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A second type of microscopic process which can result in a chemical reaction involves collision of two particles. Such a process is called a bimolecular process. A typical example of a bimolecular process is the reaction between nitrogen dioxide and carbon monoxide:
$\text{ NO}_{\text{2}} + \text{CO}\rightarrow \text{NO} + \text{ CO}_{\text{2}} \nonumber$
Here an O atom is transferred from NO2 to CO when the two molecules collide.
Several factors affect the rate of a bimolecular reaction. The first of these is the frequency of collisions between the two reactant molecules. Suppose we have a single molecule of type A (shown in black in Figure $1$a, or in blue in the animation) moving around in a gas which otherwise consists entirely of molecules of kind B (indicated in white in the figure, light blue in the animation). If we double the concentration of B molecules (Figure 1b), the number of collisions during the same time period doubles, because there are now twice as many B molecules to get in the way. Similarly, if we put twice as many A molecules into the original container, each of them collides with B molecules the same number of times, again giving twice as many A-B collisions.
The argument in the previous paragraph applies to any bimolecular process. The reaction rate must always be directly proportional to the concentration of each of the two reacting species. Thus for the general bimolecular process, A + B → products, the rate equation must be first order in A and first order in B:
$\text{Rate} = k (c_{A})(c_{B}) \nonumber$
(a)A single black molecule moving among 50 white molecules collides with 5 of them in 1 s.
A single larger particle collides 6 times with a set of 14 smaller particles over a simulated time frame of 17200 fs.
(b) If the number of white molecules is doubled, the frequency collisions rises to 10 s–1.
A single larger particle collides 14 times with a set of 28 smaller particles over a simulated time frame of 17200 fs.
(c) Two black molecules moving among 50 white ones produce 10 collisions per second. The number of collisions can thus be seen to be proportional to both the concentration of white molecules and the concentration of black molecules.
Two larger particles collide 15 times with a set of 14 smaller particles over a simulated time frame of 17200 fs.
Figure $1$ The effect of concentration on the frequency of collisions. These still images show the number of collisions between a black particle and a collection of white particles. White particles that have been hit are displayed in color. The animations display the number of collisions that occur between the dark blue particles and a set of lighter blue particles over a simulated time frame of 17200 fs, though the animation is around 8 seconds long. When hit, a lighter blue particle changes to black, then red, then white, then purple at each subsequent hit, so to quantify total collisions more easily.
The animations above serve as another way of modeling what is occurring in the still images. Notice that, as in the still image, roughly twice as many collisions occur when there are two larger blue particles or twice as many small light blue particles, than in the first initial condition. Exact count shows this is not perfectly true though, as animation (a) has 6 collisions, while animation (b) and (c) have 14 and 15 respectively. This yields a factor of 2.3 and 2.5 respectively. This highlights an important point. When we say that doubling the amount of either substance should double the collisions, we mean collisions will, on average be doubled. Notice that the exact ratio between the animations at a specific time point fluctuates. Also, it is important to note that there are never more than 30 particles of either type in any of the simulations. When dealing with the scale of moles in an actual reaction, such a difference as 2 or 3 more collisions than predicted is vanishingly small.
Collision of two molecules is a necessary but not a sufficient condition for a bimolecular process to occur. Returning to the reaction of NO2 with CO, in which an O atom is transferred from the N atom to the C atom, we can see that the orientation of the two molecules as they collide is important. This introduces what is called a steric factor. None of the collisions depicted in Figure $2$ would result in reaction, for example, because none of them involve close contact between the C atom in CO and one of the O’s in NO2. For the reaction of CO with NO2, the steric factor is estimated to be about one-sixth, meaning that only one collision in six involves an appropriate orientation. For more complex molecules, the steric factor is often much smaller. In the reaction the fraction of favorable collisions is extremely small, because OH– ions almost always hit the hydrocarbon chain at a point too far from the Br atom to cause a reaction.
The existence of a steric factor implies that there is a fairly well-defined pathway along which the reactant molecules must travel in order to produce an activated complex and then the reaction products. This pathway is called the reaction coordinate, and it applies to unimolecular as well as bimolecular reactions. In the case of cis-2-butene, for example, the reaction coordinate is the angle of rotation about the double bond, and proceeding along that pathway requires an increase in energy.
The same is true of our example of a bimolecular reaction. As the C end of a CO molecule approaches one of the O’s in NO2, the electron clouds surrounding the molecules begin to repel each other and the energy of the system rises. This is shown in Figure $3$. Only if the total energy of the two molecules exceeds the 116 kJ mol–1 activation energy can they squeeze close enough together for a C—C bond to start to form. As this occurs the N—O bond begins to break, so that the activated complex has the structure
The dashed lines indicate bonds which are just beginning to form or are in the process of breaking. As the reaction occurs, the N—O bond lengthens and the O—C distance continually gets shorter. Consequently the difference between the N—O bond length and the C—O bond length (rN—OrC—O) becomes larger and larger during the reaction. This difference, then, makes a convenient reaction coordinate. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/18%3A_Chemical_Kinetics/18.06%3A_Bimolecular_Processes.txt |
A termolecular process is one which involves simultaneous collision of three microscopic particles. In the gas phase, termolecular processes occur much less often than bimolecular processes, because the probability of three molecules all coming together at the same time is less than a thousandth the probability that two molecules will collide.
$A + B + C \rightarrow D \tag{single step}$
Consequently it is commonly found that if three molecules, A, B, and C, must combine during the course of a reaction, they will do so stepwise. That is, B and C might first combine in a bimolecular process to form BC, which would then combine with A in a second bimolecular process.
$B + C \rightarrow BC \tag{step 1}$
$BC + A \rightarrow D \tag{step 2}$
This can usually happen many times over before a successful termolecular collision would occur.
18.08: 18.7-Reaction Mechanisms
There are two main types of microscopic processes by which chemical reactions can occur. A unimolecular process involves a single molecule as a reactant, and its rate law is first order in that reactant. A bimolecular process involves collision of two molecules. Its rate law is first order in each of the colliding species and therefore second order overall. Based on this we might expect all rate laws to be first order or second order, but this conclusion does not agree with several of the experimental rate laws described earlier. (example 2 on The Rate Equation section had one rate law which was fifth order overall!)
The reason for this discrepancy is that we have not considered the possibility that an overall reaction may be the sum of several unimolecular and/or bimolecular steps. The sequence of steps by which a reaction occurs is called the mechanism of the reaction. Each unimolecular or bimolecular step in that mechanism is called an elementary process. The term elementary is used to indicate that such steps cannot be broken down into yet simpler processes. In most mechanisms some species which are produced in the earlier steps serve as reactants in later elementary processes. Such species are called intermediates.
The mechanism proposed for a given reaction must be able to account for the overall stoichiometric equation, for the rate law, and for any other facts which are known. As an example of how a mechanism can be devised to meet these criteria, consider the reaction
$\ce{2I^- + H2O2 + H3O^+ -> I2 + 4H2O}\label{1}$
When the pH of the solution is between 3 and 5, the rate law is
$\text{Rate} = k(c_{I^-})(c_{H_2O_2}) \nonumber$
We can immediately eliminate a single-step mechanism, not only because simultaneous collision of 2I, H2O2, and 2H3O+ is highly unlikely, but also because the rate law suggests a bimolecular process involving the collision of I and H2O2. The mechanism proposed for this reaction is
$\ce{HOOH + I^- <-> HOI + OH^-}\label{3}$
$\ce{HOI + I^- <-> I2 + OH^-} \nonumber$
$\ce{2OH^- +2H3O^+ <-> 4H2O} \nonumber$
You can verify for yourself that these three steps add up to the overall reaction in Equation $\ref{1}$.
The proposed mechanism can account for the rate law because the first step [Equation $\ref{3}$] is much slower than the latter two. Once HOI is produced in that first step, it is transformed almost instantaneously into I2 and OH by the second step. Similarly the OH produced by the first and second steps reacts immediately with H3O+ to form H2O. Therefore the rate of the overall reaction is limited by the rate of the first step, and the rate law must be second order since that first step is bimolecular.
What we have just said applies to any reaction mechanism. The rate of reaction is limited by the rate of the slowest step. This elementary process is called the rate-limiting step, and the rate law gives us information about the activated complex in that step. All the species whose concentrations appear in the rate law must be part of the activated complex, and the amount of each species must be given by the corresponding exponent in the rate law. It must be emphasized that any reaction mechanism is a theory about what is happening on the microscopic level and, as such, cannot be proven to be true. Thus we can say that a proposed mechanism accounts for all the known experimental facts relating to a reaction, but this does not mean it is the only mechanism which can account for those facts. A case in point is the reaction
$\text{H}_2(g) + \text{I}_2(g) \rightarrow \text{2HI}(g) \nonumber$
for which the rate law is
$\text{Rate} = k(c_{H_2})(c_{I_2})\label{7}$
This was first established experimentally in 1894, and for over 70 years chemists thought that the reaction occurred via a bimolecular collision of an H2 molecules with an I2 molecule. This agrees with the rate law since the activated complex would have the formula H2I2, containing 1H2and 1I2.
In 1967, however, it was shown that the reaction speed was increased considerably by yellow light from a powerful lamp. Such light is capable of dissociating I2 molecules into atoms:
$\ce{I2 <-> I + I} \nonumber$
The fact that the light increased the reaction rate suggested that I atoms might be involved as intermediates in the mechanism, and the currently accepted mechanism is
$\ce{I2 <-> 2I}$ fast (8a)
$\ce{I+H2 <-> H2I}$ fast (8b)
$\ce{H2I + I -> 2HI}$ slow (8c)
In this case the rate-limiting step is the last one in the mechanism. It is preceded by two rapidly established equilibria.
The rate law for the bimolecular step (8c) would be
$\text{Rate} = k'(c_{H_2})(c_{I_2})\label{9}$
but since neither H2I nor I are reactants in the overall reaction, we do not know their concentrations. These concentrations can be obtained, however, by applying the equilibrium law to Eqs. (8a) and (8b):
$\text{K}_{(8a)}= \frac{c^2_I}{c_{I_2}}$> $K={(8b)}=\frac{c_{H_2I}}{(c_{I})(c_{H_2})}$
Rearranging these equations we obtain
$c_{I^2}=\text{K}_{(8a)}(c_{I_2})$ $c_{H_2I}=\text{K}_{(8b)}(c_{I})(c_{H_2})$
These may be substituted into Equation $\ref{9}$:
$\text{Rate} = k′K_{(3b)}(c_I)(c_{H_2})(c_I) = k′K_{(3b)}(c_{H_2})(c_I)^2 \nonumber$
$= k′K_{(3b)}(c_{H_2})K_{(3a)}(c_{I_2}) \nonumber$
$= k′K_{(3b)}K_{(3a)}(c_{H_2})(c_{I_2}) \label{12}$
The rate constant k in the rate law [Equation $\ref{7}$] can he identified with the product of constants kK(8b)K(8a), and so Eqs. $\ref{12}$ and $\ref{7}$ are the same. This confirms our previous statement that the rate law tells us what species participate in the activated complex during the rate-limiting step. It also shows how more than one mechanism can lead to an activated complex having the same composition.
Example $1$: Rate Laws
The reaction between nitric oxide and oxygen:
$\ce{2NO + O2 -> 2NO2} \nonumber$
is found experimentally to obey the rate law
$\text{Rate} = k(c_{NO})^2(C_{O_2}) \nonumber$
Decide which of the following mechanisms is compatible with this rate law:
a) $\ce{NO + NO <-> N2O2}$ fast
$\ce{N2O2 + O2 -> 2NO2}$ slow
b) $\ce{NO + NO -> NO2 + N}$ slow
$\ce{N + O2 -> NO2}$ fast
c) $\ce{NO + O2 <-> NO3}$ fast
$\ce{NO3 + NO -> 2NO2}$ slow
d) $\ce{NO + O2 -> NO2 + O}$ slow
$\ce{NO + O -> NO2}$ fast
Solution
The slow step in mechanism a involves 2N atoms and 4 O atoms,i.e., an activated complex with the formula NO4. The same is true of mechanism c. Both are thus compatible with the rate law which also involves 2N and 4 O atoms. The other two mechanisms are not compatible with the measured rate law. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/18%3A_Chemical_Kinetics/18.07%3A_Termolecular_Processes.txt |
Aside from changing the concentrations of reactants or products which appear in the rate law, there are two main techniques by which reaction rates can be increased. As a rough rule of thumb, raising the temperature by an increment of 10 K usually increases the rate by a factor between 2 and 4. The effect of a catalyst can be even more striking. Certain highly efficient catalysts known as enzymes are capable of speeding up reactions to hundreds of thousands of times their normal rates. Both the effect of temperature and the effect of a catalyst are related to activation energy, and we shall consider each in turn.
18.10: The Effect of Temperature
The fact that a small increase in temperature can double the rate of a reaction is primarily dependent on an observation we made in the section on molecular speed distribution. A seemingly minor temperature rise causes a major increase in the number of molecules whose speeds (and hence molecular energies) are far above average. This happens because the distribution of molecular speeds (figure 2 in the section on molecular speed distribution) is lopsided and tails off very slowly at the high-speed end. Since only those molecules whose energy exceeds the activation energy can react, a significant increase in the fraction of molecules having high energy causes a significant increase in rate.
The fraction of molecules which have enough kinetic energy to react depends on the activation energy E, the temperature T, and the gas constant R in the following way:
Fraction of molecules having enough energy to react
$= e^{\frac {-E^ \pm}{RT}}\label{1}$
$= 10^{\frac {-E^\pm}{2.303RT}}\label{2}$
Although a derivation of this result is beyond the scope of our current discussion, we can look at this equation to see that it agrees with common sense. The larger the E, the more negative the exponent, and the smaller the fraction of molecules which can react. Thus if two similar reactions are studied under the same conditions of temperature and concentration, the one with the larger activation energy is expected to be slower. As a corollary to this, if we can find some way to reduce the activation energy of a particular reaction, then we can get that reaction to go faster. Equations $\ref{1}$ and $\ref{2}$ also shows why changing the temperature has such a large effect on reaction rates. Increasing T decreases the magnitude negative exponent in the equation. This increases the fraction of molecules which can react, as shown by the following example:
Example $1$: Mole Fractions
The reaction
$\ce{CO + NO2 -> CO2 +NO} \nonumber$
has an activation energy of 116 kJ mol. Calculate the fraction of molecules whose kinetic energies are large enough that they can collide with sufficient energy to react (a) at 298 K, and (b) at 600 K.
Solution
In both cases we use Eq. $\ref{2}$.
a) At 298 K, the power of 10 will be
$\dfrac{-E^\pm}{2.303RT}=\dfrac{-116 \frac{kJ}{mol}}{2.303 * 8.314 \dfrac{J}{K*mol} * 298K}$
$=\dfrac{-116 \dfrac{kJ}{mol}}{2.303 * 8.314 \dfrac{J}{K*mol} * 298K}=-20.3$
Then the fraction of molecules is
$Fraction = 10^{-20.3} = 5.0 * 10^{-21}$
b) At 600 k,
$Fraction = 10 ^{\dfrac{-E^{\pm}}{2.303RT}} = 10^{-10.1} = 8.0 * 10^{-11}$
Note: In an alternative way, the base e instead of 10 may be used. Using part b as an example we have
$Fraction = 10 ^{\dfrac{-E^{\pm}}{2.303RT}} = = e^{\dfrac{-E^{\pm}}{RT}} = e^{-23.25} = 8.0 * 10^{-11}$
In this case approximately doubling the temperature causes the fraction of molecular collisions which can result in reaction to increase by a factor of
that is, by about 16 billion. This means that the reaction is expected to be about 16 billion times faster at the higher temperature. This corresponds to slightly more than doubling the rate for each of the 30 intervals of 10 K between 298 and 600 K.
Equation $\ref{1}$ is also useful if we want to measure the activation energy of a reaction experimentally because it shows how the rate constant should vary with temperature. We have already seen that the rate of a unimolecular process should be proportional to the concentration of reactant. If only a fraction of those reactant molecules have enough energy to overcome the activation barrier, then the rate should also be proportional to that fraction. Thus we can write
$\text{Rate} = k(c_A) = k'10^{\frac{-E^{\pm}}{2.303RT_{(c_A)}}} \nonumber$
From this equation it is clear that
$k = k'10^{\frac{-E^{\pm}}{2.303RT}} \nonumber$
If we divide both sides by the units per second so that we can take logs, we obtain
$log \frac{k}{1 \frac{1}{s}}= log \frac{k'}{1 \frac{1}{s}} - (\frac {-E^{\pm}}{2.303RT}) \nonumber$
$log \frac{k}{1 \frac{1}{s}} = (\frac {-E^{\pm}}{2.303R}) * (\frac {1}{T}) + log \frac{k'}{1 \frac{1}{s}}$
y = m * x + b
This is the standard form for the equation of a straight line whose slope m is –E/2.303RT and whose intercept b is log (k′/s–1). Thus if we plot log (k′/s–1) versus 1/T and a straight line is obtained, we can determine E.
A similar situation applies to the second-order rate equation associated with a bimolecular process, where
$\text{Rate} = K(c_A)(c_B) = k'10^{\frac{-E^{\pm}}{2.303RT}}(c_A)(c_B) \nonumber$
In this case the rate constant has units cubic decimeter per mole per second, and so we must divide by these before taking logs. The equation that is obtained is
Again E can be obtained from the slope, provided a straight-line plot is obtained. An example of such a plot for the reaction
$\ce{2HI -> H2 + I2} \nonumber$
is shown in Figure $1$. From its slope we can determine that E = 186 kJ mol–1. This means of determining activation energy was first suggested in 1889 by Svante Arrhenius and is therefore called an Arrhenius plot
It should be pointed out that the equations just derived for Arrhenius plots apply strictly to unimolecular and bimolecular elementary processes. If a reaction mechanism involves several steps, there is no guarantee that the same elementary process will be rate limiting at several widely different temperatures. The observed rate constant may also be a product of several constants, as in the H2 + I2 reaction (Equation 5 in the section on Reaction Mechanisms). If either a different rate-limiting step or the product of constants has a different temperature dependence, then an Arrhenius plot may not be linear. This also makes it dangerous to use such a plot to predict reaction rates at temperatures quite different from the conditions under which experimental measurements were made. The slope and y-intercept of an Arrhenius plot can be used to determine the values for the Arrhenius Equation.
$k= Ae^{\frac{-E_a}{RT}} \nonumber$
where A is called the frequency factor, Ea is the activation energy, R is the gas law constant and T is the temperature in Kelvin. The frequency factor value depends on how often molecules collide and how orientation of the molecules is related to the reaction. A benefit of the Arrhenius equation is that it gives a direct method for computing the dependence of the rate constant, k, on temperature. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/18%3A_Chemical_Kinetics/18.09%3A_18.8-Increasing_the_Rate_of_a_Reaction.txt |
Altering the mechanism of a reaction so that the activation energy is lower is the second major way to speed up the reaction. A catalyst can do this by participating in the activated complex for the rate-limiting step, even though the catalyst itself is neither a reactant nor a product in the overall stoichiometric equation. A good example of catalysis is provided by the effect of I2 on the rate of isomerization of cis-2-butene. You will recall that the rate law for the catalyzed reaction involves the concentration of I2 raised to the one-half power. This implies that half a molecule of I2 (that is, an I atom) is involved in the activated complex, which probably has the structure
Since there is no double bond between the two central C atoms, one end of the activated complex can readily twist around the other.
The currently accepted mechanism for this catalyzed reaction involves three steps:
$\frac{1}{2} \ce{I2} \rightleftharpoons \ce{I} \nonumber$
$\ce{I + cis-C4H8 -> trans-C4H8 + I} \nonumber$
$\ce{I} \rightleftharpoons \ce{ \frac{1}{2} I2} \nonumber$
The first and last steps have coefficients of one-half associated with I2 because each I2 molecule that dissociates produces two I atoms, only one of which is needed to help a given cis-2-butene molecule to read. Note also that for every I2 molecule which dissociates in the first step of the mechanism, an I2 molecule is eventually regenerated by the last step. As a result, the concentration of I2 after the reaction remains exactly the same as before.
If we consider the energetics of each step in the proposed mechanism, we find that much less than the 262 kJ mol–1 activation energy of the uncatalyzed reaction is required. A complete energy profile for the catalyzed process is compared with that of the uncatalyzed one in Figure 1. The bond enthalpy is 151 kJ mol–1 (from Table 1 in the bond enthalpy section), and so 75.5 kJ mol–1 will be required for formation of the I atom in the first step. An additional increase in energy occurs as the I atom collides with cis-2-butene and bonds with it. Then about 12 kJ mol–1 is required for twisting around the C—C single bond in the activated complex. All told 115 kJ mol–1 is required to go from the initial molecules to the activated complex. When rotation to a trans structure is complete, the I atom dissociates fromtrans-2-buten and eventually reacts with another I atom to form I2. These last two processes involve an overall decrease in energy which is nearly the same as the increase required to achieve the activated complex.
The reduction in activation energy illustrated in Figure $1$ is the chief factor in speeding up the catalyzed reaction.
Example $1$: Rate Constant
Based on activation energies given, how many times larger would the rate constant be for the catalyzed rather than for the uncatalyzed isomerization of cis-2-butene at a temperature of 500 K?
Solution
The fraction of molecules which have sufficient energy to achieve the activated complex can he calculated using equation 1 from the section on the effect of temperature on catalysis:
$\text{Fraction (catalyzed)} = 10^{\dfrac{-115 \frac{kJ}{mol}}{(2.303 \times 8.314 \frac{J}{K*mol}*500K)}} = 10^{-12.0} = 1 \times 10^{-12} \nonumber$
$\text{Fraction (uncatalyzed)} = 10^{\dfrac{-262 \frac{kJ}{mol}}{(2.303 \times 8.314 \frac{J}{K*mol}*500K)}} = 10^{-27.4} = 4 \times 10^{-28} \nonumber$
Since the respective rate constants should be proportional to these fractions,
$\dfrac {k (catalyzed)}{k(uncatalyzed)} = \frac{1*10^{-12}}{4 \times 10^{-28}} = 2.5 \times 10^{15} \nonumber$
Thus at unit concentrations of cis-2-butene and iodine, the catalyzed reaction will be more than 1015 times faster.
Another example of catalysis is involved in the reaction between H2O2 and the I ion. In discussing this reaction in example 2 in the rate equation section we noted that it is first order in H2O2 and first order in I ion between pH = 3 and pH = 5. At lower pH values, however, a different mechanism takes over, and the rate law becomes
$\text{Rate}= k(c_{H_2O_2})(c_{I^-})(c_{H^+}) \nonumber$
This indicates that the activated complex contains an additional proton when compared with the uncatalyzed case. This proton apparently adds to the H2O2 molecule, forming H3O2+. Because of the positive charge, less energy increase occurs as an I ion approaches H3O2+ and the reaction
$\ce{H3O2+ + I- -> H2O + HOI} \nonumber$
has a lower activation energy than
$\ce{H2O2 + I- -> OH- + HOI} \nonumber$
The latter is the rate-limiting step in the mechanism at higher pH (Eqs. (1a), (1b), and (1c) in the reaction mechanisms section), and so protonating H2O2 results in a faster overall reaction. Measurements over a range of temperatures show that the activation energy is lowered from 56 to 43.6 kJ mol–1 by this change of mechanism.
The peroxide-iodide reaction is one example of a great many acid-catalyzed reactions. In most of these the hydrogen ion concentration appears in the rate law, indicating that the activated complex contains an extra proton. The positive charge of this proton allows negative ions to combine more readily with the protonated species. The proton may also shift electron density toward itself and away from some other site in the protonated molecule, allowing a negative species to bond to that site more readily. The net result is a lower activation energy and a more rapid reaction. The preceding examples are illustrative of the three main features of catalysis:
1. The catalyst allows the reaction to proceed via an alternative mechanism.
2. The catalyst is directly involved in this mechanism, but for every step in which a catalyst molecule is a reactant, there is another step where the catalyst appears as a product. Thus there is no net consumption of the catalyst.
3. The catalyzed mechanism results in a faster reaction, usually because the overall activation energy is lowered. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/18%3A_Chemical_Kinetics/18.11%3A_Catalysis.txt |
Our previous discussion has concentrated on catalysts which are in the same phase as the reaction being catalyzed. This kind of catalysis is called homogeneous catalysis. Many important industrial processes rely on heterogeneous catalysis, in which the catalyst is in a different phase. Usually the catalyst is a solid and the reactants are gases, and so the rate-limiting step occurs at the solid surface. Thus heterogeneous catalysis is also referred to as surface catalysis.
The detailed mechanisms of most heterogeneous reactions are not yet understood, but certain sites on the catalyst surface appear to be able to weaken or break bonds in reactant molecules. These are called active sites. One example of heterogeneous catalysis is hydrogenation of an unsaturated organic compound such as ethane (C2H4) by metal catalysts such as Pt or Ni:
The currently accepted mechanism for this reaction involves weak bonding of both H2 and C2H4 to atoms on the metal surface. This is called adsorption. The H2 molecules dissociate to individual H atoms, each of which is weakly bonded to a Pt atom:
$\ce{H2 + 2Pt (surface) -> 2H ---Pt(surface)} \nonumber$
These adsorbed H atoms can move across the metal surface, and eventually they combine with a C2H4 molecule, completing the reaction. Because adsorption and dissociation of H2 on a Pt surface is exothermic (ΔHm° >= –160 kJ mol–1), it can provide H atoms for further reaction without a large activation energy. By contrast, dissociation of gaseous H2 molecules without a metal surface would require the full bond enthalpy (ΔHm° = +436 kJ mol–1). Clearly the metal surface makes a major contribution in lowering the activation energy.
Heterogeneous catalysts are used extensively in the petroleum industry. One example is the combination of SiO2 and Al2O3 used to speed up cracking of long-chain hydrocarbons into the smaller molecules needed for gasoline. Another is the Pt catalyst used to reform hydrocarbon chains into aromatic ring structures. This improves the octane rating of gasoline, making it more suitable for use in automobile engines. Other industries also make effective use of catalysts. SO2, obtained by burning sulfur (or even from burning coal), can be oxidized to SO3 over vanadium pentoxide, V2O5. This is an important step in manufacturing H2SO4
Another important heterogeneous catalyst is used in the Haber process for synthesis of NH3 from N2 and H2. As with most industrial catalysts, its exact composition is a trade secret, but it is mainly Fe with small amounts of Al2O3 and K2O added.
The surface catalyst you are most likely to be familiar with is found in the exhaust systems of many automobiles constructed since 1976. Such a catalytic converter contains from 1 to 3 g Pt in a fine layer on the surface of a honeycomb-like structure or small beads made of Al2O3. The catalyst speeds up oxidation of unburned hydrocarbons and CO which would otherwise be emitted from the exhaust as air pollutants. It apparently does this by adsorbing and weakening the bond in the O2 molecule. Individual O atoms are then more readily transferred to CO or hydrocarbon molecules, producing CO2 and H2O. This action of the catalytic surface can be inhibited or poisoned if lead atoms (from tetraethyllead in leaded gasoline) react with the surface. Hence the prohibition of use of leaded fuel in cars equipped with catalytic converters. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/18%3A_Chemical_Kinetics/18.12%3A_Heterogeneous_Catalysis.txt |
Electrons (especially valence electrons) are the only subatomic particles which are involved in ordinary chemical changes, and we have spent considerable time describing the rearrangements they undergo when atoms and molecules combine. However, another category of reactions is possible in which the structures of atomic nuclei change. In such nuclear reactions electronic structure is incidental—we are primarily interested in how the protons and neutrons are arranged before and after the reaction. Nuclear reactions are involved in transmutation of one element into another and in natural radioactivity. In these sections we consider nuclear reactions in more detail, exploring their applications to nuclear energy, to the study of reaction mechanisms, to qualitative and quantitative analysis, and to estimation of the ages of objects as different as the Dead Sea scrolls and rocks from the moon.
19: Nuclear Chemistry
In our discussion on Radiation we described the experimental basis for the idea that each atom has a small, very massive nucleus which contains protons and neutrons. Surrounding the nucleus are one or more electrons which occupy most of the volume of the atom but make only a small contribution to its mass. Electrons (especially valence electrons) are the only subatomic particles which are involved in ordinary chemical changes, and we have spent considerable time describing the rearrangements they undergo when atoms and molecules combine. However, another category of reactions is possible in which the structures of atomic nuclei change. In such nuclear reactions electronic structure is incidental—we are primarily interested in how the protons and neutrons are arranged before and after the reaction. Nuclear reactions are involved in transmutation of one element into another and in natural radioactivity. In these sections we consider nuclear reactions in more detail, exploring their applications to nuclear energy, to the study of reaction mechanisms, to qualitative and quantitative analysis, and to estimation of the ages of objects as different as the Dead Sea scrolls and rocks from the moon.
Nuclear reactions involve rearrangements of the protons and neutrons within atomic nuclei. During naturally occurring nuclear reactions α particles, β particles, and γ rays are emitted, often in a radioactive series of successive reactions. Nuclear reactions may also be induced by bombarding nuclei with positive ions or neutrons. Artificial isotopes produced in this way may decay by positron emission or electron capture as well as by α , β or γ emission. Stability of nuclei depends on the neutron/proton ratio (usually between 1 and 1.6) and magic numbers of protons and neutrons.
Radioactive decay obeys a first-order rate law, and its rate is often reported in terms of half-life, the time necessary for half the radioactive nuclei to decompose. Known half-lives of isotopes such as ${}_{\text{6}}^{\text{14}}\text{C}$ and ${}_{\text{92}}^{\text{238}}\text{U}$ may be used to establish the ages of objects containing these elements, provided accurate measurements can be made of the quantity of radiation emitted. Geiger-Müller counters or scintillation counters are often used for such measurements. Other important applications of radioactive isotopes include tracer studies, where a particular type of atom can be labeled and followed throughout a reaction, and neutron activation analysis, which can determine extremely low concentrations of many elements.
The relative stability of a nucleus is given by the energy of formation per nuclear particle. This may be determined from the difference between the molar mass of the nucleus and the sum of the molar masses of its constituent protons and neutrons. Both fission, breaking apart of a heavy nucleus, and fusion, combining of two light nuclei, can result in release of energy. Fission usually involves ${}_{\text{92}}^{\text{238}}\text{U}$ or , ${}_{\text{94}}^{\text{239}}\text{Pu}$ and these isotopes have been used in nuclear explosives and nuclear power plants. Fission products are highly radioactive. Because of the considerable damage done to living tissue by the ability of α, β and γ radiation to break bonds and form ions, emission of radioactive materials must be carefully controlled and fission power plants are quite expensive to construct. Although it promises much larger quantities of free energy and fewer harmful by-products than fission, nuclear fusion bas not yet been shown to be feasible for use in power plants. So far its only application has been in hydrogen bombs. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/19%3A_Nuclear_Chemistry/19.01%3A_Prelude_to_Nuclear_Chemistry.txt |
When we discuss natural radioactivity in the sections under The Structure of Atoms, we mention the properties of $\alpha$ particles, $\beta$ particles, and $\gamma$ rays, but little attention is paid to the atoms which are left behind when one of these forms of radiation is emitted. Now we consider the subject of radioactivity in more detail.
$α$ Emission
An alpha particle corresponds to a helium nucleus. It consists of two protons and two neutrons, and so it has a mass number of 4 and an atomic number (nuclear charge) of 2. From a chemical point of view we would write it as He2+, indicating its lack of two electrons with the superscript 2+. In writing a nuclear reaction, though, it is unnecessary to specify the charge, because the presence or absence of electrons around the nucleus is usually unimportant. For these purposes an α particle is indicated as ${}_{\text{2}}^{\text{4}}\text{He}$ or ${}_{\text{2}}^{\text{4}}\alpha\label{1}$.
In certain nuclei a particles are produced by combination of two protons and two neutrons which are then emitted. An example of naturally occurring emission of an α particle is the disintegration of one of the isotopes of uranium, ${}_{\text{92}}^{\text{238}}\text{U}$. The equation for this process is
${}_{\text{92}}^{\text{238}}\text{U }\to \text{ }{}_{\text{90}}^{\text{234}}\text{Th + }{}_{\text{2}}^{\text{4}}\text{He} \label{2}$
Note that if we sum the mass numbers on each side of a nuclear equation, such as Equation $\ref{2}$, the total is the same. That is, 238 on the left equals 234 + 4 on the right. Similarly, the atomic numbers (subscripts) must also balance (92 on the left and 90 + 2 on the right). This is a general rule which must be followed in writing any nuclear reaction. The total number of nucleons (i.e., protons and neutrons) remains unchanged, and electrical charge is neither created nor destroyed in the process.
When a nucleus emits an a particle, its atomic number is reduced by 2 and it becomes the nucleus of an element two places earlier in the periodic table. That one element could transmute into another in this fashion was first demonstrated by Rutherford and Soddy in 1902. It caused a tremendous stir in the scientific circles of the day since it quite clearly contradicted Dalton’s hypothesis that atoms are immutable. It gave Rutherford, who was then working at McGill University in Canada, an international reputation.
The type of nucleus that will spontaneously emit an a particle is fairly restricted. The mass number is usually greater than 209 and the atomic number greater than 82. In addition the nucleus must have a lower ratio of neutrons to protons than normal. The emission of an a particle raises the neutron/proton ratio as illustrated by the nuclear equation
${}_{\text{84}}^{\text{210}}\text{Po }\to \text{ }{}_{\text{82}}^{\text{206}}\text{Pb + }{}_{\text{2}}^{\text{4}}\text{He} \nonumber$
The nucleus of ${}_{\text{84}}^{\text{210}}\text{Po}$ contains 210 – 84 = 126 neutrons and 84 protons, giving a ratio of 126:84 = 1.500. This is increased to 124:82 = 1.512 by the α-emission.
$β$ Emission
A beta particle is an electron which has been emitted from an atomic nucleus. It has a very small mass and is therefore assigned a mass number of 0. The β particle has a negative electrical charge, and so its nuclear charge is taken to be –1. Thus it is given the symbol ${}_{-\text{1}}^{\text{0}}e$ or ${}_{-\text{1}}^{\text{0}}\beta$ in a nuclear equation. Two examples of unstable nuclei which emit $β$ particles are
$\ce{ _{90}^{234}Th -> _{91}^{234}Pa + _{-1}^{0}e} \nonumber$
and
${}_{\text{6}}^{\text{14}}\text{C }\to \text{ }{}_{\text{7}}^{\text{14}}\text{N + }{}_{-\text{1}}^{\text{0}}e \nonumber$
Note that both of these equations accord with the conservation of mass number and atomic number, showing again that both the total number of nucleons and the total electrical charge remain unchanged. We can consider a β particle emitted from a nucleus to result from the transformation of a neutron into a proton and an electron according to the reaction
${}_{\text{0}}^{\text{1}}n\text{ }\to \text{ }{}_{\text{1}}^{\text{1}}p\text{ + }{}_{-\text{1}}^{\text{0}}e \nonumber$
(Indeed, free neutrons unattached to any nucleus soon decay in this way.) Thus when a nucleus emits a β particle, the nuclear charge rises by 1 while the mass number is unaltered. Therefore the disintegration of a nucleus by β decay produces the nucleus of an element one place further along in the periodic table than the original element.
β decay is a very common form of radioactive disintegration and, unlike α decay, is found among both heavy and light nuclei. Nuclei which disintegrate in this way usually have a neutron/proton ratio which is higher than normal. When a β particle is lost, a neutron is replaced by a proton and the neutron/proton ratio decreases. For example, in the decay process
${}_{\text{82}}^{\text{206}}\text{Pb }\to \text{ }{}_{\text{83}}^{\text{206}}\text{Bi + }{}_{-\text{1}}^{\text{0}}e \nonumber$
the neutron/proton ratio changes from 1.561 to 1.530.
$γ$ Radiation
Gamma rays correspond to electromagnetic radiation similar to light waves or radiowaves except that they have an extremely short wavelength—about a picometer. Because of the wave-particle duality we can also think of them as particles or photons having the same velocity as light and an extremely high energy. Since they have zero charge and are not nucleons, they are denoted in nuclear equations by the symbol ${}_{\text{0}}^{\text{0}}\gamma$ or, more simply, $\gamma$.
Virtually all nuclear reactions are accompanied by the emission of γ rays. This is because the occurrence of a nuclear transformation usually leaves the resultant nucleus in an unstable high-energy state. The nucleus then loses energy in the form of a γ-ray photon as it adopts a lower-energy more stable form. Usually these two processes succeed each other so rapidly that they cannot be distinguished. Thus when 238U decays by $α$ emission, it also emits a $γ$ ray. This is actually a two-stage process. In the first stage a high-energy (or excited) form of 234Th is produced:
${}_{\text{92}}^{\text{238}}\text{U }\to \text{ }{}_{\text{90}}^{\text{234}}\text{Th* + }{}_{\text{2}}^{\text{4}}\text{He} \label{eq7}$
This excited nucleus-then emits a $γ$ ray:
${}_{\text{90}}^{\text{234}}\text{Th* }\to \text{ }{}_{\text{90}}^{\text{234}}\text{Th} + {}_{\text{0}}^{\text{0}}\gamma \label{eq8}$
Usually when a nuclear reaction is written, the $γ$ ray is omitted. Thus Equations \ref{eq7} and \ref{eq8} are usually combined to give
${}_{\text{92}}^{\text{238}}\text{U }\to \text{ }{}_{\text{90}}^{\text{234}}\text{Th} + {}_{\text{2}}^{\text{4}}\text{He} \nonumber$ | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/19%3A_Nuclear_Chemistry/19.02%3A_Naturally_Occurring_Radioactivity.txt |
Naturally occurring uranium contains more than 99% $\ce{_{92}^{238}U}$ that decays to $\ce{_{90}^{234}Th}$ by $α$ emission:
$\ce{_{92}^{238}U -> _{90}^{234}Th + _{2}^{4}He} \nonumber$
The product of this reaction is also radioactive, however, and undergoes $β$ decay:
$\ce{ _{90}^{234}Th -> _{91}^{234}Pa + _{-1}^{0}e} \nonumber$
The $\ce{_{91}^{234}Pa}$ produced in this second reaction also emits a $β$ particle:
$\ce{ _{91}^{234}Pa -> _{92}^{234}U + _{-1}^{0}e} \nonumber$
These three reactions are only the first of 14 steps. After emission of eight $α$ particles and six $β$ particles, the isotope $\ce{ _{82}^{206}Pb}$ is produced. It has a stable nucleus which does not disintegrate further. The complete process may be written as follows:
$\text{ }{}_{\text{92}}^{\text{238}}\text{U}\xrightarrow{\alpha }{}_{\text{90}}^{\text{234}}\text{Th}\xrightarrow{\beta }{}_{\text{91}}^{\text{234}}\text{Pa}\xrightarrow{\beta }{}_{\text{92}}^{\text{234}}\text{U}\xrightarrow{\alpha }{}_{\text{90}}^{\text{230}}\text{Th}\xrightarrow{\alpha }{}_{\text{88}}^{\text{226}}\text{Ra}\xrightarrow{\alpha }{}_{\text{88}}^{\text{222}}\text{Rn}$
$\downarrow ^{\alpha }$ (2a)
${}_{\text{82}}^{\text{206}}\text{Pb}\xleftarrow{\alpha }{}_{\text{84}}^{\text{210}}\text{Po}\xleftarrow{\beta }{}_{\text{83}}^{\text{210}}\text{Bi}\xleftarrow{\alpha }{}_{\text{82}}^{\text{210}}\text{Pb}\xleftarrow{\alpha }{}_{\text{84}}^{\text{214}}\text{Po}\xleftarrow{\beta }{}_{\text{83}}^{\text{214}}\text{Bi}\xleftarrow{\beta }{}_{\text{82}}^{\text{214}}\text{Pb}\xleftarrow{\alpha }{}_{\text{84}}^{\text{218}}\text{Po }$
While the net reaction is
${}_{\text{92}}^{\text{238}}\text{U }\to \text{ }{}_{\text{82}}^{\text{206}}\text{Pb + 8}{}_{\text{2}}^{\text{4}}\text{He + 6}{}_{-\text{1}}^{\text{0}}e \nonumber$
Such a series of successive nuclear reactions is called a radioactive series. Two other radioactive series similar to the one just described occur in nature. One of these starts with the isotope $\ce{ _{90}^{232}Th}$ and involves 10 successive stages, while the other starts with $\ce{_{92}^{235}U}$ and involves 11 stages. Each of the three series produces a different stable isotope of lead.
Example $1$: Uranium-Actinium Series
The first four stages in the uranium-actinium series involve the emission of an α particle from a $\ce{_{92}^{235}U}$ nucleus, followed successively by the emission of a $β$ particle, a second $α$ particle, and then a second β particle. Write out equations to describe all four nuclear reactions.
Solution
The emission of an a particle lowers the atomic number by 2 (from 92 to 90). Since element 90 is thorium, we have
$\ce{ _{92}^{235}U -> _{90}^{231}Th + _{2}^{4}He} \nonumber$
The emission of a β particle now increases the atomic number by 1 to give an isotope of element 91, protactinium:
$\ce{_{90}^{231}Th -> _{91}^{231}Pa + _{-1}^{0}e} \nonumber$
The next two stages follow similarly:
$\ce{_{91}^{231}Pa -> _{89}^{227}Ac + _{2}^{4}He} \nonumber$
and
$\ce{_{89}^{227}Ac -> _{90}^{227}Th + _{-1}^{0}e} \nonumber$
Example $2$: Thorium Series
In the thorium series, $\ce{_{90}^{232}Th}$ loses a total of six α particles and four β particles in a 10-stage process. What isotope is finally produced in this series?
Solution
The loss of six α particles and four $β$ particles:
$\ce{6 _{2}^{4}He + 4 _{-1}^{0}e} \nonumber$
involves the total loss of 24 nucleons and 6 × 2 – 4 = 8 positive charges from the $\ce{_{90}^{232}Th}$ nucleus. The eventual result will be an isotope of mass number 232 – 24 = 208 and a nuclear charge of 90 – 8 = 82. Since element 82 is $\ce{Pb}$, we can write
$\ce{ _{90}^{232}Th -> _{82}^{208}Pb + 6 _{2}^{4}He + 4 _{-1}^{0}e} \nonumber$
19.04: Artificially Induced Nuclear Reactions
In 1919 Rutherford performed the first artificial nuclear reaction. He was able to demonstrate that when $α$ particles are introduced into a closed sample of N2 gas, an occasional collision led to the formation of an isotope of $\ce{O}$ and the release of a proton:
$\ce{_{2}^{4}He + _{7}^{14}N -> _{8}^{17}O + _{1}^{1}H} \label{1}$
Since then many thousands of nuclear reactions have been studied, most of them produced by the bombardment of stable forms of matter with a beam of nucleons or light nuclei as projectiles. Particles which have been used for this purpose include protons, neutrons, deuterons (21H) , $α$ particles, and B, C, N, and O nuclei.
19.05: Bombardment with Positive Ions
A powerful method of artificially inducing nuclear reactions is the bombardment of a sample of matter with ions. When the bombarding particle is positively charged, which is usually the case, it must have a very high kinetic energy to overcome the coulombic repulsion of the nucleus being bombarded. This is particularly necessary if the nucleus has a high nuclear charge. To give these charged particles the necessary energy, an accelerator or “atom smasher” such as a cyclotron must be used. The cyclotron was developed mainly by E. O. Lawrence (1901 to 1958) at the University of California. A schematic diagram of a cyclotron is shown in Figure $1$. Two hollow D-shaped plates (dees) are enclosed in an evacuated chamber between the poles of a powerful electromagnet. The two dees are connected to a source of high-frequency alternating current, so that when one is positive, the other is negative. Ions are introduced at the center and are accelerated because of their alternate attraction to the left- and right-hand dees. Since the magnetic field would make ions traveling at constant speed move in a circle, the net result is that they follow a spiral path as they accelerate until they finally emerge at the outer edge of one of the dees.
Some examples of the kinds of nuclear reactions which are possible with the aid of an accelerator are as follows:
${}_{\text{12}}^{\text{24}}\text{Mg + }{}_{\text{1}}^{\text{1}}\text{H}\to \text{ }{}_{\text{11}}^{\text{21}}\text{Na + }{}_{\text{2}}^{\text{4}}\text{He } \nonumber$
${}_{\text{3}}^{\text{7}}\text{Li + }{}_{\text{1}}^{\text{2}}\text{H}\to \text{ }{}_{\text{4}}^{\text{8}}\text{Be + }{}_{\text{0}}^{\text{1}}n\text{ } \nonumber$
${}_{\text{46}}^{\text{106}}\text{Pb + }{}_{\text{2}}^{\text{4}}\text{He +}\to \text{ }{}_{\text{47}}^{\text{109}}\text{Ag + }{}_{\text{1}}^{\text{1}}\text{H} \nonumber$
A particularly interesting type of nuclear reaction performed in an accelerator is the production of the transuranium elements. These elements have atomic numbers greater than that of uranium (92) and are too unstable to persist for long in nature. The heaviest of them can be prepared by bombarding nuclei which are already heavy with some of the lighter nuclei:
${}_{\text{92}}^{\text{238}}\text{U + }{}_{\text{6}}^{\text{12}}\text{C}\to \text{ }{}_{\text{98}}^{\text{246}}\text{Cf + 4}{}_{\text{0}}^{\text{1}}n \nonumber$
${}_{\text{92}}^{\text{238}}\text{U + }{}_{\text{7}}^{\text{14}}\text{N}\to \text{ }{}_{\text{99}}^{\text{247}}\text{Es + 5}{}_{\text{0}}^{\text{1}}n\text{ } \nonumber$
${}_{\text{98}}^{\text{252}}\text{Cf + }{}_{\text{5}}^{\text{10}}\text{B}\to \text{ }{}_{\text{103}}^{\text{257}}\text{Lr + 5}{}_{\text{0}}^{\text{1}}n\text{ } \nonumber$ | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/19%3A_Nuclear_Chemistry/19.03%3A_Radioactive_Series.txt |
Since a neutron has no charge, it is not electrostatically repelled by the nucleus it is bombarding. Because of this, neutrons do not need to be accelerated to high energies before they can undergo a nuclear reaction. Nuclear reactions involving neutrons are thus easier and cheaper to perform than those requiring positively charged particles.
Though neutron-bombardment reactions are often carried out in a nuclear reactor (which will be described later), they can also be very conveniently performed in a small laboratory using a neutron source. Usually a neutron source consists of an $\alpha$ emitter such as $\ce{^{222}_{86}Rn}$ mixed with $Be$, an element whose nuclei produce neutrons when bombarded by α particles:
$\ce{_4^9 Be} + \ce{_2^4He} \to \ce{^{12}_6 C} + \ce{_0^1 n} \label{1}$
This reaction was originally used in 1932 by Sir James Chadwick (1891 to 1974) to demonstrate the existence of the neutron. (Previous to this it was believed that electrons were present in the nucleus together with protons.) The neutrons produced by Equation $\ref{1}$ have a very high energy and are called fast neutrons. For many purposes the neutrons are more useful if they are first slowed down or moderated by passing them through paraffin wax or some other substance containing light nuclei in which they can dissipate most of their energy by collision. The slow neutrons produced by a moderator are then able to participate in a larger number of neutron-capture reactions of which the following two are typical:
${}_{\text{16}}^{\text{34}}\text{S + }{}_{\text{0}}^{\text{1}}n\text{ }\to \text{ }{}_{\text{16}}^{\text{35}}\text{S + }\gamma \label{2}$
${}_{\text{80}}^{\text{200}}\text{Hg + }{}_{\text{0}}^{\text{1}}n\text{ }\to \text{ }{}_{\text{80}}^{\text{201}}\text{Hg + }\gamma \label{3}$
In such a reaction a different isotope (often an unstable isotope) of the element being bombarded is produced, with the emission of a $γ$ ray. Radioactive isotopes of virtually all the elements can be produced in this way.
An important neutron-capture reaction is that undergone by the most common isotope of uranium, namely, $\ce{^{238}U}$:
${}_{\text{92}}^{\text{238}}\text{U + }{}_{\text{0}}^{\text{1}}n\text{ }\to \text{ }{}_{\text{92}}^{\text{239}}\text{U + }\gamma \label{4}$
The uranium-239 produced in this way decays by β emission to produce the first and most important of the transuranium elements, namely, neptunium:
${}_{\text{92}}^{\text{239}}\text{U }\to \text{ }{}_{\text{93}}^{\text{239}}\text{N + }{}_{-\text{1}}^{\text{0}}e \label{5}$
When nuclei are bombarded by fast neutrons, a secondary particle is emitted—usually a proton or an α particle:
${}_{\text{5}}^{\text{11}}\text{B + }{}_{\text{0}}^{\text{1}}n\text{ }\to \text{ }{}_{\text{4}}^{\text{11}}\text{Be + }{}_{\text{1}}^{\text{1}}\text{H}\label{6}$
${}_{\text{13}}^{\text{27}}\text{Al + }{}_{\text{0}}^{\text{1}}n\text{ }\to \text{ }{}_{\text{11}}^{\text{24}}\text{Na + }{}_{\text{2}}^{\text{4}}\text{He}\label{7}$
19.07: Further Modes of Decay - Positron Emission and Electron Capture
Isotopes produced by nuclear reactions, which do not occur in nature (artificial isotopes) are invariably unstable and radioactive. They exhibit two kinds of decay not found among naturally occurring radioactive elements. The first is positron emission in which a fundamental particle we have not discussed is ejected from the nucleus. The positron is identical with the electron except that it has a positive rather than a negative charge. Its symbol is $\ce{ _{+1}^{0}e^{+}}$ or $\ce{ _{+1}^{0}\beta^{+}}$. An example of a positron emission reaction is the decay of $\ce{ _6^{11}C}$
$\ce{ _6^{11}C -> _5^{11}B + _{+1}^{0}e^{+}} \nonumber$
Positron emission is common among isotopes having a low neutron-to-proton ratio.
The second new method of decay is called electron capture. The nucleus absorbs one of the electrons from its own innermost core. An example is the following reaction:
${}_{-\text{1}}^{\text{0}}e\text{ + }{}_{\text{4}}^{\text{7}}\text{Be }\xrightarrow{ec}\text{ }{}_{\text{3}}^{\text{7}}\text{Li} \nonumber$
Again this results in an increased neutron/proton ratio.
19.08: Nuclear Stability
Why is it that certain combinations of nucleons are stable in a nucleus while others are not? A complete answer to this question cannot yet be given, largely because the exact nature of the forces holding the nucleons together is still only partially understood. We can, however, point to several factors which affect nuclear stability. The most obvious is the neutron/proton ratio. As we discuss in "Further Modes of Decay", if this is too high or too low, it makes for an unstable nucleus.
If we plot the number of neutrons against the number of protons for all known stable (i.e., nonradioactive) nuclei, we obtain the result shown in Figure \(1\). All the stable nuclei lie within a definite area called the zone of stability. For low atomic numbers most stable nuclei have a neutron/proton ratio which is very close to 1. As the atomic number increases, the zone of stability corresponds to a gradually increasing neutron/proton ratio. In the case of the heaviest stable isotope, \(\ce{_{83}^{209}Bi}\) for instance, the neutron/proton ratio is 1.518. If an unstable isotope lies to the left of the zone of stability in Figure \(1\), it is neutron rich and decays by β emission. If it lies to the right of the zone, it is proton rich and decays by positron emission or electron capture.
Another factor affecting the stability of a nucleus is whether the number of protons and neutrons is even or odd. Among the 354 known stable isotopes, 157 (almost half) have an even number of protons and an even number of neutrons. Only five have an odd number of both kinds of nucleons. In the universe as a whole (with the exception of hydrogen) we find that the even-numbered elements are almost always much more abundant than the odd-numbered elements close to them in the periodic table.
Finally there is a particular stability associated with nuclei in which either the number of protons or the number of neutrons is equal to one of the so-called "magic" numbers 2, 8, 20, 28, 50, 82, and 126. These numbers correspond to the filling of shells in the structure of the nucleus. These shells are similar in principle but different in detail to those found in electronic structure. Of particular stability, and also of high abundance in the universe, are nuclei in which both the-number of protons and the number of neutrons correspond to magic numbers. Examples are 42He, 168O, 4020Ca, and 20882Pb
Example \(1\): Isotopes
Find which element has the largest number of isotopes, using Figure \(1\). Likewise find which is the number of neutrons which occurs most frequently. What do you notice about the numbers of protons and neutrons in each case?
Solution:
Tin has 10 isotopes, and its atomic number 50 is a magic number. A total of 7 stable isotopes have 82 neutrons in the nucleus, more than for any other number of neutrons. Again the number is a magic number. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/19%3A_Nuclear_Chemistry/19.06%3A_Neutron_Bombardment.txt |
We have labeled all isotopes which exhibit radioactivity as unstable, but radioactive isotopes vary considerably in their degree of instability. Some decay so quickly that it is difficult to detect that they are there at all before they have changed into something else. Others have hardly decayed at all since the earth was formed.
The process of radioactive decay is governed by the uncertainty principle, so that we can never say exactly when a particular nucleus is going to disintegrate and emit a particle. We can, however, give the probability that a nucleus will disintegrate in a given time interval. For a large number of nuclei we can predict what fraction will disintegrate during that interval. This fraction will be independent of the amount of isotope but will vary from isotope to isotope depending on its stability. We can also look at the matter from the opposite point of view, i.e., in terms of how long it will take a given fraction of isotope to dissociate. Conventionally the tendency for the nuclei of an isotope to decay is measured by its half-life, symbol t1/2.
This is the time required for exactly half the nuclei to disintegrate. This quantity, too, varies from isotope to isotope but is independent of the amount of isotope.
Figure $1$ shows how a 1-amol (attomole) sample of 12853I, which has a half-life of 25.0 min, decays with time. In the first 25 min, half the nuclei disintegrate, leaving behind 0.5 amol. In the second 25 min, the remainder is reduced by one-half again, i.e., to 0.25 amol. After a third 25-min period, the remainder is (½)3 amol, after a fourth it is (½)4 amol, and so on. If x intervals of 25.0 min are allowed to pass, the remaining amount of 12853I will be (½)x amol.
This example enables us to see what will happen in the general case. Suppose the initial amount of an isotope of half-life t1/2 is n0 and the isotope decays to an amount n in time t, we can measure the time in terms of the number of t1/2 intervals which have elapsed by defining a variable x such that
$x=\frac{t}{t_{\text{1/2}}}\label{1}$
Thus after time t our sample will have been reduced to a fraction (½)x of the original amount. In other words
$\frac{n}{n_{\text{0}}}=\left( \frac{\text{1}}{\text{2}} \right)^{x}\label{2}$
Taking natural logarithm of each side, we then have
$ln\frac{n}{n_{\text{0}}} = ln \left( \frac{\text{1}}{\text{2}} \right)^{x} = x ln \frac{\text{1}}{\text{2}} = {-{0.693}x}\,\label{3}$
Substituting from Eq. $\ref{1}$ we thus obtain
$ln \frac{n}{n_{\text{0}}} = \frac{-\text{0.693}t}{t_{\text{1/2}}}\label{4}$
Example $1$ : Decay
What amount of 12853I will be left when 3.65 amol of this isotope is allowed to decay for 15.0 min. The half-life of 12853I is 25.0 min.
Solution: Substituting in Eq. $\ref{3}$ we have
$ln \dfrac{n}{n_{\text{0}}}$ = $\dfrac{-\text{0}\text{.693}}{t_{\text{1/2}}}$t = $\dfrac{-\text{0}\text{.693 }\times \text{ 15}\text{.0 min}}{\text{25}\text{.0 min}}$ = – 0.4158
Thus
$\dfrac{n}{n_{\text{0}}}$ = e– 0.4158 = 0.6598
or
n = 0.6598n0 = 0.6598 × 3.65 amol = 2.41 amol
Equation $\ref{2}$ describes how the amount of a radioactive isotope decreases with time, but similar formulas can also be written for the mass m and also for the rate of disintegration r. This is because both the mass and the rate are proportional to the amount of isotope. Thus the rate at which an isotope decays is given by
$ln\frac{r}{r_{\text{0}}} = \frac{-\text{0}\text{.693}}{t_{\text{1/2}}} \nonumber$t
where r0 is the initial rate at time zero.
The following example uses equation $\ref{4}$ to determine the half life of the disintegration of 9038Sr using decay rates taken at two time points.
Example $2$ : Decay Rate
A sample of 9038Sr has a decay rate of 1.682×106 disintegrations per minute. A year later, the rate of decay has decreased to 1.641×106 disintigrations per minute. What is the half life of 9038Sr?
Solution
We have r0 = 1.682×106 min–1 g–1, while r = 1.641×106 min–1 g–1, and t = 1yr. Substituting into Eq. $\ref{4}$, we then have
$ln\dfrac{r}{r_{\text{0}}}$ = $\dfrac{-\text{0}\text{.693}}{t_{\text{1/2}}}$t
$ln\dfrac{1.641 \times 10^6}{1.682 \times 10^6}$ = $\dfrac{-\text{0}\text{.693}}{t_{\text{1/2}}}$1 yr
– 0.0247 = $\dfrac{-\text{0}\text{.693}}{t_{\text{1/2}}}$1 yr
t1/2 = $\dfrac{-\text{0}\text{.693}}{-0.0247}$1 yr=28.1 years
While rates of decay may be used as in example $2$, if masses of either the radioactive substance or the product formed from decay are given, they must first be converted into moles. Then equation $\ref{3}$ may be used to find any remaining unknown quantities. Example three asks for the age of the sample given masses of the radioactive isotope, product, and half life:
Example $3$ : Decay
21084Po decays to 20682Pb by emitting an alpha particle. Analysis of a sample originally containing only 21084Po is found to have 0.387 μg of 21084Po and 0.294 μg of 20682Pb. Given a half life of 138 days, how long ago was the original sample produced?
Solution:
Amount of 210Po = nPo = $\dfrac{\text{0}\text{.387 }\mu \text{g}}{\text{210 g mol}^{-\text{1}}}$ = 1.84 × 10–3 μmol
Amount of 206Pb = nPb = $\dfrac{\text{0}\text{.294 }\mu \text{g}}{\text{206 g mol}^{-\text{1}}}$ = 1.43 × 10–3 μmol
Since each mol of 206Pb was originally a mole of 210Po, the original amount of 210Po, n0, is given by
n0 = (1.84 + 1.43) × 10–3 μmol = 3.27 × 10–3 μmol
Substituting into Eq. $\ref{3}$, we have
$ln\dfrac{n}{n_{\text{0}}}$ = $\dfrac{-\text{0}\text{.693}}{t_{\text{1/2}}}$t
$ln\dfrac{\text{1}\text{.84 }\times \text{ 10}^{-\text{3}}\text{ }\mu \text{mol}}{\text{3}\text{.27 }\times \text{ 10}^{-\text{3}}\text{ }\mu \text{mol}}$ = $\dfrac{-\text{0}\text{.693}}{138\text{ days}}$ t
– 0.575 = – 5.02 × 10–3 days–1
or
t = 115 days | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/19%3A_Nuclear_Chemistry/19.09%3A_The_Rate_of_Radioactive_Decay.txt |
Because radiation is harmful to humans and other organisms, it is very important that we be able to detect it and measure how much is present. Such measurements are complicated by two factors. First, we cannot see, hear, smell, taste, or touch radiation, and so special instruments are required to measure it. Second, different types of radiation are more dangerous than others, and corrections must be made for the relative harm done by α particles as opposed to, say, γ rays.
Perhaps the most common instrument for measuring radiation levels is the Geiger-Müller counter (the same Geiger who worked with Rutherford to discover the atomic nucleus). A schematic diagram of a Geiger-Müller counter is shown in Figure \(1\). A metal tube containing Ar gas is sealed at one end with a thin glass or plastic window and contains a central wire well insulated from it. A potential difference of about 1000 V is applied between the central wire and the tube. Any incoming α, β or γ ray will ionize some of the Ar atoms. These Ar+ ions are quickly accelerated to a high velocity by the large potential difference, high enough for them in turn to start ionizing further Ar atoms. Thus, for every ray that enters the tube, a large number of ions is formed and a pulse of electrical current is produced. This pulse is amplified and allowed to drive a digital electronic counter which operates on a principle similar to that of a digital watch. The number of particles passing through the tube in a given time can thus be found. Alternatively, the tube can be made to operate a meter indicating the rate at which radiation is passing into the Geiger-Müller tube.
Another type of detector, much used for γ rays, is the scintillation counter. When a γ ray penetrates a special crystal or solution, it produces a momentary flash of light (called a scintillation) which is detected by a photoelectric cell. Again the output can be amplified and fed into a counter or a meter. A third kind of detector is used to monitor how much exposure laboratory workers have been subjected to in the course of their work. This is simply a strip of photographic film. The degree to which this film is darkened is a measure of the total quantity of radiation to which the worker has been subjected.
19.11: Units of Radiation Dose
A variety of units have been designed to measure how much radiation has been absorbed by a given sample of human or animal tissue. The simplest to understand is the radiation-absorbed dose, abbreviated rad. This corresponds to absorption of 10–5 J of energy per gram of tissue. A more useful unit is the rem (roentgen-equivalent man), which is the same as the rad except that it is corrected for the relative harmfulness of each type of radiation. For example, an α particle having a kinetic energy of 1.6 × 10–22 J can produce about 10 times as many ions as a γ ray of equal energy. Consequently 1 rad of α radiation would be corrected to 10 rem, while 1 rad of γ radiation would correspond to 1 rem. Once radiation detectors were developed, it was discovered that there is nowhere that one can be entirely free of radiation. That is, there is a natural background radiation impinging on all of us every day of our lives. This comes from natural radioactive isotopes in our surroundings and from cosmic radiation which enters the earth’s atmosphere from outer space. The average United States citizen receives just over 0.1 rem per year from natural background, although this varies from place to place. In Colorado, for example, background radiation is much higher because of the altitude (less atmosphere to block cosmic rays) and because of naturally occurring deposits of uranium.
Current estimates indicate that the actual radiation dose received by the average person is about 80 percent higher than natural background. The major portion of this increase is due to medical uses—a chest x-ray, for example, contributes about 0.2 rem. Other contributions are made by radioactive fallout from nuclear bombs (about 4 percent of background), and miscellaneous sources such as TV sets (about 2 percent).
There is evidence that the effects of small doses of radiation are cumulative, at least to some degree, and that there is no lower limit to the dose which can cause some damage. Thus even background radiation may be harmful to some extent, but it is hard to determine just how harmful because we have no way of turning it off to see how much difference it makes. In the absence of more accurate information it would seem prudent for each individual and for a society as a whole to minimize unnecessary radiation exposures. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/19%3A_Nuclear_Chemistry/19.10%3A_Instruments_for_Radiation_Detection.txt |
Tracers
A very large number of isotopes which do not occur naturally can now be made fairly readily by neutron capture using an atomic reactor or a laboratory neutron source. Many of these artificial isotopes have proved very useful in chemistry since they provide a way of identifying atoms from a particular source, a technique known as labeling or tracer study. This technique is particularly easy to use if the isotope is radioactive. Thus, for example, if a small quantity of the radioactive isotope 131I is added in the form of iodide ion to a saturated solution of lead iodide, one soon finds that the solid lead iodide in contact with the solution, as well as the solution, become radioactive. This clearly demonstrates that the solution equilibrium
$\text{PbI}_2 \rightleftharpoons \text{Pb}^{2+} + \text{2I}^{-} \nonumber$
is a dynamic process involving the constant interchange of iodide ions between the solution and the solid.
Tracer studies are also possible with isotopes which are not radioactive. The isotope 18O is often used in this way, since no convenient radioactive isotope of oxygen is available. In naturally occurring oxygen the isotope 18O is only 0.2 percent of the total. If extra 18O is added, its presence can be detected by mass spectrometry. An interesting and important example of the use of 18O is in the study of photosynthesis. If the water in this reaction is enriched with 18O, then the isotope is found in the oxygen produced:
$6 \text{CO}_{2} (g) + 6 \text{H}_{2} {\text{O}} (l) \rightarrow \text{C}_{6} \text{H}_{12} \text{O}_{6} (s) + 6 {\text{O}}_{2} (g) \nonumber$ By contrast, if the carbon dioxide is enriched with 18O, none of this enrichment appears in the oxygen produced. Another example of the use of 18O comes from inorganic chemistry. It is the reaction between the sulfite ion and the chlorate ion in aqueous solution:
By labeling the oxygens in the chlorate ion, it is found that all the 18O lost from the one species is gained by the other and none of it is transferred to the solvent water. The mechanism of this reaction must thus be a direct transfer of oxygen and is quite unrelated to the two half-equations we conventionally use when balancing the redox equation:
$\text{SO}_{3} \text{ }^{2-} + \text{H}_{2} \text{O} \rightarrow \text{SO}_{4} \text{ }^{2-} + 2\text{H}^{+} + 2e^{-} \nonumber$
$2 \text{H}^{+} + 2e^{-} + \text{ClO}_{3} \text{ }^{-} \rightarrow \text{ClO}_{2}^{-} + \text{H}_{2} \text{O} \nonumber$
Neutron Activation Analysis
An important use of radioisotopes in detecting small amounts of certain elements in a sample is neutron activation analysis. The sample being analyzed is irradiated by a neutron source. Nuclei of the element being analyzed capture neutrons, and an unstable nucleus is formed which emits a γ ray. Since the wavelength of this γ ray is characteristic of the element, it can be distinguished from other elements in the sample. This method of analysis has the advantage of being nondestructive. The sample being analyzed is scarcely altered by being irradiated. Neutron activation is also among the most sensitive of analytical techniques. As little as a pictogram (10–12 g) of arsenic, for example, can be detected. This is about 10 000 times more sensitive than Marsh’s test—so often used by the fabled detective Sherlock Holmes. Neutron activation analysis is used by many modern detectives to find evidence of air and water pollution as well as the types of crimes with which Holmes was involved. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/19%3A_Nuclear_Chemistry/19.12%3A_Uses_of_Artificial_Isotopes.txt |
In a nucleus the protons and the neutrons are held very tightly by forces whose nature is still imperfectly understood. When the nucleons are very close to each other, these "nuclear forces" are strong enough to counteract the Coulombic repulsion of the protons, but they fall off very rapidly with distance and are essentially undetectable outside the nucleus. Because the energies involved in binding the nucleons together are very large, they give rise to an effect which makes it possible to measure them. According to Einstein’s special theory of relativity, when the energy of a body increases, so does its mass, and vice versa. If the change in energy is indicated by ΔE and the change in mass by Δm, these two quantities are related by the equation
${\Delta\text{E}} = {\Delta\text{mc}}^2\label{1}$
where c is the velocity of light (2.998 × 108 m s–1).
In ordinary chemical reactions this change in mass with energy is so small as to be undetectable, but in nuclear reactions we invariably find that products and reactants have different masses. As a simple example, let us take the dissociation of a deuteron into a proton and a neutron:
${}_{\text{1}}^{\text{2}}\text{D }\to \text{ }{}_{\text{1}}^{\text{1}}p\text{ + }{}_{\text{0}}^{\text{1}}n \nonumber$
The molar mass of a deuteron is found experimentally to be 2.013 55 g mol–1 (Table $1$), but if we add the molar masses of a neutron and a proton, we obtain a somewhat higher value, namely, (1.007 28 + 1.008 67) g mol–1 = 2.015 95 g mol–1. The change in mass using the usual delta convention is thus (2.015 95 – 2.013 55) g mol–1 = 0.002 40 g mol–1. From Eq. $\ref{1}$ we then have
${\Delta\text{E}} = {\Delta\text{mc}}^2 = 0.002 40 \frac{\text{g}}{\text{mol}}\text{ }\times \text{ }\frac{\text{1 kg}}{\text{10}^{\text{3}}\text{ g}} \times \left(2.998 \times {\text{10}^{\text{8}}\frac{\text{m}}{\text{s}}} \right)^2 \nonumber$
= 2.16 × 1011 kg m2 s–2 mol–1 = 216 × 109 J mol–1 = 216 GJ mol–1
Table $1$:The Molar Masses of Some Selected Nuclei (Electrons Are Not Included in These Masses).
Nucleus M/g mol–1 Nucleus M/g mol–1
${}_{\text{0}}^{\text{1}}n$ (neutron) 1.008 67 ${}_{\text{26}}^{\text{56}}\text{Fe}$ 55.920 66
${}_{\text{1}}^{\text{1}}p$ (proton) 1.007 28 ${}_{\text{27}}^{\text{59}}\text{Co}$ 58.918 37
${}_{\text{1}}^{\text{2}}\text{D}$ (deuteron) 2.013 55 ${}_{\text{36}}^{\text{84}}\text{Kr}$ 83.8917
${}_{\text{1}}^{\text{3}}\text{T}$ (tritium) 3.015 50 ${}_{\text{50}}^{\text{120}}\text{Sn}$ 119.8747
${}_{\text{2}}^{\text{4}}\text{He}$ 4.001 50 ${}_{\text{56}}^{\text{138}}\text{Ba}$ 137.8743
${}_{\text{3}}^{\text{7}}\text{Li}$ 7.014 36 ${}_{\text{78}}^{\text{194}}\text{Pt}$ 193.9200
${}_{\text{6}}^{\text{12}}\text{C}$ 11.996 71 ${}_{\text{83}}^{\text{209}}\text{Bi}$ 208.9348
${}_{\text{8}}^{\text{15}}\text{O}$ 14.998 68 ${}_{\text{92}}^{\text{235}}\text{U}$ 234.9934
${}_{\text{8}}^{\text{16}}\text{O}$ 15.990 52 ${}_{\text{94}}^{\text{239}}\text{Pu}$ 239.0006
${}_{\text{8}}^{\text{17}}\text{O}$ 16.994 74
Since expansion work or even electronic energies are negligible compared to this change in nuclear energy, we can equate the change in nuclear energy either to the change in internal energy or the enthalpy; that is,
ΔE = ΔUm = ΔHm = 216 GJ mol–1
The energy needed to separate a nucleus into its constituent nucleons is called its binding energy. The binding energy of the 21H nucleus is thus 216 GJ mol–1. Notice how very much larger this is than the bond energy of an average molecule, which is about 200 or 300 kJ mol–1. Since a gigajoule is 1 million kJ, the energies involved in holding the nucleons together in a nucleus are something like a million times larger than those holding the atoms together in a molecule.
Since the number of nucleons in a nucleus is quite variable, it is useful to calculate the average energy of each nucleon by dividing the total binding energy by the number of nucleons, A. This gives the binding energy per nucleon. In the case of the nucleus 5626Fe, for instance, we can easily find from Table $1$ that Δm for the process
${}_{\text{26}}^{\text{56}}\text{Fe }\to \text{ 26}{}_{\text{1}}^{\text{1}}p\text{ + 30}{}_{\text{0}}^{\text{1}}n \nonumber$
has the value 0.528 72 g mol–1, giving a value for ΔHm, from Eq. (19.35)of 4750 GJ mol–1. Since A = 56 for this nucleus, the binding energy per nucleon has the value
$\frac{\Delta H_{m}}{A} = \frac{\text{4750}}{\text{56}} \nonumber$ GJ mol–1 = 848 GJ mol–1
The binding energy of a nucleus tells us not only how much energy must be expended in pulling the nuclei apart but also how much energy is released when the nucleus is formed from protons and neutrons. In the case of the 5626Fe, for instance, we have
$\text{26}{}_{\text{1}}^{\text{1}}p\text{ + 30}{}_{\text{0}}^{\text{1}}n\text{ }\to \text{ }{}_{\text{26}}^{\text{56}}\text{Fe} \nonumber$
ΔHm = –4.75 × 103 GJ mol–1
that is, the energy of formation of 5626Fe is equal to the negative of the binding energy. In Figure $1$ the energy of formation on a per nucleon basis has been against the mass number for the most stable isotope of each element. The zero energy axis in this plot corresponds to the energy of completely separated protons and neutrons, while the points on the graph correspond to the average energy of a nucleon in the nucleus in question. Obviously, the lower the energy, the more stable the nucleus.
As we can see from Figure $1$, the most stable nuclei are those of mass number close to 60, the nucleus with the lowest energy being the5626Fe nucleus. As the mass number rises above 60, the nuclei become slightly higher in energy, i.e., less stable. Decreasing the mass number below 60 also brings us into a region of high-energy nuclei. With the exception of the 42He, nucleus, the nuclei of highest energy belong to the very lightest elements.
Figure $1$ shows us that in principle there are two ways in which we can obtain energy from the nuclei of the elements. The first of these is by the splitting up or fission of a very heavy nucleus into two lighter nuclei. In such a case each nucleon will move from a situation of higher to lower energy and energy will be released. Even more energy will be released by the fusion of two very light nuclei, each containing only a few nucleons, into a single heavier nucleus. Though fine in principle, neither of these methods of obtaining energy is easy to achieve in practice in a controlled way with due respect to the environment.
19.13: Mass-Energy Relationships
Get an idea of the magnitude of binding energy by comparing the energy produced by exploding a hydrogen/oxygen mixture in a balloon with 400 mL hydrogen and 200 mL oxygen (about 0.04 g of hydrogen). Alternatively, soap bubbles with the mixture can be exploded. Caution: No bigger! Loud! Use hearing protectors.
With ΔE ~-125 kJ/g, this will give about 5 kJ. Compare to the binding energy for deuterium:
3.57 x 10-13J/atom x (1 atom / 2.14 amu) x (1 amu / 1.66 x 10-24 g)
= 1.0 x 1011 J/g
From ChemPRIME: 19.12: Mass-Energy Relationships | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/19%3A_Nuclear_Chemistry/19.13%3A_Mass-Energy_Relationships/19.13.01%3A_Lecture_Demonstrations_on_Mass_Energy_Relationships.txt |
The first time that nuclear fission was achieved in the laboratory was by the Italian physicist Enrico Fermi (1901 to 1954) in 1934. Fermi was among the first to use the neutron in nuclear reactions, following its discovery by Chadwick in 1932. He hoped, by bombarding uranium with slow neutrons, to be able to prepare the first transuranium element. Instead he obtained a product which seemed to be a group II element which he identified incorrectly as radium. It remained for the experienced German radiochemist Otto Hahn to correct Fermi’s mistake. (In the meantime Fermi had been awarded the Nobel Prize.) Somewhat reluctantly, Hahn published a paper early in 1939 showing that the element produced by the bombardment of uranium was not radium at all but the very much lighter group II element barium, 36 places earlier in the periodic table! It then became clear that instead of knocking a small chip off the uranium nucleus as had been expected, the bombarding neutron had broken the nucleus into two large fragments, one of which was barium. We now know that the initial step in this process is the formation of an unstable isotope of uranium which then fissions in a variety of ways, some of which are shown below:
Example 1: Energy of Fission
Make a rough estimate of the energy released by the fission of 1 g of uranium-235 according to the equation
${}_{\text{92}}^{\text{235}}\text{U + }{}_{\text{0}}^{\text{1}}n\text{ }\to \text{ }{}_{\text{56}}^{\text{140}}\text{Ba + }{}_{\text{36}}^{\text{93}}\text{Kr + 3}{}_{\text{0}}^{\text{1}}n$
using Figure $2$ :
Solution:
From Figure $2$ we can make the following estimates of the energies of formation per nucleon for the four species involved:
ΔHf (140Ba) = – 810 GJ mol1ΔHf (235U) = – 730 GJ mol–1
ΔHf (93Kr) = – 810 GJ mol1ΔHf (${}_{\text{0}}^{\text{1}}n$) = 0
Using these quantities in the same way as enthalpies of formation for chemical reactions, we obtain
ΔHm = [140(– 810) + 93(– 840) – 235(– 730)] GJ mol–1 = – 20 000 GJ mol–1
The enthalpy change per gram is then given by
ΔH = – 20 000 $\frac{\text{GJ}}{\text{mol}}$ × $\frac{\text{1 mol}}{\text{235 g}}$ = – 85 GJ g–1
Note: This is about the same quantity of heat energy as that produced by burning 3 tons of bituminous coal!
Calculations similar to that just performed soon persuaded scientists in 1939 that the fission of uranium was highly exothermic and could possibly be used in a super bomb. Adolph Hitler had been in power in Germany for 6 years, and Europe was teetering on the brink of World War II. The possibility that Nazi Germany might develop such a bomb and use it did not seem remote, especially to those scientists who had recently fled Nazi and Fascist Europe and come to the United States. Albert Einstein, himself one of these refugees, was persuaded to write a letter to President Franklin Roosevelt in August 1939 in which the alarming possibilities were outlined. Roosevelt heeded Einstein’s advice and established the so-called Manhattan Project, a super-secret research effort to develop an atomic bomb if at all possible. After 5 years of intense effort and the expenditure of more money than had ever been spent on a military-scientific project before, the first bomb was tested in New Mexico in July 1945. Shortly thereafter two atom bombs were dropped on the Japanese cities of Hiroshima and Nagasaki, and World War II ended almost immediately.
A crucial feature of the fission of uranium without which an atom bomb is impossible is that fission produces more neutrons than it consumes. As can be seen from Figure $2$, for every neutron captured by a 23592U nucleus, between two and four neutrons are produced. Suppose now that we have a very large sample of the pure isotope 23592U and a stray neutron enters this sample. As soon as it hits a 235U nucleus, fission will take place and about three neutrons will be produced. These in turn will fission three more 235U nuclei, producing a total of nine neutrons. A third repetition will produce 27 neutrons. a fourth 81. and so on. This process (which is called a chain reaction) escalates very rapidly. Within a few microseconds a very large number of nuclei fission, with the release of a tremendous amount of energy, and an atomic explosion results.
There are two reasons why a normal sample of uranium metal does not spontaneously explode in this way. In the first place natural uranium consists mainly of the isotope 23892U while the fissionable isotope 23592U comprises only 0.7 percent of the total. Most of the neutrons produced in a given fission process are captured by 23892U nuclei without any further production of neutrons. The escalation of the fission process thus becomes impossible. However, even a sample of pure 235U will not always explode spontaneously. If it is sufficiently small, many of the neutrons will escape into the surroundings without causing further fission. The sample must exceed a critical mass before an explosion results. In an atomic bomb several pieces of fissionable material, all of which are below the critical mass, are held sufficiently far apart for no chain reaction to occur. When these are suddenly brought together, an atomic explosion results immediately.
A great deal of the 5 years of the Manhattan Project was spent in separating the 0.7 percent of 235U from the more abundant 238U. This was done by preparing the gaseous compound UF6 and allowing it to effuse through a porous screen. (the "Kinetic Theory of Gases" sections discuss that the rate of effusion is inversely proportional to the square root of molar mass.) Each effusion resulted in a gas which was slightly richer in the lighter isotope. Repeating this process eventually produced a compound rich enough in 235U for the purposes of bomb manufacture.
Only the first bomb dropped on Japan used uranium. The second bomb used the artificial element plutonium, produced by the neutron bombardment of 238U:
${}_{\text{92}}^{\text{235}}\text{U + }{}_{\text{0}}^{\text{1}}n\text{ }\to \text{ }{}_{\text{94}}^{\text{239}}\text{Pu + 2}{}_{-\text{1}}^{\text{ 0}}e \nonumber$
Fission of 23984Po occurs in much the same way as for ${}_{\text{92}}^{\text{235}}\text{U}$, giving a variety of products; for example,
${}_{\text{94}}^{\text{239}}\text{Pu + }{}_{\text{0}}^{\text{1}}n\text{ }\to \text{ }{}_{\text{38}}^{\text{90}}\text{Sr + }{}_{\text{56}}^{\text{147}}\text{Ba + 3}{}_{\text{0}}^{\text{1}}n \nonumber$
Again this is a highly exothermic reaction yielding about the same energy per mole (20 000 GJ mol–1) as 235U.
19.14: Nuclear Fission
World War Two brought about many advances in weaponry. The Nazis used the immense science ability of their newly conquered countries to attempt to create super weapons to keep the new territory. The weapons included new tanks like the feared Tiger Tank, or the highly advanced Luftwaffe air force. However, the one weapon the Nazi's really wanted to hold was the atomic bomb. The atomic bomb would end the war. No country would be able to keep fighting after such a huge attack with a single weapon. The problem that the Nazi's began to have though was that many of the great minds they once had left the country and came to the United States. Also, in the end Allied troops got to Hitler and his scientists just before they could complete any atomic bombs. The U.S. also wanted to create an atom bomb, and for the same reasons. The Manhattan Project was started to form this new weapon.
The Manhattan Project was successful in the United States because of the vast resources that could be pulled together for the project. The great minds worked under the University of Chicago football field to try and create fission. A plant in Tennessee had the sole purpose of producing enough uranium for atomic bomb testing. Finally, New Mexico was the epicenter of the whole operation. This is where the bombs were constructed and tested. The resources needed were immense. There had to be enough fuel for the bomb, and everything required to make it. There had to be housing and food for all the scientists. Maybe most importantly, there had to be enough money to support the project. The Unites States was able to effectively use all of these resources, while the Germans did not have the resources, and didn't use them as effectively. Finally, after years of hard work an atomic bomb test took place of July 16, 1945. The bomb was held together in some parts by just tape, yet it was a huge success. The problem was that the war in Europe was already over, so President Truman had to decide whether or not to use the bomb on Japan.
The war in the Pacific was a lot different than the war in Europe. In Europe, you landed once and fought your way through to Berlin to defeat the Nazi's. However, in the Pacific Japan controlled hundreds of Islands before you got to the mainland. The United States had to battle through every little island before they were even close enough where bombers could reach Japan. Battling through every little island was extremely costly for U.S. Troops. Also, the U.S. had very little allied support, and the Japanese were ready to fight until every last soldier was killed or captured. This mentality brought about the idea of using the atomic bomb on Japan. If an invasion was used, it was estimated that there could be as many as 1 million casualties and the invasion could have lasted as long as 6 months. The invasion would have cost huge amounts of American lives and resources. As a result, when the bomb had a successful test President Truman said he hardly hesitated to approve of the usage of the atomic bomb on Japan.
The first atomic bomb was dropped on Hiroshima on August 6, 1945 by the Enola Gay. The bomb destroyed the city, killing almost 100,000 people. The city of Hiroshima was essentially leveled. Following this attack, on August 9, 1945 another atomic bomb was dropped on Nagasaki. The same effects were felt in this city. The United States was ready to drop another atomic bomb in the third week of August, 3 more in September, and 3 more in October if necessary. However, on August 15 Japan surrendered to the allies and World War Two was officially over.
The Workings of the Atomic Bomb
A crucial feature of the fission of uranium without which an atom bomb is impossible is that fission produces more neutrons than it consumes. As can be seen from Eqs. (1), for every neutron captured by a $\ce{^{235}_{92}U}$ nucleus, between two and four neutrons are produced. Suppose now that we have a very large sample of the pure $\ce{^{235}_{92}U}$ isotope and a stray neutron enters this sample. As soon as it hits a $\ce{^{235}_{92}U}$ nucleus, fission will take place and about three neutrons will be produced. These in turn will fission three more 235U nuclei, producing a total of nine neutrons. A third repetition will produce 27 neutrons. a fourth 81. and so on. This process (which is called a chain reaction) escalates very rapidly. Within a few microseconds a very large number of nuclei fission, with the release of a tremendous amount of energy, and an atomic explosion results.
)
There are two reasons why a normal sample of uranium metal does not spontaneously explode in this way. In the first place natural uranium consists mainly of the isotope $\ce{^{238}_{92}U}$, while the fissionable isotope $\ce{^{235}_{92}U}$ comprises only 0.7 percent of the total. Most of the neutrons produced in a given fission process are captured by $\ce{^{238}_{92}U}$ nuclei without any further production of neutrons. The escalation of the fission process thus becomes impossible. However, even a sample of pure 235U will not always explode spontaneously. If it is sufficiently small, many of the neutrons will escape into the surroundings without causing further fission. The sample must exceed a critical mass before an explosion results. In an atomic bomb several pieces of fissionable material, all of which are below the critical mass, are held sufficiently far apart for no chain reaction to occur. When these are suddenly brought together, an atomic explosion results immediately.
A great deal of the five years of the Manhattan Project was spent in separating the 0.7 percent of 235U from the more abundant 238U. This was done by preparing the gaseous compound UF6 and allowing it to effuse through a porous screen. (the "Kinetic Theory of Gases" sections discuss that the rate of effusion is inversely proportional to the square root of molar mass.) Each effusion resulted in a gas which was slightly richer in the lighter isotope. Repeating this process eventually produced a compound rich enough in 235U for the purposes of bomb manufacture.
Only the first bomb dropped on Japan used uranium. The second bomb used the artificial element plutonium, produced by the neutron bombardment of 238U:
$\ce{^{235}_{92} U + ^{1}_0n \rightarrow ^{239}_{94}Pu + 2 ^{0}_{-1}\beta } \nonumber$
Fission of Pu-239 occurs in much the same way as for U-235, giving a variety of products; for example,
$\ce{^{239}_{94} Pu + ^{1}_0n \rightarrow ^{90}_{38}Pu + ^{147}_{56}Ba + 3 ^{1}_{0}n } \nonumber$
Again this is a highly exothermic reaction yielding about the same energy per mole (20 000 GJ mol–1) as 235U.
This immense amount of energy released is the reason why plutonium and uranium were used. By comparison, 1 ton of TNT yields about 4 GJ mol–1 of energy. This means that the atomic bombs had about 5,000 times more energy than 1 ton of TNT, that is a huge amount of energy, and explains why the bomb was so effective and destructive.
We take a look at the size and power of explosions created from a hand grenade to the Tsar Bomba and beyond. Its not just nukes but supervolcano eruptions & meteor impacts as well. It is indeed a terrifying true scale of nuclear weapons.
Although the Fat Man atomic bomb was extremely powerful and destructive, it is not even close to the power of subsequent nuclear bombs. The atomic age was set forth by the United States on those two fateful days in August. In the subsequent years many other countries harnessed the power of nuclear weapons, including Russia, China, Britain, and France. The "highpoint" of the atomic age came with the explosion of the Tsar Bomba by Russia. This bomb was equal to 50 megatons of TNT, or 2,500 times more powerful than the Fat Man bomb. It was said that when the bomb was detonated the shock wave in the ground could be felt over 600 miles away (the bomb wasn't even detonated on the ground either, but rather in mid air!). The advance and use of nuclear weapons seemed to be growing at an extraordinarily fast pace. However, countries stepped up and signed various treaties, including the Nuclear Non-Proliferation Treaty, to stop the increase in nuclear weapons and hopefully the use of any more in the future. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/19%3A_Nuclear_Chemistry/19.14%3A_Nuclear_Fission/19.14.01%3A_Nuclear_Fission_and_WWII.txt |
Even before the atomic bomb had been produced, scientists and engineers had begun to think about the possibility of using the energy released by the fission process for the production of electrical energy. Shortly after World War II confident predictions were made that human beings would soon depend almost entirely on atomic energy for electricity. Alas, we are now years into the future from then and no such miracle has occurred. In the United States only 8.46 percent of the electrical energy in 2008 was produced by this method[1]. The proportion is a little higher in some other countries, notably Great Britain, but nowhere is nuclear power even on the verge of replacing the fossil fuels. The unfortunate truth is that producing power from atomic fission has turned out to be much more expensive than was previously expected. Even in these days of high prices for the fossil fuels it is still only barely competitive.
A schematic diagram of a typical nuclear reactor is given in Figure \(1\). The uranium is present in the form of pellets of the oxide U3O8 enclosed in long steel tubes about 2 cm in diameter. The uranium is23892U mainly, slightly enriched with the fissionable 23592U. The rate of fission can be regulated by inserting or withdrawing control rods made of cadmium, which is a very efficient neutron absorber. In addition a moderator such as graphite or water must be present to slow down the neutrons, since slow neutrons are more efficient at causing fission than fast ones.
The energy released by the fission of the uranium is carried off by a coolant, usually superheated steam at about 320°C. This steam cannot be used directly since it becomes slightly radioactive. Instead it is passed through a heat exchanger so as to produce further steam which can then be used to power a conventional steam turbine. The whole system is enclosed in a strong containment vessel which is not shown in the figure. This vessel prevents the spread of radioactivity in case of a serious accident.
Nuclear power plants have two advantages over conventional power plants using fossil fuels. First, for a given energy output they consume much less fuel. Second, they produce far smaller quantities of toxic effluents. Fossil-fueled plants produce sulfur dioxide, oxides of nitrogen, and smoke particles, all of which are injurious to health.
Despite the much lower cost for fuel, nuclear power plants are very expensive to build. This is largely because of their chief disadvantage, the extremely dangerous nature of the radioactive products of nuclear fission. Fission products consist of a great many neutron-rich, unstable nuclei, ranging in atomic number from 25 to 60. Particularly dangerous are the long-lived isotopes 9038Sr, 13755Cs, and the shorter-lived 13153I, all of which can be incorporated into the human body. Extreme precautions must be taken against accidental release of even traces of these materials into the environment. The worst case scenario is a reactor meltdown, of which the best known is the Chernobyl disaster, occurring in 1986 in Ukraine, then still a part of the Soviet Union. Recent study on the event estimates eventual deaths caused by the accident in the higher exposed populations as 4000 [2]. The overall cost of the disaster is of course difficult to quantify in terms of health, psychological, economic and environmental effects.
It should be realized that there are significant differences between a Hiroshima-type bomb and this sort of meltdown, primarily a difference in destructive explosive power. These differences arises because fuel used in nuclear reactors is not rich enough in23592U for the chain reaction to produce an atomic bomb-like explosion. It should also be noted that the safety deficiencies which caused the Chernobyl did not reflect the safety precautions taken in in nuclear plants today, or in other plants at the time. Regardless, the Chernobyl event, along with the earlier Three Mile Island incident in 1979, helped to start a decline in the use and building of nuclear power plants worldwide, and many people remain wary of nuclear power. Interest in nuclear power has renewed with discussions of national and world energy strategies, however.
Even if fission products are handled successfully during normal operation of a nuclear plant, there still remains the difficulty of their eventual disposal. Although many of the unstable nuclei produced by fission are short-lived, some, like9038Sr (25 years) and 13755Cs (30 years), have quite long half-lives. Accordingly these wastes must be stored for many hundreds of years before enough nuclei decompose to reduce their radioactivity to a safe level. At the present time, spent nuclear fuel is stored in spent fuel water pools at reaction sites, which shields the surrounding environment. Spent fuel may also be placed in dry cask storage after fuel has been stored in a spent fuel pool for at least a year. The dry casks contain inert gas, are made of steel, and may be surrounded by steel and concrete to prevent leakage of radiation [3].
Both spent fuel pools and dry cask storage are only short term solutions. One proposed long term solution is to store spent nuclear waste safely in a geological repository deep underground. In 1982, the Nuclear Waste Policy Act[4] tasked the United States Department of Energy with finding and constructing a national repository for long term storage of nuclear waste. In 1987, the Department of Energy was directed to focus solely on Yucca Mountain, Nevada, as the site to develop a national repository. Originally set to start accepting waste in 1998, the opening of the repository has been greatly delayed, often due to a large amount of unresolved debate on the topic.. Nevertheless, in 2002, the Department of Energy determined the site acceptable, and began application to use the site[5]
The Yucca Mountain Repository is currently out of favor. With shrinking budgets along with safety concerns about the site, the Department of Energy motioned to withdraw the Yucca Mountain site as a long term repository for nuclear waste in 2010[6]. The debacle over Yucca Mountain highlights both the scientific and political difficulties in finding a long term solution to spent nuclear fuels, as well as the difficulties involved in using nuclear power in general. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/19%3A_Nuclear_Chemistry/19.15%3A_Nuclear_Power_Plants.txt |
Because $\ce{_{92}^{235}U}$ is only 0.7 percent of naturally occurring uranium, its supply is fairly limited and could well only last for about 50 years of full-scale use. The other 99 percent of the uranium can also be utilized if it is first converted into plutonium by neutron bombardment
$\ce{_{92}^{238}U + _{0}^{1}n -> _{94}^{239}Pu + 2 _{-1}^{0} e} \nonumber$
$\ce{_{94}^{239}Pu}$ is also fissionable, and so it could be used in a nuclear reactor as well as $\ce{_{92}^{235}U}$.
The production of plutonium can be carried out in a breeder reactor which not only produces energy like other reactors but is designed to allow some of the fast neutrons to bombard the $\ce{_{92}^{235}U}$, producing plutonium at the same time. More fuel is then produced than is consumed.
Breeder reactors present additional safety hazards to those already outlined. They operate at higher temperatures and use very reactive liquid metals such as sodium in their cooling systems, and so the possibility of a serious accident is higher. In addition the large quantities of plutonium which would be produced in a breeder economy would have to be carefully safeguarded. Plutonium is an α emitter and is very dangerous if taken internally. Its half-life is 24,000 years, and so it will remain in the environment for a long time if dispersed. Moreover, $\ce{_{94}^{239} Pu}$ can be separated chemically (not by the much more expensive gaseous diffusion used to concentrate $\ce{_{92}^{ 235}U}$ from fission products and used to make bombs. Such a material will obviously be attractive to terrorist groups, as well as to countries which are not currently capable of producing their own atomic weapons.
19.17: Nuclear Fusion
In addition to fission, a second possible method for obtaining energy from nuclear reactions lies in the fusing together of two light nuclei to form a heavier nucleus. As we see when discussing Figure 1 from Mass-Energy Relationships, such a process results in nucleons which are more firmly bonded to each other, and hence lower in potential energy. This is particularly true if ${}_{\text{2}}^{\text{4}}\text{He}$ is formed, because this nucleus is very stable. Such a reaction occurs between the nuclei of the two heavy isotopes of hydrogen, deuterium and tritium:
${}_{\text{1}}^{\text{2}}\text{D + }{}_{\text{1}}^{\text{3}}\text{T }\to \text{ }{}_{\text{2}}^{\text{4}}\text{He + }{}_{\text{0}}^{\text{1}}n \label{1}$
For this reaction, Δm = – 0.018 88 g mol–1 so that ΔHm = – 1700 GJ mol–1. Although very large quantities of energy are released by a reaction like Equation $\ref{1}$, such a reaction is very difficult to achieve in practice. This is because of the very high activation energy, about 30 GJ mol–1, which must be overcome to bring the nuclei close enough to fuse together. This barrier is created by coulombic repulsion between the positively charged nuclei. The only place where scientists have succeeded in producing fusion reactions on a large scale is in a hydrogen bomb. Here, the necessary activation energy is achieved by exploding a fission bomb to heat the reactants to a temperature of about 108 K. Attempts to carry out fusion in a more controlled way have met only limited success. At the very high temperatures required, all molecules dissociate and most atoms ionize. A new state of matter called a plasma is formed. It is neither solid, liquid, nor gas. Plasma behaves much like the universal solvent of the alchemists by converting any solid material that it contacts into vapor.
Two techniques for producing a controlled fusion reaction are currently being explored. The first is to restrict the plasma by means of a strong magnetic field, rather than the walls of a container. This has met some success, but has not yet been able to contain a plasma long enough for usable energy to be obtained. The second technique involves the sudden compression and heating of pellets of deuterium and tritium by means of a sharply focused laser beam. Again, only limited success has been obtained.
Though these attempts at a controlled fusion reaction have so far been only partially successful, they are nevertheless worth pursuing. Because of the much readier availability of lighter isotopes necessary for fusion, as opposed to the much rarer heavier isotopes required for fission, controlled nuclear fusion would offer the human race an essentially limitless supply of energy. There would still be some environmental difficulties with the production of isotopes such as tritium, but these would be nowhere near the seriousness of the problem caused by the production of the witches brew of radioactive isotopes in a fission reactor. It must be confessed, though, that at the present rate of progress, the prospect of limitless clean energy from fusion seems unlikely in the next decade or two. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/19%3A_Nuclear_Chemistry/19.16%3A_Breeder_Reactors.txt |
It was thought that organic compounds could only be manufactured in living organisms, and chemistry was divided into the subfields of inorganic and organic on this basis. This subdivision persists today, but the definition of organic has changed in response to the discovery of numerous ways to make organic compounds from inorganic starting materials. Biochemistry is the study of chemical elements found in living systems, and how these elements combine to form molecules and collections of molecules which carry out the biological functions and behaviors that we associate with life.
• 20.1: Prelude to Biochemistry
• 20.2: The Elements of Life
• 20.3: The Building Blocks of Biochemistry
Fortunately nearly all the substances found in living cells are polymeric—they are built up by different combinations of a limited number of relatively small molecules. For example, the basic structures of all proteins in all organisms consist of covalently linked chains containing 100 or more amino acid residues. Only 20 different amino acids are commonly incorporated in proteins, but the number of ways of arranging 100 of these in a chain taking any of the amino acids at random for each place
• 20.4: Fats and Lipids
Nonpolar lipids have molecular structures which contain no electrically charged sites, few polar groups, and large amounts of carbon and hydrogen. They are similar to hydrocarbons in being almost completely insoluble in water, and so they are said to be hydrophobic (from the Greek, meaning water-hater). On the other hand, polar lipids consist of molecules which have polar groups (such as —OH) or electrically charged sites at one end, and hydrocarbon chains at the other.
• 20.5: Nonpolar Lipids
A great many nonpolar lipids can be made by combining different long- chain acids with glycerol. Because these acids were originally derived from fats, they are collectively referred to as fatty acids. This most-common form of animal fat serves as a storehouse for energy and as insulation against heat loss. On a molecular level it is constructed from three molecules of stearic acid and one of glycerol.
• 20.6: Polar Lipids
As was true of most nonpolar lipids, the structures of polar lipids are based on condensation of fatty acids with glycerol. The main difference is that only two of the three OH groups on glycerol are involved.
• 20.7: Carbohydrates
Carbohydrates are sugars and sugar derivatives whose formulas can be written in the general form: Cx(H2O)y. (The subscripts x and y are whole numbers.) Some typical carbohydrates are sucrose (ordinary cane sugar), C12H22O11; glucose (dextrose), C6H12O6; fructose (fruit sugar), C6H12O6; and ribose, C5H10O5.Since the atom ratio H/O is 2/1 in each formula, these compounds were originally thought to be hydrates of carbon, hence their general name.
• 20.8: Simple Sugars
• 20.9: Disaccharides
Disaccharides are made up by condensing two sugar units.
• 20.10: Polysaccharides
As the name suggests, polysaccharides are substances built up by the condensation of a very large number of monosaccharide units. Cellulose, for example, is a polymer of β-glucose, containing upwards of 3000 glucose units in a chain. Starch is largely a polymer of α-glucose.
• 20.11: Proteins
• 20.12: Polypeptide Chains
The backbone of any protein molecule is a polypeptide chain obtained by the condensation of a large number of amino acids with the elimination of water. You will recall that the amino acids are bifunctional organic nitrogen compounds containing an acid group, —COOH, and an amine group, —NH2. The amine group is attached to the carbon atom adjacent to the —COOH (the α carbon atom).
• 20.13: The Amino Acids
Altogether there are 20 amino acids which commonly occur in all organisms. Under most circumstances amino acids exist as zwitterions.
• 20.14: Primary Protein Structure
• 20.15: Secondary Protein Structure
• 20.16: Higher-Order Structure
• 20.17: Nucleic Acids
• 20.18: Nucleic Acid Structure
Nucleic acids were first isolated from cell nuclei (hence the name) in 1870. Since then they have been found in other portions of cells as well, especially the ribosomes, which are the sites of protein synthesis. Most nucleic acids are extremely long-chain polymers.
• 20.19: Information Storage
ow can DNA and RNA molecules act as blueprints for the manufacture of proteins? Each amino acid in a protein is determined by a specific codon of three nitrogenous bases in the DNA or RNA chain. The details of this genetic code are discussed.
• 20.20: The Double Helix
There is more to the structure of DNA than just the primary sequence of nitrogenous bases. Secondary structure also plays a crucial biochemical role. Each DNA molecule consists of two nucleotide chains wrapped around each other in a double helix and held together by hydrogen bonds. This hydrogen bonding involves only the nitrogenous bases. Each of the purine bases can hydrogen bond with one and only one of the pyrimidine bases.
• 20.21: DNA Replication
DNA is the genetically active component of the chromosomes of a cell and contains all the information necessary to control synthesis of the proteins, enzymes, and other molecules which are needed as that cell grows, carries on metabolism, and eventually reproduces. Thus when a cell divides, its DNA must pass on genetic information to both daughter cells. It must somehow be able to divide into duplicate copies. This process is called replication.
• 20.22: Transcription and Translation
20: Molecules in Living Systems
Most of us have little difficulty distinguishing living organisms from inanimate matter. The former are capable of reproducing nearly exact copies of themselves; they can appropriate both matter and energy from their surroundings, moving, growing, and repairing damage caused by external factors; and groups of them evolve and adapt in response to long-term environmental changes. On a macroscopic scale the differences are sufficiently striking that early philosophers and scientists postulated the existence of a vital force without which living organisms would be inanimate. It was thought that organic compounds could only be manufactured in living organisms, and chemistry was divided into the subfields of inorganic and organic on this basis.
This subdivision persists today, but the definition of organic has changed in response to the discovery of numerous ways to make organic compounds from inorganic starting materials. As seen in the sections on organic compounds, organic chemistry now means “chemistry of compounds containing carbon.” No restriction is placed on the origin of the compounds, and hundreds of thousands of organic compounds which are foreign to all living systems have been produced in laboratories around the world. Indeed, concern about the effects of some of these synthetic substances on the environment has led to yet another definition of organic. The general public now takes it to mean “free of substances produced as a result of human activities.”
Just as the division between organic and inorganic chemistry has become more arbitrary with the advance of knowledge, the distinction between life and nonlife has also blurred. Living organisms are made up of atoms and molecules which follow the same chemical principles as any other set of atoms and molecules. Yet there is a difference-these atoms, molecules, and even groups of molecules are organized to a much greater degree than in any of the cases we have discussed. Above a certain level of complexity a collection of chemicals begins to exhibit most of the behavior patterns that we associate with life. A virus, for example, may consist of fewer than 100 associated large molecules. It is the structures of these molecules and the ways in which they are associated that determine a virus’ behavior and make it appear to be on the threshold of life.
Biochemistry is the study of chemical elements found in living systems, and how these elements combine to form molecules and collections of molecules which carry out the biological functions and behaviors that we associate with life. Our treatment must of necessity be brief, but even if it were not, the complexity of biochemical systems would insure that it would be incomplete. Much of modern chemistry, both inorganic and organic, involves the extension to complex biological systems of principles and facts gleaned from studies of more general chemical behavior.
More than 99 percent of the atoms in most living organisms are H, O, N, or C. These are the smallest atoms which can form one, two, three, and four covalent bonds, respectively, and they are especially suited to make up the more than 1011 different, but often related, kinds of molecules estimated to exist in the biosphere. Each such molecule has a specific function to perform in a specific organism, and its molecular structure is very important in determining how that function is carried out.
Many biological molecules are formed by condensation polymerization from small building-block molecules. Reversal of such condensations (hydrolysis) breaks large molecules down into the building blocks again, allowing them to be used by another organism. Examples of condensation reactions are the formation of lipids from fatty acids and glycerol, the formation of cellulose and starch from glucose, the formation of proteins from amino acids, and the formation of nucleic acids from ribose or deoxyribose, phosphate, and nitrogenous bases.
Lipids may be divided into two categories: nonpolar (hydrophobic) and polar (hydrophilic). Both types consist of long hydrocarbon chains, but polar lipids have electrically charged or hydrogen-bonding groups at one end. Lipid bilayers, in which the hydrophilic ends of polar lipids contact an aqueous phase while the hydrophobic tails intertwine, are important components of cell walls and other membranes. Nonpolar substances often dissolve in the hydrophobic portions of lipid tissue and may be concentrated along ecological food chains, a process referred to as bioamplification.
Carbohydrates provide a storehouse for solar energy absorbed during photosynthesis. Simple sugars usually contain five or six C atoms in a ring and a large number of OH groups. Disaccharides are formed from a condensation reaction between two simple sugars. Polysaccharides, such as cellulose and starch, are condensation polymers of simple sugars. Their structures and chemical reactivities are very dependent on the exact structure of the simple sugar from which they are made.
Fibrous proteins, in which polypeptide chains are arranged parallel to one another, are fundamental components of structural tissues such as tendons, hair, etc. Globular proteins, on the other hand, have compact, nearly spherical structures in which the polypeptide chain folds back on itself. Enzymes, antibodies, hormones, and hemoglobin are examples of globular proteins. Proteins are made by polymerization of 20 different amino acids. The order of amino acid side chains along the polymer backbone constitutes primary protein structure. Secondary structure involves hydrogen bonding to form α-helix or pleated-sheet structures. The intricate folding of the polypeptide chain in a globular protein is referred to as tertiary structure. Some proteins (hemoglobin, for example) have quaternary structure— several polypeptide chains are nested together.
Nucleic acids, such as DNA and RNA and formed from nitrogenous bases, sugars and phosphate, constitute a blueprint and a mechanism for synthesizing useful proteins. Codons, each consisting of a sequence of three nitrogenous bases along a nucleic acid polymer chain, indicate which amino acid goes where. DNA also has secondary structure, the double helix. When cells divide, the double-stranded DNA molecule can replicate itself, and complementary base pairing insures that each new cell will contain identical DNA. During protein synthesis, information is transcribed from DNA to mRNA and then translated from the mRNA code into a protein. In the translation process, tRNA molecules bonded to specific amino acids, base pair their anticodon sequence with a codon sequence on the mRNA. The ribosome, itself constituted of RNA and protein performs the catalytic activity of synthesizing the protein. In this way the primary protein structure is determined, and secondary and higher order structures then follow directly. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/20%3A_Molecules_in_Living_Systems/20.01%3A_Prelude_to_Biochemistry.txt |
Hardly a year goes by without the appearance in the media of the old cliché that the average human body is worth about \$3.57 (perhaps a little more, allowing for inflation). Such figures are based on the elemental composition of the human body reported in the table.
Table \(1\) Elemental Composition of the Human Body.
Element
Weight %
Atom %
Period
H
10.2
63.5
I
O
66.0
25.6
II
C
17.5
9.1
II
N
2.4
1.06
II
Subtotal
96.1
99.3
Ca
1.6
0.25
IV
P
0.9
0.18
III
Na
0.3
0.07
III
K
0.4
0.06
IV
Cl
0.3
0.05
III
S
0.2
0.04
III
Mg
0.05
0.01
III
Subtotal
99.85
99.92
Fe
0.005
0.0006
IV
Zn
0.002
0.0002
IV
Cu
0.0004
0.000 06
IV
Mn
0.000 05
0.000 006
IV
Considering just the market value of the elements, one obtains a ridiculously low price. What has been ignored, of course, is how the atoms of those elements are put together. The raw materials for a fine watch are not worth much either-what we pay for is mainly the skill and intelligence with which they are combined.
Before considering how the elements from which we all are made have been combined, however, it is worth thinking a little about why those particular elements in the table are involved. This is especially important because the same elements in very similar ratios are found in nearly all living systems. As shown in the biological periodic table (Figure \(1\) ) a great many other elements are simply not involved in the chemistry of life-at least not to the extent that the necessity of their presence can be demonstrated experimentally.
Figure \(1\) The biological periodic table.
More than 99 percent of the atoms in the human body, or any other organism for that matter, are H, O, N, and C. This does not appear to be mere happenstance. The indications are that life as we know it could not be based on any other four elements. These elements have the lightest and smallest atoms which can form one, two, three, and four covalent bonds, respectively. Because of their small radii these atoms can approach very closely and bond very strongly to each other. The most important of these four elements is undoubtedly carbon. Carbon atoms have a special capacity for binding with each other to form long chains and rings. This capacity allows carbon to form a very large number of stable compounds whose molecular structures are different but nevertheless closely related.
No other element has this capacity. Perhaps the closest contender is silicon. Although a number of compounds containing Si—Si bonds such as
are known, none of these compounds are very stable. All are readily converted to compounds containing Si—O bonds. The bond enthalpy of the bond (368 kJ mol–1) is much larger than that of the Si—Si bond (176 kJ mol–1) and also larger than that of the Si—H bond (318 kJ mol–1). The production of Si—O bonds is thus exothermic and thermodynamically favorable. Another factor is the rapid rate at which these chemical reactions can occur. Silicon is capable of expanding its valence shell through the use of d orbitals to allow more than four bonds. This enables it to form an activated complex in which both the bond being made and the bond being broken feature, but which requires very little activation energy. Equivalent reactions involving carbon require very much higher activation energies and usually proceed so slowly as to be imperceptible at room temperature, even when thermodynamically permitted.
Carbon is not the only element with unique properties in biological molecules. Hydrogen is also special and plays two important roles. You will recall that alkanes are chemically unreactive. We can attribute this to the large C—H bond enthalpy of 413 kJ mol–1. Only fluorine forms a stronger bond than this with carbon. It is thus quite difficult to replace a hydrogen atom attached to a carbon atom with a more stable alternative. When an organic compound reacts chemically, it is almost always a functional group which undergoes a change. The presence of C—H bonds renders a large proportion of the carbon chain unreactive and restricts reaction almost entirely to those sites which include an atom other than carbon or hydrogen.
In biological molecules these other atoms are almost invariably oxygen and nitrogen, and this is no accident either. Oxygen and nitrogen are the two most electronegative elements which have a valence greater than 1. They are able to form bonds both to carbon on the one hand and to hydrogen on the other. In groups like
and
hydrogen atoms fulfill their second important role in biological molecules—forming hydrogen bonds between different molecules or between different parts of the same molecule.
This ability of different functional groups on a carbon chain to hydrogen bond with each other is a particularly important aspect of biological molecules. You will recall from figure 2 in the section on molecular equilibrium that a molecule containing a chain of carbon atoms is capable of a very large number of conformations due to free rotation around the single bonds. Since many biological molecules contain very long chains, they are capable of adopting an almost infinite number of shapes. In practice only a few of these shapes are useful, and the molecules can be "frozen" into such a useful conformation through hydrogen-bonding between various segments of the chain. A very good example of this is the enzyme trypsin. If the amino acid chain in this molecule were not held in the particular conformation shown in the figure by means of hydrogen bonds between adjacent parts of the chain, the various segments making up the active site would no longer be grouped together and the molecule would be unable to function as a catalyst. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/20%3A_Molecules_in_Living_Systems/20.02%3A_The_Elements_of_Life.txt |
It has been estimated that even a unicellular organism may contain as many as 5000 different substances, and the human body probably has well over 5 000 000. Only a few of these are exactly the same in both species, and so the total number of different compounds in the living portion of the earth (the biosphere) is approximately (105 compounds/species) × 106 species = 1011 compounds. If it were possible for chemists to synthesize one of these every second, 24 hours a day, 7 days a week, about 3000 years would be required to make all of them—obviously a hopeless task, even if we knew the composition and structure of each one.
How then can we make sense out of the chemistry of living systems? Fortunately nearly all the substances found in living cells are polymeric—they are built up by different combinations of a limited number of relatively small molecules. For example, the basic structures of all proteins in all organisms consist of covalently linked chains containing 100 or more amino acid residues. Only 20 different amino acids are commonly incorporated in proteins, but the number of ways of arranging 100 of these in a chain taking any of the amino acids at random for each place in the chain is 20100 \(\cong\) 10130, allowing an almost infinite variety of structures. Just as an understanding of the properties of atoms and their bonding characteristics was a significant aid in predicting the chemistry of molecules, a knowledge of the properties of simple molecular building blocks gives us a starting point for the study of biochemistry. Each of the building blocks and their polymeric forms has at least one major role to play in the chemistry of life. Most are quite versatile, serving several functions. Let us take a look at the building blocks of biochemistry:
Sugars, or carbohydrates are molecules that follow the form Cx(H2O)y. A simple sugar can serve as an energy source for an organism. Simple sugars can dimerize into disaccharides or polymerize into Polysaccharides. Uses for polysacharides range from energy storage, such as glycogen stored in your liver and muscles, to structural support, such as cellulose that makes up cell walls for plants.
glucose, a monosaccharide sucrose, a disaccharide glycogen a polysaccharide
Amino Acids as mentioned earlier are a class of 20 molecules that polymerize to form proteins. Proteins take on a large variety of roles, from catalyzing important reactions as enzymes, to providing structural support, such as collagen, to serving as hormone or neurotransmitter signals.
L-alanine, an amino acid A protein, a blue GPF varient, shown in stick form. Ribbon view of the same protein.
Nucleotides are a slightly more complex class of biological molecule, consisting of a phosphate group, a ribose or deoxyribiose sugar, and a nitrogenous base. Nucleotides, such as ATP serve as carriers of energy in a cell and provide energy to run processes that would otherwise be non-spontaneous. Nucleotides are able to polymerize into nucleic acids by forming a phosphodiester bond between the phosphate group of one nucleotide and a hydroxyl group on the ribose or deoxyribose of another nucleotide. The polymeric nucleic acids are able to store information for building proteins.
dATP, a nucleotide B-DNA the polymer and a nucleic acid.
Fatty Acids can already be considered somewhat polymeric, as they consist of a end carboxylic acid followed by a hydrocarbon chain of varying lengths. Three fatty acids can be attached to glycerol to form triglycerides. Other chemical groups can be added to the three hydroxyl groups of glycerol, thus yielding many chemical variations, which along with other non-polar chemicals are all included in the umbrella category of fats and lipids. Functions of these polymers again range from energy storage, to forming membranes necessary to structure, to signaling.
palmitic acid, a fatty acid. A triacylglycerol.
Salts can be thought of as a polymers when considering the ionic crystal lattice. This can be applied to extracellular structures such as shells, teeth and bones. In cells, salts serve roles as ions in solution. Effects in this "monomeric" form include Ca2+ signaling, setting up electrochemical gradients to do work, or influencing osmotic properties.
Water polymerizes in ice formation. Much of this structure still remains in liquid form, and hydrogen bonding networks still remain. So not only does water provide the solvent medium for all chemical reactions in cells, its structure dictates the way proteins fold, and how membranes form. It also maintains rigidity of cell walls, and serves as a thermoregulator. By virtue of its large polarity water is a good solvent for ionic substances and therefore provides a means of transporting inorganic nutrients such as NH4+, NO3, CO32, PO43, and monatomic ions throughout higher organisms. Its ability to dissolve a wide variety of substances also makes it useful for disposing of wastes. Many of the human body’s defense mechanisms against external toxic substances involve conversion into water-soluble forms and elimination via urine.
A water molecule. Ability to hydrogen bond allows for proton transfer, and confers structural elements to liquid water.
Vitamins and trace elements, while not as prevalent as the chemical compounds discussed above, trace elements, such as zinc, copper, and iron are highly important to biological function, often serving key structural and chemical purposes in protein function. Vitamins are organic compounds that often augment proteins to help with biological functions.
Vitamin B1, also known as thiamine. This organic molecule helps with the enzymatic activity of the protein pyruvate dehydrogenase complex.
It is clear from this overview that both monomeric and polymeric forms of the chemical components of chemicals in living systems serve important functions. Also, those functions change depending on which building blocks are used for polymerization. However, the properties of these polymers can be understood in terms of the monomers and how they combine. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/20%3A_Molecules_in_Living_Systems/20.03%3A_The_Building_Blocks_of_Biochemistry.txt |
The term lipid applies to any water-insoluble substance which can be extracted from cells by organic solvents such as chloroform, ether, or benzene. Two major categories may be identified. Nonpolar lipids have molecular structures which contain no electrically charged sites, few polar groups, and large amounts of carbon and hydrogen. They are similar to hydrocarbons in being almost completely insoluble in water, and so they are said to be hydrophobic (from the Greek, meaning water-hater). On the other hand, polar lipids consist of molecules which have polar groups (such as —OH) or electrically charged sites at one end, and hydrocarbon chains at the other. Since polar or charged groups can hydrogen bond to or electrostatically attract water molecules, one end of a polar lipid molecule is said to be hydrophilic (water-loving). Lipids with a polar and nonpolar end are sometimes called amphipathic lipids, because one end is hydrophilic, while the other is hydrophobic. Such substances often form structures which bury hydrophobic surface, while exposing hydrophilic surface to water. Some typical structures of both types of lipids are shown in Figure \(1\).
20.05: Nonpolar Lipids
A good example of a nonpolar lipid is the neutral fat glycerol tristearate. This most-common form of animal fat serves as a storehouse for energy and as insulation against heat loss. On a molecular level it is constructed from three molecules of stearic acid and one of glycerol:
(1)
A great many nonpolar lipids can be made by combining different long- chain acids with glycerol. Because these acids were originally derived from fats, they are collectively referred to as fatty acids.
Notice that for each stearic or other fatty acid molecule that combines with one of the —OH groups of glycerol, a molecule of water is given off, and so the reaction is a condensation. It turns out that a great many important biological molecules are put together by condensation reactions during which water is given off. The reverse of Eq. (1), in which water reacts with a large molecule and splits it into smaller pieces, is called hydrolysis. By carrying out hydrolysis living organisms can break down molecules manufactured by other species. The simple building blocks obtained this way can then be recombined by condensation reactions to form structures appropriate to their new host.
By contrast with the glycerol tristearate found in animals, vegetable fats contain numerous double bonds in their long hydrocarbon chains. This polyunsaturation introduces “kinks” in the hydrocarbon chains because of the barrier to rotation and the 120° angles associated with the double bonds. Consequently it is more difficult to align the chains side by side (see Figure \(1\) ), and the unsaturated fats do not pack together as easily in a crystal lattice. As was true with alkanes, chain length also determines whether a fat is liquid or solid, and where the melting point occurs.
Most unsaturated fats (like corn oil) are liquids at ordinary temperatures, while saturated fats (like butter) are solids. Vegetable oils can be converted by hydrogenation to compounds that are solids. This process involves adding H2 catalytically to the double bonds:
Hydrolysis of fats [the reverse of Eq. (1)] is important in the manufacture of soaps. It can be speeded up by the addition of a strong base like NaOH or KOH, in which case the reaction is called saponification. Since saponification requires that the pH of the reaction mixture be high, the fatty acid that is produced will dissociate to its anion. When glycerol tristearate is saponified with NaOH for example, sodium stearate, a relatively water-soluble substance and a common soap, is formed.
The ability of soaps to clean grease and oil from soiled surfaces is a result of the dual hydrophobic-hydrophilic structures of their molecules. The stearate ion, for example, consists of a long nonpolar hydrocarbon chain with a highly polar —COO group at one end.
The hydrophobic hydrocarbon chain tries to avoid contact with aqueous media, while the anionic group readily accommodates the dipole attractions and hydrogen bonds of water molecules.
The two main ways that the hydrophobic portions of stearate ions can avoid water are to cluster together on the surface or to dissolve in a small quantity of oil or grease (see Figure \(2\) ).
In the latter case the hydrophilic heads of the soap molecules contact the water outside the grease, forming a structure known as a micelle. Since the outsides of the micelles are negatively charged, they repel one another and prevent the grease droplets from recombining. The grease is therefore suspended (emulsified) in the water and can be washed away easily.
Natural soaps, such as sodium stearate, were originally made in the home by heating animal fat with wood ashes, which contained potash, K2CO3. Large quantities are still produced industrially, but to a considerable extent soaps have been replaced by detergents. This is a consequence of the undesirable behavior of soaps in hard water. Calcium, magnesium, and other hard-water cations form insoluble compounds when combined with the anions of fatty acids. This produces scummy precipitates and prevents the soap molecules from emulsifying grease unless a large excess is used.
Detergents such as alkylbenzenesulfonates (ABS) and linear alkylbenzenesulfonates (LAS) have structures very similar to sodium stearate except that the charged group in their hydrophilic heads is —SO3 attached to a benzene ring. The ABS detergents also have methyl (CH3) groups branching off their hydrocarbon chains.
Such molecules do not precipitate with hard-water cations and therefore are more suitable for machine washing of clothes. The LAS detergents replaced ABS during the mid-nineteen-sixties when it was discovered that the latter were not biodegradable. They were causing rivers and even tap water to become covered with detergent suds and foam. Apparently the enzymes in microorganisms that had evolved to break down the unbranched hydrocarbon chains in natural fats and fatty acids were incapable of digesting the branched chains of ABS molecules. LAS detergents, though manufactured by humans, mimic the structures of naturally occurring molecules and are biodegradable. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/20%3A_Molecules_in_Living_Systems/20.04%3A_Fats_and_Lipids.txt |
As was true of most nonpolar lipids, the structures of polar lipids are based on condensation of fatty acids with glycerol. The main difference is that only two of the three OH groups on glycerol are involved. The third is combined with a highly polar molecule:
In one sense the polar lipids are like the anions of fatty acids, only more so. They contain two hydrophobic hydrocarbon tails and a head which may have several electrically charged sites. As in the case of soap and detergent molecules, the tails of polar lipids tend to avoid water and other polar substances, but the heads are quite compatible with such environments.
The polar lipids are most commonly found as components of cell walls and other membranes. Nearly all hypotheses regarding membrane structure take as a fundamental component a lipid bilayer (Figure \(2\) ). Bilayers made in the laboratory have many properties in common with membranes. Ions such as Na+, K+, and Cl cannot pass through them, but water molecules can. The hydrocarbon core of such a bilayer should have large electrical resistance, as does a membrane. Certain carrier molecules can transport K+ and other ions across a bilayer, apparently by wrapping a hydrophobic cloak around them to disguise their charges. Membrane proteins in a bilayer also allow for transport of ions and other molecules across the bilayer which could not cross otherwise.
20.07: Carbohydrates
Carbohydrates are sugars and sugar derivatives whose formulas can be written in the general form: Cx(H2O)y. (The subscripts x and y are whole numbers.) Some typical carbohydrates are sucrose (ordinary cane sugar), C12H22O11; glucose (dextrose), C6H12O6; fructose (fruit sugar), C6H12O6; and ribose, C5H10O5.Since the atom ratio H/O is 2/1 in each formula, these compounds were originally thought to be hydrates of carbon, hence their general name.
Glucose, by far the most-common simple sugar, is a primary source of energy for both animals and plants. Because they contain free glucose molecules, dextrose tablets or foods such as grapes and honey can provide a noticeable “lift” for persons tired by physical exertion. The individual glucose molecules pass rapidly into the bloodstream when such foods are eaten. Glucose is also the monomer from which the polymers cellulose and starch are built up. The structural material of plants, from the woody parts of trees to the cell walls of most algae, is cellulose. Plants store energy in starch, providing a source of glucose for all but the simplest organisms. The energy in starchy foods is not as rapidly available, however, since the polymeric structure must be broken down before glucose is released. As a consequence of the ubiquity of starch and especially cellulose, carbohydrates are by far the most plentiful organic compounds in the biosphere.
20.08: Simple Sugars
A careful consideration of the geometry of this structure reveals that not one but two cyclic structures are possible. These are called α- and β-glucose and are shown in Figure \(2\). In the β form the C—O bond on carbon atom 1 (shown in dark color) is parallel to the C—O bond on carbon atom 4, while in the α form these two bonds are at an angle of 180° – 109.5° = 70.5°. This geometric difference may seem relatively trivial, but it turns out to be important when glucose molecules are used as building blocks to form larger entities. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/20%3A_Molecules_in_Living_Systems/20.06%3A_Polar_Lipids.txt |
The sugar molecules listed in Figure \(1\) are usually referred to as monosaccharides. This distinguishes them from the disaccharides which are made up by condensing two sugar units.
A familiar example of a disaccharide is ordinary cane sugar, sucrose, which may be obtained by condensing a molecule of α-glucose with one of the cyclic forms of fructose called β-fructose. The structure of sucrose is shown in Figure \(2\).
Other, less familiar, examples of disaccharides are lactose, which occurs in milk, and maltose, which are shown in Figure \(3\). In order to digest a disaccharide like sucrose or lactose, the human body must have an enzyme which can catalyze hydrolysis of the linkage between the two monosaccharide units. Many Asians, Africans, and American Indians are incapable of synthesizing lactase, the enzyme that speeds hydrolysis of lactose. If such persons drink milk, the undigested lactose makes them sick.
20.10: Polysaccharides
As the name suggests, polysaccharides are substances built up by the condensation of a very large number of monosaccharide units. Cellulose, for example, is a polymer of β-glucose, containing upwards of 3000 glucose units in a chain. Starch is largely a polymer of α-glucose.
These two substances are a classic example of how a minor difference in the monomer can lead to major differences in the macroscopic properties of the polymer. Good-quality cotton and paper are almost pure cellulose, and they give us a good idea of its properties. Cellulose forms strong but flexible fibers and does not dissolve in water. By contrast, starch has no mechanical strength at all, and some forms are water soluble. Part of the molecular structure of cellulose and starch are shown in Figure \(1\).
Cellulose and starch are different not only in overall structure and macroscopic properties. From a biochemical point of view they behave so differently that it is difficult to believe that they are both polymers of the same monosaccharide. Enzymes which are capable of hydrolyzing starch will not touch cellulose, and vice versa. From a plant’s point of view this is just as well since cellulose makes up structural material while starch serves as a storehouse for energy. If there were not a sharp biochemical distinction between the two, the need for a bit more energy by the plant might result in destruction of cell walls or other necessary structural components.
Bacteria, protozoa, termites, some cockroaches, and ruminant mammals (cattle, sheep, etc.) are capable of digesting cellulose. Ruminant mammals, termites and cockroaches themselves do not produce cellulase, the enzyme which breaks down cellulose, but rather, maintain a symbiotic relationship with bacteria in their guts which do breakdown the cellulose for digestion. Most organisms, including humans, are not capable of digesting cellulose, either through their own enzymes or through a symbiotic relationship with an organism which can. If our digestive enzymes could hydrolyze cellulose, humans would have available a much larger food supply. Quite literally we would be able to eat sawdust! It is possible to hydrolyze cellulose in the laboratory either with strong acid or with cellulase, the enzyme used by bacteria. So far, however, such processes produce more expensive (and less tasty!) food than we already have available. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/20%3A_Molecules_in_Living_Systems/20.09%3A_Disaccharides.txt |
The ability to serve a variety of functions is characteristic of most biomolecules. Nowhere is this versatility better exemplified than by the proteins. Perhaps because of their many functions, proteins are the most abundant organic molecules in living cells, constituting more than 50 percent of the mass once water is removed. It is estimated that the human body contains well over a million different kinds of protein, and even a single-cell organism contains thousands. Each of these is a polymer of amino acids which has a highly specific composition, a unique molecular weight (usually in the range from 6000 to 1 000 000) and its own sequence of different amino acids along the polymer chain.
Proteins may be divided into two major classes on the basis of their behavior when reacted with water. The products obtained upon hydrolysis of simple proteins are all amino acids. In the case of conjugated proteins other organic and/or inorganic substances are obtained. The non-amino acid portions of conjugated proteins may consist of metals, lipids, sugars, phosphate, or other types of molecules. These components are referred to as prosthetic groups
Proteins may also be subdivided on the basis of their molecular shape or conformation. In the fibrous proteins long polymer chains are arranged parallel or nearly parallel to one another to give long fibers or sheets. This arrangement results in physically tough materials which do not dissolve in water. The fibrous proteins are fundamental components of structural tissues such as tendons, bone, hair, horn, leather, claws, and feathers.
By contrast, polymer chains of the globular proteins fold back on themselves to produce compact, nearly spherical shapes. Most globular proteins are water soluble and hence are relatively mobile within a cell. Some examples are enzymes, antibodies, hormones, toxins, and substances such as hemoglobin whose function is to transport simple molecules or even electrons from one place to another. The enzyme trypsin, is a typical globular protein.
Another class of proteins are the membrane proteins, which, as the name would suggest, reside in a cell's lipid bilayer membrane. Such proteins can act as channels for ions or other molecules unable to pass through the lipid bilayer; as signal transducers, able to respond to signal molecules on one side of a membrane to begin a molecular response on the other side of the membrane; or as anchors of other molecules to the cell membrane, to name a few exemplars of membrane protein function. Because these proteins interface with non-polar portions of the lipid bilayer, they do no maintain function and structure in an aqueous solution, making them far more difficult to study than globular proteins or fibrous proteins.
Enzymes
The enzymes are the most extensive and highly specialized class of proteins. The structure and mode of action of trypsin, a typical enzyme, were described in enzymes. Most enzyme-catalyzed processes occur from 108 to 1011 times faster than the uncatalyzed reactions. Specificity is so great in some cases that only one particular molecule can serve as the enzyme’s substrate. Even closely related structures are unable to fit the active site, and so their reactions cannot be sped up. Often a enzyme will be catalytically inactive until the correct substrates enter the active site and induce a conformational switch of the enzyme to its active form.
Enzymes are of crucial importance to living organisms because they catalyze nearly every important reaction in cell metabolism. In cases where it is necessary to carry out a nonspontaneous reaction, enzymes are capable of coupling a reaction having a negative free-energy change to the desired one. When a living system reproduces, grows, or repairs damage caused by the external environment, enzymes catalyze the necessary reactions. Some enzymes are even stimulated or inhibited by the presence of smaller molecules. This permits regulation of the concentrations of certain substances because the latter can turn off enzymes which initiate their synthesis. On a molecular level enzymes are the means by which the mechanism of a cell is kept running.
20.12: Polypeptide Chains
The backbone of any protein molecule is a polypeptide chain obtained by the condensation of a large number of amino acids with the elimination of water. You will recall that the amino acids are bifunctional organic nitrogen compounds containing an acid group, —COOH, and an amine group, —NH2. The amine group is attached to the carbon atom adjacent to the —COOH (the α carbon atom). The three simplest amino acids are
In practice, though, these acids are usually in the form of their zwitterions, and we should write them
If these three amino acids are now condensed, water is eliminated and a simple polypeptide containing three amino acids is obtained:
Figure \(3\)
The two CO—NH bonds produced by this reaction are called peptide bonds. Notice that the peptide bond is an amide linkage. An important feature of such a peptide bond is that it is planar. This is because of the existence of two resonance structures
Another important aspect of the peptide bond is the opportunity it provides for hydrogen bonding. The oxygen on the carbonyl group can bond to the hydrogen on an group further along the chain:
Such a bond is somewhat stronger than a normal hydrogen bond because of the partially negative character of the oxygen atom and the partially positive character of the nitrogen atom conferred by resonance structure II. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/20%3A_Molecules_in_Living_Systems/20.11%3A_Proteins.txt |
Altogether there are 20 amino acids which commonly occur in all organisms. Under most circumstances amino acids exist as zwitterions and have the general formula
R represents a group called a side chain which varies from one amino acid to another. One aspect of amino acid structure not obvious in the above representation is the stereochemistry. Notice in the above image the carbon marked with an α, has 4 different groups attached to it, and thus amino acids are chiral. One of the most familiar chiral objects are hands, and therefore chirality is often thought of in terms of handedness. Below are the two enantiomers of the amino acid alanine, L-alanine and R-alanine, so you can see the difference.
The two enantiomers of alanine, L-alanine and R-alanine. Try as you might, it is impossible put these two molecules in the exact same position, they are mirror images. The L-form is the isomer seen in organisms.
How do you tell the difference between the two enantiomers? While most chemists use the R and S priority system to distinguish between enantiomers, many biochemical compounds, including amino acids and sugars use the D and L system. This is based upon a historical method for determining enantiomers using glyceraldehyde. Glyceraldehyde is a chiral three carbon sugar. with one form classified as D and the other as L. Amino acids and sugars can be synthesized from glyceraldehyde. The enantiomer synthesized from the D-form of glyceraldehyde are also labeled D, while those from the L-form, L. To determine if an amino acid is L or D, look at the α carbon, so that the hydrogen atom is directly behind it. This should place the three other functional groups in a circle. Follow from COOH to R to NH2, or CORN. If this is in a counterclockwise direction, the the amino acid is in the L-isomer. If this order is in the clockwise direction, the amino acid is a D-isomer. Try this trick with the two models of alanine.
If you assigned priority and used the R,S system, you will find that most amino acids are S-isomers. There is one exception however, which is cysteine. the sulfur in the R group gives it priority over the carboxylic acid group. This it is an R-isomer in the R,S system, but an L-isomer in the D,L system.
Structures of the 20 different R groups are given in Figure \(1\), where they are shown as part of a polypeptide chain. The figure also shows the three-letter code and one-letter code often used to identify amino acids in a polypeptide chain.
Although each amino acid side chain has its own individual properties, it is useful to divide them into several categories. In Figure \(1\) this has been done on the basis of how strongly hydrophilic or hydrophobic they are. On the extreme left of the figure are the six most hydrophilic side chains. All are polar, and four are actually ionic at a pH of 7. Next in the figure are the six most hydrophobic side chains. These are all large and nonpolar and contain no highly electronegative atoms like nitrogen and oxygen. The remaining eight side chains either contain small nonpolar groups or groups of fairly limited polarity, and are therefore not very strongly hydrophilic or very strongly hydrophobic.
Whether a side chain is hydrophilic or hydrophobic has considerable influence on the conformation of the polypeptide chain. Like the hydrocarbon tails of fatty acid molecules, hydrophobic amino acid side chains tend to avoid water and cluster together with other nonpolar groups. In a globular protein like trypsin for example, most hydrophobic residues are found among twisted chains deep within the molecule. Hydrophilic side chains, by contrast, tend to occupy positions on the outside surface, where they contact the surrounding water molecules. Hydrogen bonds and dipole forces attract these water molecules and help solubilize the globular protein. Were all the nonpolar R groups exposed to the aqueous medium, the protein would be much less soluble. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/20%3A_Molecules_in_Living_Systems/20.13%3A_The_Amino_Acids.txt |
Proteins which occur in nature differ from each other primarily because their side chains are different. Partly this is a matter of composition. In wool, for example, 11 percent of the side chains are cysteine, while no cysteine occurs in silk at all. To a much larger extent though, the differences between different proteins is a matter of the sequence in which the different side chains occur. This is especially true of globular proteins like enzymes. The sequence of side chains along the backbone of peptide bonds in a polypeptide is said to constitute its primary structure.
The 20 different amino acids permit construction of a tremendous variety of primary structures. Consider, for example, how many tripeptides similar to that shown in Eq. 1 on the page on polypeptide chains can be constructed from the 20 amino acids. In the example shown, the first amino acid in the chain is glycine, but it might just as well be proline or any other of the 20 amino acids. Thus there are 20 possibilities for the first place in the chain. Similarly there are 20 possibilities for the second place in the chain, making a total of 20 × 20 = 400 possible combinations. For each of these 400 structures we can again choose from among 20 amino acids for the third place in the chain, giving a grand total of 400 × 20 = 203 = 8000 possible structures for the tripeptide.
A general formula for the number of primary structures for a polypeptide containing n amino acid units is 20n—a very large number indeed when you consider that most proteins contain at least 50 amino acid residues. [2050 = (2 × 10)50 = 250 × 1050 = 1015 × 1050 = 1065] Primary structure is conventionally specified by writing the three-letter abbreviations for each amino acid, starting with the —NH3+ end of the polymer. In some cases, this is even shortened to the 1-letter abbreviation sequence. For example, the structure
which reading from the -NH3+ end is alanine, glycine, glycine would be specified as
Ala-Gly-Gly or AGG
Note that Ala-Gly-Gly is not the same as Gly-Gly-Ala. In the latter case glycine rather than alanine has the free —NH3+ group. Because its ends are different, there is a directional character in the polypeptide chain.
Both three letter and single letter abbreviations are shown in Figure \(1\) :
Determination of the primary structure of a protein is a difficult and complicated problem. It also is a rather important one—the sequence of amino acids governs the three-dimensional shape and ultimately the biological function of the protein. Consequently much effort has gone into methods by which primary structure can be elucidated.
Insulin was the first protein whose amino acid sequence was determined. This pioneering work, completed in 1953 after some 10 years of effort, earned a Nobel Prize for British biochemist Frederick Sanger (born 1918). He found the primary structure to be
Note how there are two chains in this structure, one with 21 side chains and the other with 30. These two chains are linked in two places with disulfide (—S—S—) bridges, each connecting two cysteine residues in different chains.
In order to determine the primary structure of a protein, a known mass of pure sample is first boiled in acid or base until it is completely hydrolyzed to individual amino acids. The amino acid mixture is then separated chromatographically and the exact amount of each amino acid determined. In this way, one can find that for every 3 mol serine in the insulin molecule, there are 6 mol leucine. The next step is to break down the protein into smaller fragments. Disulfide bridges are broken by oxidation after which the protein is selectively hydrolyzed by enzymes, called proteases, such as trypsin or chymotrypsin. In a favorable case one will then have several fragments each containing 10 or 20 amino acid residues. These can then be separated and analyzed individually.
Using Edman degradation, so named for Pehr Edman, the sequence of amino acids in one of these polypeptide fragments is usually determined using phenylisothiocyanate, , which selectively attacks the —NH3+ end of the polypeptide chain. This reaction is carried out under basic conditions. Addition of acid then splits off the terminal amino acid, and it can be identified. Since the rest of the polypeptide chain is left intact, this process can be repeated, and each amino acid in the sequence can be attacked, removed, and identified. By snipping off amino acids one at a time, one eventually finds the complete sequence for the fragment. This whole process can be automated and hence sped up considerably. Once the fragments have been sequenced, it becomes a matter of ordering them correctly. Since different proteases hydrolyze peptide bonds at different places in the amino acid sequence, different fragmentation patterns can be used to determine the sequence for the whole protein. The end of a fragment from a trypsin digest will be in the middle of a fragment from a chymotrypsin digest, for instance. This provides a relatively quick means to sequence unknown protein sequences.
Edman degradation is not the only method for which proteins are sequenced nowadays. Mass spectrometry can sequence polypeptides of 20 to 30 amino acids in length, using a technique called tandem mass spectrometry. In this method, a polypeptide is sent through one mass spectrometer, which ionizes the polypeptide. The charged peptide then enters a collision chamber, causing the peptide to fragment at different peptide bonds. The resulting fragments are then measured by a second mass spectrometer. The resultant spectrum can determine the peptide sequence by differences in mass of the fragments.
With the advent of DNA sequencing methods and the success of the Human Genome Project, many protein sequences are now determined indirectly, through the genetic code. When the DNA sequence for a protein is known, this can be used to determine the protein sequence. By the same token, a known protein sequence can be used to determine the gene coding for that protein. Thus there are many different ways to determine protein sequences, all of which compliment each other.[1]
As methods for determining primary structure have become more advanced, a great many proteins have been sequenced, and some interesting comparisons can be made. A particularly intriguing example is that of cytochrome c, an electron carrier which is found in all organisms that use oxygen for respiration. When samples of cytochrome c from different organisms are compared, it is found that the amino acid sequence is usually different in each case. Moreover, the more widely separated two species are in their macroscopic features, the greater the degree of difference in their protein sequences. When horse cytochrome c is compared with that of yeast, 45 out of 104 residues are different. Only two substitutions are found between chicken and duck, and cytochrome c is identical in the pig, cow, and sheep. The magnitudes of these changes coincide quite well with biological taxonomy based on macroscopically observable differences. Cytochrome c can be used to trace biological evolution from unicellular organisms to today’s diverse species, and even to estimate times at which branching occurred in the family tree of life. This makes molecular methods a powerful tool for the evolutionary biologist as well. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/20%3A_Molecules_in_Living_Systems/20.14%3A_Primary_Protein_Structure.txt |
One might expect a long-chain protein molecule to be rather floppy, adopting a variety of molecular shapes and changing rapidly from one conformation to another. In practice this seldom happens. Instead the protein chain stays more or less in the same conformation all the time. It is held in this shape by the cooperative effect of a large number of hydrogen bonds between different segments of the chain.
A particularly important conformation of the polypeptide chain is the spiral structure shown in Figure \(1\). This is called an α helix. Many fibrous proteins like hair, skin, and nails consist almost entirely of α helices. In globular proteins too, although the overall structure is more complex, short lengths of the chain often have this configuration. In an a helix the polypeptide chain is twisted into a right-hand spiral—the chain turns around clockwise as one moves along it. The spiral is held together by hydrogen bonds from the amido ( ) group of one peptide bond to the carbonyl group of a peptide bond three residues farther along the chain. Two factors contribute toward making this a particularly stable structure. One is the involvement of all the and groups in the chain in the hydrogen bonding. Spirals with slightly more or slightly less twist do not permit this. The second factor is the way in which the side chains project outward from an α helix. Bulky side chains therefore do not interfere with the hydrogen bonding, enabling a fairly rigid cylinder to be formed.
A second regular arrangement of the polypeptide chain is the β sheet, the β-keratin structure found in silk and shown in Figure \(2\). As in the α helix, this structure allows all the amido and carbonyl groups to participate in hydrogen bonds. This hydrogen bonding structure can be accomplished in two manners, either a parallel or antiparallel β sheet, which are compared in Figure \(3\). Unlike the α helix, though, the side chains are squeezed rather close together in a pleated-sheet arrangement. In consequence very bulky side chains make the structure unstable. This explains why silk is composed almost entirely of glycine, alanine, and serine, the three amino acids with the smallest side chains. Most other proteins contain a much more haphazard collection of amino acid residues.
20.16: Higher-Order Structure
Several of the amino acid side chains are difficult to fit into either the α helix or β-sheet types of structure. An obvious example is proline, in which the R group is a ring and includes nitrogen bonded to the α carbon. When proline is involved in a peptide bond, all hydrogens are gone from the nitrogen, leaving no site for hydrogen bonding.
Another problem involves side chains having the same charge, such as those of lysine and arginine or glutamic acid and aspartic acid, which repel one another considerably. When they occur close together, these groups may also destabilize an α helix or β sheet structure. The presence of several groups of the type just mentioned allows a protein chain to bend sharply instead of laying flat or curling regularly into a spiral. Such sharp bends connect sections of α helix or β sheet structures in the globular proteins. They allow the polypeptide chain to curl back upon itself, folding the protein into a very compact, nearly spherical shape. The nature of this folding is referred to as tertiary structure, since it involves a third organizational level above the primary amino acid sequence and the secondary α helix or β sheet.
An excellent example of tertiary structure in a globular protein is provided by the three-dimensional view of sperm-whale myoglobin in Figure \(1\). The molecular backbone consists of eight relatively straight segments of a helix. The longest of these contain 23 amino acids, the shortest just 7. Fewer than 50 of myoglobin's 153 amino acids are found in the bends between the helixes, but all the proline residues fall into this category, as do many of the others which destabilize the α helix or β sheet. (Although only the α helix is prominent in the structure of myoglobin, other globular proteins are found to have regions consisting of β sheets.)
Several other important generalizations may be made about the structure of myoglobin. Even in regions where the chain twists into an α helix, nearly all the nonpolar R groups point toward the interior of the molecule. Here they crowd so closely together that only four water molecules can squeeze their way in. The outside surface of the protein, however, contains all the polar R groups. These interact strongly with the many water molecules which normally surround myoglobin in muscle tissue.
Although myoglobins isolated from a variety of mammals differ slightly in their primary structure, they all seem to have nearly the same overall molecular shape. Apparently some of the amino acids are much more important than others in determining the bend points and other crucial features of tertiary structure. Substitutions at less-important positions do not cause great variations in the ability of the protein to carry out its biological function.
Finally, Figure \(1\) shows clearly that myoglobin contains a prosthetic group. In this case it is the heme group, which contains an iron atom surrounded by four nitrogens in a flat ring structure known as a porphyrin. The porphyrin ring is not covalently bonded to the protein chain, but rather fits snugly into a “pocket” surrounded by several segments of helix. A nitrogen atom in a histidine side chain on one helix does form a coordinate covalent bond with iron in the heme group, but apart from this the prosthetic group is positioned solely by the way the protein chain is folded.
The iron in the heme group marks the active site at which the oxygen-storage function of myoglobin is accomplished. Iron, like other first-row transition-metal ions, ordinarily forms six bonds directed toward the corners of an octahedron. In myoglobin only five bonds to iron are found. The sixth position (opposite the histidine nitrogen) can be occupied by an oxygen molecule, providing a convenient storage site. If the concentration of oxygen near the protein falls, this binding is reversed, releasing oxygen to replenish the supply.
If an iron ion can bond to an oxygen molecule, you may wonder why the complicated porphyrin and protein structures are necessary. It is to prevent the oxidation of the iron in the heme group from the iron(II) to the iron(III) oxidation state. Because the oxygen molecule needs to gain two electrons to be oxidized while iron(II) can only supply one, no electron transfer occurs. If the iron(II) atom in heme were not surrounded by the protein chain, a water molecule would be able to take part as an intermediate in the electron-transfer mechanism and reduction of the oxygen could occur. As matters stand, though, the oxygen molecule cannot be reduced until it is released by the myoglobin. Thus the combined effects of the tertiary structure of the protein, the prosthetic group, and a specialized active site allow the myoglobin to fulfill its biological function of storing oxygen molecules until needed, without allowing them to be reduced during storage.
In some proteins there is yet a fourth level of organization, labeled quaternary structure. This may be illustrated by hemoglobin, whose function as an oxygen carrier in the bloodstream is well known. Hemoglobin consists essentially of four myoglobin molecules packed together in a single unit. Four separate polypeptide chains are each folded as in myoglobin and then nested together. The way these four subunits fit together constitutes the quaternary structure.
Like the other type of structures, quaternary structure contributes to the function of a protein. In the case of hemoglobin, an oxygen molecule attached to one subunit causes slight shifts in tertiary and quaternary structure which make it easier for other oxygen molecules to bond to the other subunits. Consequently hemoglobin in the lungs can be loaded with its full complement of four oxygen molecules rather easily, a factor which increases its efficiency in carrying oxygen to body tissues. The converse is also true—loss of one oxygen molecule causes slight structural rearrangements which allow the remaining three to depart more readily. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/20%3A_Molecules_in_Living_Systems/20.15%3A_Secondary_Protein_Structure.txt |
Given the tremendously important role of proteins in the functioning of any organism, the question arises of how cells synthesize appropriate polypeptides. Although more than 11400 enzymes[1] are currently known, these constitute a very small fraction of the possible combinations of the 20 amino acids in chains of 100 or more. Clearly a great many of these combinations are worthless, in the sense that they are not adapted to any biological function. It would be pointless for any organism to construct them. Indeed, if they did occur, it would be useful for a living system to hydrolyze them to their individual amino acids and use these to build up proteins which did carry out necessary functions. Therefore it is reasonable to expect a living cell to contain some kind of “blueprint” which specifies the structures of those proteins which are essential for the cell’s normal functioning. Without such a guide an incredible amount of effort would be wasted in synthesizing unusable polypeptides just to get small amounts of those that worked.
The blueprint just described is contained in the molecular architecture of deoxyribonucleic acid (DNA). The structure of DNA also provides an obvious mechanism by which the information necessary to specify protein structure can be reproduced and passed from a parent cell to its progeny. The complicated task of building protein structures from the DNA blueprint involves several types of ribonucleic acids (RNA’s) whose structures are closely related to that of DNA. Hence the storage, reproduction, and application of information about protein structure depends on nucleic acids
1. ↑ PDB Statistics. RCSB Protein Data Bank. Accessed September 29, 2009. http://www.rcsb.org/pdb/statistics/h...Classification
20.18: Nucleic Acid Structure
Nucleic acids were first isolated from cell nuclei (hence the name) in 1870. Since then they have been found in other portions of cells as well, especially the ribosomes, which are the sites of protein synthesis. Most nucleic acids are extremely long-chain polymers—some forms of DNA have molecular weights greater than 109. Nucleic acids are made up from three distinct structural units. These are:
1 A five-carbon sugar. Only two sugars are involved. These are ribose (used in RNA) and deoxyribose (used in DNA). Their structures are shown in Figure \(1\). Note that deoxyribose, as its name implies, has one oxygen less than ribose in the 2 position.
2 A nitrogenous base. There are five of these bases. All are shown in Figure \(2\). Three of them, adenine, guanine, and cytosine, are common to both DNA and RNA. Thymine occurs only in DNA, and uracil only in RNA.
3 Phosphoric acid. H3PO4 provides the unit that holds the various segments of the nucleic acid chain to each other.
The combination of a sugar and a nitrogenous base is called a nucleoside. A typical example of a nucleoside is adenosine, derived by the condensation of ribose and adenine:
Nucleosides in turn can be condensed with phosphoric acid by means of the —OH group at carbon atom 5. The result is a structure called a nucleotide. Thus when adenosine condenses with phosphoric acid, the nucleotide formed is adenosine-5-monophosphoric acid, usually indicated by the acronym AMP: | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/20%3A_Molecules_in_Living_Systems/20.17%3A_Nucleic_Acids.txt |
How can DNA and RNA molecules act as blueprints for the manufacture of proteins? The exact details were unraveled in the early 1960s mainly by Marshall Nirenberg (born 1927) at the National Institutes of Health and H. G. Khorana (born 1922) at the University of Wisconsin, work which earned them the Nobel prize in 1968. They showed that each amino acid in a protein is determined by a specific codon of three nitrogenous bases in the DNA or RNA chain. The details of this genetic code are given in the table below. As an example of how this code works, let us take the section of RNA shown in Figure 3 on Nucleic Acid Structure. This has the sequence UCAUGG. This is part of the instructions for building a polypeptide chain containing the amino acid serine (UCA) followed by the amino acid tryptophan (UGG).
Table \(1\) The Genetic Code for RNA
Note
(a) A termination codon is indicated by TERM. (b) AUG, the codon for methionine is also the initiation codon. All protein synthesis begins at this codon, though this initial methionine is often removed during post-transcriptional processing.
Since each codon corresponds to three places in the nucleic acid chain and since there are four kinds of nitrogenous bases to fill each place, there are a total of 43 = 64 different possible codons. Since there are only 20 amino acids, the genetic code is degenerate—several different codons correspond to the same amino acid. This degeneracy acts as a safeguard against errors in reading the code. Thus UCU, UCC, UCA, and UCG all correspond to serine. If a mistake is made in reading the third base in this triplet, no harm is done since serine is still produced. On the molecular level transfer RNAs (tRNA), the molecules reading the codons and providing the correct amino acid, can pair with multiple codons. This only occurs in terms of the third base in the codon. For instance, G pairs with C, but is also capable of pairing with U. Some tRNAs even employ a fifth nitrogenous base, inosinate(I) which is capable of pairing with A, U or C. This use of multiple pairing with the third codon by tRNA is called the wobble hypothesis, and was first proposed by Francis Crick. Notice that while a tRNA can pair with multiple codon in the wobble hypothesis, it can only pair with codons for the same amino acid, and each codon is still specific to only one amino acid.[1]
There are three additional features of the genetic code. First, AUG, the codon for Methionine also serves as an initiation codon, and, with help from other signals, is where protein systhesis begins. A second feature is that reading RNA for protein synthesis goes from the 5' carbon end of the nucleic acid to the 3' carbon end. A final important feature of the genetic code is the existence of three termination codons. These correspond to an instruction for ending a polypeptide chain. How these features work is best illustrated by an example.
Example \(1\): RNA
Decode the RNA fragment
5' A C C U U A U G A C G C C U G U C C A U U A A C G A U 3'
Solution
First, we must decide which direction to read the RNA code. Synthesis goes from the 5' end to the 3' end, so this segment is read left to right. Had it been displayed 3' to 5', we would have needed to read it from right to left.
Second, we need to look for an initiation codon, AUG. This codon appears starting at the sixth letter in. Thus, we can divide the sequence up like this, with the start codon bold:
AC|CUU|AUG|ACG|CCU|GUC|CAU|UAA|CGA|U
Third, let us see if there is a stop codon in this sequence. Sure enough, the fifth codon after the start codon, UAA is a stop codon. Thus, the entire sequence to be translated, in bold:
AC|CUU|AUG|ACG|CCU|GUC|CAU|UAA|CGA|U
which translates to the amino acid sequence:
Met-Thr-Pro-Val-His-STOP
Notice in the example, that if we had not started with the initiation codon, an entirely different protein would have been formed. Look at what would have happened if we had simply started at the beginning of the sequence:
ACC|UUA|UGA|CGC|CUG|UCC|AUU|AAC|GAU|
a stop codon appears in a new place, and the translated protein is:
Thr-Leu-STOP
This highlights the importance of the reading frame, the place where codons start being read. Notice that, since codons are 3 bases long, any sequence has three different reading frames. Without the initiation codon, there would be no way to identify the correct reading frame. In addition to the AUG initiation codon, other element regulate initiation. In bacteria, a sequence of bases before the initiation codon, called the Shine-Dalgarno sequence precedes the AUG codon, specifying where to begin translation. A different set up occurs in eukaryotes. An initiation complex forms, but instead of having a specific sequence connected to the initiation codon, the complex slides along the mRNA strand, until it finds the AUG initiation codon.[2]
1. ↑ Nelson, D.L., Cox, M.M. Lehninger Principles of Biochemistry(5<sup>th</sup>ed). New York: W.H. Freeman and Company, 2008. pp. 1070-1072.
2. ↑ Nelson, D.L., Cox, M.M. Lehninger Principles of Biochemistry(5<sup>th</sup>ed). New York: W.H. Freeman and Company, 2008. pp. 1088-1090. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/20%3A_Molecules_in_Living_Systems/20.19%3A_Information_Storage.txt |
There is more to the structure of DNA than just the primary sequence of nitrogenous bases. Secondary structure also plays a crucial biochemical role. Each DNA molecule consists of two nucleotide chains wrapped around each other in a double helix and held together by hydrogen bonds. This hydrogen bonding involves only the nitrogenous bases. Each of the purine bases can hydrogen bond with one and only one of the pyrimidine bases.
Thus adenine can hydrogen bond with thymine and guanine with cytosine, as shown in Figure \(1\). Note that in both cases there is an exact match of hydrogen atoms on the one base with nitrogen or oxygen atoms on the other. Note also that the distance from sugar linkage to sugar linkage across each of the base pairs in Figure \(1\) is almost exactly the same. This explains why only these two combinations occur in DNA. Other combinations (i.e., adenine-cytosine) are not nearly so favorable energetically.
The overall geometry of the two nucleotide chains in the DNA molecule is in the form of the double helix shown in Figure \(2\). Each helix corresponds to a nucleotide chain, and the two chains are joined throughout their length by adenine-thymine or guanine-cytosine pairs. These base pairs are stacked one above the other with their planes perpendicular to the axis of the two spirals. This places the hydrophobic base pairs inside the structure and allows the hydrophilic sugar and phosphate groups to contact water on the exterior. The whole helix will just fit inside a cylinder 2000 pm in diameter.
The spacing between base pairs is 340 pm, and there are 10.5 base pairs in one full turn of the helix.
The two nucleotide chains in the double helix are said to be complementary to each other. Because of the exact pairing of the bases we can always tell the sequences of bases in the one chain from that in the other. Thus if the first six bases in one chain are AGATCC, we know that the first six bases in the other will be TCTAGG. Both chains are therefore alternative representations of the same information. If one or two bases become misplaced in either strand, this can be recognized because of mismatching with the complementary strand. Repair enzymes can then correct the sequence of bases along the incorrect strand. A final point to make is that the two strands are antiparallel. This means that one strand, from bottom to top is going from the 5' carbon to the 3' carbon, while the complimentary strand is going 3' to 5' from bottom to top.
This double-helix model for DNA was first suggested in 1953 by James D. Watson (born 1928) and Francis Crick (1916 to 2004). It was an important milestone in the history of science, since it marked the birth of a new field, molecular biology, in which the characteristics of living organisms could at last begin to be explained in terms of the structure of their molecules. In 1962 Crick and Watson shared the Nobel Prize with M. F. H. Wilkins, whose x-ray crystallographic data had helped them to formulate their model. Rosalind Franklin(1920-1958) who performed the x-ray crystallography experiments did not win the Nobel Prize, as they are not awarded posthumously, but should be included in any discussion on the discovery of the double helix. A fascinating account of this discovery, which does not always put the author in a favorable light, can be found in Watson’s book “The Double Helix." | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/20%3A_Molecules_in_Living_Systems/20.20%3A_The_Double_Helix.txt |
According to the central dogma of molecular genetics, DNA is the genetically active component of the chromosomes of a cell. That is, DNA in the cell nucleus contains all the information necessary to control synthesis of the proteins, enzymes, and other molecules which are needed as that cell grows, carries on metabolism, and eventually reproduces. Thus when a cell divides, its DNA must pass on genetic information to both daughter cells. It must somehow be able to divide into duplicate copies. This process is called replication. Given the complementary double strands of DNA, it is relatively easy to see how DNA as a molecule is well structured for replication, as is show in Figure \(1\). Each strand serves as a template for a new strand. Thus, after DNA is replicated, each new DNA double helix will have one strand from the original DNA molecule, and one newly synthesized molecule. This is referred to as semiconservative replication.
A rather complex mechanism exists for DNA replication, involving many different enzymes and protein factors. Let us consider some of the more important aspects of DNA replication. First, the double strand needs to be opened up to replicate each template strand. To do this, a set of proteins and enzymes bind to and open up the double helix at an origin point in the molecule. This forms replication forks, points where double stranded DNA opens up, allowing replication to occur. A helicase enzyme binds at the replication forks, with the function of further unwinding the DNA and allowing the replication fork to move along the double strand as DNA is replicated. Another enzyme, DNA gyrase, is also required to relieve stress on the duplex caused by unwinding the double strand. Further, single strand binding proteins are needed to prevent the single strands from reforming a double strand. Another essential enzyme in this initiation phase is primase, which creates an RNA primer on each single strand of DNA to begin replication from.
All of these initial functions are necessary to prepare the DNA for the main enzyme which builds then new strands, DNA polymerase. Multiple polymerase enzymes exist, but for the moment we will DNA polymerase III, the main DNA polymerase in E. coli. DNA polymerase III catalyzes the reaction by which a new nucleotide is added to a growing DNA strand. That reaction is seen in Figure \(2\). The DNA polymerase enzymes need a free 3' OH group in order to begin synthesizing a new strand, which explains the necesity of the RNA primer, which gives a 3'OH group for DNA polymerase III to start from. [2]
This leads to another constraint on DNA polymerase III. One strand, the leading strand can be polymerized continuously since the new strand being created goes 5' to 3' from the replication fork, but since the original strands are anti-parallel, the other strand, the lagging strand is going in the wrong direction for polymerization. In this case, the polymerzation reaction starts away from the replication fork and works back toward it. This means that the lagging strand is synthesized in disconnected segments, known as Okazaki fragments, instead of continuously. Later, another DNA polymerase, in the case of E. coli, DNA polymerase I, removes RNA primers and fills in the missing discontinuities. Then, another enzyme, DNA ligase, connects breaks between 3'OH groups and 5' phosphate groups in the newly synthesized strands that exist due to these discontinuities. While the enzymes of this process differ in eukaryotes, they fulfill similar mechanisms. Even with this complexity of this process, DNA polymerase III is able to add new nucleotides at a rate of 250-1,000 nucleotides per second.[3]
A number of advantages of the double-stranded structure held together by hydrogen bonds is evident in the process of replication. Complementary base pairing insures that the two new DNA molecules will be the same as the original. The large number of hydrogen bonds, each of which is relatively weak, makes complete separation of the two strands unlikely, but one hydrogen bond, or even a few, can be broken rather easily. The helicase portion of the replication complex can therefore separate the two strands in much the same way that a zipper operates. Like the teeth of a zipper, hydrogen bonds provide great strength when all work together, but the proper tool can separate them one at a time. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/20%3A_Molecules_in_Living_Systems/20.21%3A_DNA_Replication.txt |
Although DNA contains the necessary instructions for synthesizing all the proteins necessary for the functioning of a cell, it does not take part directly in the synthesis itself. Sections of the DNA chain are first copied into a type of RNA called messenger RNA, mRNA. This process is known as transcription. In transcription, RNA polymerase opens up DNA and uses the base pair sequence of one of the DNA strands to synthesis a molecule of mRNA which is complementary to the template strand. The other strand is referred to as the coding strand, as it has the same sequence as the mRNA molecule formed. The synthesis of mRNA is governed by promoter sequences in the DNA, as well as binding of protein factors that can either promote or repress transcription. Such control is necessary for a cell produce the correct proteins at the correct time.
The mRNA molecules differ from DNA in three ways:
1 RNA is generally found in a single strand form, instead of the double helix structure of DNA.
2 RNA has a hydroxyl group ( ) at the 2' carbon, wheras DNA simply has a hydrogen.
3 The base uracil (U) replaces thymine (T).
All three differences are displayed in Fig \(1\). Another aspect of mRNA molecules is that they are also considerably smaller than DNA, containing the blueprints for only a few proteins at most.
As the name implies, mRNA molecules are used to transport their coded instructions from the nucleus of the cell, where the DNA is situated, to the ribosomes, where the process of protein synthesis actually takes place.
When an mRNA molecule reaches a ribosome, a process called translation takes place in which the base sequence on the mRNA molecule is used to create a protein using the codon code. In translation, each codon on the mRNA base pairs with an anticodon base sequence on a RNA molecule called transfer RNA, tRNA. Each tRNA molecule is bound to the amino acid. Thus, the codon UUA will pair with the anticodon of the tRNA bound to leucine. The synthesis of the protein itself is performed by the ribosome, which is a protein-RNA complex, but unlike other enzymes, the catalytic activity is provided by the RNA portion, not the protein. The ribosome is thus sometimes referred to as a ribozyme, to distinguish it from the usual concept of any enzyme with protein based catalytic activity. Figure \(2\) presents this process as a cartoon.
In the cartoon, the brown structure represents the ribosome, the blue letters represent mRNA and its base sequence, the colored hooks represent tRNAs, and amino acids are represented by their three letter abbreviations. mRNA binds into the ribosome, which finds a start codon at which to begin translating. This begins with Methionine, which is bound to the tRNA which recognizes the AUG codon. The next tRNA enters in at the A site. The ribosome catalyzes the condensation reaction between the carboxyl end of methionine and the amino end of glutamate. The dipeptide is now bound to the tRNA in the A site. The tRNA for methionine shifts to the exit site and leaves the complex, while the dipeptide and bound tRNA move to the P site. The next activated tRNA, meaning the tRNA has an amino acid bound to it, enters the A site, and the process repeats until a stop codon is reached. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/20%3A_Molecules_in_Living_Systems/20.22%3A_Transcription_and_Translation.txt |
In the following sections we are going to study the way in which matter can both absorb energy and emit it in the form of electromagnetic radiation such as light. The pattern in which matter absorbs or emits radiation is called its spectrum. In the past, and still to this day, studies of the spectrum of a substance have furnished important clues to the structure of matter. At the same time, the spectrum of a substance is often a very useful way of characterizing and hence identifying and analyzing that substance.
• 21.1: Prelude to Spectroscopy
Absorption of an appropriate quantity of energy can raise the hydrogen atom from a lower to a higher energy level, while emission of electromagnetic radiation corresponds to a change from a higher to a lower energy level. Although Bohr’s theory is quantitatively accurate only for hydrogen, his idea of energy levels is useful for all other atoms and even for molecules.
• 21.2: The Nature of Electromagnetic Radiation
• 21.3: Atomic Spectra and the Bohr Theory
• 21.4: Bohr Theory of the Atom
In a classic paper published in 1913, the young Niels Bohr, then working with Rutherford in Manchester, England, proceeded to show how Rydberg’s formula could be explained in terms of a very simple model of the hydrogen atom. The model was based on the nuclear view of atomic structure which had just been proposed by Rutherford.
• 21.5: The Spectra of Molecules- Infrared
When we turn from the spectra of atoms to those of molecules, we find that the region of most interest to chemists is no longer the visible and ultraviolet but rather the infrared. As its name implies, the infrared extends beyond the red end of the visible spectrum, from the limit of visibility at roughly 0.8 μm (800 nm) up to about 100 μm where the microwave region begins.
• 21.6: The Visible and Ultraviolet Spectra of Molecules- Molecular Orbitals
When molecules absorb or emit radiation in the ultraviolet and visible regions of the spectrum, this almost always corresponds to the transition of an electron from a low-energy to a high-energy orbital, or vice versa.
• 21.7: Molecular Orbitals
In order to explain both the ground state and the excited state involved in an absorption band in the ultraviolet and visible spectra of molecules, it is necessary to look at the electronic structure of molecules in somewhat different terms. We need to look upon electrons in a molecule as occupying orbitals which belong to the molecule as a whole. Such orbitals are called molecular orbitals, and this way of looking at molecules is referred to as molecular-orbital (abbreviated MO) theory.
• 21.8: Delocalized Electrons
• 21.9: Conjugated Systems
In some molecules the delocalization of electron pairs can be very much more extensive than in ozone and benzene. This is particularly true of carbon compounds containing conjugated chains.
21: Spectra and Structure of Atoms and Molecules
In the following sections we are going to study the way in which matter can both absorb energy and emit it in the form of electromagnetic radiation such as light. The pattern in which matter absorbs or emits radiation is called its spectrum. In the past, and still to this day, studies of the spectrum of a substance have furnished important clues to the structure of matter. At the same time, the spectrum of a substance is often a very useful way of characterizing and hence identifying and analyzing that substance.
Many of the properties of electromagnetic radiation can be explained if light is thought of as periodically varying electric and magnetic fields (electromagnetic waves). Such waves can be characterized by their frequency v or their wavelength λ, and their speed of propagation is always λv = c = 2.998 × 108 m s–1. Some properties of light are more easily explained in terms of particles called photons. The energy of a photon is given by E = hv, where h = 6.626 × 10–34 J s and is called Planck’s constant.
When any element is heated to a high temperature or excited in a discharge tube, it gives a line spectrum. Niels Bohr was able to predict the wavelengths of the lines in the spectrum of hydrogen by means of a theory which assigned the single electron to specific energy levels and hence to orbits of specific radius. Absorption of an appropriate quantity of energy can raise the hydrogen atom from a lower to a higher energy level, while emission of electromagnetic radiation corresponds to a change from a higher to a lower energy level. Although Bohr’s theory is quantitatively accurate only for hydrogen, his idea of energy levels is useful for all other atoms and even for molecules.
In the case of molecules, energy levels arise because of different speeds and kinds of molecular vibrations and rotations as well as because electrons are moved farther from or closer to positively charged nuclei. In organic compounds some groups of atoms vibrate at much the same frequency no matter what molecule they are in. The energy levels of such vibrations usually differ by roughly the energies of infrared photons, and many organic functional groups can be identified by the characteristic frequencies at which they absorb infrared radiation. When molecules absorb visible or ultraviolet light, band spectra occur. Some of the energy of each absorbed photon goes to excite an electron, but varying amounts also increase vibrational and rotational energies. Thus photons are absorbed over a broad range of frequencies and wavelengths.
The most convenient theory by which the electronic energies of molecules can be predicted is the molecular-orbital theory. It assumes that electrons in a molecule occupy orbitals which are not confined to a single atom but rather extend over the entire molecule. Bonding molecular orbitals involve constructive interference between two electron waves, while antibonding molecular orbitals involve destructive interference. An electron occupying an antibonding MO is higher in energy than it would be if the atoms were not bonded together, and so antibonding electrons cancel the effect of bonding electrons. This explains why molecules such as He2 or Ne2 do not form.
Molecular-orbital theory is especially useful in dealing with molecules for which resonance structures must be drawn. Because molecular orbitals can be delocalized over several atoms, there is no need for several resonance structures in the case of molecules like O3 and C6H6. The greater the extent of electron delocalization, the smaller the separation between molecular energy levels and the longer the wavelength at which absorption of ultraviolet or visible light can occur. Thus compounds containing long chains of alternating single and double bonds or having several benzene rings connected together often absorb visible light and are colored. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/21%3A_Spectra_and_Structure_of_Atoms_and_Molecules/21.01%3A_Prelude_to_Spectroscopy.txt |
Visible light, gamma rays, x-rays, ultraviolet (black) light, infrared radiation, microwaves, and radio waves are all related. Many of their properties can be explained by a wave theory: periodically varying electric and magnetic fields (electromagnet waves). Figure $1$ indicates the relationship of these fluctuating electric and magnetic fields. It also illustrates the maximum amplitude A0 and the wavelength λ. The intensity of a wave is associated with the square of its amplitude.
Electromagnetic waves travel through a vacuum at the speed of light, c = 2.9979 × 108 m s–1. The entire wave shown in Figure 21.1 may be thought of as moving from left to right. Thus at position P, where the electric field had maximum amplitude at the instant the figure was drawn there is a progressive decrease in amplitude with time. The amplitude reaches its smallest (most negative) value when the wave has moved a distance equal to that separating points P′ and P. Eventually the amplitude increases to its maximum value again, corresponding to movement of a distance P″ to P, or one wavelength λ. A moving wave can be characterized by the frequency v at which points of maximum amplitude pass a fixed position. The speed of the wave (distance traveled per unit time) must be the product of the wavelength (distance between maxima) and the frequency (number of maxima passing per unit time):
$c= \lambda \upsilon \label{1}$
Since the speed of electromagnetic radiation in a vacuum is always the same, radiation may be characterized by specifying either λ or v. The other quantity can always be calculated from Equation $\ref{1}$.
Example $1$ : Wavelength, Frequency, & Speed
Specify the frequency, wavelength, and speed in a vacuum of each of the types of electromagnetic radiation listed below:
1. Blue-green light; λ = 500 nm.
2. Heat rays emitted from hot asphalt pavement; v = 1.5 × 1014 s–1
3. A gamma ray emitted from $^{131}_{53} I$ ; λ = 3.402 pm.
4. An FM radio transmission; v = 91.5 MHz.
Solution
In each case we make use of the relationship c = λv = 2.998 × 108 m s–1.
1. $v= \frac{c}{ \lambda } = \frac{2.998\times 10^{8}m s{-1}}{500\times10^{-9}m}=6.00\times10^{14}s^{-1}$
2. $\lambda= \frac{c}{v} = \frac{2.998\times 10^{8}m s{-1}}{1.5\times10^{14}s^{-1}}=2.00\times10^{-6}m=2.0\mu m$
3. $v= \frac{c}{\lambda} = \frac{2.998\times 10^{8}m s{-1}}{3.402\times10^{-12}m}=8.812\times10^{19}s^{-1}$
4. The unit hertz Hz is 1 s–1, therefore $\lambda=\frac{2.998\times10^{8}ms^{-1}}{91.5*10^{6}Hz} \times \frac{1 Hz}{1 s^{-1}} = 3.28 m$
The results obtained in Example $1$ indicate that the frequency and wavelength of electromagnetic radiation can vary over a wide range.
The experiments which did most to convince scientists that light could be described by a wave model are concerned with interference. In 1802 Thomas Young (1773 to 1829), an English physicist, allowed light of a single wavelength to pass through a pair of parallel slits very close to each other and then onto a screen. Young observed the interference pattern of alternating dark and bright strips, shown in Figure $2$. Instead of two strips of light, three appeared on the screen, the most prominent being in the center.
The appearance of these bright and dark strips on the screen is easy to explain if light is regarded as a wave. The bright areas are the result of constructive interference, while the dark ones result from destructive interference. Constructive interference occurs when the crests of two waves reach the same point at the same time. The amplitudes of the two waves add together, giving a resultant larger than either. In the case of destructive interference a maximum in one wave and a minimum in the other reach the same point at the same time. Thus one cancels the effect of the other, and the resultant wave is smaller. This is illustrated in Figure 3. When destructive interference occurs between two waves which have the same amplitude, the resultant wave has zero amplitude (and zero intensity). Hence the dark strips observed in the double-slit experiment.
Although the behavior of light and other forms of electromagnetic radiation can usually be interpreted in terms of wave motion, this is not always so. When radiation is absorbed or emitted by matter, it is usually more convenient to regard it as a stream of particles called photons.
Thus electromagnetic radiation has the same kind of wave-particle duality we encountered in the case of the electron. However, photons have some properties which are very different from those of electrons and other particles. Although photons have mass and energy, and although we can count them, they can travel only at the speed of light. We cannot slow down a photon or stop it without changing it into something else.
The wave-particle duality of photons and electromagnetic radiation is enshrined in an equation first proposed by the German physicist Max Planck (1858 to 1947). The energy of a photon E and the frequency of the electromagnetic radiation associated with it are related in the following way:
$E=h \upsilon \label{2}$
where h is a universal constant of nature called Planck’s constant with the value 6.6262 × 10–34 J s. The application of Equation $\ref{2}$ is best shown by an example.
Example $2$ : Energy of Photons
Calculate the energy of photons associated with each kind of electromagnetic radiation mentioned in Example 1. Compare each result with the mean bond enthalpy for a C—C single bond (348 kJ mol–1).
Solution In each case use the formula E = hv. If wavelengths are given instead of frequencies, the formula E = hc/λ may be obtained by combining Equations $\ref{1}$ and $\ref{2}$.
1. $E=\frac{hc}{\lambda}=\frac{6.626\times10^{-34}Js\times2.998\times10^{8}ms^{-1}}{500\times10^{-9}m}=3.97\times10^{-19}J=0.397aJ$
2. $E=hv=6.626\times10^{-34}Js\times1.5\times10^{14}s^{-1}=9.9\times10^{-20}J=0.099aJ$
3. $E=\frac{6.626\times10^{-34}Js\times2.998\times10^{8}ms^{-1}}{3.402\times10^{-12}m}=5.839\times10^{-14}J=58 390aJ$
4. $E=6.626\times10^{-34}Js\times91.5\times10^{-6}s^{-1}=6.06\times10^{-26}J=6.06\times10^{-8}aJ$
Since the bond enthalpy quoted refers to 1 mol C—C bonds, we must divide by the Avogadro constant to obtain a quantity which is appropriate to compare with the energy of a single quantum of radiation:
Enthalpy to dissociate one C—C bond is
$\dfrac{348kJmol^{-1}}{6.022\times10^{23}mol^{-1}} = 5.78 \times 10^{–19}\; \,J = 0.578\, aJ \nonumber$
Clearly the energies of visible photons are comparable with the energies of chemical bonds. Infrared and radio waves have far less energy per photon. Gamma-ray photons have enough energy to break open about 100 000 chemical bonds. As a consequence chemical changes often occur when gamma rays or other high-energy photons are absorbed by matter. Such changes are usually detrimental to living systems, and materials such as lead are used to shield humans from sources of high-energy radiation.
The entire spectrum of electromagnetic radiation may be characterized in terms of wavelength, frequency, or energy per photon, as shown in Figure $3$. The example calculations we have done so far and the figure both indicate the broad range covered by λ, v, and E. Electromagnetic radiation which can be detected by the human retina is but a small slice out of the total available spectrum. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/21%3A_Spectra_and_Structure_of_Atoms_and_Molecules/21.02%3A_The_Nature_of_Electromagnetic_Radiation.txt |
When any element is heated to a sufficiently high temperature, all bonds between atoms are dissociated and the element turns into a monatomic gas. At these or still higher temperatures the single atoms begin to radiate visible and ultraviolet light. The actual pattern of light emitted varies from element to element and is called its emission spectrum. Such a spectrum is commonly obtained by passing an electric arc through a powdered sample of the solid element or by applying a voltage to a discharge tube containing a gas at low pressure. Neon signs are everyday examples of discharge tubes. Neon produces a bright red glow in the tube, and other gases produce different colors.
If the emission spectrum of an element is passed through a prism and allowed to strike a photographic film, the intensity of emission as a function of wavelength can be measured. The equipment for making such measurements is called a spectroscope and is shown schematically in Figure $1$. Typical emission spectra obtained in this way are shown in Figure $2$ where the intensity is plotted against wavelength. Note from this figure how the light emitted is confined to a few very specific wavelengths. At all other wavelengths there is no emission at all. Spectra of this kind are usually referred to as line spectra. The wavelengths of the lines in a line spectrum are unique to each element and are often used, especially in metallurgy, both to identify an element and to measure the amount present. These wavelengths can often be measured to an accuracy as great as one part in a billion (1 in 109). Because the spectrum of an element is readily reproducible and can be measured so accurately, it is often used to determine lengths. For example, the meter was once defined as the distance between two marks on a platinum bar kept at Sevrés in France, but now it is taken as 1 650 763.73 wavelengths of a particular line in the spectrum of 86Kr.
At first glance the wavelengths of the lines in an emission spectrum, like those shown in Figure $2$, appear to have no pattern to them, but upon closer study regularities can be distinguished.
For example, you can see that lines in the hydrogen spectrum occur very close together in the region just above 365 nm and then are spaced farther and farther apart as λ increases. In 1885 a Swiss high-school mathematics teacher, J. J. Balmer (1825 to 1898), discovered a formula which accounted for this regularity:
$\lambda=364.6 \text{ nm}\frac{n^{2}}{n^{2}-4} ~~~~ n = 3, 4, 5, \cdots \label{1}$
When the number n is any positive integer greater than 2, the formula predicts a line in the hydrogen spectrum. For n = 3, the Balmer formula gives λ = 656.3 nm; when n = 4, λ = 486.1 nm; and so on. All spectral lines predicted by the Balmer formula are said to belong to the Balmer series.
Other similar series of lines are found in the ultraviolet (Lyman series) and infrared (Paschen, Brackett, Pfund series) regions of the spectrum. In each case the wavelengths of the lines can be predicted by an equation similar in form to Balmer’s. These equations were combined by Rydberg into the general form
$\frac{1}{\lambda}=R_{\infty}\big(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\big) ~~~~~~ n_2 \text{>} n_1 \label{2}$
where R, called the Rydberg constant , has the value 1.097 094 × 107 m–1. Thus a single equation containing only one constant and two-integer parameters is able to predict virtually all the lines in the spectrum of atomic hydrogen. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/21%3A_Spectra_and_Structure_of_Atoms_and_Molecules/21.03%3A_Atomic_Spectra_and_the_Bohr_Theory.txt |
In a classic paper published in 1913, the young Niels Bohr, then working with Rutherford in Manchester, England, proceeded to show how Rydberg’s formula could be explained in terms of a very simple model of the hydrogen atom. The model was based on the nuclear view of atomic structure which had just been proposed by Rutherford. Bohr’s model is shown in Figure $1$. An electron of charge –e and mass $m$ moves around a heavy nucleus of charge +e. Ordinarily the electron would move in a straight line, but the attraction of the nucleus bends its path so that it moves with a constant velocity $u$ in a perfect circle of radius r around the nucleus. The situation and the mathematics are very similar to that of a planet arcing round the sun. The major difference is that instead of the force of gravity there is an electrostatic force of attraction $F$ between the proton and the electron described by Coulomb’s law:
$F=k\frac{e^{2}}{r^{2}} \label{1}$
where $k$ has the value 8.9876 × 109 J m C–2.
Expressions for both the kinetic and potential energies of the electron can be derived using Equation $\ref{1}$ and the principles of elementary physics. Such a derivation can be found in most introductory physics texts. The two expressions are
$E_{k}=\frac{1}{2}mu^{2}=\frac{1}{2}k\frac{e^{2}}{r} \label{2}$
and
$E_{p}=-k\frac{e^{2}}{r} \label{3}$
If these are added together, we obtain a simple formula for the total energy of the electron:
\begin{align} E &=E_{k}+E_{p} \[4pt] &=-\frac{1}{2}k\frac{e^{2}}{r} \label{4} \end{align}
If we now insert the known values of $e$ and $k$, we have the result
\begin{align} E &= -\frac{1}{2}\times\frac{8.9876 \times 10^{9} \text{ J m C}^{-2}\times(1.6022 \times 10^{-19} \text{ C})^{2}}{r} \[4pt] &=-\frac{1.1536 \times 10^{-28} \text{ J M} }{r} \label{5} \end{align}
From Equation $\ref{5}$ we see that the total energy of the electron is very negative for an orbit with a small radius but increases as the orbit gets larger.
In addition to suggesting the planetary model just described, Bohr also made two further postulates which enabled him to explain the spectrum of hydrogen. The first of these was the suggestion that an electron of high energy circling the nucleus at a large radius can lose some of that energy and assume an orbit of lower energy closer to the nucleus. The energy lost by the electron is emitted as a photon of light of frequency $\nu$ given by Planck’s formula
$\Delta E = h\upsilon \label{6}$
where $ΔE$ is the energy lost by the electron.
Bohr’s second postulate was that only certain orbits are possible to the electron in a hydrogen atom. This enabled him to explain why it is that only light of a few particular frequencies can be emitted by the hydrogen atom. Since only a limited number of orbits are allowed, when an electron shifts from an outer to an inner orbit, the photon which emerges cannot have just any frequency but only that frequency corresponding to the energy difference between two allowed orbits.
Example $1$: Emitted Photon
According to Bohr’s theory two of the allowed orbits in the hydrogen atom have radii of 52.918 and 211.67 pm. Calculate the energy, the frequency, and the wavelength of the photon emitted when the electron moves from the outer to the inner of these two orbits.
Solution
Labeling the outer orbit 2 and the inner orbit 1, we first calculate the energy of each orbit from Equation $\ref{5}$:
$E_{2}=-\dfrac{1.1536\times10^{-28} \text{Jm}}{211.67\times10^{-12} \text{m}}=-0.54500 \text{aJ} \nonumber$
$E_{1}=-\dfrac{1.1536\times10^{-28} \text{Jm}}{52.918\times10^{-12} \text{m}}=-2.1780\, \text{aJ} \nonumber$
Thus
$\Delta E = – 0.545 00 \text{aJ} – (– 2.1780 \text{aJ}) = 1.6330\, \text{aJ} \nonumber$
Using Equation $\ref{6}$, we now have
$\upsilon=\dfrac{\Delta E}{h}=\dfrac{1.6330\times10^{-18}\text{ J}}{6.6262\times10^{-34} \text{ J s}}=2.4645\times10^{15} \text{ s}^{-1}=2.4645 \text{ PHz} \nonumber$
Finally $\lambda = \dfrac{c}{\upsilon} = 1.2164 \times 10^{-7} \text{ m} = 121.64 \text{ nm}$. In order to predict the right frequencies for the lines in the hydrogen spectrum, Bohr found that he had to assume that the quantity $mur$ (called the angular momentum by physicists) needed to be a multiple of $h/2π$. In other words the condition restricting the orbits to only certain radii and certain energies was found to be
$mur=\frac{nh}{2\pi} \qquad \label{7}$
where could have the value 1, 2, 3, etc.
By manipulating both Equations $\ref{7}$ and $\ref{2}$, it is possible to show that this restriction on the angular momentum restricts the radii of orbits to those given by the expression
$r = \frac{n^{2}h^{2}}{4\pi^{2}mke^{2}} \qquad n = 1, 2, 3, \cdots \label{8}$
If the known values of h, m, k, and e are inserted, this formula reduces to the convenient form
$r = n^2 \times 52.918 \text{ pm} \qquad n = 1, 2, 3, \cdots \label{9}$
Bohr’s postulate thus restricts the electron to orbits for which the radius is 52.9 pm, 22 × 52.9 pm, 32 × 52.9 pm, and so on.
If we substitute Equation $\ref{8}$ into Equation $\ref{4}$, we arrive at a general expression for the energy in terms of n:
$E=-\frac{1}{2}k\frac{e^{2}}{r}=-\frac{1}{2}ke^{2}\times\frac{4\pi^{2}mke^{2}}{n^{2}h^{2}} =\frac{2\pi^{2}k^{2}e^{4}m}{n^{2}h^{2}} \label{10}$ Again substituting in the known values for all the constants, we obtain $E=-\frac{2.1800 \text{ aJ}}{n^{2}} \label{11}$
The integer $n$ thus determines how far the electron is from the nucleus and how much energy it has, just as the principal quantum number $n$ described previously.
Example $2$: Ionization Energy
Using Equation $\ref{10}$ or $\ref{11}$ find the ionization energy of the hydrogen atom.
Solution
The ionization energy of the hydrogen atom corresponds to the energy difference between the electron in its innermost orbit (n = 1) and the electron when completely separated from the proton. For the completely separated electron r = ∞ (infinity) and so does n. Thus
$E_{1}=-\dfrac{2.1800 \text{aJ}}{1^{2}}=-2.1800\text{aJ} \nonumber$
and
$E_{\infty}=-\dfrac{2.1800 \text{aJ}}{\infty^{2}}=0.0000\text{aJ} \nonumber$
The energy difference is thus
$\Delta E=E_{infty}-E_{1}=2.1800 \text{aJ} \nonumber$
which is the ionization energy per atom. On a molar basis the ionization energy is the Avogadro constant times this quantity; namely,
$2.1800 \times 10^{-18} \text{J} \times 6.0221 \times 10^{23} \text{mol}^{-1} = 1312.8 \text{ kJ mol}^{-1}\ n\nonumber$
Note: In an atom, the electronic configuration of lowest energy is called the ground state while other configurations are called excited states.
We can now derive Rydberg’s experimental formula from Bohr’s theory. Suppose an electron moves from an outer orbit for which the quantum number is n2 to an inner orbit of quantum number n1. The energy lost by the electron and emitted as a photon is then given by
$\Delta E=E_{2}-E_{1}= – 2.1800 \text{ aJ} \left(\frac{1}{n_{2}^{2}}-\frac{1}{n_{1}^{2}}\right) = 2.1800 \text{ aJ} \left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right) \label{12}$
However,
$\Delta E=h\upsilon=\frac{hc}{\lambda} \label{13}$
where $λ$ is the wavelength of the photon. Combining Eqs. $\ref{2}$ and $\ref{3}$, we obtain
$\frac{1}{\lambda}=\frac{2.1800 \text{ aJ}}{hc}\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right) \nonumber$
or
$\frac{1}{\lambda}=1.0975\times10^{7}m^{-1} \left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right) \label{14}$
This expression is of exactly the same form as that found experimentally by Rydberg with a value for R of 1.0974 × 107 m–1, very close to the experimental value of 1.097 094 × 107 m–1. Even better agreement can be obtained if allowance is made for the fact that the nucleus is not stationary but that the electron and nucleus revolve around a common center of gravity.
Example $3$: Wavelength of Light in the Spectrum
Calculate the wavelength of the light emitted when the electron in a hydrogen atom drops from the n = 3 to the n = 1 orbit. In what region of the spectrum does this spectral line lie? To what series does it belong?
Solution
From Equation $\ref{14}$ we find
$\dfrac{1}{\lambda}=1.0975\times10^{7}m^{-1}\times\left(1-\dfrac{1}{9}\right)=9.7547\times10^{6}m^{-1} \nonumber$
giving $\lambda = 102.51 \text{ nm}$. This is the second line in the Lyman series and lies in the far ultraviolet. The experimentally determined wavelength is 102.573 nm.
Bohr’s success with the hydrogen atom soon led to attempts both by him and by others to extend the same model to other atoms. On the qualitative level these attempts met with some success, and a general picture of electrons occupying orbits in successive levels and sub-levels, similar to that shown in Figure 5.2, began to emerge. On the quantitative level, however, all attempts to calculate accurate values for the energies of the electrons in their quantized orbitals were dismal failures. It was not until Schrödinger’s introduction of wave mechanics in 1926 that these difficulties could be resolved. Suddenly, it seemed, everything fell into place. Since then virtually every line in the spectrum of every element has been accounted for theoretically. As a result, we now have a very exact, though mathematically rather complex, picture of the behavior of electrons in both the ground state and in excited states of atoms. In particular, the study of atomic spectra has allowed us to determine the ionization energies of all the elements very accurately.
Details of the spectra of polyelectronic atoms are complex, and so we will consider only one example: sodium. Excited states of sodium may be obtained by increasing the energy of the atom so that the 3s valence electron occupies the 3p, 3d, 4s, 4p, 4d, 4f or some other orbital. By contrast with the hydrogen atom, however, a sodium atom has other electrons which shield the valence electron from nuclear charge, and this shielding is different for each different orbital shape (s, p, d, f, etc.). Consequently the energy of an excited sodium atom whose electron configuration is 1s22s22p64s1 is not the same as that of an excited sodium atom whose configuration is 1s22s22p64p1. Different shielding of the outermost (4s or 4p) electron results in a different energy. Because of this, four formulas are needed to describe the energy of the sodium atom—one for each of the orbital shapes available to the outermost electron:
\begin{align*} E_{ns} & =\frac{2.1800 \text{ aJ}}{(n-a_{s})^{2}}\E_{np} & =\frac{2.1800 \text{ aJ}}{(n-a_{p})^{2}}\ E_{nd} &= \frac{2.1800 \text{ aJ}}{(n-a_{d})^{2}}\ E_{nf} & = \frac{2.1800 \text{ aJ}}{(n-a_{f})^{2}} \end{align*} \nonumber
In all these equations n represents the principal quantum number. It must be 3 or greater since the electron is in the 3s orbital to begin with. The different shielding requires a different correction for each type of orbital: as = 1.36; ap = 0.87; ad = 0.012; and af = 0.001.
Because there are four different sets of energy levels, the number of transitions between levels (and hence the number of lines in the spectrum) is larger for sodium than for hydrogen. Early spectroscopists were able to distinguish four different types of lines, which they labeled the sharp, principal, diffuse, and fundamental series. It is from the abbreviation of these terms that we have obtained the modern symbols s, p, d, and f.
As most readers will know, when almost any sodium compound is held in a Bunsen burner, it imparts a brilliant yellow color to the flame. This yellow color corresponds to the most prominent line in the sodium spectrum. Its wavelength is 589 nm. On the atomic scale this line is caused by the sodium atom moving from an excited state (in which the valence electron is in a 3p orbital) to the ground state (in which the electron is in a 3s orbital). Using the above equations we can obtain approximate values for the two energies involved in the transition:
$E_{3p}=\frac{2.1800 \text{ aJ}}{(3-0.87)^{2}}=-0.4805\, \text{ aJ} \nonumber$
and
$E_{3s}=\dfrac{2.1800 \text{ aJ}}{(3-1.36)^{2}}=-0.8105\, \text{ aJ}. \nonumber$
Thus $\Delta E= 0.3300 \text{ aJ}$ giving $\lambda = \dfrac{hc}{\Delta E}=602 \text{ nm}$ This agrees approximately with the experimental result. Another feature of the sodium spectrum deserves mention. Careful observation reveals that the yellow color of sodium is actually due to two closely spaced lines (a doublet). One has a wavelength of 588.995 nm, and the other is at 589.592 nm. When the electron is in a 3p orbital, its spin can be aligned in two ways with respect to the axis of the orbital. The small difference in energy between these two orientations results in two slightly different wavelengths | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/21%3A_Spectra_and_Structure_of_Atoms_and_Molecules/21.04%3A_Bohr_Theory_of_the_Atom.txt |
When we turn from the spectra of atoms to those of molecules, we find that the region of most interest to chemists is no longer the visible and ultraviolet but rather the infrared. As its name implies, the infrared extends beyond the red end of the visible spectrum, from the limit of visibility at roughly 0.8 μm (800 nm) up to about 100 μm where the microwave region begins. Another difference from the spectra discussed in the atomic spectra section is that infrared spectra are all absorption spectra rather than emission spectra. Infrared light is passed through the sample, and the intensity of light emerging is measured electronically. The energies of infrared photons are very much less than those of visible and ultraviolet photons. A photon of wavelength 10 μm has an energy of only 0.02 aJ (about 12 kJ m–1)—not even enough to break a hydrogen bond, let alone a normal covalent bond. It does have enough energy to make a molecule vibrate more strongly, however, and since vibrational energy is quantized, this can only happen at certain discrete frequencies and not at others.
Figure 1 shows the infrared spectra of two triatomic molecules, H2O and CO2, and also that of a more complex molecule, C2H5OH (ethanol). Each of the peaks in these spectra corresponds to a strong absorption of infrared radiation on the macroscopic level and a sudden increase in the amplitude with which the molecule vibrates on the microscopic level. Since a polyatomic molecule can vibrate in a variety of ways, there are several peaks for each molecule. The more complex the molecule, the larger the number of peaks. Note also that not all the vibrations correspond to the stretching and unstretching of bonds. A vibration in a polyatomic molecule is defined as any periodic motion which changes the shape or size of the molecule. In this sense bending and twisting motions also count as vibrations.
A useful feature of the vibrations which occur in polyatomic molecules is that many bonds and some small groups of atoms vibrate in much the same way no matter what molecule they are in. In Figure 1, for example, stretching of the O—H bond gives a peak between 2 and 3 μm for both H2O and C2H5OH. Because of this it is possible to identify many of the functional groups in an organic molecule merely by looking at its infrared spectrum. Figure \(2\) shows characteristic wavelengths by which some common functional groups can be identified. On the other hand, each molecule is a unique combination of chemical bonds and functional groups. Quite minor differences in molecular structure can result in noticeable differences in the infrared spectrum. Thus these spectra can be used in the same way the police use fingerprints.
When an unknown compound is prepared, one of the first things that is usually measured is its infrared spectrum. If this spectrum should happen to match that of a previously prepared compound, the unknown compound can be readily identified. If not, it may still be possible to identify some of the functional groups that are present.
In order for a molecular vibration to interact with electromagnetic radiation, the dipole moment of the molecule must change as the vibration takes place. The larger this change in dipole moment, the more strongly the substance absorbs the incident radiation. Thus very polar bonds like O—H and C==O usually produce very prominent peaks in an infrared spectrum. Conversely some vibrations do not feature in the infrared at all. In particular, diatomic molecules like N2 and O2, in which both atoms are identical, have zero dipole moment at any stage in a vibration. They produce no absorption in the infrared.
Since N2 and O2, are the chief constituents of the air, it is just as well that they do not absorb infrared radiation. The atmosphere would become intolerably hot if they did! As it is, only the minor constituents of the atmosphere, CO2 and H2O, absorb in the infrared. Nevertheless this absorption still plays an important role in maintaining the surface of the earth at its current temperature.
The earth absorbs energy from the sun by day, and radiates this energy away at night. The inflow and outflow must balance on average, otherwise the earth would heat up or cool down. Most of the sun’s radiation is in the visible region of the spectrum, but the radiation which escapes from the much cooler earth is mainly in the infrared, centered around 10 to 12 μm. As you can see from Figure \(1\), water absorbs infrared radiation between 2 to 3 μm and 6 to 7.5 μm. Water also absorbs strongly above 18 μm. Thus much of the outgoing infrared radiation is absorbed by water vapor in the earth’s atmosphere and prevented from escaping. You may have noticed that after a really humid summer day the temperature does not fall very fast at night. Excess water vapor in the atmosphere prevents radiation from escaping the earth’s surface. On the other hand, in a desert area the low humidity allows rapid heat loss. Although rocks may become hot enough to fry an egg by day, temperatures often drop to freezing overnight. While local concentrations of water vapor may vary from time to time, the total quantity in the earth’s atmosphere is buffered by the vast areas of ocean and remains nearly constant. Thus the average absorption of outgoing radiation by water seldom changes. The quantity of CO2 in the atmosphere is not so well regulated, however, and it appears that human activities are causing it to increase. (Using the data provided in Example 4 of Equations and Mass Relationships, you can calculate that about 9.4 Pg (9.4 × 1015 g) of CO2 results from the combustion of fossil fuels in the world each year.) Even in a relatively non- industrial area such as Hawaii, there has been a steady increase in CO2 concentration for many years.
Referring to Figure \(1\) again, we can clearly see that infrared absorption by CO2 occurs in just those parts of the spectrum that were not blocked by H2O absorption. Thus increasing the concentration of CO2 should decrease earth’s radiation to outer space and might increase the average surface temperature. On a global scale this is called the greenhouse effect—the CO2 and H2O act like the glass in a greenhouse, allowing visible light to pass in but blocking the loss of infrared. Climatologists have predicted that during the hundred years human beings have been using fossil fuels, the greenhouse effect should have raised surface temperatures by 0.5 to 1.0 K. Until 1950 that prediction appeared to have been borne out, but measured temperatures have since fallen back to about the 1900 level. Attempts to explain this drop on the basis of additional particulate matter in the atmosphere have met with varying degrees of success and failure. All that can be said for certain is that we know far less about the atmosphere and world climate than we would like.
21.06: The Visible and Ultraviolet Spectra of Molecules- Molecular Orbitals
When molecules absorb or emit radiation in the ultraviolet and visible regions of the spectrum, this almost always corresponds to the transition of an electron from a low-energy to a high-energy orbital, or vice versa. One might expect the spectra of molecules to be like the atomic line spectra shown in Figure \(1\), but in fact molecular spectra are very different. Consider, for example, the absorption spectrum of the rather beautiful purple-violet gas I2. This molecule strongly absorbs photons whose wavelengths are between 440 and 600 nm, and much of the orange, yellow, and green components of white light are removed. The light which passes through a sample of I2 is mainly blue and red. When analyzed with an average quality spectroscope, this light gives the spectrum shown in Figure \(2\)a. Instead of the few discrete lines typical of atoms, we now have a broad, apparently continuous, absorption band. This is typical of molecules.
Why is there this difference between atomic and molecular spectra? An answer begins to appear if we use a somewhat more expensive spectroscope. Figure \(\PageIndex{2b}\) shows a tracing of the I2 spectrum made with such an instrument. What originally appeared to he a continuous band is now shown to consist of a very large number of very narrow, closely spaced lines. Thus the broad absorption band of I2 is actually made up of discrete lines. The reason molecules give rise to such an enormous number of lines is that molecules can vibrate and rotate in a very large number of ways while atoms cannot. Furthermore both rotational levels and vibrational motion are quantized. When a molecule absorbs a photon of light and an electron is excited to a higher orbital, the molecule will not be stationary either before or after the absorption of the photon.
Because of the large number of energy possibilities both before and after the transition, a very large number of lines of slightly different wavelengths is obtained. A careful analysis of these lines yields much valuable information about the way in which the molecule rotates and vibrates. In particular, very accurate values of bond enthalpies and bond lengths can be obtained from a study of the fine structure of an absorption band like that shown in Figure \(\PageIndex{2b}\). | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/21%3A_Spectra_and_Structure_of_Atoms_and_Molecules/21.05%3A_The_Spectra_of_Molecules-_Infrared.txt |
In order to explain both the ground state and the excited state involved in an absorption band in the ultraviolet and visible spectra of molecules, it is necessary to look at the electronic structure of molecules in somewhat different terms from the description given in the sections on Chemical Bonding and Further Aspects of Covalent Bonding. In those chapters we treated electrons either as bonding pairs located between two nuclei, or as lone pairs associated with a single nucleus. Such a model of electronic structure is known as the valence-bond model. It is of very little use in explaining molecular spectra because photons are absorbed by the whole molecule, not an individual atom or bond. Thus we need to look upon electrons in a molecule as occupying orbitals which belong to the molecule as a whole. Such orbitals are called molecular orbitals, and this way of looking at molecules is referred to as molecular-orbital (abbreviated MO) theory.
The term molecular orbital is mentioned in the discussion of The Covalent Bond when we described formation of a covalent bond in an H2 molecule as a result of overlap of two 1s atomic orbitals—one from each H atom. In that section, however, we do not point out that there are two ways in which the 1s electron wave of one H atom can combine with the 1s electron wave of another. One of these involves constructive interference between the two waves and is referred to as positive overlap. This results in a bigger electron wave (and hence more electron density) between the two atomic nuclei. This attracts the positively charged nuclei together, forming a bond as described in "The Covalent Bond". A molecular orbital formed as a result of positive overlap is called a bonding MO.
It is also possible to combine two electron waves so that destructive interference of the waves occurs between the atomic nuclei. This situation is referred to as negative overlap, and it decreases the probability of finding an electron between the nuclei. In the case of two H atoms this results in a planar node of zero electron density halfway between the nuclei. Without a buildup of negative charge between them, the nuclei repel each other and no chemical bond is possible. A molecular orbital formed as a result of negative overlap is called an antibonding MO.
If one or more electrons occupy an antibonding MO, the repulsion of the nuclei increases the energy of the molecule, and so such an orbital is higher in energy than a bonding MO. This is shown in Figure $1$. Electron dot-density diagrams for the 1s electron in each of two separate H atoms are shown on the left and right sides of the figure. The horizontal lines show the energy each of these electrons would have. A dot-density diagram for a single electron occupying the bonding MO formed by positive overlap of the two orbitals is shown in the center of the diagram. This is labeled σ1s. Above it is a dot-density diagram for a single electron occupying the antibonding MO, σ1s*. (In general, antibonding MO’s are distinguished from bonding MO’s by adding an * to the label.) The energies of the molecular orbitals are indicated by the horizontal lines in the center of the diagram. The Greek letter σ in the labels for these orbitals refers to the fact that their positive or negative overlap occurs directly between the two atomic nuclei. Like an atomic orbital, each molecular orbital can accommodate two electrons. Thus the lowest energy arrangement for H2 would place both electrons in the a1s MO with paired spins. This molecular electron configuration is written (σ1s)2, and it corresponds to a covalent electron-pair bond holding the two H atoms together. If a sample of H2 is irradiated with ultraviolet light, however, an absorption band is observed between 110 and 170 nm. The energy of such an absorbed photon is enough to raise one electron to the antibonding MO, producing an excited state whose electron configuration is (σ1s)11s*)1. In this excited state the effects of the bonding and the antibonding orbitals exactly cancel each other; there is no overall bond between the two H atoms, and the H2 molecule dissociates. When the absorption of a photon results in the dissociation of a molecule like this, the phenomenon is called photodissociation. It occurs quite frequently when UV radiation strikes simple molecules.
The molecular-orbital model we have just described can also be used to explain why a molecule of He2 cannot form. If a molecule of He2 were able to exist, the four electrons would doubly occupy both the bonding and the antibonding orbitals, giving the electron configuration1s)21s*)2. However, the antibonding electrons would cancel the effect of the bonding electrons, and there would be no resultant buildup of electronic charge between the nuclei and hence no bond. Interestingly enough, an extension of this argument predicts that if He2 loses an electron to become the He2+ ion, a bond is possible. He2+ would have the structure (σ1s)21s*)1 and the single electron in the antibonding orbital would only cancel half the effect of the two electrons in the bonding orbital. This would leave the ion with a “half-bond” joining the two nuclei. The spectrum of He shows bands corresponding to He2+, and from them it can be determined that He2+ has a bond enthalpy of 322 kJ mol–1.
The molecular-orbital model can easily be extended to other diatomic molecules in which both atoms are identical (homonuclear diatomic molecules). Three general rules are followed. First, only the core orbitals and the valence orbitals of the atoms need be considered. Second, only atomic orbitals whose energies are similar can combine to form molecular orbitals. Third, the number of molecular orbitals obtained is always the same as the number of atomic orbitals from which they were derived.
Applying these rules to diatomic molecules which consist of atoms from the second row of the periodic table, such as N2, O2, and F2, we need to consider the 1s, 2s, 2px, 2py, and 2pz atomic orbitals. Since the 1s orbital of each atom differs in energy from the 2s, we can overlap the two 1s orbitals separately from the 2s. This gives a σ1s and a σ1s* MO, as in the case of H2. Similarly, the 2s orbitals can be combined to give σ2s and σ2s* before we concern ourselves with the higher energy 2p orbitals. There are three 2p orbitals on each atom, and so we expect a total of six molecular orbitals to be derived from them.
The shapes of these six molecular orbitals are shown by the boundary-surface diagrams in Figure $2$. Two of them are formed by positive and negative overlap of 2px orbitals directly between the atomic nuclei. Consequently they are labeled σ2px and σ2px*. Two more molecular orbitals are formed by sideways overlap of 2px atomic orbitals. These are labeled π2px and π2px*, because the molecular orbitals have two parts—one above and one below a nodal plane containing the nuclei. The atomic 2py orbitals also overlap sideways to form π2py and π2py* molecular orbitals. These are identical to π2px and π2px*, except for a 90° rotation around the line connecting the nuclei. Consequently π2px and π2py have the same energy, as do π2px* and π2py*.
The electron configuration for any homonuclear diatomic molecule containing fewer than 20 electrons can be built up by filling electrons into the molecular orbitals we have just derived, starting with the orbital of lowest energy.
The relative energies of the molecular orbitals at the time they are being filled are shown in Figure $3$. Like the energies of atomic orbitals given in Figure 1 from Electron Configurations, these relative molecular-orbital energies vary somewhat from one diatomic molecule to another. In particular the σ2p orbital is often lower than π2p. Nevertheless Figure $3$ gives the correct order of filling the orbitals, and we can use it to determine molecular electron configurations.
Example $1$ : Electron Configuration
Find the electronic configuration of the oxygen molecule, O2.
Solution
Starting with the lowest lying orbitals (σ1s and σ1s*) we add an appropriate number of electrons to successively higher orbitals in accord with the Pauli principle and Hund’s rule. O2 has 16 electrons, the first 14 of which are easily accommodated in the following way:
$( \sigma_{1s})^2 (\sigma_{1s}^* )^2 ( \sigma_{2s} )^2 (\sigma_{2s}^* )^2 ( \pi_{2p})^4 (\sigma_{2p})^2 \nonumber$
The remaining two electrons must now be added to the π2px* orbitals. Since both these orbitals are of equal energy, one electron must be placed in each orbital and the spins must be parallel. The total electronic structure is thus
$(\sigma_{1s})^2 (\sigma_{1s}^*)^2 (\sigma_{2s})^2 (\sigma_{2s}^*)^2 ( \pi_{2p})^4 (\sigma_{2p})^2 (\pi_{2px}^*)^1 (\pi_{2py}^*)^1 \nonumber$
As the previous problem shows, the molecular-orbital model predicts that O2 has two unpaired electrons. Substances whose atoms, molecules, or ions contain unpaired electrons are weakly attracted into a magnetic field, a property known as paramagnetism. (In a few special cases, such as iron, a much stronger magnetic attraction called ferromagnetism is also obsewed.) Most substances have all their electrons paired. Such materials are weakly repelled by a magnetic field, a property known as diamagnetism. Hence measurement of magnetic properties can tell us whether all electrons are paired or not. O2, for example, is found to be paramagnetic, an observation which agrees with the electron configuration predicted in Example 21.6. Before the advent of MO theory, however, the paramagnetism was a mystery, since the Lewis diagram predicted that all electrons should be paired.
The molecular-orbital model also allows us to estimate the strengths of bonds in diatomic molecules. We simply count each electron in a bonding orbital as contributing half a bond, while each electron in an antibonding orbital takes away half a bond. Thus if there are B electrons in bonding orbitals and A electrons in antibonding orbitals, the net bond order is given by
$\text{ Bond order} = \frac{A-B}{2} \nonumber$
The larger the bond order, the more strongly the atoms are held together.
Example $2$ : Bond Order
Calculate the bond order for the molecule N2.
Solution
There are 14 electrons, and so the electron configuration is
$( \sigma_{1s})^2 (\sigma_{1s}^* )^2 ( \sigma_{2s} )^2 (\sigma_{2s}^* )^2 ( \pi_{2p})^4 (\sigma_{2p})^2 \nonumber$
There are a total of 2 + 2 + 4 + 2 = 10 electrons in bonding MO’s and only 4 in the antibonding orbitals σ1s* and σ2s*. Thus
$\text{ Bond order} = \frac{10-4}{2} = 3 \nonumber$
The bond orders derived from the molecular-orbital model for stable molecules agree exactly with those predicted by Lewis’ theory. Not only do we find a triple bond for N2, but we also find a double bond for O2 and a single bond for F2. The results of such bond-order calculations are summarized in Table 1.
TABLE $1$ Bond Orders, Bond Enthalpies, and Bond Lengths of Some Diatomic Molecules.
Molecule* Bond Order Bond Enthalpy/kJ mol–1 Bond Length/pm
$\text{H}^{+}_{2}$ $\frac{1}{2}$ $256$
$106$
$\text{H}_2$ $432$
$74$
$\text{He}^{+}_{2}$ $\frac{1}{2}$ $322$
$108$
$\text{He}_2$
$0$
$\text{Molecule not detected}$
$\text{Li}_2$
$1$
$110$
$267$
$\text{Be}_2$
$0$
$\text{ Molecule not detected}$
$\text{B}_2$
$1$
$274$
$159$
$\text{C}_2$
$2$
$603$
$124$
$\text{N}_2$
$3$
$942$
$110$
$\text{O}_2$
$2$
$494$
$121$
$\text{F}_2$
$1$
$139$
$142$
$\text{Ne}_2$
$0$
$\text{Molecule not detected}$
* Some molecule-ions such as H2+ are included. Except fur the number of electrons involved, the MO theory is applied to them in exactly the same way as to molecules.
Some of the molecules in the table, such as C2 and B2, are only stable at high temperatures or only exist transitorily in discharge tubes, and so you are probably not familiar with them. Nevertheless, their spectra can be studied. Also included in the table are values for the bond enthalpies and bond lengths of the various species obtained from their spectra. Note the excellent qualitative agreement with the MO theory. The higher the bond order predicted by the theory, the larger the bond enthalpy and the shorter the bond length. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/21%3A_Spectra_and_Structure_of_Atoms_and_Molecules/21.07%3A_Molecular_Orbitals.txt |
One of the most useful aspects of molecular-orbital theory only becomes apparent when we consider molecules containing three or more atoms. According to molecular-orbital theory, electrons occupy orbitals which are delocalized. That is, the orbitals spread over the entire molecule. An electron is not confined only to the vicinity of one or two atomic nuclei as in the lone pairs and bond pairs of valence-bond theory. As seen in the section on Resonance, there are some molecules, such as O3 and C6H6, which cannot be described adequately by a single Lewis diagram. In such cases valence-bond theory resorts to resonance hybrids, such as
for ozone. The same molecules can be handled by MO theory without the need for several contributing structures, because electrons can occupy orbitals belonging to the molecule as a whole.
As an example of this we shall apply MO theory to the pi orbitals in ozone. The sigma bonds in this molecule may be attributed to overlap of sp2 hybrid orbitals on each of the three oxygen atoms. You will recall that sp2 hybrids are directed toward the corners of an equilateral triangle, in reasonable agreement with the 117° angle in ozone. For the sigma bonds and lone pairs, then, we have the Lewis diagram
This involves only 14 of the 18 valence electrons of the three oxygens, and so 4 electrons are left for pi bonding. Furthermore, each oxygen has a p orbital perpendicular to the plane of its sp2 hybrids (that is, perpendicular to the plane of the three oxygen atoms), which has not yet been used for bonding.
These three remaining p orbitals can overlap sideways to form molecular orbitals, which are included in the jmol shown here. Using the scroll bar, you may select which molecular orbital to display for the ozone molecule. In order to demonstrate delocalized electrons, we will focus on three of the molecular orbitals shown. The first of these orbitals to highlight is seen by selecting the orbital labeled "N7" in the scroll menu. This orbital shows a concentrated electron density between the central oxygen and each of the other two, and is therefore a bonding MO. The second to focus on is orbital "N10", which looks very similar to the original atomic p orbitals on the end oxygen atoms. An electron in this orbital neither strengthens nor weakens the attractions between the atoms, and so the MO is said to be nonbonding. Finally, there is an antibonding MO which has nodes between each pair of oxygen atoms and can be seen by selecting orbital "N13". Since there are only four electrons, only the two lower-energy MO’s (the bonding and nonbonding) will be occupied. The two electrons in the bonding MO provide a bond order of 1, but this is spread over both O—O bonds, and so it contributes a bond order of one-half between each pair of oxygens. Similarly the two electrons in the nonbonding MO correspond to a lone pair, half on the left oxygen and half on the right. Including the sigma bonding framework, this gives 1½ bonds between each pair of oxygens and 2½ lone pairs on each end oxygen. This is exactly the average of the two resonance structures already given.
The MO treatment can also be used to interpret the spectrum of ozone. Ozone in the earth’s stratosphere absorbs much solar ultraviolet radiation which would otherwise cause damage to the biosphere. This absorption is due to a band centered around 255 nm which corresponds to excitation of an electron to the unfilled antibonding pi orbital shown in Figure \(1\) c.
Another important molecule to which MO theory can be applied usefully is benzene. As described in "Resonance," benzene can be represented by the resonance hybrid
This indicates that all C—C bonds are equivalent and intermediate between a single and a double bond. As in the case of ozone, we can treat the sigma bonds of benzene in valence-bond terms, dealing only with pi bonding by the molecular-orbital, method. The sigma bonding framework is
This involves 24 electrons in bonds formed by overlap of sp2 hybrid orbitals on the carbon atoms with 1s orbitals on each hydrogen or with other sp2 hybrids on other carbons. This gives a planar framework and leaves six p orbitals (one on each carbon) which are perpendicular to the molecular plane and can overlap sideways to form six pi molecular orbitals. Only the three lowest-energy MO’s are occupied, and their electron clouds are shown in Figure \(2\). All these MO’s are bonding, while the three not shown are anti- bonding. When the six valence electrons not used for sigma bonding occupy the pi-bonding MO’s, they are evenly distributed around the ring of carbon atoms. This contributes a bond order of one-half between each pair of carbons, or a total bond order of 1½ when the sigma bond is included. Thus the C—C bond is intermediate between a single and double bond, in accord with the resonance hybrid. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/21%3A_Spectra_and_Structure_of_Atoms_and_Molecules/21.08%3A_Delocalized_Electrons.txt |
In some molecules the delocalization of electron pairs can be very much more extensive than in ozone and benzene. This is particularly true of carbon compounds containing conjugated chains, i.e., long chains of alternating single and double bonds. An example is provided by vitamin A2 which has the structure
Both the physical and chemical properties of this molecule indicate that all the bonds in the conjugated chain (shown in color) are intermediate in character between single and double bonds and that the pi electrons are free to move over the whole length of the conjugated chain.
A very simple, though approximate, method of handling delocalized electrons mathematically is to treat them as though they were particles in a one-dimensional box of the same length as the chain. In the case of vitamin A2 the distance from the carbon atom at one end of the conjugated chain to that on the other is about 1210 pm. If we feed this value and the mass of the electron (9.110 × 10–31 kg) into the particle-in-the-box formula (Equation $\ref{1}$), we can obtain approximate values of the energy levels occupied by the delocalized electrons:
$E_{k}=\dfrac{1}{2m}\left(\dfrac{nh}{2d}\right)^{2}=\dfrac{1}{2\times9.110\times10^{-31} \text{ kg}}\left(\dfrac{n\times6.626\times10^{-34} \text{J s }}{2\times 1210\times 10^{-12} \text{ m}}\right) ^{2} \label{1}$
$E_{k}=n^{2}(4.11\times10^{-20} \text{ J}) \label{2}$
The first seven energy levels derived from this formula are shown in Figure $2$. Note that since there are 12 pi electrons (one for each carbon atom on the conjugate chain), these will occupy the six lowest levels in accord with the Pauli principle.
This model allows us to calculate the wavelength of the main absorption band in the spectrum of vitamin A2. If an electron from the highest occupied level (n = 6) is excited to the lowest unoccupied level (n = 7), the energy required can be calculated from Equation $\ref{2}$:
$\Delta E = E_{7}-E_{6}=(7^{2}-6^{2})\times4.11\times10^{-20} \text{ J}=5.34\times10^{-19} \text{ J} \nonumber$
Thus
$λ = \dfrac{hc}{\Delta E} = \dfrac{6.626\times10^{-34} \text{ Js}\times2.998\times0^{8} \text{ ms}^{-1}}{5.34\times10^{-19} \text{ J}} = 327 \text{ nm} \nonumber$
This is in reasonable agreement with the observed value of 350 nm, considering the approximate nature of the model.
As a general rule, the longer a conjugated chain, the longer the wavelength at which it absorbs. Ethene (C2H4), for example, has only one double bond and absorbs at 170 nm. Hexatriene (C6H8) has three alternating double bonds and absorbs at 265 nm, while vitamin A2, with six double bonds, absorbs at 350 nm. It is not difficult to explain this effect in terms of the particle-in-a-box model. According to Equation $\ref{1}$ the energy of a level varies inversely with the square of the length of the box. Thus the longer the conjugated chain, the closer the energy levels will be to each other, and the less energy a photon need have to excite an electron. Naturally the lower the energy of a photon, the longer its wavelength will be.
A similar effect is found for molecules containing several benzene rings. Since these correspond to Lewis formulas of alternating double and single bonds, they can also be regarded as conjugated systems containing delocalized electrons. Experimentally we find that the more benzene rings a molecule contains, the longer the wavelengths at which it absorbs:
Thus increasing the extent of electron delocalization increases the wavelength at which a molecule will absorb light, whether the electron is delocalized over rings or chains.
This behavior of delocalized electrons is important in the preparation of compounds which strongly absorb visible light, i.e., in the preparation of dyes. Very few compounds which are held together by sigma bonds alone are colored. The electrons are so tightly held that a very energetic photon is needed to excite them. In order for an organic molecule to absorb in the visible region of the spectrum, it must usually contain very delocalized pi electrons. Thus most dyes and most colored compounds occurring in living organisms turn out to be large molecules with extensive systems of conjugated double bonds. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/21%3A_Spectra_and_Structure_of_Atoms_and_Molecules/21.09%3A_Conjugated_Systems.txt |
Approximately three-quarters of the known elements are metals and conduct both heat and electricity very well. Morevoer, they have shiny surfaces; they are capable of being shaped by hammering (malleable) and also of being drawn into wires (ductile). These properties can be understood in terms of metallic bonding in which valence electrons are delocalized over an entire metallic crystal. The strength of metallic bonding varies roughly as the number of electrons available in this sea. Chemical properties of the metals include a tendency to lose electrons and form positive ions, and the ability of their oxides to function as bases. The extent of these characteristics varies from one metal to another.
• 22.1: Prelude to Metals
Chemical properties of the metals include a tendency to lose electrons and form positive ions, and the ability of their oxides to function as bases. The extent of these characteristics varies from one metal to another. Several borderline cases such as B, Si, Ge, As, Sb, and Te are difficult to classify as metals or nonmetals. These elements are usually referred to as the metalloids or semimetals.
• 22.2: Metallic Bonding
Electrons can be fed into one end of a metal wire and removed from the other end without causing any obvious change in the physical and chemical properties of the metal. To account for this freedom of movement modern theories of metallic bonding assume that the valence electrons are completely delocalized; that is, they occupy molecular orbitals belonging to the metallic crystal as a whole. These delocalized electrons are often referred to as an electron gas or an electron sea.
• 22.3: Metallurgy
The processing of ores may be divided into three steps. (1) Concentration to remove worthless material (gangue) or to convert the mineral into an appropriate form for subsequent processing. (2) The most-important step is reduction of the metal from a positive oxidation state. This may involve elevated temperatures, chemical reducing agents, electrolysis, or some combination of these treatments. (3) Refining is required to achieve the purity desired in the final product.
• 22.4: Beneficiation
Beneficiation is any process which removes the gangue minerals from ore to produce a higher grade product, and a waste stream. Beneficiation may involve physical or chemical processes. Often, as in the case of panning for gold, the desired ore or metal is denser than the gangue. The latter can be suspended in a stream of water and flushed away.
• 22.5: Reduction of Metals
The ease with which a metal may be obtained from its ore varies considerably from one metal to another. Since the majority of ores are oxides or can be converted to oxides by roasting, the free-energy change accompanying the decomposition of the oxide forms a convenient measure of how readily a metal may be obtained from its ore.
• 22.6: Refining of Metals
Once a metal is reduced, it is still not necessarily pure enough for all uses to which it might be put. An obvious example is the brittleness and low tensile strength of pig iron, characteristics which make it suitable for casting, but little else. Steelmaking involves oxidation of the impurities in basic oxygen, open hearth, or electric furnaces.
• 22.7: Corrosion
An important aspect of the use of some metals, particularly of iron, is the possibility of corrosion. It is estimated that about one-seventh of all iron production goes to replace the metal lost to corrosion. Rust is apparently a hydrated form of iron(III)oxide. Rusting requires both oxygen and water, and it is usually sped up by acids, strains in the iron, contact with less-active metals, and the presence of rust itself.
• 22.8: Coordination Compounds
A characteristic feature of the transition metals is their ability to form a group of compounds called coordination compounds, complex compounds, or sometimes simply complexes.
• 22.9: Geometry of Complexes
• 22.10: Chelating Agents
Chelating agents are ligands that are able to form two or more coordinate covalent bonds with a metal ion. An important and interesting example of this is the chelating agents—ligands which are able to form two or more coordinate covalent bonds with a metal ion. One of the most common of these is 1,2-diaminoethane (usually called ethylenediamine and abbreviated en.) For metals which display a coordination number of 6, an especially potent ligand is ethylenediaminetetraacetate ion (EDTA).
• 22.11: Transitional Metal Ions in Aqueous Solutions
22: Metals
Approximately three-quarters of the known elements display the macroscopic properties characteristic of metals. They conduct both heat and electricity very well; they have shiny surfaces; they are capable of being shaped by hammering (malleable) and also of being drawn into wires (ductile). These properties can be understood in terms of metallic bonding in which valence electrons are delocalized over an entire metallic crystal. Positive metal ions formed by loss of valence electrons are held together by an electron sea. The strength of metallic bonding varies roughly as the number of electrons available in this sea. Chemical properties of the metals include a tendency to lose electrons and form positive ions, and the ability of their oxides to function as bases. The extent of these characteristics varies from one metal to another. Several borderline cases such as B, Si, Ge, As, Sb, and Te are difficult to classify as metals or nonmetals. These elements are usually referred to as the metalloids or semimetals. As you will recall from the discussion of metals on the periodic table, one can draw a zigzag line across the periodic table from B to At which separates the metals from the nonmetals and semimetals. This line is clearly indicated in most periodic tables. Periodic Table Live allows you to select metals, semi-metals, or non-metals and see how they are divided on the periodic table.
A great many metals and alloys are of commercial importance, but metals occur naturally in oxide, carbonate, or sulfide ores. Such ores must be concentrated (beneficiated) before they can be reduced to the metal, and usually the raw metal must be purified (refined) in a third step. An excellent example of these processes involves iron which can be readily beneficiated because its ore is ferromagnetic. Iron ore is then reduced in a blast furnace and purified in a steel making furnace. Since ore-reduction is a nonspontaneous process, its reverse, oxidation or corrosion of a metal, is often a problem. This is especially true in the case of iron because the oxide coating which forms on the metal surface does not protect the remaining metal from atmospheric oxidation.
Here we will be concerned mainly with the transition metals. We have already covered metals which are representative elements, such as alkali metals and alkaline earth metals. A discussion of the lanthanoid and actinoid metals is beyond the scope of general chemistry. Since transition metals contain d electrons in their valence shell, their chemistry is somewhat different from that of the representative elements. In particular they form a family of compounds called complex compounds or coordination compounds which are very different from those we have encountered up to this point. In these complexes, several ligands which can serve as Lewis bases are bonded to a metal ion which serves as a Lewis acid. The number of ligands is called the coordination number, and defines the possible geometries of a complex. For a coordination number of 2 the complex is usually linear. Both square planar and tetrahedral structures occur for coordination number 4, and coordination, number 6 usually involves an octahedral structure. Square planar and octahedral structures give rise to cis-trans isomerism.
Some ligands, called chelating agents, can coordinate-covalent bond to metal ions at more than one site. Chelate complexes are often important in biological systems because they can disguise the charge of a metal ion, stabilizing the ion in a hydrophobic environment.
In aqueous solution transition-metal ions are usually octahedrally coordinated by water molecules, but often other ligands which are stronger Lewis bases replace water. Such reactions often produce color changes, and they are usually rapid. A few metal ions, such as Cr(III), Co(III), Pt(IV), and Pt(II), undergo ligand substitution rather slowly and are said to be inert. Metal ions whose reactions are rapid are said to be labile. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/22%3A_Metals/22.01%3A_Prelude_to_Metals.txt |
Most metals have very compact crystal structures involving either the body-centered cubic, face-centered cubic, or hexagonal closest-packed lattices. Thus every atom in a metal is usually surrounded by 8 or 12 equivalent nearest neighbors. How can each atom be bonded to so many of its fellow atoms? Although there are plenty of electropositive atoms to donate electrons, there are no electronegative atoms to receive them, and so ionic bonding seems unlikely. Ordinary covalent bonding can also be ruled out. Each covalent bond would require one electron from each atom, and no metal has 12 valence electrons.
A valuable clue to the nature of bonding in metals is provided by their ability to conduct electricity. Electrons can be fed into one end of a metal wire and removed from the other end without causing any obvious change in the physical and chemical properties of the metal. To account for this freedom of movement modern theories of metallic bonding assume that the valence electrons are completely delocalized; that is, they occupy molecular orbitals belonging to the metallic crystal as a whole. These delocalized electrons are often referred to as an electron gas or an electron sea. Positive metal ions produced by the loss of these valence electrons can then be thought of as “floating” in this three-dimensional sea. Each ion is held in place by the attraction of the negatively charged electron sea and the repulsion of its fellow positive ions.
In order to see how MO theory can be applied to metals, let us first consider the simplest case, lithium. If two lithium atoms are brought together, the 1s core electrons remain essentially unchanged since there is virtually no overlap between them. The 2s orbitals, by contrast, overlap extensively and produce both a bonding and an antibonding orbital. Only the bonding orbital will actually be occupied by the two electrons, as shown in Figure 1. Somewhat higher than these two orbitals are a group of six unoccupied orbitals produced by the overlap of six 2p atomic orbitals (three on each atom). Suppose now we add a third atom to the two already considered so that we form a triangular molecule of formula Li3. As shown in the figure, the overlap of three 2s orbitals produces a lower group of three orbitals, while the overlap of three times three 2p orbitals produces a higher group of nine orbitals. Again the total number of molecular orbitals is equal to the number of atomic orbitals from which they are derived.
Continuing to add lithium atoms in this fashion, we soon attain a cluster of 25 lithium atoms. The energy-level situation for a cluster this size is a lower group of 25 MO’s, all deriving from 2s atomic orbitals, and a higher group of 75 MO’s, all deriving from 2p atomic orbitals. Note how closely spaced these energy levels have become. This is in line with the tendency for the energy levels to get closer the greater the degree of delocalization.
Finally, if we add enough lithium atoms to our cluster to make a visible, weigh-able sample of lithium, say 1020 atoms, the energy spacing between the molecular orbitals becomes so small it is impossible to indicate in the figure or even to measure. In effect an electron jumping among these levels can have any energy within a broad band from the lowest to highest. In consequence this view of electronic structure in solids is often referred to as the band theory of solids.
It should also be clear from Figure \(1\) that all the available molecular orbitals need not be completely filled with electrons. In the case of lithium, for example, a sample containing 1020 atoms would have 1020 valence electrons. Since each atom would have a single 2s orbital as well as three 2p orbitals, there would be 1 × 1020 MO’s in the 2s band and 3 × 1020 MO’s in the 2p band. If all electrons were paired, only the 0.5 × 1020 MO’s of lowest energy in the 2s band would be required to hold them. Note that there is a nice correspondence between the half-filled 2s band of the macroscopic sample and the half-filled 2s orbital of an individual Li atom.
According to the band theory, it is this partial filling which accounts for the high electrical and thermal conductance of metals. If an electric field is applied to a metallic conductor, some electrons can be forced into one end, occupying slightly higher energy levels than those already there. As a consequence of delocalization this increased electronic energy is available throughout the metal. It therefore can result in an almost instantaneous flow of electrons from the other end of the conductor.
A similar argument applies to the transfer of thermal energy. Heating a small region in a solid amounts to increasing the energy of motion of atomic nuclei and electrons in that region. Since the nuclei occupy specific lattice positions, conduction of heat requires that energy be transferred among nearest neighbors. Thus when the edge of a solid is heated, atoms in that region vibrate more extensively about their average lattice positions.
They also induce their neighbors to vibrate, eventually transferring heat to the interior of the sample. This process can be speeded up enormously if some of the added energy raises electrons to higher energy MO’s within an incompletely filled band. Electron delocalization permits rapid transfer of this energy to other atomic nuclei, some of which may be quite far from the original source.
When an energy band is completely filled with electrons, the mechanism just described for electrical and thermal conduction can no longer operate. In such a case we obtain a solid which is a very poor conductor of electricity, or an insulator. At first glance we might expect Be, Mg, and other alkaline earths to be insulators like this. Since atoms of these elements all contain filled 2s subshells, we would anticipate a filled 2s band in the solid for all of them. That this is not the case is due to the relatively small energy difference between the 2s and 2p levels in these atoms. As you can see from Figure \(1\), this small separation results in an overlap between the 2s and 2p bands. Thus electrons can move easily from the one band to the other and provide a mechanism for conduction.
Figure \(2\) shows four different possibilities for band structure in a solid. For a solid to be a conductor, a band must be either partially filled or must overlap a higher unfilled band. When there is a very large energy gap between bands and the lower band is filled, we have an insulator. If the gap is quite small, we get an intermediate situation and the solid is a semiconductor. All the semimetals found along the stairstep diagonal in the periodic table, notably germanium, have a band structure of this type.
In a semiconductor we find that collisions among atoms and electrons in the crystal are occasionally energetic enough to excite an electron into the top band. As a result there are always a small number of electrons in this band and an equal number of holes (orbitals from which electrons have been removed) in the lower-energy band. The excited electrons can carry electrical current because many different energy levels are available to them. So can the holes—other electrons from the nearly filled band can move up or down into them, a process which decreases or increases the energy of the hole.
In a metal, electrical conductivity decreases as the temperature is raised because the nuclei vibrate farther from their rest positions and therefore get in the way of moving valence electrons more often. Exactly the opposite behavior is found for semiconductors. With increasing temperature, more and more electrons are excited to the higher-energy conduction band so that more current can be carried. Excitation of electrons to the conduction band can also be accomplished by a photon, a phenomenon known as photoconduction. Selenium metal is often used in this way as a photocell in light meters and “electric eyes.”
The electron-sea model of metals not only explains their electrical properties but their malleability and ductility as well. When one layer of ions in an electron sea moves along one space with respect to the layer below it, a process we can represent pictorially as:
The final situation is much the same as the initial. Thus if we hit a metal with a hammer, the crystals do not shatter, but merely change their shape, This is very different from the behavior of ionic crystals.
The electron-sea model also enables us to explain, at least partially, why the metallic bond is noticeably stronger for some metals than others. While the alkali metals and some of the alkaline-earth metals can be cut with a knife, metals like tungsten are hard enough to scratch the knife itself. A good indication of how the strength of the metallic bonding varies with position in the periodic table is given by the melting point.
As can be seen from Figure \(3\), if the melting point of the metals is plotted against the group number for the three long periods, there is a sharp increase from group IA to group VB or VIB, after which there is a leveling off. Finally the melting point again drops to quite low values. A similar behavior is found for other properties such as boiling point, enthalpy of fusion, density, and hardness.
The initial increases in the strength of metallic bonding as we move from group IA to VIB can be explained in terms of the number of valence electrons the metal is capable of contributing to the electron sea. The more electrons an atom loses, the larger will be the charge of the positive ion embedded in the electron sea and the greater will be the electron probability density of the electron sea itself. Thus the more electrons which are lost, the more tightly the ions will be held together. Chromium with six valence electrons is thus much harder than sodium with one.
This trend cannot continue indefinitely, however. The more electrons that are removed from an atom, the more energy it takes to remove the next electron. Eventually we find that more energy is needed to remove an electron from a metal nucleus than is liberated by placing it in the electron sea. The strength of the bonding thus begins to level off and eventually to drop. It should be pointed out that metallic bonding strength is not solely dependent on the number of valence electrons (or the periodic group number) of an element. Other factors such as atomic radius and type of crystal lattice are also important. Nevertheless it is useful to remember that melting points and other properties related to metallic bond strength reach their maximum at about the middle of each transition series. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/22%3A_Metals/22.02%3A_Metallic_Bonding.txt |
Most metals are chiefly useful in elemental form, but they usually occur in compounds on the earth’s surface. An ore is a naturally occurring material from which one or more useful elements or compounds may be obtained at a cost that is economically feasible. As can be seen from the table, most metal ores are oxides, carbonates, or sulfides. A few of the least electropositive metals occur as the element.
Table \(1\): Occurrence of Metals.
Type of Ore
Examples
Native Metals Cu, Ag, Au, As, Sb, Bi, Pd, Pt
Oxides Al2O3, Fe2O3, Fe3O4, SnO2, MnO2, TiO2, FeO•Cr2O3, FeO•WO3, Cu2O, ZnO
Carbonates CaCO3, CaCO3•MgCO3,MgCO3, FeCO3, PbCO3, BaCO3, SrCO3, ZnCO3, MnCO3, CuCO3•Cu(OH) 2
Sulfides Ag2S, Cu2S, CuS, PbS, ZnS, HgS, FeS•CuS, FeS2, Sb2S3, Bi2S, MoS2, NiS, CdS
Halides NaCl, KCl, AgCl, KCl•MgCl2•6H2O, NaCl and MgCl2 in seawater
Sulfates BaSO4, SrSO4, PbSO4, CaSO4•2H2O, CuSO4•2Cu(OH)2
Silicates Be3Al2Si6O18, ZrSiO4, Sc2Si2O7
Phosphates LaPO4, LiF•AlPO4
Whether a mineral can usefully be regarded as an ore or not depends on such factors as how concentrated it is, its exact location, and whether there is a suitable process for extracting the metal. As the more accessible higher-grade ores become exhausted, less accessible and lower-grade ores are becoming increasingly utilized with a consequent shift in the meaning of the word ore. It is conceivable that many silicates could become sources of metals, notably aluminum. Currently, though, silicates are expensive decompose chemically, and silicates form ores only for relatively expensive metals like zirconium and beryllium.
The processing of ores may be divided into three steps. Often concentration or other beneficiation is required to remove worthless material (gangue) or to convert the mineral into an appropriate form for subsequent processing. The second and most-important step is reduction of the metal from a positive oxidation state. This may involve elevated temperatures, chemical reducing agents, electrolysis, or some combination of these treatments. Usually a third step, refining, is required to achieve the purity (or precise mixed composition in the case of an alloy) desired in the final product.
22.04: Beneficiation
Beneficiation is any process which removes the gangue minerals from ore to produce a higher grade product, and a waste stream. Beneficiation may involve physical or chemical processes. Often, as in the case of panning for gold, the desired ore or metal is denser than the gangue. The latter can be suspended in a stream of water and flushed away. The iron ore magnetite, Fe3O4, is ferrimagnetic. It can be separated from abundant deposits of taconite by grinding to a fine slurry in water. Passing this suspension over powerful electromagnets removes the Fe3O4, more than doubling the concentration of Fe. Both the physical beneficiation processes just described produce large quantities of tailings-water suspensions of the gangue. Usually the silicate and other particles are trapped in a settling basin, but in one case large quantities of taconite tailings have been dumped into Lake Superior for several decades.
Flotation
This is a partly physical and partly chemical beneficiation used to concentrate ores of copper, lead, and zinc. These metals have considerable affinity for sulfur. (the table on the metallurgy section lists sulfide ores for each of them.) The ore is ground and suspended with water in large tanks. Flotation agents such as the xanthates are added and a stream of air bubbles passed up through the tank. The sulfur-containing end of the xanthate molecule is attracted to the desired metal, while the hydrocarbon end tends to avoid water (like the hydrocarbon end of a soap molecule). The metal-containing particles are swept to the top of the tank in a froth much akin to soap bubbles. The concentrated ore may then be skimmed from the water surface. By means of flotation, ores containing as little as 0.3% copper can be concentrated to 20 to 30% copper, making recovery of the metal economically feasible.
Chemical Processes
Many ores can be chemically separated from their gangue by means of acid or base. Thus copper ore is often leached with acid which dissolves copper oxides and carbonates and leaves behind most silicates:
$\text{CuCo}_3 + \text{H}_2\text{SO}_4(aq) \rightarrow \text{CuSO}_4 (aq) + \text{H}_2\text{O} + \text{CO}_2 \label{1}$
In the Bayer process bauxite is dissolved in 30% NaOH:
$\text{Al}_2\text{O}_3(s) + \text{OH}^-(aq) + \text{3H}_2\text{O}(l) \rightarrow \text{2Al(OH}^-_4(aq) \label{2}$
and then reprecipitated by adding acid. Impurities such as Fe2O3 and SiO2 are eliminated in this way.
Many sulfide and carbonate ores are roasted in air in order to convert them into oxides, a form more suitable for further processing:
$\text{2ZnS}(s) + \text{3O}_2(g) \rightarrow \text{2ZnO}(s) + \text{SO}_2(g)\label{3}$
$\text{PbCO}_3 (s) \rightarrow \text{PbO}(s) + \text{Co}_2(g) \nonumber$
The production of SO2 in Equation $\ref{3}$ is a notorious source of air pollution, including acid rain. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/22%3A_Metals/22.03%3A_Metallurgy.txt |
The ease with which a metal may be obtained from its ore varies considerably from one metal to another. Since the majority of ores are oxides or can be converted to oxides by roasting, the free-energy change accompanying the decomposition of the oxide forms a convenient measure of how readily a metal may be obtained from its ore. Values of the free energy change per mol O2 produced are given in the table for a representative sample of metals at 298 and 2000 K. A high positive value of ΔGm° in this table indicates a very stable oxide from which it is difficult to remove the oxygen and obtain the metal, while a negative value of ΔGm° indicates an oxide which will spontaneously decompose into its elements. Note how the value of ΔGm° decreases with temperature in each case. This is because a gas (oxygen) is produced by the decomposition, and ΔS is accordingly positive.
Table $1$: Free-Energy Changes for the Decomposition of Various Oxides at 298 and 2000 K.
Reaction ΔGm° (298 K)/kJ mol–1 ΔGm° (2000 K)/kJ mol–1
$\frac{2}{3} \ce{Al2O3 \rightarrow \frac{4}{3} AlO_2}$
+1054
+691
$\ce{2MgO -> 2Mg + O2}$
+1138
+643
$\tfrac{2}{3}\ce{Fe2O3} \rightarrow \tfrac{4}{3}\ce{Fe + O2}$
+744
+314
$\text{SnO}_2 \rightarrow \text{Sn} + \text{O}_2$
+520
+42
$\ce{2HgO -> 2Hg + O2}$
+118
–381
$\ce{2Ag2O -> 4Ag + O2}$
+22
–331
Formation of CO2
$\text{C}(s) + \text{O}_2(g) \rightarrow \text{CO}_2(g)$
–394
–396
The two metals in the table which are easiest to obtain from their oxide ores are Hg and Ag. Since the ΔGm° value for the decomposition of these oxides becomes negative when the temperature is raised, simple heating will cause them to break up into O2 and the metal. The next easiest metals to obtain are Sn and Fe. These can be reduced by coke, an impure form of C obtained by heating coal. Coke is the cheapest readily obtainable reducing agent which can be used in metallurgy. When C is oxidized to CO2, the free-energy change is close to – 395 kJ mol–1 over a wide range of temperatures. This fall in free energy is not quite enough to offset the free-energy rise when Fe2O3 and SnO2 are decomposed at 298 K, but is more than enough if the temperature is 2000 K. Thus, for example, if Fe2O3 is reduced by C at 2000 K, we have, from Hess’ law,
${}_{\text{3}}^{\text{2}}\text{Fe}_{\text{2}}\text{O}_{\text{3}}\text{(}s\text{) }\to \text{ }{}_{\text{3}}^{\text{4}}\text{Fe(}l\text{) + O}_{\text{2}}\text{(}g\text{)}$ ΔGm° = +314 kJ mol–1
$\text{C(}s\text{) + O}_{\text{2}}\text{(}g\text{)}\to \text{ CO}_{\text{2}}\text{(}g\text{)}$ ΔGm° = –394 kJ mol–1
${}_{\text{3}}^{\text{2}}\text{Fe}_{\text{2}}\text{O}_{\text{3}}\text{(}s\text{) + C(}s\text{) }\to \text{ }{}_{\text{3}}^{\text{4}}\text{Fe(}l\text{) + CO}_{\text{2}}$ ΔGm° = –82 kJ mol–1
Thus ΔGm° for the reduction is negative, and the reaction is spontaneous.
The two metals in the table which are most difficult to obtain from their ores are Mg and Al. Since they cannot be reduced by C or any other readily available cheap reducing agent, they must be reduced electrolytically. The electrolytic reduction of bauxite to yield Al (the Hall process) is used to produce aluminum.
Reduction of Iron
Since iron is the most important metal in our industrial civilization, its reduction from iron ore in a blast furnace (Figure $1$ ) deserves a detailed description. The oxides present in most iron ores are Fe2O3 and Fe3O4. These oxides are reduced stepwise: first to FeO and then to Fe. Ore, coke, and limestone are charged to the furnace through an airlock-type pair of valves at the top. Near the bottom a blast of air, preheated to 900 to 1000 K, enters through blowpipes called tuyères. Oxygen from the air blast reacts with carbon in the coke to form carbon monoxide and carbon dioxide, releasing considerable heat. The blast carries these gases up through the ore, coke, and limestone, and they exit from the top of the furnace.
By the time the ore works its way to the lower part of the furnace, most of the Fe2O3 has already been reduced to FeO. In this region, temperatures reach 1600 to 2000 K, high enough to melt FeO and bring it into close contact with the coke. Most of the FeO is reduced by direct reaction with carbon, the latter being oxidized to carbon monoxide:
$\text{2C}(s) + \text{2FeO}(l) \rightarrow \text{2Fe}(l) + \text{2CO} \triangle G_m^o(2000 k) = -280 \frac{kJ}{mol} \nonumber$
The molten iron produced by this reaction drips to the bottom of the furnace where it is collected and occasionally tapped off.
Higher in the furnace, temperatures fall below the melting points of the iron oxides. Because there is little contact between solid chunks of ore and of coke, direct reduction by solid carbon is rather slow. Gaseous carbon monoxide contacts all parts of the ore, however, and reacts much more rapidly:
$\text{CO}(g) + \text{Fe}_2\text{O}_3(s) \rightarrow \text{CO}_2 (g) \text{2FeO}(s) \nonumber$
$\text{CO}(g) + \text{FeO}(s) \rightarrow \text{CO}_2(g) + \text{Fe}(s)$
Thus much of the “carbon reduction” in ironmaking is actually carried out by carbon monoxide.
The gangue in iron ore consists mainly of silicates and silica, SiO2. These impurities are removed in slag. Limestone added with coke and ore is calcined (decomposed to the oxide) by the high temperatures of the blast furnace:
$\text{CaCO}_3(s) \underset{\text{1100 K}}{\mathop{\rightarrow}}\, \text{CaO}(s) + \text{CO}_2(g) \nonumber$
Lime (CaO) serves as a flux, reducing the melting points (mp) of silica (SiO2) and silicates:
$\underset{\text{mp = 2853 K}}{\mathop{\text{CaO(}s\text{)}}}\,\text{ + }\underset{\text{mp = 1986 K}}{\mathop{\text{SiO}_{\text{2}}\text{(}s\text{)}}}\,\text{ }\to \text{ }\underset{\text{mp = 1813 K}}{\mathop{\text{CaSiO}_{\text{3}}\text{(}l\text{)}}} \nonumber$
The liquid silicates flow rapidly down through the hottest part of the furnace. This helps to prevent reduction of silica to silicon, hence yielding purer iron. The slag is less dense than molten iron and immiscible with it. Therefore the slag floats on the surface of the iron and can easily be tapped off.
Although most blast-furnace iron now goes directly to a steelmaking furnace in molten form, much of it used to be run into molds where it hardened into small ingots called pigs because of their shape. Consequently blast-furnace iron is still referred to as pig iron. A single large blast furnace may produce more than 106 kg iron per day. For each kilogram of iron, 2 kg iron ore, 1 kg coke, 0.3 kg limestone, 4 kg air, 63 kg water, and 19 MJ of fossil-fuel energy are required. The furnace produces 0.6 kg slag and 5.7 kg, flue gas per kg iron. Nearly 5 percent of the iron ore is lost in the form of small particles suspended in the flue gas unless, as in the furnace shown in Figure 1, air-pollution controls are installed. The latter trap FeO particles for recycling to the furnace and also make the flue gas (which contains about 12% CO and 1% H2) suitable as a fuel for preheating air fed to the tuyères. Thus control of blast-furnace air pollution (a major contributor to the one-time “smoky city” reputations of Pittsburgh, Pennsylvania and Gary, Indiana) also conserves ore supplies and energy resources. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/22%3A_Metals/22.05%3A_Reduction_of_Metals.txt |
Once a metal is reduced, it is still not necessarily pure enough for all uses to which it might be put. An obvious example is the brittleness and low tensile strength of pig iron, characteristics which make it suitable for casting, but little else. These adverse properties are due to the presence of impurities, a typical analysis of blast-furnace iron showing about 4% C, 2% P, 2.5%, Si, 2.5%Mn, and 0.1% S by weight. Further refining to remove these impurities (especially carbon) produces steel, a much stronger and consequently more useful material.
Steel making involves oxidation of the impurities in basic oxygen, open hearth, or electric furnaces. Some oxidation products (CO, CO2, and SO2) are volatile and easily separated. The others end up, along with some iron oxides, in a slag which floats on the surface of the molten steel.
In the open-hearth furnace oxidation is due to air at the surface of the molten metal. This method of steel refining was developed in the mid 1800s, contemporaneous with the industrial revolution. This method requires up to 12 hours—and natural gas or other fuel must be burned to keep the metal liquid. Thus the open hearth wastes large amounts of free energy. The use of fossil fuel does make it possible to recycle as much as 50 percent scrap iron, however, and the longer melting time allows somewhat greater control over the composition of a batch of steel.
Developed in the 1950s, the basic oxygen furnace has replaced the open-hearth furnace as the primary steel making method. In this process, pure oxygen is directed onto the surface of molten pig iron in a large crucible. Some of the iron is oxidized to Fe3O4 and Fe2O3, forming an oxidizing slag. The impurities, namely, C, P, Si, Mn, and S, are all oxidized at the same time. Since all these reactions are spontaneous and exothermic, they provide enough heat so that up to 25 percent solid scrap iron may be melted in the crucible without cooling it to the point where solid iron would remain. Oxidation of one batch of pig iron and scrap normally takes slightly more than half an hour. Thus, this method is far quicker than the open-hearth furnace, is three times more efficient.
Computers are now used to interpret spectroscopic analyses of steel in basic oxygen furnaces, indicating in a few minutes what metals must be added to obtain the desired composition. This has largely eliminated the last advantage of the open hearth and speeded up changeovers to basic oxygen. It has also decreased recycling of iron because the latter furnace cannot handle as much scrap. Much recycling of iron is now done in electric-arc furnaces which can melt a charge of 100 percent scrap.
In addition to the chemical oxidations used in steel making, electrolytic oxidation and reduction is quite important in refining metals. The electrolytic refining of copper, and aluminum production has already been described.
22.07: Corrosion
An important aspect of the use of some metals, particularly of iron, is the possibility of corrosion. It is estimated that about one-seventh of all iron production goes to replace the metal lost to corrosion. Rust is apparently a hydrated form of iron(III)oxide. The formula is approximately Fe2O3•$\tfrac{\text{3}}{\text{2}}$H2O, although the exact amount of water is variable. (Note that this is about halfway between iron(III) hydroxide, Fe(OH)3 or ½{Fe2O3•3H2O], and anhydrous Fe2O3).
Rusting requires both oxygen and water, and it is usually sped up by acids, strains in the iron, contact with less-active metals, and the presence of rust itself. In addition, observation of a rusted object, such as an iron nail from an old wooden building, shows that rust will deposit in one location (near the head of the nail) while the greatest loss of metallic iron will occur elsewhere (near the point). These facts suggest that the mechanism of rusting involves a galvanic cell. The half-equations involved are
$\text{2Fe}(S) \rightarrow \text{2Fe}^{2+}(aq) + \text{4}e^-\label{1}$
$\text{4}e^- + \text{4H}^+(aq) + \text{O}_2(g) \rightarrow \text{2H}_2\text{O}\label{2}$
yielding the full reaction:
$\text{2Fe}(s) + \text{4H}^+(aq) + \text{O}_2(g) \rightarrow \text{2Fe}^{2+}(aq) + \text{2H}_2\text{O}\label{3}$
Once Fe2+(aq) is formed, it can migrate freely through the aqueous solution to another location on the metal surface. At that point the iron can precipitate:
$\text{4Fe}(s) + \text{O}_2(g) + \text{7 H}_2\text{O}(l) \rightarrow \text{2Fe}_2\text{O}_3 \cdot \frac{3}{2} \text{H}_2\text{O}(s) + \text{8H}^+(aq) \nonumber$
Hydrogen ions liberated by this reaction are then partially consumed by Equation $\ref{2}$. The electrons required for half-equation $\ref{2}$ are supplied from Equation $\ref{1}$ via metallic conduction through the iron or by ionic conduction if the aqueous solution contains a significant concentration of ions. Thus iron rusts faster in contact with salt water than in fresh.
The mechanism proposed in the preceding paragraph implies that some regions of the iron surface become cathodic, i.e., that reduction of oxygen to water occurs there. Other locations are anodic; oxidation of Fe to Fe2+ occurs. The chief way in which such regions may be set up depends on restriction of oxygen supply, because oxygen is required for the cathodic reaction shown in Equation $\ref{2}$. In the case of the iron nail, for example, rust forms near the head because more oxygen is available. Most of the loss of metal takes place deep in the wood, however, near the point of the nail. At this location Equation $\ref{1}$ but not $\ref{2}$ can occur.
A similar situation occurs when a drop of moisture adheres to an iron surface (Figure $1$). Pitting occurs near the center of the drop, while hydrated iron(III) oxide deposits near the edge.
A second way in which anodic and cathodic regions may be set up involves the presence of a second metal which has a greater attraction for electrons (is less easily oxidized) than iron. Such a metal can drain off electrons left behind in the iron when Fe2+ dissolves. This excess of electrons makes the less-active metal an ideal site for Equation $\ref{2}$, and so a cell is set up at the intersection of the metals. Rust may actually coat the surface of the less-active metal while pits form in the iron.
The most important technique for rust prevention is simply to exclude water and oxygen by means of a protective coating. This is the principle behind oiling, greasing, painting, or metal plating of iron. The coating must be complete, however, or rusting may be accelerated by exclusion of oxygen from part of the surface. This is especially true when iron is coated with a less-active metal such as tin. Even a pinhole in the coating on a tin can will rust very quickly, since the tin becomes cathodic due to its larger electrode potential and to the oxygen exclusion from the iron beneath.
A second technique involves bringing the iron object in contact with a more active metal. This is called cathodic protection because the more active metal donates electrons to the iron, strongly inhibiting Equation $\ref{1}$. Both cathodic protection and a surface coating are provided by galvanizing, a process in which zinc is plated onto steel electrolytically or by dipping in the molten metal. Like many other metals, zinc is self-protective—it reacts with oxygen and carbon dioxide from air to form an adherent impervious coating of zinc hydroxycarbonate, Zn2(OH)2CO3. Should there be a scratch in the zinc plate, the iron still cannot rust because zinc will be preferentially oxidized. The hydroxycarbonate formed will then cover the opening, preventing further contact of oxygen with the iron or zinc.
A third technique applies to situations (such as an automobile radiator) where aqueous solutions are in contact with the iron. Corrosion inhibitors include chromate salts and organic compounds such as tributylamine, (C4H9)3N. Chromates apparently form an impervious coating of FeCrO4(s) as soon as any iron is oxidized to iron(II). Tributylamine, a derivative of ammonia, reacts with organic acids formed by decomposition of antifreeze at the high temperatures of an automobile engine. The tributylammonium salts produced are insoluble and coat the inside of the cooling system. Thus tributylamine neutralizes acid which would accelerate corrosion and provides a surface coating as well. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/22%3A_Metals/22.06%3A_Refining_of_Metals.txt |
A characteristic feature of the transition metals is their ability to form a group of compounds called coordination compounds, complex compounds, or sometimes simply complexes. A typical coordination compound is the intensely blue solid substance Cu(NH3)4SO4 which can be crystallized from solutions of CuSO4 to which a very large excess of concentrated NH3 has been added. These crystals contain two polyatomic ions, one of which is the sulfate ion, SO42, and the other of which is the complex ion Cu(NH3)42+ which is responsible for the blue color.
We can regard a complex ion such as Cu(NH3)42+ as being the result of the interaction of :NH3 acting as a Lewis base with the Cu2+ ion acting as a Lewis acid. Each NH3 molecule can be considered as donating a pair of electrons to a central Cu2+, thus forming four coordinate covalent (dative) bonds to it:
Most coordination compounds contain a complex ion similar to Cu(NH3)42+. This ion can be either positively charged like Cr(H3O)63+, or it can be negatively charged like CoCl63. Neutral complexes like Pt(NH3)2Cl2 are also known. All these species contain a central metal ion attached by coordinate covalent bonds to several ligands. These ligands are invariably Lewis bases. Some typical examples of ligands are H2O, NH3, Cl, OH, CN, Br, and SCN. The number of ligands attached to the central metal ion is said to be its coordination number and is usually 2, 4, or 6. The group of ligands bonded to the metal taken collectively is said to constitute the metal’s coordination sphere.
When writing the formula of a coordination compound containing complex ions, square brackets are usually used to enclose the coordination sphere. Examples are
• $\text{[Cu(NH}_{3} \text{)}_{4} \text{]SO}_{4}$
• $\text{[Cr(H}_{2} \text{O)}_{6} \text{]Cl}_{3}$
• $\text{[Pt(NH}_{3} \text{)}_{2} \text{Cl}_{2} ]$
• $\text{[Cu(NH}_{3} \text{)}_{4} \text{](NO}_{3} \text{)}_{2}$
• $\text{K}_{3} \text{[Fe(CN)}_{6} ]$
• $\text{[Pt(NH}_{3} \text{)}_{4} \text{][PtCl}_{4} ]$
When such compounds are dissolved in H2O, each of the ions present in the solid becomes an independent species with its own chemical and physical properties. Thus, when 1 mol [Cr(H2O)6]Cl3 crystal is dissolved in H2O the solution contains 1 mol Cr(H2O)63+ ion which can be recognized by its characteristic grayish-violet color and 3 mol Cl which can be detected by the precipitate of AgCl which forms when AgNO3 is added to the solution.
Table $1$: Observations on Complex Compounds Containing PtCl2, NH3, and KCl.
Compound Molar Conductivity/A V–1 dm2 mol–1 Moles AgCl Precipitated per Mole Compound Electrode to which Pt Migrates During Electrolysis
[Pt(NH3)4]Cl2
3.0
2 immediately
Cathode
[Pt(NH3)3Cl]Cl
1.2
1 immediately; 1 after several hours
Cathode
Pt(NH3)2Cl2
~ 0
2 after several hours
Does not migrate
K[Pt(NH3)Cl3]
1.1
3 after several hours
Anode
K2[PtCl4]
2.8
4 after several hours
Anode
An even better example of how the various ions in a coordination compound can behave independently when dissolved in water is provided by the set of Pt(II) complexes shown in the table. The first of these compounds contains the complex ion [Pt(NH3)4]2+ and in each subsequent compound one of the NH3 ligands in the coordination sphere of the Pt is replaced by a Cl ligand. As a result each compound contains a Pt complex of different composition and also of different charge, and when dissolved in H2O, it shows just the conductivity and other properties we would expect from the given formula. When 1 mol [Pt(NH3)3Cl]Cl is dissolved in H2O, it furnishes 1 mol Pt(NH3)3Cl+ ions and 1 mol Cl ions. The strongest evidence for this is the molar conductivity of the salt (1.2 A V–1 dm2 mol–1), which is very similar to that of other electrolytes like NaCl (1.3 A V–1 dm2 mol–1) which also yield a +1 ion and a –1 ion in solution, but very different from that of electrolytes like MgCl2 (2.5 A V–1 dm2 mol–1) which yield one + 2 ion and two –1 ions in solution. The conductivity behavior also suggests that the Pt atom is part of a cation, since the Pt moves toward the cathode during electrolysis. The addition of AgNO3 to the solution serves to confirm this picture. One mol AgCl is precipitated immediately, showing 1 mol free Cl ions. After a few hours a further mole of AgCl is precipitated, the Cl this time originating from the coordination sphere of the Pt atom due to the slow reaction
$\text{[Pt(NH}_{3} \text{)}_{3} \text{Cl]}^{+} (aq) + \text{Ag}^{+} (aq) + \text{H}_{2} \text{O} \rightarrow \text{[Pt(NH}_{3} \text{)}_{3} \text{H}_{2} \text{O]}^{2+} (aq) + \text{AgCl} (s) \nonumber$
It is worth noting that in all these compounds, Pt has an oxidation number of + 2. Thus the combination of Pt with one NH3 ligand and three Cl ligands yields an overall charge of 2(for Pt) – 3(for Cl) + 0(for NH3) = –1. The ion is thus the anion [PtNH3Cl3] found in compound 4.
Example $1$: Oxidation State
What is the oxidation state of Pt in the compound Ca[Pt(NH3)Cl5]2?
Solution
Since there are two complex ions for each Ca2+ ion, the charge on each must be –1. Adding the charge on each ligand, we obtain –5(for Cl) + 0(for NH3) = –5. If the oxidation number of Pt is x, then x – 5 must equal the total charge on the complex ion:
$x \text{ } – \text{ } 5 = \text{ } –1 \ ~~ \ \nonumber$
or
$x =+4 \nonumber$
The compound in question is thus a Pt(IV) complex. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/22%3A_Metals/22.08%3A_Coordination_Compounds.txt |
The geometry of a complex is governed almost entirely by the coordination number. We will consider only the most common coordination numbers, namely, 2, 4, and 6.
Coordination number = 2
Complexes with two ligands are invariably linear. The best-known examples of such compounds are Ag(I) and Au(I) complexes such as
Both of these complexes are important. The Au(CN)2 complex is used to extract minute gold particles from the rock in which they occur. The crushed ore is treated with KCN solution and air is blown through it:
$4 \text{Au} (s) + 8 \text{CN}^{-} (aq) + \text{O}_{2} (g) + 2 \text{H}_{2} \text{O} (l) \rightarrow 4 \text{[Au(CN)}_{2} \text{]}^{-} (aq) + 4 \text{OH}^{-} (aq) \label{1}$
The resultant complex is water soluble. The silver complex is also water soluble and affords a method for dissolving AgCl, which is otherwise very insoluble.
$\text{AgCl} (s) + 2 \text{NH}_{3} (aq) \rightarrow \text{[Ag(NH}_{3} \text{)}_{2} \text{]}^{+} (aq) + \text{Cl}^{-} (aq) \nonumber$
This reaction is often used in the laboratory to be sure a precipitate is AgCl(s).
Coordination number = 4
Two geometries are possible for this coordination number. Some complexes, like the [Pt(NH3)4]2+ ion shown in Figure $1$, are square planar, while others, like Cd(NH3)42–, are tetrahedral. Most of the four-coordinated complexes of Zn(II), Cd(II), and Hg(II) are tetrahedral, while the square planar arrangement is preferred by Pd(II), Pt(II), and Cu(II) complexes.
Because the square planar geometry is less symmetrical than the tetrahedral geometry, it offers more possibilities for isomerism. A well-known example of such isomerism is given by the two square planar complexes
These two isomers are called geometrical isomers. That isomer in which two identical ligands are next to each other is called the cis isomer, while that in which they are on opposite sides is called the trans isomer. Though these two isomers have some properties which are similar, no properties are identical and some are very different. For example, the cis isomer of the above complex is used as an anti-tumor drug to treat cancerous cells. The trans form, by contrast, shows no similar biological activity.
It is worth noting that cis-trans isomerism is not possible in the case of tetrahedral complexes. As you can quickly verify by examining any three-dimensional tetrahedral shape, any given corner of a tetrahedron is adjacent to the other three. Since all the corners are cis to each other, none are trans.
Coordination number = 6
When there are six ligands, the geometry of the complex is almost always octahedral, like the geometry of SF6, or of [Cr(H2O)6]3+. All ligands are equidistant from the central atom, and all ligand-metal-ligand angles are 90°. An octahedral complex may also be thought of as being derived from a square planar structure by adding a fifth ligand above and a sixth below on a line through the central metal ion and perpendicular to the plane.
The octahedral structure also gives rise to geometrical isomerism. For example, two different compounds, one violet and one green, have the formula [Co(NH3)4Cl2]Cl. The violet complex turns out to have the cis structure and the green one trans, as shown in Figure $2$. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/22%3A_Metals/22.09%3A_Geometry_of_Complexes.txt |
Although we have confined our discussion so far to simple ligands such as Cl, NH3, or H2O, much larger and more complicated molecules can also donate electron pairs to metal ions. An important and interesting example of this is the chelating agents—ligands which are able to form two or more coordinate covalent bonds with a metal ion. One of the most common of these is 1,2-diaminoethane (usually called ethylenediamine and abbreviated en.)
When both nitrogens coordinate to a metal ion, a stable five-member ring is formed. The word chelating, derived from the Greek chele, “claw,” describes the pincer-like way in which such a ligand can grab a metal ion.
A chelating agent which forms several bonds to a metal without unduly straining its own structure is usually able to replace a similar simpler ligand. For example, although both form coordinate covalent bonds via groups, ethylenediamine can readily replace ammonia from most complexes:
For metals which display a coordination number of 6, an especially potent ligand is ethylenediaminetetraacetate ion (abbreviated EDTA):
All six electron pairs marked in color are capable of coordinating to a metal ion, in which case the EDTA ion wraps completely around the metal and is very difficult to dislodge. EDTA is used to treat lead and mercury poisoning because of its ability to chelate these metals and aid their removal from the body.
Chelate complexes are often important in living systems. The coordination of iron in proteins such as myoglobin or hemoglobin involves four nitrogen of the heme group and one from a histidine side chain. Since iron normally has a coordination number of 6, this leaves one open site, to which oxygen can bond. The presence of carbon monoxide, a stronger ligand than oxygen, causes displacement of oxygen from hemoglobin. This prevents transport of oxygen from the lungs to the brain, causing drowsiness, loss of consciousness, and even death upon long exposure to carbon monoxide. Consequently operating an automobile in a closed garage, a cookstove in a tent, or burning any fossil fuel incompletely in an enclosed space may be hazardous to one’s health.
Another important application of chelates is transport of metal ions across membranes. The interior of a biological membranes contain the nonpolar, hydrophobic tails of lipid molecules. This makes it quite difficult for ionic species such as K+ and Na+ to travel from one side of a membrane to the other. One way in which this barrier may be circumvented is by carrier molecules, called ionophores. Ionophores are able to chelate an ion, but also have a hydrophobic exterior.
One such ionophore is the antibiotic nonactin, a medium-sized organic molecule with the formula
This molecule is able to transport K+ ions but not Na+ ions. Apparently the Na+ ion is too small to fit in among the eight coordinating O’s, while the K+ ion can (Figure 1). Other than these O’s, most of the nonactin molecule is a hydrocarbon chain. Therefore once K+ is chelated, the outer part of the complex is quite hydrophobic. It can easily pass through the interior of a membrane, releasing K+ on the other side. The toxic effect of nonactin and several related antibiotics is the result of their ability to transport alkali-metal ions to regions of a cell where they should not be. This breaks down ion gradients the cell has created to perform tasks and store energy. Consequently the cell wastes energy pumping K+ and other ions out again. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/22%3A_Metals/22.10%3A_Chelating_Agents.txt |
We often write transition-metal ions in aqueous solution with symbols such as Cr3+, Cu2+, and Fe3+ as though they were monatomic, but this is far from being the case. These ions are actually hydrated in solution and can be regarded as complex ions. Thus, for example, the grayish-violet color of many chromium(III) salts when dissolved in H2O is due to the species [Cr(H2O)6]3+ rather than to a bare Cr3+ ion. The same color is evident in many crystalline solids such as [Cr(H2O)6]Cl3 which are known to contain the Cr3+ ion surrounded octahedrally by six H2O molecules. In much the same way the blue color of many solutions of copper(II) salts can be attributed to the species [Cu(H2O)4]2+ and the pale violet color of some solutions of iron(III) salts to the [Fe(H2O)6]3+ ion. Because [Fe(H2O)6]3+ is capable of donating a proton, the conjugate base, [Fe(H2O)5OH]2+ is generally present when Fe3+ is dissolved in water. This imparts a yellow color to the solution. Figure $1$ shows examples of colored ion complexes in aqueous solution.
Figure $1$ Examples of colored aqueous transition metal complexes
Not all salts of transition-metal ions yield the hydrated ion when dissolved in H2O. Figure $2$ compares three aqueous copper complexes. When CuCl2 is dissolved in H2O, a beautiful green color due mainly to the complex [CuCl2(H2O)2] is produced. This is obviously different from the sky-blue color of [Cu(H2O)4]2+ which is obtained when Copper(II) sulfate or copper(II) nitrate are dissolved. This is because the Cl ion is a stronger Lewis base with respect to the Cu2+ ion than is H2O. Thus, if there is a competition between H2O and Cl to bond as a ligand to Cu2+, the Cl ion will usually win out over the H2O.
Figure $2$ The Different Colored Copper Chloride Complexes
The superior strength of the Cl as a Lewis base is easily demonstrated by adding Cl ions to a sky-blue solution of copper(II) sulfate. A green color immediately appears due to the formation of chloro complexes:
\begin{align*} \ce{[Cu(H2O)4]^{2+}} + \ce{Cl^{-}} &\rightleftharpoons \ce{[Cu(H2O)3Cl]^{+}} + \ce{H2O} \label{1} \[4pt] \ce{[Cu(H2O)3Cl]^{+}} + \ce{Cl^{-}} &\rightleftharpoons \ce{[Cu(H2O)2Cl2]} + \ce{H2O} \end{align*}
Green Complexes
If a large excess of Cl ion is added, the solution changes color again from green to yellow. This is because of even further displacement of H2O ligands by Cl ligands:
\begin{align*} \ce{[Cu(H2O)2Cl2]} + \ce{Cl^{-}} &\rightleftharpoons \ce{[Cu(H2O)2Cl3]^{-}} + \ce{H2O} \[4pt] \ce{[Cu(H2O)2Cl3]^{-}} + \ce{Cl^{-}} &\rightleftharpoons \ce{[CuCl4]^{2-}} + \ce{H2O} \end{align*} \nonumber
Yellow Complexes
Under favorable circumstances yellow crystals of salts like Cs2[CuCl4], containing the complex ion CuCl42 can be obtained from these solutions.
Because they might very possibly form complexes with it, one must be careful about what ions are added to a solution containing hydrated transition-metal ions. Not only the chloride ions, but the other halide ions are liable to complex, and the same is true of species like NH3 and CN. These ligands differ quite a lot in their affinity for a particular metal ion, but the rules governing this situation are not simple. One finds, for instance, that although NH3 will complex very readily with Cu2+ it has little or no affinity for Fe3+. In other words, a ligand which is a strong Lewis base with respect to one metal ion is not necessarily a strong base with respect to another. There are some ions, however, which almost always function as very weak Lewis bases. The perchlorate ion, ClO4 in particular, forms almost no complexes. The nitrate ion, NO3, and sulfate ion, SO42, only occasionally form complexes.
The addition of ligands to a solution in order to form a highly colored complex is often used to detect the presence or absence of a given metal in solution. The deep blue color of [Cu(NH3)4]2+ produced when excess NH3 is added to solution of Cu(II) salts is a case in point. This can be seen in the following video, where a aqueous solution of ammonia is added to a copper sulfate solution:
The initial copper sulfate solution is sky blue, due to the [Cu(H2O)4]2+ complex. When ammonia is added, a precipitate of Cu(OH)2(s) is formed. as it settles to the bottom, it can be seen that the remaining solution is a dark blue, due to the [Cu(NH3)4]2+ complex formed by copper with ammonia.
Other well-known color reactions are the blood-red complex formed between Fe(III) ions and the thiocyanate ion, SCN, as well as the pink-red complex of Ni(II) with dimethylglyoxime.
While most of the reactions we have been describing are very fast and occur just as quickly as the solutions are mixed, this is not always the case. With certain types of complexes, ligand substitution is quite a slow process. For example, if Cl ions are added to a solution containing [Cr(H2O)6]3+ ions, it is a few days before the grayish-violet color of the original ion is replaced by the green color of the chloro complexes [Cr(H2O5) Cl]2+ and [Cr(H2O)4 Cl]+. Alternatively the solution may he heated, in which case the green color will usually appear within 10 min. The reaction
$\text{[Cr(H}_{2} \text{O)}_{6} \text{]}^{3+} + \text{Cl}^{-} \rightarrow \text{[Cr(H}_{2} \text{O)}_{5} \text{Cl]}^{3+} + \text{H}_{2} \text{O} \nonumber$
is thus a slow reaction with a high activation energy. Ligand substitution reactions of other Cr(III) complexes behave similarly. In consequence Cr(III) complexes are said to be inert, as opposed to a complex like Fe(H2O)63+ which swaps ligands very quickly and is said to be labile. Other examples of inert complexes are those of Co(III), Pt(IV), and Pt(II). Almost all the compounds which were used to establish the nature and the geometry of coordination compounds were inert rather than labile. There is very little point in trying to prepare cis and trans isomers of a labile complex, for example, because either will quickly react to form an equilibrium mixture of the cis and trans forms.
A final complication in dealing with aqueous solutions of transition-metal complexes is their acid-base behavior. Hydrated metal ions like [Cr(H2O)6]3+ are capable of donating protons to water and acting as weak acids. Most hydrated ions with a charge of + 3, like Al3+ and Fe3+ behave similarly and are about as strong as acetic acid. The hydrated Hg(II) ion is also noticeably acidic in this way. Perhaps the most obvious of these cationic acids is the hydrated Fe(III) ion. When most Fe(III) salts are dissolved in water, the color of the solution is yellow or brown, though the Fe(H2O)63+ ion itself is pale violet. The yellow color is due to the conjugate base produced by the loss of a proton. The equilibrium involved is
$\text{[Fe(H}_{2} \text{O)}_{6} \text{]}^{3+} + \text{H}_{2} \text{O} \rightleftharpoons \text{[Fe(H}_{2} \text{O)}_{5} \text{OH]}^{2+} + \text{H}_{3} \text{O}^{+} \nonumber$
Pale violet Brown Complexes
If solutions of Fe(III) salts are acidified with perchloric acid or nitric acid, the brown base is protonated and the yellow color disappears from the solution entirely. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/22%3A_Metals/22.11%3A_Transitional_Metal_Ions_in_Aqueous_Solutions.txt |
According to convention there is a sweet and a bitter, a hot and a cold, and according to convention there is order. In truth there are atoms and a void. -Democritus (400 B.C.)
In the trial scene in Alice in Wonderland, the White Rabbit, asked to read a document adduced as evidence, asks, "Where shall I begin, please your Majesty?" The answer is straightforward: "Begin at the beginning, and go on till you come to the end: then stop." But we shall begin in the middle, with a description of what atoms and molecules are like, before saying anything about how we know that atoms exist. When we examine the evidence for atomic and molecular structure in later chapters, you will have at least an idea of the goal of the effort. The result, we hope, will be to make this textbook more comprehensible than most of Lewis Carroll's books. (The White Rabbit's evidence did not fare very well: "If any one of them can explain it," said Alice, "I'll give him sixpence. I don't believe there's an atom of meaning in it."
01: Atoms Molecules and Ions
An atom consists of a positively charged nucleus, surrounded by one or more negatively charged particles called electrons. The positive charges equal the negative charges, so the atom has no overall charge; it is electrically neutral. Most of an atom's mass is in its nucleus; the mass of an electron is only 1/1836 the mass of the lightest nucleus, that of hydrogen. Although the nucleus is heavy, it is quite small compared with the overall size of an atom. The radius of a typical atom is around 0.1 to 0.25 nanometres (nm), whereas the radius of a nucleus is about 10-6 nm. * If an atom were enlarged to the size of the earth, its nucleus would be only 200 feet in diameter and could easily rest inside a small football stadium.
Table 1-1. Fundamental Particles of Matter
Particle Charge Mass (amu)
Proton +1 1.00728
Neutron 0 1.00867
Electron -1 0.000549
The nucleus of an atom contains protons and neutrons. Protons and neutrons have nearly equal masses, but they differ in charge. A neutron has no charge, whereas a proton has a positive charge that exactly balances the negative charge on an electron. Table 1-1 lists the charges of these three fundamental particles, and gives their masses expressed in atomic mass units. The atomic mass unit (amu) is defined as exactly one-twelfth the mass of the nucleus of a carbon atom consisting of six protons and six neutrons. With this scale, protons and neutrons have masses that are close to, but not precisely, 1 amu each. [As a matter of information at this point, there are approximately 6.022 X 1023 amu in 1 gram (g). This number is known as Avogadro's number, N, and later in the chapter we will show one of the ways this number can be calculated.]
The number of protons in the nucleus of an atom is known as the atomic number, Z. It is the same as the number of electrons around the nucleus, because an atom is electrically neutral. The mass number of an atom is equal to the total number of heavy particles: protons and neutrons. When two atoms are close enough to combine chemically-to form chemical bonds with one another-each atom "sees" mainly the outermost electrons of the other atom. Hence these outer electrons are the most important factors in the chemical behavior of atoms. Neutrons in the nucleus have little effect on chemical behavior, and the protons are significant only because they determine how many electrons surround the nucleus in a neutral atom. All atoms with the same atomic number behave in much the same way chemically and are classified as the same chemical element. Each element has its own name and a one- or two-letter symbol (usually derived from the element's English or Latin name). For example, the symbol for carbon is C, and the symbol for calcium is Ca. The symbol for sodium is Na-the first two letters of its Latin (and German) name, natrium- to distinguish it from nitrogen, N, and sulfur, S.
Note
One nanometre equals 10-9 meters (m), or 10-7 centimeters (cm). If you are not familiar with metric units, see Appendix 1 for more information on the Système International (SI), a simplified version of the metric system adopted by scientists throughout the world. We shall generally use SI units in this book. If you are not familiar with the use of exponential numbers (scientific notation), read Appendix 4.
Example 1.1.1
What is the atomic symbol for bromine, and what is its atomic number? Why isn't the symbol for bromine just the first letter of its name? What other element preempts the symbol B?
Solution
Bromine's atomic number is 35, and its symbol is Br; B is the symbol for boron.
Contributors and Attributions
• R. E. Dickerson, H. B. Gray, and G. P. Haight, Jr. Content was used from "Chemical Principles", an introductory college-level text for General Chemistry with permission of the Caltech library and Harry B. Gray, on behalf of the authors. | textbooks/chem/General_Chemistry/Chemical_Principles_(Dickerson)/01%3A_Atoms_Molecules_and_Ions/1.1%3A_The_Structure_of_Atoms.txt |
Although all atoms of an element have the same number of protons, the atoms may differ in the number of neutrons they have (Table 1-2). These differing atoms of the same element are called isotopes. Four isotopes of helium (He) are shown in Figure 1-1. All atoms of chlorine (Cl) have 17 protons, but there are chlorine isotopes having 15 to 23 neutrons. Only two chlorine isotopes exist in significant amounts in nature, those with 18 neutrons (75.53% of all chlorine atoms found in nature), and those with 20 neutrons (24.47%). To write the symbol for an isotope, place the atomic number as a subscript and the mass number (protons plus neutrons) as a superscript to the left of the atomic symbol. The symbols for the two naturally occurring isotopes of chlorine then would be $\textstyle{\frac{35}{17}}$Cl and $\textstyle{\frac{37}{17}}$Cl. Strictly speaking, the subscript is unnecessary, since all atoms of chlorine have 17 protons. Hence the isotope symbols are usually written without the subscript: 35Cl and 37Cl. In discussing these isotopes, we use the. terms chlorine-35 and chlorine-37. For a nucleus to be stable, the number of neutrons should (for the first few elements) equal or slightly exceed the number of protons. The more protons, the greater the ratio of neutrons to protons to ensure stability. Nuclei that have too many of either kind of fundamental particle are unstable, and break down radioactively in ways that are discussed in Chapter 23.
Figure 1-1 Four isotopes of helium (He). All atoms of helium have two protons (hence two electrons), but the number of neutrons can vary. Most helium atoms in nature have two neutrons (helium-4), and fewer than one helium atom per million in nature has just one neutron (helium-3). The other helium isotopes, helium-5, helium-6, and helium-8 (not shown) are unstable and are seen only briefly in nuclear reactions (see Chapter 23). The size of the nucleus is grossly exaggerated here. If the nucleus were of the size shown, the atom would be half a kilometer across.
Example 1.2.1
How many protons, neutrons, and electrons are there in an atom of the most stable isotope of uranium, uranium-238? Write the symbol for this isotope. Refer to Figure. 1-1.
Solution
The atomic number of uranium (see the inside back cover) is 92, and the mass number of the isotope is given as 238. Hence it has 92 protons, 92 electrons,and 238 - 92 = 146 neutrons. Its symbol is $\textstyle{\frac{238}{92}}$U (or 238U).
The total mass of an atom is called its atomic weight, and this is almost but not exactly the sum of the masses of its constituent protons, neutrons and electrons. * When protons, neutrons, and electrons combine to form an atom, some of their mass is converted to energy and is given off. (This is the source of energy in nuclear fusion reactions.) Because the atom cannot be broken down into its fundamental particles unless the energy for the missing mass is supplied from outside it, this energy is called the binding energy of the nucleus.
Note: Atomic Weight vs. Atomic Mass
The terms atomic weight and molecular weight are universally used by working scientists, and will be used in this book, even though these are technically masses rather than weights.
Table 1-2. Composition of Typical Atoms and Ions
Electrons Protons Neutrons
Atomic
Number
Atomic Weight
(amu)
Total Charge
(electron units)
Hydrogen atom, 1H or H 1 1 0 1 1.008 0
Deuterium atom, 2H or D 1 1 1 1 2.014 0
Tritium atom, 3H or T 1 1 2 1 3.016 0
Hydrogen ion, H+ 0 1 0 1 1.007 +1
Helium atom, 4He 2 2 2 2 4.003 0
Helium nucleus or alpha particle, He2+ or α 0 2 2 2 4.002 +2
Lithium atom, 7Li 3 3 4 3 7.016 0
Carbon atom, 12Ca 6 6 6 6 12.000 0
Oxygen atom, 16O 8 8 8 8 15.995 0
Chlorine atom, 35Cl 17 17 18 17 34.969 0
Chlorine atom, 37Cl 17 17 20 17 36.966 0
Naturally occurring mixture of chlorine 17 17 18 or 20 17 35.453 0
Uranium atom, 234U 92 92 142 92 234.04 0
Uranium atom, 235U 92 92 143 92 235.04 0
Uranium atom, 238U 92 92 146 92 238.05 0
Naturally occurring mixture of uranium 92 92 varied 92 238.03 0
Example 1.2.2
Calculate the mass that is lost when an atom of carbon-12 is formed from protons, electrons, and neutrons.
Solution
Since the atomic number of every carbon atom is 6, carbon-12 has 6 protons and therefore 6 electrons. To find the number of neutrons, we subtract the number of protons from the mass number: 12 - 6 = 6 neutrons. We can use the data in Table 1-1 to calculate the total mass of these particles:
Protons: 6 X 1.00728 amu = 6.04368 amu
Neutrons: 6 X 1.00867 amu = 6.05202 amu
Electrons: 6 X 0.00055 amu = 0.00330 amu
Total particle mass: 12.09900 amu
But by the definition of the scale of atomic mass units, the mass of one carbon-12 atom is exactly 12 amu. Hence 0.0990 amu of mass has disappeared in the process of building the atom from its particles.
Example 1.2.3
Calculate the expected atomic weight of the isotope of chlorine that has 20 neutrons. Compare this with the actual atomic weight of this isotope as given in Table 1-2.
Solution
The chlorine isotope has 17 protons and 20 neutrons:
Protons: 17 X 1.00728 amu = 17.1238 amu
Neutrons: 20 X 1.00867 amu = 20.1734 amu
Electrons: 17 X 0.00055 amu = 0.0094 amu
Total particle mass: 37.3066 amu
Actual observed atomic weight: 36.966 amu
Mass Loss: 0.341 amu
Each isotope of an element is characterized by an atomic number (total number of protons), a mass number (total number of protons and neutrons), and an atomic weight (mass of atom in atomic mass units). Since mass losses upon formation of an atom are small, the mass number is usually the same as the atomic weight rounded to the nearest integer. (For example, the atomic weight of chlorine-37 is 36.966, which is rounded to 37.) If there are several isotopes of an element in nature, then of course the experimentally observed atomic weight (the natural atomic weight) will be the weighted average of the isotope weights. The average is weighted according to the percent abundance of the isotopes. Chlorine occurs in nature as 75.53% chlorine-35 (34.97 amu) and 24.47% chlorine-37 (36.97 amu), so the weighted average of the isotope weights is
$(0.7553 \times 34.97 \;amu) + (0.2447 \times 36.97\; amu) = 35.46\; amu$
The atomic weights given inside the back cover of this book are all weighted averages of the isotopes occurring in nature, and these are the figures we shall use henceforth-unless we are specifically discussing one isotope. All isotopes of an element behave the same way chemically for the most part. Their behavior will differ in regard to mass-sensitive properties such as diffusion rates, which we'll look at later in this book.
Example 1.2.4
Magnesium (Mg) has three significant natural isotopes: 78.70% of all magnesium atoms have an atomic weight of 23.985 amu, 10.13% have an atomic weight of 24.986 amu, and 11.17% have an atomic weight of 25.983 amu. How many protons and neutrons are present in each of these three isotopes? How do we write the symbols for each isotope? Finally, what is the weighted average of the atomic weights?
Solution
There are 12 protons in all magnesium isotopes. The isotope whose atomic weight is 23.985 amu has a mass number of 24 (protons and neutrons), so 24 - 12 protons gives 12 neutrons. The symbol for this isotope is 24Mg. Similarly, the isotope whose atomic weight is 24.986 amu has a mass number of 25, 13 neutrons, and 25Mg as a symbol. The third isotope (25.983 amu) has a mass number of 26, 14 neutrons, and 26Mg as a symbol. We calculate the average atomic weight as follows:
(0.7870 X 23.985) + (0.1013 X 24.986) + (0.1117 X 25.983) = 24.31 amu
Example 1.2.5
Boron has two naturally occurring isotopes, lOB and 11B. We know that 80.22% of its atoms are 11B, atomic weight 11.009 amu. From the natural atomic weight given on the inside back cover, calculate the atomic weight of the lOB isotope.
Solution
If 80.22% of all boron atoms are 11B, then 100.00 - 80.22, or 19.78%, are the unknown isotope. We can use W to represent the unknown atomic weight in our calculation:
(0.8022 X 11.009) + (0.1978 X W) = 10.81 amu (natural atomic weight)
W = $\textstyle{\frac{10.81-8.831}{0.1978}}$ = 10.01 amu
Contributors and Attributions
• R. E. Dickerson, H. B. Gray, and G. P. Haight, Jr. Content was used from "Chemical Principles", an introductory college-level text for General Chemistry with permission of the Caltech library and Harry B. Gray, on behalf of the authors. | textbooks/chem/General_Chemistry/Chemical_Principles_(Dickerson)/01%3A_Atoms_Molecules_and_Ions/1.2%3A_Isotopes.txt |
The formation of atoms from fundamental particles, interesting as this might be to the physicist, is far from being the ultimate stage in the organization of matter. As we mentioned earlier, when atoms are close enough to one another that the outer electrons of one atom can interact with the other atoms, then attractions can be set up between atoms, strong enough to hold them together in what is termed a chemical bond. In the simplest cases the bond arises from the sharing of two electrons between a pair of atoms, with one electron provided by each of the bonded atoms. Bonds based on electron sharing are known as covalent bonds, and two or more atoms held together as a unit by covalent bonds are known as a molecule. One of the principal triumphs of the theory of quantum mechanics in chemistry (see Chapter 8) has been its ability to predict the kinds of atoms that will bond together, and the three-dimensional structures and reactivities of the molecules that result. (A major section of this book, Chapters 8-14, is devoted to chemical bonding theories.)
Figure 1-2 Shapes and relative sizes of some simple molecules. Two bonded atoms appear to interpenetrate because their electron clouds overlap. By convention. a tapered bond in a drawing represents a bond pointing out toward the observer, with the wide end of the taper closest. and a dashed line is used for a bond that points back behind the plane of the page.
In molecular diagrams, a covalent, electron-sharing bond is represented by a straight line connecting the bonded atoms. In the water molecule, one atom of oxygen (0) is bonded to two hydrogen (H) atoms. The diagram for the molecule can be drawn two ways:
The second version acknowledges the fact that a water molecule is not linear; the two H -0 bonds make an angle of 105° with one another. Molecules of hydrogen gas, hydrogen sulfide, ammonia, methane, and methyl alcohol (methanol) have the following bond structures:
These diagrams show only the connections between atoms in the molecules. They do not show the three-dimensional geometries (or shapes) of the molecules. Figure 1-2 shows the shapes and the relative bulk of several molecules. Note that the bond angle in molecules having more than two atoms can vary. The angle in the water molecule is 105°, and the angle in hydrogen sulfide is 92°; the four atoms connected to the central carbon in methane and methyl alcohol are directed to the four corners of a tetrahedron. The bond structure in straight-chain octane, one of the components of gasoline, is
Each of the molecular diagrams shown can be condensed to a molecular formula, which tells how many atoms of each element are in the molecule, but provides little or no information as to how the atoms are connected. The molecular formula for hydrogen is H2; water, H20; hydrogen sulfide, H2S; ammonia, NH3; methane, CH4 ; methyl alcohol, CH30H or CH4O; and octane, C8H18. The formula for octane can also be written
The sum of the atomic weights of all the atoms in a molecule is its molecular weight. Using the atomic weights on the inside back cover, we can calculate molecular weights. The molecular weight of hydrogen, H2, is
2 X 1.0080 amu = 2.0160 amu
A water molecule, H2O , has two atoms of hydrogen and one atom of oxygen, so:
(2 X 1.0080 amu) + (15.9994 amu) = 18.0154 amu
Example 7
Calculate the molecular weight of methyl alcohol.
Solution
The molecular formula is CH30H or CH4O. Then:
1 carbon: 1 X 12.011 amu = 12.011 amu
4 hydrogens: 4 X 1.008 amu = 4.032 amu
1 oxygen: 1 X 15.999 amu = 15.999 amu
Total particle mass: 32.04 amu
(If you wonder why the last figure has been dropped, see the discussion of significant figures in Appendix 4.)
In Example 7 notice that the natural atomic weight of carbon is not 12.000 amu but 12.011 amu, since carbon occurs as a mixture of 98.89% carbon-12 and 1.11% carbon-1 3, with trace amounts of carbon-14.
Example 8
What is the molecular weight of pure octane?
Solution
Since the molecular formula is C8H18", the molecular weight is:
(8 X 12.011) + (18 X 1.008) = 114.23 amu
Contributors and Attributions
• R. E. Dickerson, H. B. Gray, and G. P. Haight, Jr. Content was used from "Chemical Principles", an introductory college-level text for General Chemistry with permission of the Caltech library and Harry B. Gray, on behalf of the authors. | textbooks/chem/General_Chemistry/Chemical_Principles_(Dickerson)/01%3A_Atoms_Molecules_and_Ions/1.3%3A_Molecules.txt |
Forces Between Molecules
Although the strongest attractions of an atom are for other atoms to which it is bonded in a molecule, two molecules themselves exert small but appreciable attractions on one another. Molecules are slightly "sticky." These forces, caused by momentary fluctuations in electron distributions around the atoms, are known as van der Waals attractions (after Dutch physicist Johannes van der Waals). They are responsible for the existence of three states (or phases) of matter at different temperatures: solids, liquids, and gases. Temperature is just a measure of the heat energy or energy of motion that a collection of molecules possesses. At low temperatures, the molecules have little energy of motion. The van der Waals attractions hold them together in an orderly, close-packed crystalline array or lattice (Figure 1-3c). This is the solid state. If more energy is fed into the crystal so the temperature rises, the molecules will vibrate about their average or equilibrium positions in the crystal. Enough energy will cause the ordered structure of the molecular crystal to break up, and the molecules will be free to slide past one another, although they are still touching (Figure 1-3b). This is the liquid state, and the transition temperature between solid and liquid is called the melting point, Tm. The liquid is still held together by van der Waals attractions, although the molecules have too much energy of motion to be locked into a rigid array. If still more energy is given to the liquid, the molecules will begin to move fast enough to overcome the van der Waals attractions, separate entirely from one another, and travel in independent molecular trajectories through space (Figure 1-3a). This is the gas phase, and the transition temperature between liquid and gas is called the boiling point, Tb. Changes in phase are treated in more detail in Chapter 18.
The melting and boiling points of some simple molecules are compared in Table 1-3. In general, larger molecules have higher melting and boiling points, since they have larger surface areas for van der Waals attractions. Thus at 1 atm. pressure H2 boils at - 252.5°C, CH4. boils at - 164.0 °C, but C8H18 must be heated to + 125.7°C before the molecules will separate from one another and go into the gas phase.
Table 1-3. Melting and Boiling Points of Some Simple Molecular Substances
Substance
Molecular
Formula
Tm(°C) Tb(°C)
Gases
Hydrogen H2 -259.1 -252.5
Oxygen O2 -218.4 -183.0
Methane CH4 -182.5 -164.0
Hydrogen Sulfide H2S -85.5 -60.7
Chlorine Cl2 -101.0 -34.6
Ammonia NH3 -77.7 -33.4
Liquids
Bromine Br2 -7.2 +58.8
Methanol CH3OH -93.9 +65.0
Water H2O 0 +100
n-Octane C8H18 +185 +125.7
Solids
Iodine I2 +113.5 +184.4
Sucrose (cane sugar) C12H22O11 +185 decomposes
Figure 1-4 The 0-H bonds in water and methanol (methyl alcohol) are polar because the oxygen atom has the stronger attraction for the electron pair and pulls negative charge toward itself, leaving the hydrogen with a fractional positive charge. This polarity is of great importance in interactions between molecules.
A second kind of force between molecules also influences melting and boiling points: the polarity of the molecules. If two atoms that are connected by an electron-pair covalent bond do not have the same attraction for electrons, then the electron pair will shift toward the atom with the greater electron pulling power. This will give that atom a slight excess of negative charge (represented by δ- rather than by just a minus sign, which would imply a full electron charge), and will confer a slight positive charge (δ+) on the atom that lost out in the tug-of-war for the electron pair. Because the electron-attracting power (electronegativity) of oxygen is greater than that of hydrogen, the oxygen atom in a molecule of water or methyl alcohol is slightly negative, and the hydrogen atoms are slightly positive (Figure 1-4). Such a molecule is termed polar because it behaves like a tiny electric dipole; that is, the negative charge on the oxygen attracts other nearby positive charges, and the positive charge on each hydrogen attracts other negative charges. This is another attractive force between molecules, in addition to van der Waals attractions. Because of the forces binding its molecules, methanol melts and boils at much higher temperatures than methane, which is similar to it in molecular size. Methanol is a liquid at room temperature, whereas methane is a gas. In water, the attractions between hydrogen and oxygen from different molecules are so strong that they are given the name of hydrogen bonds. Hydrogen bonds are especially' important in proteins and other giant molecules in living organisms. If it were not for polarity and hydrogen bonding, water would melt and boil at lower temperatures even than H2S (Table 1-3). It would be a gas at room temperature, rather than the Earth's most common liquid.
Contributors and Attributions
• R. E. Dickerson, H. B. Gray, and G. P. Haight, Jr. Content was used from "Chemical Principles", an introductory college-level text for General Chemistry with permission of the Caltech library and Harry B. Gray, on behalf of the authors. | textbooks/chem/General_Chemistry/Chemical_Principles_(Dickerson)/01%3A_Atoms_Molecules_and_Ions/1.4%3A_Forces_between_Molecules.txt |
So far we have talked only about individual atoms or molecules, and about masses measured in atomic mass units. But individual molecules are hard to manipulate in the laboratory, and chemists weigh their materials in grams, not in atomic mass units. To scale up from the molecular level to the laboratory level, we use a unit called a mole. A mole of a substance is equal to as many molecules of that substance as there are atoms of carbon-12 in exactly 12 g of carbon-12. This means that 1 mole of any substance is a weight, in grams, equal to that substance's molecular weight expressed in atomic mass units. Most important of all, by this definition, 1 mole of any substance contains the same number of molecules. The chemist can count atoms and molecules in the laboratory simply by weighing them. The word mole applies not just to molecules but also to atoms; in practice, we speak of a mole of helium atoms as well as of a mole of water molecules. The term gram-atom applied to a mole of atoms is no longer widely used.
Example 1.5.1
How many grams of each of the following substances are there in 1 mole of that substance: H2, H20 , CH3OH, octane (C8H18), and neon gas (Ne)?
Solution
The molecular weights (in atomic mass units) of most of these substances have been given in previous examples, and the atomic weight of neon is listed on the inside back cover. One mole of each substance is therefore:
H2 2.0160 g break C8H18 114.23 g
H2O 18.0154 g break Ne 20.179 g
CH3OH 32.04 g break
Because the weights listed in Example 9 give the correct relative weights of the molecules that are being weighed out, each of the quantities of material will contain the same number of molecules. This is what makes the concept of moles useful. It is not even necessary to know what that number is, although we know it to be 6.022 X 1023 ; it is called Avogadro's number and is given the symbol N Going from molecules to moles means a scale-up of 6.022 X 1023 times. Avogadro's number is also the conversion factor between atomic mass units and grams as units of mass: 1 g = 6.022 X 1023 amu. If we think of the molecular weight as being the mass of a mole of substance, the units for molecular weight are grams per mole; if we think of it as the actual weight of one molecule, the numerical value is unchanged but the units become atomic mass units per molecule. Both are correct.
Example 1.5.2
One molecule of H2 reacts with one molecule of Cl2 to form two molecules of hydrogen chloride gas, HCl. What weight of chlorine gas should be used in order to react completely with 1 kilogram (kg) of hydrogen gas?
Solution
The molecular weights of H2 and Cl2 are 2.0160 g mole-1 and 70.906 g mole-1, respectively. * Hence 1000 g of H2 contains:
$\textstyle\frac{1000 g}{2.0160 g mole^-1}$ = 496.0 moles of H2 molecules
Without knowing how many molecules there are in a mole, we can still be sure that 496.0 moles of Cl2 will have the same number of molecules as 496.0 moles or 1000 g of H2. How many grams of Cl2 are there in 496.0 moles? Since the molecular weight of Cl2 is 70.906 g mole-1,
496.0 moles X 70.906 g mole-1 = 35,170 g of Cl2
One kilogram equals 1000 g, so 35,170 g is 35.17 kg. If 1.00 kg H2 is made to react with 35.17 kg of C12, the reaction will be complete and none of either starting material will be left over.
* The expression "g mole-1" should be read as "grams per mole." In this notation, a speed in miles per hour is written with units of "miles hr-I."
Example 1.5.3
How many molecules of $H_2$ and $Cl_2$ would be present in the experiment of Example 1.5.2?
Solution
In 496.0 moles of any substance, there will be
$(496.0\; \cancel{mol}) \left(6.022 \times 10^{23}\; \dfrac{molecules}{\cancel{mole}}\right) = 2.99 \times 10^{26}\; molecules$
As a sobering example of just how large Avogadro's number is, 1 mole of coconuts, each 14 centimeters (cm) in diameter, would fill a volume as large as the entire planet earth. The use of moles in chemical calculations is the subject of the next chapter, but the idea has been introduced here because we need to know how to scale up from the molecular to the laboratory level.
Contributors and Attributions
• R. E. Dickerson, H. B. Gray, and G. P. Haight, Jr. Content was used from "Chemical Principles", an introductory college-level text for General Chemistry with permission of the Caltech library and Harry B. Gray, on behalf of the authors. | textbooks/chem/General_Chemistry/Chemical_Principles_(Dickerson)/01%3A_Atoms_Molecules_and_Ions/1.5%3A_Molecules_and_Moles.txt |
Figure 1-5 Common table salt (sodium chloride. NaCl) is built from closely packed sodium ions, Na+ (small spheres). and chloride ions. CI- (large. colored spheres). Each ion of one charge is surrounded by six ions of the opposite charge at the four compass points and above and below. This is a particularly stable arrangement of charges. and it occurs in many salts. From Dickerson and Geis. Chemistry, Matter. and the Universe The Benjamin / Cummings Publishing Co .. Menlo Park. Ca .. © 1976 .
The idea of a covalent bond suggests equal sharing of the electron pair by the bonded atoms, but the brief discussion of polarity in Section 1-4 indicated that the sharing is not always equal. The relative electronegativity or electron-attracting power of atoms is of great importance in explaining chemical behavior, and is treated in detail in Chapters 9 and 10. Sodium atoms (and all metals in general) have a weak hold on electrons, whereas chlorine atoms are very electronegative. Hence in common table salt (sodium chloride, NaCl), each sodium atom, Na, loses one electron (e-) to form a sodium ion, Na+. Each chlorine atom picks up one electron to become a chloride ion, Cl-:
Na → Na+ + e-wordandword$\textstyle{\frac{1}{2}}$ Cl2 + e- → Cl-
We write $\textstyle{\frac{1}{2}}$ Cl2 because free chlorine gas exists as diatomic (two-atom) molecules, not as free chlorine atoms. Solid sodium chloride (Figure 1-5) has sodium and chloride ions packed into a three-dimensional lattice in such a way that each positive Na+ ion is surrounded on four sides and top and bottom by negative Cl- ions, and each Cl- is similarly surrounded by six nearest neighbor Na+ ions. This is a particularly stable arrangement of positive and negative charges.
Metals in general lose one to three electrons easily to become positively charged ions, or cations:
Li Li+ + e- ttt lithium ion
Na Na+ + e- ttt sodium ion
K K+ + e- ttt potassium ion
Mg Mg2+ + 2e- ttt magnesium ion
Ca Ca2+ + 2e- ttt calcium ion
Al Al3+ + 3e- ttt aluminum ion
Some nonmetals, in contrast, pick up electrons to become negatively charged ions, or anions:
$\textstyle{\frac{1}{2}}$ F2 + e- F- ttt fluoride ion
$\textstyle{\frac{1}{2}}$ Cl2 + e- Cl- ttt chloride ion
$\textstyle{\frac{1}{2}}$ O2 + 2e- O2- ttt oxide ion
$\textstyle{\frac{1}{2}}$ S2 + 2e- S2- ttt sulfide ion
Table 1-4 Some Simple Ions of Elements
Other simple ions made from single atoms are shown in Table 1-4. The charge on a simple, single-atom ion such as AP+ or S2- is its oxidation state or oxidation number. It is the number of electrons that must be added to reduce (or removed to oxidize) the ion to the neutral species:
Reduction: AI3+ + 3e- Al
Oxidation: S2- S + 2e-
Pulling electrons away from an atom or removing them altogether is oxidation. Adding electrons to an atom or merely shifting them toward it is reduction.
Example 12
Is chlorine oxidized or reduced in forming the chloride ion? What is the oxidation state of the ion?
Solution
Chlorine is reduced, since one electron per chlorine atom is added to form the ion. The chloride ion, Cl- , is in the - 1 oxidation state.
Example 13
When metals are converted into their ions, are they oxidized or reduced? What is the oxidation state of the aluminum ion?
Solution
Metals are oxidized to their ions, since electrons are removed. The aluminum ion, AP+, is in the +3 oxidation state.
If two or more oxidation states for a metal ion are possible, they are differentiated by writing the oxidation state in Roman numerals after the name of the atom. An older nomenclature, still in use, identifies the higher oxidation state by the ending -ic and the lower by -ous. Hence,
Fe2+ tt iron(II) or ferrous break Fe3+ tt iron(III) or ferric
Cu+ tt copper(I) or cuprous break Cu2+ tt copper(II) or cupric
Sn2+ tt tin(II) or stannous break Sn4+ tt tin(IV) or stannic
Example 14
When the ferric ion is converted to the ferrous ion, is this an oxidation or reduction? Write the equation for the process.
Solution
The equation is Fe3+ + e- Fe2+ . The process is a reduction since an electron is added.
The modern nomenclature with Roman numerals is easier to use because it does not require you to remember what the two oxidation states of a metal are, in order to know what a compound is from its name.
A salt is a compound made up of positive and negative ions. Because a salt must be electrically neutral, the total charge on its positive and negative ions must be zero. Since each ion of Sn2+ has a charge of +2, twice as many chloride ions with -1 charge each are required to produce a zero net charge. Hence the salt of Sn2+ and Cl- ions has the overall composition SnCl2, rather than SnCl or SnCl3. It is called stannous chloride or tin (II) chloride. The formula for stannic chloride or tin(IV) chloride is SnCl4.
In addition to these simple ions, compound or complex ions can be formed between a metal or nonmetal and oxygen, chlorine, ammonia (NH3), the hydroxide ion (OH-), or other chemical groups. The sulfate ion, SO$\textstyle{\frac{2-}{4}}$, has four oxygens at the corners of a tetrahedron around the central sulfur atom, and an overall charge of -2. The nitrate ion, NO$\textstyle{\frac{}{3}}$, has three oxygen atoms in an equilateral triangle around the nitrogen, and a -1 charge. The ammonium ion, NH$\textstyle{\frac{+}{4}}$, has four hydrogens at the corners of a tetrahedron, and a +1 charge. These ions are thought of as units because they form salts the way single-atom ions do, and go through many chemical reactions unchanged. Silver nitrate, AgN03, is a salt containing equal numbers of Ag+ and NO$\textstyle{\frac{}{3}}$ ions. Ammonium sulfate is a salt with twice as many ammonium ions, NH$\textstyle{\frac{+}{4}}$, as sulfate ions, SO$\textstyle{\frac{2-}{4}}$, and the chemical formula (NH4)2S04. Other typical complex ions are shown in Table 1-5.
Table 1-5 Some Common Complex Ions
When a central atom is surrounded by several equally spaced atoms, the number of surrounding atoms is called the coordination number. The most important factor is size. Nitrogen in the nitrate ion, NO$\textstyle{\frac{ }{3}}$, has room for three oxygen atoms around it, and hence a coordination number of 3 for oxygen. The sulfur atom is larger than a nitrogen atom, and can accommodate one more oxygen atom in the sulfate ion, SO$\textstyle{\frac{2-}{4}}$. Hence the coordination number of sulfur for oxygen is 4.
The most common coordination numbers are 2, 3, 4, and 6, (See Table 1-6.) An ion or molecule with a central atom having a coordination number of 2 can be either linear, as carbon dioxide with O-C-O in a straight line, or bent, as in water, H20. Possible structures for ions or molecules with coordination numbers of 3, 4, and 6 are shown in Table 1-6.
Table 1-6 Common Coordination Numbers
Figure 1-6 Geometry of atoms around central atoms with coordination numbers 3, 4, and 6. If L is any peripheral atom and M is the central atom, then the bond angle L - M - L is 120° for trigonal planar, 109.5° for tetrahedral, and typically around 109.5° for trigonal pyramidal geometries. Square planar and octahedral geometries have two L - M - L angles, 90° and 180°.
It is not strictly correct to talk about molecular formulas and molecular weights of salts, since there are no molecules in salts-only ordered lattices of ions. No one sodium ion in the sodium chloride structure shown in Figure 1-5 to a particular chloride ion. It is correct, however, to speak of the chemical formula of a salt, and the formula weight that corresponds to it. Since the chemical formula for sodium chloride is NaCl, the formula weight of sodium chloride is the sum of the atomic weights of one atom of sodium and one atom of chlorine:
1 sodium: tt 22.990 amu
1 chlorine: tt 35.453 amu
Total: tt 58.443 amu
It is conventional to call this the "molecular weight" of sodium chloride, and no confusion results as long as you realize what a salt structure is like. A mole of sodium chloride is 58.443 g. It will contain 6.022 X 1023 sodium ions and 6.022 X 1023 chloride ions. Even though they are not paired off into molecules, the ratio is strictly one to one.
Example 15
What is the molecular weight of ammonium sulfate?
Solution
The chemical formula of ammonium sulfate is (NH4)2SO4, so the molecular weight (actually the formula weight) is
2 nitrogens: tt 2 X 14.007 amu= tt 28.014 amu
8 hydrogens: tt 8 X 1.008 amu= tt 8.064 amu
1 sulfur: tt 1 X 32.06 amu= tt 32.06 amu
4 oxygen: tt 4 X 15.999 amu= tt 63.996 amu
Total: tt tt 132.13 amu
The simple anions are named by adding -ide to the name of the element, as in the fluoride (F-), chloride (Cl-), oxide (O2-), and sulfide (S2-) ions. Where more than one complex anion of an element with oxygen can be formed, the suffixes -ate and -ite are used for the higher and lower oxidation states, respectively. Thus,
Sulfate ion: tt SO$\textstyle{\frac{2-}{4}}$ word Sulfite ion: tt SO$\textstyle{\frac{2-}{3}}$
Nitrate ion: tt NO$\textstyle{\frac{ }{3}}$ word Nitrite ion: tt NO$\textstyle{\frac{ }{2}}$
Arsenate ion: tt AsO$\textstyle{\frac{3-}{4}}$ word Arsenite ion: tt AsO$\textstyle{\frac{3-}{3}}$
If more than two such anions exist, then the prefixes hypo- ("under") and per- ("beyond") are used:
Perchlorate ion: tt ClO$\textstyle{\frac{ }{4}}$
Chlorate ion: tt ClO$\textstyle{\frac{ }{3}}$
Chlorite ion: tt ClO$\textstyle{\frac{ }{2}}$
Hypochlorite ion: tt ClO$\textstyle{\frac{ }{ }}$
Melting Points and Boiling Points of Salts
A salt crystal represents a particularly stable balance of positive and negative charges, with each type of ion being kept out of the way of others of like charge. Melting a salt crystal means upsetting this delicate balance of charges, and allowing mutually repelling ions to come closer together from time to time as the ions flow past one another. This disruption of structure requires large amounts of energy to accomplish, so the melting points of salts are higher than those of molecular solids. The melting points of two salts, sodium chloride (NaCI) and potassium sulfate (K2SO4), are compared in Table 1-7 with those of the elements from which the salts are made.
Table 1-7. Melting and Boiling Points of Two Salts and Their Component Elements
Substance
Chemical
Formula
Tm(°C) Tb(°C)
Sodium metal Na 97.8 882.9
Chlorine Gas Cl2 -101.0 -34.6
Sodium Chloride (salt) NaCl 801 1413
Potassium metal K 64 774
Sulfur S 119 445
Oxygen gas O2 -218 -183
Potassium sulfate (salt) K2SO4 1069 1689
Metallic sodium melts at 97.8°C, and solid chlorine melts at -101°C, but their combination, sodium chloride (common table salt), requires a temperature of 801°C before it will melt. Boiling or vaporizing a salt is even more difficult. The ions remain ions in the liquid state, tumbling past one another as in any other liquid; but before the gas phase can be attained, Na+ and CI- ions must pair off into neutral NaCl molecules. To accomplish this pairing, electrons have to be pulled away from CI- ions, which have a strong attraction for them, and pushed toward Na+ ions, which do not want them. The NaCl bond in sodium chloride vapor is extremely polar, with the electron pair skewed strongly toward the chlorine atom, but the separation still is not as complete as in Na+ and CI- ions. Much energy is required to push electrons where they are not wanted and to make NaCl molecules from Na+ and CI- ions, so high temperatures are required before this can happen. Hence the very high boiling points of salts in comparison with molecular compounds, as illustrated in Table 1-7.
Contributors and Attributions
• R. E. Dickerson, H. B. Gray, and G. P. Haight, Jr. Content was used from "Chemical Principles", an introductory college-level text for General Chemistry with permission of the Caltech library and Harry B. Gray, on behalf of the authors. | textbooks/chem/General_Chemistry/Chemical_Principles_(Dickerson)/01%3A_Atoms_Molecules_and_Ions/1.6%3A_Ions.txt |
Although salts are hard to melt and even harder to vaporize, many can be dissolved easily in a polar liquid such as water. The reason for this is simple. The water molecules help to dismantle the salt crystal, since the partial positive and negative charges on the polar water molecules (Figure 1-4) provide a substitute for the positive and negative charges that were present in the crystal lattice. Figure 1-7 illustrates what happens when a crystal such as sodium chloride is dissolved in water. Each positively charged Na+ is surrounded by water molecules with their negatively charged oxygens turned toward it, and each negatively charged Cl- ion is surrounded by water molecules with their positively charged hydrogens closest. The ions from the salt crystal are said to be hydrated. If the stability that hydration gives the ions in solution is greater than the stability of the crystal lattice, then the salt will dissolve. Sodium chloride is a familiar example of a soluble salt. In contrast, if the hydration energy is too small, then the crystal will be the more stable form, and it will not dissolve in water. Silver chloride (AgCl) and barium sulfate (BaS04) are examples of insoluble salts. When a salt crystal dissolves, it does not simply come apart into ions; it is taken apart by the molecules of the liquid in which it is dissolved (the solvent). This is why salts will not dissolve in nonpolar liquids such as gasoline (octane, C8H18); there are no charges on the solvent molecules to make up for the loss of charge attractions within the crystalline salt.
Figure 1-7 Breakup of a salt crystal by water molecules. with hydration of ions. Each salt ion in solution is surrounded by polar water molecules with the opposite charge to that of the ion turned toward it. This electrostatic hydration energy compensates for the loss of attractions between ions in the salt crystal. From Dickerson and Geis. Chemistry, Matter, and the Universe.
Salt solutions conduct electricity, and this property was extremely important early in the development of theories of chemical bonding. Electrical conduction in metals takes place by means of moving electrons; the metal ions remain in place. Crystalline salts do not conduct electricity at all, but if the salt is melted, then positive ions can migrate one way through the liquid and negative ions can move the other way in the presence of an electric field. This mobility of ions is even greater if the salt is dissolved in water and the ions consequently are hydrated.
Some of the first concrete ideas about the nature of chemical bonding came from the electrolysis experiments of the English scientist Michael Faraday (1791-1867). (Electrolysis means "breaking apart with electricity.") If sodium chloride is melted (above 801°C) and if two electrodes (the cathode and the anode) are inserted into the melt as shown in Figure 1-8 and an electric current is passed through the molten salt, then chemical reactions take place at the electrodes: Sodium ions migrate to the cathode, where electrons enter the melt, and are reduced to sodium metal:
Figure 1-8 A commercial electrolysis cell for the production of metallic sodium and chlorine gas from molten NaCl. Liquid sodium floats to the top of the melt above the cathode and is drained off into a storage tank. Chlorine gas bubbles out of the melt above the anode. From Dickerson and Geis. Chemistry. Matter. and the Universe.
Figure 1-9 Schematic diagram of an electrolysis cell. For current to be carried , the fluid must contain mobile ions, either as a molten salt or as hydrated ions in solution , A substance capable of carrying current by migration of ions is called an electrolyte. If the electrolyte is a solution of CuCl2 , which dissociates (breaks apart) to give Cu2+ and CI- ions, then as current is passed through the cell. Cu2+ ions migrate to the cathode and are reduced to metallic copper, and CI- ions migrate to the anode, where they are oxidized to Cl2 gas. Platinum electrodes are used because they are chemically inert and will not react.
$Na^+ + e^-(from\; cathode) → Na$
Chloride ions migrate the other way, toward the anode, give up their electrons to the anode, and are oxidized to chlorine gas:
$Cl^- → \frac{1}{2} Cl_{2} + e^-(to\; anode)$
The overall reaction is the breakdown of sodium chloride into its elements:
$Na^+ + CI^- → Na + \frac{1}{2}Cl_2$
Sodium ions are reduced and chloride ions are oxidized. Electrolysis can also be carried out by passing electric current through solutions of salts (Figure 1-9). If a solution of sodium chloride in water is electrolyzed, chlorine gas is given off at the anode as in the case of molten sodium chloride, but the cathode product is hydrogen gas rather than metallic sodium:
$Na^+ + Cl^- + H_2O \rightarrow Na + \frac{1}{2} Cl_2 + \frac{1}{2} H_2 + OH^- \label{1-1}$
This is the same result that would be obtained if liquid sodium chloride was first electrolyzed to give metallic sodium:
$Na^+ + Cl^- → Na + \frac{1}{2} Cl_2 \label{1-2}$
and the sodium was then dumped into water:
$Na + H_2O → Na^+ + \frac{1}{2}H_2 + OH^- \label{1-3}$
Equation $\ref{1-1}$ is just the sum of Equations $\ref{1-2}$ and $\ref{1-3}$, since the sodium metal that is produced in Equation $\ref{1-2}$ is used up in Equation $\ref{1-3}$. There is nothing mysterious about the different cathode products during electrolysis of sodium chloride in a melt or in solution, If water is present, some of the H2O molecules will be dissociated into H+ and OH- ions. Because H+ has a greater affinity for electrons than Na+ does, the H+ ions will take electrons away from metallic sodium, making the anode product H2 rather than Na, and leaving Na+ ions in solution. In contrast, Cu2+ ions have a greater affinity for electrons than H+ ions do, so the anode product of electrolysis of CuCl2 is metallic copper, whether the process is carried out in the melt or in solution (Figure 1-9). Typical products of electrolysis of solutions and melts are given in Table 1-8. Electrochemical reactions and cells are discussed in detail in Chapter 19. At the moment, we are focusing on what electrochemical reactions tell us about chemical bonding.
Table 1-8. Products of Electrolysis
Electrolytett Cathode Producttt Anode Product
Sulfuric acid (H2SO4) in H2Ott H2tt O2
Sodium sulfate (Na2SO4) in H2Ott H2tt O2
Sodium chloride (NaCl) in H2Ott H2tt Cl2
Potassium iodide (Kl) in H20tt H2tt I2
Copper sulfate (Cu2SO4) in H20tt Cutt O2
Silver nitrate (AgNO3) in H2Ott Agtt O2
Mercuric nitrate [Hg(NO3)2] in H20tt Hgtt O2
Lead nitrate [Pb(NO3)2] in H20tt Pbtt O2 and some PbO2
Molten lye (NaOH); not in H2Ott Natt O2
Faraday found that there was a quantitative relationship between the amount of electricity passed through an electrolytic cell and the amount of chemical change produced. He formulated Faraday's laws of electrolysis, which in terms of the modern theory of atoms and ions can be expressed as follows:
1. Passing the same quantity of electricity through a cell always leads to the same amount of chemical change for a given reaction. The weight of an element deposited or liberated at an electrode is proportional to the amount of electricity that is passed through.
2. It takes 96,485 coulombs of electricity to deposit or liberate 1 mole of a substance that gains or loses one electron during the cell reaction. If n electrons are involved in the reaction, then 96,485n coulombs of electricity are required to liberate a mole of product.
The quantity 96,485 coulombs of electricity has become known as 1 faraday in his honor, and has been given the symbol $\mathcal{F}$. Faraday's laws become self-evident when you realize that 1 $\mathcal{F}$ is simply the charge on 1 mole of electrons, or 6.022 X 1023 electrons. The scale-up factor of 6.022 X 1023 from molecules to moles is paralleled by the same scale-up factor from 1 electron charge to 1 $\mathcal{F}$ of charge. At the time, of course, Faraday knew neither the value of Avogadro's number nor the charge on an electron. His experiments did tell him, however, that charges on ions came in multiples of a fundamental unit, such that 96,485 coulombs corresponded to a mole of these units. The word electron first appeared in 1881, when the British physicist G. J. Stoney coined it to denote this fundamental unit of ionic charge. Its application to a real negatively charged particle came a decade later.
Example 16
Write equations for the reactions that occur when current is passed through molten NaCl. How many grams of sodium and chlorine are released when 1 $\mathcal{F}$ of charge is passed through the cell?
Solution
The cathode reaction is Na+ + e- Na, and the anode reaction is Cl- $\textstyle{\frac{1}{2}}$Cl2 + e-. When 1 mole of electrons (1$\mathcal{F}$) passes through molten NaCI, each electron reduces one sodium ion, so 1 mole of sodium atoms is produced. Hence 22.990 g of Na are deposited at the cathode. At the anode, 1 mole of electrons is removed from 1 mole of chloride ions, leaving 1 mole of chlorine atoms, which combine pairwise to make $\textstyle{\frac{1}{2}}$ mole of Cl2 molecules. Hence the weight of chlorine gas released is 35.453 g (the atomic weight of Cl, half the molecular weight of Cl2).
Example 17
How many grams of magnesium metal and chlorine gas are released when 1 $\mathcal{F}$ of electricity is passed through an electrolytic cell containing molten magnesium chloride, MgC12?
Solution
The cathode reaction is Mg2+ + 2e- Mg, and the anode reaction is this: 2Cl- Cl2 + 2e-. Since two electrons are required to reduce each ion of Mg2+, 1 mole of electrons will be sufficient to reduce half a mole of magnesium ions, depositing 12.153 g of magnesium. (The atomic weight of magnesium is 24.305 g mole-1.) As in Example 16, 1 mole of Cl- ions is oxidized, liberating half a mole or 35.453 g of Cl2 gas.
Example 18
The main commercial source of aluminum metal is the electrolysis of molten salts of Al3+. How many faradays of charge, and how many coulombs, must be passed through the melt to deposit 1 kg of metal?
Solution
One kilogram of aluminum is 1000 g/26.98 g mole-1, or 37.06 moles. Since each atom of aluminum deposited requires three electrons, 37.06 moles will require 3 X 37.06, or 111.2, moles of electrons. Hence 111.2$\mathcal{F}$ or 10,730,000 coulombs will be needed.
Example 19
Electron flow at the rate of 1 coulomb per second (coulomb sec-1) is a current of 1 ampere (A). Currents in industrial electrolytic production of aluminum are ordinarily in the range of 20,000 to 50,000 A. If a cell is operated at 40,000 A (40,000 coulombs sec-1), how long will it take to produce the kilogram of aluminum metal mentioned in Example 18?
Solution
The time required will be
$\textstyle{\frac{10,730,000 coulombs}{40,000 coulombs sec^-1}}$ = 268 sec or 4.5 min
Figure 1-10 Illustrations of Faraday's laws of electrolysis (a) Two electrons are required to reduce each ion of Cu2+, or 2 moles of electrons (2$\mathcal{F}$) for each mole of copper. Each faraday is enough to oxidize 1 mole of Cl- ions to $\textstyle{\frac{1}{2}}$ mole of Cl2 gas. (b) Only 1$\mathcal{F}$ of charge is required to reduce 1 mole of Ag+ ions to metallic silver. since the ionic charge on Ag+ is only +1. Chlorine gas is liberated at the same rate per faraday as before.
Faraday's laws are represented diagrammatically in Figure 1-10. We have been using these laws with a prior knowledge of the charges on different ions, and the knowledge that 96,485 coulombs is the total charge on 6.022 X 1023 electrons. History actually operated in reverse: Faraday and others used electrolysis experiments to find out what the charges on ions were. The reasoning used is illustrated in Table 1-9. If twice as much electricity is required to liberate a mole of copper as a mole of silver (assuming that you know the atomic weights of the two metals and can calculate the weights of a mole of each), then the copper ion must have twice the charge of the silver ion. In Table 1-9, the number of faradays of charge required to liberate 1 mole of an element is the same as the number of charges, positive or negative, on the ion.
Table 1-9. Deduction of Ionic Charge by Electrolysis
Product of
electrolysis
Electrode
Faradays per mole
of atoms deposited
Ion in
solution
Silver (Ag) Cathode 1a Ag+
Chlorine (Cl2) Anode 1 Cl-
Copper (Cu) Cathode 2 Cu2+
Hydrogen (H2) Cathode 1 H+
Iodine (I2) Anode 2 I-
Oxygen (O2)b Anode 2 O2-
Zinc (Zn) Cathode 2 Zn2+
a For example, electrolysis of silver nitrate solution for 1 hour by using a current of 0.5 A deposits 2.015 g of silver; 2.015/107.9 = 0.0187 mole of silver.
(0.5 coulomb sec^-1) x 3600 sec = 0.0187 $\mathcal{F}$
96.485 coulombs $\mathcal{F}$-1
b Actually, oxygen (02) is produced by a complicated electrode reaction. The species O2- can exist in molten oxides, but in water 02- becomes 2OH- by reaction with a water molecule.
Contributors and Attributions
• R. E. Dickerson, H. B. Gray, and G. P. Haight, Jr. Content was used from "Chemical Principles", an introductory college-level text for General Chemistry with permission of the Caltech library and Harry B. Gray, on behalf of the authors. | textbooks/chem/General_Chemistry/Chemical_Principles_(Dickerson)/01%3A_Atoms_Molecules_and_Ions/1.7%3A_Ions_in_Solution.txt |
It was proposed even as far back as the time of Ben Franklin and John Dalton that the forces between particles of matter must be electrical in some way. But because like charges repel one another, it was believed, wrongly, that bonds could not exist between identical atoms, whereas we now know that most common gases occur as diatomic molecules: H2, N2, 02, F2, Cl2, and so on. This one blunder led to nearly a half-century of confusion about molecular structure and atomic weights, during which it was thought that hydrogen gas was H instead of H2 water was HO instead of H2O, and oxygen had an atomic weight of 8 rather than 16. Electron pairs as the "glue" that holds atoms together in covalent bonds were not even proposed systematically until 1913, by G. N. Lewis; they were not explained theoretically for still another 20 years. Faraday's experiments showed that the charges on ions did occur naturally in fundamental pieces or units such that a mole of these charges equaled 1 $\mathcal{F}$, and Stoney named this elemental unit the electron. But Stoney's electron was not necessarily a particle that could be isolated and studied.
Figure 1-11 A Crookes tube. When a high voltage (about 10,000 volts) is applied across two electrodes in a sealed glass tube containing gas at low pressure, the voltage induces the breakdown of gas molecules into electrons and positive ions. The electrons stream toward the anode and are known as cathode rays, and the positive ions stream toward the cathode and are termed canal rays. If the cathode is perforated, the positive ions will pass through it and cause a glow where they strike the glass walls. If a lightweight pinwheel is placed in the path of the cathode rays. they can cause it to rotate . As the electrons of the cathode rays move toward the anode. they strike other gas molecules and set up a glow discharge that is familiar in neon signs.
The people who first showed that electrons were real particles that could be added to or removed from atoms were physicists studying the effects of electricity on gases. They found that if they set up an electrical potential of 10,000 volts between two electrodes in a sealed tube (a Crookes tube) containing gas at low pressure, they observed a glow discharge (Figure 1-11). This discharge is what makes neon signs glow. The electrical potential strips electrons from atoms of the gas, sending the electrons streaming toward the anode and positive ions toward the cathode. These moving electrons (the cathode rays) can be detected by watching the flashes of light on a zinc sulfide screen placed in the tube. If a lightweight pinwheel is set up in the path of the electrons inside the tube, the electrons even make the pinwheel rotate. On their way to the anode, the cathode rays strike other atoms of the gas, causing the emission of light in a glow discharge. The color of the glow discharge will vary, depending on the gas used· inside the tube.
If a metal plate with a slit is placed in front of the cathode, then the electrons in the cathode ray will be confined to a thin beam. This beam is deflected by electric and magnetic fields in a way that indicates that the particles in the beam carry a negative charge. The relative amount of bending of the canal rays (positive ions) and cathode rays (negative electrons) shows that the cathode ray particles are extremely light, whereas the positive ions are roughly as heavy as the original atoms from which they came. The exact nature of the canal rays depends on what gas is used in the tube, but the cathode rays are the same for all gases. J. J. Thomson (1856- 1940) suggested that the particles in the cathode rays might in fact be Stoney's "electrons," and in 1897 he found a way to use the deflection of the beam by electric and magnetic fields to calculate the charge-to-mass ratio (elm) of the particles. He found that
$\textstyle{\frac{e}{m}}$ = 1.76 x 108 coulombs g-1
Figure 1-12 The mass spectrometer, Electrons emitted by an ionizing source bombard gas molecules and produce positive ions . These ions are accelerated by an electric field. and they are then passed through collimating slits (51 and 52), which direct the ions into parallel beams. These beams are bent in an electric field. resulting in diverging beams of ions moving with different speeds. The collimating slits are aligned so that only ions headed straight along the tube arrive at the point of divergence. A magnetic field refocuses the beams in such a way that all ions of the same charge-to-mass ratio strike the same spot on the photographic plate.
Example 20
Assume that Thomson's cathode ray particles are in fact the same as Stoney's and Faraday's electrons, and that 1$\mathcal{F}$ is a mole of electrons. Calculate the mass of one electron.
Solution
The charge on one electron is
e = $\textstyle{\frac{1\mathcal{F}}{N}}$ = $\textstyle{\frac{96,485 coulombs mole^-1}{6.022 X 10^23 electrons mole^-1}}$
= 1.602x 1019 coulomb
m = $\textstyle{\frac{1.602x 10^19coulomb}{1.76 X 10^8 coulomb g^-1}}$ = 0.910 x 10-27 g
Figure 1-13 Millikan's oil-drop experiment. Tiny droplets of oil are introduced between two plates that can be given an electrostatic charge. A drop of oil is allowed to fall freely through the air, and its path is monitored . The radius of the drop is calculated from the terminal velocity of its fall and the viscosity of air The air is ionized by x rays. and negatively charged particles (electrons) stick to the oil drops. The charge on a drop can be determined from the voltage that must be applied across the condenser plates to make the drop hang motionless, with electrostatic and gravitational forces in balance.
A modern descendant of the Crookes tube and Thomson's apparatus is the mass spectrometer (Figure 1-12). It is a valuable research tool for measuring the mass per unit charge of any substance that can be given a positive charge. Mass spectrometry offers the most direct measurement of atomic weights of elements, and it is the method by which isotopes can be both identified and separated. By looking at the masses of the fragments into which molecules are broken down during electron bombardment in the spectrometer, organic chemists can obtain useful information about the molecular structure of a substance. (During the development of the atomic bomb in World War II, mass spectrometry was used to separate fissionable 235U from 238U, although the extremely low pressures that mass spectrometry requires were not practical for large-scale production.)
Although the ratio of electron charge to electron mass was measured in 1897 by Thomson, the charge itself was not measured until 1911 , when Robert A. Millikan (1868-1953) obtained the charge by the ingenious experiment illustrated in Figure 1-13. He used x rays to irradiate a spray of tiny oil droplets between two chargeable plates. Electrons from ionization of the air around the drops adhered to the drops, giving them one, two, or more electron charges. Millikan measured first the rate of free fall of the charged drops through air of known viscosity. Then he measured the voltage across the plates that was sufficient to suspend the drops motionless between the plates. He calculated that the charge on anyone drop was always an integral multiple of 1.6022 X 10-19 coulomb, and he concluded correctly that this was the charge on a single electron.
Example 20 actually was presented the wrong way around. We can solve it only because we know the value of Avogadro's number, whereas in fact Millikan's results furnished one way of calculating Avogadro's number.
Example 21
Assume that you do not know the value of Avogadro's number, but that you recognize that the faraday is the charge necessary to reduce 1 mole of Na+ ions, with one of Millikan's electrons combining with each ion. Calculate the number of ions in a mole, or Avogadro's number.
Solution
The charge on one electron is
N =$\textstyle{\frac{96,485 coulombs mole^-1}{1.6022 X 10^-19 coulomb ion^-1}}$ = 6.022 x 1023 ions mole-1
Contributors and Attributions
• R. E. Dickerson, H. B. Gray, and G. P. Haight, Jr. Content was used from "Chemical Principles", an introductory college-level text for General Chemistry with permission of the Caltech library and Harry B. Gray, on behalf of the authors. | textbooks/chem/General_Chemistry/Chemical_Principles_(Dickerson)/01%3A_Atoms_Molecules_and_Ions/1.8%3A_Gaseous_Ions.txt |
An atom consists of a positively charged nucleus surrounded by enough negatively charged electrons to yield zero net charge. The nucleus is constructed from positively charged protons and neutral neutrons, each of mass approximately 1 amu. The mass of an electron is approximately 1/ 1836 the mass of a proton; the charge on an electron is equal but opposite in sign to the charge on a proton. The total number of protons in the nucleus (and electrons in a neutral atom) is the atomic number, Z. The total number of both protons and neutrons is the mass number, and the mass of the atom, in atomic mass units, is its atomic weight. The atomic weight is always slightly less than the sum of masses of the particles that go into making an atom, because mass is converted to energy and lost when the atom is formed.
All atoms with the same number of protons, and therefore the same atomic number, are classified as the same element and represented by a one- or two-letter symbol. Atoms of the same element with varying numbers of neutrons are called isotopes of the element. Isotopes are identified by placing the mass number as a superscript to the left of the symbol of the element (e.g., 37Cl). The atomic number is sometimes added as a subscript (e.g., $\textstyle\frac{37}{17}$Cl), although it is actually not necessary since the element's name and atomic number are known from the symbol. Each isotope of an element has its own atomic weight, and the natural atomic weight is the weighted average of these isotopic values, the weighting being according to the natural abundance of each isotope.
A collection of atoms held together by chemical bonds is a molecule Usually, but not always, the bonding in a molecule can be explained in terms of electron pairs, each holding two atoms together. Such an electronpair bond is a covalent bond. The sum of the atomic weights of all the atoms in a molecule is its molecular weight. Although atoms in different molecules are not directly bonded to one another, all molecules are slightly "sticky," and are attracted to other molecules. These van der Waals attractions will make the molecules of a gas adhere to one another to form a liquid if the temperature falls low enough, and make the molecules of a liquid fit together in a regular crystalline array in a solid if the temperature falls lower still. The temperatures at which these two transitions occur are the boiling point, Tb , and the melting point, Tm' respectively.
If two atoms differ in their intrinsic electron-pulling power or electronegativity, then the electron pair of the bond between them will be shifted toward the atom with the greater attraction, giving it a negative charge and the other atom a positive charge. The bond, and molecules that contain such bonds, are said to be polar. Polar molecules can attract one another, and they can also attract positively and negatively charged ions. Melting and boiling points of polar molecules are higher than would be expected from van der Waals attractions alone, because their polarity provides a second type of intermolecular attraction.
Atomic and molecular weights are measured on a scale of atomic mass units (amu), where 1 amu is defined as exactly one-twelfth of the mass of a I2C atom. A quantity of a chemical substance (atoms, molecules, or ions) equal to the atomic weight expressed in grams is defined as 1 mole of that substance. One mole of any substance-atoms, molecules or ions-contains the same number of particles of that substance. This property makes the mole a useful means of counting out particles merely by weighing them. The units of atomic and molecular weights are either grams per mole or amu per molecule (or atom).
Some atoms, those of metals in particular, have a weak hold on their electrons and can lose one, two, or more electrons to become positively charged ions, or cations. Many nonmetals or groups of atoms can acquire one or more negative charges to become negatively charged ions, or anions. A salt is a compound of the relative number of cations and anions that will produce zero overall charge. Common table salt, NaCl, contains equal numbers of Na+ and Cl- ions. The pulling away or outright removal of electrons is termed oxidation, and the addition to or shifting of electrons toward an atom is reduction. Since electrons are never created or destroyed in chemical reactions, whenever one substance is oxidized, some other substance must be reduced.
Simple anions made by adding electrons to single atoms have names ending in -ide, as chloride, Cl-, and sulfide, S2-, ions. For complex ions of a nonmetal atom with oxygen, the higher and lower oxidation state ions are differentiated by the suffixes -ate and -ite. The oxidation state of a metal cation (see Chapter 10) is indicated by a Roman numeral after the name of the metal, as in Fe3+, iron(III), or by the suffixes -ic and -ous.
Although salts do not have separate molecules and, strictly speaking, cannot have molecular weights, they do have chemical formulas that express their overall composition in the simplest possibJe way. The weight of 1 mole of these atoms is the formula weight of the salt, but it is customary to refer to this as the salt's "molecular weight." Thus magnesium chloride has one Mg2+ ion for every two Cl- ions, a net charge of zero, a chemical formula of MgCl2, and a molecular weight of 95.211 g mole-1.
The coordination number in a complex ion or molecule is the number of atoms or chemical groups bonded directly to the central atom. These bonding groups can be simple ions such as O2- and Cl- or molecules such as ammonia (NH3) and water (H2O). The maximum coordination number for a given central atom depends on the size of the atom and the size of its surrounding groups. The most common coordination numbers are 2, 3, 4, and 6.
Salts have higher melting and boiling points than molecular substances, because heat energy must be supplied to break apart the stable crystal lattice, and even more heat energy is required to force positive and negative ions to pair off and share electrons in neutral molecules that can go into a gas phase. Salts often dissolve readily in water, however, because polar attractions by the water molecules can compensate for the attractions of other ions in the crystal. Ions surrounded by polar water molecules in solution are said to be hydrated. Gasoline and similar nonpolar liquids cannot dissolve salts because they cannot hydrate (or solvate, if the solvent is other than water) the ions.
If a current of electricity is passed through molten salt or a salt solution, the current is carried by ions migrating in opposite directions. At the cathode, where electrons enter the salt medium, metal cations can be reduced to pure metal. At the anode, where electrons flow out of the salt and back into the external circuit, anions can be oxidized to liberate pure nonmetallic elements. This is the process of electrolysis. Faraday found a quantitative relationship between the amount of charge passed through a cell and the amount of chemical change produced: 96,485 coulombs of charge will bring about 1 mole of a change that involves one electron per ion. The quantity, 96,485 coulombs, is simply the charge on 1 mole of electrons and is called 1 faraday ($\mathcal{F}$) of charge.
Electrons as separate particles were studied by physicists interested in low-pressure gas discharges under high voltages. Cathode rays consist of a beam of electrons stripped away from the gas atoms. J. J. Thomson showed, by means of deflecting magnetic and electrostatic fields, that the cathode rays were made of negatively charged particles, and he measured the charge-to-mass ratio of the particles. R. A. Millikan completed the process, in his oil-drop experiment, by successfully measuring the charge on the electron. This, combined with Faraday's results, led to the calculation of Avogadro's number, the number of electrons in a faraday of charge, or the number of particles in a mole of any substance. The mass spectrometer, a descendant of Thomson's gas-discharge tubes, is a modern analytical tool and a means of finding the charge-to-mass ratio for any atomic or molecular species that can be given a charge.
Contributors and Attributions
• R. E. Dickerson, H. B. Gray, and G. P. Haight, Jr. Content was used from "Chemical Principles", an introductory college-level text for General Chemistry with permission of the Caltech library and Harry B. Gray, on behalf of the authors. | textbooks/chem/General_Chemistry/Chemical_Principles_(Dickerson)/01%3A_Atoms_Molecules_and_Ions/1.9%3A_Chapter_Summary.txt |
1. When you can measure what you are
speaking about, and express it in numbers,
you know something about it; but when
you cannot measure it, when you cannot
express it in numbers your knowledge is of
a meagre and unsatisfactory kind; it may
be the beginning of knowledge, but you
have scarcely, in your thoughts, advanced
to the stage of science.''
William Thomson, Lord Kelvin (1891)
Introduction
The concept of atoms goes back to the Greek philosophers. Democritus (470-360? b.c.) proposed that all matter is made up of separate, indestructible atoms, that different kinds of atoms have different structures and behavior, and that the observed properties of substances arise because of the way their individual atoms arrange themselves and combine with one another. His theories are essentially a primitive version of the material in Chapter 1. Why, then, did the ancient Greeks not use the theories of Democritus and go on to develop atomic energy? Why did 2000 years pass before modem science began to develop?
The answer, in large part, is that the Greeks did not think quantitatively about atoms, and they were not experimentalists. Their science was a philosophical explanation of the universe, rather than a pragmatic tool with which to manipulate the world around them. Cheap human labor kept them from having to worry about developing a scientific technology.
The Greek scientist Heron of Alexandria invented several steam-driven mechanisms that could have led directly to the steam engine, but he used them only as toys and novelties.
The atomic theory of Democritus was sterile because it did not lead to quantitative predictions that could be tested. It failed to develop beyond abstract concepts because it did not have the feedback from successful and unsuccessful experiments in the real world to challenge and improve it.
A scientific theory, to be useful, must be quantitative. It must predict: "If I do this, then that will happen, and to an extent that I can calculate in advance." Such a prediction is testable. It can be seen to be correct, increasing our confidence in the theory behind it; or what frequently is more important, it can be seen to be incorrect, causing us to revise and improve the theory.
Scientific theories grow by continual destruction and rebuilding. A theory that predicts nothing that could possibly be tested conveys no information, and is worthless.
The importance of precise measurement of mass in chemical reactions escaped the Greek philosophers. It also escaped the medieval European alchemists, metallurgists, and iatrochemists (medicinal chemists). The great French chemist Antoine Lavoisier (1743-1794) was the first to realize that mass was the fundamental quantity conserved during chemical reactions.
The total mass of all products formed must be precisely the same as the total mass of the starting materials. With this principle Lavoisier demolished the long-accepted phlogiston theory of heat (see Chapter 6) by showing that when a substance burns, it combines with another element, oxygen, rather than decomposing and giving off a mysterious universal substance called phlogiston. The principle of the conservation of mass is the cornerstone of all chemistry. More than just total mass is conserved; the same number of each type of atom must be present both before and after a chemical reaction, no matter how intricately these atoms may combine and rearrange into molecules.
Energy also must be conserved in chemical reactions. To the chemist, this means that the heat that is absorbed or given off in a particular chemical reaction (the heat of reaction) must be the same no matter how that reaction is carried out—in one step or in several. For example, the heat given off when hydrogen gas and graphite (a form of carbon) are burned must be the same as that given off when hydrogen and carbon are used to make synthetic gasoline, and this gasoline is then used as fuel for an automobile engine.
If the heat given off in the two variations of the reaction were not the same, then the more efficient reaction could be run in the forward direction and the less efficient one could be run in reverse. The result would be a cyclical no-fuel furnace that would pour out endless quantities of heat at no cost to the operator. Perpetual-motion schemes of all types vanish as soon as one becomes quantitative about heat, energy, and work. This is the basis of thermodynamics, which is covered in detail in Chapters 15-17.
In this chapter we shall look at the consequences, for chemistry, of two principles:
1. Atoms are neither created nor destroyed during chemical reactions (conservation of mass).
2. Heats of reaction are additive. If two reactions can be added to give a third, then the heat of the third reaction is equal to the sum of the heats of the first two reactions (conservation of energy).
Both these principles may seem obvious at first, but they are also quite powerful tools in explaining chemical behavior.
Atomic Weights, Molecular Weights, and Moles
Figure 2-1. Two molecules of hydrogen gas combine with one molecule of oxygen to yield two molecules of water. Avogadro's principle tells us that equal volumes of different gases contain equal numbers of molecules, at a specified temperature and pressure. Hence two volumes of H2 gas will combine with one volume of 02, to produce two volumes of water vapor, and two moles of H2 will combine with one mole of 02 to produce two moles of water vapor. From Dickerson and Geis, Chemistry, Matter, and the Universe
As soon as chemists realized that mass — not volume, density or some other measurable property — was the fundamental property that was conserved during chemical reactions, they began to try to establish a correct scale of atomic masses (atomic weights) for all the elements. How they did this is described in Chapter 6; the result of their years of work is the table of natural atomic weights on the inside back cover of this book. As we saw in Chapter 1, the molecular weights of molecular compounds and the formula weights of nonmolecular compounds (such as salts) are found by adding the atomic weights of all the constituent atoms.
Central to all chemical calculations is the concept of the mole. As defined in Chapter 1, a mole of any substance is the quantity that contains as many particles of the substance as there are atoms in exactly 12 g of carbon-12. Thus a mole of a substance is a quantity in grams that is numerically equal to its molecular weight expressed in atomic mass units. The number of particles in a mole is called Avogadro's number, and the experiments of Millikan and Faraday described at the end of Chapter 1 are one means of establishing its value:
N = 6.022 X 1023 particles mole-1
Moles are a way of manipulating atoms or molecules in bundles of 6.022 X 1023. The molecular weights of H2, 02, and H20 were worked out in Chapter 1. If we know that two molecules of hydrogen gas, H2, react with one molecule of oxygen gas, 02, to produce two molecules of water, H20, then we can predict that 2 moles of H2, or 4.032 g, will react with 1 mole of 02, or 31.999 g, to yield 2 moles of water, or 36.031 g (Figure 2-1). The check addition, 4.032 + 31.999 = 36.031, verifies the conservation of mass during the reaction. The chemist measures substances in grams, by weighing them.
Yet it is more meaningful to convert these quantities from grams to moles, because then one is working with relative molecular proportions, scaled up by a uniform factor of N.
Chemical Analyses. Percent Composition and Empirical Formulas
Chemical analysis involves breaking a substance down into its elements and then measuring the relative amount of each element present, either in grams per 100 g of original compound, or as a percent by weight. One way of doing this, if the compound is a hydrocarbon (made up only of carbon and hydrogen), is to burn a known amount of the substance in oxygen, and measure the quantities of C02 (carbon dioxide) and H20 that result.
Example 1
When 25.00 g of an unknown hydrocarbon is burned, 68.58 g of CO2 and 56.15 g of H20 are produced.
How many grams of carbon and hydrogen did the original sample contain?
Solution
The atomic weight of carbon is 12.011 g mole-1, and the molecular weight of CO2 is 44.010 g mole-1.
First we find the percentage of carbon in carbon dioxide:
$\textstyle{(\frac{12.011}{44.010})} \times 100% = 27.29%\mbox{ carbon}$
If 27.29% of CO2 is carbon, then the quantity of carbon in 68.58 g of CO2 will be
$27.29% \times 68.58\mbox{ g} = 18.72\mbox{ g carbon.}$
A similar calculation for hydrogen in water gives:
$\textstyle{(\frac{2 \times 1.008}{18.015})} \times 100% = 11.19%\mbox{ hydrogen}$
$11.19% \times 56.15\mbox{ g} = 6.283\mbox{ g hydrogen.}$
As a check: 18.72 g + 6.283 g = 25.00 g.
Example 2
How many grams of carbon and hydrogen per 100.0 g of sample are there
in the hydrocarbon of Example 1?
Solution
$\textstyle{(\frac{100\mbox{ g}}{25\mbox{ g}})} \times 18.72\mbox{ g carbon} = 74.88\mbox{ g carbon per 100 g sample}$
$\textstyle{(\frac{100\mbox{ g}}{25\mbox{ g}})} \times 6.28\mbox{ g carbon} = 25.12\mbox{ g hydrogen per 100 g sample}$
Figure 2-2. Seven different molecules with the empirical formula CH. A simple elemental analysis could not distinguish between them. An approximate molecular weight could distinguish between C2H2, C4H4, and C6H6, but even more information would be required to identify the particular C6H6 molecule present.
Example 3
What is the percent composition by weight of the hydrocarbon in Example 1?
Solution
$\left ( \frac{18.72\mbox{ g carbon}}{25\mbox{ g total}} \right ) \times 100% = 74.88%\mbox{ carbon}$
$\left ( \frac{6.28\mbox{ g hydrogen}}{25\mbox{ g total}} \right ) \times 100% = 25.12%\mbox{ hydrogen}$
Example 4
Calculate the relative number of carbon and hydrogen atoms in the hydrocarbon of Example 3.
Solution
It is easiest to work with 100.0 g of the substance, so the percent elemental composition figures become grams of the respective elements. First we divide each amount of carbon and hydrogen by their atomic weights:
$\frac{74.88\mbox{ g carbon}}{12.011\mbox{ g/mole}} = 6.234\mbox{ moles carbon}$
$\frac{25.12\mbox{ g hydrogen}}{1.008\mbox{ g/mole}} = 24.92\mbox{ moles hydrogen}$
These are the relative numbers of moles of carbon and hydrogen, and it is here that the concept of moles becomes useful. These numbers must also be the relative numbers oí atoms of carbon and hydrogen. For every 6.234 atoms of carbon in the unknown hydrocarbon, there are 24.92 atoms of hydrogen. If we look for a common factor for these two numbers, we see that they are in a 1:4 ratio. Dividing both numbers by the lower number, 6.234, we find that for every one atom of carbon there are $\frac{24.92}{6.234} = 3.997\mbox{ atoms of hydrogen.}$
Example 5
A common liquid is 11.19% hydrogen and 88.81% oxygen by weight. What are the relative numbers of hydrogen and oxygen atoms?
Solution
Again working with 100.0 g of the substance, we calculate the number of moles of each element:
$\frac{74.88\mbox{ g hydrogen}}{1.008\mbox{ g/mole}} = 11.10\mbox{ moles hydrogen}$
$\frac{25.12\mbox{ g oxygen}}{15.999\mbox{ g/mole}} = 5.551\mbox{ moles oxygen}$
Dividing both numbers by the smaller, to search for a common factor, we find that there are two atoms of hydrogen for every atom of oxygen.
Example 6
A common laboratory solvent, a hydrocarbon, is made up of 92.26% carbon and 7.74% hydrogen. What are the relative numbers of carbon and hydrogen atoms in the substance?
Solution
The answer is that one carbon is found for each hydrogen.
An elemental analysis, by itself, is not enough to decide the correct molecular formula of a compound. The formula for methane is CH4, which would fit the results of the calculation in Example 4. But the analytical results would also be compatible with the molecules C2H8, C3H12, or C4H16, if they could exist. The substance in Example 5 might be water, H20, but it could also be H402 or some higher multiple. If you recognize, correctly, that only CH4 and H20 are chemically sensible, then you are bringing to bear new chemical information that is not present in the analytical data alone. Most chemists would assume that the molecule in Example 6 was benzene, C6H6. But it could also be acetylene, C2H2 (except for the fact that acetylene is a gas at room temperature, and the unknown hydrocarbon was said to be a common laboratory solvent, which would exclude acetylene) or any of the five other less common hydrocarbon molecules shown in Figure 2-2.
A chemical formula that gives the relative number of each type of atom, as integers with no common factor, is called the empirical formula of the substance. It is the empirical formula that results from an elemental analysis of a substance, not the molecular formula, which could be the same as the empirical formula or could be some integral multiple of it. The empirical formula is the same as the molecular formula for methane, CH4, and for water, H2O; the empirical formulas of acetylene and benzene are both CH, but the molecular formulas are C2H2 and C6H6, respectively. It frequently happens that some simple physical measurement can give a rough approximation of the molecular weight of a substance. Gas densities (Chapter 3), freezing-point depression, and osmotic pressure measurements (Chapter 18) are useful in this regard. If such an approximate molecular weight is available, then it can be used along with the empirical formula to decide the true molecular formula.
Example 7
Glucose is 40.00% carbon by weight, 6.71% hydrogen, and 53.29% oxygen. What is its empirical formula, and what is its molecular formula?
Solution
Working with 100.0 g of glucose, we first find the number of moles of each element:
$\frac{40.00\mbox{ g carbon}}{12.011\mbox{ g/mole}} = 3.330\mbox{ moles carbon}$
$\frac{6.71\mbox{ g hydrogen}}{1.008\mbox{ g/mole}} = 6.66\mbox{ moles hydrogen}$
$\frac{53.29\mbox{ g hydrogen}}{15.999\mbox{ g/mole}} = 3.331\mbox{ moles hydrogen}$
This is obviously a molar ratio of one carbon to two hydrogens to one oxygen, so the empirical formula is CH2O. With the information provided we have no way of knowing whether this, or some multiple of this, is the true molecular formula.
Example 8
From other experiments we know that glucose has a molecular weight of approximately 175 g mole-1. Use this information and the results of Example 7 to find the molecular formula and the exact molecular weight of glucose.
Solution
The weight corresponding to the empirical formula is
$12.011 + (2 \times 1.008) + 15.999 = 30.026\mbox{ g/mole}$
The approximate molecular weight is roughly six times this value, so the precise molecular weight is
$6 \times 30.026\mbox{ g/mole} = 180.16\mbox{ g/mole,}$
and the molecular formula is C6H1206.
Chemical Equations. Balancing Equations and Conservation of Mass
When propane gas, C3H8, is burned in oxygen, the products are carbon dioxide and water. This can be written as a chemical equation:
C3H8 + O2 → CO2 + H2O (2-1)
If chemistry were not a quantitative science, then this description of the reaction, identifying both the reactants and the products, would be adequate. But we expect more from a chemical equation. How many molecules of oxygen are required per molecule of propane, and how many molecules of carbon dioxide and water result? Equation 2-1 is an unbalanced equation. When we add numerical coefficients (placed to the left of the formula) that tell how many of each kind of molecule are involved, then there will be the same number of each kind of atom on the left and right sides of the equation, since atoms are neither created nor destroyed in a chemical reaction. The result will be a balanced equation.
To balance equation 2-1, we note first that the 3 carbon atoms on the left will lead to 3 molecules of CO2 as products, each requiring 2 oxygen atoms, or 6 oxygens in all. Similarly, the 8 hydrogen atoms in propane will produce 4 molecules of water, requiring 4 more oxygen atoms. This total of 10 oxygens on the right must come from 5 molecules of 02. The correct coefficients for the four substances in equation 2-1 are therefore 1,5,3, and 4:
C3H8 + 5 O2 → 3 CO2 + 4 H2O (2-2)
Each side of this balanced equation contains 3 carbon atoms, 8 hydrogen atoms, and 10 oxygen atoms.
Example 9
Trinitrotoluene (TNT), C7H5N3O6, combines violently with oxygen to produce CO2, water, and N2. Write a balanced chemical equation for the explosion.
Solution
The unbalanced equation is
C7H5N306 + O2 → C02 + H2O + N2
Since there are odd numbers of hydrogen and nitrogen atoms on the left, and even numbers on the right, it will be easier to balance the equation on the basis of two molecules of TNT:
2 C7H5N3O6 + O2 → CO2 + H2O + N2
The 14 carbons, 10 hydrogens, and 6 nitrogens then will result in 14 carbon dioxide, 5 water, and 3 nitrogen molecules:
2 C7H5N306 + O2 → 14 CO2 + 5 H2O + 3 N2
Now all atoms are balanced on left and right sides of the equation except for oxygen. Of the 33 oxygens on the right, 12 are provided on the left by the 2 starting molecules of TNT, and 21 must be supplied by $\textstyle{10\tfrac{1}{2}}$ 02. The final, balanced equation is
2 C7H5N306 + 10$\tfrac{1}{2}$ O2 → 14 CO2 + 5 H2O + 3 N2
Example 9 led to an equation with a fractional coefficient for oxygen. This can be removed by multiplying all coefficients on both sides by 2:
4 C7H5N3O6 + 21 O2 -> 28 CO2 + 10 H2O + 6 N2 (2-3)
but this is not necessary, since there is no reason why all coefficients must be integers. It would even be correct to base the equation on a single molecule of TNT:
C7H5N3O6 + 21/4 O2 → 7 CO2 + 5/2 H2O + 3/2 N2 (2-4)
A balanced chemical equation such as equation 2-3 has several levels of meaning. Most simply, it describes the starting materials and the products. It also tells us that the number of each kind of atom entering the reaction is the same as the number leaving. Each type of atom individually is conserved during the reaction. Equation 2-3 is also a statement that for every 4 molecules of TNT, 21 molecules of oxygen are required, and the products are 28 molecules of CO2, 10 molecules of water, and 6 molecules of N2. Scaling the reaction up by a factor of 6.022 X 1023 to go from molecules to moles, 4 moles of TNT react with 21 moles of O2 to produce 28 moles of CO2, 10 moles of H2O, and 6 moles of N2. The individual molecular weights are
C7H5N3O6 227.13 g mole-1
O2 31.999 g mole-1
CO2 44.010 g mole-1
H2O 18.015 g mole-1
N2 28.013 g mole-1
Hence equation 2-3 also tells us that 4 X 227.13 g = 908.52 g of TNT requires 21 X 31.999 g = 671.98 g of oxygen for complete reaction. It also tells us that the products will be
28 X 44.010 g = 1232.3 g of CO2 10 X 18.015 g = 180.15 g of H2O 6 x 28.013 g = 168.08 g of N2
We can verify that mass is indeed conserved:
Reactants: 908.52 g + 671.98 g = 1580.5 g Products: 1232.3 g + 180.15 g + 16.08 g = 1580.5 g
What a balanced chemical equation does not tell us is the molecular mechanism or course of events by which the reaction takes place. Equation 2-3 should not be construed as suggesting that 4 TNT molecules must collide simultaneously with 21 oxygen molecules. Even three-body collisions are so much rarer than two-body collisions that they can be dismissed from consideration in most chemical reactions. An elaborate series of individual steps could take place, as long as the overall net reaction was described correctly by equation 2-3 or equation 2-4.
The reactants and products need not be molecules:
CaCO3 + 2 HCl → CaCl2 + CO2 + H2O (2-5)
Equation 2-5 describes the reaction of CaCO3, calcium carbonate (limestone), and HCl, hydrochloric acid, to produce an aqueous solution of calcium chloride, CaCl2, and carbon dioxide. The equation is balanced, because the number of each type of atom is the same on both sides. The molar meaning is clear: 1 mole or 100.09 g of calcium carbonate requires 2 moles or 72.92 g of hydrochloric acid for complete reaction, and the products will be 1 mole each of calcium chloride (110.99 g mole-1), carbon dioxide (44.01 g mole-1), and water (18.02 g mole-1). You can verify from these figures that mass is conserved during the reaction. The molecular interpretation is less straightforward, since calcium carbonate is a salt and not a molecular compound, Equation 2-5 should not be taken as meaning that one molecule of calcium carbonate reacts with two molecules of HCl. Although HCl exists as discrete molecules in the gas phase, in solution the molecules dissociate into H+ and Cl- ions. A better approximation of what actually happens at the molecular level is
CaCO3(s) + 2 H+(aq) → Ca2+(aq) + CO2(g) + H2O(l) (2-6)
The letters in parentheses describe the physical state of each species (s, solid; aq, a hydrated ion in aqueous solution; g, gas; l, liquid). This equation says that solid calcium carbonate reacts with two hydrated protons (hydrogen ions) in aqueous solution to produce hydrated calcium ions, gaseous carbon dioxide, and liquid water. The chloride ions remain as hydrated chloride ions in solution before and after the reaction; hence they are omitted from the equation. Equation 2-5, like other balanced chemical equations, is most useful in describing the amounts of materials involved, rather than the molecular mechanism of reaction. Equation 2-6, although a better description of what is going on at the level of atoms and ions, is less useful in keeping track of the quantities of matter involved.
Example 10
Metallic sodium reacts with water to produce hydrogen gas and sodium hydroxide solution (a mixture of Na+ and OH- ions). Write (a) a balanced equation for the overall reaction and (b) an equation that more accurately describes the actual atomic or ionic species present.
Solution
The balanced equation is
Na + H2O → 1/2 H2 + NaOH
or
2 Na + 2 H2O → H2 + 2 NaOH
A better description of what is actually present would be
2 Na(s) + 2 H2O(l) → H2(g) + 2 Na+(aq) + 2 OH-(aq)
Calculations of Reaction Yields
Balanced chemical equations are primarily used to calculate the expected yield (quantity of product) from a reaction, and to determine whether any of the reactants will remain unused when the other reactants are depleted.
Example 11
How many grams of hydrogen are needed to combine with 100.0 g of carbon
to make benzene, C6H6? How many moles and how many grams of benzene will be produced?
Solution
The balanced equation for the reaction is
6 C + 3 H2 → C2H6
The number of moles of carbon present is
$\frac{100.0\mbox{ g carbon}}{12.011\mbox{ g/mole}}=8.326\mbox{ moles carbon}$
The balanced equation tells us that half as many moles of H2 are needed as moles of C, so we need 4.163 moles of H2
$4.163\mbox{ moles hydrogen}\times2.016\mbox{ g/mole}=8.393\mbox{ g hydrogen}$
The molecular weight of benzene is
(6 X 12.011 g) + (6 X 1.008 g) = 78.11 g mole-1
One-sixth as many moles of benzene are produced as moles of carbon used up, or 8.326/6 = 1.388 moles of benzene. Hence the amount of benzene produced is
$1.388\mbox{ moles} \times 78.11\mbox{ g/mole} = 108.4\mbox{ g benzene}$
As a check on arithmetic, note that 100.0 g of carbon and 8.4 g of hydrogen combine to produce 108.4 g of benzene. Again, mass is conserved during a chemical reaction.
Example 12
How many grams of silver sulfide (Ag2S) can be formed by the reaction
2Ag + S → Ag2S
if we start with 10.00 g of silver (Ag) and 1.00 g of sulfur (S)? Which starting material, if any, will be left over, and how much?
Solution
Reaction: 2 Ag + S → Ag2S
Masses: 215.7 g + 32.06 g = 247.8 g
The quantity of sulfur required to react with 10.00 g of silver is
$\frac{32.06\mbox{ g S}}{215.7\mbox{ g Ag}}\times10.00\mbox{ g Ag}=1.486\mbox{ g S}$
But we only have 1.00 g sulfur, so not all of the silver can react. Turning the problem around, the amount of silver needed to react with 1.00 g sulfur is
$\frac{215.7\mbox{ g Ag}}{32.06\mbox{ g S}}\times1.00\mbox{ g S}=6.73\mbox{ g Ag}$
Hence, there will be $10.00 - 6.73 = 3.72\mbox{ g Ag}$ left over. The quantity of silver sulfide produced will be
$\frac{247.8\mbox{ g silver sulfide}}{32.06\mbox{ g S}}\times1.00\mbox{ g S}=7.73\mbox{ g silver sulfide}$
Notice that Example 12 was worked a different way: First the total masses of all reactants and products were written under the balanced chemical equation, and then the ratios of these masses were worked out to find the desired answers. The problem can also be solved using moles, and the choice is one of convenience.
Alternative Solution
First we find the number of moles of silver and sulfur:
$\frac{10.00\mbox{ g Ag}}{107.9\mbox{ g/mole}}=0.0927\mbox{ mole Ag}$
$\frac{1.00\mbox{ g S}}{32.06\mbox{ g/mole}}=0.0312\mbox{ mole S}$
Since 2 moles of silver are required for every mole of sulfur, and there are more than twice as many moles of silver as moles of sulfur, some of the silver must be left behind when all of the sulfur has been used. The 0.0312 mole of sulfur will combine with 0.0624 mole of silver, adn form 0.0312 mole of Ag2S. Left behind are 0.0927 - 0.0624 = 0.0303 mole of silver. Translating these quantities back to grams, we have
$0.0303\mbox{ mole Ag}\times107.9\mbox{ g/mole}=3.27\mbox{ g Ag left over}$
$0.0312\mbox{ mole Ag}_2\mbox{S}\times247.8\mbox{ g/mole}=7.73\mbox{ g Ag}_2\mbox{S produced}$
We arrived at the same answers in the first solution to the example. The mole method is surer, but slower. The ratio method is faster, but you can go astray more easily if you are not absolutely sure of what you are doing. Use the mole method until you are proficient in chemical calculations.
Solutions as Chemical Reagents
Liquid solutions are convenient media for chemical reactions. Rapid mixing of the liquid means that potential reactants are brought close to one another frequently, so collisions and chemical reactions can take place much faster than they would in a crystalline solid. Moreover, a given number of molecules in a liquid is confined to a smaller space than the same number of molecules in a gas, so reactant molecules in a liquid have more of a chance to come in contact. Water is an especially good solvent for chemical reactions because its molecules are polar. The H2O molecules, and the H+ and OH- ions into which water dissociates to a small extent, can help to polarize bonds in other molecules, weaken bonds, and encourage chemical attack. It is no accident that life evolved in the oceans rather than in the upper atmosphere or on dry land. If life had been forced to evolve using solid-state crystal reactions, the 4.5 billion years of earth's history to date might barely have been time enough for the process to begin.
Concentration Units: Molarity and Molality
In solutions involving a liquid and a gas or solid, the liquid component is called the solvent, and the other component is called the solute. If the solution is made up of two liquids, the distinction is less clear, but the substance present in the greater amount is usually considered as the solvent. The most common way of expressing concentration in solution is molarity, or the number of moles of solute per liter of solution. [1] The symbol M is read as "moles per liter of solution," as in 1.5M NaCl. The symbol c is used to denote concentration in moles per liter, as is the chemical symbol in brackets [H], although such brackets are sometimes used to represent concentration in any units. Hence the expression cNaC1 would be read as "the concentration of sodium chloride in moles per liter of solution." This is not the solution that would result from adding 1 mole of NaCl to a liter of water, since the total volume after mixing would be a little more than 1 liter. Sodium and chloride ions take up room, even when dissolved in water. The proper procedure in making a 1.0M solution would be to dissolve the salt in less than a liter of water, and then slowly add more water, with mixing, until the total volume reached 1.00 liter.
For many salts, we can use the approximation that volumes are additive, or that the volume of a solution will be equal to the original volume of the solvent plus that of the crystals that were dissolved.
Example 13
If 264 g of ammonium sulfate, (NH4)2SO4, are dissolved in 1.000 liter of water, what will the approximate final volume and the approximate molarity of the solution be, assuming additivity of volumes? The density of crystalline (NH4)2SO4 is 1.76 g ml-1.
Solution
The volume of solid ammonium sulfate added is
$\frac{264\mbox{ g}}{1.76\mbox{ g/mL}}=150\mbox{ mL or 0.150 L}$
The final solution volume then will be 1.000 + 0.150 = 1.150 liters. The number of moles of solute is $\frac{264\mbox{ g}}{132\mbox{ g/mole}}=2.00\mbox{ moles ammonium sulfate}$
The molarity of ammonium sulfate then is $\frac{2.00\mbox{ moles}}{1.15\mbox{ liters}}=1.74\mbox{ moles/liter or 1.74}M$
The approximation of additivity of volumes must be used with care. In this example, the true molarity of such a solution is 1.80M, so the approximation is only 3.3% in error. But for liquids whose molecules interact strongly, such as ethyl alcohol and water, the total volume may shrink after mixing because of molecular attractions. Additivity of volumes should be used only as a rough guide to molarity.
An alternative expression of concentration, molality, is based on the amount of solvent used rather than the solution that results. The molality of a solute is the number of moles of solute in 1 kg of solvent (not of solution). The density of water is 1.00 g mL-1, so 1 kg of water occupies a volume of 1 liter. Hence the ammonium sulfate solution of Example 13 is a 2.00 molal solution, since it was made up from 2.00 moles of solute in a kilogram (1 liter) of water. For solvents other than water, we must use the density of the liquid to convert from kilograms to liters.
Example 14
Suppose 5.00 g of acetic acid, C2H4O2, are dissolved in 1 liter of ethanol.
Calculate the molality of the resulting solution. The density of ethanol is 0.789 g mL-1. Can you calculate the molarity from the information given?
Solution
The molecular weight of acetic acid is 60.05 g mole-1, so the number of moles is
$\frac{5.00\mbox{ g}}{60.05\mbox{ g/mole}}=0.0833\mbox{ mole acetic acid}$
The number of kilograms of solvent used is
$1.00 \mbox{ liter}\times0.789\mbox{ kg/liter}=0.789\mbox{ kg ethanol}$
Notice that 1 g mL-1 is the same as 1 kg liter-1, since there are 1000 g in a kilogram and 1000 mL in a liter. The molality then is
$\frac{0.0833\mbox{ mole solute}}{0.789\mbox{ kg solvent}}=0.106\mbox{ mole/kg}$
The solution is therefore 0.106 molal. The molarity of the solution cannot be calculated because we know neither the volume of the acetic acid nor whether volumes are additive when acetic acid is dissolved in ethanol.
The symbol m is used for concentration expressed as molality. We would write the results of Example 14 as
macetic acid = 0.106 mole kg-1
Dilution Problems
If we dilute a solution (add more solvent), the number of moles of solute does not change. If c is the molarity (not molality) of the solution and V is the volume in liters, then the number of moles of solute is
c (moles liter-1) X V(liters) = cV(moles)
If we use the subscript 1 to represent a solution before it is diluted with more solvent, and the subscript 2 for the diluted solution, then
Moles of solute = cl Vl = c2 V2
Example 15
To what volume must 5.00 mL of 6.00M HCl be diluted to make the concentration 0.100M?
Solution
$V_2=\frac{c_1}{c_2}\times V_1=\frac{6.00M}{0.100M}\times 5.00\mbox{ mL}=300\mbox{ mL}$
This does not mean that 300 mL of water must be added, but that the total volume of solution must be brought up to 300 mL
Example 16
If 175 ML of a 2.00M solution are diluted to 1.00 liter, what will the molarity be?
Solution
$c_2=\frac{V_1}{V_2}\times c_1=\frac{175\mbox{ mL}}{1000\mbox{ mL}}\times 2.00M=0.350M$
Acid-Base Neutralization
Probably the most familiar definition of acids and bases is that by the Swedish physicist and chemist Svante Arrhenius (1859-1927): An acid is a substance that increases the hydrogen ion concentration, [H+], when added to water, and a base is a substance that increases the hydroxide ion concentration, [OH-], when added to water. Some of the more common acids and bases are listed in Tables 2-1 and 2-2. The first 11 acids in Table 2-1, from HF to HNO3, dissociate in aqueous solution to release one proton or hydrogen ion:
HNO3 → H+(aq) + NO3-(aq)
nitric
acid
nitrate
ion
Table 2-1. Common Acids
HF Hydrofluoric
HCl Hydrochloric
HClO Hypochlorous
HClO2 Chlorous
HClO3 Chloric
HClO4 Perchloric
HBr Hydrobromic
HBrO3 Bromic
HI Hydroiodic
HNO2 Nitrous
HNO3 Nitric
H2CO3 Carbonic
H2SO3 Sulfurous
H2SO4 Sulfuric
H3PO2 Hypophosphorous
H3PO3 Phosphorous
H3PO4 Phosphoric
H3BO3 Boric
HCOOH Formic
CH3COOH Acetic
Table 2-2. Common Bases
LiOH Lithium hydroxide
NaOH Sodium hydroxide
Mg(OH)2 Magnesium hydroxide
Ca(OH)2 Calcium hydroxide
Ba(OH)2 Barium hydroxide
NH3 Ammonia
The abbreviation (aq) is a reminder that the ions are hydrated, but it is really not necessary since every ion in aqueous solution is hydrated, and we shall omit it in the rest of this discussion. Remember that the water molecules are always present, surrounding each ion and helping to stabilize it in solution. Carbonic, sulfurous, and sulfuric acids release two protons in two stages, and the three phosphorus-containing acids produce three protons:
H2CO3 → H+ + HCO3- → 2 H+ + CO32-
carbonic acid
H2SO4 → H+ + HSO4- → 2 H+ + SO42-
sulfuric acid
H3PO4 → H+ + H2PO4- → 2 H+ + HPO42- → 3 H+ + PO43-
phosphoric acid
Carbonic acid is classed as a weak acid because its loss of protons is only partial; the species present in aqueous solution are a mixture of carbonate and bicarbonate ions and a small amount of undissociated carbonic acid. In contrast, sulfuric acid is a strong acid because the loss of the first of the two H+ is complete in aqueous solution. (Acid-dissociation equilibria are considered in detail in Chapter 5.) Nitric and hydrochloric acids are common strong acids, and phosphoric acid is weak. Organic acids such as formic and acetic release a proton from their —COOH carboxyl groups:
CH3COOH CH3COO- + H+
acetic acid acetate ion
It is common to use the abbreviation HOAc for acetic acid and OAC- for the acetate ion.
Hydroxide bases such as sodium hydroxide and magnesium hydroxide dissolve in water to release hydroxide ions:
NaOH → Na+ + OH-
Mg(OH)2 Mg2+ + 2 OH-
The hydroxide ions are already present in solid NaOH, just as chloride ions are present in NaCl. Ammonia, NH3, is also a base, but it has no hydroxide ions of its own. Instead, it produces them by reacting with water molecules:
NH3 + H2O → NH4+ + OH-
ammonia ammonium
ion
hydroxide
ion
Ammonia is sometimes written as ammonium hydroxide, NH4OH, to make it resemble the metal hydroxide bases (such as sodium hydroxide, NaOH). But this is incorrect; there is no such substance as ammonium hydroxide; there is only ammonia.
Acids and bases are useful because the H+ and OH- ions that they produce can attack molecules in solution and bring about chemical changes that would be difficult or slow in their absence. When acids and bases react with one another, the H+ and OH- ions combine to form water molecules. This is called neutralization:
H+ + OH- → H2O
The easiest way to determine how much of an acid or base is present is to find out how much of a base or acid of known concentration is required to neutralize it completely. This is the process of acid-base titration. One equivalent (equiv) of an acid is the quantity of acid that will release 1 mole of protons or H+ in neutralizing a base, and 1 equiv of a base is the quantity that will produce 1 mole of OH- ions. Complete neutralization occurs when the same number of equivalents of acid and base react with one another. For acids that release one proton per molecule, such as HCl and HNO3, the equivalent is the same as the mole, and 1 equivalent weight is the same as the molecular weight. But since H2SO4 is capable of releasing two H+ ions, 1 mole of H2SO4 corresponds to 2 equiv, and the equivalent weight of sulfuric acid in acid-base neutralizations is half the molecular weight. The equivalent weight of phosphoric acid, H3PO4, or the quantity that will produce 1 mole of H+ ions, is one-third the molecular weight. Similarly, the mole and the equivalent are identical for NaOH, KOH, and NH3, but the equivalent weight of Ca(OH)2 is half its molecular weight. We can appreciate the usefulness of the concept of equivalents by looking at the neutralization of phosphoric acid by magnesium hydroxide:
2 H3PO4 + 3 Mg(OH)2 Mg3(PO4)2 + 6 H2O
Molecular weight: 98.0 g 58.3 g 262.9 g 18.0 g
Equivalent weight: 32.7 g 29.2 g
One mole or 98.0 g of phosphoric acid will not neutralize 1 mole or 58.3 g of magnesium hydroxide, but 1 equiv or 32.7 g of phosphoric acid will neutralize 1 equiv or 29.2 g of magnesium hydroxide. This is the same answer that would be obtained by using the balanced equation shown. Since 2 moles of acid react with 3 moles of base as shown, 2 X 98.0 = 196 g of phosphoric acid will neutralize 3 X 58.3 = 175 g of magnesium hydroxide. These numbers are just the numbers obtained by using equivalents, but scaled up by a factor of 6.
Example 17
Use equivalents to find the number of grams of nitric acid, HNO3, needed to neutralize 100.0 g of barium hydroxide, Ba(OH)2.
Solution
The molecular weight of HNO3 is 63.01 g mole-1; of Ba(OH)2, 171.34 g mole-1. The corresponding equivalent weights are 63.01/1 = 63.01 g equiv-1 for HNO3, and 171.34/2 = 85.67 g equiv-1 for Ba(OH)2. The number of equivalents of barium hydroxide is
$\frac{100.0\mbox{ g}}{85.67\mbox{g/equiv}} = 1.167\mbox{ equiv of Ba(OH)}_2$
The same number of equivalents of nitric acid is needed:
$1.167\mbox{ equiv}\times63.01\mbox{g/equiv}=73.53\mbox{ g nitric acid}$
Alternative Solution
This example could also be solved by using the balanced chemical equation:
2 HNO3 + Ba(OH)2 → Ba(NO3)2 + 2 H2O
The number of moles of barium hydroxide at the start is
$\frac{100.0\mbox{ g}}{171.3\mbox{g/mole}}=0.5838\mbox{ mole Ba(OH)}_2$
The balanced equation tells us that twice as many moles of nitric acid are required: 1.167 moles HNO3. In grams, this is
$1.167\mbox{ moles}\times63.01\mbox{ g/mole}=73.53\mbox{ g nitric acid}$
The use of equivalents eliminates the need to work out a balanced equation for the reaction.
The normality of a solution, represented by N is the number of equivalents of solute per liter of solution. A 1.00 M solution of phosphoric acid is 3.00N, and a 0.010M solution of Ca(OH)2 is 0.020N.
Example 18
If 4.00 g of sodium hydroxide are dissolved in water and the volume is
brought up to 500 mL, find the molarity and the normality of the solution.
Solution
Since the molecular weight of NaOH is 40.0 g mole-1,
$\frac{4.00\mbox{ g}}{40.0\mbox{ g/mole}}=0.010\mbox{ mole NaOH}$
$\frac{0.100\mbox{ mole NaOH}}{0.500\mbox{ liter solution}}=0.200\mbox{ mole/liter, or }0.200M\mbox{ NaOH}$
Because 1 mole of NaOH releases 1 mole of OH- ions, the molarity and normality are the same. The solution is 0.200N.
Example 19
If 10.0 g of sulfuric acid (H2SO4) are mixed slowly with enough water to make a final volume of 750 mL, what are the molarity and normality of the resulting solution?
Solution
Since the molecular weight of NaOH is 40.0 g mole-1,
$\frac{10.0\mbox{ g}}{98.1\mbox{ g/mole}}=0.102\mbox{ mole sulfuric acid}$
$\frac{0.102\mbox{ mole}}{0.750\mbox{ liter}}=0.136M\mbox{ sulfuric acid}$
Since each mole of sulfuric acid contributes 2 equiv, the solution is
$2\times0.136M=0.272N\mbox{ H}_2\mbox{SO}_4$
Acid-Base Titration
Figure 2-3. An acid—base titration. The solution in the flask contains an unknown number of equivalents of base (or acid). The burette is calibrated to show volume to the nearest 0.001 cm3. It is filled with a solution of strong acid (or base) of known concentration. Small increments are added from the burette until, at the end point, one drop or less changes the indicator color permanently. (An indication that the equivalence point is being approached is the appearance — and disappearance on stirring — of the color that the indicator assumes beyond neutralization.) At the equivalence point, the total amount of acid (or base) is recorded from the burette readings. The number of equivalents of acid and base must be equal at the equivalence point.
Chemists frequently use titrations to compare relative concentrations of chemical equivalents in acid—base solutions (Figure 2-3). When enough acid solution from a burette (shown in the figure) has been added to neutralize the base in the sample being analyzed, the number of equivalents of acid and base involved must be the same. The point of neutralization is called the equivalence point. An acid—base indicator such as litmus or Phenolphthalein can be used to determine the equivalence point. From the volume of acid solution used and its normality, we can calculate the number of equivalents of base in the unknown sample. If NA and NB are the normalities of acid and base solutions, and VA and VB are the volumes of each at neutrality, then
Number of equivalents = NAVA = NBVB (2-7)
Example 20
If 25.00 ml of phosphoric acid (H3PO4) are just enough to neutralize 30.25 mL of a sodium hydroxide solution, what is the ratio of the normalities of
the two solutions? What is the ratio of molarities?
Solution
$\frac{N_A}{N_B}=\frac{V_B}{V_A}=0.102\mbox{ mole sulfuric acid}$
$\frac{0.102\mbox{ mole}}{0.750\mbox{ liter}}=0.136M\mbox{ sulfuric acid}$
Since each mole of sulfuric acid contributes 2 equiv, the solution is
$2\times0.136M=0.272N\mbox{ H}_2\mbox{SO}_4$
Since the normality of the acid is three times its molarity, and the normality of the base is the same as its molarity, the molarity ratio is
$\frac{c_A}{c_B}=\frac{N_A/3}{N_B}=0.403$
Example 21
In a titration, 25.00 mL of a solution of calcium hydroxide, Ca(OH)2, require 10.81 mL of 0.100N HCl for neutralization. Calculate (a) the normality of the Ca(OH)2 solution, (b) the molarity, and (c) the number of grams of Ca(OH)2 present in the sample.
Solution
The normality of the solution of Ca(OH)2 is
$N_B=\frac{V_A}{V_B}\times N_A=\frac{10.81\mbox{ mL}}{25.00\mbox{ mL}}\times0.100N=0.0432N$
Since 1 mole of calcium hydroxide yields 2 equiv OH-, the molarity is half the normality, or 0.0216M Ca(OH)2.
The number of moles of Ca(OH)2 is
$0.0216\mbox{ mole/liter}\times0.02500\mbox{ liter}=0.000541\mbox{ mole}$
Since the molecular weight of Ca(OH)2 is 74.1 g mole-1, the mass present is
$0.000541\mbox{ mole}\times74.1\mbox{ g/mole}=0.0401\mbox{g Ca(OH)}_2$
Example 22
An organic chemist synthesizes a new acid. She dissolves 0.500 g in a convenient volume of water and finds that it requires 15.73 mL of 0.437N NaOH for neutralization. What is the equivalent weight of the new compound as an acid? If it is known that the acid contains three ionizing -COOH groups, what is the molecular weight?
Solution
The number of equivalents of the base are
$0.01573\mbox{ liter}\times0.437\mbox{ eqiv/liter}=0.00687\mbox{ equiv}$
The equivalent weight is found from
$\frac{0.500\mbox{ g}}{0.00687\mbox{ equiv}}=72.8\mbox{ g/equiv}$
If the equivalent weight of the acid is 72.8 g, and each mole yields 3 equiv, then the molecular weight is
$3\times72.8=218\mbox{ g}$
Heats of Reaction
So far this chapter has been devoted to the consequences of the conservation of mass, and little has been said about energy. But the principle that heats of reaction are additive, that energy is conserved in a process whether the process is carried out in one step or in several, is an important one. Heat and work are both forms of energy, and are measured in the same units. If you do work on an object or collection of objects, you can increase the energy or make the system heat up, depending on how the work is done. Lifting a heavy object is a conversion of work to potential energy, and friction is a conversion of work to heat. Conversely, energy can be reconverted to work when a heavy object falls, and heat is converted to work in an automobile engine. Of these three — heat, work, and energy — the chemist usually is more concerned with heat: the heat that may be absorbed or given off when a chemical reaction takes place.
By Newton's laws of motion, the force on an object is the product of its mass and acceleration:
Force = mass × acceleration
F = m × a
The force that must be applied to a 1.00 kg mass to give it an acceleration of 1 meter per second per second (1 m s-2) is defined as a force of 1 newton (N). Hence 1 N = 1 kg m s-2. (SI units are based on length in meters and mass in kilograms.)
Example 23
When a pitcher whips a 5.00 ounce (oz) baseball around an arc 5.00 m in circumference in order to accelerate it from zero to 90 miles hr-1, what average acceleration does he give to the ball during the pitch, and what average force does he exert on it during his windup?
Solution
Assume uniform acceleration on the ball from the time the windup begins until the ball leaves the pitcher's hand. For uniform acceleration from rest, v = at and s = at2, where v is velocity, a is acceleration, t is time, and s is distance. Eliminating time from the two expressions yields a = v2/2s. If v = 40.2 m s-1 (90 mph) and s = 5.00 m, then
$a=\frac{40.2^2}{2\times5.00}=162\mbox{ m s}^{-2}$
Since the mass (m) is 0.142 kg (5.00 oz), the average force applied to the ball during the swing is
$F=m\times a=0.142\times162\mbox{ kg m s}^{-2}=23.0N$
The work done when 1 N of force is exerted on an object for a distance of 1 m is defined as 1 joule (J). Hence 1 J = 1 N m = 1 kg m2 s-2. For a thrown object, all the work is converted into kinetic energy (energy of motion); in other circumstances, part or all the work can end as heat.
Example 24
How much work is done on the baseball in the pitch described in Example 23? How much kinetic energy does the ball have as it leaves the pitcher's hand?
Solution
The work done on the baseball is
$W=F\times s=23.0\mbox{ N} \times 5.00\mbox{ m}=115\mbox{ N m} = 115\mbox{ J}$
The ball ends with a kinetic energy of 115 J.
As a check on these results, we can calculate the kinetic energy directly:
$E = \tfrac{1}{2} mv^2 = \tfrac{1}{2}\times0.142\times(40.2)^2\mbox{ kg m}^2\mbox{s}^{-2}=115\mbox{ J}$
where E is energy. The advantage of the joule as a unit of heat is that it makes immediately apparent the connection between heat, work, and energy. An older unit of energy that arose from heat measurements is the calorie. One calorie (cal) is defined as the quantity of heat required to raise the temperature of 1 g of pure water by 1°C (from 14.5°C to 15.5°C, to be exact). This definition had no obvious connection with work, and in fact the calorie was defined in the nineteenth century, before anyone realized that heat and work were alternative forms of the same thing: energy. We will use only joules in this book, but you should be aware of calories since most of the preexisting literature uses that unit. The calorie is approximately four times as big as a joule: 1 cal = 4.184 J. Heats of reaction of mole quantities of substances are typically in the range of kilojoules (kJ) or kilocalories (kcal), where 1 kJ = 1000 J and 1 kcal = 1000 cal.
As an illustration of heats of reaction and the principle of additivity of heat, let us look at the decomposition of hydrogen peroxide, H2O2. When an aqueous solution of hydrogen peroxide reacts to form oxygen gas and liquid water, heat is given off. The amount of heat will vary somewhat with the temperature at which the reaction occurs, but at 25°C, the commonly accepted standard "room temperature" for measuring and tabulating heats of reaction, each mole of H2O2 that decomposes produces 94.7 kJ of heat. (If this energy could be used with perfect efficiency, it would be enough to accelerate 823 baseballs as described in Example 24.)
A schematic representation of a bomb calorimeter used for the measurement of heats of combustion. The weighed sample is placed in a crucible, which in turn is placed in the bomb. The sample is burned completely in oxygen under pressure. The sample is ignited by an iron wire ignition coll that glows when heated. The calorimeter is filled with fluid, usually water, and insulated by means of a jacket. The temperature of the water is measured with the thermometer. From the change in temperature, the heat of reaction can be calculated.
The heat involved in a chemical reaction carried out at constant pressure (or at least with the final pressure brought back to the starting value) is known as the change in the enthalpy of the reacting system, ΔH (read as "delta H"). As we shall see in Chapter 15, the energy change, ΔE, corresponds to the heat of the reaction if the reaction is carried out at constant volume, as in the bomb calorimeter shown in Figure 2-4. Enthalpy can be regarded as a "corrected" energy, the correction being for any work that the chemicals might do in pushing against the atmosphere if they expand. The difference between ΔE and ΔH is small but significant, but it is not important to us now. If heat is given off during the reaction, then the enthalpy of the reacting system of chemical falls; ΔH, the change in enthalpy, is negative. Such a reaction is called exothermic. In an endothermic reaction heat is absorbed and the enthalpy of the reaction mixture rises. For the hydrogen peroxide reaction, we can write
H2O2(aq) → H2O(l) + 1/2 O2 ΔH = -94.7 kJ (2-8)
This heat is released when 1 mole of hydrogen peroxide decomposes to 1 mole of water and 1/2 mole of oxygen gas, or for 1 mole of the reaction as just written. If all the coefficients of the reaction are doubled, then the heat of reaction must be doubled also, since it then refers to twice as much reaction:
2 H2O2(aq) → 2 H2O(l) + O2 ΔH = -189.4 kJ (2-9)
"One mole of the reaction as written" now means 2 moles of hydrogen peroxide decomposing to 2 moles of water and 1 mole of oxygen, because the coefficients in the equation have all been doubled. The heat of a reaction also depends on the physical state of the reactants and products. If hydrogen peroxide were to decompose to give oxygen gas and water vapor instead of liquid, part of the 94.7 kJ would be diverted to evaporating H2O:
H2O(l) → H2O(g) ΔH= +44.0 kJ (2-10)
and less heat would be given off by decomposition of peroxide:
H2O2(aq) → H2O(g) + 1/2 O2(g) ΔH= -50.7 kJ (2-11)
An important assumption is hidden here: that heats of reaction are additive (Figure 2-5). Equation 2-9 plus equation 2-10 gives equation 2-11, and so we have assumed that the heat of the third reaction will be the sum of the first two:
ΔH= -94.7 + 44.0 kJ = -50.7 kJ
The change in enthalpy for the reaction
H2O2(aq) → H2O(g) + 1/2 O2(g) can be obtained without actually measuring the enthalpy change for the reaction by adding ΔH for the two reactions
H2O2(aq) → H2O(l) + 1/2 O2(g) ΔH = -94.7 kJ/mol
H2O(l) → H2O(g) ΔH = +44.0 kJ/mol
The additivity of reaction heats follows directly from the first law of thermodynamics (Chapter 15): The change in energy or enthalpy between two
states depends only on the nature of those states, and not on how the change is carried out. A collection of chemicals in a given state has a certain energy and a certain enthalpy, neither of which depends in any way on how the chemicals were brought to that state. (That is, the past history of the chemicals can affect their present energy and enthalpy, but we do not need to know that history to measure the values of E and H.) Hence the difference between enthalpies of reactants and products, or the heat of reaction, can depend only on the nature of the starting and ending states, and not on the particular way that the reaction is carried out. This is sometimes called Hess' law of heat summation, which is a rather dignified name for a natural consequence of the first law of thermodynamics.
The additivity of heats of reaction makes a great amount of experimentation in thermochemistry (the chemistry of heat and energy) unnecessary. We need not measure and tabulate the enthalpy change of every conceivable chemical reaction. For example, if we know the heat of vaporization of liquid water (equation 2-10) and the heat of decomposition of hydrogen peroxide to liquid water (equation 2-9), then we never need to measure the heat of decomposition of hydrogen peroxide to water vapor; the answer can be calculated ahead of time. If a certain reaction is inconvenient to carry out, there may be a set of easier reactions whose sum is the reaction in question. After we have carried out the individual experiments, we can add the enthalpy changes in the same way as the chemical equations, to find the heat of the difficult-to-measure reaction.
Suppose that someone proposed a scheme for making diamonds by oxidizing methane:
CH4(g) + O2(g) → C(di) + 2 H2O(l)
[The notation (s) is not sufficient for carbon, since diamond (di) must be differentiated from graphite (gr).] You would like to find whether the reaction will liberate heat that must be allowed for in the design of the reaction vessel. This particular synthesis has never been carried out (and probably never will be), yet you can give your misguided friend his answer from a knowledge of the heats of easier reactions. The heat of combustion of a substance containing C, N, O, and H is the heat, per mole of substance, of the reaction with enough oxygen to produce CO2, N2, and liquid H2O. Heats of combustion are easy to measure, and were among the first reaction heats to be measured and tabulated systematically. Extensive tables of heats of combustion can be found in books such as the CRC Handbook of Chemistry and Physics or Lange's Handbook of Chemistry. The heats of combustion of methane and diamond are
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l) ΔH = -890 kJ/mol (2-12)
C(di) + O2(g) → CO2(g) ΔH = -395 kJ/mol (2-13)
The desired diamond-synthesizing reaction is produced by subtracting the second reaction from the first, or by adding the first reaction to the reverse of the second, and the heat of reaction is found in the same way:
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l) ΔH = -890 kJ/mol (2-12)
CO2(g) → C(di) + O2(g) ΔH = +395 kJ/mol (2-14)
CH4(g) + O2(g) → C(di) + 2 H2O(l) ΔH = -495 kJ/mol (2-15)
Notice that when a reaction is turned around and run in reverse, the heat of reaction changes sign, since a process that gave off 395 kJ in one direction must absorb 395 kJ in reverse.
Heats of Formation
Because of the additivity of heats of reaction, not all heats have to be tabulated — only those for the minimum set of reactions from which all others can be obtained. The set that has been agreed upon by scientists and engineers is made up of the heats of formation of compounds from their pure elements in standard states. For solids and liquids this standard state is the most common form of the element at 25 °C or 298 K and 1 atmosphere (atm)[2] external pressure; gases are defined similarly but at 1 atm partial pressure.[3] The standard state for thermodynamic measurements involving carbon is graphite (gr), not diamond (di). The heats of formation for all the compounds involved in the diamond synthesis are
C(gr) + 2 H2(g) → CH4(g) ΔH = -74.8 kJ/mol (2-16)
C(gr) → C(di) ΔH = +1.9 kJ/mol (2-17)
H2(g) + 1/2 O2(g) → H2O(l) ΔH = -285.8 kJ/mol (2-18)
A table of standard heats of formation of compounds from their pure elements is given in Appendix 3. In that table, the subscript 298 refers to the temperature (298 K), and the zero superscript signifies that reactants and products are all in their standard states. To illustrate how heats of general reactions are found from heats of formation, let us look again at the diamond synthesis, equation 2-15. That reaction can be obtained by adding equation 2-17 to twice equation 2-18 and the reverse of equation 2-16:
C(gr) → C(di) ΔH = +1.9 kJ (2-17)
2 H2(g) + O2(g) → 2 H2O(l) ΔH = -571.6 kJ 2x(2-18)
CH4(g) → C(gr) + 2 H2(g) ΔH = +74.8 kJ -(2-16)
CH4(g) + O2(g) → C(di) + 2 H2O(l) ΔH = -494.9 kJ (2-15)
The heat of reaction is found in exactly the same manner:
ΔH = (+1.9) + 2(-285.8) - (-74.8) = -494.9 kJ
As you can see, the consequences of not keeping track of signs and coefficients can be disastrous. The surest method is to write out each equation, with its heat of reaction, in such a way that the sum of the individual equations will be the desired reaction. If all the coefficients of an equation are multiplied by an arbitrary number n, then the heat of formation must also be multiplied by n, and if a formation equation is turned around and run in reverse, then the sign ΔH must be changed. If the individual equations add to give the desired reaction, then the individual heats add to give the corresponding overall heat of reaction.
A convenient shortcut is to think of the heat of formation of a compound as if it were, in a sense, the enthalpy of the compound itself. (Warning: This is possible only because the heats of formation of the elements are zero by definition.) Then the heat of a reaction becomes the sum of heats of formation of all the products, minus the heats of formation of all the reactants, each being multiplied by the coefficient of that substance in the balanced equation.
Example 25
What is the standard heat of the reaction by which ferric oxide is reduced by carbon to iron and carbon monoxide in a blast furnace?
Solution
The reaction is as follows, with the standard heat of formation per mole written below each compound:
Fe2O3(s) + 3 C(gr) 2 Fe(s) + 3 CO(g)
ΔH (kJ/mole) -822.1 0.0 0.0 -110.5
The standard heat of formation of the elements from themselves, of course is zero, by definition. For the reaction as written,
ΔH = 2(0.0) + 3(-110.5) - (-822.1) - 3(0.0) = +490.6 kJ
These results are consistent with the fact that much heat must be supplied to reduce iron ore to iron. Note, however, that 490.6 kJ is the net heat that would be absorbed if the reaction were run at 298 K, not the 1800 K of a blast furnace. Yet this calculated figure is also the heat absorbed if ferric oxide and carbon are heated from 298 K to 1800 K, allowed to react, and the products are cooled again to room temperature. The enthalpy change or heat of a reaction depends only on the initial and final states of the participants, and not on whether the temperature remained constant or went to blast-furnace levels in between. All that matters is that the temperature is brought back down to 298 K at the end.
As another example of the principle, the net heat evolved when water is made from hydrogen and oxygen will be the same whether a mixture of H2 and 02 at 298 K explodes violently and the resulting water is cooled back to 298 K, or whether the same mixture reacts slowly in the presence of finely divided platinum as a catalyst, never increasing its temperature. So, in referring to heats of reaction, when we say that the values are correct for the process carried out "at 1 atm pressure and 298 K," we require only that the reactants begin at these conditions, and that the products end there. This is why tables of heats of formation under standard conditions (Appendix 3) are useful.
References and Notes
2. The SI unit of length is the meter (m), divided into 10 decimeters (dm) or 100 centimeters (cm). The unit of volume is the cubic meter (m3). For laboratory work the cubic meter is too large to be convenient, so it is customary to use the liter, which in the SI is defined as 1 dm3, and the milliliter (mL), which equals 1 cm3 (or sometimes cc). By strict logic the liter is an extraneous unit in SI, but it is too convenient, and its use too deeply ingrained, to be eliminated. Scientists in the past tended to use milliliters for liquid volumes, and cubic centimeters for volumes of solids. Hence the volume of a sodium chloride solution would be measured in milliliters, but the density of rock salt (sodium chloride crystals) would be reported in grams per cubic centimeter, or g cm-3. We shall use only milliters in this chapter, but thereafter shall feel free to use cubic centimeters wherever that unit seems more natural. Remember that 1 m3 = 1000 liters, 1 liter = 1000 mL, and 1 mL = 1 cm3. For more information on SI, see Appendix 1.
3. The conversion to the absolute or Kelvin temperature scale is considered in Chapter 3, as is the atmosphere as a unit of pressure. (By SI convention, no degree sign is used for the Kelvin scale.)
4. The partial pressure of a gas in a mixture is the pressure that the gas would show if all the other gases were removed and it were the only gas present. | textbooks/chem/General_Chemistry/Chemical_Principles_(Dickerson)/02%3A_Conservation_of_Mass_and_Energy/5.01%3A_Hydrogen_Oxygen_and_Water.txt |
Scientific research consists in seeing what
everyone else has seen, but thinking what
no one else has thought.
A. Szent-Gyorgyi (b. 1893)
IntroductionEdit
The word gas comes from gaos, a Dutch form of the word chaos. Gases were the last substances to be understood chemically. Solids and liquids were easy to identify and differentiate, but the idea of different kinds of "ayres" came only slowly. Carbon dioxide was not prepared from limestone until 1756. Hydrogen was discovered in 1766; nitrogen, in 1772; and oxygen, in 1781. Although gases were late in being identified, they were the first substances whose physical properties could be explained in terms of simple laws. It is fortunate that when matter in this most elusive state is subjected to changes in temperature and pressure, it behaves according to rules much simpler than those that solids and liquids follow. Moreover, one of the best tests of the atomic theory is its ability to account for the behavior of gases. This is the story of the present chapter.
Given any trapped sample of gas, we can measure its mass, its volume, its pressure against the walls of a container, its viscosity, its temperature, and its rate of conducting heat and sound. We can also measure the rate at which it effuses through an orifice into another container, and the rate at which it diffuses through another gas. In this chapter we shall show that these properties are not independent of one another, that they can all be related by a simple theory which assumes that gases consist of moving and colliding particles.
Avogadro's LawEdit
One of the most important hypotheses in the development of atomic theory was made by Amedeo Avogadro (1776-1856) in 1811. He proposed that equal volumes of all gases, at a specified temperature and pressure, contain equal numbers of molecules. This means that the density of a gas-its weight per unit volume-in grams per milliliter must be proportional to the molecular weight of the gas. If Avogadro's ideas had not been ignored for another 50 years, the process of arriving at a dependable set of atomic weights for the elements would have taken much less time. (The entire story is told in Chapter 6.) It was a belated tribute to an unjustly ignored scientist to call the number of molecules per mole Avogadro's number.
If we accept Avogadro's principle, then the number of molecules, n, and also the number of moles, is proportional to V, the volume of the gas:
n = number of moles = kV (at constant P and T).
In this equation, k is a proportionality constant; it changes with temperature and pressure. We shall be looking for other such relationships for gases -relationships connecting the pressure, P, the volume, V, the temperature, T, and the number of moles in a sample, n.
The Pressure of GasEdit
Figure 3-1 Measuring gas pressure . (a) Torricellian barometer . When a mercury-filled tube is inverted in a dish of mercury. the level in the tube falls, thereby leaving a vacuum at the top of the tube . Only a trace of mercury vapor is present in the space at the top. The height of the column is determined by the pressure of the atmosphere on the mercury in the reservoir. (b) In a gas-handling system, pressure P (in millimeters of Hg) is determined by measuring the difference in heights of the two mercury columns of a manometer. If the system is evacuated completely. the levels are equal.
If a glass tube, closed at one end, is filled with mercury (Hg), and the open end is inverted in a pool of mercury as in Figure 3-1a, the level of mercury in the tube will fall until the mercury column stands about 760 millimeters (mm) above the surface of the pool. The pressure produced at the pool surface by the weight of the mercury column is balanced exactly by the pressure of the surrounding atmosphere. Because there is a balance of opposing pressures, more mercury will not flow into or out of the tube. A device such as this (called a barometer) can measure atmospheric pressure, as Italian mathematician and physicist Evangelista Torricelli (1608-1647) first realized. He showed that it was the pressure at the bottom of the mercury column that mattered, and not the total weight of mercury; thus, the height of mercury in a barometer tube is independent of the size or shape of the tube. Atmospheric pressure at sea level supports a column of mercury 760 mm high. Because mercury columns were used so often in early pressure experiments, the "millimeter of mercury" became a common unit of pressure. Pressure is force per unit area (P = F / A), and the SI unit of pressure is the pascal (Pa), defined as 1 newton per square meter (N m-2). (The newton, as you may recall from Chapter 2, is the force that will impart an acceleration of 1 m sec-2 to a 1-kg mass; 1 N = 1 kg m sec-2 .)
Example 1
The density of liquid mercury is 13.596 g cm-3. How would you express a pressure of 1 mm Hg in pascals?
Solution
Imagine a sheet of mercury 1 m square and 1 mm thick. Converting first to centimeters for convenience) we would find its volume to be
0.100 cm X 100 cm X 100 cm = 1000 cm3
and its mass to be
1000 cm3 X 13.596 g cm-3 = 13,596 g or 13.596 kg
Since force is equal to mass times acceleration, and the acceleration of gravity at sea level, g, is 9.806 m sec-2, the force exerted by the mercury on its supporting table would be
F = mg
F = 13.596 kg X 9.806 m sec-2 = 133.32 kg m sec-2, or 133.32 N
Since the area of the sheet is I m2, the pressure on the table would be
P = $\textstyle\frac{F}{A}$
P = $\textstyle\frac{133.32 N}{1 m^2}$ = 133.32 N m-2 = 133.32 pascals (Pa)
Example 2
Standard sea level pressure is exactly 760 mm Hg. Express this in pascals.
Solution
From Example I , we know that a pressure of I mm Hg is equal to 133.3 Pa. Therefore,
760 mm Hg X 133.32 Pa mm-1 = 101,323 Pa
The pascal is too small a unit to be convenient for measuring gas pressures, just as the cubic meter is too large to be convenient for measuring liquid volumes. Hence we shall follow a long-established tradition of measuring gas pressures in standard atmospheres, where
I atmosphere (atm) = 101,325 Pa = 760 mm Hg
The atmosphere then becomes an auxiliary or secondary unit to the strict S1, like the liter for volume, and the charge on the electron for ionic charges.
Example 3
At 8000 ft in the Colorado Rockies, the pressure of the atmosphere is approximately three-quarters what it is at sea level. Express this pressure in standard atmospheres, in pascals, and in millimeters of mercury.
Solution
The pressure is 0.750 atm, 76,000 Pa, or 570 mm Hg.
Boyle's Law Relating Pressure and VolumeEdit
Figure 3-2 Dependence of volume of gas sample on pressure. (a) The simple J·tube apparatus used by Boyle to measure pressure and volume. When the height of the column is equal in the open and closed parts of the tube, the pressure exerted on the gas sample is equal to atmospheric pressure . (b) The pressure on the gas is increased by adding mercury to the tube. (e) The gas burette. a device employing the same principle as the J-tube apparatus. The gas is at atmospheric pressure . (d) The pressure on the gas is increased by raising the mercury reservoir. In (a) and (b) the cross section of the J-tube is assumed constant so the height of the gas sample is a measure of volume . In (c) and (d) the volume of the gas is measured by the calibrated burette.
Robert Boyle (1627-1691), who gave us the first operational definition of an element (see Chapter 6), was also interested in phenomena occurring in evacuated spaces. When devising vacuum pumps for removing air from vessels, he noticed a property familiar to anyone who has used a hand pump for inflating a tire or football, or who has squeezed a balloon without breaking it: As air is compressed, it pushes back with increased vigor. Boyle called this the "spring of the air," and measured it with the simple device shown in Figure 3-2a and b.
Boyle trapped a quantity of air in the closed end of the J-tube as in Figure 3-2a, and then compressed it by pouring increasing amounts of mercury into the open end (b) . At any point in the experiment, the total pressure on the enclosed gas is the atmospheric pressure plus that produced by the excess mercury, which has height h in the open tube. Boyle's original pressure-volume data on air are given in Table 3-1. Although he did not take special pains to keep the temperature of the gas constant, it probably varied only slightly. Boyle did note that the heat from a candle flame produced a drastic alteration in the behavior of air.
Figure 3-3 Plots of Boyle's data of Table 3 1 on various scales (a) P versus V. which gives a hyperbola. (b) P versus 1/ V. (c) V versus 1/ P (d) log V versus log P. (e) PV versus P. A and B mark the same end data points in each plot . If the data are plotted as P versus 1/ V (or V versus 1/ P), the curve obeys the linear equation y = ax + b , in which P is the y-coordinate and 1/ V is the x coordinate. The proportionality constant a can be determined from the slope of the straight line in (b) , or from the height of the horizontal line in (e) Note how sensitive plot (e), with its expanded vertical scale. is to errors in the data (and possibly unsuspected trends) .
Table 3-1 Boyle's Original Data Relating Pressure and Volume for Atmospheric Aira
Volume (index markstt Pressure (inches tt P x V
along uniform bore tubing)btt of mercury)ctt
A word48 29 $\textstyle\frac{2}{16}$ 1400
word46 30 $\textstyle\frac{9}{16}$ 1406
word44 31 $\textstyle\frac{15}{16}$ 1408
word42 33 $\textstyle\frac{8}{16}$ 1410
word40 35 $\textstyle\frac{5}{16}$ 1412
word38 37 1408
word36 39 $\textstyle\frac{5}{16}$ 1416
word34 41 $\textstyle\frac{10}{16}$ 1420
word32 44 $\textstyle\frac{3}{16}$ 1416
word30 47 $\textstyle\frac{1}{16}$ 1414
word28 50 $\textstyle\frac{5}{16}$ 1410
word26 54 $\textstyle\frac{5}{16}$ 1412
word24 58 $\textstyle\frac{13}{16}$ 1414
word23 61 $\textstyle\frac{5}{16}$ 1411
word22 64 $\textstyle\frac{1}{16}$ 1411
word21 67 $\textstyle\frac{1}{16}$ 1410
word20 70 $\textstyle\frac{11}{16}$ 1415
word19 74 $\textstyle\frac{2}{16}$ 1410
word18 77 $\textstyle\frac{14}{16}$ 1403
word17 82 $\textstyle\frac{12}{16}$ 1410
word16 87 $\textstyle\frac{14}{16}$ 1407
word15 93 $\textstyle\frac{1}{16}$ 1398
word14 100 $\textstyle\frac{7}{16}$ 1408
word13 107 $\textstyle\frac{13}{16}$ 1395
B word12 111 $\textstyle\frac{9}{16}$ 1342
aReprinted by permission from J. B. Conant. Harvard Case Histories In Experimental Science, Harvard University Press. Cambridge, 1957. Vol. 1. p. 53.
bEnd data points A and B correspond to those labels on Figure 3-3.
cThe height h, in Figure 3-2b. plus 29$\textstyle\frac{1}{8}$ inches for atmospheric pressure.
Analysis of DataEdit
After a scientist obtains data such as those in Table 3-1, he then attempts to infer a mathematical equation relating the two mutually dependent quantities that he has measured. One technique is to plot various powers of each quantity against one another until a straight line is obtained. The general equation for a straight line is
y= ax + b (3-2)
in which x and y are variables and a and b are constants. If b is zero, the line passes through the origin.
Figure 3-3 shows several possible plots of the data for pressure, P, and volume, V, given in Table 3-1. The plots of P versus 1/V and V versus 1/P are straight lines through the origin. A plot of the logarithm of P versus the logarithm of V is also a straight line with negative slope of -1. From these plots the equivalent equations are deduced:
P = $\textstyle\frac{a}{V}$ (3-3a)
V = $\textstyle\frac{a}{P}$ (3-3b)
and
log V = log a - log P (3-3c)
These equations represent variants of the usual formulation of Boyle's law: For a given number of moles of gas molecules, the pressure is inversely proportional to the volume if the temperature is held constant.
When the relationship between two measured quantities is as simple as this one, it can be deduced numerically as well. If each value of P is multiplied by the corresponding value of V, the products all are nearly the same for a single sample of gas at constant temperature (Table 3-1 ). Thus,
PV = a$\textstyle\frac{~}{}$ 1410 (3-3d)
Equation 3-3d represents the hyperbola obtained by plotting P versus V (Figure 3-3a). This experimental function relating P; and V now can be checked by plotting PV against P to see if a horizontal straight line is obtained (Figure 3-3e).
Boyle found that for a given quantity of any gas at constant temperature, the relationship between P and V is given reasonably precisely by
PV = constant word(at constant T and n) (3-4)
For comparing the same gas sample at constant temperature under different pressure and volume conditions, Boyle's law can be written conveniently as
P1V=1 = P2V2 (3-5)
with the subscripts 1 and 2 representing the different conditions.
Example 3
Plastic bags of peanuts or potato chips purchased in Aspen, Colorado, are frequently puffed up because the air sealed inside at sea level has expanded under the lower surrounding pressure at the 8000-ft elevation. If 100 cm3 of air are sealed inside a bag at sea level, what volume will the air occupy at the same temperature in Aspen? (Assume that the bag is so wrinkled that it does not limit gas expansion, and see Example 3 for missing data.)
Solution
Use Boyle's law in the form of equation 3-5, with subscript 1 representing sea level and subscript 2 representing 8000 ft. Then P1 = 1.000 atm, V1 = 100 cm3, P2 = 0.750 atm, and V2 is to be calculated:
1.000 atm X 10cP1V1 = P2V2
1.000 atm X 100 cm3 = 0.750 atm X V2
1.000 atm X 10000V2 = 133 cm3
Charles' Law Relating Volume and TemperatureEdit
Figure 3-4 Experimental determination of the relationship between volume and temperature of a gas. The apparatus consists of a small capillary tube and a thermometer mounted on a ruled scale and Immersed In a hot oil bath As the system cools. the oil rises in the tube, and the length of air space and the temperature are measured at intervals. For a tube of constant bore, the length of the air space is a measure of the gas volume. So long as the bottom of the air space In the capillary is maintained at the same depth below the surface of the oil bath. the pressure in the capillary Will be constant.
Figure 3-5 A plot of data obtained with the apparatus In Figure 3-4 , showing that volume is proportional to the absolute temperature . Just such a plot employing the Celsius scale of temperature was originally used to locate the absolute zero of temperature , Notice how easily a small error in the slope of the line through the data points could produce a large error in the value of absolute zero . It should be clear that. if at all possible, such long extrapolations should be avoided.
We know that air expands on heating, thereby decreasing its density. For this reason, balloons rise when inflated with warm air. About 100 years after Boyle derived his law, Jacques Charles (1746-1823), in France, measured the effect of changing temperature on the volume of an air sample. This measurement can be made quite easily with the device shown in Figure 3-4.
Some sample data are plotted in Figure 3-5; these show that a graph of V versus T is a straight line with an extrapolated intercept of - 273 ° on the Celsius scale of temperature, or - 460° on the Fahrenheit scale. Charles expressed his law as
V = c(t + 273)
where V is the volume of gas sample, t is the temperature on the Celsius scale, and c is a proportionality constant.
Later, Lord Kelvin (1824- 1907) suggested that the intercept of -273°C represented an absolute minimum of temperature below which it is not possible to go. Scientists now use Kelvin's absolute scale of temperature with 0 K = -273.15°C and O°C = 273.15 K. Charles' law is expressed as
V = cT word(at constant P and n) (3-6)
where T is the absolute Kelvin temperature (i.e., T = t + 273.15). Equation 3-6 indicates that at constant pressure the volume of a given number of moles of gas is directly proportional to the absolute temperature. For light gases such as hydrogen and helium, Charles' law is so accurate that gas thermometers often replace mercury thermometers for precise temperature measurement (Figure 3-6). A mercury thermometer calibrated to read O°C in a water-ice mixture and 100°C in boiling water is inaccurate by as much as 0.1 degree (deg) at intermediate points, whereas a hydrogen thermometer is much more accurate throughout this region.
If the same gas sample is being compared at constant pressure but different temperatures and volumes, then Charles' law can be written as
$\frac{V_{1}}{T_{1}}$ = $\frac{V_{2}}{T_{2}}$ word or word $\frac{V_{1}}{V_{2}}$ = $\frac{T_{1}}{T_{2}}$ (3-7)
This way of writing the expression emphasizes the fact that the ratio of volumes matches the ratio of absolute temperatures, if pressure and the number of moles are constant.
Figure 3-6 A simple gas thermometer. The gas volume is a measure of the absolute temperature. The scale can be calibrated with the freezing point (O°C) and boiling point (100°C) of water. The mercury is injected into or removed from the apparatus to maintain constant atmospheric pressure.
Example 5
The same plastic bag of peanuts mentioned in Example 4 is laid on a windowsill in the sun, where its temperature increases from 20°C to 30°C. If the original volume is 100.0 cm3, what is the final volume after warming?
Solution
V1 = 100.0 cm3; T1 = 20°C, or 293.15 K; and T2 = 30°C, or 303.15 K. To calculate V2 we use equation 3-7:
$\frac{V_{1}}{T_{1}}$ = $\frac{V_{2}}{T_{2}}$
Substituting our data and solving for V2 we get
$\textstyle\frac{100.0 cm^3}{293.15}$ = $\textstyle\frac{V_{2}}{303.15}$
V2 = $\textstyle\frac{303.15}{293.15}$ x 100.0 cm3 = 103.4 cm3
Notice that the absolute Kelvin temperature (K) must be used, not the Celsius temperature.
The Combined Gas LawEdit
The three gas equations that we have encountered so far all may be written in terms of the proportionality of volume to another quantity:
Avogadro's law: V$\propto \!\,$n (at constant P and T)
Boyle's law: V$\propto \!\,$\textstyle\frac{1}{P}$ (at constant T and n)
Charles' law: V$\propto \!\,$T (at constant P and n)
(The symbol $\propto \!\,$ means "is proportional to.") Therefore, the volume must be proportional to the product of these three terms, or
V$\propto \!\,$\textstyle\frac{nT}{P}$ = R$\textstyle\frac{nT}{P}$
or
PV = nRT (3-8)
where R is the proportionality constant. Equation 3-8 is known as the ideal gas law. It contains all of our earlier laws as special cases and, in addition, predicts more relationships that can be tested. For example, the French chemist and physicist Joseph Gay-Lussac verified the prediction that, at constant volume, the pressure of a fixed amount of gas is proportional to its absolute temperature. (In effect, equation 3-8 is a definition of the ideal gas; the differences between real gases and the hypothetical ideal gas are discussed in Section 3-8.)
The gas law is often useful when expressed in the form of ratios of starting and final variables. Suppose a fixed amount of gas at constant temperature is compressed from P1 to P2 with volumes V2 and V2. Then P1 = nRT/V1, P2 = nRT/V2, and the pressure ratio and volume ratio are related by
$\frac{P_2}{P_1}$ = $\frac{V_1}{V_2}$ (T, n constant) (3-9)
This is Boyle's law in ratio form. In a similar way, as we have seen, Charles' law states that the ratio of starting volume to final volume matches the temperature ratio at constant pressure:
$\frac{V_2}{V_1}$ = $\frac{T_2}{T_1}$ (P, n constant) (3-7)
Increasing the number of moles of gas at constant temperature and pressure by a certain factor increases the volume by the same factor:
$\frac{V_2}{V_1}$ = $\frac{n_2}{n_1}$ (P, T constant) (3-10)
And increasing the number of moles of gas at constant temperature in a tank of fixed volume by a certain factor increases the pressure inside the tank by the same factor:
$\frac{P_2}{P_1}$ = $\frac{n_2}{n_1}$ (T, V constant) (3-11)
You should be able to derive these equations and also the analogous equation that expresses Gay-Lussac's observations about pressure and temperature at constant volume easily from the ideal gas law.
Example 6
An old-fashioned diving bell is simply a cylinder closed at the top and open
at the bottom, like an inverted drinking glass, with benches around the inside for the divers to sit on. Air pressure alone keeps the water out. A diver sitting in a bell that has an air volume of 8000 liters wants to drive the water level down by increasing the volume to 10,000 liters, because his feet are getting wet. If he filled the bell with 650 moles of an 02-N2 mixture to begin with, how many more moles of gas will he have to release into the bell to obtain his desired volume increase?
Solution
This is a problem in changes of volume with number of moles; pressure and temperature remain constant. Therefore we use equation 3-10, with V1 = 8000 liters, V2 = 10,000 liters, and n1 = 650 moles. We need to calculate n2. Substituting our data in equation 3-10 and solving for n2 , we get
$\textstyle\frac{10,000 liters}{8,000 liters}$ = $\textstyle\frac{n_2}{650 moles}$
n2 = $\textstyle\frac{10,000 liters}{8,000 liters}$ x 650 moles = 813 moles
Hence 813-650 = 163 moles of gas must be added
Notice that it makes no difference to the problem whether the gas is pure or a mixture of N2 and O2. Within the limits of validity of the ideal gas expression, all gases behave the same way with respect to pressure, volume, and temperature, if measured in moles rather than in grams. The ideal gas expression itself can be written as a ratio, in a form useful for considering simultaneous pressure, temperature, and volume changes in a fixed quantity of gas:
$\frac{P_1V_1}{T_1}$ = $\frac{P_2V_2}{T_2}$ (3-12)
Example 7
When a weather balloon is filled with hydrogen gas at 1.000 atm pressure and 25°C, it has a diameter of 3.00 m and a volume of 14,100 liters. At high altitude the atmospheric pressure drops to half its sea-level value; the temperature is -40°C. What then is the volume of the balloon? What is its diameter?
Solution
We have P1 = 1.000 atm; V1 = 14,100 liters; T1 = 25°C, or 298.15 K; P2 = 0.500 atm; T2 = -40°C, or 233.15 K. Rearranging equation 3-12 to solve for V2 , we get
V2 = $\textstyle\frac{T_2}{T_1}$ x $\textstyle\frac{P_1}{P_2}$ x V1 = $\textstyle\frac{233.15 K}{298.15 K}$ x $\textstyle\frac{1.000 atm}{0.500 atm}$ x 14,100 liters
= 22,100 liters
Assuming that the balloon is spherical, we can find its diameter, d, by using the formula V = $\textstyle\frac{4}{3}$ πr3 or V = 4πr3/3. The diameter then is 3.48 m.
Notice that in Example 7 the temperature drop alone would have brought about a volume decrease to 233.15/298.15 = 0.782 of the initial volume, and the pressure drop alone would have brought about a twofold volume increase. The actual increase, by a factor of 1.56, is the product of these two effects.
The numerical value of the gas constant, R) in the ideal gas law (equation 3-8) depends on the units in which pressure and volume are measured -assuming that only the absolute, or Kelvin, temperature scale is used. If pressure is in atmospheres and volume is in liters, then R = 0.082054 liter atm K-1 mole-1. But, as you can see from Appendix 1, R also can be expressed as 8.3143 J K-1 mole-1. We shall show in Chapter 15 that the product PV has the units of work or energy.
Example 8
How much volume will 75.0 g of hydrogen gas occupy at 1.000 atm pressure and 298 K?
Solution
To answer this question we must use the full ideal gas equation, equation 3-8, with P = 1.000 atm, R = 0.08205 liter atm K-1 mole-1, T = 298 K, and n = 75.0 g/2.016 g mole-1 = 37.2 moles. Thus
V = $\textstyle\frac{nRT}{P}$ = $\textstyle\frac{37.2 moles \times 0.08205 liter atm K-^1 mole-^1 \times 298 K}{1.000 atm}$
= 910 liters
Example 9
A 1000-liter tank is filled to a pressure of 10.00 atm at 298 K, requiring 11.5 kg of gas. How many moles of gas are present? What is the molecular weight of the gas? Assuming the gas to be a pure element, can you identify it?
Solution
V = 1000 liters, P = 10.00 atm, and T = 298 K. Hence,
n = $\textstyle\frac{PV}{RT}$=$\textstyle\frac{10.00 atm \times 1000 liters}{0.08205 liter atm K-^1 mole-^1 \times 298 K}$
= 409 moles
We convert the weight of the gas in kilograms to grams (11.5 kg = 11,500 g) and then calculate the molecular weight (mol wt):
Mol wt = $\textstyle\frac{11,500 g}{409 moles}$ = 28.1 g mole-1
Since we are assuming that the gas is an element, it must be N2
Example 10
An 8-liter boiler is designed to withstand pressures up to 1000 atm. If 1.50 kg of water vapor is in the boiler, to what temperature can it be heated before the boiler explodes?
Solution
P = 1000 atm, V = 8.00 liters. To find the number of moles, we divide the number of grams of water (1.50 kg = 1500 g) by the molecular weight of water. Hence, n = 1500 g/18.0 g mole-1 = 83.3 moles. Therefore,
T = $\textstyle\frac{PV}{nR}$=$\textstyle\frac{1000 \times 8.00}{83.3 \times 0.08205}$ K = 1170 K or 897°C
Standard Temperature and PressureEdit
It is frequently useful to compare volumes of gases involved in physical and chemical processes. Such comparisons are interpreted most easily if the gases are at the same temperature and pressure, although generally it is inconvenient to make all measurements under such carefully controlled conditions; O°C (273 K) and 1.000 atm have been designated arbitrarily as standard temperature and pressure (STP). If we know the volume of a sample of gas at any condition, we can easily calculate the volume it would have as an ideal gas at STP by employing the combined gas law. This calculated volume is useful even if the substance itself becomes a liquid or solid at STP.
Example 11
In an experiment, 300 cm3 of steam are at 1.000 atm pressure and 150°C. What is the ideal volume at STP?
Solution
We can use equation 3-7, substituting the subscript STP for the subscript 2. Rearranging the equation to solve for VSTP, we get
VSTP = V1 $\textstyle\frac{T_{STP}}{T_1}$
= 300 cm3 $\times$ $\textstyle\frac{273 K}{423 K}$ = 194 cm3
This is the volume that the steam would occupy at STP if it behaved like an ideal gas instead of condensing.
At STP, 1 mole of an ideal gas occupies 22.414 liters, as can be seen from the ideal gas law:
Volumes per mole = $\textstyle\frac{V}{n}$ = $\textstyle\frac{RT}{P}$
= $\textstyle\frac{0.082054 liter atm K^{-1} mole^{-1} X 273.15 K}{1.0000 atm}$
= 22.414 liters mole-l
This volume is often called a standard molar volume.
Ideality and NonidealityEdit
The equations describing the various gas laws are exact mathematical expressions. Measurements of volume, pressure, and temperature more accurate than those of Boyle and Charles show that gases only approach the behavior that the equations express. Gases depart radically from so-called ideal behavior when under high pressure or at temperatures near the boiling point of the corresponding liquids. Thus, the gas laws, or more precisely the ideal gas laws, accurately describe the actual behavior of a real gas only at low pressures and at temperatures far above the boiling point of the substance in question. In Section 3-8 we shall return to the problem of how to correct the simple ideal gas law for the behavior of real gases.
The Kinetic Molecular Theory of GasesEdit
Figure 3-7 An experiment to test whether gas molecules move in straight lines . Two flasks are joined by a straight tube with a side arm and stopcock, The bottom flask contains material such as iodine, that can be vaporized by heating . The flasks are evacuated and the material heated. thereby producing vapor Molecules leave the solid in random directions. and condensation occurs uniformly over the entire surface of the bottom flask. However, only molecules moving vertically can pass through the collimating holes in the connecting tube and into the top flask. These molecules pass straight through and form a single spot directly opposite the source material. A high vacuum (low pressure) is required to prevent molecular collisions from randomizing molecular motion in the connection tube and upper flask.
Figure 3-8 An experiment demonstrating collisions of gas molecules with large solid particles. Particles of ground glass are kept suspended like dust particles in air by bombardment with moving mercury molecules . The heavy molecules (mainly monatomic Hg) leaving the surface of the boiling mercury have high kinetic energy, some of which is transferred to the glass particles on collision
At STP, 1 mole of carbon dioxide gas occupies 22.2 liters (ideally 22.4 liters), whereas the same amount of dry ice (solid CO2) has a volume of only about 28 cm3 (assuming a density of dry ice of 1.56 g cm-3). This greater volume of the gas, plus the fact that a gas is compressed or expanded so easily, suggests strongly that much of a gas is empty space. But how does a system that is mostly empty space exert pressure on its surroundings? Experiments such as the one in Figure 3-7 indicate that gas molecules move, and move in straight lines. They also collide with the walls of the container, with one another, and with any other objects that may be in the container with the gas (Figure 3-8). As we shall see, the collision with the container walls produces pressure. It is unnecessary to assume any forces between molecules and container to account for pressure.
Figure 3-9 (a) The velocity of a molecule of gas resolved into components. We determine the components of the velocity vector v by dropping perpendiculars from the head and tail of the vector to the coordinate axes. (b) Collision of molecule with wall showing change in direction of x component of velocity.
We can explain many observed properties of gases by a simple theory of molecular behavior that was developed in the latter half of the nineteenth century by Ludwig Boltzmann (1844-1906), James Clerk Maxwell (1831 - 1879), and others. This kinetic molecular theory has three assumptions:
1. A gas is composed of molecules that are extremely far from one
another in comparison with their own dimensions. They can be considered as essentially shapeless, volumeless points, or small, hard spheres.
2. These gas molecules are in a state of constant random motion, which
is interrupted only by collisions of the molecules with each other and with the walls of the container.
3. The molecules exert no forces on one another or on the container
other than through the impact of collision. Furthermore, these collisions are elastic; that is) no energy is lost as friction during collision.
Our experience with colliding bodies such as a tennis ball bouncing on pavement is that some kinetic energy is lost on collision: The energy is transformed into heat as a result of what we call friction. A bouncing tennis ball gradually "dies down" and comes to rest because its collisions with the pavement are subject to friction and are therefore inelastic. If molecular collisions involved friction, the molecules gradually would slow down and lose kinetic energy, thereby hitting the walls with decreasing ,change of momentum, so the pressure would drop slowly to zero. This process does not take place; therefore, we must postulate that molecular collisions are frictionless, that is, perfectly elastic. In other words, the total kinetic energy of colliding molecules remains constant.
The Phenomenon of Pressure and Boyle's LawEdit
This simple model is adequate to explain pressure and to provide a molecular explanation of Boyle's law. Consider a container, which we will make cubical for simplicity, with a side of length l (Figure 3-9b). Suppose that the container is evacuated completely except for one molecule of mass m that moves with a velocity v having components vx, vy, and vz parallel to the x, y, and z edges of the box (see Figure 3-9a).*
Let us look first at what happens when the molecule rebounds from a collision with one of the YZ walls, which are perpendicular to the x axis.
Pressure is force per unit area, and force is the rate of change of momentum (mass times velocity) with time. When a molecule bounces off the shaded wall in Figure 3-9b, it exchanges momentum of 2mv. with the wall; for the particle begins with momentum in the x direction of -mvx and ends with momentum +mvx. The velocity components in the y and z directions are not changed during a collision with the YZ wall and do not enter into the calculation. No matter how many collisions the molecule has with an XY or an XZ wall along the way, if the x component of velocity is vx the molecule will return to collide with the original YZ wall in a time 2l/vx. If the molecule transfers momentum of 2mvx. every 2l/vx sec, then the rate of change of momentum with time, or the force, Fx, is
Fx = $\textstyle\frac{2mv_x}{2l/v_x}$ = $\textstyle\frac{mv^2_x}{l}$
The force per unit area, or the pressure, is
Px = $\textstyle\frac{mv^2_x}{l \cdot \!\ l^2}$ = $\textstyle\frac{mv^2_x}{l^3}$ = $\textstyle\frac{mv^2_x}{V}$ (3-13)
*If the idea of the breakdown of a vector such as velocity into it's three components, vx,vy and vz, is unfamiliar, there is another explanation that, although less exact, leads to the same answer. This is to assume that since the motions of a molecule in the x, y, and z directions are unrelated, we can think of the molecules as being divided into three groups: one third moving in the x direction, one third in the y direction, and one third in the z direction. The pressure from one molecule on the YZ wall is then Px = mv2/V (analogous to equation 3-13). The pressure from all the molecules moving in a direction perpendicular to that wall is N/3 times this value (where N is the total number of molecules), or
Px = $\textstyle\frac{N}{3}$ $\textstyle\frac{m\overline{v^2}}{V}$
as in equation 3-21. The rest of the proof is the same. (The bar over v2 indicates an average over all molecules.)
since the area of the wall is l2, and the total volume of the box is V = 13 .Similarly, for the other walls,
Py = $\textstyle\frac{mv^2_y}{V}$ (3-14)
Pz = $\textstyle\frac{mv^2_z}{V}$ (3-15)
If the box now contains N molecules rather than just one,
Px = N $\textstyle\frac{m\overline{v^2_x}}{V}$ (3-16)
Py = N $\textstyle\frac{m\overline{v^2_y}}{V}$ (3-17)
Pz = N $\textstyle\frac{m\overline{v^2_z}}{V}$ (3-18)
in which the quantities $\overline{v^2}$ are the averages over all molecules of the squares of the velocity components, since we cannot assume that all molecules have the same velocity.
The total velocity of a molecule is related to its velocity components by
v2 = v${_x^2}$ + v${_y^2}$ + v${_z^2}$ (3-19)
If the motions of the individual molecules are truly random and unrelated, the average of the square of the velocity component in each direction will be the same. There will be no preferred direction of motion in the gas:
$\overline{v{_x^2}}$ = $\overline{v{_y^2}}$ = $\overline{v{_z^2}}$ = ${_3^1}\overline{v^2}$ (3-20)
(As before, the bars over v2 indicate averages over all molecules.) An immediate consequence of this randomness of motion is that the pressure will be the same on all walls, a fact that certainly agrees with our observations of real gases. Rewriting equations 3-16, 3-17, and 3-18 in terms of v2 gives
Px = $\textstyle\frac{N}{3}\frac{m\overline{v^2}}{V}$ Py = $\textstyle\frac{N}{3}\frac{m\overline{v^2}}{V}$ Pz = $\textstyle\frac{N}{3}\frac{m\overline{v^2}}{V}$
and
Px = Py = Pz = P = $\textstyle\frac{N}{3}\frac{m\overline{v^2}}{V}$ (3-21)
or
PV = $\textstyle\frac{N}{3}$ m$\overline{v^2}$ (3-22)
This last expression looks very much like Boyle's law. Boyle's law maintains that the product of pressure and volume for a gas is constant at constant temperature; our derivation from the simple kinetic molecular theory states that the PV product is constant for a given mean velocity of gas molecules. If the theory is correct, the mean velocity of the molecules of a gas cannot depend on either pressure or volume, but only on temperature. The mean molecular kinetic energy is represented by the symbol $\overline{\epsilon}$ (where ε is the Greek letter epsilon) and is expressed as $\textstyle\frac{1}{2}$mv2; furthermore, if N is Avogadro's number, the kinetic energy, Ek , of 1 mole of molecules is N$\overline{\epsilon}$. For a mole of gas, the PV product of Boyle's law is proportional to the kinetic energy per mole:
Ek = N$\overline{\epsilon}$ = $\textstyle\frac{1}{2}$N m$\textstyle\overline{v^2}$ (3-23)
Multiplying and dividing the right-hand term by 3 and rearranging gives
Ek = $\textstyle({3 \over 2}) ({1 \over 3})N m\overline{v^2}$ (3-24)
Comparison with equation 3-22 shows that
PV = $\textstyle\frac{2}{3}$ Ek (3-25)
The combination of this derivation from the kinetic theory and the observed ideal gas law (equation 3-8) tells us that the kinetic energy per mole is directly proportional to the temperature. Or, reversing the statement:
Absolute temperature,T; is an indication of the kinetic energy of gas molecules and ultimately of their mean square velocity. For 1 mole of an ideal gas, PV = RT. Substituting the value for PV given in equation 3-25 gives
Ek = $\textstyle\frac{3}{2}$ RT (3-26)
But Ek = N$\overline{\epsilon}$, in which $\overline{\epsilon}$ =$\textstyle\frac{1}{2}$ m$\textstyle\overline{v^2}$; therefore,
T = $\textstyle\frac{2}{3}\frac{N}{R}\frac{1}{2}m\overline{v^2}$ = $\textstyle\frac{M\overline{v^2}}{3R}$ (3-27)
in which the molecular weight is M = Nm. In short, temperature is a measure rifthe motion of molecules. If we heat a gas and raise its temperature, we do so by increasing the velocity of its molecules. When a gas (or any other substance) cools, its molecular motion diminishes. This molecular motion need not be confined to movement of whole molecules from one place to another, which is the picture that we have drawn for an ideal gas. It also can include rotations of entire molecules or of groups on a molecule, and vibrations of molecules.
We now can see more clearly what happens when the kinetic energy of macroscopic objects is dissipated as heat. When a speeding car skids to a halt, its braking is achieved by converting its energy of motion into frictional heat. But this conversion means changing the motion of the large object-the automobile-into increased relative motion of the molecules of the brake shoes and drum, the tires, and the pavement. Instead of having rubber molecules in the tires vibrating relatively slowly but moving rapidly as a unit, we have a heated tire with molecules moving more rapidly relative to one another but without a net direction of motion. The motions of the molecules have become less directional and more randomized.
This behavior is typical of all real processes. It is easy to go from coherent motion (the rolling tire) to incoherent motion (the hot but stationary tire); it is not possible to go the other way without paying a price. As we shall see in Chapter 16, in any real process the disorder of the object under examination, plus all of the surroundings with which it interacts, always will increase. In other words, in this world things always get messier. This notion is simply the second law of thermodynamics. The quantity that measures this disorder, and which we shall learn to use later in chemical situations, is called entropy, S.
Predictions of the Kinetic Molecular TheoryEdit
The test of any theory is not its beauty or its internal consistency, but its usefulness in predicting the behavior of real systems correctly. By this criterion, the kinetic molecular theory is a good one, as we shall see.
Molecular SizeEdit
Some simple calculations with solid and gaseous carbon dioxide illustrate an important difference between molecular environments in gases and in condensed phases of matter.
Example 12
The density of dry ice, or solid carbon dioxide, is 1.56 g cm-3. Find the volume per mole and the volume per molecule.
Solution
The molecular weight of carbon dioxide is 44.01 amu, so 1 mole of dry ice weighs 44.01 g. To find the volume of 1 mole, divide weight by density:
V = $\textstyle\frac{44.01 g}{1.56 g cm^{-3}}$ = 28.2 cm3
Dividing the volume per mole by Avogadro's number yields the volume per molecule:
V = $\textstyle\frac{28.2 cm^3}{6.022 \times \!\, 10^{23} molecules}$ = 4.68 $\textstyle\times \!\,$ 10-23 cm3 molecule -1
Since 1 cm = 108 Å, the volume per molecule is 46.8 Å3.
The cube root of 46.8 is 3.6, so we should expect the carbon dioxide molecule to fit inside a cube approximately 3.6 Å on edge; therefore the carbon dioxide molecule is about 3.6 Å in diameter. This is roughly what would be expected from other experiments for the size of the carbon dioxide molecule, giving us confidence in the correctness of the picture of dry ice as built up from closely packed molecules of carbon dioxide. Now let's try a comparable calculation for gaseous carbon dioxide.
Example 13
The density of carbon dioxide gas at STP is 1.977 g liter-1. Find the volume
per mole and the volume per molecule.
Solution
One mole or 44.01 g of carbon dioxide gas has a volume of
V = $\textstyle\frac{44.01 g}{1.977 g liter^{-1}}$ = 22.26 liters or 22,260 cm3
(Notice the small deviation from ideal gas behavior.) The volume per molecule is
V = $\textstyle\frac{22,260 cm^3}{6.022 \times \!\, 10^{23} molecules} \times \!\,$ $10^{24}Å^3}{1 cm^3$
= 36,900 Å3 molecule-1
Figure 3-10 The relative size of CO2 molecule and the volume per molecule available to it in carbon dioxide gas at STP. Of course, one molecule is not confined to this volume, nor are other molecules excluded from it.
Carbon dioxide gas has a molar volume that is 790 times the volume of the solid. The volume per molecule in the gas phase corresponds to a cube that is 33.4 Å on a side (Figure 3- 10). Only one part in 790 of the gas volume is actually filled by molecules.
Molecular SpeedsEdit
With nothing more than the elementary kinetic theory presented here, we can calculate the root-mean-square (rms) speed, vrms which is the square root of the average of the squares of the speeds of individual molecules. From equation 3-27, vrms is
$\sqrt{\overline{v^2}}$ = vrms = $\textstyle\sqrt\frac{3RT}{M}$ (3-28)
where R is the gas constant, T is the absolute temperature, and M is the molecular weight. This equation is a good example of the absolute necessity of keeping track of units. The gas constant, R, must be expressed as 8.314 J K-1 mole-1, not as 0.08205 liter atm K-1 mole-1, if the speed is to be expressed in m sec-1. Since 1 J = 1 kg m2 sec-2, the units of 3RT/M are
$\textstyle\frac{(kg m^2 sec^{-2}K^{-1}mole^{-1})(K)}{kg mole^{-1}}$ = m2 sec-2
and vrms is in the desired units. At STP the expression is
vrms = $\textstyle\frac{2610}{M^{1/2}}$ m sec-1 (3-29)
Figure 3-11 The distribution of speeds among molecules in nitrogen gas at three different temperatures. At higher temperatures the average speed is greater, there are fewer molecules that have precisely this average speed. and there is a broader distribution of speed among molecules.
Example 14
Use equation 3·29 to calculate the root-mean-square speeds at STP of molecules of H2, N2, 02 and HBr.
Solution
The molecular weights of these four gases are 2.02, 28.01, 32.00, and 80.91 g mole-1, respectively. Hence the rms speeds are 1840 m sec-1 for H2, 493 m sec-1 for N2, 461 m sec-1 for 02, and 290 m sec-1 for HBr. In more familiar units, these are speeds of 4140, 1109, 1037, and 653 miles hr-1, respectively.
Notice that heavier molecules are slower moving at a given temperature. Molecules that have greater mass do not have to move as rapidly as lighter molecules do to have the same kinetic energy, and it is kinetic energy that is directly related to temperature.
Although the root-mean-square speed of nitrogen molecules at STP may be 493 m sec-1 , this does not mean that all nitrogen molecules travel at this speed. There is a distribution of speeds, from zero to values considerably above 493 m sec-1. As individual gas molecules collide and exchange energy, their speeds will vary. The actual distribution of speeds in nitrogen gas at 1 atm pressure and three different temperatures is shown in Figure 3-11. These curves portray a Maxwell- Boltzmann distribution of speeds. The equations for these curves can be derived from the kinetic theory of gases by using statistical or probability arguments. At higher temperatures, the root-mean- square speed increases, as expected from equation 3-28. But Figure 3-11 shows that the distribution of speeds also becomes more diffuse. There is a greater range of speeds, and fewer molecules have a speed close to the average value.
From the size of a molecule, the speed with which it travels, and the number of other molecules per unit of volume around it, we can calculate the mean free path (the distance a molecule travels between two successive collisions) and the collision frequency. Molecules such as 02 or N2 travel an average of 1000 Å between collisions, and they experience approximately 5 billion collisions per second at STP (Figure 3-12).
Dalton's Law of Partial PressuresEdit
Figure 3-12 The mean free path, or distance traveled between collisions. for gas molecules that show ideal behavior at STP, is about 1000 Å or 10-5 cm. We can imagine the size of a molecule in this drawing by comparing the relative sizes of the cube 33.4 Å on a side in this figure and in Figure 3-10.
If each molecule in a gas travels independently of every other except at moments of collisions, and if collisions are elastic, then in a mixture of different gases the total kinetic energy of all the different gases will be the sum of the kinetic energies of the individual gases:
E = E1 + E2 + E3 + E4 + • • •
Since each gas molecule moves independently, the pressure that each gas exerts on the walls of the container can be derived separately (equation 3-25). For example,
p1 = $\textstyle\frac{2E_1}{3V}$ p2 = $\textstyle\frac{2E_2}{3V}$ p3 = $\textstyle\frac{2E_3}{3V}$ (3-30)
This pressure exerted by one component of a gas mixture is called its partial pressure, p . Each of these equations can be rewritten to give kinetic energy in terms of pressure:
E1 = $\textstyle\frac{3}{2}p_1V$ E2 = $\textstyle\frac{3}{2}p_2V$ E3 = $\textstyle\frac{3}{2}p_3V$
Substituting in the energy expression and canceling the $\textstyle\frac{3}{2}$V terms from both sides of the equation produces
P = p1 + p2 + p3 + p4 + • • • = $\sum_{j}^{}{p_j}$ (3-31)
The special sign at the right is a summation sign, which is a shorthand way of writing the instructions: Sum all the terms of the type Pj for all the different values of j. It will be used frequently.
The total pressure, P, then, is the sum of the partial pressures of the individual components of the gas mixture, each considered as if it were the only gas present in the given volume. John Dalton (1766-1844) proposed his law of partial pressures during the gas investigations that eventually led him to the theory of atoms.
An important measure of concentrations in a mixture of gases (and in solutions and solids as well) is the mole fraction, X. The mole fraction of the j th component in a mixture of substances is defined as the number of moles (n) of the given substance divided by the total number of moles of all substances:
Xj = $\textstyle\frac{n_j}{n_1 + n_2 + n_3 + n_4 + . . .}$ = $\frac{n_j}{\sum_{i}^{}{n_i}}$ (3-32)
Another version of Dalton's law is the statement that the partial pressure of one component in a mixture of gases is its concentration in mole fraction times the total pressure. If there are nj moles of gas j present in a mixture, the partial pressure of that gas is calculable from the ideal gas law:
pj = nj$\textstyle\frac{RT}{V} = \frac{n_j}{n} \times n \times \frac{RT}{V}$ $n = (\ n_1 + n_2 + n_3 + n_4 + . . . = \sum_{i}^{}{n_i}\ ) \!\,$
Since nj/n = Xj is the mole fraction, and nRT/V = P is the total pressure, Dalton's law becomes
pj = XjP (3-33)
Example 15
A gas mixture at 100°C and 0.800 atm pressure contains 50% helium, HE, and 50% xenon, Xe, by weight. What are the partial pressures of the individual gases?
Solution
First find the number of moles of helium and xenon in any given sample. A- convenient sample choice is 100 g. Then the number of moles of each gas is
nHe = $\textstyle\frac{50.0 g}{4.00 g mole^{-1}}$ = 12.85 moles He
nXe = $\textstyle\frac{50.0 g}{131.3 g mole^{-1}}$ = 0.381 mole Xe
The next step is to calculate the mole fraction, Xj of each component:
XHe = $\textstyle\frac{12.5}{12.5 + 0.381}$ = 0.970
XXe = $\textstyle\frac{0.381}{12.5 + 0.381}$ = 0.030
According to Dalton's law, the partial pressure of each component is expressed as pj = XjP. Thus, we have
pHe = 0.970 P = 0.970(0.800) = 0.776 atm
pXe = 0.030 P = 0.030(0.800) = 0.024 atm
Notice that no total volume was specified, and a convenient but arbitrary sample size was used for calculation purposes. Why is the answer independent of volume? Will the answer change if the temperature is changed?
Often gases are collected over liquids such as water or mercury, as in Figure 3-13. Dalton's law must be applied in such cases to account for partial evaporation of the liquid into the space occupied by the gas.
Figure 3-13 When oxygen gas is collected by displacing water from an inverted bottle, the presence of water vapor in the collecting bottle must be recognized when calculating the amount of oxygen collected. The correction is made easily by using Dalton's law of partial pressures.
=
Example 16
Oxygen gas generated in an experiment is collected at 25°C in a bottle inverted in a trough of water (Figure 3-13). The external laboratory pressure is 1.000 atm. When the water level in the originally full bottle has fallen to the level in the trough, the volume of collected gas is 1750 ml. How many moles of oxygen gas have been collected?
Solution
If the water levels inside and outside the bottle are the same, then the total pressure inside the bottle equals 1.000 atm. But at 25°C the vapor pressure of water (or the pressure of water vapor in equilibrium with the liquid) is 23.8 mm Hg or 0.0313 atm, so the partial pressure of oxygen gas is only 1.000 - 0.031, or 0.969 atm. The mole fraction of oxygen gas in the bottle is 0.969 and not 1.000, and the partial pressure of oxygen also is 0.969 atm. The number of moles is
n = $\textstyle\frac{PV}{RT} = \textstyle\frac{0.969 atm \times 1750 cm^3}{82.054 cm^3 atm K^{-1}mole^{-1}\times 298 K}$
= 0.0694 mole
What would the answer have been had the pressure of water vapor been neglected?
Example 17
On a humid day at 43.3°C in Galveston, Texas, the vapor pressure of water is 0.087 atm. What is the water content of the atmosphere, expressed as a mole fraction? Assuming that dry air is 20 mole percent 0 and 80 mole percent N2 what is the water content in percent by weight?
Solution
The answers are 0.087 and 5.62%
Other Predictions of the Kinetic Molecular TheoryEdit
Figure 3-14 Effusion is the flow of gas from a small hole in a container into an outside region of equal pressure. According to Graham's law, the rates of effusion of two gases at equal temperature are inversely proportional to the square roots at their molecular weights or, by kinetic molecular theory, proportional to their velocities.
Derivations from the kinetic molecular theory that are not much more complicated in principle than the ones we have seen for the gas pressure furnish us with a host of other predictions about the behavior of gases. These predictions have been tested by many scientists and have encouraged confidence in the theory. A derivation of the probability of a molecule hitting a hole in the wall of a container leads to Graham's law of effusion, which predicts that the rate of leakage of a gas from a small hole in a tank will be inversely proportional to the square root of the molecular weight (Figure 3-14).
Thomas Graham (1805-1869) observed, in 1846, that the rates of effusion of gases are inversely proportional to the square roots of their densities. Since, by Avogadro's hypothesis, the density of a gas is proportional to its molecular weight, Graham's observation agrees with the kinetic theory, which predicts that the rate of escape is proportional to molecular velocity or inversely proportional to the square root of the molecular weight (equation 3-29):
$\frac{Rate_2}{Rate_1} = \frac{v_2}{v_1} = (\tfrac{M_1}{M_2})^{1/2}$ (3-34)
However, the law begins to fail at high densities, in which molecules collide several times with one another as they escape through the hole. The law also fails when there are holes large enough so the gas has a hydrodynamic flow toward the hole, thereby leading to the formation of a jet of escaping gas. But so long as isolated molecules escape by going through the hole during their random motions through a stationary gas, the kinetic molecular theory prediction is exact.
Example 18
A given volume of oxygen gas effuses through a small orifice into a vacuum in 1 min. The same volume of an unknown gas takes 1 min and 34 sec to effuse at the same temperature. What is the approximate molecular weight of the unknown gas? If its empirical formula is CH, what is its molecular formula, and what is its molecular weight?
Solution
From equation 3-34 we can derive
$\frac{t_2}{t_1} = \frac{Rate_1}{Rate_2} = (\tfrac{M_1}{M_2})^{1/2}$
the subscript 1 representing oxygen and the subscript 2 representing the unknown gas. Substituting the observed data, and rearranging the equation to solve for M2, we get
M2 = 32.00 g mole-1 $\times (\tfrac{1.57}{1.00})^2$ = 78.9 g mole-1
This is an approximate value, subject to errors in measuring gas flow. Since the formula weight of the unknown gas is 13.02 g (we know this from the empirical formula, CH), there must be six formula weights in the true molecular weight. Since 6 X 13.02 = 78.12, the gas is C6H6 .
Figure 3-15 Deviations from the ideal gas law for several gases at 273 K, in terms of the compressibility factor Z = PV/RT. The dip of Z below 1.0 at low pressures is caused by intermolecular attractions: the rise above 1 .0 at high pressures is produced by the shorter range intermolecular repulsions as the molecules. of finite bulk. are crowded closely together.
Figure 3-16 PV/RT for 1 mole of methane gas at several temperatures. Note that PV/RT is less than 1.0 at low pressures and greater than 1.0 at high pressures. Ideal gas behavior is approached at high temperatures.
Figure 3-17 Pressure-volume curves for nitrogen and an ideal gas at constant temperature. At low pressures, the molar volume of N2 gas is less than that of an ideal gas because of intermolecular attraction . At high pressures. the nonzero volume of individual N2 molecules makes the gas volume greater than ideal.
Figure 3-18 Reduction of pressure of a real gas as a result of intermolecular attractions. (a) Gas at low density. (b) Gas at high density. A molecule M in a high-density gas hits the wall with a smaller impact than in a low-density gas because the attractions of its nearest neighbors reduce the force of its impact.
Figure 3-19 This center of no other molecule can come closer than a molecular diameter to the center of a given molecule. The volume around each molecule from which other molecules are excluded then is $\textstyle\frac{4}{3}\pi d^3$ or eight times the molecular volume of $\textstyle\frac{\pi}{6} d^3$.
Example 19
Calculate the relative effusion rates of the two isotopic forms of uranium hexafluoride, 238UF6 and 235UF6 . All fluorine is 19F.
Solution
The molecular weight of 238UF6 is 352.0 amu, and that of 235UF6 is 349.0 amu. The ratio of effusion rates then is
$\textstyle\frac{Rate(235)}{Rate(238)} = (\tfrac{352.0}{349.0})^{1/2} = 1.0043$
Although there is only 0.43% difference in effusion rates of the hexafluorides of the two isotopes of uranium, scientists used this difference to separate fissionable 235U from 238U during the construction of the first atomic bombs at the end of World War II. No other separation method proved workable at the time. The scientists used UF6 because it is a gaseous compound of uranium, but the small separation ratio meant that the gas had to be passed through many thousands of porous barriers in the special gas diffusion plant at Oak Ridge, Tennessee, to achieve a useful degree of separation.
The kinetic molecular theory also allows us to predict gaseous diffusion, viscosity, and thermal conductivity, the three so-called transport properties. Each phenomenon can be treated mathematically as the diffusion of some molecular property down a gradient. In gaseous diffusion, mass diffuses from regions of high to low concentration, or down a concentration gradient. Viscosity of a fluid arises because slowly moving molecules diffuse into (and retard) rapidly moving fluid layers, and faster molecules diffuse into (and accelerate) the slow regions. This is a transport of momentum down a velocity gradient. Thermal conductivity is the scattering of rapidly moving molecules into regions of slower ones. It can be described as a transport of kinetic energy down a temperature gradient. In all three cases, the kinetic molecular theory predicts the diffusion coefficient, with best accuracy at low gas pressures and high temperatures. These are just the conditions for which the simple ideal gas law is most applicable.
In summary, the elementary kinetic molecular theory, as outlined here, provides a correct explanation for the behavior of ideal gases. It gives us confidence in the reality of molecules, and encourages us to look for molecular modifications of the simple theory that will account for deviations from ideal gas behavior.
Real Gases and Deviations from the Ideal Gas LawEdit
If gases were ideal, the quotient PV/RT would always equal 1 for 1 mole of gas. Actually all real gases deviate, to some extent, from ideal behavior; the quantity Z = PV/ RT, called the compressibility coefficient, is one measure of this deviation. Z is plotted against pressure for several gases at 273 K in Figure 3-15, and for one gas at several temperatures in Figure 3-16. We can interpret the behavior of real gases as a combination of intermolecular attractions (which are effective over comparatively long distances) and repulsions caused by the finite sizes of molecules (which become significant only when molecules are crowded together at high pressures). At low pressures- but still too high for ideal behavior-intermolecular attractions make the molar volume unexpectedly low, and the compressibility coefficient is less than 1. However, at sufficiently high pressures the crowding of molecules begins to predominate, and the molar volume of the gas is greater than it would be if the molecules were point masses. The higher the temperature (Figure 3-16), the less significant the intermolecular attraction will be in comparison with the kinetic energy of the moving molecules, and the lower will be the pressure at which the bulk factor dominates and Z rises above 1.
An equation such as the ideal gas law, PV = nRT, is known as an equation of state because it describes the state of a system in terms of the measurable variables P, V, T, and n (Figure 3-17). Other equations of state that have been proposed describe the behavior of real gases better than the ideal gas law. The best known of these equations is the one introduced, in 1873, by Johannes van der Waals. Van der Waals assumed that, even for a real gas, there is an ideal pressure, p* , and an ideal volume, V* , that would apply to the ideal expression P*V* = nRT; but because of the imperfections of the gas, these were not the same as the measured pressure, P, and measured volume, V. The ideal volume, he reasoned, should be less than the measured volume because the individual molecules have a finite volume instead of being point masses, and the portion of the container's volume that is occupied by other molecules is unavailable to any given molecule. Therefore, the "ideal" volume should be less than the measured volume by a constant, b, that is related to molecular size by V* = V - b.
Moreover, a gas molecule subject to attractions from other gas molecules strikes the walls with less force than it would if these attractions were absent. For as the molecule approaches the wall, there are more gas molecules behind it in the bulk of the gas than there are between it and the wall (Figure 3-18). The number of collisions with the wall in a given time is proportional to the density of the gas, and each collision is softened by a back-attraction factor, which itself is proportional to the density of molecules doing the attracting. Therefore, the correction factor to P is proportional to the square of the gas density, or inversely proportional to the square of the volume: P* = P + a/ V2, where a is related to the attractions between molecules. The complete van der Waals equation is
$\textstyle(P + \tfrac{a}{\overline{V}^2}) (\overline{V} - b)$ = RT
Here $\overline{V}$ is the volume per mole, or: $\overline{V}$ = V/n. The equation is simpler when written this way. Similarly, the ideal gas law can be written P$\overline{V}$ = RT as easily as PV = nRT. The constants a and b are chosen empirically to provide the best relationship of the equation to the actual PVT behavior of a gas. Even so, the molecular size calculated from this purely experimental b agrees well with the ones obtained by other means (Table 3-2), and gives us confidence that we have the right explanation for deviations from ideality.
Experimentally obtained values of a and b for several gases are given in Table 3-2, along with several calculations of molecular diameters. We might suppose that the constant b is simply the excluded volume per mole, Vm (as in Figure 3-19): b = 8NVm = $\textstyle\frac{4}{3}\pi Nd^3$. However, collision is a two-molecule process and this calculation overcounts the excluded volume by a factor of 2. The molecular diameters in Table 3-2 were obtained from the b values by having b equal to 4NVm, in which Vm = $\pi d^3$ /6 is the volume of one molecule.
The van der Waals equation is applicable over a much wider range of temperatures and pressures than is the ideal gas law; it is even compatible with the condensation of a gas to a liquid.
SummaryEdit
We have seen four experimentally derived principles or laws of gas behavior, which all gases obey approximately, especially under conditions of low pressure and high temperature.
1. Avogadro's law: At fixed pressure and temperature, the volume of any gas is proportional to the number of moles present.
2. Boyle's law: At constant temperature, the volume of a sample of gas is inversely proportional to the pressure on the gas.
3. Charles' law: At constant pressure, the volume of a sample of gas is proportional to the temperature of the gas on the absolute, or Kelvin, scale.
4. Gay-Lussac's law: At constant volume, the pressure exhibited by a sample of gas is proportional to the temperature of the gas on the absolute scale.
Although real gases only approximate this behavior, we can define an ideal gas as one that follows the preceding laws exactly under all conditions. All the foregoing observations can be combined in one expression, the ideal gas law:
PV = nRT
If P is pressure in atmospheres, V is volume in liters, T is the absolute Kelvin temperature, and n is the number of moles, then the proportionality constant R, known as the gas constant, has the value
R = 0.08205 liter atm K-1 mole-1
In practical calculations, the ideal gas law is often most useful in one of the various ratio forms given in equations 3-9 through 3-12.
Standard temperature and pressure, or STP, is defined as 273.15 K (O°C) and exactly 1 atm. Gas properties are frequently converted to STP conditions for comparison purposes, even for gases such as H2O that liquefy at STP. One mole of any ideal gas at STP has a volume of 22.414 liters, and this is called a standard molar volume.
The kinetic molecular theory of gases successfully explains the behavior of ideal gases with a minimum of starting assumptions, and also provides a framework for understanding the deviations of real gases from ideal gas behavior. In its simplest form, the kinetic theory assumes that a gas is made up of noninteracting point molecules in a state of constant motion, colliding elastically with one another and with the walls of a container. In extending the theory to cover real gases, we recognize that molecules have finite volume and exert attractive forces on one another.
In the kinetic theory, pressure is simply the result of collision of molecules with the container walls, and transfer of momentum. The product of pressure and volume is equal to two-thirds the kinetic energy of motion of the molecules (equation 3-25). Combining this fact with the observed ideal gas law, we come to the important conclusion that the kinetic energy of motion of the molecules is directly proportional to abso1ute temperature (equation 3-26), or that temperature is simply a consequence of molecular motion.
Comparing the densities of substances in the gaseous and condensed phases, we find that the average space available to a gas molecule at STP is roughly three orders of magnitude (or 103) greater than the volume of the molecule itself. The root-mean-square (rms) speed of a molecule is inversely proportional to the square root of its molecular weight (equation 3-27), and this speed is on the order of several thousand miles per h,our at STP. The heavier the molecule, the more slowly it moves.
The actual molecular velocities in a gas vary in a distribution around this rms value, with some velocities nearly zero and others very much faster than average. Velocities of individual molecules vary as the molecules collide with one another and rebound. Nevertheless, the distribution of molecular speeds remains constant at constant temperature. An ideal gas at STP has a mean free path, or average distance between collisions, on the order of 1000 Å, and a collision rate of around 5 X 109 collisions sec-1.
Dalton's law of partial pressures says that each component of a gas mixture behaves as if it were the only gas present. The mole fraction, Xj, is the number of moles of gas j present divided by the total number of moles of all gases. The partial pressure of gas j is the mole fraction of j times the total pressure: pj = XjP. The sum of partial pressures of all components is the total pressure.
The kinetic theory of gases predicts that the rate of effusion of a gas through a small orifice will be inversely proportional to the square root of molecular velocity (equation 3-34), and this prediction is borne out by experiment. The theory is also successful in accounting quantitatively for diffusion of gases, viscosity, and conductivity of heat.
Real gases deviate from ideal behavior because molecules are not volumeless, shapeless points, and because real molecules attract one another. Molecular attractions become impossible to neglect when the molecules are moving more slowly, at low temperatures; molecular volumes become significant when the gas is compressed. Hence the gases approximate ideal behavior most closely at high temperature and low pressure.
Van der Waals modified the ideal gas law to take both of the preceding factors into account. The van der Waals equation, (P + a/$\overline{V}^2$)($\overline{V}$ - b) = RT for 1 mole of gas, has an experimental constant, b, that is related to molecular volume, and another constant, a, that is related to molecular attractions or "stickiness." From the van der Waals constant b we can obtain approximate molecular diameters, and these values agree roughly with estimates of diameters obtained from densities of solids or from gas viscosities. | textbooks/chem/General_Chemistry/Chemical_Principles_(Dickerson)/03%3A_Gas_Laws_and_the_Kinetic_Theory/5.01%3A_Hydrogen_Oxygen_and_Water.txt |
And so, nothing that to our world appears,
Perishes completely, for nature ever
Upbuilds one thing from another's ruin;
Suffering nothing yet to come to birth
But by another's death.
Lucretius (95-55 B.C.)
Introduction
The main question asked in Chapter 2 was "If a given set of substances will react to give a desired product, how much of each substance is needed?" Our basic assumptions were that matter cannot be arbitrarily created or destroyed, and that atoms going into a reaction must come out again as products.
In this chapter we ask a second question: "Will a reaction occur, eventually?" Is there a tendency or a drive for a given reaction to take place, and if we wait long enough will we find that reactants have been converted spontaneously into products? This question leads to the ideas of spontaneity and of chemical equilibrium. A third question, "Will a reaction occur in a reasonably short time?" involves chemical kinetics, which will be discussed in Chapter 22. For the moment, we will be satisfied if we can predict which way a chemical reaction will go by itself, ignoring the time factor.
Spontaneous Reactions
A chemical reaction that will occur on its own, given enough time, is said to be spontaneous. In the open air, and under the conditions inside an automobile engine, the combustion of gasoline is spontaneous:
C7H16 + 11 O2 → 7CO2 + 8 H2O
(The reaction is exothermic, or heat emitting. The enthalpy change, which was defined in Chapter 2, is large and negative: $\vartriangle \!\,H$ = -4812 kJ mole-1 of heptane at 298 K. The heat emitted causes the product gases to expand, and it is the pressure from these expanding gases that drives the car.) In contrast, the reverse reaction under the same conditions is not spontaneous:
7CO2 + 8H2O $\nrightarrow$ C7H16 + 11 O2
No one seriously proposes that gasoline can be obtained spontaneously from a mixture of water vapor and carbon dioxide.
Explosions are examples of rapid, spontaneous reactions, but a reaction need not be as rapid as an explosion to be spontaneous. It is important to understand clearly the difference between rapidity and spontaneity. If you mix oxygen and hydrogen gases at room temperature, they will remain together without appreciable reaction for years. Yet the reaction to produce water is genuinely spontaneous:
2H2 + O2 → 2H2O
We know that this is true because we can trigger the reaction with a match, or catalyst of finely divided platinum metal.
The preceding sentence suggests why a chemist is interested in whether a reaction is spontaneous, that is, whether it has a natural tendency to occur. If a desirable chemical reaction is spontaneous but slow, it may be possible to speed up the process. Increasing the temperature will often do the trick, or a catalyst may work. We will discuss the functions of a catalyst in detail in Chapter 22. But in brief, we can say now that a catalyst is a substance that helps a naturally spontaneous reaction to go faster by providing an easier pathway for it. Gasoline will burn rapidly in air at a high enough temperature. The role of a spark plug in an automobile engine is to provide this initial temperature. The heat produced by the reaction maintains the high temperature needed to keep it going thereafter. Gasoline will combine with oxygen at room temperature if the proper catalyst is used, because the reaction is naturally spontaneous but slow. But no catalyst will ever make carbon dioxide and water recombine to produce gasoline and oxygen at room temperature and moderate pressures, and it would be a foolish chemist who spent time trying to find such a catalyst. In short, an understanding of spontaneous and nonspontaneous reactions helps a chemist to see the limits of what is possible. If a reaction is possible but not currently realizable, it may be worthwhile to look for ways to carry it out. If the process is inherently impossible, then it is time to study something else.
Equilibrium and the Equilibrium Constant
The speed with which a reaction takes a place ordinarily depends on the most concentrations of the reacting substances. This is common sense, since most reactions take place when molecules collide, and the more molecules there are per unit of volume, the more often collisions will occur.
The industrial fixation of atmospheric nitrogen is very important in the manufacture of agricultural fertilizers (and explosives). One of the steps in nitrogen fixation, in the presence of a catalyst, is
N2 + O2 → 2NO (4-1)
If this reaction took place by simple collision of one molecule of N2 and one molecule of O2, then we would expect the rate of collision (and hence the rate of reaction) to be proportional to the concentrations of N2 and O2:
Rate of NO production $\textstyle\propto \!\, [N_2][O_2]$
or
R1 = k1[N2][O2] (4-2)
The proportionality constant k1 is called the forward-reaction rate constant, and the bracketed terms [N2][O2] represent concentrations in moles per liter. This rate constant, which we will discuss in more detail in Chapter 22, usually varies with temperature. Most reactions go faster at higher temperatures, so k1 is larger at higher temperatures. But k1 does not depend on the concentrations of nitrogen and oxygen gases present. All of the concentration dependence of the overall forward reaction rate, R1, is contained in the terms [N2] and [O2]. If this reaction began rapidly in a sealed tank with high starting concentrations of both gases, then as more N2 and O2 were consumed, the forward reaction would become progressively slower. The rate of reaction would decrease because the frequency of collision of molecules would diminish as fewer N2 and O2 molecules were left in the tank.
The reverse reaction can also occur. If this reaction took place by the collision of two molecules of NO to make one molecule of each starting gas,
2NO → N2 + O2 (4-3)
then the rate of reaction again would be proportional to the concentration of each of the colliding molecules. Since these molecules are of the same compound, NO, the rate would be proportional to the square of the NO concentration:
Rate of NO removal $\textstyle\propto \!\,$[NO][NO]
or
R2 = k2[NO]2 (4-4)
where R2 is the overall reverse reaction rate and k2 is the rev~rse-reaction rate constant. If little NO is present when the experiment begins, this reaction will occur at a negligible rate. But as more NO accumulates by the forward reaction, the faster it will be broken down by the reverse reaction.
Thus as the forward rate, R1, decreases, the reverse rate, R2 , increases. Eventually the point will be reached at which the forward and reverse reactions exactly balance (4-5):
R1 = R2
[N2][O2]k1 = k2[NO]2
This is the condition of equilibrium. Had you been monitoring the concentrations of the three gases, N2 O2, and NO, you would have found that the composition of the reacting mixture had reached an equilibrium state and thereafter ceased to change with time. This does not mean that the individual reactions had stopped, only that they were proceeding at equal rates; that is, they had arrived at, and thereafter maintained, a condition of balance or equilibrium.
Figure 4-1 Illustration of dynamic equilibrium: two fish tanks connected by a channel (a) Start of experiment, with 10 goldfish in the left tank and 10 guppies in the right. (b) Equilibrium state. with 5 of each kind of fish in each tank. (c) If we were to observe one single fish (here a guppy among goldfish). we would find that it spends half its time in each tank . The equilibrium of the tank in (b) is a dynamic. averaged state and not a static condition . The fish do not stop swimming when they have become evenly mixed.
The condition of equilibrium can be illustrated by imagining two large fish tanks, connected by a channel (Figure 4-1). One tank initially contains 10 goldfish, and the other contains 10 guppies. If you watch the fish swimming aimlessly long enough, you will eventually find that approximately 5 of each type of fish are present in each tank. Each fish has the same chance of blundering through the channel into the other tank. But as long as there are more goldfish in the left tank (Figure 4-la), there is a greater probability that a goldfish will swim from left to right than the reverse. Similarly, as long as the number of guppies in the right tank exceeds that in the left, there will be a net flow of guppies to the left, even though there is nothing in the left tank to make the guppies prefer it. Thus the rate of flow of guppies is proportional to the concentration of guppies present. A similar statement can be made for the goldfish.
At equilibrium (Figure 4-1b), on an average there will be 5 guppies and 5 goldfish in each tank. But they will not always be the same 5 of each fish. If 1 guppy wanders from the left tank into the right, then it or a different guppy may wander back a little later. Thus at equilibrium we find that the fish have not stopped swimming, only that over a period of time the total number of guppies and goldfish in each tank remains constant. If we were to fill each tank with 9 goldfish and then throw in 1 guppy, we would see that, in its aimless swimming, it would spend half its time in one tank and half in the other (Figure 4-1 c).
In the NO reaction we considered, there will be a constant concentration of NO molecules at equilibrium, but they will not always be the same NO molecules. Individual NO molecules will react to re-form N2 and O2, and other reactant molecules will make more NO. As with the goldfish, only on a head-count or concentration basis have changes ceased at equilibrium.
The equilibrium condition for the NO-producing reaction, equation 4-1, can be rewritten in a more useful form:
$\textstyle\frac{[NO]^2}{[N_2][O_2]} = \textstyle\frac{k_1}{k_2} = K_eq$ (4-6)
in which the ratio of forward and reverse rate constants is expressed as a simple constant, the equilibrium constant, Keq. This equilibrium constant will vary as the temperature varies, but it is independent of the concentrations of the reactants and products. It tells us the ratio of products to reactants at equilibrium, and is an extremely useful quantity for determining whether a desired reaction will take place spontaneously.
General Form of the Equilibrium Constant
We derived the equilibrium-constant expression for the NO reaction by assuming that we knew the way that the forward and reverse steps occurred at the molecular level. If the NO reaction proceeded by simple collision of two molecules, the derivation would be perfectly correct. The actual mechanism of this reaction is more complicated. But it is important, and fortunate for chemists, that we do not have to know the reaction mechanism to write the proper equilibrium constant. The equilibrium-constant expression can always be written from the balanced chemical equation, with no other information, even when the forward and reverse rate expressions are more complicated than the balanced equation would suggest. (We shall prove this in Chapter 16.) In our NO example, the forward reaction actually takes place by a series of complicated chain steps. The reverse reaction takes place by a complementary set of reactions, so that these complications cancel one another in the final ratio of concentrations that gives us the equilibrium constant. The details of the mechanism are "invisible" to the equilibrium-constant expression, and irrelevant to equilibrium calculations.
A general chemical reaction can be written as
aA + bB cC + dD (4-7)
In this expression, A and B represent the reactants; C and D, the products. The letters a, b, c, and d represent the number of moles of each substance involved in the balanced reaction, and the double arrows indicate a state of equilibrium. Although only two reactants and two products are shown in the general reaction, the principle is extendable to any number. The correct equilibrium-constant expression for this reaction is
$K_{eq} = \textstyle\frac{[C]^c[D]^d}{[A]^a[B]^b}$ (4-8)
It is the ratio of product concentrations to reactant concentrations, with each concentration term raised to a power given by the number of moles of that substance appearing in the balanced chemical equation. Because it is based on the quantities of reactants and products present at equilibrium, equation 4-8 is called the law of mass action.
Example 1
Give the equilibrium-constant expression for the reaction
CO + H2O CO2 + H2
Solution
The equilibrium constant is given by
$K_{eq} = \textstyle\frac{[CO_2][H_2]}{[CO][H_2O]}$
Since all four substances have a coefficient of 1 in the balanced equation, their concentrations are all raised to the first power in the equilibrium-constant expression.
Example 2
What is the equilibrium-constant expression for the formation of water from hydrogen and oxygen gases? The reaction is
2H2 + O2 2H2O
Solution
$K_{eq} = \textstyle\frac{[H_2O]^2}{[H_2]^2[O_2]}$
Since two moles of hydrogen and water are involved in the chemical equation, their concentrations are squared in the Keq expression.
Example 3
Give the equilibrium-constant expression for the dissociation (breaking up) of water into hydrogen and oxygen. The reaction is
2H2O 2H2 + O2
Solution
$K_{eq} = \textstyle\frac{[H_2]^2[O_2]}{[H_2O]^2}$
An important general point emerges here. This reaction is the reverse of that of Example 2, and the equilibrium-constant expression is the inverse, or reciprocal, of the earlier one. If a balanced chemical reaction is reversed, then the equilibrium-constant expression must be inverted, since what once were reactants now are products, and vice versa.
Example 4
The dissociation of water can just as properly be written as
H2O H2 + $\textstyle\frac{1}{2}O_2$
What then is the equilibrium-constant expression?
Solution
$K_{eq} = \textstyle\frac{[H_2][O_2]^{1/2}}{[H_2O]}$
Notice that when the reaction from Example 3 is divided by 2, resulting in the Example 4 reaction, the equilibrium constant is the square root of the old value, or the old Keq to the one-half power. Similarly, if the reaction is doubled, the Keq must be squared. In general, it is perfectly proper to multiply all the coefficients of a balanced chemical reaction by any positive or negative number, n, and the equation will remain balanced. (Multiplying all the coefficients of an equation by - 1 is formally the same as writing the equation in reverse. Write out a simple equation and prove to yourself that this is so.) But if all the co1ficients of an equation are multiplied by n, then the new equilibrium-constant expression is the old one raised to the nth power. Hence, when working with equilibrium constants, one must keep the corresponding chemical reactions clearly in mind.
Example 5
The reaction for the formation or the breakdown of ammonia can be written in a number of ways:
a) N2 + 3H2 2NH3
b) $\textstyle\frac{1}{2}N + 3H_2$ NH3
c) $\textstyle\frac{1}{3}N + H_2$ $\textstyle\frac{2}{3}$NH3
d) NH3 $\textstyle\frac{1}{2}$N2 + $\textstyle\frac{3}{2}H_2$
(Each of these expressions might be appropriate, depending on whether you were focusing on nitrogen, ammonia, hydrogen, or the dissociation of ammonia.) What are the equilibrium-constant expressions for each formulation, and how are the equilibrium constants related?
Solution
a) $K_{a} = \textstyle\frac{[NH_3]^2}{[N_2][H_2]^3}$ c) $K_{c} = \textstyle\frac{[NH_3]^{2/3}}{[N_2]^{1/3}[H_2]}$
b) $K_{b} = \textstyle\frac{[NH_3]^2}{[N_2]^{1/2}[H_2]^{3/2}}$ d) $K_{d} = \textstyle\frac{[N_2]^{1/2}}{[H_2]^{3/2}[NH_3]}$
$K_a = K_b^2 = K_c^3 = K_d^{-2} = \textstyle\frac{1}{K_d^2}$
Notice that there is nothing wrong with fractional powers in the equilibrium-constant expression.
Using Equilibrium Constants
Equilibrium constants have two main purposes:
1. To help us tell whether a reaction will be spontaneous under specified conditions.
2. To enable us to calculate the concentration of reactants and products that will be present once equilibrium has been reached.
We can illustrate how equilibrium constants can be used to achieve these ends, and also the fact that an equilibrium constant is indeed constant, with real data from one of the most intensively studied of all reactions, that between hydrogen and iodine to yield hydrogen iodide:
H2(g) + I2(g) 2HI(g) (4-9)
If we mix hydrogen and iodine in a sealed flask and observe the reaction, the gradual fading of the purple color of the iodine vapor tells us that iodine is being consumed. This reaction was studied first by the German chemist Max Bodenstein in 1893. Table 4-1 contains the data from Bodenstein's experiments. The experimental data are in the first three columns. In the fourth column, we have calculated the simple ratio of product and reactant concentrations, [HI]/[H2][I2], to see if it is constant. It clearly is not, for as the hydrogen concentration is decreased and the iodine concentration is increased, this ratio varies from 2.60 to less than 1. The law of mass action (Section 4-3) dictates that the equilibrium-constant expression should contain the square of the HI concentration, since the reaction involves 2 moles of HI for every mole of H2 and I2 , The fifth column shows that the ratio [HI]2/[H2][I2] is constant within a mean deviation of approximately 3%. * Therefore, this ratio is the proper equilibrium-constant expression, and the average value of Keq for these six runs is 50.53.
The equilibrium constant can be used to determine whether a reaction under specified conditions will go spontaneously in the forward or in the reverse direction. The ratio of product concentration to reactant concentration, identical to the equilibrium constant in form but not necessarily at equilibrium conditions, is called the reaction quotient, Q:
Q = $\textstyle\frac{[HI]^2}{[H_2][I_2]}$ (not necessarily at equilibrium) (4-10)
If there are too many reactant molecules present for equilibrium to exist, then the concentration terms in the denominator will make the reaction quotient, Q, smaller than Keq. The reaction will go forward spontaneously to make more product. However, if an experiment is set up so that the reaction quotient is greater than Keq, then too many product molecules are present for equilibrium and the reverse reaction will proceed spontaneously. Therefore, a comparison of the actual concentration ratio or reaction quotient with the equilibrium constant allows us to predict in which direction a reaction will go spontaneously under the given set of circumstances:
Q < Keq (forward reaction spontaneous)
Q > Keq (reverse reaction spontaneous) (4-11)
Q = Keq (reactants and products at equilibrium)
• These are Bodenstein's original numbers. Modern data can be much more accurate, with
less deviation in Keq. The mean deviation is the average of the deviations of individual calculated Keq from the average Keq.
Example 6
If 1.0 X 10-2 mole each of hydrogen and iodine gases are placed in a I-liter flask at 448°C with 2.0 X 10-3 mole of HI, will more HI be produced?
Solution
The reaction quotient under these conditions is
$Q = \textstyle\frac{(2.0 \times 10^{-3})}{(1.0 \times 10^{-2})^2}$ = 0.040
This is smaller than the equilibrium value of 50.53 in Table 4-1, which tells us that excess reactants are present. Hence, equilibrium will not be reached until more HI has been formed.
Example 7
If only 1.0 X 10-3 mole each of H2 and I2 had been used, together with 2.0 X 10-3 mole of HI, would more HI have been produced spontaneously?
Solution
You can verify that the reaction quotient is Q = 4.0. Because this is less than Keq, the forward reaction is still spontaneous.
Example 8
If the conditions of Example 7 are changed so that the HI concentration is increased to 2.0 X 10-2 mole liter-1 , what happens to the reaction?
Solution
The reaction quotient now is Q = 400. This is greater than Keq- There are now too many product molecules and too few reactant molecules for equilibrium to exist. Thus the reverse reaction occurs more rapidly than the forward reaction. Equilibrium is reached only by converting some of the HI to H2 and 12, so the reverse reaction is spontaneous.
Example 9
If the conditions of Example 7 are changed so that the HI concentration is 7.1 X 10-3 mole liter-1 , in which direction is the reaction spontaneous?
Solution
Under these conditions,
Q = $\textstyle\frac{(7.1 \times 10^{-3})^2}{(1.0 \times 10^{-3})^2} = 50.4 = K_{eq}$
Since Q equals Keq within the limits of accuracy of the data, the system as described is at equilibrium, and neither the forward nor the backward reaction is spontaneous. (Both reactions are still taking place at the molecular level, of course, but they are balanced so their net effects cancel.)
The second use for equilibrium constants is to calculate the concentrations of reactants and products that will be present at equilibrium.
Example 10
If a 1-liter flask contains 1.0 X 10-3 mole each of H2 and I2 at 448°C, what amount of HI is present when the gas mixture is at equilibrium?
Solution
The Keq expression is treated as an ordinary algebraic equation, and solved for the HI concentration:
$\textstyle\frac{[HI]^2}{(1.0 \times 10^{-3})^2} = K_{eq}$ = 50.53
$[HI]^2 = 50.53 \times 1.0 \times 10^{-6}$
$\textstyle [HI] = 7.1 \times 10^{-3} \textstyle mole liter^{-1}$
You can verify that in Example 7 the HI concentration was less than this equilibrium value; in Example 8 it was more; and in Example 9 it was just this value.
Example 11
One-tenth of a mole, 0.10 mole, of hydrogen iodide is placed in an otherwise empty 5.0 liter flask at 448°C. When the contents have come to equilibrium, how much hydrogen and iodine will be in the flask?
Solution
From the stoichiometry of the reaction, the concentrations of H2 and I2 must be the same. For every mole of H2 and I2 formed, 2 moles of HI must decompose. Let y equal the number of moles of H2 or I2 per liter present at equilibrium. The initial concentration of HI before any dissociation has occurred is
[HI]0 = $\textstyle\frac{0.10 mole}{(5.0 liters} = 0.020 mole liter^{-1}$
Begin by writing a balanced equation for the reaction, then make a table of concentrations at the start and at equilibrium:
H2 + I2 2HI
Start (moles Liter-1): 0 0 0.020
Equilibrium: y y 0.020 - 2y
The HI concentration of 0.020 mole liter-1 has been decreased by 2y for every y moles of H2 and I2 that are formed. The equilibrium-constant expression is
50.53 = $\textstyle\frac{(0.020 - 2y)^2}{y^2}$
We immediately see that we can take a shortcut by taking the square root of both sides:
7.11 = $\textstyle\frac{0.020 - 2y}{y}$
9.11 y = 0.020
y = 0.0022 mole liter-1
For 5 liters, 5 $\times$ 0.0022 = 0.011 mole of H2 and of I2 will be present at equilibrium. Only (0.020 - 0.0044) $\times$ 5 = 0.080 mole of HI will be left in the 5-liter tank, and the fraction of HI dissociated at equilibrium is
$\textstyle\frac{2y}{[HI]_0} = \frac{0.0044}{0.020}$ = 0.22, or 22% dissociation
Shortcuts such as taking the square root in the preceding example are not always possible, yet part of the skill of solving equilibrium problems lies in recognizing shortcuts when they occur and using them. The key is often a good intuition about what quantities are large and small relative to one another, and this intuition comes from thoughtful practice and understanding of the chemistry involved. You should remember that these are chemical problems, not mathematical ones.
In many cases a quadratic equation must be solved.
Example 12
If 0.00500 mole of hydrogen gas and 0.0100 mole of iodine gas are placed in a 5.00 liter tank at 448°C, how much HI will be present at equilibrium?
Solution
The initial concentrations of H2 and I2 are
[H2]0 = $\textstyle\frac{0.00500 mole}{5.00 liters}$ = 0.00100 mole liter-1
[I2]0 = $\textstyle\frac{0.0100 mole}{5.00 liters}$ = 0.00200 mole liter-1
This time, let the unknown variable y be the moles per liter of H2 or I2 that have reacted at equilibrium:
H2 + I2 2HI
Start (moles Liter-1): 0.00100 0.00200 0.0
Equilibrium: 0.00100 - y 0.00200 -y 2y
The quilibrium expression is
50.53 = $\textstyle\frac{(2y)^2}{(0.00100 - y)(0.00200 - y)}$
The square-root shortcut is now impossible because the starting concentrations of H2 and I2 are unequal. Instead we must reduce the equation to a quadratic expression:
46.53y2 - 0.1516y + 1.011 $\times$ 10-4 = 0
A general quadratic equation of the form ay2 + by + c = 0 can be solved by the quadratic formula,
y = $\textstyle\frac{-b \pm \!\, \sqrt{b^2 -4ac}}{2a}$
Thus for this problem
y = $\textstyle\frac{0.1516 \pm \!\, \sqrt{0.02298 - 0.01881}}{93.06}$
y = 2.32 $\times$ 10-3 and 0.935 $\times$ 10-3 mole liter-1
The first solution is physically impossible since it shows more H2 reacting than was originally present. The second solution is the correct answer: y=0.935 $\times$ 10-3 mole liter-1. Therefore, the equilibrium concentrations are
[H2] = 0.00100 - 0.000935 = 0.065 $\times$ 10-3 mole liter-1
[I2] = 0.00200 - 0.000935 = 1.065 $\times$ 10-3 mole liter-1
[HI] = 2(0.0935 $\times$ 10-3) = 1.87 $\times$ 10-3 mole liter-1
Units and Equilibrium Constants
As we have seen, the square brackets around a chemical symbol, as in [N2], represent concentrations, usually but not exclusively in units of moles liter-1. Concentrations expressed as moles liter-1 are often given the special symbol c, as in cN2, the concentrations measured in these units is denoted by Kc.
An equilibrium constant as we have defined it thus far may itself have units. In Example 1, Keq is unitless since the moles2 1iter-2 of the numerator and denominator cancel. In Example 2, the units of Keq are moles-1 liter since concentration occurs to the second power in the numerator and to the third power in the denominator. In Example 3 the units of Keq are the inverse: moles liter-1. The units demanded by Example 4, moles1/2 liter-1/2, may seem strange but they are perfectly respectable.
Example 13
What are the units for the equilibrium constants in the four reactions of Example 5?
Solution
The Keq expression is treated as an ordinary algebraic equation, and solved for the HI concentration:
Constant Units
Ka break break moles-2 liter2
Kb break break moles-1 liter
Kc break break moles-2/3 liter2/3
Kd break break moles liter-1
The question of units for Keq becomes important as soon as we realize that we can measure concentration in units other than moles liter-1. The partial pressure in atmospheres is a convenient unit when dealing with gas mixtures, and the equilibrium constant then is identified by Kp. Since the numerical values of Kp and Kc in general will be different, one must be sure what the units are when using a numerical constant.
Example 14
One step in the commercial synthesis of sulfuric acid is the reaction of sulfur dioxide and oxygen to make sulfur trioxide:
2SO2 + O2 2SO3
At 1000 K, the equilibrium constant for this reaction is Kp = 3.50 atm-1. If the total pressure in the reaction chamber is 1.00 atm and the partial pressure of unused 02 at equilibrium is 0.10 atm, what is the ratio of concentrations of product (S03) to reactant (S02)?
Solution
kp = $\textstyle\frac{p_{SO_3}^2}{p_{SO_2}^2 p_{O_2}}$ (pj = partial pressure of j)
Ratio = $\textstyle\frac{p_{SO_3}}{p_{SO_2}} = \textstyle\sqrt{K_p\times p_{O_2}} = \sqrt{3.50 \times 0.10} = 0.59$
The equilibrium mixture has 0.59 mole of S03 for every 1 mole of S02.
The ideal gas law permits us to convert between atmospheres and moles liter-1, and between Kp and Kc:
PV = nRT (3-8)
$P = \textstyle\frac{n}{V}RT = cRT$ (4-12)
In the general chemical reaction written earlier,
aA + bB cC + dD (4-7)
Δn (read "delta n"), the increase in number of moles of gas during the reaction, is
$\delta$n = c + d - a - b (4-13)
The equilibrium-constant expression in terms of partial pressures is
$K_p = \textstyle\frac{p_C^c p_D^d}{p_A^a p_B^b}$ (4-14)
With the ideal gas law applied to each gas component, we can convert this expression to Kc:
$K_p = \textstyle\frac{(cRT)^c (cRT)^d}{(cRT)^a (cRT)^b} = \textstyle\frac{c^c c^d}{c^a c^b}$(RT)Δn = Kc(RT)Δn (4-15)
(Do not confuse the two uses of the symbol c in equation 4-15: one is for concentration in moles liter-1 and the other for the number of moles of substance C.)
Example 15
What is the numerical value of Kc for the reaction of Example 14?
Solution
Three moles of reactant gases are converted into only 2 moles of product, so Δn = - 1. Hence at 1000 K,
Kp = 3.50 atm-1 = Kc(RT)-1
Kc = Kp $\times$ RT
= 3.50 atm-1 $\times$ 0.08205 liter atm K-1 mole-1 $\times$ 1000 K
= 287 moles-1 liter
Although the numerical answers that result when different units are used may differ, the physical reality must be the same.
Example 16
What is the concentration of oxygen in Example 14, in moles liter-1? Solve Example 14 again using Kc from Example 15.
Solution
Three moles of reactant gases are converted into only 2 moles of product, so Δn = - 1. Hence at 1000 K,
Kc = $\textstyle\frac{c_{SO_3}^2}{c_{SO_2}^2 c_{O_2}} =$ 287 moles-1 liter
cO2 $= \textstyle\frac{p_{O_2}}{RT} = \textstyle\frac{0.10 atm}{82.05 liter atm mole^{-1}} =$ 0.00122 mole liter-1
Ratio$= \textstyle\frac{c_{SO_3}}{c_{SO_2}} = \textstyle\sqrt{K_c \times c_{O_2}} = \textstyle\sqrt{287 \times 0.00122} = 0.59$
This is the same ration of SO_3 to SO_4 as was obtained when atmospheres were used. The choice is one of convenience.
Equilibrium Involving Gases with Liquids or Solids
Figure 4-2 Solid CaC03 and CaO are placed in a cylinder with a movable piston, which is initially pressed against the solids to ensure that no extraneous gases are present . At equilibrium, enough CaC03 will have decomposed so that the pressure of CO2 gas above the solid phases is a fixed value that varies with temperature but is 'Independent of how much of each of the solids is present.
All the examples considered so far have involved only one physical state, a gas, and are examples of homogeneous equilibria. Equilibria that involve two or more physical states (such as a gas with a liquid or a solid) are called hetergenous equilibria. If one or more of the reactants or products are solids or liquids, how does this affect the form of the equilibrium constant?
The answer, in short, is that any pure solids or liquids that may be present at equilibrium have the same effect on the equilibrium no matter how much solid or liquid is present. The concentration of a pure solid or liquid can be considered constant, and for convenience all such constant terms are brought to the left side of the equation and incorporated into the equilibrium constant itself. As an example, limestone (calcium carbonate, CaCO3), breaks down into quicklime (calcium oxide, CaO) and carbon dioxide, CO2:
CaCO3 CaO(s) + CO2(g)
The simple equilibrium-constant expression is
K'eq = $\textstyle\frac{[CaO(s)][CO_2(g)]}{[CaCO_3(s)]}$
As long as any solid limestone and quicklime are in contact with the gas, their effect on the equilibrium is unchanging. Hence the terms [CaCO3] and [CaO] remain constant and can be merged with K'eq:
Keq = K'eq $\textstyle\frac{[CaCO_3(s)}{CaO(s)} =$[CO2(g)]
This form of the equilibrium-constant expression tells us that, at a given temperature, the concentration of carbon-dioxide gas above limestone and calcium oxide is a fixed quantity. (this is true only as long as both solid forms are present.) Measuring concentration in units of atmospheres, we get
Kp = pCO2
with the experimental value 0.236 atm at 800°C.
We can see what this means experimentally by considering a cylinder to which CaCO, and CaO have been added. The cylinder has a movable piston, as shown in Figure 4-2. If the piston is fixed at one position, then CaCO3 will decompose until the pressure of CO2 above the solids is 0.236 atm (if the temperature is 800°C). If you try to decrease the pressure by raising the piston, then more CaCO3 will decompose until the pressure again rises to 0.236 atm. Conversely, if you try to increase the pressure by lowering the piston, some of the CO2 gas will react with CaO and become CaCO3 decreasing the amount of CO2 gas present until the pressure once more is 0.236 atm. The only way to increase pCO2, is to raise the temperature, which increases the value of Kp itself to 1 atm at 894°C and to 1.04 atm at 900°C.
An even simpler example is the vaporization of a liquid such as water:
H2O(l) H2O(g)
This process can be treated as a chemical reaction in a formal sense even though bonds within molecules are not made or broken. Imagine that the cylinder shown in Figure 4-2 is half-filled with water rather than with CaCO3 and CaO, and that the piston is initially brought down to the surface of the water. As the piston is raised, liquid will evaporate until the pressure of water vapor is a constant value that depends only on the temperature. This is the equilibrium vapor pressure of water at that temperature. At 25°C, the vapor pressure of water is 0.0313 atm. At 100°C, the vapor pressure reaches 1 atm and, as we shall see in Chapter 18, this is just the definition of the normal boiling point of water. The pressure of water vapor above the liquid in the cylinder does not depend on whether the water in the cylinder is 1 cm or 10 cm deep; the only requirement is that some water be present and capable of evaporating to make up any decrease in vapor pressure. Only when the piston is raised to the point where no more liquid exists can the pressure of water vapor fall below 0.0313 atm, if the cylinder is at 25°C. Similarly, if the piston is lowered, some of the vapor condenses, keeping the pressure at 0.0313 atm. Only when all vapor has condensed and the piston is resting on the surface of the liquid can the pressure inside the cylinder be raised above 0.0313 atm.
The formal equilibrium treatment of the evaporation of water would be
K'eq = $\textstyle\frac{[H_2O(g)]}{[H_2O(l)]}$
[H2O(l)] = constant , as long as liquid is present
Keq = K'eq[H2O(l)] = [H2O(g)]
In pressure units, the expression would be
Kp = pH2O(g)
From a practical standpoint, what the preceding discussion means is that the concentration terms for pure solids and liquids are simply eliminated from the equilibrium-constant expression. (They are present, implicitly, in the Keq.)
Example 17
If the hydrogen iodide reaction previously discussed in this chapter is carried out at room temperature, then iodine is present as deep purple crystals rather than as vapor. What then is the form of the equilibrium-constant expression, and does the equilibrium depend on the amount of iodine crystals present?
Solution
The reaction is
H2(g) + I2(s) 2HI(g)
and the equilibrium-constant expression is:
Keq = $\textstyle\frac{[HI]^2}{[H_2]}$
As long as some I2(s) crystals are present, the quantity is immaterial as far as equilibrium is concerned.
Example 18
Tin(IV) oxide reacts with carbon monoxide to form metallic tin and CO2 by the reaction
SnO2(s) + 2CO(g) Sn(s) + 2CO2(g)
What is the equilibrium-constant expression?
Solution
Keq = $\textstyle\frac{[CO_2]^2}{[CO]^2}$
Example 19
What is the equilibrium-constant expression for the following reaction leading to liquid water?
CO2(g) + H2(g) CO(g) + H2O(l)
What would the expression be if the product were water vapor?
Solution
If the product is H2O(l), the equilibrium-constant expression is
Keq = $\textstyle\frac{[CO]}{[CO_2][H_2]}$
If the product is H2O(g), the equilibrium-constant expression is
Keq = $\textstyle\frac{[CO][H_20]}{[CO_2][H_2]}$
The preceding example shows that as long as liquid water is present the gas-phase concentration is fixed at the vapor pressure of water at that temperature. Hence the water contribution, being constant, can be lumped into Keq.
Factors Affecting Equilibrium: Le Chatelier's Principle
Equilibrium represents a balance between two opposing reactions. How sensitive is this balance to changes in the conditions of a reaction? What can be done to change the equilibrium state? These are very practical questions if, for example, one is trying to increase the yield of a useful product in a reaction.
Under specified conditions, the equilibrium-constant expression tells us the ratio of product to reactants when the forward and backward reactions are in balance. This equilibrium constant is not affected by changes in concentration of reactants or products. However, if products can be withdrawn continuously, then the reacting system can be kept constantly off-balance, or short of equilibrium. More reactants will be used and a continuous stream of new products will be formed. This method is useful when one product of the reaction can escape as a gas, be condensed or frozen out of a gas phase as a liquid or solid, be washed out of the gas mixture by a spray of a liquid in which it is especially soluble, or be precipitated from a gas or solution.
For example, when solid lime (CaO) and coke (C) are heated in an electric furnace to make calcium carbide (CaC2),
CaO(s) + 3C(s) CaC2(s) + CO(g)
the reaction, which at 2000-3000°C has an equilibrium constant of close to 1.00, is tipped toward calcium carbide formation by the continuous removal of carbon monoxide gas. In the industrial manufacture of titanium dioxide for pigments, TiCl4 and O2 react as gases:
TiCl4 (g) + O2 (g) TiO2 (s) + 2Cl2(g)
The product separates from the reacting gases as a fine powder of solid Ti02 , and the reaction is thus kept moving in the forward direction. When ethyl acetate or other esters used as solvents and flavorings are synthesized from carboxylic acids and alcohols,
CH2COOH + HOCH2CH3 CH3COOCH2CH3 + H2O
acetic acid ethanol ethyl acetate
the reaction is kept constantly off-balance by removing the water as fast as it is formed. This can be done by using a drying agent such as Drierite (CaS04), by running the reaction in benzene and boiling off a constant-boiling benzene-water mixture, or by running the reaction in a solvent in which the water is completely immiscible and separates as droplets in a second phase. A final example: Since ammonia is far more soluble in water than either hydrogen or nitrogen is, the yield of ammonia in the reaction
N2(g) + 3H2(g) 2NH3(g)
can be raised to well over 90% by washing the ammonia out of the equilibrium mixture of gases with a stream of water, and recycling the nitrogen and hydrogen.
Temperature
Figure 4-3 Le Chatelier's principle and temperature. The dissociation of ammonia. 2NH3(g') 3H2(g) + N2(g) is endothermic or heat-absorbing . (a) Ammonia equilibrium at room temperature . (b) The temperature increase produced by adding heat is partially counteracted by using some of the heat to dissociate NH3 molecules and form N2 and H2. From Dickerson and Geis. Chemistry. Matter. and the Universe.
All the preceding methods will upset an equilibrium (in our examples, in favor of desired products) without altering the equilibrium constant. A chemist can often enhance yields of desired products by increasing the equilibrium constant so that the ratio of products to reactants at equilibrium is larger. The equilibrium constant is usually temperature dependent. In general, both forward and reverse reactions are speeded up by increasing the temperature, because the molecules move faster and collide more often. If the increase in the rate of the forward reaction is greater than that of the reverse, then Keq. increases with temperature and more products are formed at equilibrium. If the reverse reaction is favored, then Keq. decreases. Thus Keq for the hydrogen- iodine reaction at 448°C is 50.53, but at 425°C it is 54.4, and at 357°C it increases to 66.9. Production of HI is favored to some extent by an increase in temperature, but its dissociation to hydrogen and iodine is favored much more.
The hydrogen iodide-producing reaction is exothermic or heat emitting:
H2(g) + I2(g) 2HI(g)
ΔH298 = -10.2 kJ per 2 moles of HI
(If you check this figure against Appendix 3, remember that this reaction involves gaseous iodine, not solid.) If the external temperature of this reaction is lowered, the equilibrium is shifted in favor of the heat-emitting or forward reaction; conversely, if the temperature is increased, the reverse reaction, producing H2 and I2 is favored. The equilibrium shifts so as to counteract to some extent the effect of adding heat externally (raising the temperature) or removing it (lowering the temperature).
The temperature dependence of the equilibrium point is one example of a more general principle, known as Le Chatelier's principle: If an external stress is applied to a system at chemical equilibrium, then the equilibrium point will change in such a way as to counteract the effects of that stress. If the forward half of an equilibrium reaction is exothermic, then Keq will decrease as the temperature increases; if it is endothermic, Keq will increase. Only for a heat-absorbing reaction can the equilibrium yield of products be improved by increasing the temperature. A good way to remember this is to write the reaction explicitly with a heat term:
H2(g) + I2(g) 2HI(g) + heat(given off)
Then it is clear that adding heat, just like adding HI, shifts the reaction to the left. (see Figure 4-3.)
Pressure
Le Chatelier's principle is true for other kinds of stress, such as pressure changes. The equilibrium constant, Keq, is not altered by a pressure change at constant temperature. However, the relative amounts of reactants and products will change in a way that can be predicted from Le Chatelier's principle.
The hydrogen- iodine reaction involves an equal number (2) of moles of reactants and product. Therefore, if we double the pressure at constant temperature, the volume of the mixture of gases will be halved. All concentrations in moles liter-1 will be doubled, but their ratio will be the same. In Example 12, doubling the concentrations of the reactants and product does not change the equilibrium constant:
Keq = $\textstyle\frac{(1.87 \times 10^{-3} mole liter^{-1})^2}{(0.065 \times 10^{-3}mole liter^{-1})(1.065 \times 10^{-3} mole liter^{-1})}$
= $\textstyle\frac{(3.74 \times 10^{-3})^2}{(0.13\times 10^{-3})(2.13 \times 10^{-3})} =$ 50.51
Thus the hydrogen- iodine equilibrium is not sensitive to pressure changes. Notice that in this case Keq does not have units, since the concentration units in the numerator and denominator cancel.
In contrast, the dissociation of ammonia is affected by changes in pressure because the number of moles (2) of reactant does not equal the total number of moles (4) of products:
2NH3(g) N2(g) + 3H2(g)
The equilibrium constant for this reaction at 25°C is
Keq = $\textstyle\frac{[N_2][H_2]^3}{[NH_3]^2} =$ 2.5 $\times$ 10-9 mole2 liter -2
One set of equilibrium conditions is
N2 = 3.28 $\times$ 10-3 mole liter-1
H2 = 2.05 $\times$ 10-3 mole liter-1
NH3 = 0.106 mole liter-1
(Can you verify that these concentrations satisfy the equilibrium condition?) If we now double the pressure at constant temperature, thereby halving the volume and doubling each concentration,
N2 = 6.56 $\times$ 10-3 mole liter-1
H2 = 4.10 $\times$ 10-3 mole liter-1
NH3 = 0.212 mole liter-1
the ratio of products to reactants, the reaction quotient, is no longer equal to Keq:
Q = $\textstyle\frac{(6.56 \times 10^{-3})(4.10 \times 10^{-3})^3}{(0.212)^2} =$ 1.0 $\times$ 10-8 mole2 liter-2
Figure 4-4Le Chatelier's principle and pressure. (a) At initial equilibrium 17 molecules (moles) of gas are present: 12 of H2 , 4 of N2, and 1 of NH3. (b) When the gas is compressed Into a smaller volume a stress is created, which is evidenced by a higher pressure. (c) This stress can be relieved and the pressure reduced if some of the molecules of H2 and N2 combine to form more NH3, since the total number of gas molecules is thereby reduced. From Dickerson and Geis, Chemistry, Matter, and the Universe.
Since Q is greater than Keq, too many product molecules are present for equilibrium. The reverse reaction will run spontaneously, thereby forming more NH3 and decreasing the amounts of H2 and N2. Consequently, part of the increased pressure is offset when the reaction shifts in the direction that lowers the total number of moles of gas present. In general, a reaction that reduces the number of moles of gas will be favored by an increase in pressure, and one that produces more gas will be disfavored. (See Figure 4-4.)
Example 20
If the hydrogen iodide reaction were run at a temperature at which the iodine was a solid, would an increase in pressure shift the equilibrium reaction toward more HI, or less? What would be the effect of pressure on Keq?
Solution
Since the reaction of 2 moles of gaseous HI now yields 1 mole of gaseous H2 and 1 mole of solid I2 the stress of increased pressure is relieved by dissociating HI to H2 and I2. However, Keq will be unchanged by the pressure increase.
Figure 4-5 The ammonia perpetual-motion engine. A mixture of NH3 , H2, and N2 is contained in a chamber by the piston at the left and the hatched cylinder suspended from the left end of the rocker arm contains a mythical catalyst that would shift the equilibrium point of the reaction 2NH3(g) N2(g) + 3H2(g) to the right. In (a) and (b), as the catalyst is introduced, ammonia dissociates to nitrogen and hydrogen. The total volume of gas increases and the piston is pushed to the right. In (c) and (d), as the catalyst is withdrawn, N2 and H2 reassociate to form ammonia; hence the volume shrinks and the piston is driven to the left. This self-contained, two-stage process provides an unlimited supply of power at the flywheel on the right without an external input of energy. For practical difficulties, see the text.
Catalysis
What effect does a catalyst have on a reaction at equilibrium? None. A catalyst cannot change the value of Keq, but it can increase the speed with which equilibrium is reached. This is the main function of a catalyst. It can take the reaction only to the same equilibrium state that would be reached eventually without the catalyst.
Catalysts are useful, nevertheless. Many desirable reactions, although spontaneous, occur at extremely slow rates under ordinary conditions. In automobile engines, the main smog-producing reaction involving oxides of nitrogen is
N2 + O2 2NO
(Once NO is present, it reacts readily with more oxygen to make brown N02.) At the high temperature of an automobile engine, Keq for this reaction is so large that appreciable amounts of NO are formed. However, at 25°C, Keq= 10-30. (Using only the previous two bits of information and Le Chatelier's principle, predict whether the reaction as written is endothermic or exothermic. Check your answer using data from Appendix 3.) The amount of NO present in the atmosphere at equilibrium at 25°C should be negligible. NO should decompose spontaneously to N2 and O2 as the exhaust gases cool. But any Southern Californian can verify that this is not what happens. Both NO and N02 are indeed present, because the gases of the atmosphere are not at equilibrium.
The rate of decomposition of NO is extremely slow, although the reaction is spontaneous. One approach to the smog problem has been to search for a catalyst for the reaction
2NO N2 + O2
that could be housed in an exhaust system and could break down NO in the exhaust gases as they cool. Finding a catalyst is possible; a practical problem arises from the gradual poisoning of the catalyst by gasoline additives, such as lead compounds. This is the reason why new cars with catalytic converters only use lead-free gasoline.
A proof of the assertion that a catalyst cannot change the equilibrium constant is illustrated in Figure 4-5. If a catalyst could shift the equilibrium point of a reacting gas mixture and produce a volume change, then this expansion and contraction could be harnessed by mechanical means and made to do work. We would have a true perpetual-motion machine that would deliver power without an energy source. From common sense and experience we know this to be impossible. This "common sense" is stated scientifically as the first law of thermodynamics, which will be discussed in Chapter 15. A mathematician would call this a proof by contradiction: If we assume that a catalyst can alter Keq, then we must assume the existence of a perpetual-motion machine. However, a perpetual-motion machine cannot exist; therefore our initial assumption was wrong, and we must conclude that a catalyst cannot alter Keq.
In summary, Keq is a function of temperature, but it is not a function of reactant or product concentrations, total pressure, or the presence or absence of catalysts. The relative amounts of substances at equilibrium can be changed by applying an external stress to the equilibrium mixture of reactants and products, and the change is one that will relieve this stress. This last statement, Le Chatelier's principle, enables us to predict what will happen to a reaction when external factors are changed, without having to make exact calculations.
Summary
A spontaneous reaction is one that will take place, given enough time, without outside assistance. Some spontaneous reactions are rapid, but time is not an element in the definition of spontaneity. A reaction can be almost infinitely slow and still be spontaneous.
The net reaction that we observe is the result of competition between forward and reverse steps. If the forward process is faster, then products accumulate, and we say that the reaction is spontaneous in the forward direction. If the reverse process is faster, then reactants accumulate, and we say that the reverse reaction is the spontaneous one. If both forward and reverse processes take place at the same rate, then no net change is observed in any of the reaction components. This is the condition of chemical equilibrium.
The ratio of products to reactants, each concentration term being raised to a power corresponding to the coefficient of that substance in the balanced chemical equation, is called the equilibrium constant, Keq. (See equation 4-8.) It can be used to predict whether a given reaction under specified conditions will be spontaneous, and to calculate the concentrations of reactants and products at equilibrium. The reaction quotient, Q, has a form that is identical with that of the equilibrium constant, Keq, but Q applies under nonequilibrium conditions as well. For a given set of conditions, if Q is smaller than Keq, the forward reaction is spontaneous; if Q is greater than Keq, the reverse reaction is spontaneous; and if Q = Keq, the system is at equilibrium.
The equilibrium constant can be used with any convenient set of concentration units: moles liter-1 , pressure in atmospheres, or others. Its numerical value will depend on the units of concentration, so one must be careful to match the proper values of Keq and units when solving problems. If gas concentrations are expressed in moles liter-1, the equilibrium constant is designated by Kc; if in atmospheres, by Kp. Just as partial pressure of the jth component of a gas mixture is related to moles per liter by pj = cjRT, so Kp and Kc are related by Kp = Kc(RT)Δn, in which Δn is the net change in number of moles of gas during the reaction.
When some of the reactants or products are pure solids or liquids, they act as infinite reservoirs of material as long as some solid or liquid is left. Their effect on equilibrium depends only on their presence, not on how much of the solid or liquid is present. Their effective concentrations are constant, and can be incorporated into Keq. In practice, this simply means omitting concentration terms for pure solids and liquids from the equilibrium-constant expression. Evaporation of a liquid can be treated formally as a chemical reaction with the liquid as reactant and vapor as product. These conventions for writing concentration terms for a liquid permit us to write the equilibrium constant for evaporation as Kp = pj where pj is the equilibrium vapor pressure of substance j.
Le Chatelier's principle states that if stress is applied to a system at equilibrium the amounts of reactants and products will shift in such a manner as to minimize the stress. This means that for a heat-absorbing, or endothermic, reaction, Keq increases as the temperature is increased, since carrying out more of the reaction is a way of absorbing some of the added heat. Similarly, cooling increases Keq for a heat-emitting or exothermic reaction. Although the equilibrium constant Keq is independent of pressure, and changing the total pressure on a reacting system does not alter Keq directly, an increase in pressure does cause the reaction to shift in the direction that decreases the total number of moles of gas present.
A catalyst has no effect at all on Keq or the conditions of equilibrium. All that a catalyst can do is to make the system reach equilibrium faster than it would have done otherwise. Catalysts can make inherently spontaneous but slow reactions into rapid reactions, but they cannot make nonspontaneous reactions take place of their own accord. | textbooks/chem/General_Chemistry/Chemical_Principles_(Dickerson)/04%3A_Will_It_React_An_Introduction_to_Chemical_Equilibrium/5.01%3A_Hydrogen_Oxygen_and_Water.txt |
The words "acid" and "base" are
functional terms, and not labels.
They describe what a substance does,
rather than what it is.
R. von Handler (b. 1931)
IntroductionEdit
Almost all the reactions that a chemist is concerned with take place in solution rather than in gaseous or solid phases. Most of these reactions occur in aqueous solution, where water is the solvent. There are good reasons for this preference for liquid media. Molecules must come into contact to react, and the rates of migration of atoms or molecules within crystals usually are too slow to be useful. In contrast, molecules of gases are mobile, but gas volumes are inconveniently large, and many substances cannot be brought into the gas phase without decomposing. Solutions of reacting molecules in liquids offer an optimum combination of compactness, ease of handling, and rapidity of mixing of different substances.
As we saw in Chapter 1, water has special virtues as a solvent. It is polar, in the sense illustrated in Figure 5-1. The oxygen atom draws the electrons of the 0 - H bonds toward itself, acquiring a slight negative charge and leaving small positive charges on the two hydrogen atoms. Water therefore can interact with other polar molecules. Moreover, water molecules dissociate to a small extent into H+ and OH- ions, a property that is important in acid-base reactions. This chapter is concerned with reactions and equilibria in aqueous solution, especially those involving acids and bases.
Equilibria in Aqueous SolutionsEdit
Figure 5-1 Water is a polar molecule, with excess electrons and a partial negative charge on the oxygen atom, and an electron deficiency and a partial positive charge on each hydrogen atom. (b) The methane molecule, CH4, is nonpolar: Its electrons are distributed evenly over the molecule. It has no local regions of positive and negative charge to attract water molecules, so water is a poor solvent for methane, (c) Methanol, CH3OH, is polar, although less so than water. It has an excess of electrons and a small negative charge on the oxygen atom. and a small positive charge on the attached hydrogen atom. Methanol interacts well with water molecules by electrostatic forces. making it soluble in water. (d) Sodium hydroxide. NaOH. dissociates into positive and negative ions. These ions interact strongly with the polar water molecules, so NaOH is extremely soluble in water. Each Na+ and OH- ion has a cluster of water molecules surrounding it, with their negative charges closest to the sodium ions and their positive charges closest to the hydroxide ions. The ions are said to be hydrated.
If reactants and products in a chemical reaction are in solution, the form of the equilibrium-constant expression is the same as for gas reactions, but the logical units of concentration are moles per liter of solution (units of molarity).
aA + bB cC + dD (4-7)
Keq = $\textstyle\frac{[C]^c[D]^d}{[A]^a[B]^b}$ (4-8)
Some reactions in aqueous solution involve water as a participant. A well-studied example is the hydrolysis ("splitting by water") of the ethyl acetate molecule to yield acetic acid and ethyl alcohol (ethanol):
Because all the other participant molecules themselves are polar, they dissolve well in water, which is therefore a good dispersing agent. In addition, water plays a direct role as a reactant molecule.
The equilibrium-constant expression for this reaction, in principle, is
K'eq = $\textstyle\frac{[CH_3COOH][C_2H_5OH]}{[CH_3COOC_2H_5][H_2O]}$ (5-2)
However, since water is present in such excess in its role as solvent, the water concentration is virtually unchanged during the reaction. In dilute solutions this is approximately the concentration of water in its pure state.
H2O = $\textstyle\frac{1000 g liter^{-1}}{18.0 g mole^{-1}} =$ 55.6 moles liter-1 (5-3)
This constant water concentration can be brought over to the left side of equation 5-2 and incorporated into the equilibrium constant, as we saw for condensed phases in Chapter 4, so the equilibrium-constant expression becomes
Keq = K'eq[H2O] = $\textstyle\frac{[CH_3COOH][C_2H_5OH]}{[CH_3COOC_2H_5]}$ (5-4)
Other reactions in aqueous solution involve ions; an example is the precipitation of silver ions with chloride ions, in the form of insoluble silver chloride:
Ag+ + Cl- AgCl(s)
In this process water is not a direct reactant or product, but it does interact with the ions to keep them in solution. Any ion in aqueous solution is hydrated, or surrounded by polar water molecules as in Figure 5-1d. If the central ion is positive (a cation), then the negatively charged oxygen atoms of the water molecules are pointed toward it; if the central ion is negative (an anion), the positively charged hydrogen atoms of the water molecules are closest.
Each hydrated ion thus is stabilized by an immediate environment of charges opposite in sign to its own charge. When a salt crystal dissolves in water, the attractions between ions of opposite charge in the crystal are broken. In compensation, similar attractions are set up between ions and the hydrating water molecules. Solubility of salt crystals is the result of a balance or competition between crystal forces and hydration forces. This is why salts do not dissolve in nonpolar solvents such as benzene, which cannot offer hydrating attractions.
Ionization of Water and the pH ScaleEdit
Water itself ionizes to a small extent:
H2O(l) H+ + OH- (5-5)
Each ion is surrounded with polar water molecules (as Na+ and OH- are in Figure 5-1d). The hydrated state of the proton, H+, is sometimes represented as H3O+, meaning H+ • H20. But this is an unnecessary and even misleading notation. A more accurate representation of a hydrated proton would be H9O$_4^+$, or H+ • (H20)4, to represent the cluster:
We will assume that H+ and OH-, like all other ions, are hydrated in aqueous solution, and we will therefore represent them simply as H+ and OH-.
Table 5-1. lon-Product Constant for Water, Kw
Kw = [H+][OH-]
T (°C): 0 25 40 60
Kw 0.115 $\times$ 10-14 1.008 $\times$ 10-14 2.95 $\times$ 10-14 9.5 $\times$ 10-14
The equilibrium-constant expression for the dissociation of water is
K'eq $= \textstyle\frac{[H^+][OH^-]}{[H_2O]}$ (5-6)
The constant [H2O] form can be combined with K'eq as before, producing
Kw = 55.6K'eq = [H+][OH-] (5-7)
This new equilibrium constant, Kw, is called the ion-product constant for water. Like most equilibrium constants, Kw varies with temperature. Some experimental values of the ion-product constant are given in Table 5-1.
Example 1
From the data in Table 5-1 and Le Chatelier's principle, predict whether the dissociation of water liberates or absorbs heat.
Solution
Since a higher temperature favors dissociation, dissociation is an endothermic or heat-absorbing process. From Appendix 3, ΔH(diss of H20) = +55.90 kJ mole-1. This is the energy required to break one O—H bond, thereby leaving both electrons with the oxygen atom.
It is customary to take Kw,/sub> = 1.00 $\times$ 10,sup>-14 as being accurate enough for room-temperature equilibrium calculations. (It is also customary in acid base equilibrium calculations to write Kw as if it were an exact number, 10,sup>-14 rather than 1.00 $\times$ 10-14.) This means that in pure water, where the concentrations of hydrogen and hydroxide ions are equal,
[H+] = [OH-] = 10-7 mole liter-1 (5-8)
The pH scale. From Dickerson and Geis, Chemistry, Matter, and the Universe.
Since large powers of 10 are clumsy to deal with, a logarithmic notation has been devised, called the pH scale (Figure 5-2). (The symbol pH stands for "negative power of hydrogen ion concentration.") The pH is the negative logarithm of [H+]:
pH = -log10[H+] (5-9)
If the hydrogen ion concentration is 10-7 mole liter-1, then
pH = -log10(10-7) = +7
By an analogous definition,
pOH = -log10[OH-] (5-10)
and the pOH of pure water is also +7. The equilibrium constant Kw also can be expressed in logarithmic terms:
pKw = -log10Kw = +14 (5-11)
Finally, the equilibrium expression for dissociation of water,
[H+][OH-] = Kw = 10-14 (5-12)
can be written
pH + pOH = 14 (5-13)
In an acid solution, [H+] is greater than 10-7, so the pH is less than 7. The ion-product equilibrium still holds, and [OH-] can be found from the expression
[OH-] = $\textstyle\frac{K_w}{[H^+]} = \frac{10^{-14}}{[H^+]}$ (5-14)
or
pOH = pKw - pH = 14 - pH (5-15)
The approximate pH values of some common solutions are given in Table 5-2.
Table 5-2. Acidity of Some Common Solutions
Substance pH
Commercial concentrated HCl (37% by weight) ~ -1.1
1 M;; HCl solution 0.0
Gastric Juice 1.4
Lemon Juice 2.1
Orange Juice 2.8
Wine 3.5
Tomato Juice 4.0
Black Coffee 5.0
Urine 6.0
Rainwater 6.5
Milk 6.9
Pure Wtaer at 24°C 7.0
Blood 7.4
Baking soda solution 8.5
Borax solution 9.2
Limewater 10.5
Household ammonia 11.9
1M NaOH solution 14.0
Saturated NaOH solution ~15.0
Example 2
From Table 5-2, what is the hydrogen ion concentration of orange juice? What is the hydroxide ion concentration?
Solution
Since the pH of orange juice is 2.8, the hydrogen ion concentration is
[H+] = 10-2.8 = 10+0.2 $\times$ 10-3 = 1.6 $\times$ 10-3
= 0.0016 mole liter-1
The hydroxide ion concentration can be obtained by either of two equivalent methods:
[OH-] = $\textstyle\frac{10^{-14}}{1.6 \times 10^{-3}} =$ 6.3 $\times$ 10-12 mole liter-1
or
pOH = 14 - pH = 11.2
[OH-] = 10-11.2 = 10+0.8 $\times$ 10-12 = 6.3 $\times$ 10-12 mole liter-1
Example 3
What is the ratio of hydrogen ions to hydroxide ions in pure water? In orange juice?
Solution
In pure water the ratio is 10-7 to 10-7 or 1 to 1. In orange juice, from Table 5-2, the ratio is 1.6 $\times$ 10-3 to 6.3 $\times$ 10-12 or 250,000,000 to 1.
To maintain equilibrium, the added H+ ions from the juice have pushed the water dissociation reaction in the direction of undissociated H20, thereby removing OH- ions from the solution. Orange juice is not a particularly strong acid, and the enormous fluctuation of ionic ratios even in this example illustrates the usefulness of power-of-ten and logarithmic (pH, pOH, pK) notation.
Strong and Weak AcidsEdit
Figure 5-3 Bronsted-Lowry acids and bases. In the theory of Bronsted and Lowry, an acid is any substance that releases protons in solution, and a base is any substance that removes protons by combining with the. HCl is a strong acid because it readily releases H+,/sup> ions. Cl- is a weak base because it has a small tendency to combine with H+, HCl, and Cl- are termed a conjugate pair of acid and base. From Dickerson and Geis, Chemistry, Matter and the Universe.
Figure 5-4 Comparison of relative strengths of solvation of a hydrogen ion in (a) liquid ammonia, (b) water, and (c) diethyl ether. The binding between proton and solvent ammonia molecules is extremely strong, and liquid ammonia will take protons from and make strong acids of substances that in aqueous solution are only weak acids. In contrast, diethyl ether is such an ineffectual proton-solvating molecule that many substances that are strong aC ids in water can retain their proton and be only partially dissociated weak acids in ethyl ether. The + and - represent partial charges arising from local deficiencies and excesses of electrons.
Arrhenius defined an acid (Chapter 2) as a substance that increases the hydrogen ion concentration of an aqueous solution, and a base as a substance that increases the hydroxide ion concentration. A more general definition was proposed in 1923 by Johannes Bronsted and T. M. Lowry. The Bronsted-Lowry definition can be applied to nonaqueous solutions as well: An acid is any substance that is capable of giving up a hydrogen ion, or proton, and a base is any substance that can combine with and therefore remove a hydrogen ion. Now that we understand that water molecules exist in equilibrium with their dissociated H+ and OH- ions, we can see that the two definitions are equivalent when water is the solvent. Arrhenius and Br~nsted acids are both hydrogen-ion-releasing substances. If a Br~nsted base combines with hydrogen ions, it shifts the equilibrium of equation 5-5 in favor of dissociation until balance is restored. More hy~ droxide ions are formed in the process, so in water a Br~nsted base is an Arrhenius base as well.
In aqueous solution, acids are classified as either strong or weak. Strong acids are completely dissociated or ionized, and they include hydrogen acids such as hydrochloric acid (HCl) and hydroiodic acid (HI), and oxyacids (oxygen-containing acids) such as nitric acid (HN03), sulfuric acid (H2S04), and perchloric acid (HCI04). Each of these acids loses one proton in solution, and the acid-dissociation constant, Ka, is so large (> 103) that too little undissociated acid remains to be measured. (HS0$_4^-$ loses a second proton and is a weak acid.)
Weak acids have measurable ionization constants in aqueous solution, because they do not dissociate completely. Examples (at 25°C) are
Sulfuric: HSO$_4^-$ H+ + SO$_4^2-$ Ka = $\textstyle\frac{[H^+][SO_4^{2-}]}{[HSO_4^-]}$
(2nd Ionization) = 1.2 $\times$ 10-2 (5-16)
Hydrofluoric: HF H+ + F$^-$ Ka = $\textstyle\frac{[H^+][F^-]}{[HF]}$
= 3.5 $\times$ 10-4 (5-17)
Acetic: CH3COOH CH3COO- + H$^+$ Ka = $\textstyle\frac{[H^+][CH_3COO^-]}{[CH_3COOH]}$
= 1.76 $\times$ 10-5 (5-18)
Hydrocyanic: HCN H+ + CN$^-$ Ka = $\textstyle\frac{[H^+][CN^-]}{[HCN]}$
= 4.9 $\times$ 10-10 (5-19)
The distinction between strong and weak acids is somewhat artificial. The ionization of HCl is not simply a dissociation; it is, rather, the result of successful competition of H20 molecules with Cl- ions for the proton, H+:
HCl + xH2O H+ • (H2O)x + Cl- (5-20)
In the Bronsted-Lowry theory, any proton donor is an acid, and any proton acceptor is a base (Figure 5-3). Therefore, HCl is an acid, and Cl- is its conjugate base. Since HCl loses a proton readily it is a strong acid, and since Cl- has so little affinity for the proton it is a weak base. In contrast, HCN is a very weak acid, because relatively few HCN molecules release their proton. Its conjugate base, CN-, is a strong base by virtue of its high affinity for a proton.
Water is a somewhat stronger base than Cl-, and when it is present in excess, as in an aqueous solution of HCl, it takes virtually all the protons from HCl, leaving it completely ionized. CN,sup>- is a much stronger base than H2O, so only a small fraction of the protons from HCN become bound to the water molecules. In other words, HCN is only slightly ionized in aqueous solution, as its Ka of 4.9 $\times$ 10-10 indicates.
Because water is present in great excess, any acid whose conjugate base is weaker than H2O (i.e., has a lesser affinity for protons than has H2O) will be ionized almost completely in aqueous solution. We cannot distinguish between the behavior of HCl and of HCl04 (perchloric acid) in water solution. Both are completely dissociated and are therefore strong acids. However, for a solvent with a lesser attraction for protons than water, we do find differences between HCl and HCl04. With diethyl ether as a solvent, perchloric acid is still a strong acid, but HCl is only partially ionized and hence is a weak acid. Diethyl ether does not solvate a proton as strongly as water does (Figure 5-4). (Solvation is a generalization of the concept of hydration, which applies to solvents other than water.) The equilibrium point in the reaction
HCl + xC2H5OC2H5 H+ • (C2H5OC2H5)x + Cl- (5-21)
lies far to the left, so HCl is only partially dissociated in ether. Only in an extremely strong acid, such as perchloric acid, does the anion have so little attraction for the proton that it will release it to ether as an acceptor solvent. Clearly, by using solvents other than water, we can see differences in acidity (or proton affinity) that are masked in aqueous solution. This masking of relative acid strengths by solvents such as water is known as the leveling effect.
The dissociation constants for a number of acids in aqueous solution are listed in Table 5-3, with estimates of the Ka for strong acids that are "leveled" by the solvent in aqueous solution. The dissociation of protonated solvent, H3O+, into hydrated protons and H20, represents merely a shuffling of protons from one set of water molecules to another, and must have a Keq of 1.00. In liquid ammonia as a solvent, all acids whose conjugate bases are weaker than NH3 would be leveled by the solvent and would be totally ionized strong acids. Thus hydrofluoric acid and acetic acid are both strong acids in liquid ammonia.
The leveling effect of solvent and the origin of strong and weak acids are summarized in Figure 5-4. T he distinction between strong and weak acids depends on the solvent as much as it does on the inherent properties of the acids themselves. Nevertheless, in aqueous solution the distinction is real. As long as the discussion is confined to aqueous solutions (as ours will be from now on), we shall find it useful to think about and to treat the two classes of acids separately.
Strong and Weak BasesEdit
In Arrhenius' terminology a base is a substance that decreases the hydrogen ion concentration of a solution. Sodium hydroxide, potassium hydroxide, and similar compounds are bases because they dissolve and dissociate completely in aqueous solution to yield hydroxide ions:
NaOH Na + + OH- (5-22)
KOH K+ + OH-
These excess hydroxide ions then disturb the water dissociation equilibrium, and combine with some of the protons normally found in pure water:
H+ + OH- H2 [H+] = $\textstyle\frac{K_w}{OH^-}<$ 10-7 (5-23)
In the more generalized Bronsted-Lowry defnition, the hydroxide ion itself is the base, because it is the substance that combines with the proton. The Na+ and K+ ions merely provide the positive ions that are necessary for overall electrical neutrality for the chemical compound.
The commonly encountered hydroxides of alkali metals (Li, Na, K) all dissolve and dissociate completely to produce the same Bronsted-Lowry base, OH-. These hydroxides all are strong bases, analogous to strong acids such as HCl and HNO3. Other substances such as ammonia and many organic nitrogen compounds also can combine with protons in solution and act as Bronsted-Lowry bases. These compounds are generally weaker bases than the hydroxide ion, because they have smaller attraction for protons. For example, when ammonia competes with OH- for protons in an aqueous solution, it is only partially successful. It can combine with only a portion of the H+ ions, thus will have a measurable equilibrium constant.
NH3 + H+ NH$_4^+$ (5-24)
There is no logical reason why this reaction cannot be described by an acd-dissociation constant, as in Table 5-3. The ammonium ion, NH$_4^+$, is the Bronsted-Lowry conjugate acid of the base NH3. There is no reason why, in an acid-base pair, it is the acid that must be neutral and the base charged, as in HCl/Cl- and HCN/CN-. The NH$_4^+$ ion is just as respectable an acid as HCl or HCN, and although weaker than HCl, it is actually stronger than HCN. Thus, we can describe the ammonia reaction as an acid dissociation:
NH$_4^+$ NH3 + H+ Ka= 5.6 $\times$ 10-10 (from Table 5-3) (5-25)
or, if we want to focus on the basic behavior of NH3,
NH3 + H+ NH$_4^+$ Keq = $\textstyle\frac{1}{K_a}$ = 1.8 $\times$ 10+9 (5-26)
However, chemical language has become trapped by the older acid-base terminology introduced by Arrhenius, and you should be aware of this. Arrhenius thought of a base as a substance that releases OH- ions into aqueous solution. For alkali metal hydroxides such as NaOH the process was straightforward:
NaOH Na+ + OH- (5-27)
But what about NH3? Where do the hydroxide ions come from? Arrhenius assumed that when ammonia dissolved in water the reaction was
NH3 + H2 NH4OH NH$_4^+$ + OH- (5-28)
This brought NH3 into line by postulating an intermediate-ammonium hydroxide base that dissociates completely; ammonium hydroxide would be a weak base that dissociates only partially. Arrhenius defined a base-dissociation constant, Kb, as
BOH B+ + OH<sup- Kb = $\textstyle\frac{[B^+][OH^-]}{[BOH]}$ (5-29)
where B usually represents a metal. For ammonia, Ka and Kb would be related by
Ka = $\textstyle\frac{[NH_4^+][OH^-]}{[NH_3]} = \frac{[NH_4^+][OH^-][H^+]}{[NH_3][H^+]} = \frac{K_w}{K_a}$ (5-30)
Kb = $\textstyle\frac{10^{-14}}{5.6\times 10^{-10}} =$ 1.8 $\times$ 10-5 (5-31)
Unfortunately for Arrhenius' theory, there is no evidence that ammonium hydroxide, NH40H, exists as a real compound. It is more accurate to say that the polar ammonia molecule is hydrated like any other polar molecule: NH3 • (H20)x. Ammonia, NH3, combines directly with a proton and with water molecules:
NH3 + H+O + xH2 NH$_4^+$ (in acid solutions) (5-32)
NH3 + xH2O NH$_4^+$ + OH- (in acid solutions)
Nevertheless, Arrhenius' notation is too deeply embedded in the fabric of chemistry to dislodge, and we often will use Kb for weak bases rather than Ka for their conjugate acids. In general, the completely dissociated strong bases that we shall encounter will be hydroxide compounds, and the weak bases will be ammonia and organic nitrogen compounds such as those listed in Table 5-4. Kb always can be found from Ka and Kw and the expression
Ka $\times$ Kb = Kw (5-33)
Solutions of Strong Acids and Bases: Neutralization and TitrationEdit
When an amount of strong acid is added to water, the effect is that of adding the same amount of hydrogen ions, since the acid is totally dissociated,
Example 4
What is the hydrogen ion concentration of a 0.0l00M nitric acid solution? What is the pH?
Solution
[H+] = 0.010 mole liter-1
pH = -log10(10-2) = 2.00
The solution is quite acidic.
Example 5
What are the hydrogen ion concentration and the pH of a 0.0050M sodium hydroxide solution?
Solution
The hydroxide ion contribution from completely dissociated NaOH is
[OH-] = 0.0050 mole liter-1
This large quantity of hydroxide ions will repress the normal dissociation of water and enhance the reaction to the left:
H2O H+ + OH-
The hydrogen ion concentration is found from the water equilibrium expression:
[H+] = $\textstyle\frac{K_w}{[OH^-]} = \frac{10^{-14}}{0.0050} =$ 2.0 $\times$ 10-12 mole liter-1
pH = -log10(2.0) - log10(10-12) = -0.30 + 12.0 = 11.7
The solution is quite basic.
Example 6
If we mix equal volumes of the solutions of the previous two examples, what will be the pH of the resulting solution?
Solution
If equal volumes are mixed, then the concentration of each solute will be halved, since the final volume is twice the volume of each starting solution. The final solution would be 0.0050M in nitric acid and 0.0025M in sodium hydroxide. But acid and base will react and neutralize one another until one or the other is used up:
H+ + NO$_3^-$ + Na+ + OH- H2O + NO$_3^-$ + Na+
or simply
H+ + OH- H2O
since sodium and nitrate ions take no part in the neutralization reaction. In this case, sodium hydroxide is in shorter supply. When all the base has been neutralized, we still have
0.0050 - 0.0025 = 0.0025 mole liter-1 excess nitric acid
[H+] = 0.0025 = 2.5 $\times$ 10-3 mole liter-1
pH = -log10(2.5) + 3.0 = 2.6
Example 7
How many milliliters of 0.10M HCl must we add to 200 ml of 0.0050M KOH to bring the pH down to 10.0?
Solution
Without HCl, the pH of the potassium hydroxide solution would be 11.7, as in Example 5. Let y equal the number of milliliters of HCl solution needed to yield a pH of 10.0. Since 0.0050 mole liter-1 is the same as 0.0050 millimoles ml-1, the total number of millimoles (m moles) of KOH is
nKOH = 0.0050 m mole ml-1 $\times$ 200 ml = 1.00 m mole
The total number of millimoles of HCl that must be added is
nHCl = 0.10 m mole ml-1 $\times$ y ml = 0.10y m mole
Since the final solution is basic, nKOH - nHCl. The net amount of hydroxide ions left over after partial neutralization by HCl is
nbase = nKOH - nHCl = 1.0 -0.10y
The final volume is
V = 200 + y ml
and therefore the final hydroxide in concentration is
[OH-] = $\textstyle\frac{n_{base}}{V} = \frac{1.0 - 0.10y}{200 + y}$
A pH of 10.0 means a pOH of 4.0 and [OH-] = 10-4 mole liter-1, thus
$\textstyle\frac{1.0 - 0.10y}{200 + y}$ = 10-4
and
y = 9.8 ml of 0.10 M HCl to be added.
Titration and Titration CurvesEdit
If we add equal numbers of equivalents of a strong acid and a strong base, they will neutralize one another completely, and the pH will be 7.0. As we saw in Chapter 2, this makes possible the titration method of measuring quantities of acid or base.
Example 8
One hundred fifty milliliters of HCl solution of unknown concentration are
titrated with 0.10M NaOH. Eighty milliliters of base solution are required to neutralize the acid. How many moles of HCl were present originally, and what was the acid-solution concentration?
Solution
The number of millimoles of base used is
nNaOH = 0.10 m mole ml-1 $\times$ 80 ml = 8.0 m mole
This must be the same as the number of millimoles of acid originally present, if neutralization was complete.
Thus the original concentration of HCl was
[HCl]0 = $\textstyle\frac{8.0 m moles}{150 ml} =$ 0.053 m moles ml-1 or mole liter-1
A common way of determining the equivalence point of titration (the point at which neutralization occurs) is with an acid-base indicator. Indicators are weak organic acids or bases that have different colors in their ionized and neutral states (or in two ionized states). If their color change occurs in the neighborhood of pH 7, and if we add a few drops of indicator solution to the solution being titrated, we see this color change at the end point of the titration. We will discuss some common indicators in the section on weak acids. The matching of indicator color-change point and the end point of a titration does not have to be very exact, because the pH swings drastically through several units as neutralization becomes complete. This can make life easy for the analytical chemist, and it is worth looking more closely at the behavior of pH during titration. To illustrate what we have just said, let us calculate the titration curve for a typical strong acid and strong base.
Example 9
Fifty milliliters of 0.10M HN03 are titrated with 0.10M KOH, in an experimental arrangement such as that shown in Figure 2-3. Calculate the pH of the solution as a function of the volume of KOH solution added (v, in milliliters).
Solution
It is easiest to treat this calculation in three parts: before neutralization, at neutralization (equivalence point), and after neutralization. Before the equivalence point, calculate how much base has been added, assume that all of this base was used to neutralize some of the acid, and calculate how much acid would remain unneutralized, as a function of the volume of base solution added.
Figure 5-5 Titration curve for typical strong acid and base Fifty milliliters of 0.1 OM HNO3 are titrated with increasing amounts of 0.10M KOH. Data are given in Table 5-5. Notice how rapidly the pH changes in the region of the end point, or of exact neutralization of acid by base. Any acid-base indicator that changes color between pH 4 and pH 10 could be used to detect the equivalence point in this titration.
Original : nHNO3 = 50 ml $\times$ 0.10 mmole ml-1 = 5.0 mmoles
Added : nKOH = v ml $\times$ 0.10 mmole ml-1
Net Acid : nacid = 5.0 mmoles - 0.10v mmole
Total Volume : V = 50 + v ml
Hydrogen ion
concentration : [H+]net = $\textstyle\frac{5.0 - 0.10v}{50 + ''v''} = \frac{50-v}{50 + v}$(0.10) mmole ml-1
The calculation of [H+] for various values of v is shown in Table 5-5, and these calculations are plotted with open circles at the left of Figure 5-5. At the equivalence point, the amounts of acid and base are equal and the pH is 7.0. After the equivalence point, we only need to calculate how much base was added in excess of that required to neutralize the acid, and use this to find [OH-j, pOH, and pH:
Original : nHNO3 = 50 mmoles (as before)
Added : nKOH = v ml $\times$ 0.10 mmole ml-1
Net Acid : nbase = 0.10v - 5.0 mmoles
Total Volume : V = 50 + v ml
Hydrogen ion concentration :
[OH-] $\textstyle\frac{0.10v - 5.0}{50 + v} = \frac{v - 50}{v + 50}$(0.10) mmole ml-1
This calculation for several values of v and the corresponding pH values are listed in Table 5-5 and are plotted with solid circles on the right of Figure 5-5. It now is obvious why the choice of an indicator is not too critical in such a titration. Any indicator that changes color between pH 4 and pH 10 will do.
Titrating a weak acid with a strong base, or a weak base with a strong acid, is more complicated because the weak component is only partially dissociated. Dissociation equilibria of the type discussed in the next section must be used. We will not be concerned in this chapter with such titrations, but they are treated in Appendix 5, with an example of a titration curve corresponding to Figure 5-5.
Equilibria with Weak Acids and BasesEdit
Because weak acids are only partially dissociated in water, the contribution of a weak acid such as acetic acid to the hydrogen ion concentration is less than the total concentration of added acid. The equilibrium-constant expression for dissociation of the acid must be used explicitly. These general principles can be illustrated with a concrete example, that of calculating the pH of a solution of 0.0l00M acetic acid. As we saw in Example 5-4 for nitric acid, a strong acid, a 0.0100M solution has a pH of 2.00. Because acetic acid is a weak acid and only partially dissociated, a 0.0 100M solution will have a hydrogen ion concentration of less than 0.0100M, and a pH greater than 2.0.
It is common to represent the acetate ion, CH3COO-, simply by Ac-, and the undissociated acetic acid molecule, CH3COOH, by HAc as if it were a simple inorganic acid. (The forms OAc- and HOAc also are used, to indicate that acetic acid is an oxyacid with the dissociating proton attached to an oxygen atom.) The dissociation of HAc is incomplete:
HAc H+ + Ac-
and the equilibrium expression describing dissociation is
Ka = $\textstyle\frac{[H^+][Ac^-]}{[HAc]}$ = 1.76 $\times$ 10-5 (from Table 5-3)
We know the initial overall concentration, c0, of acetic acid:
c0 = 0.0100 mole liter-1
and we know that at equilibrium some of this acetic acid remains undissociated and some of it has ionized to acetate ions, Ac-:
c0 = [HAc] + [Ac-] (mass-balance equation)
This is called a mass-balance equation, because it states that total acetate is neither created nor destroyed during dissociation. We also know that the concentrations of hydrogen ions and acetate ions are equal, since dissociation of HAc is the only source of H+. (It is legitimate to neglect H+ from the dissociation of water, since acetic acid represses water dissociation even below its normal small extent.) Thus
[H+] = [Ac-] (charge-balance equation)
This is known as a charge-balance equation, because it states that the total positive charge in the solution must equal the total negative charge. We now can use these data about conservation of acetate and neutrality of the solution to simplify the equilibrium-constant expression. Let the hydrogen ion concentration that we are seeking be [H+] = y, and eliminate [Ac-] at once using the charge-balance equation:
Ka = $\textstyle\frac{y^2}{[HAc]}$ (eequilibrium equation)
c0 = [HAc] + y (mass-balance equation)
The second equation tells us that the concentration od undissociated HAc equals the original overall concentration, c0, minus the amount that has dissociated y:
[HAc] = c0 - y
The equilibrium expression then is
Ka = $\textstyle\frac{y^2}{c_0 -y}$ (5-34)
Substituting the value of Ka from Table 5-3, we get
1.76 $\times$ 10-5 = $\textstyle\frac{y^2}{0.0100 - y}$
or
y2 + 1.76 $\times$ 10-5y - $\times$ 10-7 = 0
This is a quadratic equation, which can be solved with the quadratic formula. If ay2 + by + c = 0, then
y = $\textstyle\frac{-b\pm \!\, \sqrt{b^2-4ac}}{2a}$ (5-35)
For this problem, a = 1, b = 1.76 $\times$ 10-5, and c = -1.76 $\times$ 10-7.
y = $\textstyle\frac{-1.76\times10^{-5}\pm \!\,\sqrt{3.10\times10^{-10} + 7.04 \times 10^{-7}}}{2}$
or
y = $\textstyle\frac{-1.76 \times 10^{-5}\pm \!\, 8.39 \times 10^{-4}}{2}$
Only the positive answer is reasonable, because one cannot have a negative concentration. Thus the answer is
y = 4.11 $\times$ 10-4 mole liter-1
Under certain physical conditions you can take a shortcut to avoid the quadratic formula. In this example, since you know that the acid is only slightly dissociated, you can try neglecting y in the denominator of the equilibrium expression for Ka, thereby assuming that it is small in comparison with 0.0100 mole liter-1, and that the concentration of undissociated acetic acid is virtually the same as the total acetic acid present. This assumption gives
1.76 $\times$ 10-5 = $\textstyle\frac{y^2}{0.0100}$
and an approximate answer of
y = 4.2 $\times$ 10-4 = 0.00042 mole liter-1
This is close to the correct answer of 0.000411 mole liter-1. You can make a quick improvement by using this approximate value in the undissociated acetate concentration in the denominator:
1.76 $\times$ 10-5 = $\textstyle\frac{y^2}{0.0100 - 0.00042}$
= 4.11 $\times$ 10-4 mole liter-1
Repetition of the foregoing process until the answer remains constant from one cucle to the next is called the method of successive approximation. If your intuition for how much dissociation the acid undergoes is good enough, you can often solve equilibrium problem by an approximate solution and a quick correction in less time than it takes to solve the quadratic formula. If your original guess is not so good, two or three cycles of approximation may be required before you arrive at an unchanging value for y.
As our results show, acetic acid is indeed only slightly dissociated at 0.0100M concentration. Of the initial 0.0100 mole liter-1, 0.000411 mole has dissociated, and 0.0096 mole remains as dissolved but undissociated HAc molecules. The percent dissociation is
$\textstyle\frac{4.11 \times 10^{-4} mole}{0.0100 mole} \times$ 100 = 4.11%
Since the hydrogen ion concentration is [H+] = 4.11 $\times$ 10-4 M, the pH of this solution is 3.39.
What happens if we dilute the acetic acid solution? Does a greater or lesser percent of the acetic acid then dissociate? Does the pH increase or decrease?
Example 8
What are the pH and percent dissociation in a solution of 0.00100M acetic acid?
Solution
The equilibrium expression is as before:
KaNaOH = $\textstyle\frac{y^2}{c_0 - y}$
1.76 $\times$ 10-5 = $\textstyle\frac{y^2}{0.00100 - y}$
Neglecting y in comparison with co, the approximate solution is
1.76 $\times$ 10-5 = $\textstyle\frac{y^2}{0.00100}$
y = 1.33 $\times$ 10-4 mole liter -1
and the solution obtained by using this value to correct the undissociated HAc concentration is
y = 1.24 $\times$ 10-4 mole liter-1
Using this second value to correctc0 in another cycle of approximation makes no change in y, so the process can be halted. Now the pH is 3.91 instead of 3.39, and the percent dissociation is
$\textstyle\frac{1.24 \times 10^{-4} mole}{0.00100 mole} \times 100 =$ 12.4%
Although the actual hydrogen ion concentration is lower (witness the larger pH), a greater fraction of the HAc present is dissociated into ions. This is Le Chatelier's principle again. If a solution containing HAc, H+, and Ac- is diluted, thereby lowering its total concentration of all ions and molecules, the equilibrium will attempt to reestablish itself, as reactions change, in the direction that will increase the total concentration of solute particles of one kind or another. Compare this behavior with the effect of increasing the pressure on the ammonia gas equilibrium in Chapter 4.
IndicatorsEdit
Figure 5-6 The basic form (a) and acid form (b) of the indicator methyl orange. The different colors of the two structures, yellow and red, give methyl orange its usefulness in displaying the pH of a solution into which it has been introduced. The complex structure can be symbolized by an ion, In-, which can combine with a proton as shown at the bottom of the figure.
An indicator is a weak acid (or a weak base) that has sharply different colors in its dissociated and undissociated states. Methyl orange (Figure 5-6) is a complex organic compound that is red n its neutral, un-ionizes form and yellow when ionized. It can be represented as the weak acid HIn:
HIn H+ + In-
red yellow
Adding acid shifts the indicator equilibrium to the left, and adding base shifts it to the right. Hence methyl orange is red in acids and yellow in bases.
The intensity of color from indicators such as methyl orange is so great that the colors can be seen easily even when the amount added to a solution is too small to have an appreciable influence on the pH of the solution. Nevertheless, the ratio of dissociated to undissociated indicator depends on the hydrogen ion concentration
Ka = $\textstyle\frac{[H^+][In^-]}{[HIn]}$ (5-36)
and
$\textstyle\frac{[In^-]}{[HIn]} = \frac{K_a}{[H^+]}$ (5-37)
log10($\textstyle\frac{[In^-]}{[HIn]}$) = pH - pka (5-38)
For methyl orange, Ka = 1.6 X 10-4 and pKa = 3.8. The neutral (red) and dissociated (yellow) forms of the indicator are present at equal concentrations when the pH = 3.8. The eye is sensitive to color changes over a range of concentration ratios of approximately 100, or over two pH units. Below pH 2.8, a solution containing methyl orange is red, and above approximately 4.8 it is clearly yellow. As you can see from Figure 5-5, an indicator change over two pH units is quite satisfactory for strong acid-base titrations.
Methyl orange could be used for the titration in Figure 5-5, even though its pKa is far from the titration equivalence point of 7.0, only because the change in pH at the equivalence point is so large. For titrations of weak acids, this would not be true, and it would be better to pick an indicator with a pKa closer to the expected equivalence point. Other indicators are shown in Figure 5-7, along with the pH range in which their color changes occur. Phenolphthalein is a particularly convenient and common indicator, which changes from colorless to pink in the range of pH 8 to 10.
Contribution to [H+] from Dissociation of WaterEdit
Figure 5-7 Some common acid-base indicators, with the pH ranges in which their color changes occur. The choice of an indicator for an acid-base titration depends on the expected pH at the equivalence point of the titration and the width of swing of pH values as the equivalence point is passed.
Nothing has been said in the discussions of either strong acids or weak acids about a contribution to the hydrogen ion concentration from the dissociation of water itself. It has been tacitly assumed that all H+ comes from the acid. This is a valid assumption for all but the most dilute solutions of very weak acids such as HCN. The correction for water dissociation seldom is necessary, so it will not be covered in this chapter. A complete treatment is given in Appendix 5.
Weak Acids and Their SaltsEdit
What will happen to a weak acid such as acetic acid if we add some sodium acetate (NaAc), which is the salt of a strong base (NaOH) and acetic acid? The salt will dissolve and dissociate completely into sodium and acetate ions. From Le Chatelier's principle, we would expect these added acetate ions to force the weak acetic acid equilibrium system in the direction of less dissociation. This is exactly what happens. The acid-equilibrium expression is the same:
Ka = $\textstyle\frac{[H^+][Ac^-]}{[HAc]}$ (5-39)
However, two sources of acetate ions now exist: NaAc and HAc. The acetate ion supplied by sodium acetate is measured by cs , the total molarity of the salt, since dissociation is complete. Acetate concentration from acetic acid is measured by the hydrogen ion concentration, since every dissociation of HAc to produce Ac- also produces a proton. Therefore, the total acetate ion concentration is
[Ac-]total = [Ac-]NaAc + [Ac-]HAc = cs + [H+] (5-40)
(Again, we have neglected any protons contributed by the dissociation of water.) The concentration of un-ionized acetic acid is the overall acid concentration, ca, less the acetate from dissociation:
[HAc] = ca - [Ac-]HAc = ca - [H+] (5-41)
If we represent the hydrogen ion concentration by y, we have
Ka = $\textstyle\frac{y(c_s + y)}{(c_a - y)}$ (5-42)
When the added salt concentration, cs, is zero, this is the simple weak acid-dissociation equilibrium expression that we have seen previously in equation 5-34.
Figure 5-8 The effect of added sodium acetate on the dissociation of acetic acid. Data plotted here are listed in Table 5-6, and were calculated as explained in the text. The first salt added represses acetic acid dissociation to a great extent and causes a rapid increase in pH. Later additions are not as effective.
Example 11
What are the pH and percent dissociation of a solution of 0.0l0M acetic
acid in the presence of (a) no NaAc, (b) 0.0050M NaAc, and (c) 0.010M NaAc?
Solution
From Le Chatelier's principle, we would expect the dissociation of HAc to be repressed as more NaAc is added. The pH should increase and the percent dissociation should decrease. (a) This problem was already solved in Section 5-6, yielding pH 3.39 and 4.11% dissociation. (b) For cs = 0.0050 mole liter-1,
1.76 $\times$ 10-5 = $\textstyle\frac{y(0.0050 + y)}{0.010 - y}$ (from equation 5-42)
This is most easily solved by successive approximations. As a first approximation we can assume that y will be smaller than 0.0050 or 0.010, and we can therefore neglect it when it is added to or subtracted from these quantities:
y1 = 1.76 $\times$ 10-5 $\times \textstyle\frac{0.010}{0.0050} =$ 3.52 $\times$ 10-5 = 0.000035 mole liter-1
As a second approximation, we can use this trial value of y tpo "correct" 0.0050 to 0.005035, and 0.010 to 0.009965, and solve the equation again:
y2 = 1.76 $\times$ 10-5 $\times\textstyle\frac{0.009965}{0.005035} =$ 3.48 $\times$ 10-5 mole liter-1
A third approximation is unnecessary, and the answer should be rounded to 3.5 ,math>\times[/itex] 10-5 mole liter-1:
pH = 5 - log10 3.5 = 5 - 0.54 = 4.46
Percent dissociation = $\textstyle\frac{3.5 \times 10^{-5}}{0.010} \times$ 100 = 0.35%
(c) For cs = 0.010 mole liter-1
y = [H+] = 1.76 $\times$ 10-5 mole liter-1
pH= 4.75
Percent dissociation = 0.18%
Notice that the acetic acid now dissociates so little that even the first approximation is adequate.
Results for these and a few other sodium acetate concentrations are listed in Table 5-6 and are plotted in Figure 5-8. The first salt added has a large effect on the degree of dissociation and pH; later additions of salt cause less change. When acid and salt are present in equal concentrations, the pH is equal to the pKa of the acid.
Table 5-6. Effect of Adding Sodium Acetate to 0.010M Acetic Acid Solution
cs = concentration of sodium acetate in moles liter-1
cs: 0.0 0.001 0.002 0.005 0.010 0.020
break
pH: 3.4 3.8 4.1 4.5 4.8 5.1
break
Percent dissociation 4.1 1.5 0.84 0.35 0.18 0.09
of acetic acid:
BuffersEdit
If the concentrations of a solution of a weak acid and a salt of the acid anion are reasonably high, then the solution is resistant to changes in hydrogen ion concentration.
Example 12
A solution is 0.050M in HAc and 0.050M in NaAc. Calculate the change in
pH when 0.0010 mole of hydrochloric acid (HCl) is added to a liter of solution, assuming that the volume increase upon adding the HCl is negligible. Compare this to the pH if the same amount of HCI is added to a liter of pure water.
Solution
Before adding HCl the acetic acid equilibrium is
Ka = $\textstyle\frac{[H^+][Ac^-]}{[HAc]} = \frac{y(0.050)}{(0.050)}$
Thus
y = Ka = 1.76 $\times$ 10-5 mole liter-1
pH = pKa = 4.75
(Again, we were justified in ignoring y in the [Ac-] and [HAc] terms because the value is small compared to 0.050.)
The added protons from HCl combine with the acetate ions to form more acetic acid:
Ac- + H+ (from HCl) HAc
Thus to a good approximation, all the added protons are used up, and the new acetic acid and acetate concentrations are
[HAc] = 0.050 + [H+]HCl = 0.051 mole liter<sup.-1
[Ac-1] = 0.050 - [H+]HCl = 0.049 mole liter-1
Ka = $\textstyle\frac{y(0.049)}{(0.051)}$
y = 1.76 $\times$ 10-5 $\times \textstyle\frac{0.051}{0.049} =$ 1.83 $\times$ 10-5 mole liter-1
pH = 5 - 0.26 = 4.74
The pH changes from 4.75 to 4.74, a difference of only 0.01 unit. In the absence of HAc and NaAc, the same concentration of HCl would produce a pH of 3.0.
This resistance to pH change is called buffering action, and the solution of HAc and NaAc is an acetate buffer. Buffers are used widely for pH control in laboratory chemistry, in the; chemical industry, and in living organisms. A carbonate buffer system in your bloodstream, involving the reaction
H+ + HCO$_3^-$ H2CO3 CO2 + H2O (5-43)
maintains the blood pH around 7.4. When a biochemist studies enzyme activity in the laboratory, he must use a buffer system to maintain a constant pH during the experiments, otherwise his results may have little meaning. One of the sillier disputes in commercial advertising is that between two pharmaceutical companies as to whether buffers added to aspirin to combat an acid reaction in the stomach are a benefit or an adulterant.
In general, if the concentration of strong acid added to a buffer solution is y moles liter-1, the equilibrium equation becomes
Ka = $\textstyle\frac{[H^+][A^-]}{[HA]} = \frac{[H^+](c_s - y)}{c_a + y}$ (5-44)
in which cs and ca are the salt and buffering acid concentrations, respectively. After addition of the foreign acid, the hydrogen ion concentration is
[H+] = Ka$\textstyle\frac{(c_a + y)}{(c_s - y)}$ (5-45)
and the pH is
pH = pKa + log10$\textstyle\frac{(c_s - y)}{(c_a + y)}$ (5-46)
If base is added, hydrogen ions are removed, and the same expressions can be used with a negative value of y.
Example 13
A formic acid buffer is prepared with 0.010 mole liter-1 each of formic acid
(HCOOH) and sodium formate (HCOONa). What is the pH of the solution? What is the pH if 0.0020 mole liter-1 of solid sodium hydroxide (NaOH) is added to a liter of buffer? What would be the pH of the sodium hydroxide solution without buffer? What would the pH have been after adding sodium hydroxide if the buffer concentrations had been 0.10 mole liter-1 instead of 0.0l0?
Solution
The answers are
Buffer: pH = 3.75
After adding NaOH: pH = 3.92
Without buffer: pH = 11.30
Without stronger buffer: pH = 3.77
In the preceding example, you can see the dramatic effect of the formate buffer in keeping the solution acidic in spite of the added base, and the importance of reasonably high buffer concentrations if the buffering capacity of the solution is not to be exceeded.
Salts of Weak Acids and Strong Bases : HydrolysisEdit
A sodium chloride solution is neutral, with a pH of 7.0. This is reasonable, because sodium hydroxide is a strong base and hydrochloric acid is a strong acid, and if equal amounts of each were added, neutralization would be complete. In contrast, sodium acetate is the salt of a strong base and a weak acid. Intuitively we would expect a sodium acetate solution to be somewhat basic, and it is. Some of the acetate ions from the salt combine with water to form undissociated acetic acid and hydroxide ions:
Ac- + H2O HAc + OH- (5-47)
This sometimes is called a hydrolysis reaction, the implication being that H2O breaks up crystals of sodium acetate. It does, when the salt crystal dissolves in water, but this is not the point. In solution the acetate ion acts as a base. It is as good a Bronsted base as ammonia, and the ammonium ion is a perfectly good acid, like HAc.
NH3 + H2O NH$_4^+$ + OH-
We should not let the different charges on the acetate ion (-1) and ammonia (0) obscure the similarity of their acid-base behavior.
The equilibrium constant for acetate hydrolysis has the same form as any other base dissociation:
Kb = $\textstyle\frac{[HAc][OH^-]}{[Ac^-]}$ (5-48)
Kb = $\textstyle\frac{[NH_4^+][OH^-]}{[NH_3]}$ (5-30)
where, as usual, the virtually unchanging water concentration is incorporated into the equilibrium constant. This constant sometimes is written Kh for "hydrolysis constant," but the added nomenclature is unnecessary. It is a simple base-equilibrium constant of the kind we have seen before, except that acetate ion is the base.
As always, Kb is related to the cooresponding acid-dissociation constant, Ka, by
Kb = $\textstyle\frac{[HAc][OH^-]}{[Ac^-]} = \frac{[HAc][OH^-][H^+]}{[Ac^-][h^+]} = \frac{K_w}{K_a}$ (5-49)
(Recall the ammonia-water equilibrium expressions at the end of Section 5-4.) This value is all we need to calculate the pH of a sodium acetate solution.
Example 14
What is the pH of a solution of 0.010M NaAc?
Solution
Acetate ions from NaAc combine with H20 to produce undissociated HAc molecules and OH- ions (equation 5-47). The equilibrium expression is
5.68 $\times$ 10-10 = $\textstyle\frac{[HAc][OH^-]}{[Ac^-]}$
Let the hydroxide ion concentration be y. Since every reaction of an acetate ion with water produces one hydroxide ion and one undissociated HAc molecule, the concentration of each of the latter two species must bey moles liter-1. The remaining acetate ions are those originally present from NaAc minus those that have combined with water:
[Ac-] = 0.010 - y
and we arrive at the familiar expression
Kb = $\textstyle\frac{y^2}{c_s - y} = \frac{y^2}{0.010 - y} =$ 5.68 $\times$ 10-10 (5-50)
This is even easier to solve than the weak-acid problems. Since the equilibrium constant is so small, y will be correspondingly small and can be neglected in the denominator in comparison to 0.010. The result is
y2 = 0.010 $\times$ 5.68 $\times$ 10-10 = 5.7 $\times$ 10-12
y = 2.4 $\times$ 10-6 mole liter-1 = [OH-]
[H+] = $\textstyle\frac{K_w}{[OH^-]} = \frac{10^{-14}}{2.4 \times 10^{-6}}$ = 4.2 $\times$ 10-9
pH = 9 - 0.6 = 8.4
(As before, we have neglected any contribution to the hydrogen ion concentration from water molecules. Our procedure is accurate enough for most situations, including the purposes of this chapter. The full derivation is found in Appendix 5.)
Polyprotic Acids: Acids That Liberate More Than One Hydrogen IonEdit
If water is the solvent, sulfuric acid, H2SO4, loses one proton as a strong acid with an immeasurably large dissociation constant.
H2SO4 H+ + HSO$_4^-$
It also can lose a second proton as a weak acid with a measurable dissociation constant. Acids that can liberate more than one proton are called polyprotic acids.
HSO$_4^-$ H+ + SO$_4^{2-}$Ka2 = 1.20 $\times$ 10-2 pKa2 = 1.92
For carbonic acid, H2C03, both dissociations are weak:
H2CO3 H+ + HCO$_3^-$Ka1 = 4.3 $\times$ 10-7 pKa1 = 6.37
HCO$_3^-$ H+ + CO$_3^{2-}$Ka2 = 5.61 $\times$ 10-11 pKa2 = 10.25
The relative values of Ka1 and Ka2 for a given acid are intuitively reasonable. One would expect HCO$_3^-$, which already has a negative charge, to be less ready than neutral H2C03 to lose another proton.
Phosphoric acid, H3 PO4, has three dissociations:
H3PO4 H+ + H2PO$_4^-$ pKa1 = 2.12
H2PO$_4^-$ H+ + HPO$_4^{2-}$ pKa2 = 7.21
HPO$_4^{2-}$ H+ + PO$_4^{2-}$ pKa3 = 12.67
Thus, in an aqueous solution of phosphoric acid there will be seven ionic and molecular species present: H3PO4 , H2PO$_4^-$, HPO$_4^{2-}$, PO$_4^{3-}$, H2O, H+, and OH-. Life might appear impossibly complicated, were we not able to make some approximations.
At a pH equal to the pKa for a particular dissociation, the two forms of the dissociating species are present in equal concentrations. For the second dissociation of phosphoric acid, for which pKa2 = 7.21,
Ka2 = $\textstyle\frac{[H^+][HPO_4^{2-}]}{[H_2PO_4^-]}$
log$\textstyle\frac{[HPO_4^{2-}]}{[H_2PO_4^-]}$ = pH - pKa2
When pH = pKa2, we have the ratio
$\textstyle\frac{[HPO_4^{2-}]}{[H_2PO_4^-]}$ = 1.00
Hence, in a neutral solution, H2PO$_4^{2-}$ and HPO$_4^-$ are present in about the same concentrations. Very little undissociated H3PO4 will be found, since from the first dissociation constant,
Ka1 = $\textstyle\frac{[H^+][H_2PO_4^-]}{[H_3PO_4]}$
log$\textstyle\frac{[H_2PO_4^-]}{[H_3PO_4]}$ = pH - pKa1 = 7.00 - 2.12 = 4.88
$\textstyle\frac{[H_2PO_4^-]}{[H_3PO_4]}$ = 104.88 = 7.6 $\times$ 104 = 76,000
Similarly, little PO$_4^{3-}$ will exist:
log$\textstyle\frac{[PO_4^{3-}]}{[HPO_4^{2-}]} =$ pH - pKa3 = 7.00 - 12.67 = -5.67
$\textstyle\frac{[PO_4^{3-}]}{[HPO_4^{2-}]} =$ 10-5.67 = 2.1 $\times$ 10-6 = $\textstyle\frac{1}{480,000}$
The only phosphate species that we have to consider near pH = 7 are H2PO$_4^{2-}$ and HPO$_4^{2-}$. Similarly, in strong acid solutions near pH = 3, only H3P04 and H2P0$_4^-$ are important. As long as the pKa's of successive dissociations are separated by three or four units (as they almost always are), matters are simplified.
There is still another simplification. When a polyprotic acid such as carbonic acid, H2CO3, dissociates, most of the protons present come from the first dissociation:
H2CO,sub>3 H+ + HCO$_3^-$ pKa1 = 6.37
Since the second dissociation constant is smaller by four orders of magnitude (and the pKa2 larger by four units), the contribution of hydrogen ions from the second dissociation will be only one ten-thousandth as large. Correspondingly, the second dissociation has a negligible effect on the concentration of the product of the first dissociation, HCO$_3^-$.
Example 15
At room temperature and 1 atm CO2 pressure, water saturated in CO2 has a
carbonic acid concentration of approximately 0.040 mole liter-1. Calculate the pH and the concentrations of all carbonate species for a 0.040 M H2CO3 solution.
Solution
Considering initially only the first dissociation:
Ka1 = 4.3 $\times$ 10-7 = $\textstyle\frac{y^2}{0.0040 -y}$ in which y = [H^+]
From our experience with acetic acid, which has an even larger Ka, we should expect to be able to neglect y in the denominator. The extent of dissociation of an acid with such a small Ka will be very small:
y2 = 4.3 $\times$ 10-7 $\times$ 0.040 = 1.7 $\times$ 10-8
y = 1.3 $\times$ 10-4 mole liter-1
This is the concentration of both hydrogen ion and bicarbonate ion, HCO$_3^-$:
[H+] = 1.3 $\times$ 10-4 mole liter-1
[HCO$_3^-$] = 1.3 $\times$ 10-4 mole liter-1
[H2CO3] = 0.040 - 0.00013 = 0.040 mole liter-1
pH = 4 - 0.12 = 3.88
Consequently, carbonated beverages have an acidity somewhere between those of wine and tomato juice (see Table 5-2). For the second dissociation:
HCO$_3^-$ H+ + CO$_3^{2-}$
Ka2 = 5.6 $\times$ 10-11 = $\textstyle\frac{[H^+][CO_3^{2-}]}{[HCO_3^-]}$
Since this second dissociation has only a minor effect on the first one, we can assume that the hydrogen ion and bicarbonate ion concentrations are effectively the same:
[CO$_3^{2-}$] = $\textstyle\frac{[HCO_3^-]}{[H^+]} \times$ Ka2 = Ka2 = 5.6 $\times$ 10-11 mole liter-1
Note the rather surprising result that the concentration of the second dissociation product is equal to the second dissociation constant!
Example 16
Calculate the sulfide ion concentration in a solution saturated in H2S
(0.10 mole liter-1, which may be gotten from the application of Henry's Law for the solubility of gases in water) (a) if the solution is made from distilled water and (b) if the solution is made pH = 3.0 with HCl. Use Ka values in Table 5-3.
Solution
In distilled water, the first dissociation is
Ka1 = 9.1 $\times$ 10-8 = $\textstyle\frac{y^2}{0.10}$
The dissociation constant is so small that y in the denominator can be neglected immediately. Dissociation will be extremely slight:
y = [H+] = [HS-] = 9.5 $\times$ 10-5 mole liter-1
pH = 5 - 0.98 = 4.02
From the second dissociation:
[S^{2-}] = $\textstyle\frac{[HS^-]}{[H^+]} \times$ Ka2 = Ka2 = 1.1 $\times$ 10-12 mole liter-1
As with the H2CO3 example, the anion produced by the second dissociation has a concentration equal to the second dissociation constant.
In contrast, in HCl solution at pH = 3.0:
Ka1 = $\textstyle\frac{[H^+][HS^-]}{[H_2S]} = \frac{1.0 \times 10^{-3} [HS^-]}{0.10} =$ 9.1 $\times$ 10-8
[HS-] = 9.1 $\times$ 10-6 mole liter-1
Ka2 = $\textstyle\frac{[H^+][S^{2-}]}{[HS^-]} = \frac{1.0 \times 10^{-3}[S^{2-}]}{9.1 \times 10^{-6}}$ = 1.1 $\times$ 10-12
[s2-] = $\textstyle\frac{9.1 \times 10^{-6} \times 1.1 \times 10^{-12}}{1.0 \times 10^{-3}} =$ 1.0 $\times$ 10-14
The acid has repressed the dissociation of H2S, making the sulfide ion concentration only one-hundredth of what it is in pure water. As we shall see in the next section, we can use acids to exert a fine control on sulfide concentration in analytical methods by controlling the pH.
Equilibria With Slight Soluble SaltsEdit
When most solid salts dissolve in water, they dissociate almost completely into hydrated positive and negative ions. The solubility of a salt in water represents a balance between the attraction of the ions in the crystal lattice and the attraction between these ions and the polar water molecules. This balance may be a delicate one, easily changed in going from one compound to an apparently similar one, or from one temperature to another. It is not possible to give hard-and-fast rules as to whether a compound is soluble, or even to account for all observed behavior.
One important factor certainly is the electrostatic attraction between ions. Crystals made up of small ions that can be packed closely together are generally harder to pull apart than crystals made up of large ions. Therefore, for a given cation, fluorides (F-) and hydroxides (OH-) are less soluble than nitrates (NO$_3^-$) and perchlorates (ClO$_4^-$). Chlorides are intermediate in size, and their behavior is difficult to predict from general principles.
The charge on the ions also is important. More highly charged ions such as phosphates (PO$_4^{3-}$) and carbonates (CO$_3^{2-}$) interact strongly with cations and are less soluble than the singly charged nitrates and perchlorates.
The terms soluble and insoluble are relative, and the degree of solubility can be related to an equilibrium constant. For a "slightly soluble" salt such as silver chloride, an equilibrium exists between the dissociated ions and the solid compound:
AgCL(s) Ag+ + Cl- (5-51)
The equilibrium expression for this reaction is
Keq = $\textstyle\frac{[Ag^+][Cl^-]}{[AgCl(s)]}$ (5-52)
As long as solid AgCl remains, its effect on the equilibrium does not change. As with the H2O concentration in the water dissociation equilibrium, the concentration of the solid salt can be incorporated into the equilibrium constant:
Ksp = Keq[AgCl(s)] = [Ag+][Cl-] (5-53)
This new equilibrium constant, Ksp, is called the solubility-product constant. For substances in which the ions are not in a 1: 1 ratio, the form of the solubility-product expression is analogous to our previous equilibrium expression:
PbCl2 Pb2+ + 2Cl- Ksp = [Pb2+][Cl-]2
Al(OH)3 Al3+ + 3OH- Ksp = [Al3+][OH-]3
Ag2CrO4 2Ag+ + CrO$_4^{2-}$ Ksp = [Ag+]+[CrO$_4^{2-}$]
Ba3(PO4)2 3Ba2+ + 2PO$_4^{3-}$ Ksp = [Ba2+]3[PO$_4^{3-}$]2
Solubility equilibria are useful in predicting whether a precipitate will form under specified conditions, and in choosing conditions under which two chemical substances in solution can be separated by selective precipitation.
The solubility-product constant of a slightly soluble compound can be calculated from its solubility in moles liter-1.
Example 17
The solubility of AgCl in water is 0.000013 mole liter-1 at 25°C. What is its solubility-product constant, Ksp?
Solution
The equilibrium expression is
AgCl Ag+ + Cl-
The concentrations of Ag+ and Cl- are equal because for each mole of solid AgCl that dissolves, 1 mole each of Ag+ and Cl- ions is produced. Hence the concentration of each ion is equal to the overall solubility, s, of the solid in moles liter-1:
[Ag+] = [Cl-] = s = 1.3 $\times$ 10-5 mole liter-1
Ksp = [Ag+][Cl-] = s2 = 1.7 $\times$ 10-10
Example 18
At a certain temperature the solubility of Fe(OH)2 in water is 7.7 $\times$ 10-6 mole liter-1. Calculate its Ksp at that temperature.
Solution
The equilibrium equation is
Fe(OH)2 Fe2 + 2OH-
and the solubility-product expression is
Ksp = [Fe2+][OH-]2
Since one mole of dissolved Fe(OH)2 produces one mole of Fe2+ and two moles of OH-,
[Fe2+] = s = 7.7 $\times$ 10-6 mole liter-1
[OH-] = 2s = 1.54 $\times$ 10-5 mole liter-1
Ksp = 7.7 $\times$ 10-6 $\times$ (1.54 $\times$ 10-5)2 = 1.8 $\times$ 10-15
The solubility-product constants of a number of substances are listed in Table 5-7. Substances are grouped by anion and listed in the order of decreasing Ksp; anions are listed roughly in the order of decreasing solubility. Once the solubility-product constant is known, it can be used to calculate the solubility of a compound at a specified temperature.
Example 19
What is the solubility of lead sulfate, PbS04 , in water at 25°C?
Solution
The dissociation reaction is
PbSO4 Pb2+ + SO$_4^{2-}$
Let the unknown solubility be s moles liter-1. Then since each mole of dissolved PbS04 produces 1 mole of each ion,
[Pb2+] = [SO$_4^{2-}$] = s
The solubility-product equation is
Ksp = [Pb2+][SO$_4^{2-}$] = s2 = 1.3 $\times$ 10-8 (from Table 5-7)
s = 1.1 $\times$ 10-4 mole liter-1
Example 20
In Table 5-7 we see that cadmium carbonate, CdC03 , and silver carbonate,
Ag2CO3, have approximately the same solubility-product constants. Compare their molar solubilities in water (at 25°C).
Solution
For cadmium carbonate,
Ksp = [Cd2+][CO$_3^{2-}$] = s2 = 5.2 $\times$ 10-12
s = 2.3 $\times$ 10-6 mole liter-1
For Ag2CO3 the expression is slightly different. If the solubility again is s moles liter-1, since each mole of salt produces 2 moles of Ag+ ions,
[Ag+] = 2s
[CO$_3^{2-}$] = s
Ksp = [Ag+]2[CO$_3^{2-}$] = (2s)2 $\times$ s = 4s3 = 8.2 $\times$ 10-12
s = 1.3 $\times$ 10-4 mole liter-1
Although cadmium carbonate and silver carbonate have nearly the same solubility-product constants, their solubilities in moles liter-1 differ by a factor of 100 because the form of the solubility-product expression is different. The solubility of Ag2CO3 is sensitive to the square of the metal-ion concentration, because two silver ions per carbonate ion are necessary to build the solid crystal.
Common-Ion EffectEdit
In the preceding example, the solubility of silver carbonate in pure water was calculated to be 1.3 $\times$ 10-4 mole liter-1. Will silver carbonate be more soluble or less soluble in silver nitrate solution? Le Chatelier's principle leads us to predict that a new, outside source of silver ions would shift the silver carbonate equilibrium reaction in the direction of less dissociation:
Ag2 CO3 2Ag+ + CO$_3^{2-}$ (5-54)
or that silver carbonate would be less soluble in a silver nitrate solution than in pure water. This decrease in the solubility of one salt in a solution of another salt that has a common cation or anion is called the common-ion effect.
Example 21
What is the solubility at 25°C of calcium fluoride, CaF2, (a) in pure water, (b) in 0.10M calcium chloride, CaCl2, and (c) in 0.10M sodium fluoride, NaF?
Solution
(a) If the solubility in pure water is s, then
{Ca2+] = s
[F-] = 2s
Ksp = s $\times$ 4s2 = 4s3 = 3.9 $\times$ 10-11
s = 2.1 $\times$ 10-4 mole liter-1
(b) In 0.01M CaCl2, the calcium ion concentration is the sum of the concentration of calcium ions from calcium chloride and from calcium fluoride, whose solubility we are seeking:
[Ca2+] = 0.10 + s
[F-] = 2s
Ksp = (0.10 + s)(2s)2 = 3.9 $\times$ 10-11
This is a cubic equation, but a moment's thought about the chemistry involved will eliminate the need to solve it as such. With such a small solubility-product constant, you can predict that the solubility of calcium fluoride will be very small in comparison with 0.10 mole liter-1. (You already should realize from (a) and Le Chatelier's principle that in this problem s will be less than 2.1 $\times$ 10-4 mole liter-1.) If our prediction is valid, we can simplify the solubility-product equation and calculate the approximate solubility:
0.10 $\times$ (2s)2 = 3.9 $\times$ 10-11
s2 = $\textstyle\frac{3.9 \times 10^{-11}}{4 \times 0.10}$ = 9.75 $\times$ 10-11
s = 0.99 $\times$ 10-5 = 9.9 $\times$ 10-6 mole liter-1
Therefore the approximation is justified. Only 4.7% as much CaF2 will dissolve in 0.10M CaCl2 as in pure water:
$\textstyle\frac{9.9 \times 10^{-6}}{2.1 \times 10^{-4}} \times$ 100 = 4.7%
(c) In 0.10M NaF,
[Ca2+] = s and [F-] = 0.10 + 2s
since fluoride ions come from NaF as well as from CaF2. The solubility-product equation is
Ksp = s(2s + 0.10)2 = 3.9 $\times$ 10-11
Again, thinking about the chemical meaning will avoid the necessity of solving a cubic equation. The 2s term will be very small compared to 0.10 mole liter-1, therefore,
s(0.10)2 = 3.9 $\times$ 10-11
s = 3.9 $\times$ 10-9 mole liter-1
This approximation is even more valid than the previous one, since from the calculation
$\textstyle\frac{3.9 \times 10^{-9}}{2.1 \times 10^{-4}} \times$ 100 = 0.0019%
only 0.0019% as much CaF2 will dissolve in 0.10M NaF as in pure water. Fluoride is more effective than calcium as a common ion because it has a second-power effect on the solubility equilibrium.
The common-ion method of controlling solubility often is used with solutions of sulfide ion, S2-, because many metals form insoluble sulfides, and the sulfide ion concentration can be controlled by adjusting the pH.
Example 22
What is the maximum possible concentration of Ni2+ ion in water at 25°C that is saturated with H2S and maintained at pH 3.0 with HCl?
Solution
From the solubility-product equilibrium equation we predict that too much nickel ion will cause the precipitation of nickel sulfide, NiS:
Ksp = [Ni2+][S2-] = 3 $\times$ 10-21
The only new twist to this problem is finding the sulfide ion concentration from the H2S equilibrium. Hydrogen sulfide dissociates in two steps, each with an equilibrium constant:
H2S H+ + HS- Ka1 = 9.1 $\times$ 10-8
HS- H+ + S2- Ka1 = 1.1 $\times$ 10-12
H2S 2H+ + S2- Ka1.2 = Ka1 $\times$ Ka2
Because the overall dissociation is the sum of two dissociation steps, the overall equilibrium constant, Ka1.2, is the product of Ka1 and Ka2:
Ka1.2 = $\textstyle\frac{[H^+][HS^-]}{[H_2S]} \times \frac{[H^+][S^{2-}]}{[HS^-]} = \frac{[H^+]^2[S^{2-}]}{[H_2S]}$
Ka1.2 = 9.1 $\times$ 10-8 $\times$ 1.1 $\times$ 10-12 = 1.0 $\times$ 10-19
Saturated H2S is approximately 0.10M at 25°C (which can be gotten from Henry's Law for the solubility of gases in water), and the very small value of Ka1.2 means that dissociation of H2S is very slight. Hence we can write
[H2S] = 0.10 mole liter-1 and [H+]2[S2-] = 1.0 $\times$ 10-20
in a saturated H2S solution. This "ion product" for saturated H2S is a useful relationship to remember.
In this problem, the pH has been adjusted to 3.0 with hydrochloric acid, so
[H+] = 1.0 $\times$ 10-3 mole liter-1
Therefore, the sulfide ion concentration can be calculated from
[S2-] = Ka1.2 $\times \textstyle\frac{[H_2S]}{[H^+]^2} =$ 1.0 $\times$ 10-19 $\times \textstyle\frac{0.10}{(1.0 \times 10^{-3})^2}$
which gives
[S2-] = 1.0 $\times$ 10-14 mole liter-1
Since NiS will precipitate if the solubility product is exceeded, the highest value that the nickel ion concentration can have is
[Ni2+] = $\textstyle\frac{K_{sp}}{S^{2-}} = \frac{3 \times 10^{-21}}{1 \times 10^{-14}}$ = 3 $\times$ 10-7 mole liter-1
Separation of Compounds by PrecipitationEdit
Solubility-product constants can be used to devise methods for separating ions solution by selective precipitation. The entire traditional qualitative-analysis scheme is based on the use of these equilibrium constants to determine the correct precipitating ions and the correct strategy.
Example 23
A solution is 0.010M in barium chloride, BaCl2, and 0.020M in strontium chloride, SrCl2. Can either Ba2+ or Sr2+ be precipitated selectively with concentrated sodium sulfate, Na2SO4, solution? Which ion will precipitate first? When the second ion just begins to precipitate, what is the residual concentration of the first ion, and what fraction of the original amount of the first ion is left in solution? (For simplicity, assume that the Na2SO4 solution is so concentrated that the volume change in the Ba-Sr solution can be neglected.)
Solution
The upper limit on barium sulfate solubility is given by
Ksp = [Ba2+][SO$_4^{2-}$] = 1.5 $\times$ 10-9
With 0.010 mole liter-1 of Ba2+, precipitation of barium sulfate will not occur until the sulfate ion concentration increases to
[SO$_4^{2-}$] = $\textstyle\frac{1.5 \times 10^{-9}}{0.010}$ = 1.5 $\times$ 10-7 mole liter-1
Strontium sulfate will precipitate when the sulfate concentration is
[SO$_4^{2-}$] = $\textstyle\frac{K_{sp(SrSO_4)}}{[Sr^{2+}]} = \frac{7.6 \times 10^{-7}}{0.020} =$ 3.8 $\times$ 10-5 mole liter-1
Therefore, barium will precipitate first. When the sulfate concentration has risen to 3.8 $\times$ 10-5 mole liter-1 and strontium sulfate just begins to precipitate, the residual barium concentration left in solution will be
[Ba2+] = $\textstyle\frac{1.5 \times 10^{-9}}{3.8 \times 10^{-5}}$ = 3.9 $\times$ 10-5 mole liter-1
The quantity is
$\textstyle\frac{3.9 \times 10^{-5}}{0.010} \times$ 100 = 0.39%
or 0.39% of the original Ba2+ present. Thus 99.6% of the barium has been precipitated before any strontium begins to precipitate.
SummaryEdit
In this chapter we have applied the concepts of chemical equilibrium to ions in aqueous solution, especially to acid-base and precipitation reactions. We have used the equilibrium-constant expression from Chapter 4, with concentrations expressed in units of moles per liter (moles liter-1). Since the concentration of water is effectively constant, especially in dilute solutions, we have incorporated all water concentration terms, [H20], into the equilibrium constants.
Water itself ionizes with an equilibrium or ion-product constant at room temperature of Kw = [H+][OH-] = 10-14. To avoid the inconvenience of large exponential numbers, a negative exponent notation is used, whereby pH = -log10[H+], pOH = -log10[OH-], and pKeq = -log10 Keq. In this notation, the dissociation of water can be represented by pH + pOH = pKw = 14. For pure water, [H+] and [OH-] must be the same, and equal to 10-7 mole liter-1, so the pH and pOH each are equal to 7. The pH is a convenient measure of acidity, since in acid solutions the pH is less than 7, and in basic solutions it is greater than 7.
According to the Bronsted-Lowry theory of acids and bases, any substance that gives up a proton is an acid, and any substance that can combine with a proton and remove it from solution is a base. When an acid loses its proton, it becomes the conjugate base. A strong acid such as HCl has a weak conjugate base, Cl-, and a weak acid such as HAc or NH$_4^+$ has a relatively strong conjugate base, Ac- or NH3. Any acid whose conjugate base is sufficiently weaker than H2O (has a lesser affinity for H+) will be dissociated completely in aqueous solution, hence it is classified as a strong acid. Acids that dissociate only partially in aqueous solution are weak acids.
Strong acids and bases are simple to deal with, since their dissociation is complete in aqueous solution. When a strong acid is added to water, the increase in hydrogen ion concentration equals the concentration of added acid. Neutralization occurs when H+ from an acid combines with OH- from a base to form water molecules. The amount of acid present in a sample can be determined by finding out how much base of known strength is required to make the solution neutral as measured by an acid-base indicator. This is called titration, and it is a useful analytical procedure.
The equilibrium expression for a weak acid, equation 5-34, is obtained with the help of two conservation expressions: a mass-balance equation that says the total amount of acid anion is constant, and a charge-balance equation that says the solution must remain neutral as a whole. This simple expression can be solved as a quadratic equation or by the method of successive approximations, and it is valid as long as the solution is so acid that the contribution to [H+] from the dissociation of water can be neglected. If this is not the case, then a more complete expression (Appendix 5) must be used. Acid-base indicators themselves are weak acids or weak bases whose dissociated and undissociated forms have different colors.
A buffer is a mixture of a weak acid and its salt with a strong base, or alternatively, of a weak base and its salt with a strong acid. The equilibrium between the acid and salt form of the substances shifts to counteract the effect of added acid or base, making the buffered solution resistant to pH change. The pH in such solutions can be calculated from equations 5-42 and 5-46.
Hydrolysis is the interaction of the salt of a weak acid (or weak base) with water to form undissociated acid (or base) and OH- (or H+) ions. What is sometimes described as a hydrolysis constant is actually nothing more than the dissociation constant for the conjugate of the weak acid or base. The base constant, Kb, and the acid constant, Ka, are related by KaKb = Kw.
Some acids can release more than one proton in successive dissociations. These are called polyprotic acids. As long as the successive dissociation constants, K1, K2 , and so on, differ by factors of 10-4 or 10-5, the successive dissociations can be treated as separate events.
Most of the general comments just made about solving acid-base equilibrium problems are applicable to solubility equilibria, for situations in which ions combine to form an insoluble salt. Solubility-product calculations are more useful to indicate whether precipitation will occur under certain conditions, what the upper limit on concentration of an ion in solution may be, and whether two ions can be separated in solution by selective precipitation. | textbooks/chem/General_Chemistry/Chemical_Principles_(Dickerson)/05%3A_Solution_Equilibria%3A_Acids_and_Bases/5.01%3A_Hydrogen_Oxygen_and_Water.txt |
The atomic hypothesis provides a convenient
form of speech, which succesifully
describes many of the facts in a metaphorical
manner. But the handy way in
which the atomic hypothesis lends itself
to the representation of the characteristic
features of a chemical change falls
short of constituting a proof that atoms
have any real existence.
Alexander Smith, Professor of
Chemistry, University of Chicago
(1910)
Introduction
In the first five chapters you encountered some of the most fundamental ideas in chemistry: atoms, molecules, moles, conservation of mass and energy, behavior of gases, kinetic theory, equilibrium, and acid-base chemistry in solution. All these ideas have been presented in a very dogmatic way, without proof of any kind. It is time now to stop being a believer and to become a skeptic. How do we know that the material in the first five chapters is true? How do we know, for example, that the molecular formula for water is H20? After all, the best chemists in the world thought it was HO for a full 58 years after the atomic theory was proposed in 1802 by John Dalton. Why did they change their minds? What gives us the right to assert that one atom of carbon is approximately 12 times as heavy as one atom of hydrogen? It is not easy to think of ways to weigh out equal numbers of atoms without the mole concept, and this concept depends on the existence of a reliable set of atomic weights, which brings us back in a circle to the relative weights of carbon and hydrogen. How can the circle be broken?
How are the atomic numbers for elements obtained? Why should atoms with the same atomic number but different atomic weights (isotopes) have so nearly identical chemical properties that they are given the same symbol and classed as one element? What evidence is there that the negative charges in an atom are on the outside, and the positive charges are grouped in a tiny central nucleus that contains virtually all the mass of the atom? And what do we mean by the radius of an atom? Is not the size of an atom as difficult to measure as its weight? What laboratory measurements can be related to such microscopic dimensions, and how can we be sure that the relationship is correct?
How, in fact, do we know that atoms exist at all? How do we really know that everything said so far is not the product of the chemist's hyperactive imagination? Perhaps Professor Smith, the author of our chapter opening quote, was right. Alchemists explained chemical reactions in terms of mythological figures or planets (the distinction was not clear in their own minds) that they associated with the reagents: gold with the sun, copper with Venus, iron with Mars, tin with Jupiter, and lead with Saturn. In what way are atoms more successful models than Greek gods? And how are hydrogen, helium, lithium, beryllium, and so on really more satisfactory as "fundamental materials" than the earth, air, fire and water of Empedocles in ancient Greece?
We have already mentioned Faraday's experiments with ions and electrolysis, and Thomson's and Millikan's measurements of electron charge and mass, in Chapter 1. The tremendous achievement of Mendeleev and Meyer in building the periodic table of the elements is the subject of Chapter 7. The work of Rutherford, Bohr, Schrödinger, and others in developing the modern theory of atomic structure and bonding is described in Chapter 8. In this chapter we shall go back even farther, and focus on two men who revolutionized chemistry: Antoine Lavoisier (1743-1794), who demonstrated that the fundamental quantity in any chemical reaction is mass) and John Dalton (1766-1844), who proposed that the fundamental units in chemical reactions are atoms. Dalton was not the first to propose the idea of atoms in principle, but he was the first to show in a convincing way that atoms do exist, and that they are a useful basis for understanding chemical reactions.
This chapter is an exercise in both chemical history and chemical understanding-the two frequently go hand in hand. One of the guiding principles of this book is that knowing how chemical concepts evolved helps to make them more comprehensible and more interesting. Such historical material is usually presented in postscripts at the ends of chapters. In effect, this chapter is one long postscript to the first five chapters. As you travel through the rest of this book, study and learn the material in the chapters, and relax and enjoy the postscripts.
The Concept of An Element
Figure 6-1 The Greeks of the fifth century B.C. pictured all material substances as composed of different proportions of the four basic elements: earth, air, fire, and water. These elements shared, in pairs, the properties of heat or cold and wetness or dryness: Earth was cold and dry, water was cold and wet, air was hot and wet, and fire was hot and dry.
One of the oldest ideas in science is that of fundamental materials out of which everything else is made. Empedocles (500 B.C.), in Greece, performed what may be the first recorded chemical analysis. He noted that when wood burns, smoke or air rises first and is followed by flame or fire. Water vapor will condense on a cool surface held near the flame. After combustion, the remains are ash or earth. Empedoc1es interpreted combustion as a breaking down of the burning substance into its four elements: earth, air, fire, and water. He and later writers generalized these into the four elements of which all substances were composed in varying proportions (Figure 6-1). Originally, at least, these ideas were not meant to be flights of metaphysical invention, but were attempts to explain observations. Later, among the Greek, Arabic, and medieval alchemists, the ideas become imbued with mysticism. Then earth, air, fire, and water were abandoned as fundamental elements, but varying sets of what we now would call elements or simple compounds were chosen by different alchemists as the fundamental materials of nature.
Aristotle (384 -322 B.C.) gave a definition of an element that, even now, can hardly be improved:
"Everything is either an element or composed of elements.... An element is that into which other bodies can be resolved, and which exists in them either potentially or actually, but which cannot itself be resolved into anything simpler, or different in kind."
However, this definition doesn't answer the question of how to recognize an element when you encounter one. Robert Boyle (1627- 1691) gave a more practical definition: An element is a substance that will always gain weight when undergoing chemical change. This statement must be understood in the sense in which it was intended. For example, when iron rusts, the iron oxide produced weighs more than the original iron. Yet the weigh t of the iron and the oxygen that combines with it is exactly the same as the weight of the iron oxide. Conversely, when the red powder of mercuric oxide is heated, oxygen gas is emitted, and the silvery liquid mercury that remains weighs less than the original red powder. But if this decomposition takes place in an enclosed flask, one sees that there is no overall loss of weight during the reaction. (It was a century after Boyle that Lavosier made careful weighing experiments demonstrating the conservation of mass in such reactions. )
By Boyle's definition, mercuric oxide could not be an element, because it can be decomposed into parts, each of which is lighter than the original substance. Mercury could provisionally be called an element, at least until the day when someone else succeeded in separating it into components. Until the present century of spectroscopy and other laboratory techniques, it was easy to prove that a substance was not an element, but impossible to prove that one was. As the famous German chemist Justus von Liebig wrote, in 1857, "The elements count as simple substances not because we know that they are so, but because we do not know that they are not."
The elements called the rare earths provide an example of the difficulties of proving by purely chemical means that a substance is an element. In 1839, the Swedish chemist Carl Mosander extracted a new element from cerium nitrate and named it lanthanum (from the Greek lanthanein, "to lie hidden"). Two years later he showed that his lanthanum-containing preparation contained a second element which he christened didymium (from the Greek didymos, or "twin"). In 1879, Fran~ois Lecoq de Boisbaudran isolated another substance, samarium, from the didymium preparation, and all these were accepted as chemical elements. But didymium vanished from the rolls of chemistry in 1885, when the Austrian Carl von Welsbach separated it into two new elements, neodymium ("new twin") and praseodymium ("green twin"). It is only because we now have the periodic table, and understand the principles behind its construction, that we can say that there can be no other new elements between hydrogen, 1H , and element 105.
What kinds of substances are elements? The first to be recognized correctly as such were the metals. Gold, silver, copper, tin, iron, platinum, lead, zinc, mercury, nickel, tungsten, and cobalt all are metals. In fact, all but 22 of the 105 known elements have metallic properties. Five of the nonmetals (helium, neon, argon, krypton, and xenon) were discovered in the mixture of gases that remained when all the nitrogen and oxygen in air were removed. Chemists thought that these "noble" gases were inert until 1962, when it was shown that xenon combines with fluorine, the most chemically active nonmetal. The other chemically active nonmetals are either gases (such as hydrogen, nitrogen, oxygen, and chlorine) or brittle, crystalline solids (such as carbon, sulfur, phosphorus, arsenic, and iodine). Only one nonmetallic element, bromine, is liquid under ordinary conditions.
Compounds, Combustion, and the Conservation of Mass
Figure 6-2 Priestley's apparatus for preparing oxygen gas. Mercuric oxide in a small pan floating on the surface of the mercury bath is decomposed to liquid mercury and oxygen by solar heat. The arrangement of the mercury bath and bell jar prevents loss of the gas evolved.
Most eighteenth-century chemists were devoted to preparing and describing pure compounds, and to decomposing them to the elements from which they are formed. The great advances of the time were in the chemistry of gases. In 1756, Joseph Black completely changed chemists' ideas about gases when he showed, in his M.D. thesis at Edinburgh, that marble (which we know to be primarily calcium carbonate, CaCO3) could be decomposed to quicklime (calcium oxide, CaO) and a gas (carbon dioxide, CO2), and that the process could be reversed. This demonstration proved that there were different kinds of gases, and that they could take part in chemical reactions just as well as liquids and solids could. One of Black's contemporaries, John Robinson, wrote the following:
"He had discovered that a cubic inch of marble consisted of about half its weight of pure lime and as much air as would fill a vessel holding six wine gallons.... What could be more singular than to find so subtile a substance as air existing in the form of a hard stone, and its presence accompanied by such a change in the properties of the stone?"
In the following years, Henry Cavendish discovered hydrogen (1766), Daniel Rutherford found nitrogen (1772), and Joseph Priestley invented carbonated water and identified nitrous oxide ("laughing gas"), nitric oxide, carbon monoxide, sulfur dioxide, hydrogen chloride, ammonia, and oxygen. In 1781 , Cavendish proved that water is a combination of only hydrogen and oxygen, after he had witnessed Priestley explode the two gases in what Priestley later recalled as "a random experiment to entertain a few philosophical friends." The discovery of oxygen (Figure 6-2) led Antoine Lavoisier to overthrow the predominant idea of eighteenth-century chemistry, the phlogiston theory. The process by which this theory was shattered illustrates the great importance of quantitative measurements in chemistry.
Phlogiston
When Empedocles watched wood burn, he was impressed with the idea that something left the wood; only a light fluffy ash remained. It became generally accepted that combustion was the decomposition of a substance accompanied by a loss of weight. Metal oxides are usually less dense and less compact than the metals from which they come. Even when it became known that the oxide was heavier than the original metal, a confusion between density (weight per unit volume) and weight itself compounded the error. The Germans Johann Becher and Georg Stahl proposed, in 1702, that all combustible material contains an element called phlogiston, which escapes when the material burns. According to their theory,
1. Metals, when heated, lose phlogiston and become calces. (A calx is a crumbly residue.)
2. Calces, when heated with charcoal, reabsorb phlogiston and become metals again. The charcoal is necessary because the original phlogiston has become scattered through the surrounding atmosphere and lost.
3. Charcoal must therefore be very rich in phlogiston.
By this theory, a lit match goes out when it is placed in a closed bottle because the air in the bottle becomes saturated with phlogiston; respiration in living organisms is a purification process in which phlogiston is removed; a mouse under a bell jar eventually dies when the air around him has absorbed all the phlogiston it can.
Think about these ideas for a while. So long as you make no weighing experiments, this theory explains combustion as well as our present ideas do, and seems to agree with common-sense observations about the appearance of metals and calces. Jean Rey, in France, had demonstrated that tin gains weight when it burns, but chemists, unaccustomed to attaching much importance to weight, overlooked the significance of Rey's work. In 1723, Stahl gave a clever explanation for Rey's finding:
"The fact that metals when transformed into their calces increase in
weight, does not disprove the phlogiston theory, but, on the contrary, confirms it, because phlogiston is lighter than air, and, in combining with substances, strives to lift them, and so decrease their weight; consequently, a substance which has lost phlogiston must be heavier than before."
It is no wonder that hydrogen, when it was discovered, was hailed as the first preparation of pure phlogiston! Again there was a confusion between the two ideas of weight and of density (in terms of buoyancy).
Conservation of Mass
Lavoisier discovered that mercuric calx lost weight when it was heated and free mercury and a gas were produced. He measured the volume of gas released. Then he showed that when mercury was reconverted to calx, the same volume of this gas was reabsorbed and there was a weight increase equal to the earlier loss. On the basis of careful weighing experiments such as these, Lavoisier proposed that combustible materials burn by adding oxygen, thus increasing in weight. (Oxygen was his name for the gas. Priestley called it dephlogisticated air since it could apparently absorb even more phlogiston than atmospheric air could.) Lavoisier demonstrated that the products obtained from burning wood, sulfur, phosphorus, charcoal, and other substances were gases whose weight always exceeded that of the solids that burned. His rebuttal to the metallurgical explanations of Becher and Stahl was as follows:
1. Metals combine with oxygen from the air to form calces, which are oxides.
2. Hot charcoal removes oxygen from calces to form a metal and a gas, 02 (at that time called fixed air).
3. Charcoal, therefore, does not combine with the metal)· rather, it removes the oxygen that had previously been combined with the metal in the calx.
The key to this theory was the chemical balance. Lavoisier was the first chemist to realize the importance of the principle of the conservation of mass. In his Traite Etementaire de Chimie) he wrote:
"We must lay it down as an incontestable axiom, that in all the operations of art and nature, nothing is created; an equal quantity of matter exists both before and after the experiment .... Upon this principle, the whole art of performing chemical experiments depends."
Lavoisier was a businessman first and a chemist second. His full-time occupation was as a member of the Ferme Generale) an agency that collected taxes on a commission basis for the French government before the revolution. One of his biographers has called his conservation of mass dictum the "principle of the balance sheet," and has claimed to see its origin in his role as tax collector. Be that as it may, in 1794, his connection with the Ferme Generale cost him his life. *
*Hailed before a revolutionary tribunal because of his past aristocratic associations, Lavoisier heard Coffinhal, president of the tribunal, reject a plea for clemency: "The Republic has no need of chemists and savants. The course of justice shall not be interrupted." This was surely one of the most serious governmental cutbacks in the history of science.
Lavoisier published his textbook, Traite Etementaire de Chimie) in 1789, and it would be difficult to overemphasize the impact that it had on chemistry. In addition to setting forth the principle of conservation of mass in chemical reactions and overthrowing the phlogiston theory, the book contained in an appendix what is essentially our present system of nomenclature. For a generation, therefore, chemistry became "the French science" (the phrase lingered longer in France than elsewhere).
Does a Compound Have a Fixed Composition?
After Lavoisier, chemists began an intensive study of quantities in chemical reaction, that is, masses. The distinction between compounds and mixtures or solutions gradually became clear. A feud developed between those who claimed that the ratios of elements in compounds were fixed and those who believed that a continuous range of proportions was possible. The French chemist Berthollet cited alloys of metals in support of the idea of variable composition. But]. L. Proust, in Madrid, insisted that compounds had fixed composition, and correctly recognized alloys as solid solutions, not compounds:
"The properties of true compounds are invariable as is the ratio of their constituents. Between pole and pole, they are found identical in these two respects; their appearance may vary owing to the manner of aggregation, but their [chemical] properties never. No differences have yet been observed between the oxides of iron from the South and those from the North. The cinnabar of Japan is constituted according to the same ratio as that of Spain. Silver is not differently oxidized or muriated in the muriate of Peru than in that of Siberia."
This principle has been called the law of constant composition. The dispute between Berthollet and Proust had the good effect of sending chemists to the laboratory to prove the ideas of one or the other camps, and incidentally to compile rapidly a body of knowledge about chemical composition. *Of course, Proust was right; yet there are solid crystalline materials in which, because of defects in the crystal structure, the same ratio of atoms is not quite that predicted by the ideal chemical formula. For example, iron sulfide can vary from Fe1.1S to FeS1.1, depending on how the sample is prepared. Such substances are called nonstoichiometric solids, although it has been suggested that they be called "berthollides" after the loser in the debate just discussed.
*The orthodox viewpoint is that they went to the laboratory to decide between two conflicting theories. Let us be honest: Scientists are people, and science is seldom conducted in such a nonpartisan vacuum.
Equivalent Proportions
Between 1792 and 1802, an obscure German chemist named Jeremias Richter made an important discovery that was almost completely ignored by his contemporaries. His idea was the one of the equivalent proportions: The same relative amounts of two elements that combine with on another will also combine with a third element (assuming that the reactions are possible at all). This concept is easy to understand from a few examples:
1 g oh hydrogen combines with 8 g of oxygen to form water.
1 g of hydrogen combines with 3 g of carbon to form methane.
1 g of hydrogen combines with 35.5 g of chlorine to form hydrogen chloride.
1 g of hydrogen combines with 25 g of arsenic to form arsine.
The chemical reactions and formulas (which were not known at the time) are, in fact,
2H2 + O2 2H2O
2H2 + O2 2H2O
H2 + O2 2H2O
3H2 + O2 2H2O
Using modern atomic weights, verify that the preceding statements about weights involved in the reactions are true.
Richter's law of equivalent proportion states that if carbon and oxygen combine they should do so in the ratio of 3 to 8 by weight. This is true for what we now know to be CO2. If they react, carbon and chlorine should do so in the ratio of 3 to 35.5, and this is true for the liquid that we now know as carbon tetrachloride, CCl4. In a similar way, arsenic forms AsCl3, and As2O3 and chlorine and oxygen form Cl2O.
Combining Weights
Figure 6-3 Weights of elements that combine with one another in form ing the compounds indicated. We can pred ict from the diagram that 25.0 g of arsenic, for example, will combine with 35.5 9 of chlorine or 16.0 9 of sulfur. This, in fact, does occur. Arsenic and chlorine react In the predicted weight ratio to give arsenic trichloride (AsCl3). whereas arsenic and sulfur react in the predicted ratio to give the compound arsenous sulfide (As2S3).
A combining weight can be defined for each element as the weight of the element that combines with 1 g of hydrogen. If no hydrogen compound exists, it is the weight that combines with 8 g of oxygen or with one combining weight of some other element that does form a hydrogen compound. In this way a branching network of reactions can lead to a table of combining weights for all the elements. Richter's principle, if true, assures us that there will be no contradictions within the table. Such a set of combining weights for all the elements. Richter's principle, if true, assures us that there will be no contradictions within the table. Such a set of combining weights is shown in Figure 6-3 and Table 6-1.
There is one serious flaw to this scheme, which is why no one took Richter very seriously. The flaw is that many elements appear to have more than one combining weight. Carbon forms a second oxide (we know it now as carbon monoxide, CO), in which the ratio of carbon to oxygen is only 3 to 4. Either the combining weight of carbon has risen to 6, or that of oxygen has fallen to 4. In ethane (C2H6) the combining weight of carbon is 4, in ethylene (C2H4) it is 6, and in acetylene (C2H2) it is 12. The expected oxide of sulfur, SO, does not appear, and in the two most common oxides (SO2 and SO3) sulfur has combining weights of 8 and 5$\textstyle\frac{1}{3}$(Figure 6-4).
Figure 6-4 Variable combining weights for sulfur and carbon. Note that in this figure and in Figure 6-3 there are three combining weights for sulfur: 5.33 g, 8.00 g, and 16.0 g. These weights are in the ratios of 2:3:6. The two combining weights for carbon are in the ratio of 1:2.
Nitrogen is particularly troublesome. In ammonia it has a combining weight of 4$\textstyle\frac{2}{3}$, and in the three oxides known since Priestley's time its combining weights are 3$\textstyle\frac{1}{2}$, 7, and 14. If you know chemical formulas, the combining weights are easy to calculate, and you should be able to check them. But if you know only the combining weights, could you deduce the formulas? The significance of the ratios of elements in compounds was obscured even more by the habit of reporting composition in percent by weight; it was John Dalton who developed the trick of writing them as ratios to one common element and setting up combining weight tables, . which we still do. When Humphry Davy reported that the three oxides of nitrogen contained 29.50%, 44.05%, and 63.30% nitrogen by weight, no one noticed that the nitrogen was combining in the relative ratios of 1 to 2 to 4. (These percentages are Davy's experimental values. What are the correct percentages?) By 1802, it was established that compounds had fixed compositions, and that there could be several such definite compositions between the same two elements. Yet no one knew why, or where to go from there.
John Dalton and the Theory of Atoms
Figure 6-5 Dalton's oriinal symbolism for reactions that form simple compounds. Modern symbols appear beneath them. The namees to the right are Dalton's. When either the formula or the name of the product is different today, the modern formula or name is given in parentheses.
John Dalton, a science (or "natural philosophy") teacher in the Manchester, England, schools was compelled by such data as those in Section 6-3 to propose a theory of atoms, which he presented to the Literary and Philosophical Society of Manchester in 1802 and published three years later. His theory was as follows:
1. All matter is made up of atoms. These are the ultimate particles, and are indivisible and indestructible.
2. All atoms of a given element are identical, both in weight and in chemical properties.
3. Atoms of different elements have a different weights and different chemical properties.
4. Atoms of different elements can combine in simple whole numbers to form compounds.
5. When a compound is decomposed, the recovered atoms are unchanged and can form the same or new compounds.
Dalton also emphasized weights, as had Lavoisier; furthermore, Dalton invented a convenient symbolism for atoms, shown in Figure 6-5. Dalton's symbol for hydrogen represents more than merely an unspecified amount of hydrogen. It represents either one atom of hydrogen or some standard weight of hydrogen containing a standard number of atoms (such as the atomic weight containing Avogadro's number of atoms). Chemical formulas and equations are therefore not merely symbolic, but quantitative.
The Greek Atomic Theory
The idea of atoms was far from new. Democritus and the Epicureans in Greece had proposed an atomic theory, about 400 B.C., that contained virtually all of Dalton's ideas on the subject. The original writings are lost, but we know of this theory from attacks by its opponents and from a long poem written, in 55 B.C., by a Roman Epicurean, Lucretius. (The poem is entitled De Rerum Natura, "On the Nature of Things.") After Lucretius, the ideas of atomism drifted in and out of alchemy for nearly 1900 years without making a significant impact on it. Isaac Newton and Lavoisier both believed in atoms, but more as philosophical concepts or figures of speech that helped in thinking about reactions than as a theory requiring experiment.
There is an important point here that cannot be overstressed. A theory in science is important if, and only if, it makes the understanding of the behavior of the real world clearer. Describing bronze as a substitutional alloy of tin and copper is superior to describing it as the confluence of Jupiter and Venus, in alchemical terminology, because the tin-copper theory suggests experiments by which the properties of bronze might be explained, predicted, and even improved, whereas the "celestial marriage" theory leads nowhere. But perhaps it is less apparent that Democritus' atomic theory, and even Newton's, was not much of an improvement on this celestial marriage idea; it was Dalton's measurements, explanations, and predictions that made atomic theory valuable.
Fixed Ratios
Dalton took the table of combining weights as his point of departure and asked why the ratios of elements in compounds should be fixed. His answer was that a compound consists of a large number of identical molecules, each of which is built up from the same small number of atoms, arranged in the same way. Yet Dalton still needed to know how many atoms of carbon and oxygen combined in each molecule of an oxide of carbon, and how many hydrogen and oxygen atoms combined in a water molecule. Lacking any other guide, he proposed a "rule of simplicity" that started him off well but eventually landed him in serious trouble. The most stable two-element molecule, he reasoned, would be the simple diatomic one, AB. If only one compound of two elements were known, it would be an AB compound. Next most stable would be the triatomic molecules, AB2 and A2B. If only two or three compounds of two elements were known, they would be of these three types. This rule was one of those principles of economy, like the minimization of energy in mechanics or the principle of least action in physics, which are sometimes right, and sometimes wrong. Dalton was wrong.
Dalton began by mistakenly assuming from his rule of simplicity that water had a diatomic formula, HO. This made the atomic weight of oxygen equal to its combining weight of 8 (all relative to 1 for hydrogen). He then turned to the oxides of carbon and nitrogen; the possible choices are shown in Table 6-2. (All atomic weights in this discussion are based on the true numerical values, not on Dalton's values. He was a notoriously poor experimentalist. The atomic weight of oxygen, even on his own terms, began at 6.5 and slowly worked up to 8.) One oxide of carbon had a carbon-to-oxygen ratio of 0.75, and the other had a ratio of 0.375. If the first oxide were CO- he assumed that one of them had to be-then, as Table 6-2 shows, the other would be CO2 , Thus, the atomic weight of carbon would be 6. If the second oxide were CO, the first would have to be C20. (Can you prove this?) Then carbon would have an atomic weight of 3. Since oxide A was more stable to decomposition, he argued that this one must be CO, and correctly chose possibility 1. For the oxides of nitrogen, he similarly ruled out possibilities 1 and 3 because the five-atom molecules clashed with his rule of simplicity; and he again made the correct assignment of an atomic weight of 7 for nitrogen. (Correct, that is, relative to 8 for oxygen.)
Dalton should have sensed trouble as soon as he came to ammonia. He assumed by the rule of simplicity that the molecular formula for ammonia was NH. However, since 4$\textstyle\frac{2}{3}$ g of nitrogen combines with 1 g of hydrogen, this assumption would have meant an atomic weight of 4$\textstyle\frac{2}{3}$ for nitrogen, a value in conflict with the number 7 calculated from the oxides. As an alternative, he could have kept the atomic weight of 7 and worked out the formula for ammonia:
Hydrogen: $\textstyle\frac{1 g of hydrogen}{1 g mole^{-1}}$ = 1 mole of hydrogen atoms
Nitrogen: $\textstyle\frac{4\frac{2}{3}g of nitrogen}{7 g mole^{-1}}$ = 0.667 mole of nitrogen atoms
With the molar ratio of hydrogen to nitrogen (and therefore the ratio of atoms as well) being 1:0.667, or 3:2, the chemical formula would have to be N2H3, N4H6, or some higher multiple. Such a result would have shaken Dalton's faith in the rule of simplicity, and might have forced him to go back and find the right track. Yet he was undone by the poor quality of his experimental data. His initial value for the combining weight of oxygen was 6.5, which he raised to 7 in 1808. Davy increased it 7.5, and Proust finally arrived at the correct figure (given Dalton's assumptions) of 8. Dalton refused to believe their values (a stubborn attitude for such a poor experimentalist), and all the nitrogen calculations described here were carried out by Dalton with a nitrogen atomic weight of 5 rather than 7.
Law of Multiple Proportions
It is easy to be critical of a man who has gone astray because of bad data. But the real achievement of the atomic theory, which made people accept it almost at once, was not the calculation of atomic weights. It was the atomic theory explained perfectly a principle that had lain unnoticed in he published literature for over 15 years, relating elements that combine to form more than one compound. This was Dalton's law of multiple proportions.
The law of multiple proportions states that if two elements combine to form more than one compound, then the amounts of one element that combine with a fixed amount of the other will differ by factors that are the ratios of small whole numbers. (Or, that you can multiply the amounts by a suitable constant and produce a set of integers.) Since we have been using combining weights, perhaps a more meaningful statement is that if an element shows several combining weights, these weights will differ among themselves by ratios of small whole numbers. For example, the combining weights of carbon in Table 6-1 differ in the ratios of 3 to 4 to 6 to 12, or, more revealingly, in the ratio of 1/4 to 1/3 to 1/2 to 1. The combining weights for sulfur are in the ratios of 1 to 1/2 to 1/3, and the ones for nitrogen are 1/3 to 1/4 to 1/2 to 1 in NH3, NO2, NO, and N2O. Dalton's explanation of these simple fractions was that one, or two, or small number of atoms could combine with one of another kind, but that a molecule with 1.369... atoms combined with 1 atom of another was physically impossible according to atomic theory. The combining weights differ by small whole number fractions because the atoms combine in small whole numbers.
A search through the chemical literature showed that this law was the universal rule. It is one thing to prove your theory with new data that you have collected, but it is much more impressive to prove it with everyone else's; this is what Dalton did. The acceptance of the atomic theory was rapid and almost unanimous.
Equal Numbers In Equal Volumes: Gay-Lussac and Avogadro
As chemists tried to deduce formulas for more and more compounds, the flaws in Dalton's atomic weights and in his rule of simplicity became more and more obvious. No one could come up with dependable method of deciding on chemical formulas. Of the three pieces of molecular information -combining weights of the elements, atomic weights of the elements, and molecular formulas- any one could be calculated if the other two were known. Yet only one, the combining weight, was directly measurable. Dalton's wrong assumptions about formulas led to wrong atomic weights, which led back again to wrong formulas were assigned to acetic acid, the common acid of vinegar. The confusion was so great that some chemists despaired for the atomic theory. Jean Dumas wrote:
"If I were in charge, I would efface the word atom from science, for I am persuaded that it goes far beyond experience, and chemistry must never go beyond experience."
The great German chemist Friedrich Wöhler complained, even as early as 1835, that
"...organic chemistry just now is enough to drive one mad. It gives me the impression of a primeval tropical forest, full of the most remarkable things, a monstrous and boundless thicket, with no way to escape, into which one may well dread to enter."
However, the key to dilemma was already in the chemical literature, and had been since 1811. The first step was provided by Gay-Lussac, and the second by Avogadro.
Gay-Lussac
Figure 6-6 Gay-Lussac's results on the combining volumes of gases and the explanations by (a) Dalton and (b) Avogadro. Gay-Lussac found that one volume of hydrogen and one of chlorine produce two volumes of HCl gas, and that two volumes of hydrogen react with one of oxygen to produce two volumes of steam. (a) Dalton agreed that if the volume of HCl is twice the volume of either hydrogen or chlorine there must be half as many molecules per volume unit in HCL. Similarly, if there are n molecules of hydrogen per unit of volume, and if each of these molecules produces a water molecule in the same total volume, there will also be n molecules per volume unit in water. But only half the volume of oxygen is required, so the density of oxygen must be 2n molecules per volume unit. Thus. in hydrogen chloride, hydrogen, chlorine, water, and oxygen, the numbers of molecules per volume unit are n/2, n, n, and 2n. (b) Avogadro proposed that each molecule of hydrogen, chlorine. or oxygen contains two atoms. All the participants in the HCl reaction would therefore have the same number of molecules per volume unit of gas. Applying this same assumption to the water reaction led to a new formula for water, H2O, and ultimately to a complete revision of Dalton's atomic weight scale.
In 1808, Joseph Gay-Lussac (1778-1850) began a series of experiments with the volumes of reacting gases. He found that equal volumes of HCL gas and ammonia form neutral, solid ammonium chloride. An initial excess of either gas is left over at the end of the reaction. Two volumes of hydrogen react with one of oxygen to form two volumes of steam; three volumes of hydrogen react with one of nitrogen to yield two volumes of ammonia; and one volume of hydrogen reacts with one of chlorine to produce two volumes of HCl gas. In these and other experiments, in which the gas reactions were usually explosions triggered by a spark in an enclosed container, Gay-Lussac always found that gases react [in simple whole number units of volume units 1 provided that after the explosion the products are brought back to the temperature and pressure of the initial gases.
Gay-Lussac was a cautious man and a protégé of Berthollet, who as we have seen did not believe in compounds with fixed compositions. Gay-Lussac drew no conclusions in his Memoire, but the possibility of a connection with Dalton's atomic theory was apparent.
Avogadro
Dalton used Gay-Lussac's data to "prove" that equal volumes of gas do not have equal numbers of molecules, another wrong turn, like his rule of simplicity. Dalton's argument is illustrated in Figure 6-6a. The Italian physicist Amedeo Avogadro (1776- 1856) saw another path. He began by assuming that equal volumes of gas (at the same temperature and pressure) contain equal numbers of molecules. As Figure 6-6b shows, this assumption requires that gases of the reactive elements such as hydrogen, oxygen, chlorine, and nitrogen be composed of two-atom molecules instead of single, isolated atoms. If Avogadro had been believed when he published his ideas in 1811, a half-century of confusion in chemistry would have been avoided. To many people, though, his ideas seemed like one shaky assumption (equal numbers in equal volumes) buttressed by an even shakier one (diatomic molecules). At that time, ideas of chemical bonding were based almost entirely on electrical attraction and repulsion, and it was difficult for scientists to understand how two identical atoms could do anything but repel each other. And if they did attract, why did they not form larger molecules, such as H3 and H4? Jöns Jakob Berzelius (1779- 1848) used data on the vapors of sulfur and phosphorus to undercut Avogadro. Yet Berzelius did not realize that these were examples of just such higher aggregates (S8 and P4). Avogadro himself did not help matters; he mixed terminology so much that it sometimes appeared as if he were splitting hydrogen atoms ("elementary molecules") rather than separating atoms in a diatomic molecule ("integral molecules").
Cannizzaro and A Rational Method Of Calculating Atomic Weights
By 1860, the confusion about atomic weights was so widespread that nearly every chemist of any repute had his own private method of writing chemical formulas. August Kekulé (the inventor of the Kekulé structure for benzene) called a conference in Karlsruhe, Germany, to try to reach some kind of an agreement. The man who settled the entire issue was the Italian Stanislao Cannizzaro (1826-1910), who based a rigorous method for finding atomic weights on the long-ignored work of his countryman Avogadro.
Cannizzaro's reasoning, based on Avogadro's principle that equal volumes of gas contain equal numbers of molecules, was as follows:
1. Assume that the atomic weight of hydrogen is 1.0, and that hydrogen gas is made up of diatomic molecules, as Gay-Lussac's gas volume experiments suggest.
2. Assume that Avogadro was correct in deducting that oxygen gas is also diatomic, O2, and hence that the molecular formula for water is H2O, not HO. (See Figure 6-6b.) Since the combining weight of oxygen in water is 8.0, the atomic weight of oxygen must be 16.0, and the molecular weight of O2 must be 32.0.
3. If equal volumes of all gases contain equal numbers of molecules, then the molecular weight (m) of a gas is proportional to the density (D) of the gas: M = k D. Use H2 and O2 to evaluate the proportionality constant, k:
Density, D Molecular Weight, Constant, k
Gas (g liter-1) M (g mole-1 (liters mole-1)
H2 0.0894 2.0 22.37
O2 1.427 32.0 22.42
Average Value: 22.4
(The fact that k is the same or both H2 and O2 indicated that Cannizzaro was on the right track.)
4. Evaluate the molecular weights of a series of compounds containing the elements whose atomic weights are to be determined. Starting with percent composition by weight from chemical analyses, and molecular weights calculated from gas densities, calculate the weight of each element per molecular unit. Look over these weights for a given element to see if the numbers can be interpreted as integral multiples of some common factor, which may then be the atomic weight.
In the data given in Table 6-3, carbon occurs only in multiples of 12, hydrogen in multiples of 1, and chlorine in multiples of 35.3. Hence the atomic weight of carbon cannot be greater than 12, although it could be an integral submultiple of 12, such as 6, 4, or 3.
The atomic weights obtained by Cannizzaro's method are either the true atomic weights or integral multiples of them. If just ethane, benzene, and ethyl chloride had been included in the table, then the conclusion might have been drawn that the atomic weight of carbon was 24. If information from other carbon compounds had been added to the table, and just one analysis gave a weight of 6 for carbon, then the lower value would have to have been accepted, making the probable formulas C2H4, C4H6, C12H6, C2HCl3, and so on. However, no matter how many carbon compounds were analyzed by Cannizzaro's method, the weight per molecular unit always came out to be an integral multiple of 12. Hence this value was accepted as the atomic weight of carbon.
Cannizzaro's achievement was the last link in the chain of logic that began with Proust and the law of constant composition. The battle was over, save for the computing. Scientists could find accurate atomic weights for any element that appeared in compounds having measurable vapor densities. With these atomic weights, the percent composition of a new compound would lead unambiguously to the chemical formula. The mole was defined as we have stated in Chapter 1, that is, as the number of grams of a compound equal to its molecular weight on Cannizzaro's scale (which is the one we use today, with improvements in accuracy). It was realized that a mole of any compound would have the same number of molecules. Although the value of that number was not then known, it was named Avogadro's number, N, in belated recognition of his contribution.
With the hindsight that comes from knowledge of the ideal gas law, we can see that Cannizzaro's constant k is simply RT/P:
PV = nRT = $\textstyle\frac{wRT}{M}$
PM = DRT
k = $\textstyle\frac{M}{D} = \frac{RT}{P}$
where P = pressure w = weight in grams
V = volume M = molecular weight
n = number of molecules D = density = w/V
R = constant k = constant = M/D
T = temperature
The gas density values that were used in the previous demonstration of Cannizzaro's argument are those at standard temperature and pressure or STP: 1.00 atm and 273 K. Thus,
k = $\textstyle\frac{0.08205 \times 273}{1.00}$ liters mole-1 = 22.4 liters mole-1
Example 1
At STP, the following vapor densities and percent compositions are observed for three compounds of C, H, and S:
Density Percent by weight
Compound (g liter-1) C H S
x 3.40 16.0 0.0 84.0
y 2.14 25.0 8.4 66.6
z 2.77 38.7 9.7 51.6
Assuming atomic weights of 12 for C and 1 for H, as just found, calculate the probable atomic weight of sulfur, S, and the probable formulas for molecules x, y, and z.
Solution
For each compound, first calculate the molecular weight (mol wt) from the gas density, and the weight of each element per molecule:
Weight per molecule Probable
Compound Mol wt C H S formula
x 76.1 12.2 0.0 63.9 CS2
y 48.0 12.0 4.0 32.0 CH4S
z 62.1 24.0 6.0 32.0 C2H6S
Greatest common fa ctor for sulfur = 32.0
The probable atomic weight of sulfur is therefore 32.0 g mole-1.
Example 2
Suppose that another compound, w, has a vapor density of 1.38 g liter-1, and an analysis of 38.7% C, 9.4% H, and 51.6% S by weight. How would this force you to revise your conclusions for Example 1?
Solution
The molecular weight of compound w would be 31.0 g mole-1, and the percent composition would indicate 12.0 g or 1 mole of C per mole, 2.9 g or 3 moles of H, and only 16.0 g of sulfur. Hence the revised atomic weight of S would be 16.0 g mole-1 and the formula for molecule w would be CH3S. The formula for x then would be CS4; for y, CH4S2 ; and for z, C2H6S2. Needless to say, a compound such as w has never been observed, and the atomic weight of 32.0 for sulfur is valid.
By calculations such as these, a consistent set of atomic weights was obtained for the lighter elements that can be found in gaseous molecules.
Atomic Weights for the Heavy Elements: Dulong and Petit
One problem remained: What does one do about the heavy elements, especially metals, that cannot be prepared readily in gaseous compounds? The problem can be illustrated by considering lead and silver.
Example 3
The combining weight of lead (amount per 8.00 g of oxygen) in lead oxide is 51.8 g. What is the atomic weight of lead?
Solution
We know that 103.6 of lead combine with 16 g, or 1 mole, of oxygen atoms, but we can go no further without knowing the chemical formula for lead oxide. Hence, we are caught in the same vicious circle from which Cannizzaro escaped for the lighter atoms. If the formula is PbO, then the atomic weight oflead is 103.6. But if the formula is Pb2O, the atomic weight is 51.8, and if PbO2, 207.2. Can you show that, in general, if the formula for lead oxide is PbxOy, the atomic weight of lead will be 103.6 (y/x)? The problem has several solutions.
Example 4
Silver oxide is 93.05% silver by weight. What is the atomic weight of silver?
Solution
If we take, for simplicity, a specimen sample of 100 g, there will be 93.05 g of silver for every 6.95 g of oxygen. The combining weight of silver (amount per 8.00 g of oxygen) is then 93.05 g $\times$ (8.00 g/6.95 g) = 107.1 g. One mole of oxygen atoms combines with twice this amount, or 214.2 g, of silver. The choice of atomic weight now is limited to a set of multiples or fractions of 107.1 g, depending on what we assume the formula to be.
Formula: Ag2O Ag3O2 AgO Ag2O3 AgO2
Atomic Weight: 107.1 142.8 214.2 321.3 428.4
We need some means of deciding among these values. Without further information a choice cannot be made.
Pierre Dulong (1785-1838) and Alexis Petit (1791-1820) had discovered a method of estimating atomic weights of the heavier elements in 1819, but it had been overlooked in the general confusion that attended chemistry at that time. They made a systematic study of all physical properties that could possibly have a correlation with atomic weight, and found a good one in the heat capacities of solids. The gram heat capacity of a substance is the number of joules of heat needed to raise the temperature of 1 g of the substance by 1°C. It is an easily measured property. The product of the gram heat capacity and the atomic weight of an element is the heat required to raise the temperature of 1 mole by 1°C, or the molar heat capacity. Dulong and Petit noticed that for many solid elements whose atomic weight was known the molar heat capacity was very close to 25 J deg-1 mole-1 (Table 6-4). This indicates that the process of heat absorption must be related more strongly to the number of atoms of matter present than to the mass of matter. Later work on the theory of heat capacities of solids has shown that there should be such a constant molar heat capacity for simple solids. Dulong and Petit gave no explanation, however.
Since they advanced no reason for this phenomenon, at the time it was regarded by most chemists as being as dubious as Dalton's rule of simplicity (which was wrong) or Avogadro's principle of equal volumes/equal number of molecules (which was right). It was not until Cannizzaro prepared the way with light atoms that the method of Dulong and Petit was appreciated for heavy atoms.
We now can choose among the possible precise values of atomic weight derived from analytical data by using an approximate value obtained by assuming, with Dulong and Petit, that the molar heat capacity is approximately constant for all solids, 25 J deg-1 mole-1.
Example 5
The heat capacities of lead and silver, as tabulated by Dulong and Petit, are 0.123 and 0.233 J deg-1 g-1, respectively. Use this information to choose the proper atomic weights for Examples 3 and 4.
Solution
Approximate atomic weight of lead = $\textstyle\frac{25}{0.123}$ = 203
Approximate atomic weight of silver = $\textstyle\frac{25}{0.233}$ = 107
The correct choices from the previous examples must be 207.2 for lead and 107.1 for silver; the chemical formulas then are PbO2 and Ag2O.
Example 6
A common coba g-l. Assuming that you know the atomic weight of sulfur to be 32.06, compute the atomic weight of cobalt and write the correct empirical formula for linnaeite.
Solution
The correct answers are 59 and Co3S4.
Combining Capacities, "Valence," and Oxidation Number
With Dalton's atomic theory, and with the contributions of Avogadro, Dulong and Petit, and Cannizzaro, it became possible to deduce atomic weights for elements from chemical analyses and physical data such as vapor densities and heat capacities. These deductions have given us the table of atomic weights shown on the inside back cover. The next great task of chemistry was to explain the formulas that could be derived.
The most primitive concept in chemical bonding is probably the idea of combining capacity, sometimes called "valence." The combining capacity of an element in a given compound is defined as the ratio of its true atomic weight to its combining weight in that compound:
Combining capacity = $\textstyle\frac{atomic weight }{combining weight}$
Hydrogen has a combining capacity of 1, by definition. Oxygen has a combining capacity of 2 in H2O and most other compounds, but a combining capacity of 1 in hydrogen peroxide, H2O2. Using the data in Table 6-1, we can see that Cl and Br have combining capacities of 1; Ca, 2; and As, 3; carbon shows several combining capacities: 4, 3, 2, and 1. Sulfur has a combining capacity of 2 in H2S, 4 in S02, and 6 in SO3. Nitrogen has a combining capacity of 3 in ammonia, 4 in NO2, 2 in NO, and 1 in N2O. Notice that in these two-element compounds the total combining capacity of one element exactly balances the total combining capacity of the other. In SO3 one sulfur atom with a combining capacity of 6 balances three oxygen atoms having a capacity of 2 each. The formulation of the concept of combining capacity or "valence" was the first step toward a theory of chemical bonding. The second step was to assign plus and minus signs to these combining capacities so that the sum of the signed capacities for a molecule is zero. Hydrogen was assigned the value +1; therefore the value for oxygen had to be -2 so the sum for water, H2O, would be zero. The formula for sulfuric acid, H2SO4, requires that sulfur be associated with the value +6:
H: 2 $\times$ +1 = +2
O: 4 $\times$ -2 = -8
S: 1 $\times$ +6 = +6
Sum: 0
These signed combining capacities are just the oxidation numbers that we encountered in Chapter 1. They are important in theories of chemical bonding because they describe how electrons are shifted toward or away from atoms in a molecule.
Summary
This chapter has chronicled the process by which scientists deduced that chemical compounds are made up of specific numbers of atoms of different kinds having specific atomic weights, and slowly worked out a set of reliable atomic weights. The theory of atoms originated as a philosophical concept rather than as a means of manipulating substances and reactions. Antoine Lavoisier laid the foundation by establishing that mass was the fundamental quantity in chemical reactions. John Dalton turned the philosophy into reality by showing that the atomic theory would account for the experimental observations that were summarized in the laws of equivalent proportions and multiple proportions.
The task of deciding upon a set of consistent atomic weights was not an easy one, and Dalton himself went astray. The circular argument involving assumed atomic weights and assumed molecular formulas was not broken until 1860, when Cannizzaro applied a principle that had been discovered in 1811 by Avogadro but had been ignored: Under the same conditions of temperature and pressure, equal volumes of all gases contain equal numbers of molecules. Since this meant that gas density was proportional to molecular weight, the way was open for establishing the standard atomic weight scale that we still use today. The quantitative foundations of modern chemistry had been laid. | textbooks/chem/General_Chemistry/Chemical_Principles_(Dickerson)/06%3A_Are_Atoms_Real_From_Democritus_to_Dulong_and_Petit/5.01%3A_Hydrogen_Oxygen_and_Water.txt |
It often matters vastly with what others,
In what arrangements the primordial germs
Are bound together, and what motions, too,
They exchange among themselves,
for these same atoms
Do put together sky, and sea, and lands,
Rivers and suns, grains, trees and
breathing things.
have any real existence.
But yet they are commixed in different ways
With different things, with motions each its own.
—Lucretius (55 B.C.)
Introduction
In this chapter we shall examine the correlations that exist between the physical and chemical properties of the elements and their compounds. These correlations lead directly to a fundamental classification scheme for matter, the periodic table. To Ernest Rutherford, who once remarked that there are two kinds of science-physics and stamp collecting-the periodic table would be the ultimate stamp album. If this were the final chapter of our book, his impression would be confirmed. But we organize the elements of the universe into the periodic table so chemistry can begin, not end. Once we have established the classification scheme, we must set out to explain it, in terms of electrons and the other subatomic particles from which atoms are constructed. This explanation is the task of later chapters. But before we begin to theorize about the world, let's see what it's really like.
Early Classification Schemes
Very early in the development of chemistry, chemists recognized that certain elements have similar properties. In the earliest classification scheme, there were only two divisions, metals and nonmetals. Metallic elements have a certain lustrous appearance, they are malleable (can be hammered into thin sheets) and ductile (can be drawn into wires), they conduct heat and electricity, and they form compounds with oxygen that are basic (alkaline). Nonmetallic elements have no one characteristic appearance, they generally do not conduct heat and electricity, and they form acidic oxides.
Döbereiner's Triads
In 1829, the German chemist Johann Dobereiner observed several groups of three elements (triads) with similar chemical properties. In every case the atomic weight of one element in the triad was nearly the average of the other two. For example, each element in the triad chlorine, bromine, and iodine forms colored vapors containing diatomic molecules. Each element combines with metals and has a combining weight equal to its atomic weight. Each element forms ions with oxygen that have a single negative charge: ClO-, ClO$_3^-$, BrO$_3^-$, and IO$_3^-$. The atomic weight of bromine (80) is approximately the average of those of chlorine (35.5) and iodine (127). Table 7-1 lists the similarities of elements in this and other triads.
In addition to recognizing the triads given in Table 7-1, Döbereiner observed a peculiar triad of the metals iron, cobalt, and nickel, all of which have similar properties and almost the same atomic weights. The metals are used in structural materials (steel) and may be ferromagnetic like iron; in their +2 and +3 states they form complex ions that are colored.
This discovery of families of elements (the number 3 per family proved to be insignificant) provided an incentive to those who were attempting to find a rational means of classifying the elements.
Newlands' Law of Octaves
Between 1850 and 1865 many new elements were discovered, and chemists made considerable progress in the determination of atomic weights. Thus, more accurate atomic weights were made available for old elements, and reasonably accurate values were presented for new elements. In 1865, the English chemist John Newlands (1839–1898) explored the problem of the periodic recurrence of similar behavior of elements. He arranged the lightest of the known elements in order of increasing atomic weight as follows:
H Li Be B C N O
F Na Mg Al Si P S
Cl K Ca Cr Ti Mn Fe
Newlands noticed that the eighth element (fluorine, F) resembled the first (hydrogen, H), the ninth resembled the second, and so forth. His observation that every eighth element had similar properties led him to compare his chemical octaves with musical octaves, and he himself called it his law of octaves. Periodicity by octaves in chemistry suggested to him a fundamental harmony like the one in music. The comparison, although appealing, is invalid. Had Newlands known of the noble gases, his periodicity of properties would have been by nines rather than by eights. He never would have used his musical analogy, and he might have been spared some of the ridicule and indifference that he suffered. (See the Postscript for more on Newlands.)
File:Chemical Principles Table 7.1.png
Newlands' effort was admittedly a step in the right direction. However, three serious criticisms can be directed at his classification scheme:
1. There were no places in his table for the new elements, which were being discovered rapidly. Moreover, in the later parts of the table, there were several places where two elements were forced into the same position. (See the Postscript.)
2. There was no scholarly evaluation of the work on atomic weights and no selection of probable best values.
3. Certain elements did not seem to belong where they were placed in the scheme. For example, chromium (Cr) is not sufficiently similar to aluminum (AI), nor is manganese (Mn), a metal, to phosphorus (P), a nonmetal. Iron (Fe), a metal, and sulfur (S), a nonmetal, do not resemble each other either.
The Basis for Periodic Classification
The development of the periodic table as we now know it is credited mainly to the Russian chemist Dmitri Mendeleev (1834–1907), although the German chemist Lothar Meyer worked out essentially the same system independently and almost simultaneously. So far as we know, neither man was aware of Newlands' work. Mendeleev's periodic table (Figure 7-1), presented in 1869, followed Newlands' plan of arranging the elements in order of increasing atomic weight, but with the following substantial improvements:
1. Long periods were instituted for the elements now known as transition
metals. These long periods are shown folded in half in his original table, with each full period taking two lines. This innovation removed the necessity of placing metals such as vanadium (V), chromium, and manganese under nonmetals such as phosphorus, sulfur, and chlorine.
2. If the properties of an element suggested that it did not fit in the
arrangement according to atomic weight, a space was left in the table. For example, no element existed that would fit in the space below silicon (Si). Thus, a space was left for a new element, which was named ekasilicon.
3. A scholarly evaluation of atomic weight data was made. For example,
as a result of this work the combining capacity of chromium in its highest oxide was changed from 5 to the correct value of 6. The combining weight of chromium was known to be 8.66 g. Hence, instead of 43.3 (5 $\times$ 8.66), the revised atomic weight of chromium became 52.0 (6 $\times$ 8.66).
File:Chemical Principles Fig 7.1.png
Indium (In), with a combining weight of 38.5, had been assigned a combining capacity of 2 and therefore an atomic weight of 77, and had been placed between arsenic (As) and selenium (Se). Since their properties were consistent with placement below phosphorus and sulfur, which were next to one another, arsenic and selenium also had to be side by side in Mendeleev's scheme. A reevaluation showed that indium had an atomic weight of 114.8 and a combining capacity of 3, which is consistent with its position below aluminum and gallium (Ga) in the present table.
The atomic weight of platinum (Pt) had been thought to be greater than that of gold (Au). Mendeleev thought otherwise, because of the chemistry of the two metals and the places that they should occupy in his table. New determinations inspired by Mendeleev showed 198 for platinum and 199 for gold, thereby placing platinum ahead of gold and under palladium (Pd), which of all the other elements most resembles platinum.
4. On the basis of the known periodic behavior summarized in the
table, predictions of the properties of the undiscovered elements were made. These predictions later proved to be amazingly accurate, as one can see by comparing the predicted properties of ekasilicon and the properties reported for the element called germanium (Ge), which now occupies the ekasilicon space. This comparison is given in Table 7 -2.
File:Chemical Principles Table 7.2.png
From the table it is evident how Mendeleev was able to predict accurately the physical and chemical properties of the missing element. Its position in the periodic table was below silicon and above tin (Sn). The physical properties of germanium are just about the average between those observed for silicon and for tin. To predict the chemical properties for ekasilicon Mendeleev also used information from the known relative properties of phosphorus, arsenic, and antimony (Sb) in the column to the right in the periodic table.
Correlations such as this guided the search for new elements and compounds and stimulated investigation when known data did not conform with other correlations. One consequence of this research was that we gained improved values for atomic weights and densities.
The Periodic Law
Mendeleev summarized his discoveries in the periodic law: The properties of chemical elements are not arbitrary, but vary with the atomic weight in a systematic way.
After most of the elements had been discovered and their atomic weights carefully determined, several discrepancies persisted. For example, the order of increasing atomic weight within Mendeleev's Group VIII (Figure 7-1) was found to be Fe, Co, Ni, Cu in the fourth period (row 4), Ru, Rh, Pd, Ag in the fifth (row 6), and Os, Ir, Pt, Au in the sixth (row 10). Vet Ni resembles Pd and Pt more than Co does. Again, Te has a higher atomic weight than I, but I clearly belongs with Br and CI, and Te resembles Se and S in chemical properties. When the noble gases were discovered, it was revealed that Ar had a higher atomic weight than K, whereas all the other noble gases had lower atomic weights than the adjacent alkali metals. In these three instances, increasing atomic weight clearly is not acceptable as a means of placing elements in the periodic table. Therefore, the elements were assigned atomic numbers from 1 to 92 (now 105). (The atomic numbers of the elements approximately increase with their atomic weights.) When the elements are arranged according to increasing atomic number, chemically similar elements lie in vertical columns (families or groups) of the periodic table.
In 1912, Henry G. J. Moseley (1887–1915) observed that the frequencies of x rays emitted from elements could be correlated better with atomic numbers than with atomic weights. The relationship between an element's atomic number and the frequency (or energy) of x rays emitted from the element is a consequence of atomic structure. As we shall see in Chapter 8, the electrons in an atom are arranged in energy levels. When an element is bombarded by a powerful beam of electrons, electrons from the innermost levels or shells (closest to the nucleus) can be ejected from the atoms. When outer electrons drop into these shells to fill the vacancies, energy is emitted as x radiation. The x-ray spectrum of an element (the collection of frequencies of x rays emitted) contains information about the electronic energy levels of the atom. The important point for our present purpose is that the energy of a level varies with the charge on the nucleus of the atom. The greater the nuclear charge, the more tightly the innermost electrons are bound. More energy is required to knock off one of these electrons; consequently, there is more energy emitted when an electron falls back into a vacancy in the shells. Moseley discovered that the frequency of x rays emitted (designated by the Greek letter P, nul varies with atomic number, Z, according to
$\textstyle v = c(Z - b)^2$
in which c and b are characteristic of a given x-ray line and are the same for all elements.
File:Chemical Principles Fig 7.2.png
In April 1914, Moseley published the results of his work on 39 elements from 13Al to 79Au. (Recall that the atomic number is indicated by a subscript to the left of the symbol of an element.) A portion of his data [s plotted in Figure 7 -2. Moseley wrote the following:
"The spectra of the elements are arranged on horizontal lines spaced at equal distances. The order chosen for the elements is the order of the atomic weights, except in the cases of Ar, Co, and Te, where this clashes with the order of the chemical properties. Vacant lines have been left for an element between Mo and Ru, an element between Nd and Sm, and an element between W and Os, none of which are yet known. . . . This is equivalent to assigning to successive elements a series of successive characteristic integers . ... Now if either the elements were not characterized by these integers, or any mistake had been made in the order chosen or in the number of places left for unknown elements, these regularities (the straight lines) would at once disappear. We can therefore conclude from the evidence of the x-ray spectra alone, without using any theory of atomic structure, that these integers are really characteristic of the elements.... Now Rutherford has proved that the most important constituent of an atom is its central positively charged nucleus, and van den Broek has put forward the view that the charge carried by this nucleus is in all cases an integral multiple of the charge on the hydrogen nucleus. There is every reason to suppose that the integer which controls the x-ray spectrum is the same as the number of electrical units in the nucleus, and these experiments therefore give the strongest possible support to the hypothesis of van den Broek."*
The three undiscovered elements mentioned by Moseley were later found to be elements 43 (technetium, Tc), 61 (promethium, Pm), and 75 (rhenium, Re). A confusing "double element" was cleared up in 1923, when D. Coster and G. Hevesy showed that one of the unoccupied horizontal lines of Moseley's chart belonged to the new element hafnium (Hf, 72). Moseley's work was perhaps the most fundamental single step in the development of the periodic table. It proved that atomic number (or the charge on the nucleus), and not atomic weight, was the essential property in explaining chemical behavior.
The Modern Periodic Table
The easiest way to understand the periodic table is to build it. Although this may seem to be a difficult task, surprisingly little knowledge of chemistry is required to understand that the form on the inside front cover of this book is inevitable. If we arrange elements by atomic number, as Moseley did, then certain chemical properties repeat at definite intervals (Figure 7-3, top). The chemically inert gases (at least thought to be inert until 1962, when chemists produced compounds containing xenon bound to fluorine and oxygen), He, Ne, Ar, Kr, Xe, and Rn, have atomic numbers 2, 10, 18, 36, 54, and 86, or numerical intervals of 2, 8, 8, 18, 18, and 32. Each of these gases precedes an extremely reactive, soft metal that tends to form a + 1 ion: the alkali metals Li, Na, K, Rb, Cs, and Fr. And each gas is preceded by a reactive element that can gain an electron to form a - 1 ion: hydrogen and the halogens F, CI, Br, I, and At. These key elements are shown in color in the row at the top of Figure 7-3.
These chemical similarities are best represented by dividing the list of 105 elements into seven rows or periods (Figure 7-3). However, the first period has only 2 elements, the next two have 8, the next two, 18, and the sixth and probably the seventh periods have 32. How can we align 8 entries over 18, and 18 over 32?
File:Chemical Principles Fig 7.3.png
*By the time these lines were published, Moseley was in the British army, and less than a year later he was dead, at the age of 27, on a hillside on Gallipoli.
The alkaline earth metals, Be, Mg, Ca, Sr, and Ba, are so similar in chemical properties that we need little imagination to place them as shown. Nonmetals are at the right end of each period, and O, S, Se, and Te constitute a series of elements with a combining capacity of 2 and an increase in metallic behavior from O to Te: O is a nonmetal, and Te exists in the unspecific intermediate zone known as the semimetals or metalloids. Elements N, P, As, Sb, and Bi comprise a group whose characteristics are the ability to gain three electrons in certain compounds and a gradation from the nonmetallic N and P, to the semimetallic As, to the metallic Sb and Bi. T he elements C, Si, Ge, Sn, and Pb all have a combining capacity of 4. For these elements, the border zone between metals and nonmetals is located at an earlier period ; C is a nonmetal, Si and Ge are semimetals, and Sn and Pb are meta ls. Finally, the series B, Al, Ga, In, and Tl form +3 ions; B is semimetallic and the others are metallic. The properties of Al and Ga are more similar than those of Al and Sc. To bring Al above Ga, it is necessary to shift the S-element periods to the extreme right above the 18-element period below.
The "superfluous" elements in Periods 4 and 5 c,. Sc to 30Zn, and 39 Y to "Cd) constitute a series of metals, all of which exhibit a great variety of ionic states, the + 2 and + 3 states seeming to be the most common. Their properties do not change from one element to another nearly as much as the properties in the series B, C, N, 0, and F change. We call these "superfluous" elements transition metals. (We defer the question of what is in transition to Chapters 9 and 10.) When we look for chemical parallels between the fifth and sixth periods, we find that 40Zr and 72Hf are virtually identical in behavior. Again, our preferred arrangement is to place the elements in Period 5 beyond 38Sr as far as possible to the right atop Periods 6 and 7. The extra elements in Period 6, 57La to 70Yb, are practically identical in chemical behavior. These elements are called the rare earths, or lanthanides. Their partners in the seventh period (89Ac to 102No) are known as the actinides. Because the lanthanides are so similar in chemical properties, they are found together in nature and are extremely difficult to separate.
In summary, the elements can be classified into three groups (Figure 7-3): the representative elements, with diverse properties; the transition metals, more similar but yet clearly distinguishable; and the inner transition metals (lanthanides and actinides), with very similar properties. The representative elements are called representative because they show a broader range of properties than are found in the other elements, and because they are the elements with which we are most familiar.
(The radioactivity and nuclear instability of the actinides, especially uranium, have given them a historical significance that their chemical properties perhaps would not have justified. An old-time chemist is a person who still thinks of uranium as an obscure heavy element used in yellow pottery glazes and stained glass. It is ironic that a nuclear war would be fought with the raw material of stained-glass windows.)
There is a more compact form of the periodic table that indicates more clearly the relative variability of properties of neighboring elements (Figure 7-4). Trends in chemical behavior are often easier to understand if only the representative elements are examined, with the transition metals set to one side as a special case and the inner transition metals virtually ignored. In this table, the vertical columns are called groups, and those of the representative elements are numbered IA through VIIA and O. The groups of the transition elements are numbered in a way to remind you that they should be inserted in the representative element table. The numbering includes Groups IIIB to VIIB, then three columns all labeled collectively Group VIIIB, then Groups IB and lIB. Group HIB follows Group IIIA in the representative elements, and Group lIB precedes Group IlIA. This kind of numbering is clearer in the standard, "long form" of the periodic table on the inside front cover. We can see that the standard form is a compromise between the compactness of Figure 7-4 and the completeness of Figure 7-3. The lanthanides and actinides have been of so little relative importance that they have not been given group numbers.
File:Chemical Principles Fig 7.4.png
Periodicity of Chemical Properties as Illustrated by Binary Hydrides and Oxides
In this section we shall see how the periodic table enables us to predict the molecular formulas and the chemical properties of compounds of metals and nonmetals with hydrogen and oxygen.
Binary Hydrides
The number of hydrogen atoms that combine with one atom of a given representative element in the first three periods of the periodic table varies (as shown in Figure 7-5) from 1 to 4 and back to 1 again across each period. This number is equal to either the group number or 8 minus the group number, whichever is smaller. This face alone offers a clue to the way in which H is bonded in each hydrogen compound.
File:Chemical Principles Fig 7.5.png File:Chemical Principles Fig 7.6.png
Compounds of metals with hydrogen-called hydrides-are mostly ionic. In alkali hydrides such as KH or NaH, there is a transfer of negative charge to each hydrogen atom. Alkali hydrides have the NaCl crystal structure (Chapter 1), but BeH2 MgH2 and AlH3, manifest a new phenomenon, "bridging" hydrogen. In this arrangement each H atom in the crystal is equidistant between two metal atoms and appears to form a hydrogen bridge between them. Whenever H has a net negative charge, this extra charge can apparently be used to make a second bond to another atom, if there is enough potential bonding capacity in the other atom. The negatively charged H is present in NaH, but not the capacity for multiple bonds. However, Be, Mg, and Al satisfy both demands, and bridge structures are formed. The boron- hydrogen compound B2H6 (Figure 7-6) is an example of hydrogen bridging within a molecule, and the other known boron hydrides (such as B4H10, B5H9, B5H11, B6H10, and B10H14) all make extensive use of such hydrogen bridges.
Compounds of hydrogen with elements in the right half of the periods are small molecular compounds in which the number of hydrogen atoms in a molecule is dictated by the number of covalent bonds that the other atom can form. Molecules of such compounds are held together in crystals only by weak forces between molecules; thus their melting and boiling points are very low (Figure 7 -6).
Ionic hydrides react with water to produce basic solutions:
Na+H- + H2 H2 + OH- + Na+
Conversely, the halogen compounds are acidic:
HCl + H2O H3O+ + Cl-
Example 1
What are the formulas of the hydrides of cesium (Cs) and selenium (Se)? Which hydride has the higher melting point? Write a balanced equation for the reaction of cesium hydride with water.
Solution
Cesium is in Group IA. Therefore the formula of its hydride is CsH. Selenium is in Group VIA, and its hydride is H2Se. Cesium hydride is an ionic hydride with a much higher melting point than that of H2Se, which is a molecular substance with a low melting point (and boiling point). The reaction of CsH with H2O yields H2, Cs+, and OH-:
CsH(s) + H2O(l) H2(g) + Cs+ + OH-
Binary Oxides
The representative elements form oxides with the formulas expected from the elements' positions in the periodic table; in the third period these oxides are Na2O. MgO, Al2O3, SiO2, P2O5, SO3, and Cl2O7. Oxides of elements at the lower left of the table are strong bases. They have a large negative charge on the O atom, and are ionic. The melting points of these ionic oxides are typically around 2000°C, and many decompose before melting. They react with water to make basic solutions:
Na2O + H2O 2Na+ + 2OH-
At the other extreme, oxides of elements at the upper right of the table are strong acids:
Cl2O7 + 3H2O 2H3O+ + 2ClO$_4^-$ (perchloric acid)
SO3 + 3H2O 2H3O+ + SO$_4^{2-}$ (sulfric acid)
Cl2O7 is explosively unstable, and SO3 reacts vigorously with water to produce acidic solutions. The acids have been represented here as completely ionized or dissociated, but this can be as misleading as writing them in their undissociated forms: HClO4 and H2SO4. As we saw in Chapter 5, H2SO4 is only partially dissociated in water.
Between acidic and basic oxides lies a diagonal band of oxides that are amphoteric: BeO, Al2O3, and Ga2O3; GeO2 through PbO2; and Sb2O5 and Bi2O5. (Amphoteric oxides show both acidic and basic behavior.) They are virtually insoluble in water but can be dissolved by either acids or bases:
BeO + 2H3O+ Be2+ + 3H2O
BeO + 2OH- H2 + Be(OH)$_4^{2-}$
The notation in the first equation is conventional but inconsistent. The hydration of the proton is denoted by the symbol H3O+. However, the Be2+ cation is also very strongly hydrated, especially so because of its small size. It should be written as Be(H2O)$_n^{2+}$, or at least Be2+(aq). But so long as the hydration of cations is understood, it need not be spelled out every time.
The amphoteric and basic oxides are solids with high melting points. For instance, Al203 is the abrasive known as corundum, or emery; SiO2 is quartz. Only the oxides of C, N, S, and the halogens are normally liquids or gases. The contrast between C and Si in carbon dioxide and quartz is analogous to the contrast between C and N in diamond and nitrogen gas. The difference between C and Si arises because C can make double bonds to O and therefore form a molecular compound of limited size. However, Si must make single bonds with four different O atoms; hence, it must assume a three-dimensional network structure in which tetrahedrally arranged Si atoms are connected by bridging O atoms.
In all the oxides discussed so far, the chemical formula can be predicted from the group number. But there are other oxides whose formulas cannot be predicted from group numbers. For example, C can form CO as well as CO2. The compound N2O5 is not the only nitrogen oxide: NO2, N2O3, and NO are others. Sulfur can form SO2, S2O3, S2O, and S2O7, as well as SO3. But in these compounds the element does not make full use of its potential combining capacity. Thus, the general trends in properties are best illustrated by the oxides that we have been examining.
Example 2
Write a balanced equation for the reaction of calcium oxide with water.
Solution
Calcium (Ca) is in Group IIA. Its oxide is therefore CaO. The reaction is
CaO(s) + H2O(l) Ca2+ + 2OH-
The equation shows that CaO is a strong base, as its reaction with water yields OH- ions.
Summary
Physical and chemical properties of elements are periodic functions, not of atomic weight, but of atomic number. Moseley suggested, and it later was verified, that the atomic number is the total positive charge on the nucleus, equal to the total number of electrons around the nucleus in a neutral atom.
Particularly stable, inert elements occur at intervals of 2, 8, 8, 18, 18, and 32 in atomic number. These intervals, and only the most basic knowledge of similarities among elements, led to the formulation of a periodic table, in which similar elements are in vertical columns or groups, and in which chemical properties change in an orderly manner along horizontal rows or periods. The full, extended periodic table can be folded into a compact form that illustrates the division of the elements into three categories: the diversified representative elements, the more similar transition metals, and the virtually identical inner transition metals.
The correlation between combining capacity and group number in the periodic table may be illustrated by the hydrides and oxides of the representative elements. Elements at the lower left of the periodic table are metals. Their hydrides and oxides are ionic, and aqueous solutions of these compounds are basic. Elements at the upper right corner of the table are nonmetals. Their hydrogen compounds and oxides are small molecules with covalent bonds; they are gases or liquids, and are acidic. Between the two extremes at upper right and lower left in the table there is a gradation of properties. As the elements pass from nonmetals through semimetals to metals, their hydrogen compounds go from acidic, to neutral or inert, to basic (although there are many complications to this overall trend), and the oxides proceed in a more regular manner from acidic, to amphoteric, to basic. | textbooks/chem/General_Chemistry/Chemical_Principles_(Dickerson)/07%3A_The_Periodic_Table/5.01%3A_Hydrogen_Oxygen_and_Water.txt |
Contents
• 1 Introduction
• 2 Rutherford and The Nuclear Atom
• 3 The Quantization of Energy
• 3.1 The Ultraviolet Catastrophe
• 3.2 The Photoelectric Effect
• 3.3 The Spectrum of the Hydrogen Atom
• 4 Bohr's Theory of the Hydrogen Atom
• 4.1 Energy Levels of a General One-Electron Atom
• 4.2 The Need for a Better Theory
• 5 Particles of Light and Waves of Matter
• 6 The Uncertainty Principle
• 7 Wave Equations
• 8 The Hydrogen Atom
• 9 Many-Electron Atoms
• 10 Summary
The continuity if all dynamical effects was
formerly taken for granted as the basis of all
physical theories and, in close correspondence
with Aristotle, was condensed in the well-
known dogma-Natura non facit saltus-
nature makes no leaps. However, present-day
investigation has made a considerable breach
even in this venerable stronghold if physical
science. This time it is the principle if thermo-
dynamics with which that theorem has been
brought into collision by new facts, and unless
all signs are misleading, the days if its
validity are numbered. Nature does indeed seem
to make jumps-and very extraordinary ones.
Max Planck(1914)
IntroductionEdit
Physics seemed to be settling down quite satisfactorily in the late nineteenth century. A clerk in the U.S. Patent Office wrote a now-famous letter of resignation in which he expressed a desire to leave a dying agency, an agency that would have less and less to do in the future since most inventions had already been made. In 1894, at the dedication of a physics laboratory in Chicago, the famous physicist A. A. Michelson suggested that the more important physical laws all had been discovered, and "Our future discoveries must be looked for in the sixth decimal place." Thermodynamics, statistical mechanics, and electromagnetic theory had been brilliantly successful in explaining the behavior of matter. Atoms themselves had been found to be electrical, and undoubtedly would follow Maxwell's electromagnetic laws.
Then came x rays and radioactivity. In 1895, Wilhelm Rontgen (1845- 1923) evacuated a Crookes tube (Figure 1-11) so the cathode rays struck the anode without being blocked by gas molecules. Rontgen discovered that a new and penetrating form of radiation was emitted by the anode. This radiation, which he called x rays, traveled with ease through paper, wood, and flesh but was absorbed by heavier substances such as bone and metal. Rontgen demonstrated that x rays were not deflected by electric or magnetic fields and therefore were not beams of charged particles. Other scientists suggested that the rays might be electromagnetic radiation like light, but of a shorter wavelength. The German physicist Max von Laue proved this hypothesis 18 years later when he diffracted x rays with crystals.
In 1896, Henri Becquerel (1852–1908) observed that uranium salts emitted radiation that penetrated the black paper coverings of photographic plates and exposed the photographic emulsion. He named this behavior radioactivity. In the next few years, Pierre and Marie Curie isolated two entirely new, and radioactive, elements from uranium ore and named them polonium and radium. Radioactivity, even more than x rays, was a shock to physicists of the time. They gradually realized that radiation occurred during the breakdown of atoms, and that atoms were not indestructible but could decompose and decay into other kinds of atoms. The old certainties, and the hopes for impending certainties, began to fall away.
The radiation most commonly observed was of three kinds, designated alpha (α), beta (β), and gamma (γ). Gamma radiation proved to be electromagnetic radiation of even higher frequency (and shorter wavelength) than x rays. Beta rays, like cathode rays, were found to be beams of electrons. Electric and magnetic deflection experiments showed the mass of α radiation to be 4 amu and its charge to be +2; α particles were simply nuclei of helium, $_2^4$He.
The next certainty to slip away was the quite satisfying model of the atom that had been proposed by J. J. Thomson.
Rutherford and The Nuclear AtomEdit
File:Chemical Principles Fig 8.1.png
Figure 8-1
File:Chemical Principles Fig 8.2.png
Figure 8-2
File:Chemical Principles Fig 8.3.png
Figure 8-3
In Thomson's model of the atom all the mass and all the positive charge were distributed uniformly throughout the atom, with electrons embedded in the atom like raisins in a pudding. Mutual repulsion of electrons separated them uniformly. The resulting close association of positive and negative charges was reasonable. Ionization could be explained as a stripping away of some of the electrons from the pudding, thereby leaving a massive, solid atom with a positive charge.
In 1910, Ernest Rutherford (1871–1937) disproved the Thomson model, more or less by accident, while measuring the scattering of a beam of α particles by extremely thin sheets of gold and other heavy metals. (His experimental arrangement is shown in Figure 8-1.) He expected to find a relatively small deflection of particles, as would occur if the positive charge and mass of the atoms were distributed throughout a large volume in a uniform way (Figure 8-2a). What he observed was quite different, and wholly unexpected. In his own words:
"In the early days I had observed the scattering of α particles, and Dr. Geiger in my laboratory had examined it in detail. He found in thin pieces of heavy metal that the scattering was usually small, of the order of one degree. One day Geiger came to me and said, 'Don't you think that young Marsden, whom I am training in radioactive methods, ought to begin a small research?' Now I had thought that too, so I said, 'Why not let him see if any α particles can be scattered through a large angle?' I may tell you in confidence that I did not believe they would be, since we knew that the α particle was a very fast massive particle, with a great deal of energy, and you could show that if the scattering was due to the accumulated effect of a number of small scatterings, the chance of an α particle's being scattered backwards was very small. Then I remember two or three days later Geiger coming to me in great excitement and saying, 'We have been able to get some of the α particles coming backwards.' ... It was quite the most incredible event that has ever happened to me in my life. It was almost as incredible as if you fired a 15-inch shell at a piece of tissue paper and it came back and hit you."
Rutherford, Geiger, and Marsden calculated that this observed backscattering was precisely what would be expected if virtually all the mass and positive charge of the atom were concentrated in a dense nucleus at the center of the atom (Figure 8-2b). They also calculated the charge on the gold nucleus as 100 ± 20 (actually 79), and the radius of the gold nucleus as something less than 10−12 cm (actually nearer to 10−13 cm).
The picture of the atom that emerged from these scattering experiments was of an extremely dense, positively charged nucleus surrounded by negative charges-electrons. These electrons inhabited a region with a radius 100,000 times that of the nucleus. The majority of the α particles passing through the metal foil were not deflected because they never encountered the nucleus. However, particles passing close to such a great concentration of charge would be deflected; and those few particles that happened to collide with the small target would be bounced back in the direction from which they had come.
The validity of Rutherford's model has been borne out by later investigations. An atom's nucleus is composed of protons and neutrons (Figure 8-3). Just enough electrons are around this nucleus to balance the nuclear charge. But this model of an atom cannot be explained by classical physics. What keeps the positive and negative charges apart? If the electrons were stationary, electrostatic attraction would pull them toward the nucleus to form a miniature version of Thomson's atom. Conversely, if the electrons were moving in orbits around the nucleus, things would be no better. An electron moving in a circle around a positive nucleus is an oscillating dipole when the atom is viewed in the plane of the orbit; the negative charge appears to oscillate up and down relative to the positive charge, By all the laws of classical electromagnetic theory, such an oscillator should broadcast energy as electromagnetic waves. But if this happened, the atom would lose energy, and the electron would spiral into the nucleus. By the laws of classical physics, the Rutherford model of the atom could not be valid. Where was the flaw?
The Quantization of EnergyEdit
Other flaws that were just as disturbing as Rutherford's impossibly stable atoms were appearing in physics at this time. By the turn of the century scientists realized that radio waves, infrared, visible light, and ultraviolet radiation (and x rays and y rays a few years later) were electromagnetic waves with different wavelengths. These waves all travel at the same speed, c, which is 2.9979 × l08 m sec-l or 186,000 miles sec-l (This speed seems almost instantaneous until you recall that the slowness of light is responsible for the 1.3-sec delay each way in radio messages between the earth and the moon.) Waves such as these are described by their wavelength (designated by the Greek letter lambda, λ), amplitude, and frequency (designated by the Greek letter nu, ν), which is the number of cycles of a moving wave that passes a given point per unit of time (Figure 8-4). The speed of the wave, c, which is constant for all electromagnetic radiation, is the product of the frequency (the number of cycles per second or hertz, Hz) and the length of each cycle (the wavelength):
c = νλ (8-1)
The reciprocal of the wavelength is called the wave number, $\textstyle\overline{\nu}$:
$\textstyle\overline{\nu}$ = 1/λ
Its units are commonly waves per centimeter, or cm−1.
The electromagnetic spectrum as we know it is shown in Figure 8-5a. The scale is logarithmic rather than linear in wavelength; that is, it is in increasing powers of 10. On this logarithmic scale, the portion of the electromagnetic radiation that our eyes can see is only a small sector halfway between radio waves and gamma rays. The visible part of the spectrum is shown in Figure 8-5b.
Example 1
Light of wavelength 5000 Å (or 5 × 10−5 cm) falls in the green region of the visible spectrum. Calculate the wave number, $\textstyle\overline{\nu}$, corresponding to this wavelength.
Solution
The wave number is equal to the reciprocal of the wavelength, so
$\textstyle\overline{\nu}$ = 1/λ
= 1/5 × 10-5 cm = 0.2 × 105 cm-1
= 2 × 104 cm-1
The Ultraviolet CatastropheEdit
Classical physics gave physicists serious trouble even when they used it to try to explain why a red-hot iron bar is red. Solids emit radiation when they are heated. The ideal radiation from a perfect absorber and emitter of radiation is called blackbody radiation. The spectrum, or plot of relative intensity against frequency, of radiation from a red-hot solid is shown in Figure 8-6a. Since most of the radiation is in the red and infrared frequency regions, we see the color of the object as red. As the temperature is increased, the peak of the spectrum moves to higher frequencies, and we see the hot object as orange, then yellow, and finally white when enough energy is radiated through the entire visible spectrum.
The difficulty in this observation is that classical physics predicts that the curve will keep rising to the right rather than falling after a maximum. Thus there should be much more blue and ultraviolet radiation emitted than is actually observed, and all heated objects should appear blue to our eyes. This complete contradiction of theory by facts was called the ultraviolet catastrophe by physicists of the time.
In 1900, Max Planck provided an explanation for this paradox. To do this he had to discard a hallowed tenet of science-that variables in nature change in a continuous way (nature does not make jumps). According to classical physics, light of a certain frequency is emitted because charged objects- atoms or groups of atoms- in a solid vibrate or oscillate with that frequency. We could thus theoretically calculate the intensity curve of the spectrum if we knew the relative number of oscillators that vibrate with each frequency.All frequencies are thought to be possible, and the energy associated with a particular frequency depends only on how many oscillators are vibrating with that frequency . There should be no lack of high-frequency oscillators in the blue and ultraviolet regions.
Planck made the revolutionary suggestion that the energy of electromagnetic radiation comes in packages, or quanta. The energy of one package of radiation is proportional to the frequency of the radiation:
E = hν (8-2)
The proportionality constant, h, is known as Planck's constant and has the value 6.6262 × 10−34 J sec. By Planck's theory, a group of atoms cannot emit a small amount of energy at a high frequency; high frequencies can be emitted only by oscillators with a large amount of energy, as given by E = . The probability of finding oscillators with high frequencies is therefore slight because the probability of finding groups of atoms with such unusually large vibrational energies is low. Instead of rising, the spectral curve falls at high frequencies, as in Figure 8-6.
Was Planck's theory correct, or was it only an ad hoc explanation to account for one isolated phenomenon? Science is plagued with theories that explain the phenomenon for which they were invented, and thereafter never explain another phenomenon correctly. Was the idea that electromagnetic energy comes in bundles of fixed energy that is proportional to frequency only another one-shot explanation?
The Photoelectric EffectEdit
Albert Einstein (1879–1955) provided another example of the quantization of energy, in 1905, when he successfully explained the photoelectric effect, in which light striking a metal surface can cause electrons to be given off. (Photocells in automatic doors use the photoelectric effect to generate the electrons that operate the door-opening circuits.) For a given metal there is a minimum frequency of light below which no electrons are emitted, no matter how intense the beam of light. To classical physicists it seemed nonsensical that for some metals the most intense beam of red light could not drive off electrons that could be ejected by a faint beam of blue light.
Einstein showed that Planck's hypothesis explained such phenomena beautifully. The energy of the quanta of light striking the metal, he said, is greater for blue light than for red. As an analogy, imagine that the low-frequency red light is a beam of Ping-Pong balls and that the high-frequency blue light is a beam of steel balls with the same velocity. Each impact of a quantum of energy of red light is too small to dislodge an electron; in our analogy, a steady stream of Ping-Pong balls cannot do what one rapidly moving steel ball can. These quanta of light were named photons. Because of the successful explanation of both the blackbody and photoelectric effects, physicists began recognizing that light behaves like particles as well as like waves.
Example 2
Consider once again the green light in Example 1. The relationship E = allows us to calculate the energy of one green photon. What is this energy in kilojoules? In kilojoules per mole of green photons?
Solution
Let us assume that we know the wavelength to two significant digits, 5.0 X 10−5 cm. The frequency, ν, of this green light is
c = λν = 3.0 × 1010 cm sec-1 = (5.0 × 10-5 cm) ν
ν = $\textstyle\frac{3.0 \times 10^{10} cm sec^{-1}}{5.0 \times 10^{.5} cm}$
= 0.60 × 1015 sec-1 (or 0.60 × 1015 Hz)
The energy of one green photon, then, is
E =
(6.63 × 10-34 J sec)(0.60 × sec-1)
4.0 × 10-19 J, or 4.0 × 10-22 kJ
This is the energy of one green photon. To obtain the energy of a mole of green photons, we must multiply by Avogadro's number:
E = (4.0 × 10-22 kJ photon-1)(6.02 × 10-23 photons mole-1)
= 2.4 × 102 kJ mole-1
The Spectrum of the Hydrogen AtomEdit
The most striking example of the quantization of light, to a chemist, appears in the search for an explanation of atomic spectra. Isaac Newton (1642–1727) was one of the first scientists to demonstrate with a prism that white light is a spectrum of many colors, from red at one end to violet at the other. We know now that the electromagnetic spectrum continues on both sides of the small region to which our eyes are sensitive; it includes the infrared at low frequencies and the ultraviolet at high frequencies.
All atoms and molecules absorb light of certain characteristic frequencies. The pattern of absorption frequencies is called an absorption spectrum and is an identifying property of any particular atom or molecule. The absorption spectrum of hydrogen atoms is shown in Figure 8-7. The lowest-energy absorption corresponds to the line at 82,259 cm−1. Notice that the absorption lines are crowded closer together as the limit of 109,678 cm−1 is approached. Above this limit absorption is continuous.
If atoms and molecules are heated to high temperatures, they emit light of certain frequencies. For example, hydrogen atoms emit red light when they are heated. An atom that possesses excess energy (e.g., an atom that has been heated) emits light in a pattern known as its emission spectrum. A portion of the emission spectrum of atomic hydrogen is shown in Figure 8-8. Note that the lines occur at the same wave numbers in the two types of spectra.
If we look more closely at the emission spectrum in Figure 8-8, we see that there are three distinct groups of lines. These three groups or series are named after the scientists who discovered them. The series that starts at 82,259 cm−1 and continues to 109,678 cm−1 is called the Lyman series and is in the ultraviolet portion of the spectrum. The series that starts at 15,233 cm−1 and continues to 27,420 cm−1 is called the Balmer series and covers a large portion of the visible and a small part of the ultraviolet spectrum. The lines between 5332 cm−1 and 12,186 cm−1 are called the Paschen series and fall in the near-infrared region. The Balmer spectra of hydrogen from several stars are shown in Figure 8-9.
J. J. Balmer proved, in 1885, that the wave numbers of the lines in the Balmer spectrum of the hydrogen atom are given by the empirical relationship
$\textstyle\overline{\nu} =$ RH $\textstyle\times (\frac{1}{4} - \frac{1}{n^2})$ n = 3, 4, 5, . . . (8-3)
Later, Johannes Rydberg formulated a general expression that gives all of the line positions. This expression, called the Rydberg equation, is
$\textstyle\overline{\nu}$ = RH $\textstyle\times (\frac{1}{n{_1^2}} - \frac{1}{n^{_2^2}})$ (8-4)
In the Rydberg equation n1 and n2 are integers, with n2 greater than n1; RH is called the Rydberg constant and is known accurately from experiment to be 109,677.581 cm−1.
Example 3
Calculate $\textstyle\overline{\nu}$ for the lines with n1 = 1 and n2 = 2, 3, and 4.
Solution
n1 = 1, n2 = 2 line:
$\textstyle\overline{\nu} = 109,678 \textstyle (\frac{1}{1^2} - \frac{1}{2^2}) = 109,678 (1 - \frac{1}{4}) = 82,259 cm^{-1}$
n1 = 1, n2 = 3 line:
$\textstyle\overline{\nu} = 109,678 (\frac{1}{1^2} - \frac{1}{3^2}) = 109,678 (1 - \frac{1}{9}) = 97,492 cm^{-1}$
n1 = 1, n2 = 4 line:
$\textstyle\overline{\nu} = 109,678 (\frac{1}{1^2} - \frac{1}{4^2}) = 109,678 (1 - \frac{1}{16}) = 102,823 cm^{-1}$
We see that the wave numbers obtained in Example 3 correspond to the first three lines in the Lyman series. Thus we expect that the Lyman series corresponds to lines calculated with n1 = 1 and n2 = 2, 3, 4, 5, . . . . We can check this by calculating the wave number for the line with n2 = 1 and n2 = . n1 = 1, n2 = line:
$\textstyle\overline{\nu} = 109,678 (1 - 0) = 109,678 cm^{-1}$
The wave number 109,678 cm-1 corresponds to the highest emission line in the Lyman series.
The wave number for n1 = 2 and n2 = 3 is
$\textstyle\overline{\nu} = 109,678 (\frac{1}{4} - \frac{1}{9}) = 15,233 cm^{-1}.$
This corresponds to the first line in the Balmer series. Thus, the Balmer series corresponds to the n1 = 2, n2 = 3, 4, 5, 6, . . . lines. You probably would expect the lines in the Paschen series to correspond to n1 = 3, n2 = 4, 5, 6, 7, . . . They do. Now you should wonder where the lines are with n1 = 4, n2 = 5, 6, 7, 8, . . . , and n1 = 5, n2 = 6, 7, 8, 9, . . . . They are exactly where the Rydberg equation predicts they should be. The n = 4 series was discovered by Brackett and the n = 5 series was discovered by Pfund. The series with n = 6 and higher are located at very low frequencies and are not given special names.
The Rydberg formula, equation 8-4, is a summary of observed facts about hydrogen atomic spectra. It states that the wave number of a spectral line is the difference between two numbers, each inversely proportional to the square of an integer. If we draw a set of horizontal lines at a distance RH/n' down from a baseline, with n = 1, 2, 3, 4, . . ., then each spectral line in any of the hydrogen series is observed to correspond to the distance between two such horizontal lines in the diagram (Figure 8-10). The Lyman series occurs between line n = 1 and those above it; the Balmer series occurs between line n = 2 and those above it; the Paschen series occurs between line n = 3 and those above it; and the higher series are based on lines n = 4, 5, and so on. Is the agreement between this simple diagram and the observed wave numbers of spectral lines only a coincidence? Does the idea of a wave number of an emitted line being the difference between two "wave-number levels" have any physical significance, or is this just a convenient graphical representation of the Rydberg equation?
Bohr's Theory of the Hydrogen AtomEdit
In 1913, Niels Bohr (1885–1962) proposed a theory of the hydrogen atom that, in one blow, did away with the problem of Rutherford's unstable atom and gave a perfect explanation of the spectra we have just discussed.
There are two ways of proposing a new theory in science, and Bohr's work illustrates the less obvious one. One way is to amass such an amount of data that the new theory becomes obvious and self-evident to any observer. The theory then is almost a summary of the data. This is essentially the way Dalton reasoned from combining weights to atoms. The other way is to make a bold new assertion that initially does not seem to follow from the data, and then to demonstrate that the consequences of this assertion, when worked out, explain many observations. With this method, a theorist says, "You may not see why, yet, but please suspend judgment on my hypothesis until I show you what I can do with it." Bohr's theory is of this type.
Bohr answered the question of why the electron does not spiral into the nucleus by simply postulating that it does not. In effect, he said to classical physicists: "You have been misled by your physics to expect that the electron would radiate energy and spiral into the nucleus. Let us assume that it does not, and see if we can account for more observations than by assuming that it does." The observations that he explained so well are the wavelengths of lines in the atomic spectrum of hydrogen.
Bohr's model of the hydrogen atom is illustrated in Figure 8-11: an electron of mass me moving in a circular orbit at a distance r from a nucleus. If the electron has a velocity of v, it will have an angular momentum of mevr. (To appreciate what angular momentum is, think of an ice skater spinning on one blade like a top. The skater begins spinning with his arms extended. As he brings his arms to his sides, he spins faster and faster. This is because, in the absence of any external forces, angular momentum is conserved. As the mass of the skater's arms comes closer to the axis of rotation, or as r decreases, the velocity of his arms must increase in order that the product mvr remain constant.) Bohr postulated, as the first basic assumption of his theory, that in a hydrogen atom there could only be orbits for which the angular momentum is an integral multiple of Planck's constant divided by 2$\pi$:
mevr $= \textstyle n \frac{h}{2\pi}$
There is no obvious justification for such an assumption; it will be accepted only if it leads to the successful explanation of other phenomena. Bohr then showed that, with no more new assumptions, and with the laws of classical mechanics and electrostatics, his principle leads to the restriction of the energy of an electron in a hydrogen atom to the values
E = $\textstyle - \frac{k}{n^2}$ n = 1, 2, 3, 4, . . . (8-5)
The integer n is the same integer as in the angular momentum assumption, mevr = n(h/ 2[/itex]\pi[/itex]); k is a constant that depends only on Planck's constant, h, the mass of an electron, me, and the charge on an electron, e:
k = $\textstyle\frac{2\pi^2m_ee^4}{h^2} =$13.595 electron volts (eV)* atom-1
= 1312 kJ mole-1
The radius of the electron's orbit also is determined by the integer n:
r = n2a0 (8-6)
The constant, a0, is called the first Bohr radius and is given in Bohr's theory by
a0 = $\textstyle\frac{h^2}{4\pi^2m_ee^2} =$ 0.529Å
The first Bohr radius is often used as a measure of length called the atomic unit, a.u.
The energy that an electron in a hydrogen atom can have is quantized, or limited to certain values, by equation 8-5. The integer, n, that determines these energy values is called the quantum number. When an electron is removed (ionized) from an atom, that electron is described as excited to the quantum state n = . From equation 8-5, we see that as n approaches , E approaches zero. Thus, the energy of a completely ionized electron has been chosen as the zero energy level. Because energy is required to remove an electron from an atom, an electron that is bound to an atom must have less energy that this, and hence a negative energy. The relative sizes of the first five hydrogen-atom orbits are compared in Figure 8-12.
Example 4
For a hydrogen atom, what is the energy, relative to the ionized atom, of the ground state, for which n = 1? How far is the electron from the nucleus
in this state? What are the energy and radius of orbit of an electron in the first excited state, for which n = 2?
Solution
The answers are
E1 = $-\textstyle\frac{k}{1^2}$ = -1312 kJ mole-1
E2 = $-\textstyle\frac{k}{2^2}$ = -328.0 kJ mole-1
r1 = 12 × 0.529 Å = 0.529 Å
r2 = 22 × 0.529 Å = 2.12 Å
• An electron volt is equal to the amount of energy an electron gains as it passes from a point of low potential to a point one volt higher in potential (1 eV = 1.6022 × 10−19 J).
Example 5
Using the Bohr theory, calculate the ionization energy of the hydrogen atom.
Solution
The ionization energy, IE, is that energy required to remove the electron, or to go from quantum state n = 1 to n = . This energy is
IE = E - E1 = 0.00 - (- 1312 kJ mole-1)
= + 1312 kJ
Example 6
Diagram the energies available to the hydrogen atom as a series of horizontal lines. Plot the energies in units of k for simplicity. Include at least the first eight quantum levels and the ionization limit. Compare your result with Figures 8-10 and 8-13.
Try this one yourself.
In the second part of his theory, Bohr postulated that absorption and emission of energy occur when an electron moves from one quantum state to another. The energy emitted when an electron drops from state n2 to a lower quantum state n1 is the difference between the energies of the two states:
ΔE = E1 - E2 = -k$\textstyle ( \frac{1}{n_1^2} - \frac{1}{n_2^2})$ (8-7)
The light emitted is assumed to be quantized in exactly the way predicted from the blackbody and photoelectric experiments of Planck and Einstein:
|ΔE| = hv = hc$\textstyle\overline{v}$ (8-8)
If we divide equation 8-7 by hc to convert from energy to wave number units, we obtain the Rydberg equation,
$R_H = \textstyle\frac{k}{hc} = \frac{2\pi^2m_ee^4}{h^3c} = 109,737.3 cm^{-1}$
Recall that the experimental value of RH is 109,677.581 cm-1.
The graphic representation of the Rydberg equation, Figure 8-10, now is seen to be an energy-level diagram of the possible quantum states of the hydrogen atom. We can see why light is absorbed or emitted only at specific wave numbers. The absorption of light, or the heating of a gas, provides the energy for an electron to move to a higher orbit. Then the excited hydrogen atom can emit energy in the form of light quanta when the electron falls back to a lower-energy orbit. From this emission come the different series of spectral lines:
1. The Lyman series of lines arises from transitions from the n = 2, 3, 4, . . . levels to the ground state (n = 1)
2. The Balmer series arises from transitions from the n = 3, 4, 5, . . . levels to the n = 2 level.
3. The Paschen series arises from transitions from the n = 4, 5, 6, . . . levels to the n = 3 level.
An excited hydrogen atom in quantum state n = 8 may drop directly to the ground state and emit photon in the Lyman series, or it may drop first to n = 3, emit a photon in the Paschen series, and then drop to n = 1 and emit a photon in the Lyman series. The frequency of each photon depends on the energy difference between levels:
ΔE = Ea - Eb = hv
By cascading down the energy levels, the electron in one excited hydrogen atom can successively emit photons in several series. Therefore, all series are present in the emission spectrum from hot hydrogen. However, when measuring the absorption spectrum of hydrogen gas at lower temperatures we find virtually all the hydrogen atoms in the ground state. Therefore, almost all the absorption will involve transitions from n = 1 to higher states, and only the Lyman series will be observed.
Energy Levels of a General One-Electron AtomEdit
Bohr's theory can also be used to calculate the ionization energy and spectral lines of any atomic species possessing only one electron (e.g., He+, Li2+, Be3+ ). The energy of a Bohr orbit depends on the square of the charge on the atomic nucleus (Z is the atomic number):
E = $-\frac{Z^2k}{n^2}$
where
k = 13.595 eV or 1312 kJ mole-1
n = 1, 2, 3, 4, . . .
The equation reduces to equation 8-5 in the case of atomic hydrogen (Z = 1).
Example 7
Calculate the third ionization energy of a lithium atom.
Solution
A lithium atom is composed of a nucleus of charge +3 (Z = 3) and three electrons. The first ionization energy, IE1, of an atom with more than one electron is the energy required to remove one electron. For lithium,
Li(g) Li+(g) + e- ΔE = IE1
The energy needed to remove an electron from the unipositive ion, Li+, is defined as the second ionization energy, IE2, of lithium,
Li(g) Li2+(g) + e- ΔE = IE2
and the third ionization energy, IE3, of lithium is the energy required to remove the one remaining electron from Li2+. For lithium, Z = 3 and IE3 = (3)2(13.595 eV) = 122.36 eV. (The experimental value is 122.45 eV.)
The Need for a Better TheoryEdit
The Bohr theory of the hydrogen atom suffered from a fatal weakness: It explained nothing except the hydrogen atom and any other combination of a nucleus and one electron. For example, it could account for the spectra of He+ and Li2+, but it did not provide a general explanation for atomic spectra. Even the alkali metals (Li, Na, K, Rb, Cs), which have a single valence electron outside a closed shell of inner electrons, produce spectra that are at variance with the Bohr theory. The lines observed in the spectrum of Li could be accounted for only by assuming that each of the Bohr levels beyond the first was really a collection of levels of different energies, as in Figure 8-13: two levels for n = 2, three levels for n = 3, four for n = 4, and so on. The levels for a specific n were given letter symbols based on the appearance of the spectra involving these levels: s for "sharp," p for "principal," d for "diffuse," and f for "fundamental."
Arnold Sommerfeld (1868–1951) proposed an ingenious way of saving the Bohr theory. He suggested that orbits might be elliptical as well as circular. Furthermore, he explained the differences in stability of levels with the same principal quantum number, n, in terms of the ability of the highly elliptical orbits to bring the electron closer to the nucleus (Figure 8-14). For a point nucleus of charge +1 in Hydrogen, the energies of all levels with the same n would be identical. But for a nucleus of +3 screened by an inner shell of two electrons Li, an electron in an outer circular orbit would experience a net attraction of +1, whereas one in a highly elliptical orbit would penetrate the screening shell and feel a charge approaching +3 for part of its traverse. Thus, the highly elliptical orbits would have the most additional stability illustrated in Figure 8-13. The s orbits, being the most elliptical of all in Sommerfeld's model, would be much more stable than the others in the set of common n.
The Sommerfeld scheme led no further than the alkali metals. Again an impasse was reached, and an entirely fresh approach was needed.
Particles of Light and Waves of MatterEdit
At the beginning of the twentieth century, scientists generally believed that all physical phenomena could be divided into two distinct and exclusive classes. The first class included all phenomena that could be described by laws of classical, or Newtonian, mechanics of motion of discrete particles.
The second class included all phenomena showing the continuous properties of waves.
One outstanding property of matter, apparent since the time of Dalton, is that it is built of discrete particles. Most material substances appear to be continuous: water, mercury, salt crystals, gases. But if our eyes could see the nuclei and electrons that constitute atoms, and the fundamental particles that make up nuclei, we would discover quickly that every material substance in the universe composed of a certain number of these basic units and therefore is quantized. Objects appear continuous only because of the minuteness of the individual units.
In contrast, light was considered to be a collection of waves traveling through space at a constant speed; any combination of energies and frequencies was possible. However, Planck, Einstein, and Bohr showed that light when observed under the right conditions, also behaves as though it occurs in particles, or quanta.
In 1924, the French physicist Louis de Broglie (b. 1892) advanced the complementary hypothesis that all matter possesses wave properties. De Broglie pondered the Bohr atom, and asked himself where, in nature, quantization of energy occurs most naturally. An obvious answer is in the vibration of a string with fixed ends. A violin string can vibrate with only a selected set of frequencies: a fundamental tone with the entire string vibrating as a unit, and overtones of shorter wavelengths. A wavelength in which the vibration fails to come to a node (a place of zero amplitude) at both ends of the string would be an impossible mode of vibration (Figure 8-15). The vibration of a string with fixed ends is quantized by the boundary conditions that the ends cannot move.
Can the idea of standing waves be carried over to the theory of the Bohr atom? Standing waves in a circular orbit can exist only if the circumference of the orbit is an integral number of wavelengths (Figure 8-15c, d). If it is not, waves from successive turns around the orbit will be out of phase and will cancel. The value of the wave amplitude at 10° around the orbit from a chosen point will not be the same as at 370° or 730°, yet all these represent the same point in the orbit. Such ill-behaved waves are not single-valued at any point on the orbit: Single-valuedness is a boundary condition on acceptable waves.
For single-valued standing waves around the orbit, the circumference is an integer, n, times the wavelength:
2$\pi$r = nλ
But from Bohr's original assumption about angular momentum,
2$\pi$r = n$\textstyle ( \frac{h}{m_ev})$
Therefore, the idea of standing waves leads to the following relationship between the mass of the electron me its velocity, v, and its wavelength, λ:
λ = $\textstyle\frac{h}{m_ev}$ (8-10)
De Broglie proposed this relationship as a general one. With every particle, he said, there is associated a wave. The wavelength depends on the mass of the particle and how fast it is moving. If this is so, the same sort of diffraction from crystals that von Laue observed with x rays should be produced with electrons.
In 1927, C. Davisson and L. H. Germer demonstrated that metal foils diffract a beam of electrons exactly as they diffract an x-ray beam, and that the wavelength of a beam of electrons is given correctly by de Broglie's relationship (Figure 8-16). Electron diffraction is now a standard technique for determining molecular structure.
Example 8
A typical electron diffraction experiment is conducted with electrons accelerated through a potential drop of 40,000 volts, or with 40,000 eV of
energy. What is the wavelength of the electrons?
Solution
First convert the energy, E, from electron volts to joules:
E = 40,000 eV $\times \textstyle\frac{1.6022 \times 10^{-19} J}{1 eV}$ = 6.409 × 10-15 J
(This and several other useful conversion factors, plus a table of the values of frequently used physical constants, are in Appendix 2.) Since the energy is E = $\textstyle\frac{1}{2}$mev2, the velocity of the electrons is
v = $\textstyle(\frac{2E}{m_e}^{1/2} = \frac{2 \times 6.409 \times 10^{-15}kg m^2 sec^-2}{9.110 \times 10^{-31} kg})^{1/2}$
(1.407 × 1016 m2 sec-2)1/2 = 1.186 × 108 m sec-1
(In the expression E = $\textstyle\frac{1}{2}$mev2, if the mass is in kilograms and the velocity is in m sec−1, then the energy is in joules: 1 J equals 1 kg m2 sec−2 of energy. We used this conversion of units in the preceding step. The mass of the electron, me = 9.110 × 10−31 kg, is found in Appendix 2.) The momentum of the electron, mev, is
mev = 9.110 × 10-31 kg × 1.186 × 108 m sec-1
= 10.80 × 10-23 kg m sec-1
Finally, the wavelength of the electron is obtained from the de Broglie relationship:
λ = $\textstyle\frac{h}{m_ev} = \frac{6.6262 \times 10^{-34}J sec}{10.80 \times 10 ^{-23} kg m sec ^{-1}}$
= 0.06130 × 10-10 $\textstyle\frac{kg m^2 sec^{-2} sec}{kg m sec^{-1}} =$ 0.06130 × 10-10 m
= 0.06130 Å
So 40-kilovolt (kV) electrons produce the diffraction effects expected from waves with a wavelength of six-hundredths of an angstrom.
Such calculations are all very well, but the question remains: Are electrons waves or are they particles? Are light rays waves or particles? Scientists worried about these questions for years, until they gradually realized that they were arguing about language and not about science. Most things in our everyday experience behave either as what we would call waves or as what we would call particles, and we have created idealized categories and used the words wave and particle to identify them. The behavior of matter as small as electrons cannot be described accurately by these large-scale categories. Electrons, protons, neutrons, and photons are not waves, and they are not particles. Sometimes they act as if they were what we commonly call waves, and in other circumstances they act as if they were what we call particles. But to demand, "Is an electron a wave or a particle?" is pointless.
This wave-particle duality is present in all objects; it is only because of the scale of certain objects that one behavior predominates and the other is negligible. For example, a thrown baseball has wave properties, but a wavelength so short we cannot detect it.
Example 9
A 200-g baseball is thrown with a speed of 30 m sec−1 Calculate its de Broglie wavelength.
Solution
The answer is λ = 1.1 × 10−34 m = 1.1 × 10−24 Å.
Example 10
How fast (or rather, how slowly) must a 200-g baseball travel to have the same de Broglie wavelength as a 40-kV electron?
Solution
The wavelength of a 40-kV electron is 0.0613 Å.
v = $\textstyle\frac{h}{\lambda m}$ = $\textstyle \frac{6.6262 \times 10^{-34} kg m^2 sec^{-2} sec}{0.06130 \times 10^{-10}m \times 0.200 kg}$
= 0.540 × 10-21 m sec-1 = 1.70 × 10-4 Å year-1
Such a baseball would take over 10,000 years to travel the length of a carbon-carbon bond, 1.54 Å. This sort of motion is completely outside our experience with baseballs; thus we never regard baseballs as having wave properties.
The Uncertainty PrincipleEdit
One of the most important consequences of the dual nature of matter is the uncertainty principle, proposed in 1927 by Werner Heisenberg (1901–1976). This principle states that you cannot know simultaneously both the position and the momentum of any particle with absolute accuracy. The product of the uncertainty in position, $\delta x$, and in momentum, $\delta(mv)$, will be equal to or greater than Planck's constant divided by 4$\pi$:
[Δx][Δ(mvx)] ≥ $\textstyle\frac{h}{4\pi}$ (8-11)
We can understand this principle by considering how we determine the position of a particle. If the particle is large, we can touch it without disturbing it seriously. If the particle is small, a more delicate means of locating it is to shine a beam of light on it and observe the scattered rays. Yet light acts as if it were made of particles - photons - with energy proportional to frequency: E = . When we illuminate the object, we are pouring energy on it. If the object is large, it will become warmer; if the object is small enough, it will be pushed away and its momentum will become uncertain. The least interference that we can cause is to bounce a single photon off the object and watch where the photon goes. Now we are caught in a dilemma. The detail in an image of an object depends on the fineness of the wavelength of the light used to observe the object. (The shorter the wavelength, the more detailed the image.) But if we want to avoid altering the momentum of the atom, we have to use a low-energy photon. However, the wavelength of the low-energy photon would be so long that the position of the atom would be unclear. Conversely, if we try to locate the atom accurately by using a short-wavelength photon, the energy of the photon sends the atom ricocheting away with an uncertain momentum (Figure 8-17). We can design an experiment to obtain an accurate value of either an atom's momentum or its position, but the product of the errors in these quantities is limited by equation 8-11.
Example 11
Suppose that we want to locate an electron whose velocity is 1.00 × 106 m sec−1 by using a beam of green light whose frequency is 0.600 × 1015 sec−1. How does the energy of one photon of such light compare with the energy of the electron to be located?
Solution
The energy of the electron is
E = $\textstyle\frac{1}{2}m_ev^2 = \frac{9.110 \times 10^{-31} kg \times 1.00 \times 10^{12} m^2 sec^{-2}}{2}$
= 4.56 × 10-19 J
But the energy of the photon is almost as large:
Ep = = 6.6262 × 10-34 J sec × 0.600 × 1015 sec-1
= 3.97 × 10-19 J
Finding the position and momentum of such an electron with green light is as questionable a procedure as finding the position and momentum of one billiard ball by striking it with another. In either case, you detect the particle at the price of disturbing its momentum. As a final difficulty, green light is a hopelessly coarse yardstick for finding objects of atomic dimensions. An atom is about 1 Å in radius, whereas the wavelength of green light is around 5000 Å. Shorter wavelengths make the energy quandary worse.
We do not see the uncertainty limitations in large objects because of the sizes of the masses and velocities involved. Compare the following two problems.
Example 12
An electron is moving with ~ velocity of 106 m sec−1 Assume that we can measure its position to 0.01 Å, or 1% of a typical atomic radius. Compare the uncertainty in its momentum, p, with the momentum of the electron itself.
Solution
The uncertainty in position is Δx ≅ 0.01 Å = 0.01 × 10−10m. The momentum of the electron is approximately
p = mev ≅ 10-30 kg × 106 m sec-1 = 10-24 kg m sec-1
By the Heisenberg uncertainty principle, the uncertainty in the knowledge of the momentum is
Δp = $\textstyle\frac{h/4\pi}{\vartriangle x}$\textstyle\frac{0.5 \times 10^{-34} kg m^2 sec^{-1}}{0.01 \times 10^{-22} kg m sec^{-1}}$
≅ 0.5 × 10-22 kg m sec-1
The uncertainty in the momentum of the electron is 50 times as great as the momentum itself!
Example 13
A baseball of mass 200 g is moving with a velocity of 30 m sec−1. If we can
locate the baseball with an error equal in magnitude to the wavelength of light used (e.g., 5000 Å), how will the uncertainty in momentum compare with the total momentum of the baseball?
Solution
The momentum, p, of the baseball is 6 kg m sec−1, and Δp = 1 × 10−28 kg m sec−1. The intrinsic uncertainty in the momentum is only one part in 1028, far below any possibility of detection in an experiment.
Wave EquationsEdit
In 1926, Erwin Schrödinger (1887–1961) proposed a general wave equation for a particle. The mathematics of the Schrödinger equation is beyond us, but the mode of attack, or the strategy of finding its solution, is not. If you can see how physicists go about solving the Schrödinger equation, even though you cannot solve it yourself, then quantization and quantum numbers may be a little less mysterious. This section is an attempt to explain the method of solving a differential equation of motion* of the type that we encounter in quantum mechanics. We shall explain the strategy with the simpler analogy of the equation of a vibrating string.
The de Broglie wave relationship and the Heisenberg uncertainty principle should prepare you for the two main features of quantum mechanics that contrast it with classical mechanics:
1. Information about a particle is obtained by solving an equation for a wave.
2. The information obtained about the particle is not its position;
rather, it is the probability of finding the particle in a given region of space.
We can't say whether an electron is in a certain place around an atom, but we can measure the probability that it is there rather than somewhere else.
Wave equations are familiar in mechanics. For instance, the problem of the vibration of a violin string is solved in three steps:
1. Set up the equation of motion of a vibrating string. This equation will involve the displacement or amplitude of vibration, A (x), as a function of position along the string, x.
2. Solve the differential equation to obtain a general expression for amplitude. For a vibrating string with fixed ends, this general expression is a sine wave. As yet, there are no restrictions on wavelength or frequency of vibration.
3. Eliminate all solutions to the equation except those that leave the ends of the string stationary. This restriction on acceptable solutions of the wave equation is a boundary condition. Figure 8-15a shows solutions that fit this boundary condition of fixed ends of the string; Figure 8-15b shows solutions that fail. The only acceptable vibrations are those with λ = 2a/n, or $\textstyle\overline{\nu}$ = n/2a, in which n = 1, 2, 3, 4, .... The boundary conditions and not the wave equation are responsible for the quantization of the wavelengths of string vibration.
*Equations of motion are always differential equations because they relate the change in one quantity to the change in another, such as change in position to change in time.
Exactly the same procedure is followed in quantum mechanics:
1. Set up a general wave equation for a particle. The Schrödinger equation is written in terms of the function ψ(x,v,z) (where ψ is the Greek letter psi), which is analogous to the amplitude, A(x), in our violin-string analogy. The square of this amplitude, |ψ|2, is the relative probability density of the particle at position (x, y, z). That is, if a small element of volume, dv, is located at (x, y, z), the probability of finding an electron within that element of volume is If |ψ|2 dv.
2. Solve the Schrödinger equation to obtain the most general expression for ψ(x, y, z).
3. Apply the boundary conditions of the particular physical situation. If the particle is an electron in an atom, the boundary conditions are that |ψ|2 must be continuous, single-valued, and finite everywhere. All these conditions are only common sense. First, probability functions do not fluctuate radically from one place to another; the probability of finding an electron a few thousandths of an angstrom from a given position will not be radically different from the probability at the original position. Second, the probability of finding an electron in a given place cannot have two different values simultaneously. Third, since the probability of finding an electron somewhere must be 100%, or 1.000, if the electron really exists, the probability at anyone point cannot be infinite.
We now shall compare the wave equation for a vibrating string and the Schrödinger wave equation for a particle. In this text you will not be expected to do anything with either equation, but you should note the similarities between them.
Vibrating string. The amplitude of vibration at a distance x along the string is A(x). The differential equation of motion is
$\textstyle\frac{d^2A_{(x)}}{dx^2} + 4\pi^2\overline{\nu}^2A_{x} = 0$ (8-12)
The general solution to this equation is a sine function
$A_{(x)} = A_{max}sin(2\pi\overline{\nu}x + \alpha)$
and the only acceptable solutions (Figure 8-15a) are those for which ,math>\textstyle\overline{\nu}[/itex] = n/2a, where n = 1, 2, 3, 4, ... , and for which the phase shift, α, is zero:
$A_{(x)} = A_{max}sin$ $\textstyle n(\frac{\pi}{a})x$
Schrödinger equation. The square of the amplitude If(x.v.z>i' is the probability density of the particle at (x,), z). The differential equation is |ψ(x, y, z)|2 is the probability density of the particle at (x, y,z). The differential equation is
$\textstyle\frac{\partial^2\psi}{\partial x^2} + \frac{\partial^2\psi}{\partial y^2} + \frac{\partial^2\psi}{\partial z^2} + \frac{8\pi^2m_e}{h}(E - V_{(x,y,z)}\psi_{(x,y,z)} = 0$ (8-13)
V is the potential energy function at (x, y, z), and me is the mass of the electron.
Although solving equation 8-13 is not a simple process, it is purely a mathematical operation; there is nothing in the least mysterious about it. The energy, E, is the variable that is restricted or quantized by the boundary conditions on |ψ|2. Our next task is to determine what the possible energy states are.
The Hydrogen AtomEdit
The sine function that is the solution of te equation for the vibrating string is characterized by one integral quantum muber: n = 1, 2, 3, 4, . . . . The first few acceptable sine functions are
These are the first four curves in Figure 8-15a.
An atom is three-dimensional, whereas the string has only length. The solutions of the Schrödinger equation for the hydrogen atom are characterized by three integer quantum numbers: n, l, and m. These arise when solving the equation for the wave function, Ψ, which is analogous to the function An(x) in the vibrating string analogy. In solving the Schrödinger equation, we divide it into three parts. The solution of the radial part describes how the wave function, Ψ, varies with distance from the center of the atom. If we borrow the customary coordinate system of the earth, an azimuthal part produces a function that reveals how Ψ varies with north or south latitude, or distance up or down from the equator of the atom. Finally, an angular part is a third function that suggests how the wave function varies with east-west longitude around the atom. The total wave function, Ψ, is the product of these three functions. The wave functions that are solutions to the Schrödinger equation for the hydrogen atom are called orbitals.
In the process of separating the parts of the wave function, a constant, n, appears in the radial expression, another constant, l, occurs in the radial and azimuthal expressions, and m appears in the azimuthal and angular expressions. The boundary conditions that give physically sensible solutions to these three equations are that each function (radial, azimuthal, and angular) be continuous, single-valued, and finite at all points. These conditions will not be met unless n, l, and m are integers, l is zero or a positive integer less than n, and m has a value from -l to + l. From a one-dimensional problem (the vibrating string) we obtained one quantum number. With a three-dimensional problem, we obtain three quantum numbers.
The principal quantum number, n, can be any positive integer: n = 1, 2, 3, 4, 5, . . . . The azimuthal quantum number, l, can have any integral value from 0 to n - 1. The magnetic quantum number, m, can have any integral value from -l to +l. The different quantum states that the electron can have are listed in Table 8-1. For one electron around an atomic nucleus, the energy depends only on n. Moreover, the energy expression is exactly the same as in the Bohr theory:
En = $\textstyle - \frac{Z^2k}{n^2}$ ::k$= \textstyle\frac{2\pi^2m_ee^4}{h^2}$
For Z = 1 (the hydrogen atom), we have simply:
En = $-\textstyle\frac{k}{n^2}$
where k = 13.595 eV or 1312 kJ mole−1.
Quantum states, with l = 0, 1, 2, 3, 4, 5, . . ., are called the s, p, d, f, g, h, . . . states, in an extension of the old spectroscopic notation (Figure 8-13). The wave functions corresponding to s, p, d, . . . states are called s, p, d, . . . orbitals. All of the l states for the same n have the same energy in the hydrogen atom; the energy-level diagram is as in Figure 8-10.
Example 14
An electron in atomic hydrogen has a principal quantum number of 5. What are the possible values of l for this' electron? When l = 3, what are the possible values of m? What is the ionization energy (in electron volts) of this electron? What would it be in the same n state in He+?
Solution
With n = 5, l may have a value of 4, 3, 2, 1, or 0. For l = 3, there are seven possible values of m: 3, 2, 1, 0, -1, -2, -3. The ionization energy of the electron depends only on n, according to:
IE = -En
En = -$\textstyle\frac{k}{n^2}$
IE = $\textstyle\frac{k}{n^2}$
Since k = 13.6eV. the IE of an electron with n = 5 is
IE = $\textstyle\frac{13.6 eV}{5^2} =$ 0.544 eV
In general, for one-electron atomic species:
IE = -En
En = $\textstyle\frac{Z^2k}{n^2}$
IE = $\textstyle\frac{Z^2k}{n^2}$
For HE+, Z = 2:
IE = $\textstyle\frac{2^2(13.6eV}{n^2}$
For a He+ electron with n = 5, we have
IE =$\textstyle\frac{4(13.6 eV)}{25} =$ 4 × 0.544 eV = 2.18 eV
Each of the orbitals for the quantum states differentiated by n, l, and m in Table 8-1 corresponds to a different probability distribution function for the electron in space. The simplest such probability functions, for s orbitals (l = 0), are spherically symmetrical. The probability of finding the electron is the same in all directions but varies with distance from the nucleus. The dependence of Ψ and of the probability density |Ψ|2 on the distance of the electron from the nucleus in the 1s orbital is plotted in Figure 8-18. You can see the spherical symmetry of this orbital more clearly in Figure 8-19. The quantity |Ψ|2dv can be thought of either as the probability of finding an electron in the volume element dv in one atom, or as the average electron density within the corresponding volume element in a great many different hydrogen atoms. The electron is no longer in orbit in the Bohr-Sommerfeld sense; rather, it is an electron probability cloud. Such probability density clouds are commonly used as pictorial representations of hydrogenlike atomic orbitals.
The 2s orbital is also spherically symmetrical, but its radial distribution function has a node, that is, zero probability, at r = 2 atomic units (I atomic unit is a0 = 0.529 Å). The probability density has a crest at 4 atomic units, which is the radius of the Bohr orbit for n = 2. There is a high probability of finding an electron in the 2s orbital closer to or farther from the nucleus than r = 2, but there is no probability of ever finding it in the spherical shell at a distance r = 2 from the nucleus (Figure 8-20). The 3s orbital has two such spherical nodes, and the 4s has three. However, these details are not as important in explaining bonding as are the general observations that s orbitals are spherically symmetrical and that they increase in size as n increases.
There are three 2p orbitals: 2px, 2py, 2pz. Each orbital is cylindrically symmetrical with respect to rotation around one of the three principal axes x, y, z, as identified by the subscript. Each 2p orbital has two lobes of high electron density separated by a nodal plane of zero density (Figures 8-21 and 8-22). The sign of the wave function, Ψ, is positive in one lobe and negative in the other. The 3p, 4p, and higher p orbitals have one, two, or more additional nodal shells around the nucleus (Figure 8-23); again, these details are of secondary importance. The significant facts are that the three p orbitals are mutually perpendicular, strongly directional, and increasing size as n increases.
The five d orbitals first appear for n = 3. For n = 3, l can be 0, 1, or 2, thus s, p, and d orbitals are possible. The 3d orbitals are shown in Figure 8-24. Three of them, dxy, dyz, and dxz, are identical in shape but different in orientation. Each has four lobes of electron density bisecting the angles between principal axes. The remaining two are somewhat unusual: The dx2-y2 orbital has lobes of density along the x and y axes, and the dz2 orbital has lobes along the z axis, with a small doughnut or ring in the xy plane/ However, there is nothing sacrosanct about the z axis. The proper combination of wave functions of these five d orbitals will give us another set of five d orbitals in which the dz2 -like orbital points along the x axis, or the y axis. We could even combine the wave functions to produce a set of orbitals, all of which were alike but differently oriented. However, the set of orbitals, all of which were alike but differently oriented. However, the set that we have described, dxy, dyz, dxz, dx2-y2 and dz2, is convenient and is used conventionally in chemistry. The sign of the wave function, Ψ, changes from lobe to lobe, as indicated in Figure 8-24.
The azimuthal quantum number l is related to the shape of the orbital, and is referred to as the orbital-shape quantum number: s orbitals with l = 0 are spherically symmetrical, p orbitals with l = 1 have plus and minus extensions along one axis, and d orbitals with l = 2 have extensions along two mutually perpendicular directions (Figure 8-25). The third quantum number, m, describes the orientation of the orbital in space. It is sometimes called the magnetic quantum number because the usual way of distinguishing between orbitals with different spatial orientations is to place the atoms in a magnetic field and to note the differences in energy produced in the orbitals. We will use the more descriptive term, orbital-orientation quantum number.
There is a fourth quantum number that has not been mentioned. Atomic spectra, and more direct experiments as well, indicate that an electron behaves as if it were spinning around an axis. Each electron has a choice of two spin states with spin quantum numbers, s = + $\textstyle\frac{1}{2}$ or - $\textstyle\frac{1}{2}$. A complete description of the state of an electron in a hydrogen atom requires the specification of all four quantum numbers: n, l, m, and s.
Many-Electron AtomsEdit
It is possible to set up the Schrödinger wave equation for lithium, which has a nucleus and three electrons, or uranium, which has a nucleus and 92 electrons. Unfortunately, we cannot solve the differential equations. There is little comfort in knowing that the structure of the uranium atom is calculable in principle, and that the fault lies with mathematics and not with physics. Physicists and physical chemists have developed many approximate methods that involve guesses and successive approximations o solutions of the Schrödinger equation. Electronic computers have been of immense value in such successive approximations. But the advantage of Schrödinger's theory of the hydrogen atom is that it gives us a clear qualitative picture of the electronic structure of many-electron atoms without such additional calculations. Bohr's theory was too simple and could not do this, even with Sommerfeld's help.
The extension of the hydrogen-atom picture to many-electron atoms is one of the most important steps in understanding chemistry, and we shall reserve it for the next chapter. We shall begin by assuming that electronic orbitals for other atoms are similar to the orbitals for hydrogen and that they can be described by the same four quantum numbers and have analogous probability distributions. If the energy levels deviate from the ones for hydrogen (which they do), then we shall have to provide a persuasive argument, in terms of the hydrogenlike orbitals, for these changes.
SummaryEdit
Rutherford's scattering experiments showed the atom to be composed of an extremely dense, positively charged nucleus surrounded by electrons. The nucleus is composed of protons and neutrons. A proton has one positive charge and a mass of 1.67 × 10−27 kg. A neutron is uncharged and has a mass of 1.67 × 10−27 kg.
Radio waves, infrared, visible, and ultraviolet light, x rays and γ rays are electromagnetic waves with different wavelengths. The speed of light, c, equal to 2.9979 × 1010 cm sec−1, is related to its wavelength (λ) and frequency (ν) by c = νλ. The wave number, $\overline{\nu}$, is the reciprocal of the wavelength, $\overline{\nu}$ = 1/λ. Hot objects radiate energy (blackbody radiation). Planck proposed that the energy of electromagnetic radiation is quantized. The energy of a quantum of electromagnetic radiation is proportional to its frequency, E = , in which h is Planck's constant, 6.6262 × 10−34 J sec. Electron ejection caused by light striking a metal surface is called the photoelectric effect. Photon is the name given to a quantum of light. The energy of a photon is equal to , in which ν is the frequency of the electromagnetic wave. The pattern of light absorption by an atom or molecule as a function of wavelength frequency or wave number is called absorption spectrum. The related pattern of light emission from an atom or molecule is called emission spectrum. The emission spectrum of atomic hydrogen is composed of several series of lines. The positions of these lines are given accurately by a single equation, the Rydberg equation,
$\overline{\nu}$ = $R_H \times\textstyle(\frac{1}{n_1^2}- \frac{1}{n_2^2})$
in which $\overline{\nu}$ is the wave number of a given line, RH is the Rydberg constant, 109,677.581 cm−1, and n1 and n2 are integers (n2 is greater than n1). The Lyman series is that group of lines with n1 = 1 and n2 = 2, 3, 4, . . . . The Balmer series has n1 = 2 and n2 = 3, 4, 5, . . ., and the Paschen series has n1 = 3 and n2 = 4, 5, 6, . . . .
Bohr pictured the hydrogen atom as containing an electron moving in a circular orbit around a central proton. He proposed that only certain orbits were allowed, corresponding to the following energies:
$E = - \textstyle\frac{k}{n^2}$
in which E is the energy of an electron in the atom (relative to an ionized state, H+ + e-), k is a constant, equal to 13.595 eV atom−1 or 1312 kJ mole−1, and n is a quantum number that can take only integer values from 1 to ∞. The radius of a Bohr orbit is r = n2a0, where a0 is called the first Bohr radius; a0 = 0.529 Å. One atomic unit of length equals a0. The ground state of atomic hydrogen is the lowest energy state, where n = 1. Excited states correspond to n = 2, 3, 4, . . . . The energy levels in a general one-electron atomic species, such as He+ and Li2+, with atomic number Z, are given by
$E = -\textstyle\frac{Z^2k}{n^2}$
The wave nature of electrons was established when Davisson and Germer showed that metal foils diffract electrons in the same way that they diffract a beam of x rays. The wave-particle duality exhibited by electrons is present in all objects. For large objects (such as baseballs), particle behavior predominates to such an extent that wave properties are unimportant.
Heisenberg proposed that we cannot know both the position and the momentum of a particle with absolute accuracy. The product of the uncertainty in position, Δx, and momentum, Δ(mv), must be at least as large as h/4$\pi$:
x][Δ(mvx)] ≥ $\textstyle\frac{h}{4\pi}$
The wave equation for a particle is called the Schrödinger equation. The solutions to the Schrödinger equation, |Ψ(x,y,z)|2, is the relative probability density of the particle at position (x,y,z). A place where the amplitude of a wave is zero is called a node.
Solution of the Schrödinger equation for the hydrogen atom yields wave functions Ψ(x,y,z) and discrete energy levels for the electron. The wave functions Ψ(x,y,z) are called orbitals. An orbital is commonly represented as a probability density cloud, that is, a three-dimensional picture of |Ψ(x,y,z)|2. Three quantum numbers are obtained from solving the Schrödinger equation: the principal quantum number, n, can be any positive integer (n = 1, 2, 3, 4, . . . ); the azimuthal (or orbital-shape) quantum number, l, can have any integral value from 0 to n - 1; the magnetic (or orbital-orientation) quantum number, m, takes integral values from -l to + l. The energy levels depend only on n,
$E = - \textstyle\frac{k}{n^2}$
Wave functions with l = 0 are called s orbitals; those with l = 1 are called p orbitals; those with l = 2 are called d orbitals; those with l = 3, 4, 5, . . ., are called f, g, h, . . ., orbitals. A fourth quantum number is needed to interpret atomic spectra. It is the spin quantum number,s, which can be $+ \textstyle\frac{1}{2}$ or $- \textstyle\frac{1}{2}$ | textbooks/chem/General_Chemistry/Chemical_Principles_(Dickerson)/08%3A_Quantum_Theory_and_Atomic_Structure/5.01%3A_Hydrogen_Oxygen_and_Water.txt |
I have been asked sometimes how one can
be sure that elsewhere in the universe there
may not be further elements) other than
those in the periodic system. I have tried to
answer by saying that it is like asking how
one knows that elsewhere in the universe
there may not be another whole number
between 4 and 5. Unfortunately) some
persons think that is a good question) too.
George Wald, 1964
Introduction
We now know the wave functions and energy levels for a hydrogen atom. With this information and the aufbau (or buildup) process, we can go on to determine the electronic structures for atoms of all the elements. These structures lead directly to the periodic table of Figures 7-3 and 7-4. As we shall see, the structures explain the stability of eight-electron shells in noble gases and the trends in ionization energies and electron affinities of the elements.
Buildup of Many-Electron Atoms
Although we cannot solve the Schrödinger equation exactly for many electron atoms, we can show that no radical new features are expected as the atomic number increases. There are the same quantum states, the same four quantum numbers (n, l, m, and s), and virtually the same electronic probability functions or electron-density clouds. The energies of the quantum levels are not identical for all elements; rather, they vary in a regular fashion from one element to the next.
In studying the electronic structure of a many-electron atom, we shall assume the existence of a nucleus and the required number of electrons. We shall assume that the possible electronic orbitals are hydrogenlike, if not identical to the hydrogen orbitals. Then we shall build the atom by adding electrons one at a time, placing each new electron in the lowest-energy orbital available. In this way we shall build a model of an atom in its ground state, or the state of the lowest electronic energy. Wolfgang Pauli (1900 - 1958) first suggested this treatment of many-electron atoms, and called it the aufbau, or buildup, process.
The aufbau process involves three principles:
1. No two electrons in the same atom can be in the same quantum state. This principle is known as the Pauli exclusion principle. It means that no two electrons can have the same n, l, m, and s values. Therefore, one atomic orbital, described by n, l, and m, can hold a maximum of two electrons: one of spin $+ \textstyle\frac{1}{2}$ and one of spin $- \textstyle\frac{1}{2}$. We can represent an atomic orbital by a circle and an electron by an arrow:
File:Chemical Principles Equation 9.1.png
When two electrons occupy one orbital with spins $+ \textstyle\frac{1}{2}$ and one of spin $- \textstyle\frac{1}{2}$, we say that their spins are paired. A paired spin is represented as follows:
File:Chemical Principles Equation 9.2.png
2. Orbitals are filled with electrons in order of increasing energies. The s orbital can hold a maximum of 2 electrons. The three p orbitals can hold a total of 6 electrons, the five d orbitals can hold 10, and the seven f orbitals can hold 14. We must decide on the order of increasing energies of the levels before we can begin the buildup process. For atoms with more than one electron, in the absence of an external electric or magnetic field, energy depends on n and l (the size and shape quantum numbers) and not on m, the orbital-orientation quantum number.
3. When electrons are added to orbitals of the same energy (such as the five 3d orbitals), one electron will enter each of the available orbitals before a second electron enters any one orbital. This follows Hund's rule, which states that in orbitals of identical energy electrons remain unpaired if possible. This behavior is understandable in terms of electron-electron repulsion. Two electrons, one in px orbital and one in a py orbital, remain farther apart than two electrons paired in the same px orbital (Figure 8-22). A consequence of Hund's rule is that a half-filled set of orbitals (each orbital containing a single electron) is a particularly stable arrangement. The sixth electron in a set of five d orbitals is forced to pair with another electron in a previously occupied orbital. The mutual repulsion of negatively charged electrons means that less energy is required to remove this sixth electron than to remove one of the five in a set of five half-filled d orbitals. Similarly, the fourth electron in a set of three p orbitals is held less tightly than the third.
Relative Energies of Atomic Orbitals
File:Chemical Principles Fig 9.1.png
Figure 9-1 Radial distribution fnctions for electrons in the 3s, 3p, and 3d atomic orbitals of hydrogen. These curves are obtained by spinning the orbital in all directions around the nucleus to smear out all details that depend on direction away from the nucleus, and then by measuring the smeared electron probabil ity as a function of distance from the nucleus. The 35 orbital, which is already spherically symmetrical without the smearing operation. has a most probable radius at 13 atomic units and two minor peaks close to the nucleus. The 3p orbital has a maximum density near r = 12 atomic units, one spherical node at r = 6 atomic units and a density peak close to the nucleus. The 3d orbital has only one density peak, which occurs very close to the Bohr orbit radius of 9 atomic units. The shapes of the three orbitals before the spherical smearing process are to the right of each curve. An electron in the hydrogen atom with n = 2 will be in the neighborhood of r = 4 atomic units. The scale of distances changes in many electron atoms, but relative distances in different orbitals in the atom are the same as in H. An electron in a 3s orbital is more stable than one in a 3p or 3d orbital because it has a greater probability of being inside the orbital of n = 2 electrons, in which it experiences a greater attraction from the nucleus. The 3p orbital is similarly more stable than the 3d.
File:Chemical Principles Fig 9.2.png
Figure 9-2
File:Chemical Principles Fig 9.3.png
Figure 9-3
The 3s, 3p, and 3d orbitals in the hydrogen atom have the same energy but differ in the closeness of approach of the electron to the nucleus (Figure 9-1).
The energy of an electron in an orbital depends on the attraction exerted on it by the positively charged nucleus. Electrons with low principal quatum numbers will lie close to the nucleus and will screen some of this electrostatic attraction from electrons with higher principal quantum numbers. In the Li+ ion, the effective nuclear charge beyond 1 or 2 atomic units from the nucleus is not the true nuclear charge of +3, but a net charge of +1 produced by the nucleus plus the two 1s electrons. Similarly, the lone n = 3 electron in sodium experiences a net nuclear charge of approximately +1 rather than the full nuclear charge of +11.
If the net charge from the nucleus and the filled inner orbitals were concentrated at a point at the nucleus, then the energies of 3s, 3p, and 3d orbitals would be the same. But the screening electrons extend over an appreciable volume of space. The net attraction that an electron with a principal quantum number of 3 experiences depends on how close it comes to the nucleus, and whether it penetrates the lower screening electron clouds. As in Sommerfeld's elliptical-orbit model, the s orbital comes closer to the nucleus and is somewhat more stable than the p, and the p is more stable than the d. This is the reason for the variation of the l energy levels in the lithium energy-level diagram in Figure 8-13.
For a given value of the principal quantum number, n, the order of increasing energy is s, p, d, f, g, . . . . It is less easy to decide whether and when the high l-value orbitals of one n overtake the low-l orbitals of the next: for example, whether a 4f orbital has a higher energy than a 5s, or a 3d a higher energy than a 4s. The question was originally settled empirically by choosing the order of overlap that accounted for the observed structure of the periodic table. The energies have since been calculated theoretically, and (fortunately for quantum mechanics) they agree with the observed order of levels. The sequence of energy levels is shown in Figure 9-2.
Orbital Configurations and First Ionization Energies
File:Chemical Principles Fig 9.2.png
Figure 9-2 Idealized diagram of the energy levels of the hydrogenlike atomic orbitals during the buildup of many-electron atoms. On each level are written the symbols of those elements that are completed with the addition of electrons on that level. Note the nearly equal energies of 4s and 3d, of 5s and 4d, of 6s, 4f, and 5d, and finally of 7s, 5f, and 6d. The near equivalence of energies is reflected in some irregularity in the order of filling levels in the transition metals and inner transition metals. Elements with such irregularities are circled. For example, after the 6s and 7s orbitals fill in lanthanum and actinium, the next electron goes into a d rather than an f orbital. See Figure 9-3 for details.
File:Chemical Principles Fig 9.3.png
Figure 9-3 "Superlong" form of the periodic table. Column head indicates last electron to be added in the Pauli buildup process. Those elements whose electronic structures in the ground state differ from this simple buildup model are in color. In Gd, Cm, Cr, Mo, Cu, Ag, and Au, this difference arises from the extra stability of half-filled (f7, d5) or completely filled (d10) shells. Other deviations arise from the extremely small energy differences between d and s, or d and f levels. These deviations are less important to us now than the overall patterns of buildup and the way they account for the structure of the periodic table. He is placed over Be in Group IIA since the second electron is added to complete the s orbital in each of these elements. In the usual periodic table (inside front cover), He is placed over Ne, Ar, and the other noble gases to indicate that the entire valence shell is filled in these elements.
We shall build up the electronic structures of the atoms in the periodic table by adding electrons to the hydrogenlike orbitals in order of increasing energy, and by increasing the nuclear charge by one at each step. During this process we shall pay particular attention to the relationship between the orbital electronic configurations of atoms and their first ionization energies. The first ionization energy (IE1) of an atom is the energy required to remove one electron:
atom(g) + energy(IE1) positive ion(g) + e-
Numerical IE1 values are given in Table 9-1.* Use the periodic table in Figure 9-3 as an aid as you follow this building process.
A hydrogen (H) atom has only one electron, which in the ground state must go in the 1s orbital. So we write 1s1 (the superscript represents the number of electrons in the orbital), and illustrate the electronic configuration as follows:
*It is common to refer to IE1 simply as IE: This is done in Table 9-1.
File:Chemical Principles Table 9.1.png
File:Chemical Principles Equation 9.3.png
In helium (He), the second electron also can be in the 1s orbital if its spin is paired with that of the first electron. In spite of electron - electron repulsion , this electron is more stable in the 1s orbital than in the higher-energy 2s orbital: File:Chemical Principles Equation 9.4.png
Because of the electron - electron repulsion, the first ionization energy of He is less than we might might have expected for an atom with a nuclear charge of +2. A simple calculation illustrates this point. If electron - electron repulsion were not important, each electron would feel the full force of the +2 nuclear charge, and the first ionization energy could be calculated from the one-electron formula:
IE1 = -E1 = $\textstyle\frac{Z^2k}{n^2} = \frac{(2)^2(1312 kJ mole^{-1})}{(1)^2}$
= 5248 kJ mole-1
However, the experimental value of IE1 for He is much less, 2372 kJ mole-1. Although the strong attraction of a 1s electron to the +2 He nucleus is partially counterbalanced by the electron - electron repulsion, the IE1 is still very large, showing how tightly each electron is bound in He.
Lithium (Li) begins the next period in the periodic table. Two electrons fill the 1s orbital; the third electron in Li must, by the Pauli exclusion principle, occupy the next lowest-energy orbital, namely, the 2s: File:Chemical Principles Equation 9.5.png
The fourth electron in beryllium (Be) fills the 2s orbital, and the fifth electron in boron (B) must occupy one of the higher-energy 2p orbitals: File:Chemical Principles Equation 9.6.png
For B, the first ionization energy is less than that of Be because its outermost electron is in a less-stable (higher-energy) orbital. In carbon (C), two of the three 2p orbitals contain an electron. As Hund's rule predicts, in nitrogen (N) the three p electrons are found in all three 2p orbitals, instead of two being paired in one: File:Chemical Principles Equation 9.7.png
The fourth 2p electron in an oxygen (O) atom is held less tightly than the first three because of the electron - electron repulsion with the other electron in one of the 2p orbitals. The first ionization energy of O is accordingly low.
The general trend across this period is for each new electron to be held more tightly because of the increased charge on the nucleus. Because the others 2s and 2p electrons are approximately the same distance from the nucleus, they do not shield the new electron from the steadily increasing charge. This increased charge overcomes the electron repulsion as the fifth 2p electron is added in fluorine (F). Therefore, the fifth electron is held very tightly in F, and the first ionization energy increases again. The most stable configuration results when the sixth 2p electron is added to complete the n = 2 shell with the noble gas neon (Ne): File:Chemical Principles Equation 9.8.png
The complete n = 1 shell of two electrons is often given the symbol K, and the complete n = 2 shell of eight electrons is given the symbol l. A briefer representation of the Ne atom then is
Ne: KL
For all but a few atoms, writing the complete orbital electronic structure is a tedious procedure. It is also unnecessary because only the outer electrons are important in chemical reactions. We call the chemically important or outer electrons the valence electrons. The valence electrons of an atom are the electrons in the s and p orbitals beyond the closed-shell configurations. For example, in Li the two 1s electrons in He, they are chemically unreactive. Thus we say that the valence electronic structure of Li is 2s<sup.1,/sup>. Similarly, the valence electronic structure of Be is 2s2; of B, 2s22p1; of C, 2s22p2; of N, 2s22p3; of O, 2s22p4; and of F, 2s22p5.
The buildup of the third period of the periodic table proceeds exactly as that of the preceding period did. Each new electron is bound more firmly because of the increasing nuclear charge, except for the fluctuations at aluminum (Al) and sulfur (S) produced by the filling of 3s in magnesium (Mg) and the half-filling of 3p in phosphorus (P): File:Chemical Principles Equation 9.9.png
The outermost electron for each element in this period is bound less firmly than the outermost electron in the corresponding element of the previous period because the n = 3 electrons are farther from the nucleus. Therefore, the first ionization energies for the n = 3 elements are smaller than for the corresponding n = 2 elements. With the completion of the 3s and 3p orbitals, we have again reached a particularly stable electronic configuration with the noble gas argon (Ar).
Something unusual happens in the fourth period. The 4s orbital penetrates closer to the nucleus than does the 3d orbital, and at this point in the buildup process the 4s has slightly lower energy than the 3d. Hence, the one and two electrons that are added to form potassium (K) and calcium (Ca) go into the 4s orbital before the 3d orbital is filled in the elements scandium (Sc) through zinc (Zn). If we assume a constant inner electronic configuration of KL 3s23p6, the valence electronic configurations for the 4s and 3d elements are
K 3d04s1 Mn 3d54s2
Ca 3d04s2 Fe 3d64s2
Sc 3d14s2 Co 3d74s2
Ti 3d24s2 Ni 3d84s2
V 3d34s2 Cu 3d104s1
Cr 3d54s1 Zn 3d104s2
There are two anomalies in this order of filling. The half-filled (d5) and filled (d10) levels are particularly stable, therefore the chromium (Cr) and copper (Cu) atoms have only one 4s electron each.
Example 1
The valence electronic configuration for the ground state of chromium is 3d54s1. Predict the configuration of the first (i.e., lowest-energy) excited state of chromium.
Solution
The aufbau process predicts 3d44s2 for the ground state, but the extra stability of a half-filled level makes the 3d54s1 configuration slightly lower in energy than 3d44s2. The latter configuration thus becomes the first excited state.
For those elements whose ground-state configurations differ from those predicted by the aufbau process, the predicted configuration is that of an excited state usually only slightly higher in energy than the ground state.
Although the 4s orbital penetrates closer to the nucleus than the 3d and therefore has a lower energy, the majority of the probability density of the 4s orbital is farther from the nucleus than in the 3d. An electron in a 4s orbital is simultaneously farther from the nucleus, on the average, than a 3d electron and more stable because of the small but not negligible probability that it will be very close to the nucleus. In chemical bonding, the energies of electrons in such a closely spaced levels in atoms are not as significant as distances of the electrons from the nucleus. Therefore, the 4s electrons have more of an effect on chemical properties than the relatively buried 3d electrons. With the exception of Cr and Cu, all the elements from Ca through Zn have the same outer electronic structure: two 4s electrons. The chemical properties of this series of elements will vary less rapidly than those in a series in which s or p electrons are being added. This is the reason for the relatively unchanging properties of the transition metals.
After the 3d orbitals are filled, the 4p orbitals fill, in a straightforward manner, to form the representative elements from gallium (Ga), 3d104s24p1, to the noble gas krypton (Kr), 3d104s24p6. The first ionization energy, which had risen with increasing nuclear charge in the transition metals, plummets at Ga when the next electron is placed in the less stable 4p orbital.
The fifth period repeats the same pattern: first the filling of the 5s orbitals, then an interruption while the buried 4d orbitals are filled in another series of transition metals, and finally the filling of the 5p orbitals, ending with the noble gas xenon (Xe), 4d105s25p6. The common feature of all noble gases is the outermost electronic arrangement s2p6. This is the origin of the stable eight-electron shells that we mentioned in Chapter 7. The late filling of the d orbitals (and f orbitals) produces the observed lengths of the periods of the periodic table: first 2, then 8, then only 8 instead of 18 for n = 3, then only 18 instead of 32 for n = 4.
According to the energy diagram in Figure 9-2, the 6s orbital is more stable than the 5d, although the difference is small and there are exceptions. The idealized filling pattern is for the 6s orbital to fill in cesium (Cs) and barium (Ba), followed by the deeply buried 4f orbitals in the 14 inner transition elements lanthanum (La) through ytterbium (Yb). There are minor deviations from this pattern, as shown in Figure 9-3. The most important of these deviations is that the first electron after Ba goes into the 5d orbital in La and not into the 4f. Lanthanum is more properly a transition metal than an inner transition metal. It is more relevant to understand the idealized filling pattern, however, than to worry about the individual exceptions to it.
The chemical properties of the inner transition metals cesium (ce) to lutetium (Lu) vary even less than the properties of the transition metals, because successive electrons are in the deeply buried 4f orbitals. After the 4f orbitals are filled, the balance of the third transition-metal series, hafnium (Hf) to mercury (Hg), occurs with the filling of the 5d orbitals. The representative elements thallium (Tl) through radon (Rn) are formed as the 6p orbitals fill.
The seventh and last period begins in the same way. First the 7s orbital fills, then the inner transition metals from actinium (Ac) to nobelium (No) - with the irregularities shown in Figure 9-3 - and finally the beginning of a fourth transition-metal series with lawrencium (Lr). There are more deviations from this simple f-first, d-next filling pattern in the actinides than in the lanthanides (Figure 9-3), and consequently the first few actinide elements show a greater diversity of chemical properties than do the lanthanides.
In summary, the idealized sequence of filling of orbitals across a period is as follows:
1. For period n, the ns orbital is filled first with two electrons. These elements are the alkali metals (Group IA) and the alkaline earths (Group IIA) and are classed with the representative elements.
2. The very deeply buried (n - 2) f orbitals are filled next. They exist only for (n - 2) greater than 3, or for Periods 6 and 7. These elements, which have virtually identical outer electronic structure and therefore virtually identical chemical properties, are the inner transition metals.
3. The less deeply buried (n -1)d orbitals are then filled if they exist. They exist only for (n -1) greater than 2, or for Period 4 and greater. These elements are similar to one another, but not as similar as the inner transition metals. They are called the transition metals (B groups).
4. Finally, the three np orbitals are filled to form the remaining representative elements (Groups IIIA-VIIA) and to conclude in each period with the outermost s2p6 configuration of the noble gases.
We can now explain many of the facts that we presented in Chapter 7. The structure of the periodic table, with its groups and periods, can be seen to be a consequence of the order of energy levels (Figure 9-2). Elements in the same group have similar chemical properties because they have the same outer electronic structure in the s and p orbitals. The outer valence electrons that are so important in chemistry are these s and p electrons. The closed, inert shell of the noble gases is the completely filled s2p6 configuration. We can understand the mechanism of formation of the transition metals and the inner transition metals in terms of the filling of inner d and f orbitals. We can see the reasons for general trends across a period or down a group, and for local fluctuations within a period.
Electron Affinities
Another atomic property that depends strongly on the orbital electronic configuration is the electron affinity (EA), which is the energy change that accompanies the addition of an electron to a gaseous atom to form a negative ion:
Atom (g) + e- negative ion (g)
If energy is released when an atom adds an electron to form a negative ion, the EA has a positive value. If energy is required, the EA is negative. (Values for the known atomic electron affinities are given in Table 9-1.)
Within a period, the halogens have the highest electron affinities because, after the effect of screening electrons in lower quantum levels has been accounted for, the net nuclear charge is greater for a halogen than for any other element in the period. The noble gases have negative electron affinities because the new electron must be added to the next higher principal quantum level in each atom. Not only would the added electron be farther from the nucleus than the other electrons, it also would receive the full screening effect from all the others.
Lithium and sodium have moderate electron affinities; beryllium has a negative electron affinity, and magnesium has a near zero electron affinity. In Be and Mg the valence s orbital is full and the added electron must go into a higher-energy p orbital. Nitrogen and phosphorus have low electron affinities because an added electron must pair with an electron in one of the half-filled p orbitals.
Example 2
Write the ground-state orbital electronic configurations of Cl+, Cl, and Cl-. What are the valence electronic configurations of these species?
Solution
The atomic number of Cl is 17. Therefore the positive ion Cl+ has 16 electrons, Cl has 17, and Cl- has 18. The ground-state orbital electronic configurations are as follows:
Cl+: 1s22s22p63s23p4 or KL 3s23p4
Cl: 1s22s22p63s23p5 or KL 3s23p5
Cl-: 1s22s22p63s23p6 or KL 3s23p6
The valence electronic configurations are 3s23p4 for Cl+, 3s23p5 for Cl, and 3s23p4 for Cl-.
Example 3
The first ionization energy of p, 1063 kJ mole-1, is greater than that of S, 1000 kJ mole-1. Explain this difference in terms of the valence orbital electronic configurations of P and S atoms.
Solution
The valence orbital electronic configuration of P is 3s23p3; of S, 3s23p4. The 3p shell is exactly half-filled in a P atom, whereas there is one extra electron in S that is forced to pair with another 3p electron:
Chemical Principles Equation 9.10.png
As a result of the added electron-electron repulsion of the paired 3p electrons in atomic s, the normal trend of increasing first ionization energies with increasing atomic number in a given period is reversed, with the IE1 of P being grater than the IE1 of S. This effect illustrates the special stability associated with a half-filled p shell. After the half-filled p shell is disrupted (p3 to p4), the electron-electron repulsions associated with the addition of the fifth and sixth p electrons in Cl and Ar are not large enough to override the attractive effect of the increasing positive nuclear charge. Thus the ionization energies of S, Cl, and Ar increase in the usual order (S < Cl < Ar).
Example 4
In each of the following orbital electronic configurations, does the configuration
represent a ground state, an excited state, or a forbidden state (that is, a configuration that cannot exist): (a) 1s22s22p24s1; (b) 1s12s22p1; (c) 1s22s22p6; (d) 1s22s22p53s3 ?
Solution
(a) The ground-state configuration for an atom or ion with 7 electrons is 1s22s22p3. If it is an atom, then it must be N (atomic number 7). The configuration 1s22s22p24s1 represents a state in which a 2p electron has been excited to a 4s orbital. Therefore the configuration represents an excited state. (b) The configuration 1s22s22p1 represents an excited state (for a 4-electron atom or ion, 1s22s2 is the ground state; if it is an atom, it is Be). (c) The configuration 1s22s22p6 represents a ground state (F-, Ne, Na+, Mg2+). (d) The ground-state configuration for an atom or ion with 12 electrons is 1s22s22p63s2 (magnesium atom). In the configuration 1s22s22p53s3, 3 electrons are placed in the 3s orbital, which can take only 2 (one with s = $+\textstyle\frac{1}{2}$ and one with s = $-\textstyle\frac{1}{2}$). The configuration with three 3s electrons violates the Pauli principle and cannot exist. It represents a forbidden state.
Example 5
The electron affinity of Si, 138 kJ mole-1, is much larger than that of P, 75 kJ mole-1. Explain why this is so in terms of valence-orbital electronic configurations.
Solution
(The valence-orbital configuration of Si is 3s23p2; of P, 3s23p3. Adding one electron to Si to give Si<sup.- gives a half-filled 3p shell (Si- is 3s23p3); addition of one electron to P disrupts a half-filled 3p shell (P- is 3s23p4). The special stability of the half-filled shell of Si-, taken with the extra electron-electron repulsion in the 3s23p4 configuration of P-, accounts for the fact that the EA of Si is larger than the EA of P.
Types of Bonding
File:Chemical Principles Fig 9.4.png
Figure 9-4 The two hydrogen 1s orbitals overlap to form an electron-pair covalent bond in H2
A covalent bond forms between combining atoms that have electrons of similar, or equal, valence-orbital energies. For example, two atoms of hydrogen are joined by a covalent bond in the H2 molecule. The energy required to separate two bonded atoms is known as the bond energy. For H2, the bond energy (corresponding to the process H2 H + H) is 432 kJ mole-1.
The two electrons in H2 are shared equally by the two hydrogen 1s orbitals. This, in effect, gives each hydrogen atom a stable, closed-shell (helium-type configuration. An orbital representation of the covalent electron-pair bond in H2 is shown in Figure 9-4.
An ionic bond is formed between atoms with very different ionization energies and electron affinities. This situation allows one atom in a two-atom pair to transfer one or more valence electrons to its partner. An atom of Na is so different from an atom of Cl, for example, that it is not possible for atoms in NaCl to share their electrons equally. The Na atom has a relatively low IE1 of 498 kJ mole-1 and a small EA of 117 kJ mole-1. Therefore, it will readily form Na+ in the presence of an atom with a high EA. The chlorine atom has an EA of 356 kJ mole-1 and an IE1 of 1255 kJ mole-1. Rather than lose an electron, a Cl atom has a strong tendency to gain one. The result is that in diatomic NaCl an ionic bond is formed, Na+Cl-, in which the 3s valence electron in Na is transferred to the one vacancy in the 3p orbitals of Cl.
Atomic Radii
The separation of the nuclei of two atoms that are bonded together (such as H2 or Na+Cl-) is called the bond distance. In the hydrogen molecule, H2, the bond distance is 0.74 Å. Each hydrogen atom in H2 may be assigned an atomic radius of 0.37 Å. The average radii of atoms o some representative elements shown in the periodic-table arrangement of Figure 9-5 were determined from experimentally observed bond distances in many molecules. The atomic radius in most cases is compared with the size of the appropriate closed-shell positive or negative ion.
You will notice (Figure 9-5) that the atomic radii become smaller across a given row (or period) of the periodic table. This shrinkage occurs because in any given period s and p orbitals acquire additional electrons, which are not able to shield each other effectively from the increasing positive nuclear charge. Thus an increase in the effective nuclear charge, thereby decreasing the effective atomic radius. This is why a Be atom, for example, is smaller than a Li atom.
From H to Li there is a large increase in effective atomic radius; the third electron in a Li atom is in an orbital that has a much larger effective radius than the H 1s orbital has. According to the Pauli principle, the third electron in Li must be in an orbital with a larger principal quantum number, namely the 2s orbital. Seven more electrons can be added to the 2s and 2p orbitals, which have approximately the same radii. However, these electrons do not effectively shield each other from the positive nuclear charge as it increases, and the result is an increase in the effective nuclear charge and a corresponding decrease in radii in the series Li (Z = 3) through Ne (Z = 10). After Ne, additional electrons cannot be accommodated by the n = 2 level. Thus an eleventh electron must go into the n = 3 level, specifically, into the 3s orbital. Since the effective radii increase from the n = 1 to n = 2 to n = 3 valence orbitals, the effective size of an atom also increases with increasing atomic number within each group in the periodic table.
File:Chemical Principles Fig 9.5.png
Figure 9-5 Relative atomic radii of some elements compared with the radii of the appropriate closed-shell ions. Radii are in angstroms. Solid spheres represent atoms and dashed circles represent ions. Notice that positive ions are smaller than their neutral atoms and negative ions are larger. Why is this so?
Example 6
Predict the order of decreasing atomic radii for S, Cl, and Ar. Predict the order for Ca2+, Cl-, Ar, K+, and S2-.
Solution
The order S > Cl > Ar is correct for the radii of these atoms, because the nuclear charge increases by one unit from S to Cl and by one unit from Cl to Ar. The valence electrons are attracted more strongly to the nuclei with higher positive charges in any given period, so the atomic radii decrease correspondingly. For isoelectronic atomic and ionic species (those having the same number of electrons), radii decrease as the nuclear charge (atomic number) increases, again because of increasing electron-nucleus attraction. Thus the correct order for the isoelectronic species is S2- > Cl- > Ar > K+ > Ca2+.
Electronegativity
Most bonds in molecules fall somewhere between the extremes of the covalent and ionic types. The bond in the HF molecule, for example, is neither purely covalent nor purely ionic. Just how unequal is the sharing of electrons in HF? And which atom in HF is able to attract the greater share of the bonding electrons, H or F? To anser the second question, and to provide a qualitative guide to the first, Linus Pauling (b. 1901) defined a quantity called electronegativity, χ (the Greek letter chi), in 1932: two years later R. S. Mulliken (b. 1896) showed that electronegativity could be related to the average of the electron affinity and the ionization energy of an atom.
Pauling obtained electronegativity values by comparing the energy of a bond between unlike atoms, AB, with the average energies of the A2 and B2 bonds. If HF formed a covalent bond as in H2 and F2, then we would expect the bond energy in HF to be close to the average (say, the arithmetic mean or the geometric mean) of the bond energies in H2 and F2. However, in molecules such as HF, the bonds are stronger than predicted from such averages. The bond energy of HF is 565 kJ mole-1, whereas the bond energies of H2 and F2 are 432 and 139 kJ mole-1, respectively. The geometric mean of the last two values is (139 $\times$ 432)1/2 = 245 kJ mole-1, which is much less than the observed bond energy of HF. This "extra" bond energy (designated Δ) in an AB molecule is assumed to be a consequence of the partial ionic character of the bond due to the electronegativity difference between atoms A and B. The electronegativity difference between A and B may be defined as
χA - χB = 0.102Δ1/2 (9-1)
in which χA and χB are the electronegativities of atoms A and B, and Δ is the extra bond energy in kilojoules per mole. The extra bond energy is calculated from the equation
Δ = DEAB - [(DEA2)(DEB2)]1/2
in which DE is the particular bond dissociation energy.
In equation 9-1 the square root of Δ is used because it gives a more consisten set of atomic electronegativity values. Since only differences are obtained from equation 9-1, one atom must be assigned a specific electronegativity value, and then the values for the other atoms can be calculated easily. In a widely adopted electronegativity scale, the most electrnegative atom, F, is assigned a value of 3.98. (Electronegativity values based on this assignment are given in Table 9-1.)
10: Oxidation-Reduction and Chemical Properties
The original source material for this chapter can be found at caltechbook.library.caltech.e...Chapter_10.pdf
Unfortunately, the PDF version is a scanned image of the textbook pages, so the text cannot be cut and pasted into this wikibook directly. However, Google automatically generates html versions of documents as it crawls the web. Therefore, if you perform a Google search for the PDF file, you will frequently (but not always) find a link to an poorly formatted HTML version that was created using optical character recognition (OCR) software.
An easy way to work on this chapter is to cut and paste the text from the HTML version, and then capture the images from the original PDF version. Otherwise, you may have to retype the text from scratch (ugh). | textbooks/chem/General_Chemistry/Chemical_Principles_(Dickerson)/09%3A_Electronic_Structure_and_Atomic_Properties/5.01%3A_Hydrogen_Oxygen_and_Water.txt |
Chemical phenomena must be treated as if
1. Lothar Meyer (1868)
Introduction
The goal of every chemist, no matter what types of chemical compounds he or she works with, is to understand how and why chemicals react and change. Yet this is the most difficult task of all. It is not enough to know the structures of all the reactants and products, although such knowledge is a vital starting point. We must also know how the molecules approach one another and with what energies and with what orientations they interact. The concepts of energy and entropy are important in understanding chemical reactions. In this chapter we shall look at some of the problems that face us because we cannot examine individual molecular events. We shall examine two theories for predicting the rates of such simple reactions and compare their success or lack of it. We shall look at the two factors that often make reactions slow - energy and entropy - and see how catalysts can overcome these factors and accelerate chemical changes. Although we cannot present a complete theory of chemical reaction (no one can do this yet), we can outline the foundations on which this theory will someday be constructed.
What Happens When Molecules React?
Mechanisms of Reaction
Let’s suppose that we can watch what happens when two molecules react. We can take as an example the reaction of a molecule of thioacetamide, CH3—CS—NH2, with water to yield acetamide, CH3—CO—NH2, and H2S (Figure 22-1).
thioacetamide, with water to make acetamide, and H2S, (a) The thioacetamide molecule has S, C and N in one plane w around a central C. The orbitals of the double bond are distorted toward S to represent its greater elecironegativity. Orbitals that play no part in the reaction are not drawn.]]
state, with partial bonds from C to both S and O. The former 0—H bonding electron pairs are becoming lone pairs. (c) Products of the reacuon acetamide and H2S. The tetrahedral geometry of the transition state has reverted to trigonal planar geometry as the S atom leaves. The bond angle in H2S is 92° in contrast to the 105° in (a)]]In the original thioacetamide molecule, the central carbon atom is bound to carbon and to nitrogen by σ bonds, and by a σ and a π double bond to sulfur (Figure 22-1a). Since sulfur and nitrogen are more electronegative than carbon, electrons are more attracted to these atoms. Thus the electron pairs of the carbon-sulfur and carbon-nitrogen bonds their bonds are slightly displaced toward sulfur and nitrogen, causing these two atoms to bear a small negative charge and the central carbon to bear a small positive charge.
The most favorable direction of approach of a water molecule is perpendicularly from either side of the plane of the four heavy atoms. The most favorable orientation of the incoming water molecule is as shown in Figure 22-1a. Here a lone electron pair from the water is attracted to the positive charge on the central carbon. As the water molecule approaches this carbon atom, the lone-pair electrons are drawn to it and begin to form a partial bond. This partial bond formation has two effects: It weakens the bond between carbon and sulfur by repelling the electrons even more toward sulfur, and it simultaneously weakens the O—H bonds in water by pulling electrons from these bonds toward oxygen as the oxygen lone-pair electrons are attracted toward carbon. This intermediate state appears in Figure 22-1b. The central carbon atom now has two single bonds to carbon and nitrogen, and two partial bonds to sulfur and oxygen.
This intermediate state is not stable. If the water molecule falls away again and the situation reverts to that of Figure 22-1a (and there is no reason why this could not happen), then we see no net reaction. The water molecule rebounds from a collision with thioacetamide and goes its separate way. It also could happen that the sulfur atom falls away, as in Figure 22-1c. In this reaction, the two protons released by oxygen as it makes a double bond with carbon are attracted by the sulfur atom with four electron pairs, and a molecule of H2S results. The reaction
$\ {CH_3-CS-NH_2\ +\ H_2O\ \longrightarrow\ CH_3-CO-NH_2\ +\ H_2S}$
is complete. The reverse reaction also can occur; a molecule of H2S can collide with one of acetamide and produce water and thioacetamide. We would be less likely to see such an event if we could watch reactions at the molecular level, simply because there are very few H2S molecules in comparison to the number of water molecules.
Factors of Reactions
Geometry of approach is a major factor in reactions . If the water molecule approached in the plane of the thioacetamide molecule, it would find its entry blocked by sulfur lone electron pairs and hydrogen atoms (to a greater extent than is apparent in the skeletal drawings of Figure 22-1). Moreover, if the water molecule approached with a hydrogen atom, instead of a lone electron pair, pointed at the central carbon, it would not be as attracted to the thioacetamide molecule and would be more likely to rebound without reacting. If we could watch every collision, we might see that only 1 collision in 10, or in 100, had both molecules properly oriented for reaction.
A second factor is the energy of the two molecules. In the simplest theories this is expressed only as the relative speed of the two molecules upon collision. If the relative speed of the two molecules is small upon impact, the intermediate state will be more likely to revert to the starting molecules. A slowly moving water molecule can bounce harmlessly off the thioacetamide. In contrast, a water molecule that slams forcefully against the thioacetamide has more of a chance of driving away the sulfur atom, thereby producing acetamide and H2S. We might find that we could plot a curve of the probability of reaction as a function of the velocity of approach of the two molecules along a line connecting their centers.
Unfortunately, the kinds of observations we have been describing for a reaction are an unattainable dream. We must try to find out what is happening during a reaction in a more indirect way. Frequently the most that we can say about a proposed mechanism of reaction is that it is not incompatible with the data. There is always the lingering possibility that some other mechanism of reaction might explain the same data just as well. A classic example of this ambiguity is the reaction of H2 and I2. In 1893, Max Bodenstein, in Germany, studied the reaction:
$\ {H_2\ +\ I_2\ \longrightarrow\ 2HI}$
This was the first comprehensive kinetic study of a reaction occurring in the gas phase. From that time until 1967, virtually every kinetics text and treatise used this reaction as the ideal example of a two-body collision mechanism. One gas molecule of H2 collides with a molecule of I2, they reshuffle atoms, and two molecules of HI are the result. But in 1967, J. H. Sullivan showed that this reaction does not take place by a two-body collision at all, but by a complicated chain reaction. We shall see later why the data measured before 1967 could be explained with equal ease with the two-body or a three-body model.
Not only are we unable to watch individual molecules, we cannot choose the orientation of the molecules upon collision. The best we can do is to estimate the probability of the molecules’ being suitably oriented and then modify our calculations of rates of reaction by a suitable factor. Such a correction sometimes is used and is called a steric factor.
Discovering the Process of Reactions and Crossed Molecular Beams
In a gas reaction or a reaction in solution, we cannot even choose the velocity of approach of the reacting molecules. The molecules in a sample of gas will have a distribution of velocities. We can shift the distribution of velocities by varying the temperature of the gas. As illustrated in Figure 3-11, in nitrogen gas the fraction of all the molecules having a velocity greater than some value such as 1000 m sec-1 increases as the temperature increases. At 273 K, only 0.44% of the N2 molecules have velocities of 1000 m sec-1 or greater; at 1273 K, 35% have this velocity or greater; at 2273 K, this fraction increases to 55%. However, nothing that we can do to the system will give us one specific velocity.
We can remove the velocity distribution for certain reactions by using the method of crossed molecular beams (Figure 22-2). Instead of reactions occurring between molecules dispersed in a solution or a gas, beams of molecules or ions are passed through one another in an evacuated chamber with negligible amounts of other molecules present. are used. A pair of wheels with spaced openings control the velocity of molecules. The beam sources typically are ovens that emit a stream of gas molecules, and electrostatic fields that accelerate ions]] The molecules in the crossed beams react with one another and are scattered from the beams. The products of the reaction, and the unreacted initial molecules, can be observed as a function of angle of scattering by using a movable detector mounted inside the chamber. This arrangement has the great advantage that the velocity selectors can limit the beam to molecules with velocities in a chosen small range. A knowledge of the products of the reaction as a function of angle of deflection or scatter provides much more information about the process of reaction. The orientation problem remains in a molecular-beam experiment, but one can imagine experiments in which this factor is controlled as well. Intense magnetic or electric fields placed just before the beams intersect might give the majority of the molecules in the beam one preferred orientation in space if the molecules had magnetic moments or dipole moments.
Some of the reactions that have been studied with crossed molecular beams are
$\ {K\ +\ HBr\ \longrightarrow\ KBr\ +\ H}$
$\ {K\ +\ CH_3I\ \longrightarrow\ CH_3\ +\ KI}$
$\ {K\ +\ C_2H_5I\ \longrightarrow\ C_2H_5\ +\ KI}$
The reactants, beams of K atoms, HBr, CH3I, and C2H5I molecules, are emitted from heated ovens within the evacuated chamber. The detector is a heated wire filament called a surface ionization detector, which is sensitive to alkali metals or compounds of alkali metals.
Yuan T. Lee, a winner of the 1986 Nobel Prize in Chemistry, has extended molecular beam experiments to include larger and more complex molecules. He studied various reactions between organic molecules and fluorine or oxygen atoms. His most recent work has focused on basic reactions related to ozone depletion. Yuan Lee has utilized crossed-molecular beams to further explore the depletion of ozone by sending a beam of Helium seeded with ozone (O3) through a pulsed laser beam. The laser beam gives rise to high powered UV light and promotes the dissociation of O3 into a neutral O and O2. This research suggests that we could potentially control this sequence of reactions in order to create stratospheric O3 which would then be released into the environment, thereby repairing the Earth's ozone layer.[1]
Lee's group at the Lawrence Berkeley Laboratory is also using molecular beams to research photochemical processes. The researchers use a laser beam to excite molecules or atoms after they have been accelerated but before they collide, giving them more control over the type of chemical reaction that occurs. They are also studying the use of laser excitation during molecular beam experiments to promote the removal of one or more specific atoms from larger molecules as well as the combustion of hydrocarbons. [2]
Crossed-molecular beam research has facilitated a greater understanding of what intermediates (the short-lived molecules that form first in a molecular collision but soon decay into a more stable molecule) are formed in a reaction and how. By varying the velocities and angles of the incoming reactants, and by measuring and determining the distributions of the resulting products, researchers were able to infer the necessary and sufficient conditions under which the reaction would take place. In addition, the molecular cross beam has provided insight on how catalyses interact in chemical reactions. As more research is performed using crossed-molecular beam experiments, the ability to control the products of chemical reactions is becoming a reality. [3]
The disadvantage of molecular-beam experiments is that not all chemical reactions are suitable for study with molecular beams in evacuated chambers. Molecular-beam methods remain a special tool for making complete studies of certain special reactions. The majority of chemical reactions must be studied by bulk methods, in gas mixtures, solutions, and (less frequently) solids.
Measurement of Reaction Rates
The rate of reaction is usually followed in bulk methods by watching the disappearance of a reactant or the appearance of a product in a given time. If the chemical reaction is
A + 2 B → 3 C
then the rate of appearance of product C in a time interval Δt is
$rate = \frac{\Delta [C]}{\Delta t}\,$ (22-1)
in which the concentration of C, [C], is usually expressed in moles liter-1. This is the average rate of appearance of C during the time interval Δt. The limit of this average rate as the time interval becomes smaller is called the rate of appearance of C at time t. It is the slope of the curve of [C] versus t at time t. This instantaneous slope or rate is written
$\frac{d[C]}{dt}.$
Since one molecule of A disappears for every three of C that are produced, and two molecules of B disappear during the same process, the rates of disappearance and appearance of chemical species are related by the expression
$-\frac{d[A]}{dt}\,=-\frac{1}{2}\,\frac{d[B]}{dt}\,=+\frac{1}{3}\,\frac{d[C]}{dt}\,$
The rate of a chemical reaction will depend on the concentrations of the reactants, although not always in the way that might be expected from the overall chemical equation. For the reaction of hydrogen gas with gaseous iodine to produce HI,
H2 + I2 → 2 HI
the relationship between rates is
$\frac{d[HI]}{dt}\,=-2\frac{d[H_2]}{dt}\,=-2\frac{d[I_2]}{dt}\,$
and as you might intuitively expect, the rate equation is
$\frac{d[HI]}{dt}\,=k[H_2][I_2]$ (22-2)
The rate of reaction is proportional to the concentrations of H2 and I2, and dependent on the first power of each concentration. This does not mean that the reaction proceeds by a collision of one H2 molecule with one I2; since 1967 we have had evidence that it does not. We must distinguish clearly between the order of a reaction and the molecularity of the reaction. The constant of proportionality in equation 22-2 is called the rate constant.
The order of a reaction is the sum of all the exponents of the concentration terms in the rate equation. The HI reaction is first order in each of the reactant concentrations and is second order overall. Order is a purely experimental parameter and describes what is observed about the rate equation rather than implying anything about the mechanism of reaction.
The molecularity of a simple one-step reaction is the number of individual molecules that interact in the reaction. Molecularity requires a knowledge of the reaction mechanism. A reaction such as that of hydrogen and iodine actually may take place as a series of half a dozen individual reactions for which we could specify the molecularity of each. The concept of the molecularity of an overall reaction that occurs in a series of steps has no meaning. Most simple one-step reactions are unimolecular (spontaneous decay) or bimolecular (collision). True trimolecular reactions are rare, as three-body collisions are unlikely. Tetramolecular and higher reactions are virtually unheard of. Reactions that appear to be trimolecular or higher (from their stoichiometry) are, after careful study, usually seen to be the sum of a series of simple unimolecular and bimolecular steps. One of the challenges of chemical kinetics is to determine the true set of reactions in such a case.
The reaction of hydrogen gas with bromine is in complete contrast to that with iodine. The overall reaction is similar:
H2 + Br2 → 2 HBr
but the experimental dependence of the rate of production of HBr upon concentrations of reactants and products is utterly different from equation 22-2:
$\frac{d[HBr]}{dt}\,=\frac{k[H_2][Br_2]^{1/2}}{1+k'([HBr]/[Br_2])}\,$ (22-3)
This expression has two experimental rate constants, k and k’. We cannot talk about the molecularity of the reaction, because the overall process is the result of an elaborate chain of reactions that we shall come back to later. Even the order is a puzzle. At the start of a reaction of H2 with Br2, when little HBr is present, the second term in the denominator can be neglected. Then the reaction is effectively 11/2 order: first order in H2 and one-half order in Br2. As the product, HBr, accumulates, it slows down the rate of production of more HBr. Therefore, HBr is called an inhibitor of the reaction.
The formation of HCl is even more complicated. The production of HCl is accelerated by light of intensity I and is inhibited by the presence of oxygen gas, even at low oxygen concentrations. For years the difficulty of purifying the H2 and Cl2 gases and eliminating all traces of O2 led to erroneous conclusions about the kinetics of this reaction. The best experimental rate equation for the appearance of HCl is
$\frac{d[HCl]}{dt}\,=\frac{k_1I[H_2][Cl_2]}{k_2[Cl_2]+[O_2]([H_2]+k_3[Cl_2])}\,$ (22-4)
Notice that, in the limit of the complete absence of oxygen gas, the rate is proportional to the concentration of H2 gas and not dependent on the concentration of Cl2 gas at all! (The second term in the denominator of equation 22-4 is zero, and the remaining Cl2 concentrations in the numerator and denominator cancel.) The reaction is complicated by side reactions that take place on the surfaces of the reacting vessels. The results obtained sometimes depend on the size and shape of the reaction container. All of this is a far cry from the simplicity of the HI system. There are side reactions in the HI system, too, but they are not important below 800 K.
Following the Course of a Reaction
How do we measure concentrations of reactants and products during a reaction to find rate equations such as the ones we have been examining? If the total number of moles of gas changes during a gas reaction, the course of the reaction can be measured in two ways: (1) the change in pressure at constant volume or (2) the change in volume at constant pressure. These are examples of physical measurements that can be performed on the system while it is reacting. They have the advantage of not disturbing the reacting system, and they are usually rapid. With automatic recording devices, we can monitor a physical quantity continuously during the reaction.
Other physical measurements often used in kinetic studies include optical methods such as the rotation of light by a solution (useful if reactants and products have different abilities to rotate polarized light), changes in refractive index of a solution, color, and absorption spectra. Common electrical methods include the electrical conductivity of a solution (especially useful when ions are being produced or consumed), electrical potential in a cell, and mass spectrometry. Thermal conductivity, viscosity of a polymerizing solution, heats of reaction, and freezing points also have been used. The disadvantage of all these methods is that they are indirect. The property observed must be calibrated in terms of concentrations of reactants and products. The calibration is subject to systematic errors, especially if there are side reactions occurring.
Several methods are used to prepare a system for observation in a kinetics experiment. One of today's common methods, called the 'temperature-jump' or 'T-jump' method, was developed in the 1950s by German physicist Manfred Eigen, who later shared the 1967 Nobel Prize in Chemistry for his work. In the T-jump method, a small sample of the system being studied is contained and allowed to react to equilibrium. Then the temperature of the system is raised several degrees in as little as 100 nanoseconds and the reaction is observed as it adjusts to its new equilibrium point. 100 nanoseconds turns out to be too slow for some reactions, but the T-jump method is adequate for a great number of chemical processes. The men with whom Eigen shared his prize, R.G.W. Norrish and George Porter, developed the flash-photolysis method, which was similar to the T-jump method in that it provided a way for making observations of high-speed reactions. Often scientists who excite reactions using the T-jump method observe using spectrometry; a 2009 study by Michael Frunzi, Hai Xu, R. James Cross, and Martin Saunders used NMR to determine the kinetics of the reaction of C60 with 9,10-dimethylantracene.
Chemical methods are more straightforward and yield concentrations directly. With such methods, a small sample is extracted from the reacting mixture, and the reaction is halted by dilution or cooling the mixture long enough to measure concentrations. A serious disadvantage is that we are removing a part of the reacting system and thereby gradually changing it. Moreover, if the reaction cannot be stopped in the sample that is removed for analysis, then the analysis is that much less accurate. In the gas-phase reactions between H2 and Cl2, Br2, or I2, there is no change in the number of moles of gas before and after reaction, so pressure- or volume-change methods cannot be used. To study these reactions, samples are taken, and the gas mixtures are analyzed chemically for their compositions.
A First-Order Rate Equation and the Decay of 14C
Wikipedia article: Carbon-14
In a first-order process, the rate of disappearance of a reactant is proportional to the amount of the reactant present. Each reactant molecule has the same probability of breakdown in a given time interval; the total rate of breakdown simply depends on how many molecules are present. The expression is
$\frac{dn}{dt}\,=-kn$
with n being the total number of molecules present. This rate expression can be integrated to yield the concentration as a function of time:
$n=n_0e^{-kt}\,$
This rate equation is used in the example of dating with 14C in section 23.5, where the expressions in terms of the concentration of 14C are
$\frac{d[^{14}C]}{dt}=-k[^{14}C]\,$ (22-5)
$[^{14}C]=[^{14}C]_0e^{-kt}\,$ (22-6)
The integrated equation 22-6 is plotted in figure 23-3. Taking the logarithm of both sides of equation 22-6 yields
$ln[^{14}C]=[^{14}C]_0-kt\,$ (22-7)
This is the equation of figure 23.8. The plot is a straight line, with a negative slope equal to the rate constant, k. The slope of the plot of equation 22-6 at any time t, as shown in figure 23-3, is proportional to the concentration of 14C remaining at that time. This, in words, is what equation 22-5 means.
An alternative view While this is the traditional solution for the integration of the first order process, the use of Eulers constant in this situation introduces a systematic error in the reported kinetic constants.
As previously stated, in a first-order process, the rate of disappearance of a reactant is proportional to the amount of the reactant present.
Therefore the kinetic constant must represent the fraction of the population of reactant present that will breakdown in a given time period and the fraction must be less than one.
For rates that are very small in comparison to the total population the traditional equation works fairly well.
For example, for simplicity, if the initial population is assumed to be 1 (no=1) and we restrict the time period to the first interval (t=1), and we examine a rate of 5% conversion per time period we would assume a remainder of 95% of the original reactant after the first time period. However inserting these terms we get
$n=n_0e^{-kt}\,$
$n=e^{-0.05}\,$
$n=95.12%\,$
To get the expected remainder of 95% this equation requires the kinetic constant to be increased to 5.129%.
$n=e^{-0.05129}\,$
$n=0.95\,$
The problem becomes more apparent the higher the rate observed, for example if the rate was 95%, then the remaining reactant after one time period would be expected to be 5% of the initial starting population however,
$n=e^{-0.95}\,$
$n=0.3867\,$
To produce the expected 5% remainder the rate constant must be increased to k=2.99573 a rate of approximately 300%, which may be difficult to observe.
$n=e^{-2.99573}\,$
$n=0.05\,$
The problems introduced by the use of this equation can be overcome by recognizing the artificial spliting of the constant as both eulers constant and the kinetic constant are constant and do not change, so a constant raised to a constant is a constant. This constant represents the fraction of the reactant population remaining (%RP) per time period so can also be rewritten to incorporate the rate of the fraction of the population that will breakdown (%BD) per time period as well.
$e^{-k}=%RP=1-%BD\,$
Therefore the kinetic equation for first order kinetics can be rewritten as
$n=n_0(1-(%BD))^{t}\,$
Where the fraction of the population that will breakdown (%BD) can be expressed as the observed reaction rate (r) divided by the initial reactant concentration no.
$n=n_0(1-(r/n_0))^{t}\,$
This notation relates the kinetic constant directly to the observed rate and recognizes the kinetic constant can not exceed 1 as the rate can never be a value greater than the number in initial starting reactants.[4]
Decomposition of N2O5
Table 22-1
In section 16.5, we encountered the decomposition of solid N2O5 as an example of a reaction that is spontaneous yet strongly endothermic. Now let's look at the decomposition of N2O5 dissolved in carbon tetrachloride as an example of a first-order chemical reaction. Solid N2O5 and one product, NO2, are soluble in CCl4; the other product, O2, is not. The reaction
N2O5 → 2 NO2 + O2(g)
can be followed by measuring the total volume of oxygen gas that bubbles out of the solution.
The data for this reaction are given in Table 22-1, after O2 volume measurements have been converted to concentrations of N2O5 left in the solution. These data are plotted in Figure 22-3 as an example of the way in which concentration data are treated. The figure shows the concentration of N2O5 at any time, the rate of change in this concentration, and this rate of change divided by the concentration. This last quantity is equal to the rate constant. That the rate of change divided by the concentration is constant (within the limits of experimental error in the data in Table 22-1) demonstrates that the reaction is indeed first order.
Stoichiometry and Rate Expressions
Figure 22-3
The reaction
2 NO(g) + O2(g) → 2 NO2(g)
has an observed rate equation of the form
$\frac{d[NO_2]}{dt}=k[NO]^2[O_2]\,$
The reaction is second order in NO and first order in O2, and is third order overall. The rate equation happens to agree with the stoichiometry of the chemical reaction; this agreement suggests (but does not prove) that this may be a simple one-step reaction involving three molecules. In contrast, ethanol and decaborane react in solution according to the equation
30 C2H5OH + B10H14 → 10 B(OC2H5)3 + 22 H2
One might naively expect this to have a thirty-first order rate expression. In fact, the reaction is second order, first order in each of the two reactants. For the rate of disappearance of ethanol,
$-\frac{d[C_2H_5OH]}{dt}=k[C_2H_5OH][B_{10}H_{14}]\,$
The Goals of Chemical Kinetics
Some chemical processes are simple one-step reactions involving one, two, or occasionally three molecules. Many more processes are the combination of several such simple reactions. One of the goals of chemical kinetics is to find out what the true molecular mechanism of a complex process is. Why do HI, HBr, and HCl have such different experimental rate equations for a reaction that looks superficially the same in all three cases? To a kineticist the question, “What is the mechanism of the reaction?” means this: “What is the sequence of simple reactions that produces the observed kinetics and stoichiometry of the overall reaction?” To this question organic and inorganic chemists have added, “What is the geometry of the reaction for each simple step in the overall process?” The goal of this inquiry is to predict why the simple reactions proceed as they do and to predict the rates at which they occur. The theories that have been developed to calculate the rate constants for simple unimolecular and bimolecular reactions are our next topic.
Calculating Rate Constants from Molecular Information
Let us assume a simple bimolecular reaction,
$\ {A\ +\ B\harr\ C\ +\ D}$
with a rate expression,
$\frac{-d[A]}{dt} = k[A][B]$
How far can we go in calculating k from the molecular properties of A, B,C, and D? One of the earliest observations was that k varies with temperature; the rate constant is larger, and the rate of reaction is faster, at higher temperatures.
Arrhenius’ Activation Energy
Arrhenius’ activation energy by definition is the energy that must be overcome in order for a chemical reaction to occur. The activation energy of a reaction is usually denoted in units of kilojoules per mole. The theory and history of Arrhenius activation energy will now be discussed. If we plot the logarithm of the rate constant against the reciprocal of temperature, we usually obtain a straight line. Although Arrhenius was not the first person to do this, he developed the idea and gave it an explanation. Therefore, such a plot is called an Arrhenius plot. What does it mean in terms of reaction mechanisms? Van’t Hoff and others had been working, in the late 1800s, on the variation of free energy change of reaction and of the equilibrium constant with temperature. They discovered that the equilibrium constant, Keq, varies with absolute temperature, T, and with the heat of reaction in the following way:
$\frac{dln{k_{eq}}}{dT}\ =\frac{\Delta H}{dT^2}$
This expression can be derived from the Gibbs—Helmholtz equation and ultimately can be derived rigorously from thermodynamics. During the same period, G. M. Guldberg and P. Waage found that they could derive the equilibrium constant from kinetic arguments. If the forward reaction in the above general bimolecular reaction has the rate
$Rate1\ = k_1[A][B]$
and the reverse reaction has the rate
$Rate2\ = k_2[C][D]$
they assumed equilibrium was the state in which forward and reverse rates are equal, so no net change in the reacting system is occurring with time:
$[A][B]k_1\ = k_2[C][D]$
$K_{eq}\ = \frac{k_1}{k_2}\ = \frac{[C][D]}{[A][B]}$
The equilibrium constant, in this argument, is the ratio of the rate constants for the forward and reverse reactions. This is an erroneous derivation. It is valid only when the reaction is a simple one-step process in which the stoichiometry of the reaction is reflected in the coefficients of the concentration terms in the rate equation. Nevertheless, it is valid for the kind of reactions we are considering here: simple bimolecular reactions. If the equilibrium constant is the ratio of forward and reverse rate constants, the above equation suggests that the enthalpy of reaction might be the difference between two energies, Eforward and Ereverse: G°=Eforward—Ereverse If the Arrhenius energy of activation is not a function of temperature, this equation predicts that a plot of In k against the reciprocal of the absolute temperature will generate a straight line. This is true for many reactions, but not all, and the activation energy is one of the standard experimental parameters by which a chemical reaction is described. A simple, but accurate, representation of the temperature dependence of the rate of a chemical reaction can be expressed in the following formula.
$\ E_a\ = -RT\ ln{\frac{k}{A}}$
This expression can be rearranged to yield the more common Arrhenius equation.
$k = A e^{{-E_a}/{RT}}$
A historically useful generalization supported by the Arrhenius equation is that, for many common chemical reactions at room temperature, the reaction rate doubles for every 10 degree Celsius increase in temperature. The activation energy is a barrier that the colliding molecules must surmount if they are to react rather than recoil from one another.
Using the reaction of water with thioacetamide we can postulate that if thioacetamide and water molecules do not collide head-on with sufficient energy the redistribution of bonds and subsequent reaction will never occur. Water will recoil from the thioacetamide molecule and no reaction will take place. This recoil is due to the fact that insufficient energy was provided to overcome the needed Arrhenius activation energy. With this established we have experimental evidence, in the form of the temperature dependence of k, that some such threshold energy is involved in chemical reactions.
Arrhenius’ explanation of activation energies assumes that every pair of molecules with energy less than Ea will not react, and every pair with energy greater than Ea will react. Nothing is changed in the thermodynamics of the overall reaction. The activation barriers to forward and reverse reactions, Eforward and Ereverse, are such that their difference, ΔH° = Eforward — Ereverse, is the thermodynamic heat of reaction. The higher the barrier, Eforward, the slower the forward reaction will be. However, since Ereverse must rise by the same amount as Eforward if their difference is fixed, the reverse reaction is slowed by the same amount.
The point of equilibrium is affected not by the individual numerical values of the activation energies for forward and reverse reactions, but only by the difference between them, which is ΔH°.
Collision Theory of Bimolecular Gas Reactions
The next logical step is to construct a collision theory for gas reactions. A reaction between two molecules occurs, in this theory, when the molecules collide with energy in excess of Ea. A theory could hardly be simpler. There are two questions to be answered before the rate constant can be calculated:
1. How often do two molecules collide per cubic centimeter of gas mixture?
2. In what fraction of these collisions does the combined energy of the two molecules exceed Ea?
The collision frequency can be calculated from the simple kinetic theory of gases with the methods that were introduced in Chapter 3. The frequency depends on the concentrations of the two reacting gases, and also on their molecular weights, the distance between the molecular centers on collision, and on the square root of temperature. Since the molecules move more rapidly at higher temperatures, they collide more often. The fraction of pairs of molecules having energy equal to or greater than Ea upon collision is
$e^{\frac{-E_a}{RT}}$
According to simple collision theory, the rate of reaction then is
$\ Rate\ =\ (collision\ frequency)(probability\ that\ E\ge\ E_a)$
$-\frac{d[A]}{dt}=(Z[A][B])(e^{\frac{-E_a}{RT}})$
$-\frac{d[A]}{dt}=Ze^{\frac{-E_a}{RT}}[A][B]$
The rate of a reaction is greater at higher temperatures because collisions are more frequent and because the probability that a colliding pair will have an energy greater than E is also higher. The constant, Z, can be calculated from the molecular weights and the diameters of the reacting molecules by approximating them with spheres. The bimolecular rate constant, k, then is
$\ k\ = Ze^{\frac{-E_a}{RT}}$
This theory is tested in the data in Table 22-2. The Arrhenius activation energy is tabulated for six bimolecular gas reactions, along with the observed pre-exponential factor, pz, and its theoretical value as calculated from the collision theory and the absolute rate theory that we will discuss in the next sections.
Table 22-2
Keep in mind that these are the logarithms of Z that are tabulated, so a disagreement between theory and experiment of 1.0 means an error by a factor of 10 in the rate constant. The agreement is generally encouraging for so simple a theory that has no assumptions other than those of the kinetic theory of gases. There are discrepancies; for example, the ClO reaction rate is incorrect by a factor of 400. When discrepancies occur, the absolute rate theory usually does a better job of predicting Z than the collis ion theory does.
In the right column of Table 22-2 are the standard enthalpies or heats of reaction. The relative enthalpy of reactants and products, and the activ ation barrier between them, are plotted for these reactions in Figure 22-5. Some reactions, such as NO2 + CO, must surmount a considerable activat ion barrier. For other reactions, the barrier is nonexistent, as with 2Cl0. For others such as 2NO2, the barrier to reaction is only the heat of reaction itself, and the reverse reaction has a zero activation energy. The most general case is diagrammed at the bottom of Figure 22-5.
Figure 22-5
Molecular Table
Figure 22-6
Activated Complexes
Before we consider the absolute rate theory, we must look more closely at the state of the reactant molecules as they cross the activation barrier. In the reaction
$2ClO \rarr Cl_2 + O_2$
the Cl and O atoms are bonded at the start, and the two ClO molecules are too far apart to exert any influence on one another. At the end of the reaction, the Cl atoms are 1.99 A apart in a Cl2 molecule, the O atoms are 1.21 A apart in an O2 molecule, and these two molecules are far apart. What is the intermediate, activated state?
The activated complex is diagrammed in Figure 22-6b. All four atoms are an unspecified distance apart, somewhat farther away than if they were bonding in the stable sense of the term. We can be sure that the activated complex is not one in which all four atoms are so far apart that they exert no influence on one another; some sort of a loose complex must exist. The basis for this assertion is a knowledge of the bond energies of the three molecules, and that the activation energy for the 2ClO reaction is zero. The bond energies, or the energies required to separate completely the atoms in a diatomic molecule, are shown in the molecular table.
If during the reaction the two ClO molecules were first pulled apart, and then the isolated atoms were combined into Cl2 and O2, the activation energy for this reaction would be twice the bond energy of ClO, or 540 kJ per 2 moles of ClO. Instead, the activation energy is zero. The activated complex must be a combination of the four atoms such that whatever instability created as Cl and O separate is immediately compensated by the stabilizing influence of associations between Cl and Cl and between O and O.
We can think of the activated complex as an unstable “molecule,” with many of the properties of a molecule, except that it decomposes spontaneously either to reactants or to products. The thioacetamide and water molecules in Figure 22-lb are in an activated complex, and the energy of this complex is greater than either that of thioacetamide and water or that of acetamide and hydrogen sulfide.
Potential Energy Surfaces
The 2ClO reaction suggests that in principle we should be able to calculate the total potential energy of a collection of atoms as a function of their positions in space. This calculation would produce a potential energy surface with hills and plateaus of high energy, and valleys of low energy. Any region of a minimum of potential energy in this plot will represent a stable molecule. Even with four atoms as in the 2Cl0 reaction we would need, unfortunately, six variables to describe the arrangement of atoms: the bond lengths from each atom to the other three, for example. Our potential energy plot would have to be in seven-dimensional space. This is difficult to visualize and impossible to construct. We need an example with only two variables so the map can be plotted in three-dimensional space. One of the first maps to be calculated, by Henry Eyring in 1935, is the potential energy surface for the reaction
$H+H_2\harr H_2+H$
in which all three atoms are constrained to lie on a straight line. The only variables are the distances from the central hydrogen atom to the other two, r1 and r2; r1 being the distance between the first two atoms, while r2 is the distance between the second and third atoms.
$\ H \frac{}{r_1} H \frac{}{r_2}H$
The potential energy of the three-atom system as a function of r1 and r2 is shown in Figure 22-7a.
Caption 22-7
Figure 22-7
Figure 22-8
The actual arrangements of the three atoms at the six numbered points marked in color are drawn in Figure 22-8. Sections through this potential energy surface at fixed values of r1 are shown in Figure 22-7b. If either r1 or r2 is large, the three hydrogen atoms exist as an H2 molecule and an iso1ated atom. The potential energy section at constant r1, for r1 greater than 3 A in Figure 22-7b, is the same as that for an isolated H2 molecule in Figure 12-2. As an atom approaches an H2 molecule from the right (points 1 and 2 of Figures 22-7 and 22-8), the first noticeable effect is an increase in the potential energy of the system of three atoms.
The incoming atom is repelled by the molecule, and a more stable situation results if the atom rebounds and moves away again. If the atom has enough kinetic energy to keep approaching the H2 molecule, it will begin to weaken the H—H bond in the H2 molecule.
At point 3, both outer atoms are slightly farther from the central one than a normal H — H bond length, but the potential energy of the system of atoms is 25 kJ /mole higher than that of isolated H2 and H. Point 3 is the activated complex for the reaction. The activated complex can decompose either to products or to reactants. There is no reason why the three atoms in the state of point 3 cannot return to point 1 as well as proceed to point 5. What is certain is that the activated complex is unstable and must decompose. Points 1 through 5 in Figure 22-7 are connected by a colored dashed line called a reaction pathway. If we plot potential energy along this pathway, an activation energy barrier curve, such as those in Figure 22-5, results. Notice that the reaction pathway at all times is a path along a valley between steep walls. It may take 25 kJ of energy to build the activated complex of point 3, but it takes over 400 kJ to separate the atoms as at point 6. Even with reactions of molecules so complicated that we cannot calculate or even plot their complete multidimensional potential energy surface such profiles are still useful.
Absolute Rate Theory
In the absolute rate theory, also known as transition-state theory, a reaction takes place when an activated complex breaks down into products. Therefore, the rate of reaction is the product of three factors:
1. The concentration of activated complexes per cubic centimeter
2. The rate of breakdown of individual complexes or their rate of passage over the activation energy barrier
3. The probability that a breakdown will form products and not reactants again
Since the activated complex represents an unstable state of transition between reactants and products, it is often called a transition state. The transition-state theory assumes an equilibrium between reactants and the activated complex, usually represented by a superscript double dagger:
$A+B\harr AB^{\Dagger}$
$K^{\Dagger} =\frac{[AB^{\Dagger]}}{[A][B]}$
Hence, the concentration of the activated complex is given by:
$[AB^{\Dagger}]=K^{\Dagger} [A][B]$
The rate of decomposition is more complicated to calculate, but it turns out to be a universal constant for all bimolecular reactions at a given temperature:
$\ Rate\ of \ decomposition =\frac{kT}{h}$
in which k = Boltzmann’s constant, and h = Planck’s constant. The probability that a breakdown will be to products and not to reactants is the transmission coefficient, m also often written as κ. It can be estimated only as having a value between 0.5 and 1.0 in most reactions. Therefore, the overall rate of reaction is
$-\frac{d[A]}{dt}=\kappa\frac{kT}{h}K^{\Dagger}[A][B]$
and the rate constant, k2, is
$k_2=\kappa\frac{kT}{h}K^{\Dagger}$
(The symbol k2 is used here instead of k to represent a bimolecular rate constant to avoid confusion with Boltzmann’s constant.) It is possible to calculate the equilibrium constant, K , between reactants and the activated complex from molecular properties by looking at the thermodynamic interpretation of this rate-constant expression. The equilibrium constant is related to the standard free energy of formation of the activated complex from reactants, and this in turn is related to the standard enthalpy and entropy of the formation of the activated complex:
$-RTln{K}^{\Dagger}=\Delta G^{0\Dagger}=\Delta H^{0\Dagger}-T\Delta S^{0\Dagger}$
Thus, the bimolecular rate constant, k2, can be written
$k_2=\kappa\frac{kT}{h}e^{\frac{-\Delta G^{0\Dagger}}{RT}}=\kappa\frac{kT}{h}e^{\frac{\Delta S^{0\Dagger}}{R}}e^{\frac{-\Delta H^{0\Dagger}}{RT}}$
The enthalpy of activation, ∆Ho, is nearly the same quantity as the activation energy, Ea. The difference is irrelevant is this discussion. The equation above indicates that the rate of reaction is slower if the activation energy is large. This result was already obtained in the collision theory; if the activation energy is large, only a small fraction of the molecules will have enough energy to surmount the barrier and to react rather than rebound upon collision. The equation also suggests that the reaction rate is faster if the entropy of activation is large. If the activated complex is much more disordered than the reactants, the reaction is enhanced because the equilibrium constant for formation of the complex is large, and more of the complex is present. In contrast, if the reactants are severely constrained when they combine to make the activated complex, then the reaction is inhibited. We might guess that the entropy of activation in the thioacetamide-plus-water reaction is negative since the two molecules combine to form one unit in the complex. Both molecules are limited severely in their initial orientations if they are to build the activated complex successfully.
The entropy of activation in bimolecular reactions is almost always large and negative because the two reactants lose entropy when they combine in the complex. Often the most useful application of absolute rate theory is not to calculate the rate constant directly, but to use the observed rate constant and equation above to calculate the entropy of activation. The entropy of activation provides information about the structure of the activated complex. For example, if the calculated entropy of activation is positive, then any mechanism that leads through a tightly organized activated complex must be rejected. As an example, in the next section we will consider two reactions of the type:
$R_3C-Br+OH^{-}\longrightarrow R_3C-OH+Br^{-}$
which proceed by different mechanisms, depending on the nature of the R groups. In one mechanism, the Br- is driven away as OH- approaches, in the same fashion as the thioacetamide reaction. The activated complex is then a combination of R3C-Br and OH-. This is called an associative or SN2 reaction, meaning that it is a substitution of one group for another, that the groups are nucleophilic (donating electrons and attracting nuclei: Lewis bases, in fact), and that two molecules are involved in the reaction.
The other mechanism is for the R3C—Br molecule to dissociate spontaneously into Br and what is known as a carbonium ion, R3C, and for the OH- ions to react rapidly in a separate step with any free carbonium ions. The activated complex or transition state in this mechanism will be the reactant R3C—Br just before dissociation. This is called a dissociative or SN1 mechanism since it is a nucleophilic substitution in which the slowest step involves dissociation of a single molecule. One should be able to distinguish between these two mechanisms by their entropies of activation, which can be calculated from the equation above and the measured rate constants. The SN2 mechanism will have a large negative entropy of activation since the activated complex is formed by combining two molecules. In contrast, the SN1 mechanism will have virtually a zero entropy of activation because the activated complex differs only slightly from the reactant molecule.
Laser Control of Chemical Reactions
Given a set of molecules that can combine in two possible ways, scientists wonder how they can most effectively use a laser to prod the chemical reaction one way or the other. Because interference of molecular pathways is the key to governing reactions, any laser scenario that induces such interference may serve as a means of controlling reactions. Instead of shining two steady beams on a target, one might use ultrashort pulses of laser light. Modern lasers can generate bursts as short as 10-15 second. Unlike continuous-wave radiation, a light pulse is made up of a collection of distinct frequencies and, hence, of a collection of photons with different energies. Such light also has a perhaps counterintuitive property. The briefer the pulse, the broader the range of energies within it.
This property plays a major role in pulsed-laser methods for controlling the outcomes of chemical reactions. By delivering a range of energies, a pulse can induce motion (such as vibration or rotation) in a molecule, which in turn affects the way it interacts with other light pulses. Ordinarily a molecule exists at a specific energy value. A system at one of these fixed energies resides in a so-called stationary state and does not move over time. For a molecule to undergo the dynamics, it must live in several energy levels at once. Such an assemblage of energy levels is called a superposition state. The wave function describing the superposition state is the sum of wave functions representing stationary states of different energies. To construct it, researchers shine a pulse of coherent laser light on the molecule. The way the molecule then moves depends on the nature of the light pulse and its interaction with the molecule. Thus, we can affect dynamic changes in the molecule by shifting the relative contribution of the frequencies that compose the pulse--that is, by shaping the pulse.
Several researchers have developed these ideas. They include Stuart A. Rice of the University of Chicago, Robert J. Gordon of the University of Illinois at Chicago, Herschel Rabitz of Princeton University, Ronnie Kosloff of the Hebrew University of Jerusalem and Kent R. Wilson of the University of California at San Diego. Their results show that pulses built out of a complicated mixture of frequencies are required to control molecular dynamics optimally, but simple approximations often suffice to break apart molecules in a controlled way.
Complex Reactions
Most chemical reactions are not simple unimolecular or bimolecular reactions, but combinations of these. This is why such complicated rate equations as equations 22-3 or 22-4 arise. Even the hydrogen—iodine reaction, which has been used for over half a century as the classic example of a simple bimolecular reaction (equation 22-2), is complex.
The Hydrogen—Iodine Reaction
For the reaction
H2 + I2→ 2HI
(22-19)
the observed rate equation is
$-\frac{d[H_2]}{dt}\,=k[H_2][I_2]$
(22-20)
However, both N. N. Semenov and Henry Eyring have suggested that the true mechanism might be not that of equation 22-19, but a two-step mechanism involving the reversible dissociationof I2 to 2I, followed by the trimolecular reaction of I and H2:
I2 ↔ 2I
H2 + 2I → 2HI
(22-21)
The rate expression for the reaction of one H2 molecule with two I atoms is
$\frac{d[H_2]}{dt}\,=k'[H_2][I]^{2}$
(22-22)
and if the dissociation of I2 is reversible and at equilibrium, we can write an equilibrium-constant expression:
$K = \frac{[I]^2}{[I_2]}\,$
[I]2=K[I2]
(22-23)
Then substituting the concentration of I atoms in equation 22-22 with the equilibrium equation 22-23 it produces
$-\frac{d[H_2]}{dt}\,=k'K[H_2][I_2]$
(22-24)
which is the same rate expression as if the mechanism were really one of bimolecular collision. We thus have two different mechanisms with the same rate expression. How can we choose between them?.(Above 800 K, side reactions with different mechanisms occur, yet these reactions can be neglected at moderate temperatures.)
The two mechanisms have the same rate equation so long as the dissociation of I2 is at thermal equilibrium, and the amount of I atoms present is given by the thermal equilibrium constant of equation 22-23. At higher temperatures, more I2 dissociates, thereby producing the same effect that would have resulted from the greater bimolecular rate constant in the bimolecular mechanism. J. H. Sullivan decided to test the two mechanisms by making the concentration of iodine atoms different from what it is normally in the thermal dissociation of I2. He did this by dissociating I2 with 578-nm light from a mercury vapor lamp. This light should have had little effect if the reaction was bimolecular, aside from a slight decrease in the I2 concentration. Conversely, if the trimolecular reaction was correct, the rate of reaction should have increased with the intensity of irradiating light since more I atoms were being produced.
Sullivan calculated the concentration of I atoms present at several intensities of irradiating light and found that the rate of appearance of HI is proportional to the square of the I atom concentration. Therefore, the mechanism of equation 22-21 is the correct one. The classical H2 + I2 reaction is a trimolecular reaction imitating a simpler bimolecular reaction because of the thermal equilibrium that normally exists between I2 and 2I. (At least, until someone even more ingenious designs an experiment that proves that it is a more complicated reaction imitating a trimolecular reaction.) As Sullivan points out [j Chem. Phys. 46, 73 (1967)], the trimolecular reaction
H2+2I→2HI
can be replaced by two bimolecular steps:
H2+2I→H2I
H2I+I2↔HI
If the first of these is fast and reversible, so that reactants and products are in equilibrium, then the rate expression is the same as for the trimolecular process, and the two mechanisms cannot be distinguished by reaction rates. This example makes a point that must be kept in mind at all times - we can never prove that a proposed mechanism is right; we can only prove that it has not yet been shown to be wrong. There is always the chance that a more subtle experiment, such as Sullivan’s upsetting of thermal equilibrium with light, may uncover the weaknesses in an accepted mechanism. When two theories are presented on the same topic, the temptation (and usually the wiser choice) is to opt for the simpler one, until data compels you to do otherwise. But you should always be prepared to change your mind when new data demands it.
Rates and Mechanisms of Substitution Reactions
The reaction of tert-butyl bromide with OH,
(CH3)3CBr + OH- ↔ (CH3)3COH + Br-
(22-25)
has the experimental rate expression
- $\frac{d[(CH_3)_3CBr]}{dt}\,=k[(CH_3)_3CBr]$
(22-26)
The rate does not appear to depend on the OH- concentration at all. In contrast, the similar reaction with a less highly substituted carbon atom in ethyl bromide,
CH3CH2Br + OH- ↔ CH3CH2OH + Br
(22-27)
has the rate expression that we might expect from the chemical equation:
$-\frac{d[CH_3CH_2Br]}{dt}\,=k[CH_3CH_2Br][OH^-]$
(22-28)
Why should these two similar reactions proceed by different mechanisms and have different rate equations? And how is it that the rate in equation 22-26 can be independent of concentration of one of the reactants?
Reaction 22-25 takes place by the SN1 mechanism the equation
$R_3C-Br+OH^{-}\longrightarrow R_3C-OH+Br^{-}$
The tert-butyl bromide first dissociates in a slow reaction, and than the carbonium ion that is formed reacts immediately with OH. Whenever a process takes place by a series of rapid steps with one slower step, the overall rate of reaction will be controlled by the slow step. The rate in this SN1 reaction depends entirely on how fast the molecules of (CH3)3CBr decompose. The capacity of reacting carbonium ions with OH far exceeds the amount of carbonium ions supplied by the dissociation of tert-butyl bromide. The total amount of OH- present is unimportant.
Why do reactions go with different mechanisms? TheSN2 scheme is possible for ethyl bromide because there is room for three substituents of the C atom (CH3 and two H), plus OH and Br. Because of this the activated complex
-
is sterically possible.
In tert-butyl bromide,C4H9Br, on the other hand, the groups attached to the carbon atom (three CH3) are large enough that OH- and Br- cannot bind at the same time. The activated complex of the SN2 reaction is impossible, and no reaction can occur until a molecule of tert-butyl bromide spontaneously dissociates. The dissociated Carbonium Ion is then the ion to attack, either by Br- to form the reactants again or by OH to form the product. If Br is present only as a result of a previous reaction of tert-butyl bromide, its concentration is probably much smaller than that of OH, and most of the carbonium ions will be converted to tert-butyl alcohol, (CH3)3COH.
In general, a rate expression that disagrees with the stoichiometry of the overall reaction is an indication that the reaction is proceeding by a series of steps. Then the problem is to find a set of steps including a slow step that accounts for the observed rate law.
In the reaction mechanisms of tert-butyl bromide and ethyl bromide the rate difference is also encountered in the octahedral and square planar
complexes of the transition metals. The square planarcomplexes of Pt(II) and other metals can react with new ligands by associative (SN2) mechanisms because the metal atom is accessible from either side of the plane. The SN2 mechanism of the reaction
Pt(NH3)3Cl+ + Br- → Pt(NH3)3Br+ + Cl
can be written
-
The activated complex is a five-coordinated platinum, which breaks down rapidly to products. The rate of the overall reaction depends on the rate of formation of the activated complex. This rate is influenced strongly by the nature of the entering group (Br- in this example). Ligands capable of forming strong bonds with the central atom are the best entering groups because they displace the leaving group (C1 in this example) rapidly. The ions CN and I- are good entering groups for Pt(II) complexes, whereas NH3 and H20 are relatively poor.
It is much more difficult for the six-coordinated octahedral complexes to react by an SN2 mechanism because six ligands around a central metal, such as Co(III), leave little or no room for the attachment of an entering group in a transition state. Studies of substitution reactions of octahedral Co(III) complexes have established that the rate-determining step involves the dissociation of the bond between the Co(III) and the leaving group (the entering group is not involved in this dissociation step.) In aqueous solution, for example, H20 displaces Cl -in the complex Co(NH3)5Cl2+, thereby producing Co(NH3)5H203+. The mechanism that is consistent with the experimental rate of this and similar reactions is the dissociative or SN1 mechanism, which can be written.
-[[
For such mechanisms, the entering group plays no significant role in the creation of the Transition state and the rate of the overall reaction. A characteristic of most octahedralsubstitution reactions is the lack of influence of entering groups on the rate of reaction.
Chain Reactions
The reaction H2 + Br2 → 2HBr has the strange rate equation that we have already seen,
$\frac{d[HBr]}{dt}\,=\frac{k[H_2][Br_2]^{1/2}}{1+k'([HBr]/[Br_2])}\,$
(22-3)
For 13 years after this rate law was discovered, no one could account for it. Then, three groups scientist those groups of Henry Eyring, K. F. Herzfeld,and Michael Polanyidid so simultaneously. They proposed that the reaction proceeds by a chain mechanism involving two chain-propagating steps:
(1) H2+Br→ H+HBr (k1)
(2) H+ Br2 → HBr+Br (k2)
When a molecule breaks apart into uncharged fragments having unpaired electrons, the unpaired electrons (e.g., in H and Br) make the fragments called radicals,chemically reactive. The atomic product of each of these steps is a reactant for the other step, and they both produce HBr. Thus, HBr results, not from a bimolecular collision, but from an endless chain of reactions 1 and 2. The first of these two steps is the reaction of Figure 22-9. But where do these atoms of Br and H come from? The Br atoms are postulated to come initially from a chain-initiating step:
(3) Br2 → Br+ Br (k3)
Why is the dissociation of H2 not included also? The real reason is that the explanation of the reaction rate (equation 22-3) does not require it, and if we add it, we obtain the wrong rate expression. We can justify this omission in another way: The dissociation energy of H2 is 432 kJ mole-1, whereas that of Br2 is only 190 kJ mole-1. A large concentration of HBr inhibits the reaction ,as we can see from the HBr term in the denominator of equation 22-3, and a large concentration of Br2 counteracts this inhibition. From this it is evident that HBr and Br2 are competing for the same chemical substance. What might that substance be? The most likely candidate is hydrogen atoms and the inhibiting reaction would be
(4) H+HBr→H2+Br (k4)
This is a chain-inhibiting reaction and is counteracted if an excess of Br2 makes reaction 2 go rapidly, as the rate equation 22-3 predicts. Finally, the chain is terminated by the recombination of Br:
(5) Br+Br → Br2(k5)
If we can obtain equation 22-3 from these five reactions this will be a strong argument for the correctness of this chain mechanism (although not an absolute proof, as we have seen with HI.) The rate of appearance of HBr is given by
$\frac{d[HBr]}{dt}\,=+k_1[H_2][Br]+k_2[H][Br_2]-k_4[H][HBr]$
(22-29)
since HBr appears as a result of reactions 1 and 2, and disappears in reaction 4. The rates of production of H and Br atoms are given by
$\frac{d[H]}{dt}\,=k_1[H_2][Br]+k_2[H][Br_2]-k_4[H][HBr]$
(22-30)
$\frac{d[Br]}{dt}\,=-k_1[H_2][Br]+k_2[H][Br_2]+k_4[H][HBr]+2k_3[Br_2]-2k_5[Br]^{2}$
(22-31)
The coefficients of 2 in front of k3 and k5 arise because each unit of reaction 3 produces two Br atoms, and each unit of reaction 5 removes two Br atoms.
At this point an essential simplification must be made. It is assumed that the actual amount of H and Br atoms present at any time must be small because they are consumed at almost the same rate that they are produced. Soon after the reaction begins, the concentrations of H and Br will reach a steady state and will remain constant so long as the reaction continues with a plentiful supply of reactants. Because of this each of the rate equations in 22-28 and 22-29 can be set equal to zero:
$0=k_1[H_2][Br]-k_2[H][Br_2]-k_4[H][HBr]$
(22-32)
$0=-k_1[H_2][Br]+k_2[H][Br_2]+k_4[H][HBr]+2k_3[Br_2]-2k_5[Br]^{2}$
(22-33)
Adding these two equations yields
$2k_5[Br]^{2}=2k_3[Br_2]$
with the rate equation
$[Br]=(\frac{k_3}{k_5}\,)^{1/2}[Br_2]^{2}$
(22-34)
This calculation gives us a steady-state concentration for Br atoms in terms of the concentration of Br2 molecules. The HBr rate equation can be rewritten as
$\frac{d[HBr]}{dt}\,=k_1[H_2][Br]+(k_2[Br_2]-k_4[HBr])[H]$
(22-35)
We can eliminate the H concentration by expressing it in terms of Br concentration from equation 22-32:
$[H]=\frac{k_1[H2]}{k_2[Br_2]+k_4[HBr]}\,[Br]$
(22-36)
Substituting equation 22-36 into 22-35, placing everything over a common denominator, and canceling terms yields
$\frac{d[HBr]}{dt}\,=\frac{2k_1k_2[H_2][Br_2][Br]}{k_2[Br_2]+k_4[HBr]}\,$
(22-37)
Dividing top and bottom by [Br2] and then eliminating [Br] with equation 22-34 yields
$\frac{d[HBr]}{dt}\,=\frac{1}{1+(k_4/k_2)([HBr]/[Br_2])}\,$ * (2k1(k3/k5)1/2[H2][Br2]1/2)
(22-38)
This is exactly the experimental rate law, in which the experimental rate constants are related to those for the individual reactions in the chain by
k = 2k1 (k3/k5)1/2
k'= k4/k2
Now that we know what these two experimental constants mean in terms of the individual reactions, we can give a much fuller interpretation to the rate law, equation 22-38. Suppose that we could vary the individual rate constants, k1 to k3, at will. What effects would these changes have on the overall rate? The overall rate of production of HBr is accelerated if rate constants k1, k2, and k3 are large, or if reactions 1, 2, and 3 are fast. The first two of these reactions produce HBr; the third prepares the way by making more Br atoms. The production of HBr is slowed if k4 and k5 are large, or if the chain-inhibiting and chain-terminating reactions are fast.
So long as k3 and k5 change together, there is no change in the overall rate of reaction. Reactions 3 and 5 are the opposing initiating and terminating steps. Similarly, so long as k2 and k4 change together, the rate is unaffected. This, too, is sensible; for reactions 2 and 4 are similar in that they both consume an H and produce a Br, but differ in producing HBr in reaction 2 and removing it in reaction 4. Inhibition by HBr occurs because reaction 4 is enhanced, and inhibition is lessened by Br2 because reaction 2 is enhanced.
Practical uses of Complex Reactions
Although the chemistry behind complex reaction is not always understood, or even thought about, these reactions are used very often in the world around us.
A prime example of this is Hydrogen-Iodide which is used as a reactant in the complex reaction resulting in the production of methamphetamine. The production of methamphetamine utilizes the reduction of Ephedrine with Red Phosphorous, and Hydrogen-Iodide (aka Hydriodic Acid) [5]. The reaction mechanism for the reduction of Ephedrine with Red Phosphorous, and Hydrogen-Iodide is summarized as follows: Ephedrine reacts with Hydrogen-Iodide to form iodoephedrine (iodomethamphetamine) which is then reduced to methamphetamine [6]. The reduction of Ephedrine with Red Phosphorous, and Hydrogen-Iodide to create methamphetamine involves a cyclic oxidation of the iodide anion to iodine and reduction of iodine back to the anion by the red phosphorus [7]. This multi-step process is a classic complex reaction in which two reactants undergo an oxidation reaction to from an intermediate product. This intermediate chemical undergoes a reduction reaction with a third chemical, and transforms one of the original reactants back to its original state as well as changing the intermediate into the final product.
Classically, methamphetamine has been thought of as an illicit substance that used harsh chemicals to release a cascade of dopamine from the brain [8]. Although it is true that methamphetamine can be abused in this way, methamphetamine is also commonly used in medicine as well. In controlled dosages, and in specially designed time-release pills methamphetamine and Amphetamine are used to treat conditions such as Attention Deficit Hyperactivity Disorder, Traumatic Brain Injury, and the daytime drowsiness symptoms of Narcolepsy and Chronic Fatigue Syndrome. [9]. A typical daily dose of oral methamphetamine for the treatment of attention-deficit hyperactivity disorder in children is 20–25 mg. In the brain methamphetamine elevates the levels of extracellular monoamine neurotransmitters (dopamine, serotonin, norepinephrine) by promoting their release from the nerve endings. It is not completely understood how methamphetamine causes neurotransmitter release, but it appears to involve changing the distribution of the monoamine neurotransmitters from synaptic vesicles to the neuronal cytoplasm. It also reverses the transport of neurotransmitters from within the cells plasma membrane into the extracellular space [10]. This change in body chemistry results in the following symptoms and effects: alertness, wakefulness, energy, well-being, euphoria (at high doses) and suppression of appetite. Methamphetamine also activates the cardiovascular system (increased heart rate and blood pressure) and, for this reason, can cause death at high doses [11].
Another way in which complex reactions, and more specifically chain reactions, are used is in Nuclear Fission reactions. In these reactions an atom is bombarded by a neutron which splits the atom into two atoms which have a combined weight less than that of the original atom.[12] This reaction releases several products: 1) two smaller atoms; 2) heat; and 3) more neutrons. This allows the reaction to not only continue, but to grow in size on an exponential scale, thus becoming a chain reaction. This is the concept behind nuclear reactors and nuclear explosions; a sustained nuclear chain reaction of molecules that produces energy. For this reaction to occur and be efficient, atoms with high molecular weight, such as Plutonium and Uranium, must be used because their Binding Energy is low. As a result, when the atom is split, the amount of energy released is large. Also there is a specific mass, known as the critical mass, which is the mass the reactants must cumulatively be for the reaction to remain sustainable. If all of these criteria are met then a fission reaction will occur and it will result in a release of energy. The difference between a fission reactor and a fission bomb is the rate at which the fission reaction takes place.[13] In a nuclear reactor the metal is in a moderator, which slows the speed of the reaction, and the reaction is dispersed by control rods which absorb a portion of the free neutrons and slow the reaction further.[14] In these fission reactions the moderator, usually water, is used to make steam which powers a turbine and produces electrical power.[15]
References
2. ALSNews,Ozone Photodissociation Probed Using Undulator Light,http://www.als.lbl.gov/als/als_news/...35_090595.html
3. Gale Group Chemistry:probing reaction dynamics,en.wikibooks.org/w/index.php?...edit§ion=5
4. Yuan T. Lee's Crossed Molecular Beam Experiment , http://www.osti.gov/accomplishments/YuanLee_Exp.pdf
5. Walsh R, Martin E, Darvesh S. A method to describe enzyme-catalyzed reactions by combining steady state and time course enzyme kinetic parameters. Biochim Biophys Acta. 2010, 1800 (1), pp. 1-5
6. Skinner, Henry F., "Methamphetamine Synthesis Via HI/Red Phosphorous Reduction of Ephedrine," Forensic Science International, 48 128-134 (1990)
7. Skinner, Henry F., "Methamphetamine Synthesis Via HI/Red Phosphorous Reduction of Ephedrine," Forensic Science International, 48 128-134 (1990)
8. Skinner, Henry F., "Methamphetamine Synthesis Via HI/Red Phosphorous Reduction of Ephedrine," Forensic Science International, 48: 128-134 (1990)
9. Kish, Stephen J., "Pharmacologic mechanisms of crystal meth," CMAJ 13: 178 (June 17, 2008)
10. Kish, Stephen J., "Pharmacologic mechanisms of crystal meth," CMAJ 13: 178 (June 17, 2008)
11. Kish, Stephen J., "Pharmacologic mechanisms of crystal meth," CMAJ 13: 178 (June 17, 2008)
12. Kish, Stephen J., "Pharmacologic mechanisms of crystal meth," CMAJ 13: 178 (June 17, 2008)
13. Hill, John W., and Ralph H. Petrucci. General chemistry an integrated approach. Ed. Paul F. Corey. Upper Saddle River, N.J: Prentice Hall, 1999.
14. Hill, John W., and Ralph H. Petrucci. General chemistry an integrated approach. Ed. Paul F. Corey. Upper Saddle River, N.J: Prentice Hall, 1999.
15. Hill, John W., and Ralph H. Petrucci. General chemistry an integrated approach. Ed. Paul F. Corey. Upper Saddle River, N.J: Prentice Hall, 1999.
1. Hydroformylation is the process of producing aldehydes from alkenes by adding a formyl group (CHO) and a hydrogen atom to a carbon-carbon double bond using homogeneous catalysis. It is particularly useful because the aldehydes that are formed can be converted into useful products such as detergents and specialty chemicals.[3]
2. Ziegler-Natta catalysts are catalysts often based on titanium compounds and organometallic aluminium compounds. They are mostly used to polymerize terminal 1-alkenes.[4]
3. The majority of today's knowledge about homogeneous catalysis comes from earlier studies of hydrogenation. Hydrogenation (as shown in the figure below) is a chemical reaction that results from bonding hydrogen to organic compounds through the use of catalysts.
1. Carvone hydrogenation
The homogeneous catalysts involved in this reaction include Wilkinson’s catalyst, a rhodium-based compound, and Crabtree’s catalyst, an iridium-based compound. Moreover, many modern applications of hydrogenation can be found in petrochemical, pharmaceutical and food industries. [5]
2. Carbon-hydrogen bond activation or CH activation may be defined as a reaction that forms a carbon-hydrogen bond. Although they are traditionally unreactive, CH bonds can form by coordination using a catalyst. A significant role of CH activation is the ability to convert inexpensive and abundant alkanes into valuable functionalized organic compounds. [6]
1. In competitive reversible inhibition the substrate and the inhibitor compete for the enzyme's active site since they cannot bond to it at the same time. This type of inhibition can be overcome by high concentrations of substrate.[14] Competitive reversible inhibition can also occur when the competition is for the allosteric site instead of the active site.[15]
2. In uncompetitive reversible inhibition the inhibitor binds to the complex created by the substrate binding to the enzyme's active site. Because the active site is no longer available, the binding efficiency decreases as does the maximum velocity of the enzyme.[16]
3. In mixed inhibition the inhibitor can bind to the enzyme and its substrate simultaneously. While this effect can be reduced by a high concentration of substrate is cannot be completely overcome. This type of simultaneous binding can occur at the enzyme's active site, but generally occurs via the allosteric effect (The inhibitor binds to a site other than the active site.) which changes the shape of the enzyme and reduces the affinity of the substrate for the active site.[17]
1. Non-competitive inhibition, a type of mixed reversible inhibition, reduces the activity of the enzyme without affecting the binding capabilities of the substrate. The extent to which inhibition occurs is directly dependent on the concentration of the inhibitor.[18] The inhibitor always binds to an allosteric site.[19]
16. Bryant, Charles W. and Karim Nice. "How Catalytic Converters Work". http://auto.howstuffworks.com/catalytic-converter2.htm.
17. Brittanica Online Encyclopedia, "Homogeneous Catalysis", http://www.britannica.com/EBchecked/topic/270491/homogeneous-catalysis
18. Wikipedia, "Hydroformylation", en.Wikipedia.org/wiki/Hydroformylation
19. Wikipedia, "Ziegler-Natta Catalysts",en.Wikipedia.org/wiki/Ziegler-Natta
20. Wikipeida, "Hydrogenation", en.Wikipedia.org/wiki/Hydrogenation
21. Wikipedia, "C-H Activation", en.Wikipedia.org/wiki/C-H_activation
22. R. E. Dickerson, H. B. Gray, and G. P. Haight, Jr., Chemical Principles, 3rd edition, The Benjamin/Cummings Publishing Company, Menlo Park, CA, 1979, http://resolver.caltech.edu/CaltechBOOK:1979.001
23. R. E. Dickerson, H. B. Gray, and G. P. Haight, Jr., Chemical Principles, 3rd edition, The Benjamin/Cummings Publishing Company, Menlo Park, CA, 1979, http://resolver.caltech.edu/CaltechBOOK:1979.001
24. Accepta: Advanced Environmental Technologies, "Cooling Tower Scale & Corrosion Inhibitors" http://www.accepta.com/cooling-towers-water-treatment/cooling-tower-scale-corrosion-inhibitors.asp
25. NCI Cancer Bulletin, "Aromatase Inhibitors Come of Age," vol. 4/no. 10, 6 March 2007 www.cancer.gov/cancertopics/t...inhibitors0307
26. Protease Inhibitors: A Simple FactSheet from the AIDS Treatment Data Network, 15 Aug. 2006 http://atdn.org/simple/protease.html
27. Consumer Reports: Best Buy Drugs www.consumerreports.org/healt...ager_ACEIs.pdf
28. Wikipedia, Enzyme inhibitor, en.Wikipedia.org/wiki/Enzyme_inhibitor
29. Wikipedia, Enzyme inhibitor, en.Wikipedia.org/wiki/Enzyme_inhibitor
30. Wikipedia, Non-competitive inhibition, en.Wikipedia.org/wiki/Non-competitive_inhibition
31. Wikipedia, Enzyme inhibitor, en.Wikipedia.org/wiki/Enzyme_inhibitor
32. Wikipedia, Enzyme inhibitor, http://en.Wikipedia.org/wiki/Enzyme_inhibitor
33. Wikipedia, Enzyme inhibitor, http://en.Wikipedia.org/wiki/Enzyme_inhibitor
34. Wikipedia, Non-competitive inhibition, http://en.Wikipedia.org/wiki/Non-competitive_inhibition | textbooks/chem/General_Chemistry/Chemical_Principles_(Dickerson)/22%3A_Rates_and_Mechanisms_of_Chemical_Reactions/5.01%3A_Hydrogen_Oxygen_and_Water.txt |
Most everything you do and encounter during your day involves chemistry. Making coffee, cooking eggs, and toasting bread involve chemistry. The products you use—like soap and shampoo, the fabrics you wear, the electronics that keep you connected to your world, the gasoline that propels your car—all of these and more involve chemical substances and processes. Whether you are aware or not, chemistry is part of your everyday world. In this course, you will learn many of the essential principles underlying the chemistry of modern-day life.
01: Essential Ideas of Chemistry
Learning Objectives
• Outline the historical development of chemistry
• Provide examples of the importance of chemistry in everyday life
• Describe the scientific method and informs the scientific process
• Differentiate among hypotheses, theories, and laws
• Provide examples illustrating macroscopic, microscopic, and symbolic domains
Throughout human history, people have tried to convert matter into more useful forms. Our Stone Age ancestors chipped pieces of flint into useful tools and carved wood into statues and toys. These endeavors involved changing the shape of a substance without changing the substance itself. But as our knowledge increased, humans began to change the composition of the substances as well—clay was converted into pottery, hides were cured to make garments, copper ores were transformed into copper tools and weapons, and grain was made into bread.
Humans began to practice chemistry when they learned to control fire and use it to cook, make pottery, and smelt metals. Subsequently, they began to separate and use specific components of matter. A variety of drugs such as aloe, myrrh, and opium were isolated from plants. Dyes, such as indigo and Tyrian purple, were extracted from plant and animal matter. Metals were combined to form alloys—for example, copper and tin were mixed together to make bronze—and more elaborate smelting techniques produced iron. Alkalis were extracted from ashes, and soaps were prepared by combining these alkalis with fats. Alcohol was produced by fermentation and purified by distillation.
Attempts to understand the behavior of matter extend back for more than 2500 years. As early as the sixth century BC, Greek philosophers discussed a system in which water was the basis of all things. You may have heard of the Greek postulate that matter consists of four elements: earth, air, fire, and water. Subsequently, an amalgamation of chemical technologies and philosophical speculations were spread from Egypt, China, and the eastern Mediterranean by alchemists, who endeavored to transform “base metals” such as lead into “noble metals” like gold, and to create elixirs to cure disease and extend life (Figure \(1\)).
From alchemy came the historical progressions that led to modern chemistry: the isolation of drugs from natural sources, metallurgy, and the dye industry. Today, chemistry continues to deepen our understanding and improve our ability to harness and control the behavior of matter. This effort has been so successful that many people do not realize either the central position of chemistry among the sciences or the importance and universality of chemistry in daily life.
Chemistry: The Central Science
Chemistry is sometimes referred to as “the central science” due to its interconnectedness with a vast array of other STEM disciplines (STEM stands for areas of study in the science, technology, engineering, and math fields). Chemistry and the language of chemists play vital roles in biology, medicine, materials science, forensics, environmental science, and many other fields (Figure \(2\)). The basic principles of physics are essential for understanding many aspects of chemistry, and there is extensive overlap between many subdisciplines within the two fields, such as chemical physics and nuclear chemistry. Mathematics, computer science, and information theory provide important tools that help us calculate, interpret, describe, and generally make sense of the chemical world. Biology and chemistry converge in biochemistry, which is crucial to understanding the many complex factors and processes that keep living organisms (such as us) alive. Chemical engineering, materials science, and nanotechnology combine chemical principles and empirical findings to produce useful substances, ranging from gasoline to fabrics to electronics. Agriculture, food science, veterinary science, and brewing and wine making help provide sustenance in the form of food and drink to the world’s population. Medicine, pharmacology, biotechnology, and botany identify and produce substances that help keep us healthy. Environmental science, geology, oceanography, and atmospheric science incorporate many chemical ideas to help us better understand and protect our physical world. Chemical ideas are used to help understand the universe in astronomy and cosmology.
What are some changes in matter that are essential to daily life? Digesting and assimilating food, synthesizing polymers that are used to make clothing, containers, cookware, and credit cards, and refining crude oil into gasoline and other products are just a few examples. As you proceed through this course, you will discover many different examples of changes in the composition and structure of matter, how to classify these changes and how they occurred, their causes, the changes in energy that accompany them, and the principles and laws involved. As you learn about these things, you will be learning chemistry, the study of the composition, properties, and interactions of matter. The practice of chemistry is not limited to chemistry books or laboratories: It happens whenever someone is involved in changes in matter or in conditions that may lead to such changes.
The Scientific Method
Chemistry is a science based on observation and experimentation. Doing chemistry involves attempting to answer questions and explain observations in terms of the laws and theories of chemistry, using procedures that are accepted by the scientific community. There is no single route to answering a question or explaining an observation, but there is an aspect common to every approach: Each uses knowledge based on experiments that can be reproduced to verify the results. Some routes involve a hypothesis, a tentative explanation of observations that acts as a guide for gathering and checking information. We test a hypothesis by experimentation, calculation, and/or comparison with the experiments of others and then refine it as needed.
Some hypotheses are attempts to explain the behavior that is summarized in laws. The laws of science summarize a vast number of experimental observations, and describe or predict some facet of the natural world. If such a hypothesis turns out to be capable of explaining a large body of experimental data, it can reach the status of a theory. Scientific theories are well-substantiated, comprehensive, testable explanations of particular aspects of nature. Theories are accepted because they provide satisfactory explanations, but they can be modified if new data become available. The path of discovery that leads from question and observation to law or hypothesis to theory, combined with experimental verification of the hypothesis and any necessary modification of the theory, is called the scientific method (Figure \(3\)).
The Domains of Chemistry
Chemists study and describe the behavior of matter and energy in three different domains: macroscopic, microscopic, and symbolic. These domains provide different ways of considering and describing chemical behavior.
Macro is a Greek word that means “large.” The macroscopic domain is familiar to us: It is the realm of everyday things that are large enough to be sensed directly by human sight or touch. In daily life, this includes the food you eat and the breeze you feel on your face. The macroscopic domain includes everyday and laboratory chemistry, where we observe and measure physical and chemical properties, or changes such as density, solubility, and flammability.
The microscopic domain of chemistry is almost always visited in the imagination. Micro also comes from Greek and means “small.” Some aspects of the microscopic domains are visible through a microscope, such as a magnified image of graphite or bacteria. Viruses, for instance, are too small to be seen with the naked eye, but when we’re suffering from a cold, we’re reminded of how real they are.
However, most of the subjects in the microscopic domain of chemistry—such as atoms and molecules—are too small to be seen even with standard microscopes and often must be pictured in the mind. Other components of the microscopic domain include ions and electrons, protons and neutrons, and chemical bonds, each of which is far too small to see. This domain includes the individual metal atoms in a wire, the ions that compose a salt crystal, the changes in individual molecules that result in a color change, the conversion of nutrient molecules into tissue and energy, and the evolution of heat as bonds that hold atoms together are created.
The symbolic domain contains the specialized language used to represent components of the macroscopic and microscopic domains. Chemical symbols (such as those used in the periodic table), chemical formulas, and chemical equations are part of the symbolic domain, as are graphs and drawings. We can also consider calculations as part of the symbolic domain. These symbols play an important role in chemistry because they help interpret the behavior of the macroscopic domain in terms of the components of the microscopic domain. One of the challenges for students learning chemistry is recognizing that the same symbols can represent different things in the macroscopic and microscopic domains, and one of the features that makes chemistry fascinating is the use of a domain that must be imagined to explain behavior in a domain that can be observed.
A helpful way to understand the three domains is via the essential and ubiquitous substance of water. That water is a liquid at moderate temperatures, will freeze to form a solid at lower temperatures, and boil to form a gas at higher temperatures (Figure 1.1.4) are macroscopic observations. But some properties of water fall into the microscopic domain—what we cannot observe with the naked eye. The description of water as comprised of two hydrogen atoms and one oxygen atom, and the explanation of freezing and boiling in terms of attractions between these molecules, is within the microscopic arena. The formula H2O, which can describe water at either the macroscopic or microscopic levels, is an example of the symbolic domain. The abbreviations (g) for gas, (s) for solid, and (l) for liquid are also symbolic.
Key Concepts and Summary
Chemistry deals with the composition, structure, and properties of matter, and the ways by which various forms of matter may be interconverted. Thus, it occupies a central place in the study and practice of science and technology. Chemists use the scientific method to perform experiments, pose hypotheses, and formulate laws and develop theories, so that they can better understand the behavior of the natural world. To do so, they operate in the macroscopic, microscopic, and symbolic domains. Chemists measure, analyze, purify, and synthesize a wide variety of substances that are important to our lives.
Glossary
chemistry
study of the composition, properties, and interactions of matter
hypothesis
tentative explanation of observations that acts as a guide for gathering and checking information
law
statement that summarizes a vast number of experimental observations, and describes or predicts some aspect of the natural world
macroscopic domain
realm of everyday things that are large enough to sense directly by human sight and touch
microscopic domain
realm of things that are much too small to be sensed directly
scientific method
path of discovery that leads from question and observation to law or hypothesis to theory, combined with experimental verification of the hypothesis and any necessary modification of the theory
symbolic domain
specialized language used to represent components of the macroscopic and microscopic domains, such as chemical symbols, chemical formulas, chemical equations, graphs, drawings, and calculations
theory
well-substantiated, comprehensive, testable explanation of a particular aspect of nature | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/01%3A_Essential_Ideas_of_Chemistry/1.1%3A_Chemistry_in_Context.txt |
Learning Objectives
• Describe the basic properties of each physical state of matter: solid, liquid, and gas.
• Define and give examples of atoms and molecules.
• Classify matter as an element, compound, homogeneous mixture, or heterogeneous mixture with regard to its physical state and composition.
• Use symbolic, particulate, or macroscopic representations to describe or classify the different types of matter.
• Distinguish between mass and weight.
• Apply the law of conservation of matter.
Matter is defined as anything that occupies space and has mass, and it is all around us. Solids and liquids are more obviously matter: We can see that they take up space, and their weight tells us that they have mass. Gases are also matter; if gases did not take up space, a balloon would stay collapsed rather than inflate when filled with gas.
Solids, liquids, and gases are the three states of matter commonly found on earth (Figure $1$). A solid is rigid and possesses a definite shape. A liquid flows and takes the shape of a container, except that it forms a flat or slightly curved upper surface when acted upon by gravity. (In zero gravity, liquids assume a spherical shape.) Both liquid and solid samples have volumes that are very nearly independent of pressure. A gas takes both the shape and volume of its container.
A fourth state of matter, plasma, occurs naturally in the interiors of stars. A plasma is a gaseous state of matter that contains appreciable numbers of electrically charged particles (Figure $2$). The presence of these charged particles imparts unique properties to plasmas that justify their classification as a state of matter distinct from gases. In addition to stars, plasmas are found in some other high-temperature environments (both natural and man-made), such as lightning strikes, certain television screens, and specialized analytical instruments used to detect trace amounts of metals.
Video $1$: In a tiny cell in a plasma television, the plasma emits ultraviolet light, which in turn causes the display at that location to appear a specific color. The composite of these tiny dots of color makes up the image that you see. Watch this video to learn more about plasma and the places you encounter it.
Some samples of matter appear to have properties of solids, liquids, and/or gases at the same time. This can occur when the sample is composed of many small pieces. For example, we can pour sand as if it were a liquid because it is composed of many small grains of solid sand. Matter can also have properties of more than one state when it is a mixture, such as with clouds. Clouds appear to behave somewhat like gases, but they are actually mixtures of air (gas) and tiny particles of water (liquid or solid).
The mass of an object is a measure of the amount of matter in it. One way to measure an object’s mass is to measure the force it takes to accelerate the object. It takes much more force to accelerate a car than a bicycle because the car has much more mass. A more common way to determine the mass of an object is to use a balance to compare its mass with a standard mass.
Although weight is related to mass, it is not the same thing. Weight refers to the force that gravity exerts on an object. This force is directly proportional to the mass of the object. The weight of an object changes as the force of gravity changes, but its mass does not. An astronaut’s mass does not change just because she goes to the moon. But her weight on the moon is only one-sixth her earth-bound weight because the moon’s gravity is only one-sixth that of the earth’s. She may feel “weightless” during her trip when she experiences negligible external forces (gravitational or any other), although she is, of course, never “massless.”
The law of conservation of matter summarizes many scientific observations about matter: It states that there is no detectable change in the total quantity of matter present when matter converts from one type to another (a chemical change) or changes among solid, liquid, or gaseous states (a physical change). Brewing beer and the operation of batteries provide examples of the conservation of matter (Figure $4$). During the brewing of beer, the ingredients (water, yeast, grains, malt, hops, and sugar) are converted into beer (water, alcohol, carbonation, and flavoring substances) with no actual loss of substance. This is most clearly seen during the bottling process, when glucose turns into ethanol and carbon dioxide, and the total mass of the substances does not change. This can also be seen in a lead-acid car battery: The original substances (lead, lead oxide, and sulfuric acid), which are capable of producing electricity, are changed into other substances (lead sulfate and water) that do not produce electricity, with no change in the actual amount of matter.
Although this conservation law holds true for all conversions of matter, convincing examples are few and far between because, outside of the controlled conditions in a laboratory, we seldom collect all of the material that is produced during a particular conversion. For example, when you eat, digest, and assimilate food, all of the matter in the original food is preserved. But because some of the matter is incorporated into your body, and much is excreted as various types of waste, it is challenging to verify by measurement.
Atoms and Molecules
An atom is the smallest particle of an element that has the properties of that element and can enter into a chemical combination. Consider the element gold, for example. Imagine cutting a gold nugget in half, then cutting one of the halves in half, and repeating this process until a piece of gold remained that was so small that it could not be cut in half (regardless of how tiny your knife may be). This minimally sized piece of gold is an atom (from the Greek atomos, meaning “indivisible”) (Figure 1.2.4). This atom would no longer be gold if it were divided any further.
The first suggestion that matter is composed of atoms is attributed to the Greek philosophers Leucippus and Democritus, who developed their ideas in the 5th century BCE. However, it was not until the early nineteenth century that John Dalton (1766–1844), a British schoolteacher with a keen interest in science, supported this hypothesis with quantitative measurements. Since that time, repeated experiments have confirmed many aspects of this hypothesis, and it has become one of the central theories of chemistry. Other aspects of Dalton’s atomic theory are still used but with minor revisions (details of Dalton’s theory are provided in the chapter on atoms and molecules).
An atom is so small that its size is difficult to imagine. One of the smallest things we can see with our unaided eye is a single thread of a spider web: These strands are about 1/10,000 of a centimeter (0.0001 cm) in diameter. Although the cross-section of one strand is almost impossible to see without a microscope, it is huge on an atomic scale. A single carbon atom in the web has a diameter of about 0.000000015 centimeter, and it would take about 7000 carbon atoms to span the diameter of the strand. To put this in perspective, if a carbon atom were the size of a dime, the cross-section of one strand would be larger than a football field, which would require about 150 million carbon atom “dimes” to cover it. (Figure $5$) shows increasingly close microscopic and atomic-level views of ordinary cotton.
An atom is so light that its mass is also difficult to imagine. A billion lead atoms (1,000,000,000 atoms) weigh about $3 \times 10^{−13}$ grams, a mass that is far too light to be weighed on even the world’s most sensitive balances. It would require over 300,000,000,000,000 lead atoms (300 trillion, or 3 × 1014) to be weighed, and they would weigh only 0.0000001 gram.
It is rare to find collections of individual atoms. Only a few elements, such as the gases helium, neon, and argon, consist of a collection of individual atoms that move about independently of one another. Other elements, such as the gases hydrogen, nitrogen, oxygen, and chlorine, are composed of units that consist of pairs of atoms (Figure $6$). One form of the element phosphorus consists of units composed of four phosphorus atoms. The element sulfur exists in various forms, one of which consists of units composed of eight sulfur atoms. These units are called molecules. A molecule consists of two or more atoms joined by strong forces called chemical bonds. The atoms in a molecule move around as a unit, much like the cans of soda in a six-pack or a bunch of keys joined together on a single key ring. A molecule may consist of two or more identical atoms, as in the molecules found in the elements hydrogen, oxygen, and sulfur, or it may consist of two or more different atoms, as in the molecules found in water. Each water molecule is a unit that contains two hydrogen atoms and one oxygen atom. Each glucose molecule is a unit that contains 6 carbon atoms, 12 hydrogen atoms, and 6 oxygen atoms. Like atoms, molecules are incredibly small and light. If an ordinary glass of water were enlarged to the size of the earth, the water molecules inside it would be about the size of golf balls.
Classifying Matter
We can classify matter into several categories. Two broad categories are mixtures and pure substances. A pure substance has a constant composition. All specimens of a pure substance have exactly the same makeup and properties. Any sample of sucrose (table sugar) consists of 42.1% carbon, 6.5% hydrogen, and 51.4% oxygen by mass. Any sample of sucrose also has the same physical properties, such as melting point, color, and sweetness, regardless of the source from which it is isolated.
We can divide pure substances into two classes: elements and compounds. Pure substances that cannot be broken down into simpler substances by chemical changes are called elements. Iron, silver, gold, aluminum, sulfur, oxygen, and copper are familiar examples of the more than 100 known elements, of which about 90 occur naturally on the earth, and two dozen or so have been created in laboratories.
Pure substances that can be broken down by chemical changes are called compounds. This breakdown may produce either elements or other compounds, or both. Mercury(II) oxide, an orange, crystalline solid, can be broken down by heat into the elements mercury and oxygen (Figure $7$). When heated in the absence of air, the compound sucrose is broken down into the element carbon and the compound water. (The initial stage of this process, when the sugar is turning brown, is known as caramelization—this is what imparts the characteristic sweet and nutty flavor to caramel apples, caramelized onions, and caramel). Silver(I) chloride is a white solid that can be broken down into its elements, silver and chlorine, by absorption of light. This property is the basis for the use of this compound in photographic films and photochromic eyeglasses (those with lenses that darken when exposed to light).
The properties of combined elements are different from those in the free, or uncombined, state. For example, white crystalline sugar (sucrose) is a compound resulting from the chemical combination of the element carbon, which is a black solid in one of its uncombined forms, and the two elements hydrogen and oxygen, which are colorless gases when uncombined. Free sodium, an element that is a soft, shiny, metallic solid, and free chlorine, an element that is a yellow-green gas, combine to form sodium chloride (table salt), a compound that is a white, crystalline solid.
A mixture is composed of two or more types of matter that can be present in varying amounts and can be separated by physical changes, such as evaporation (you will learn more about this later). A mixture with a composition that varies from point to point is called a heterogeneous mixture. Italian dressing is an example of a heterogeneous mixture (Figure $\PageIndex{1a}$). Its composition can vary because we can make it from varying amounts of oil, vinegar, and herbs. It is not the same from point to point throughout the mixture—one drop may be mostly vinegar, whereas a different drop may be mostly oil or herbs because the oil and vinegar separate and the herbs settle. Other examples of heterogeneous mixtures are chocolate chip cookies (we can see the separate bits of chocolate, nuts, and cookie dough) and granite (we can see the quartz, mica, feldspar, and more).
A homogeneous mixture, also called a solution, exhibits a uniform composition and appears visually the same throughout. An example of a solution is a sports drink, consisting of water, sugar, coloring, flavoring, and electrolytes mixed together uniformly (Figure $\PageIndex{1b}$). Each drop of a sports drink tastes the same because each drop contains the same amounts of water, sugar, and other components. Note that the composition of a sports drink can vary—it could be made with somewhat more or less sugar, flavoring, or other components, and still be a sports drink. Other examples of homogeneous mixtures include air, maple syrup, gasoline, and a solution of salt in water.
Although there are just over 100 elements, tens of millions of chemical compounds result from different combinations of these elements. Each compound has a specific composition and possesses definite chemical and physical properties by which we can distinguish it from all other compounds. And, of course, there are innumerable ways to combine elements and compounds to form different mixtures. A summary of how to distinguish between the various major classifications of matter is shown in (Figure 1.2.8).
Eleven elements make up about 99% of the earth’s crust and atmosphere (Table $1$). Oxygen constitutes nearly one-half and silicon about one-quarter of the total quantity of these elements. A majority of elements on earth are found in chemical combinations with other elements; about one-quarter of the elements are also found in the free state.
Table $1$: Elemental Composition of Earth
Element Symbol Percent Mass Element Symbol Percent Mass
oxygen O 49.20 chlorine Cl 0.19
silicon Si 25.67 phosphorus P 0.11
aluminum Al 7.50 manganese Mn 0.09
iron Fe 4.71 carbon C 0.08
calcium Ca 3.39 sulfur S 0.06
sodium Na 2.63 barium Ba 0.04
potassium K 2.40 nitrogen N 0.03
magnesium Mg 1.93 fluorine F 0.03
hydrogen H 0.87 strontium Sr 0.02
titanium Ti 0.58 all others - 0.47
Decomposition of Water / Production of Hydrogen
Water consists of the elements hydrogen and oxygen combined in a 2 to 1 ratio. Water can be broken down into hydrogen and oxygen gases by the addition of energy. One way to do this is with a battery or power supply, as shown in (Figure $9$).
The breakdown of water involves a rearrangement of the atoms in water molecules into different molecules, each composed of two hydrogen atoms and two oxygen atoms, respectively. Two water molecules form one oxygen molecule and two hydrogen molecules. The representation for what occurs, $\ce{2H2O}(l)\rightarrow \ce{2H2}(g)+\ce{O2}(g)$, will be explored in more depth in later chapters.
The two gases produced have distinctly different properties. Oxygen is not flammable but is required for combustion of a fuel, and hydrogen is highly flammable and a potent energy source. How might this knowledge be applied in our world? One application involves research into more fuel-efficient transportation. Fuel-cell vehicles (FCV) run on hydrogen instead of gasoline (Figure $10$). They are more efficient than vehicles with internal combustion engines, are nonpolluting, and reduce greenhouse gas emissions, making us less dependent on fossil fuels. FCVs are not yet economically viable, however, and current hydrogen production depends on natural gas. If we can develop a process to economically decompose water, or produce hydrogen in another environmentally sound way, FCVs may be the way of the future.
Chemistry of Cell Phones
Imagine how different your life would be without cell phones (Figure $11$) and other smart devices. Cell phones are made from numerous chemical substances, which are extracted, refined, purified, and assembled using an extensive and in-depth understanding of chemical principles. About 30% of the elements that are found in nature are found within a typical smart phone. The case/body/frame consists of a combination of sturdy, durable polymers comprised primarily of carbon, hydrogen, oxygen, and nitrogen [acrylonitrile butadiene styrene (ABS) and polycarbonate thermoplastics], and light, strong, structural metals, such as aluminum, magnesium, and iron. The display screen is made from a specially toughened glass (silica glass strengthened by the addition of aluminum, sodium, and potassium) and coated with a material to make it conductive (such as indium tin oxide). The circuit board uses a semiconductor material (usually silicon); commonly used metals like copper, tin, silver, and gold; and more unfamiliar elements such as yttrium, praseodymium, and gadolinium. The battery relies upon lithium ions and a variety of other materials, including iron, cobalt, copper, polyethylene oxide, and polyacrylonitrile.
Summary
Matter is anything that occupies space and has mass. The basic building block of matter is the atom, the smallest unit of an element that can enter into combinations with atoms of the same or other elements. In many substances, atoms are combined into molecules. On earth, matter commonly exists in three states: solids, of fixed shape and volume; liquids, of variable shape but fixed volume; and gases, of variable shape and volume. Under high-temperature conditions, matter also can exist as a plasma. Most matter is a mixture: It is composed of two or more types of matter that can be present in varying amounts and can be separated by physical means. Heterogeneous mixtures vary in composition from point to point; homogeneous mixtures have the same composition from point to point. Pure substances consist of only one type of matter. A pure substance can be an element, which consists of only one type of atom and cannot be broken down by a chemical change, or a compound, which consists of two or more types of atoms.
Glossary
atom
smallest particle of an element that can enter into a chemical combination
compound
pure substance that can be decomposed into two or more elements
element
substance that is composed of a single type of atom; a substance that cannot be decomposed by a chemical change
gas
state in which matter has neither definite volume nor shape
heterogeneous mixture
combination of substances with a composition that varies from point to point
homogeneous mixture
(also, solution) combination of substances with a composition that is uniform throughout
liquid
state of matter that has a definite volume but indefinite shape
law of conservation of matter
when matter converts from one type to another or changes form, there is no detectable change in the total amount of matter present
mass
fundamental property indicating amount of matter
matter
anything that occupies space and has mass
mixture
matter that can be separated into its components by physical means
molecule
bonded collection of two or more atoms of the same or different elements
plasma
gaseous state of matter containing a large number of electrically charged atoms and/or molecules
pure substance
homogeneous substance that has a constant composition
solid
state of matter that is rigid, has a definite shape, and has a fairly constant volume
weight
force that gravity exerts on an object | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/01%3A_Essential_Ideas_of_Chemistry/1.2%3A_Phases_and_Classification_of_Matter.txt |
Learning Objectives
• Describe the difference between physical and chemical properties or changed.
• Identify a property or transformation as either physical or chemical using symbolic, particulate, or macroscopic representations.
• Identify the properties of matter as extensive or intensive.
• Recognize and describe the parts of the NFPA hazard diamond.
The characteristics that enable us to distinguish one substance from another are called properties. A physical property is a characteristic of matter that is not associated with a change in its chemical composition. Familiar examples of physical properties include density, color, hardness, melting and boiling points, and electrical conductivity. We can observe some physical properties, such as density and color, without changing the physical state of the matter observed. Other physical properties, such as the melting temperature of iron or the freezing temperature of water, can only be observed as matter undergoes a physical change. A physical change is a change in the state or properties of matter without any accompanying change in its chemical composition (the identities of the substances contained in the matter). We observe a physical change when wax melts, when sugar dissolves in coffee, and when steam condenses into liquid water (Figure \(1\)). Other examples of physical changes include magnetizing and demagnetizing metals (as is done with common antitheft security tags) and grinding solids into powders (which can sometimes yield noticeable changes in color). In each of these examples, there is a change in the physical state, form, or properties of the substance, but no change in its chemical composition.
The change of one type of matter into another type (or the inability to change) is a chemical property. Examples of chemical properties include flammability, toxicity, acidity, reactivity (many types), and heat of combustion. Iron, for example, combines with oxygen in the presence of water to form rust; chromium does not oxidize (Figure \(2\)). Nitroglycerin is very dangerous because it explodes easily; neon poses almost no hazard because it is very unreactive.
To identify a chemical property, we look for a chemical change. A chemical change always produces one or more types of matter that differ from the matter present before the change. The formation of rust is a chemical change because rust is a different kind of matter than the iron, oxygen, and water present before the rust formed. The explosion of nitroglycerin is a chemical change because the gases produced are very different kinds of matter from the original substance. Other examples of chemical changes include reactions that are performed in a lab (such as copper reacting with nitric acid), all forms of combustion (burning), and food being cooked, digested, or rotting (Figure \(3\)).
Properties of matter fall into one of two categories. If the property depends on the amount of matter present, it is an extensive property. The mass and volume of a substance are examples of extensive properties; for instance, a gallon of milk has a larger mass and volume than a cup of milk. The value of an extensive property is directly proportional to the amount of matter in question. If the property of a sample of matter does not depend on the amount of matter present, it is an intensive property. Temperature is an example of an intensive property. If the gallon and cup of milk are each at 20 °C (room temperature), when they are combined, the temperature remains at 20 °C. As another example, consider the distinct but related properties of heat and temperature. A drop of hot cooking oil spattered on your arm causes brief, minor discomfort, whereas a pot of hot oil yields severe burns. Both the drop and the pot of oil are at the same temperature (an intensive property), but the pot clearly contains much more heat (extensive property).
Hazard Diamond
You may have seen the symbol shown in Figure \(4\) on containers of chemicals in a laboratory or workplace. Sometimes called a “fire diamond” or “hazard diamond,” this chemical hazard diamond provides valuable information that briefly summarizes the various dangers of which to be aware when working with a particular substance.
The National Fire Protection Agency (NFPA) 704 Hazard Identification System was developed by NFPA to provide safety information about certain substances. The system details flammability, reactivity, health, and other hazards. Within the overall diamond symbol, the top (red) diamond specifies the level of fire hazard (temperature range for flash point). The blue (left) diamond indicates the level of health hazard. The yellow (right) diamond describes reactivity hazards, such as how readily the substance will undergo detonation or a violent chemical change. The white (bottom) diamond points out special hazards, such as if it is an oxidizer (which allows the substance to burn in the absence of air/oxygen), undergoes an unusual or dangerous reaction with water, is corrosive, acidic, alkaline, a biological hazard, radioactive, and so on. Each hazard is rated on a scale from 0 to 4, with 0 being no hazard and 4 being extremely hazardous.
While many elements differ dramatically in their chemical and physical properties, some elements have similar properties. We can identify sets of elements that exhibit common behaviors. For example, many elements conduct heat and electricity well, whereas others are poor conductors. These properties can be used to sort the elements into three classes: metals (elements that conduct well), nonmetals (elements that conduct poorly), and metalloids (elements that have properties of both metals and nonmetals).
The periodic table is a table of elements that places elements with similar properties close together (Figure \(5\)). You will learn more about the periodic table as you continue your study of chemistry.
Summary
All substances have distinct physical and chemical properties, and may undergo physical or chemical changes. Physical properties, such as hardness and boiling point, and physical changes, such as melting or freezing, do not involve a change in the composition of matter. Chemical properties, such flammability and acidity, and chemical changes, such as rusting, involve production of matter that differs from that present beforehand.
Measurable properties fall into one of two categories. Extensive properties depend on the amount of matter present, for example, the mass of gold. Intensive properties do not depend on the amount of matter present, for example, the density of gold. Heat is an example of an extensive property, and temperature is an example of an intensive property.
Glossary
chemical change
change producing a different kind of matter from the original kind of matter
chemical property
behavior that is related to the change of one kind of matter into another kind of matter
extensive property
property of a substance that depends on the amount of the substance
intensive property
property of a substance that is independent of the amount of the substance
physical change
change in the state or properties of matter that does not involve a change in its chemical composition
physical property
characteristic of matter that is not associated with any change in its chemical composition | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/01%3A_Essential_Ideas_of_Chemistry/1.3%3A_Physical_and_Chemical_Properties.txt |
Learning Objectives
• Explain the process of measurement and describe the three basic parts of a quantity.
• Describe the properties and units of length, mass, volume, density, temperature, and time.
• Recognize the common unit prefixes and use them to describe the magnitude of a measurement.
• Describe and calculate the density of a substance.
• Perform basic unit calculations and conversions in the metric and other unit systems.
Measurements provide the macroscopic information that is the basis of most of the hypotheses, theories, and laws that describe the behavior of matter and energy in both the macroscopic and microscopic domains of chemistry. Every measurement provides three kinds of information: the size or magnitude of the measurement (a number); a standard of comparison for the measurement (a unit); and an indication of the uncertainty of the measurement. While the number and unit are explicitly represented when a quantity is written, the uncertainty is an aspect of the measurement result that is more implicitly represented and will be discussed later.
The number in the measurement can be represented in different ways, including decimal form and scientific notation. For example, the maximum takeoff weight of a Boeing 777-200ER airliner is 298,000 kilograms, which can also be written as 2.98 $\times$ 105 kg. The mass of the average mosquito is about 0.0000025 kilograms, which can be written as 2.5 $\times$ 10−6 kg.
Units, such as liters, pounds, and centimeters, are standards of comparison for measurements. When we buy a 2-liter bottle of a soft drink, we expect that the volume of the drink was measured, so it is two times larger than the volume that everyone agrees to be 1 liter. The meat used to prepare a 0.25-pound hamburger is measured so it weighs one-fourth as much as 1 pound. Without units, a number can be meaningless, confusing, or possibly life threatening. Suppose a doctor prescribes phenobarbital to control a patient’s seizures and states a dosage of “100” without specifying units. Not only will this be confusing to the medical professional giving the dose, but the consequences can be dire: 100 mg given three times per day can be effective as an anticonvulsant, but a single dose of 100 g is more than 10 times the lethal amount.
We usually report the results of scientific measurements in SI units, an updated version of the metric system, using the units listed in Table $1$. Other units can be derived from these base units. The standards for these units are fixed by international agreement, and they are called the International System of Units or SI Units (from the French, Le Système International d’Unités). SI units have been used by the United States National Institute of Standards and Technology (NIST) since 1964.
Table $1$: Base Units of the SI System
Property Measured Name of Unit Symbol of Unit
length meter m
mass kilogram kg
time second s
temperature kelvin K
electric current ampere A
amount of substance mole mol
luminous intensity candela cd
Sometimes we use units that are fractions or multiples of a base unit. Ice cream is sold in quarts (a familiar, non-SI base unit), pints (0.5 quart), or gallons (4 quarts). We also use fractions or multiples of units in the SI system, but these fractions or multiples are always powers of 10. Fractional or multiple SI units are named using a prefix and the name of the base unit. For example, a length of 1000 meters is also called a kilometer because the prefix kilo means “one thousand,” which in scientific notation is 103 (1 kilometer = 1000 m = 103 m). The prefixes used and the powers to which 10 are raised are listed in Table $2$.
NG, equals 4 times ten to the negative 9, or 0.000000004 g. The prefix micro has the greek letter mu as its symbol and a factor of 10 to the negative sixth power. Therefore, 1 microliter, or mu L, is equal to one times ten to the negative 6 or 0.000001 L. The prefix milli has a lowercase M as its symbol and a factor of 10 to the negative third power. Therefore, 2 millimoles, or M mol, are equal to two times ten to the negative 3 or 0.002 mol. The prefix centi has a lowercase C as its symbol and a factor of 10 to the negative second power. Therefore, 7 centimeters, or C M, are equal to seven times ten to the negative 2 meters or 0.07 M O L. The prefix deci has a lowercase D as its symbol and a factor of 10 to the negative first power. Therefore, 1 deciliter, or lowercase D uppercase L, are equal to one times ten to the negative 1 meters or 0.1 L. The prefix kilo has a lowercase K as its symbol and a factor of 10 to the third power. Therefore, 1 kilometer, or K M, is equal to one times ten to the third meters or 1000 M. The prefix mega has an uppercase M as its symbol and a factor of 10 to the sixth power. Therefore, 3 megahertz, or M H Z, are equal to three times 10 to the sixth hertz, or 3000000 H Z. The prefix giga has an uppercase G as its symbol and a factor of 10 to the ninth power. Therefore, 8 gigayears, or G Y R, are equal to eight times 10 to the ninth years, or 800000000 G Y R. The prefix tera has an uppercase T as its symbol and a factor of 10 to the twelfth power. Therefore, 5 terawatts, or T W, are equal to five times 10 to the twelfth watts, or 5000000000000 W." data-quail-id="64" data-mt-width="1076">
Table $2$: Common Unit Prefixes
Prefix Symbol Factor Example
femto f 10−15 1 femtosecond (fs) = 1 $\times$ 10−15 s (0.000000000000001 s)
pico p 10−12 1 picometer (pm) = 1 $\times$ 10−12 m (0.000000000001 m)
nano n 10−9 4 nanograms (ng) = 4 $\times$ 10−9 g (0.000000004 g)
micro µ 10−6 1 microliter (μL) = 1 $\times$ 10−6 L (0.000001 L)
milli m 10−3 2 millimoles (mmol) = 2 $\times$ 10−3 mol (0.002 mol)
centi c 10−2 7 centimeters (cm) = 7 $\times$ 10−2 m (0.07 m)
deci d 10−1 1 deciliter (dL) = 1 $\times$ 10−1 L (0.1 L )
kilo k 103 1 kilometer (km) = 1 $\times$ 103 m (1000 m)
mega M 106 3 megahertz (MHz) = 3 $\times$ 106 Hz (3,000,000 Hz)
giga G 109 8 gigayears (Gyr) = 8 $\times$ 109 yr (8,000,000,000 Gyr)
tera T 1012 5 terawatts (TW) = 5 $\times$ 1012 W (5,000,000,000,000 W)
SI Base Units
The initial units of the metric system, which eventually evolved into the SI system, were established in France during the French Revolution. The original standards for the meter and the kilogram were adopted there in 1799 and eventually by other countries. This section introduces four of the SI base units commonly used in chemistry. Other SI units will be introduced in subsequent chapters.
Length
The standard unit of length in both the SI and original metric systems is the meter (m). A meter was originally specified as 1/10,000,000 of the distance from the North Pole to the equator. It is now defined as the distance light in a vacuum travels in 1/299,792,458 of a second. A meter is about 3 inches longer than a yard (Figure $1$); one meter is about 39.37 inches or 1.094 yards. Longer distances are often reported in kilometers (1 km = 1000 m = 103 m), whereas shorter distances can be reported in centimeters (1 cm = 0.01 m = 10−2 m) or millimeters (1 mm = 0.001 m = 10−3 m).
Mass
The standard unit of mass in the SI system is the kilogram (kg). A kilogram was originally defined as the mass of a liter of water (a cube of water with an edge length of exactly 0.1 meter). It is now defined by a certain cylinder of platinum-iridium alloy, which is kept in France (Figure 1.4.2). Any object with the same mass as this cylinder is said to have a mass of 1 kilogram. One kilogram is about 2.2 pounds. The gram (g) is exactly equal to 1/1000 of the mass of the kilogram (10−3 kg).
Temperature
Temperature is an intensive property. The SI unit of temperature is the kelvin (K). The IUPAC convention is to use kelvin (all lowercase) for the word, K (uppercase) for the unit symbol, and neither the word “degree” nor the degree symbol (°). The degree Celsius (°C) is also allowed in the SI system, with both the word “degree” and the degree symbol used for Celsius measurements. Celsius degrees are the same magnitude as those of kelvin, but the two scales place their zeros in different places. Water freezes at 273.15 K (0 °C) and boils at 373.15 K (100 °C) by definition, and normal human body temperature is approximately 310 K (37 °C). The conversion between these two units and the Fahrenheit scale will be discussed later in this chapter.
Time
The SI base unit of time is the second (s). Small and large time intervals can be expressed with the appropriate prefixes; for example, 3 microseconds = 0.000003 s = 3 $\times$ 10−6 and 5 megaseconds = 5,000,000 s = 5 $\times$ 106 s. Alternatively, hours, days, and years can be used.
Derived SI Units
We can derive many units from the seven SI base units. For example, we can use the base unit of length to define a unit of volume, and the base units of mass and length to define a unit of density.
Volume
Volume is the measure of the amount of space occupied by an object. The standard SI unit of volume is defined by the base unit of length (Figure $3$). The standard volume is a cubic meter (m3), a cube with an edge length of exactly one meter. To dispense a cubic meter of water, we could build a cubic box with edge lengths of exactly one meter. This box would hold a cubic meter of water or any other substance.
A more commonly used unit of volume is derived from the decimeter (0.1 m, or 10 cm). A cube with edge lengths of exactly one decimeter contains a volume of one cubic decimeter (dm3). A liter (L) is the more common name for the cubic decimeter. One liter is about 1.06 quarts. A cubic centimeter (cm3) is the volume of a cube with an edge length of exactly one centimeter. The abbreviation cc (for cubic centimeter) is often used by health professionals. A cubic centimeter is also called a milliliter (mL) and is 1/1000 of a liter.
Density
We use the mass and volume of a substance to determine its density. Thus, the units of density are defined by the base units of mass and length.
The density of a substance is the ratio of the mass of a sample of the substance to its volume. The SI unit for density is the kilogram per cubic meter (kg/m3). For many situations, however, this as an inconvenient unit, and we often use grams per cubic centimeter (g/cm3) for the densities of solids and liquids, and grams per liter (g/L) for gases. Although there are exceptions, most liquids and solids have densities that range from about 0.7 g/cm3 (the density of gasoline) to 19 g/cm3 (the density of gold). The density of air is about 1.2 g/L. Table $3$ shows the densities of some common substances.
Table $3$: Densities of Common Substances
Solids Liquids Gases (at 25 °C and 1 atm)
ice (at 0 °C) 0.92 g/cm3 water 1.0 g/cm3 dry air 1.20 g/L
oak (wood) 0.60–0.90 g/cm3 ethanol 0.79 g/cm3 oxygen 1.31 g/L
iron 7.9 g/cm3 acetone 0.79 g/cm3 nitrogen 1.14 g/L
copper 9.0 g/cm3 glycerin 1.26 g/cm3 carbon dioxide 1.80 g/L
lead 11.3 g/cm3 olive oil 0.92 g/cm3 helium 0.16 g/L
silver 10.5 g/cm3 gasoline 0.70–0.77 g/cm3 neon 0.83 g/L
gold 19.3 g/cm3 mercury 13.6 g/cm3 radon 9.1 g/L
While there are many ways to determine the density of an object, perhaps the most straightforward method involves separately finding the mass and volume of the object, and then dividing the mass of the sample by its volume. In the following example, the mass is found directly by weighing, but the volume is found indirectly through length measurements.
$\mathrm{density=\dfrac{mass}{volume}} \nonumber$
Example $1$
Calculation of Density Gold—in bricks, bars, and coins—has been a form of currency for centuries. In order to swindle people into paying for a brick of gold without actually investing in a brick of gold, people have considered filling the centers of hollow gold bricks with lead to fool buyers into thinking that the entire brick is gold. It does not work: Lead is a dense substance, but its density is not as great as that of gold, 19.3 g/cm3. What is the density of lead if a cube of lead has an edge length of 2.00 cm and a mass of 90.7 g?
Solution
The density of a substance can be calculated by dividing its mass by its volume. The volume of a cube is calculated by cubing the edge length.
$\mathrm{volume\: of\: lead\: cube=2.00\: cm\times2.00\: cm\times2.00\: cm=8.00\: cm^3} \nonumber$
$\mathrm{density=\dfrac{mass}{volume}=\dfrac{90.7\: g}{8.00\: cm^3}=\dfrac{11.3\: g}{1.00\: cm^3}=11.3\: g/cm^3} \nonumber$
(We will discuss the reason for rounding to the first decimal place in the next section.)
Exercise $1$
1. To three decimal places, what is the volume of a cube (cm3) with an edge length of 0.843 cm?
2. If the cube in part (a) is copper and has a mass of 5.34 g, what is the density of copper to two decimal places?
Answer a
0.599 cm3;
Answer b
8.91 g/cm3
Example $2$: Using Displacement of Water to Determine Density
This PhET simulation illustrates another way to determine density, using displacement of water. Determine the density of the red and yellow blocks.
Solution
When you open the density simulation and select Same Mass, you can choose from several 5.00-kg colored blocks that you can drop into a tank containing 100.00 L water. The yellow block floats (it is less dense than water), and the water level rises to 105.00 L. While floating, the yellow block displaces 5.00 L water, an amount equal to the weight of the block. The red block sinks (it is more dense than water, which has density = 1.00 kg/L), and the water level rises to 101.25 L.
The red block therefore displaces 1.25 L water, an amount equal to the volume of the block. The density of the red block is:
$\mathrm{density=\dfrac{mass}{volume}=\dfrac{5.00\: kg}{1.25\: L}=4.00\: kg/L} \nonumber$
Note that since the yellow block is not completely submerged, you cannot determine its density from this information. But if you hold the yellow block on the bottom of the tank, the water level rises to 110.00 L, which means that it now displaces 10.00 L water, and its density can be found:
$\mathrm{density=\dfrac{mass}{volume}=\dfrac{5.00\: kg}{10.00\: L}=0.500\: kg/L} \nonumber$
Exercise $1$
Remove all of the blocks from the water and add the green block to the tank of water, placing it approximately in the middle of the tank. Determine the density of the green block.
Answer
2.00 kg/L
Summary
Measurements provide quantitative information that is critical in studying and practicing chemistry. Each measurement has an amount, a unit for comparison, and an uncertainty. Measurements can be represented in either decimal or scientific notation. Scientists primarily use the SI (International System) or metric systems. We use base SI units such as meters, seconds, and kilograms, as well as derived units, such as liters (for volume) and g/cm3 (for density). In many cases, we find it convenient to use unit prefixes that yield fractional and multiple units, such as microseconds (10−6 seconds) and megahertz (106 hertz), respectively.
Key Equations
• $\mathrm{density=\dfrac{mass}{volume}}$
Glossary
Celsius (°C)
unit of temperature; water freezes at 0 °C and boils at 100 °C on this scale
cubic centimeter (cm3 or cc)
volume of a cube with an edge length of exactly 1 cm
cubic meter (m3)
SI unit of volume
density
ratio of mass to volume for a substance or object
kelvin (K)
SI unit of temperature; 273.15 K = 0 ºC
kilogram (kg)
standard SI unit of mass; 1 kg = approximately 2.2 pounds
length
measure of one dimension of an object
liter (L)
(also, cubic decimeter) unit of volume; 1 L = 1,000 cm3
meter (m)
standard metric and SI unit of length; 1 m = approximately 1.094 yards
milliliter (mL)
1/1,000 of a liter; equal to 1 cm3
second (s)
SI unit of time
SI units (International System of Units)
standards fixed by international agreement in the International System of Units (Le Système International d’Unités)
unit
standard of comparison for measurements
volume
amount of space occupied by an object | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/01%3A_Essential_Ideas_of_Chemistry/1.4%3A_Measurements.txt |
Learning Objectives
• Compare and contrast exact and uncertain numbers.
• Correctly represent uncertainty in quantities using significant figures.
• Identify the number of significant figures in value.
• Solve problems that involve various calculations and report the results with the appropriate number of significant figures.
• Apply proper rounding rules to computed quantities
• Define accuracy and precision, and use accuracy and precision to describe data sets.
Counting is the only type of measurement that is free from uncertainty, provided the number of objects being counted does not change while the counting process is underway. The result of such a counting measurement is an example of an exact number. If we count eggs in a carton, we know exactly how many eggs the carton contains. The numbers of defined quantities are also exact. By definition, 1 foot is exactly 12 inches, 1 inch is exactly 2.54 centimeters, and 1 gram is exactly 0.001 kilogram. Quantities derived from measurements other than counting, however, are uncertain to varying extents due to practical limitations of the measurement process used.
Significant Figures in Measurement
The numbers of measured quantities, unlike defined or directly counted quantities, are not exact. To measure the volume of liquid in a graduated cylinder, you should make a reading at the bottom of the meniscus, the lowest point on the curved surface of the liquid.
Refer to the illustration in Figure $1$. The bottom of the meniscus in this case clearly lies between the 21 and 22 markings, meaning the liquid volume is certainly greater than 21 mL but less than 22 mL. The meniscus appears to be a bit closer to the 22-mL mark than to the 21-mL mark, and so a reasonable estimate of the liquid’s volume would be 21.6 mL. In the number 21.6, then, the digits 2 and 1 are certain, but the 6 is an estimate. Some people might estimate the meniscus position to be equally distant from each of the markings and estimate the tenth-place digit as 5, while others may think it to be even closer to the 22-mL mark and estimate this digit to be 7. Note that it would be pointless to attempt to estimate a digit for the hundredths place, given that the tenths-place digit is uncertain. In general, numerical scales such as the one on this graduated cylinder will permit measurements to one-tenth of the smallest scale division. The scale in this case has 1-mL divisions, and so volumes may be measured to the nearest 0.1 mL.
This concept holds true for all measurements, even if you do not actively make an estimate. If you place a quarter on a standard electronic balance, you may obtain a reading of 6.72 g. The digits 6 and 7 are certain, and the 2 indicates that the mass of the quarter is likely between 6.71 and 6.73 grams. The quarter weighs about 6.72 grams, with a nominal uncertainty in the measurement of ± 0.01 gram. If we weigh the quarter on a more sensitive balance, we may find that its mass is 6.723 g. This means its mass lies between 6.722 and 6.724 grams, an uncertainty of 0.001 gram. Every measurement has some uncertainty, which depends on the device used (and the user’s ability). All of the digits in a measurement, including the uncertain last digit, are called significant figures or significant digits. Note that zero may be a measured value; for example, if you stand on a scale that shows weight to the nearest pound and it shows “120,” then the 1 (hundreds), 2 (tens) and 0 (ones) are all significant (measured) values.
Whenever you make a measurement properly, all the digits in the result are significant. But what if you were analyzing a reported value and trying to determine what is significant and what is not? Well, for starters, all nonzero digits are significant, and it is only zeros that require some thought. We will use the terms “leading,” “trailing,” and “captive” for the zeros and will consider how to deal with them.
Starting with the first nonzero digit on the left, count this digit and all remaining digits to the right. This is the number of significant figures in the measurement unless the last digit is a trailing zero lying to the left of the decimal point.
Captive zeros result from measurement and are therefore always significant. Leading zeros, however, are never significant—they merely tell us where the decimal point is located.
The leading zeros in this example are not significant. We could use exponential notation (as described in Appendix B) and express the number as 8.32407 $\times$ 10−3; then the number 8.32407 contains all of the significant figures, and 10−3 locates the decimal point.
The number of significant figures is uncertain in a number that ends with a zero to the left of the decimal point location. The zeros in the measurement 1,300 grams could be significant or they could simply indicate where the decimal point is located. The ambiguity can be resolved with the use of exponential notation: 1.3 $\times$ 103 (two significant figures), 1.30 $\times$ 103 (three significant figures, if the tens place was measured), or 1.300 $\times$ 103 (four significant figures, if the ones place was also measured). In cases where only the decimal-formatted number is available, it is prudent to assume that all trailing zeros are not significant.
When determining significant figures, be sure to pay attention to reported values and think about the measurement and significant figures in terms of what is reasonable or likely—that is, whether the value makes sense. For example, the official January 2014 census reported the resident population of the US as 317,297,725. Do you think the US population was correctly determined to the reported nine significant figures, that is, to the exact number of people? People are constantly being born, dying, or moving into or out of the country, and assumptions are made to account for the large number of people who are not actually counted. Because of these uncertainties, it might be more reasonable to expect that we know the population to within perhaps a million or so, in which case the population should be reported as 317 million, or $3.17 \times 10^8$ people.
Significant Figures in Calculations
A second important principle of uncertainty is that results calculated from a measurement are at least as uncertain as the measurement itself. We must take the uncertainty in our measurements into account to avoid misrepresenting the uncertainty in calculated results. One way to do this is to report the result of a calculation with the correct number of significant figures, which is determined by the following three rules for rounding numbers:
1. When we add or subtract numbers, we should round the result to the same number of decimal places as the number with the least number of decimal places (the least precise value in terms of addition and subtraction).
2. When we multiply or divide numbers, we should round the result to the same number of digits as the number with the least number of significant figures (the least precise value in terms of multiplication and division).
3. If the digit to be dropped (the one immediately to the right of the digit to be retained) is less than 5, we “round down” and leave the retained digit unchanged; if it is more than 5, we “round up” and increase the retained digit by 1; if the dropped digit is 5, we round up or down, whichever yields an even value for the retained digit. (The last part of this rule may strike you as a bit odd, but it’s based on reliable statistics and is aimed at avoiding any bias when dropping the digit “5,” since it is equally close to both possible values of the retained digit.)
The following examples illustrate the application of this rule in rounding a few different numbers to three significant figures:
• 0.028675 rounds “up” to 0.0287 (the dropped digit, 7, is greater than 5)
• 18.3384 rounds “down” to 18.3 (the dropped digit, 3, is less than 5)
• 6.8752 rounds “up” to 6.88 (the dropped digit is 5, and the retained digit is even)
• 92.85 rounds “down” to 92.8 (the dropped digit is 5, and the retained digit is even)
Let’s work through these rules with a few examples.
Example $1$: Rounding Numbers
Round the following to the indicated number of significant figures:
1. 31.57 (to two significant figures)
2. 8.1649 (to three significant figures)
3. 0.051065 (to four significant figures)
4. 0.90275 (to four significant figures)
Solution
1. 31.57 rounds “up” to 32 (the dropped digit is 5, and the retained digit is even)
2. 8.1649 rounds “down” to 8.16 (the dropped digit, 4, is less than 5)
3. 0.051065 rounds “down” to 0.05106 (the dropped digit is 5, and the retained digit is even)
4. 0.90275 rounds “up” to 0.9028 (the dropped digit is 5, and the retained digit is even)
Exercise $1$
Round the following to the indicated number of significant figures:
1. 0.424 (to two significant figures)
2. 0.0038661 (to three significant figures)
3. 421.25 (to four significant figures)
4. 28,683.5 (to five significant figures)
Answer a
0.42
Answer b
0.00387
Answer c
421.2
Answer d
28,684
Example $2$: Addition and Subtraction with Significant Figures Rule:
When we add or subtract numbers, we should round the result to the same number of decimal places as the number with the least number of decimal places (i.e., the least precise value in terms of addition and subtraction).
1. Add 1.0023 g and 4.383 g.
2. Subtract 421.23 g from 486 g.
Solution
(a)
\begin{align*} &\mathrm{1.0023\: g}\ +\: &\underline{\mathrm{4.383\: g}\:\:}\ &\mathrm{5.3853\: g} \end{align*} \nonumber
Answer is 5.385 g (round to the thousandths place; three decimal places)
(b)
\begin{align*} &\mathrm{486\: g}\ -\: &\underline{\mathrm{421.23\: g}}\ &\mathrm{\:\:64.77\: g} \end{align*} \nonumber
Answer is 65 g (round to the ones place; no decimal places)
Exercise $2$
1. Add 2.334 mL and 0.31 mL.
2. Subtract 55.8752 m from 56.533 m.
Answer a
2.64 mL
Answer b
0.658 m
Example $3$: Multiplication and Division with Significant Figures
Rule: When we multiply or divide numbers, we should round the result to the same number of digits as the number with the least number of significant figures (the least precise value in terms of multiplication and division).
1. Multiply 0.6238 cm by 6.6 cm.
2. Divide 421.23 g by 486 mL.
Solution
(a)
$\mathrm{0.6238\: cm\times6.6\:cm=4.11708\:cm^2\rightarrow result\: is\:4.1\:cm^2}\:\textrm{(round to two significant figures)} \nonumber$
$\textrm{four significant figures}\times \textrm{two significant figures}\rightarrow \textrm{two significant figures answer} \nonumber$
(b)
$\mathrm{\dfrac{421.23\: g}{486\: mL}=0.86728...\: g/mL\rightarrow result\: is\: 0.867\: g/mL} \: \textrm{(round to three significant figures)} \nonumber$
$\mathrm{\dfrac{five\: significant\: figures}{three\: significant\: figures}\rightarrow three\: significant\: figures\: answer} \nonumber$
Exercise $3$
1. Multiply 2.334 cm and 0.320 cm.
2. Divide 55.8752 m by 56.53 s.
Answer a
0.747 cm2
Answer b
0.9884 m/s
In the midst of all these technicalities, it is important to keep in mind the reason why we use significant figures and rounding rules—to correctly represent the certainty of the values we report and to ensure that a calculated result is not represented as being more certain than the least certain value used in the calculation.
Example $4$: Calculation with Significant Figures
One common bathtub is 13.44 dm long, 5.920 dm wide, and 2.54 dm deep. Assume that the tub is rectangular and calculate its approximate volume in liters.
Solution
\begin{align*} V&=l\times w\times d\ &=\mathrm{13.44\: dm\times 5.920\: dm\times 2.54\: dm}\ &=\mathrm{202.09459...dm^3}\:\textrm{(value from calculator)}\ &=\mathrm{202\: dm^3,}\textrm{ or 202 L (answer rounded to three significant figures)} \end{align*} \nonumber
Exercise $4$: Determination of Density Using Water Displacement
What is the density of a liquid with a mass of 31.1415 g and a volume of 30.13 cm3?
Answer
1.034 g/mL
Example $4$
A piece of rebar is weighed and then submerged in a graduated cylinder partially filled with water, with results as shown.
1. Use these values to determine the density of this piece of rebar.
2. Rebar is mostly iron. Does your result in (a) support this statement? How?
Solution
The volume of the piece of rebar is equal to the volume of the water displaced:
$\mathrm{volume=22.4\: mL-13.5\: mL=8.9\: mL=8.9\: cm^3}\nonumber$
(rounded to the nearest 0.1 mL, per the rule for addition and subtraction)
The density is the mass-to-volume ratio:
$\mathrm{density=\dfrac{mass}{volume}=\dfrac{69.658\: g}{8.9\: cm^3}=7.8\: g/cm^3}\nonumber$
(rounded to two significant figures, per the rule for multiplication and division)
The density of iron is 7.9 g/cm3, very close to that of rebar, which lends some support to the fact that rebar is mostly iron.
Exercise $4$
An irregularly shaped piece of a shiny yellowish material is weighed and then submerged in a graduated cylinder, with results as shown.
1. Use these values to determine the density of this material.
2. Do you have any reasonable guesses as to the identity of this material? Explain your reasoning.
Answer a
19 g/cm3
Answer b
It is likely gold; it has the right appearance for gold and very close to the density given for gold.
Accuracy and Precision
Scientists typically make repeated measurements of a quantity to ensure the quality of their findings and to know both the precision and the accuracy of their results. Measurements are said to be precise if they yield very similar results when repeated in the same manner. A measurement is considered accurate if it yields a result that is very close to the true or accepted value. Precise values agree with each other; accurate values agree with a true value. These characterizations can be extended to other contexts, such as the results of an archery competition (Figure $2$).
Suppose a quality control chemist at a pharmaceutical company is tasked with checking the accuracy and precision of three different machines that are meant to dispense 10 ounces (296 mL) of cough syrup into storage bottles. She proceeds to use each machine to fill five bottles and then carefully determines the actual volume dispensed, obtaining the results tabulated in Table $2$.
Table $2$: Volume (mL) of Cough Medicine Delivered by 10-oz (296 mL) Dispensers
Dispenser #1 Dispenser #2 Dispenser #3
283.3 298.3 296.1
284.1 294.2 295.9
283.9 296.0 296.1
284.0 297.8 296.0
284.1 293.9 296.1
Considering these results, she will report that dispenser #1 is precise (values all close to one another, within a few tenths of a milliliter) but not accurate (none of the values are close to the target value of 296 mL, each being more than 10 mL too low). Results for dispenser #2 represent improved accuracy (each volume is less than 3 mL away from 296 mL) but worse precision (volumes vary by more than 4 mL). Finally, she can report that dispenser #3 is working well, dispensing cough syrup both accurately (all volumes within 0.1 mL of the target volume) and precisely (volumes differing from each other by no more than 0.2 mL).
Summary
Quantities can be exact or measured. Measured quantities have an associated uncertainty that is represented by the number of significant figures in the measurement. The uncertainty of a calculated value depends on the uncertainties in the values used in the calculation and is reflected in how the value is rounded. Measured values can be accurate (close to the true value) and/or precise (showing little variation when measured repeatedly).
Glossary
uncertainty
estimate of amount by which measurement differs from true value
significant figures
(also, significant digits) all of the measured digits in a determination, including the uncertain last digit
rounding
procedure used to ensure that calculated results properly reflect the uncertainty in the measurements used in the calculation
precision
how closely a measurement matches the same measurement when repeated
exact number
number derived by counting or by definition
accuracy
how closely a measurement aligns with a correct value | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/01%3A_Essential_Ideas_of_Chemistry/1.5%3A_Measurement_Uncertainty_Accuracy_and_Precision.txt |
Learning Objectives
• Explain the dimensional analysis (factor label) approach to mathematical calculations involving quantities.
• Describe how to use dimensional analysis to carry out unit conversions for a given property and computations involving two or more properties.
• Convert between the three main temperature units: Fahrenheit, Celsius, and Kelvin.
It is often the case that a quantity of interest may not be easy (or even possible) to measure directly but instead must be calculated from other directly measured properties and appropriate mathematical relationships. For example, consider measuring the average speed of an athlete running sprints. This is typically accomplished by measuring the time required for the athlete to run from the starting line to the finish line, and the distance between these two lines, and then computing speed from the equation that relates these three properties:
$\mathrm{speed=\dfrac{distance}{time}} \nonumber$
An Olympic-quality sprinter can run 100 m in approximately 10 s, corresponding to an average speed of
$\mathrm{\dfrac{100\: m}{10\: s}=10\: m/s} \nonumber$
Note that this simple arithmetic involves dividing the numbers of each measured quantity to yield the number of the computed quantity (100/10 = 10) and likewise dividing the units of each measured quantity to yield the unit of the computed quantity (m/s = m/s). Now, consider using this same relation to predict the time required for a person running at this speed to travel a distance of 25 m. The same relation between the three properties is used, but in this case, the two quantities provided are a speed (10 m/s) and a distance (25 m). To yield the sought property, time, the equation must be rearranged appropriately:
$\mathrm{time=\dfrac{distance}{speed}} \nonumber$
The time can then be computed as:
$\mathrm{\dfrac{25\: m}{10\: m/s}=2.5\: s} \nonumber$
Again, arithmetic on the numbers (25/10 = 2.5) was accompanied by the same arithmetic on the units (m/m/s = s) to yield the number and unit of the result, 2.5 s. Note that, just as for numbers, when a unit is divided by an identical unit (in this case, m/m), the result is “1”—or, as commonly phrased, the units “cancel.”
These calculations are examples of a versatile mathematical approach known as dimensional analysis (or the factor-label method). Dimensional analysis is based on this premise: the units of quantities must be subjected to the same mathematical operations as their associated numbers. This method can be applied to computations ranging from simple unit conversions to more complex, multi-step calculations involving several different quantities.
Conversion Factors and Dimensional Analysis
A ratio of two equivalent quantities expressed with different measurement units can be used as a unit conversion factor. For example, the lengths of 2.54 cm and 1 in. are equivalent (by definition), and so a unit conversion factor may be derived from the ratio,
$\mathrm{\dfrac{2.54\: cm}{1\: in.}\:(2.54\: cm=1\: in.)\: or\: 2.54\:\dfrac{cm}{in.}} \nonumber$
Several other commonly used conversion factors are given in Table $1$.
Table $1$: Common Conversion Factors
Length Volume Mass
1 m = 1.0936 yd 1 L = 1.0567 qt 1 kg = 2.2046 lb
1 in. = 2.54 cm (exact) 1 qt = 0.94635 L 1 lb = 453.59 g
1 km = 0.62137 mi 1 ft3 = 28.317 L 1 (avoirdupois) oz = 28.349 g
1 mi = 1609.3 m 1 tbsp = 14.787 mL 1 (troy) oz = 31.103 g
When we multiply a quantity (such as distance given in inches) by an appropriate unit conversion factor, we convert the quantity to an equivalent value with different units (such as distance in centimeters). For example, a basketball player’s vertical jump of 34 inches can be converted to centimeters by:
$\mathrm{34\: \cancel{in.} \times \dfrac{2.54\: cm}{1\:\cancel{in.}}=86\: cm} \nonumber$
Since this simple arithmetic involves quantities, the premise of dimensional analysis requires that we multiply both numbers and units. The numbers of these two quantities are multiplied to yield the number of the product quantity, 86, whereas the units are multiplied to yield
$\mathrm{\dfrac{in.\times cm}{in.}}. \nonumber$
Just as for numbers, a ratio of identical units is also numerically equal to one,
$\mathrm{\dfrac{in.}{in.}=1} \nonumber$
and the unit product thus simplifies to cm. (When identical units divide to yield a factor of 1, they are said to “cancel.”) Using dimensional analysis, we can determine that a unit conversion factor has been set up correctly by checking to confirm that the original unit will cancel, and the result will contain the sought (converted) unit.
Example $1$: Using a Unit Conversion Factor
The mass of a competition Frisbee is 125 g. Convert its mass to ounces using the unit conversion factor derived from the relationship 1 oz = 28.349 g (Table $1$).
Solution
If we have the conversion factor, we can determine the mass in kilograms using an equation similar the one used for converting length from inches to centimeters.
$x\:\mathrm{oz=125\: g\times unit\: conversion\: factor}\nonumber$
We write the unit conversion factor in its two forms:
$\mathrm{\dfrac{1\: oz}{28.349\: g}\:and\:\dfrac{28.349\: g}{1\: oz}}\nonumber$
The correct unit conversion factor is the ratio that cancels the units of grams and leaves ounces.
\begin{align*} x\:\ce{oz}&=\mathrm{125\:\cancel{g}\times \dfrac{1\: oz}{28.349\:\cancel{g}}}\ &=\mathrm{\left(\dfrac{125}{28.349}\right)\:oz}\ &=\mathrm{4.41\: oz\: (three\: significant\: figures)} \end{align*} \nonumber
Exercise $1$
Convert a volume of 9.345 qt to liters.
Answer
8.844 L
Beyond simple unit conversions, the factor-label method can be used to solve more complex problems involving computations. Regardless of the details, the basic approach is the same—all the factors involved in the calculation must be appropriately oriented to insure that their labels (units) will appropriately cancel and/or combine to yield the desired unit in the result. This is why it is referred to as the factor-label method. As your study of chemistry continues, you will encounter many opportunities to apply this approach.
Example $2$: Computing Quantities from Measurement Results
What is the density of common antifreeze in units of g/mL? A 4.00-qt sample of the antifreeze weighs 9.26 lb.
Solution
Since $\mathrm{density=\dfrac{mass}{volume}}$, we need to divide the mass in grams by the volume in milliliters. In general: the number of units of B = the number of units of A $\times$ unit conversion factor. The necessary conversion factors are given in Table 1.7.1: 1 lb = 453.59 g; 1 L = 1.0567 qt; 1 L = 1,000 mL. We can convert mass from pounds to grams in one step:
$\mathrm{9.26\:\cancel{lb}\times \dfrac{453.59\: g}{1\:\cancel{lb}}=4.20\times 10^3\:g}\nonumber$
We need to use two steps to convert volume from quarts to milliliters.
1. Convert quarts to liters.
$\mathrm{4.00\:\cancel{qt}\times\dfrac{1\: L}{1.0567\:\cancel{qt}}=3.78\: L}\nonumber$
1. Convert liters to milliliters.
$\mathrm{3.78\:\cancel{L}\times\dfrac{1000\: mL}{1\:\cancel{L}}=3.78\times10^3\:mL}\nonumber$
Then,
$\mathrm{density=\dfrac{4.20\times10^3\:g}{3.78\times10^3\:mL}=1.11\: g/mL}\nonumber$
Alternatively, the calculation could be set up in a way that uses three unit conversion factors sequentially as follows:
$\mathrm{\dfrac{9.26\:\cancel{lb}}{4.00\:\cancel{qt}}\times\dfrac{453.59\: g}{1\:\cancel{lb}}\times\dfrac{1.0567\:\cancel{qt}}{1\:\cancel{L}}\times\dfrac{1\:\cancel{L}}{1000\: mL}=1.11\: g/mL}\nonumber$
Exercise $2$
What is the volume in liters of 1.000 oz, given that 1 L = 1.0567 qt and 1 qt = 32 oz (exactly)?
Answer
$\mathrm{2.956\times10^{-2}\:L}$
Example $3$: Computing Quantities from Measurement Results
While being driven from Philadelphia to Atlanta, a distance of about 1250 km, a 2014 Lamborghini Aventador Roadster uses 213 L gasoline.
1. What (average) fuel economy, in miles per gallon, did the Roadster get during this trip?
Answer a
51 mpg
Answer b
\$62
Conversion of Temperature Units
We use the word temperature to refer to the hotness or coldness of a substance. One way we measure a change in temperature is to use the fact that most substances expand when their temperature increases and contract when their temperature decreases. The mercury or alcohol in a common glass thermometer changes its volume as the temperature changes. Because the volume of the liquid changes more than the volume of the glass, we can see the liquid expand when it gets warmer and contract when it gets cooler.
To mark a scale on a thermometer, we need a set of reference values: Two of the most commonly used are the freezing and boiling temperatures of water at a specified atmospheric pressure. On the Celsius scale, 0 °C is defined as the freezing temperature of water and 100 °C as the boiling temperature of water. The space between the two temperatures is divided into 100 equal intervals, which we call degrees. On the Fahrenheit scale, the freezing point of water is defined as 32 °F and the boiling temperature as 212 °F. The space between these two points on a Fahrenheit thermometer is divided into 180 equal parts (degrees).
Defining the Celsius and Fahrenheit temperature scales as described in the previous paragraph results in a slightly more complex relationship between temperature values on these two scales than for different units of measure for other properties. Most measurement units for a given property are directly proportional to one another (y = mx). Using familiar length units as one example:
$\mathrm{length\: in\: feet=\left(\dfrac{1\: ft}{12\: in.}\right)\times length\: in\: inches} \nonumber$
where
• y = length in feet,
• x = length in inches, and
• the proportionality constant, m, is the conversion factor.
The Celsius and Fahrenheit temperature scales, however, do not share a common zero point, and so the relationship between these two scales is a linear one rather than a proportional one ($y = mx + b$). Consequently, converting a temperature from one of these scales into the other requires more than simple multiplication by a conversion factor, m, it also must take into account differences in the scales’ zero points ($b$).
The linear equation relating Celsius and Fahrenheit temperatures is easily derived from the two temperatures used to define each scale. Representing the Celsius temperature as $x$ and the Fahrenheit temperature as $y$, the slope, $m$, is computed to be:
\begin{align*} m &=\dfrac{\Delta y}{\Delta x} \[4pt] &= \mathrm{\dfrac{212\: ^\circ F - 32\: ^\circ F}{100\: ^\circ C-0\: ^\circ C}} \[4pt] &= \mathrm{\dfrac{180\: ^\circ F}{100\: ^\circ C}} \[4pt] &= \mathrm{\dfrac{9\: ^\circ F}{5\: ^\circ C} }\end{align*} \nonumber
The y-intercept of the equation, b, is then calculated using either of the equivalent temperature pairs, (100 °C, 212 °F) or (0 °C, 32 °F), as:
\begin{align*} b&=y-mx \[4pt] &= \mathrm{32\:^\circ F-\dfrac{9\:^\circ F}{5\:^\circ C}\times0\:^\circ C} \[4pt] &= \mathrm{32\:^\circ F} \end{align*} \nonumber
The equation relating the temperature scales is then:
$\mathrm{\mathit{T}_{^\circ F}=\left(\dfrac{9\:^\circ F}{5\:^\circ C}\times \mathit{T}_{^\circ C}\right)+32\:^\circ C} \nonumber$
An abbreviated form of this equation that omits the measurement units is:
$\mathrm{\mathit{T}_{^\circ F}=\dfrac{9}{5}\times \mathit{T}_{^\circ C}+32} \nonumber$
Rearrangement of this equation yields the form useful for converting from Fahrenheit to Celsius:
$\mathrm{\mathit{T}_{^\circ C}=\dfrac{5}{9}(\mathit{T}_{^\circ F}+32)} \nonumber$
As mentioned earlier in this chapter, the SI unit of temperature is the kelvin (K). Unlike the Celsius and Fahrenheit scales, the kelvin scale is an absolute temperature scale in which 0 (zero) K corresponds to the lowest temperature that can theoretically be achieved. The early 19th-century discovery of the relationship between a gas's volume and temperature suggested that the volume of a gas would be zero at −273.15 °C. In 1848, British physicist William Thompson, who later adopted the title of Lord Kelvin, proposed an absolute temperature scale based on this concept (further treatment of this topic is provided in this text’s chapter on gases).
The freezing temperature of water on this scale is 273.15 K and its boiling temperature 373.15 K. Notice the numerical difference in these two reference temperatures is 100, the same as for the Celsius scale, and so the linear relation between these two temperature scales will exhibit a slope of $\mathrm{1\:\dfrac{K}{^\circ\:C}}$. Following the same approach, the equations for converting between the kelvin and Celsius temperature scales are derived to be:
$T_{\ce K}=T_{\mathrm{^\circ C}}+273.15 \nonumber$
$T_\mathrm{^\circ C}=T_{\ce K}-273.15 \nonumber$
The 273.15 in these equations has been determined experimentally, so it is not exact. Figure $1$ shows the relationship among the three temperature scales. Recall that we do not use the degree sign with temperatures on the kelvin scale.
Although the kelvin (absolute) temperature scale is the official SI temperature scale, Celsius is commonly used in many scientific contexts and is the scale of choice for nonscience contexts in almost all areas of the world. Very few countries (the U.S. and its territories, the Bahamas, Belize, Cayman Islands, and Palau) still use Fahrenheit for weather, medicine, and cooking.
Example $4$: Conversion from Celsius
Normal body temperature has been commonly accepted as 37.0 °C (although it varies depending on time of day and method of measurement, as well as among individuals). What is this temperature on the kelvin scale and on the Fahrenheit scale?
Solution
$\mathrm{K= {^\circ C}+273.15=37.0+273.2=310.2\: K}\nonumber$
$\mathrm{^\circ F=\dfrac{9}{5}\:{^\circ C}+32.0=\left(\dfrac{9}{5}\times 37.0\right)+32.0=66.6+32.0=98.6\: ^\circ F}\nonumber$
Exercise $4$
Convert 80.92 °C to K and °F.
Answer
354.07 K, 177.7 °F
Example $5$: Conversion from Fahrenheit
Baking a ready-made pizza calls for an oven temperature of 450 °F. If you are in Europe, and your oven thermometer uses the Celsius scale, what is the setting? What is the kelvin temperature?
Solution
$\mathrm{^\circ C=\dfrac{5}{9}(^\circ F-32)=\dfrac{5}{9}(450-32)=\dfrac{5}{9}\times 418=232 ^\circ C\rightarrow set\: oven\: to\: 230 ^\circ C}\hspace{20px}\textrm{(two significant figures)}\nonumber$
$\mathrm{K={^\circ C}+273.15=230+273=503\: K\rightarrow 5.0\times 10^2\,K\hspace{20px}(two\: significant\: figures)}\nonumber$
Exercise $5$
Convert 50 °F to °C and K.
Answer
10 °C, 280 K
Summary
Measurements are made using a variety of units. It is often useful or necessary to convert a measured quantity from one unit into another. These conversions are accomplished using unit conversion factors, which are derived by simple applications of a mathematical approach called the factor-label method or dimensional analysis. This strategy is also employed to calculate sought quantities using measured quantities and appropriate mathematical relations.
Key Equations
• $T_\mathrm{^\circ C}=\dfrac{5}{9}\times T_\mathrm{^\circ F}-32$
• $T_\mathrm{^\circ F}=\dfrac{9}{5}\times T_\mathrm{^\circ C}+32$
• $T_\ce{K}={^\circ \ce C}+273.15$
• $T_\mathrm{^\circ C}=\ce K-273.15$
Glossary
dimensional analysis
(also, factor-label method) versatile mathematical approach that can be applied to computations ranging from simple unit conversions to more complex, multi-step calculations involving several different quantities
Fahrenheit
unit of temperature; water freezes at 32 °F and boils at 212 °F on this scale
unit conversion factor
ratio of equivalent quantities expressed with different units; used to convert from one unit to a different unit | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/01%3A_Essential_Ideas_of_Chemistry/1.6%3A_Mathematical_Treatment_of_Measurement_Results.txt |
1.1: Chemistry in Context
Q1.1.1
Explain how you could experimentally determine whether the outside temperature is higher or lower than 0 °C (32 °F) without using a thermometer.
S1.1.1
Place a glass of water outside. It will freeze if the temperature is below 0 °C.
Q1.1.2
Identify each of the following statements as being most similar to a hypothesis, a law, or a theory. Explain your reasoning.
1. Falling barometric pressure precedes the onset of bad weather.
2. All life on earth has evolved from a common, primitive organism through the process of natural selection.
3. My truck’s gas mileage has dropped significantly, probably because it’s due for a tune-up.
Q1.1.3
Identify each of the following statements as being most similar to a hypothesis, a law, or a theory. Explain your reasoning.
1. The pressure of a sample of gas is directly proportional to the temperature of the gas.
2. Matter consists of tiny particles that can combine in specific ratios to form substances with specific properties.
3. At a higher temperature, solids (such as salt or sugar) will dissolve better in water.
S1.1.3
(a) law (states a consistently observed phenomenon, can be used for prediction); (b) theory (a widely accepted explanation of the behavior of matter); (c) hypothesis (a tentative explanation, can be investigated by experimentation)
Identify each of the underlined items as a part of either the macroscopic domain, the microscopic domain, or the symbolic domain of chemistry. For any in the symbolic domain, indicate whether they are symbols for a macroscopic or a microscopic feature.
1. (a) The mass of a lead pipe is 14 lb.
2. (b) The mass of a certain chlorine atom is 35 amu.
3. (c) A bottle with a label that reads Al contains aluminum metal.
4. (d) Al is the symbol for an aluminum atom.
Identify each of the underlined items as a part of either the macroscopic domain, the microscopic domain, or the symbolic domain of chemistry. For those in the symbolic domain, indicate whether they are symbols for a macroscopic or a microscopic feature.
1. (a) A certain molecule contains one H atom and one Cl atom.
2. (b) Copper wire has a density of about 8 g/cm3.
3. (c) The bottle contains 15 grams of Ni powder.
4. (d) A sulfur molecule is composed of eight sulfur atoms.
(a) symbolic, microscopic; (b) macroscopic; (c) symbolic, macroscopic; (d) microscopic
According to one theory, the pressure of a gas increases as its volume decreases because the molecules in the gas have to move a shorter distance to hit the walls of the container. Does this theory follow a macroscopic or microscopic description of chemical behavior? Explain your answer.
The amount of heat required to melt 2 lbs of ice is twice the amount of heat required to melt 1 lb of ice. Is this observation a macroscopic or microscopic description of chemical behavior? Explain your answer.
Macroscopic. The heat required is determined from macroscopic properties.
1.2: Phases and Classification of Matter
Questions
1. Why do we use an object's mass, rather than its weight, to indicate the amount of matter it contains?
2. What properties distinguish solids from liquids? Liquids from gases? Solids from gases?
3. How does a heterogeneous mixture differ from a homogeneous mixture? How are they similar?
4. How does a homogeneous mixture differ from a pure substance? How are they similar?
5. How does an element differ from a compound? How are they similar?
6. How do molecules of elements and molecules of compounds differ? In what ways are they similar?
7. How does an atom differ from a molecule? In what ways are they similar?
8. Many of the items you purchase are mixtures of pure compounds. Select three of these commercial products and prepare a list of the ingredients that are pure compounds.
9. Classify each of the following as an element, a compound, or a mixture:
1. copper
2. water
3. nitrogen
4. sulfur
5. air
6. sucrose
7. a substance composed of molecules each of which contains two iodine atoms
8. gasoline
10. Classify each of the following as an element, a compound, or a mixture:
1. iron
2. oxygen
3. mercury oxide
4. pancake syrup
5. carbon dioxide
6. a substance composed of molecules each of which contains one hydrogen atom and one chlorine atom
7. baking soda
8. baking powder
11. A sulfur atom and a sulfur molecule are not identical. What is the difference?
12. How are the molecules in oxygen gas, the molecules in hydrogen gas, and water molecules similar? How do they differ?
13. We refer to astronauts in space as weightless, but not without mass. Why?
14. As we drive an automobile, we don't think about the chemicals consumed and produced. Prepare a list of the principal chemicals consumed and produced during the operation of an automobile.
15. Matter is everywhere around us. Make a list by name of fifteen different kinds of matter that you encounter every day. Your list should include (and label at least one example of each) the following: a solid, a liquid, a gas, an element, a compound, a homogenous mixture, a heterogeneous mixture, and a pure substance.
16. When elemental iron corrodes it combines with oxygen in the air to ultimately form red brown iron(III) oxide which we call rust. (a) If a shiny iron nail with an initial mass of 23.2 g is weighed after being coated in a layer of rust, would you expect the mass to have increased, decreased, or remained the same? Explain. (b) If the mass of the iron nail increases to 24.1 g, what mass of oxygen combined with the iron?
17. As stated in the text, convincing examples that demonstrate the law of conservation of matter outside of the laboratory are few and far between. Indicate whether the mass would increase, decrease, or stay the same for the following scenarios where chemical reactions take place:
1. Exactly one pound of bread dough is placed in a baking tin. The dough is cooked in an oven at 350 °F releasing a wonderful aroma of freshly baked bread during the cooking process. Is the mass of the baked loaf less than, greater than, or the same as the one pound of original dough? Explain.
2. When magnesium burns in air a white flaky ash of magnesium oxide is produced. Is the mass of magnesium oxide less than, greater than, or the same as the original piece of magnesium? Explain.
3. Antoine Lavoisier, the French scientist credited with first stating the law of conservation of matter, heated a mixture of tin and air in a sealed flask to produce tin oxide. Did the mass of the sealed flask and contents decrease, increase, or remain the same after the heating?
18. Yeast converts glucose to ethanol and carbon dioxide during anaerobic fermentation as depicted in the simple chemical equation here: $\ce{glucose\rightarrow ethanol + carbon\: dioxide}$
1. If 200.0 g of glucose is fully converted, what will be the total mass of ethanol and carbon dioxide produced?
2. If the fermentation is carried out in an open container, would you expect the mass of the container and contents after fermentation to be less than, greater than, or the same as the mass of the container and contents before fermentation? Explain.
3. If 97.7 g of carbon dioxide is produced, what mass of ethanol is produced?
Solutions
2 Liquids can change their shape (flow); solids can’t. Gases can undergo large volume changes as pressure changes; liquids do not. Gases flow and change volume; solids do not.
4.The mixture can have a variety of compositions; a pure substance has a definite composition. Both have the same composition from point to point.
6 Molecules of elements contain only one type of atom; molecules of compounds contain two or more types of atoms. They are similar in that both are comprised of two or more atoms chemically bonded together.
8. Answers will vary. Sample answer: Gatorade contains water, sugar, dextrose, citric acid, salt, sodium chloride, monopotassium phosphate, and sucrose acetate isobutyrate.
11. (a) element; (b) element; (c) compound; (d) mixture, (e) compound; (f) compound; (g) compound; (h) mixture
12. In each case, a molecule consists of two or more combined atoms. They differ in that the types of atoms change from one substance to the next.
14. Gasoline (a mixture of compounds), oxygen, and to a lesser extent, nitrogen are consumed. Carbon dioxide and water are the principal products. Carbon monoxide and nitrogen oxides are produced in lesser amounts.
16. (a) Increased as it would have combined with oxygen in the air thus increasing the amount of matter and therefore the mass. (b) 0.9 g
18. (a) 200.0 g; (b) The mass of the container and contents would decrease as carbon dioxide is a gaseous product and would leave the container. (c) 102.3 g
1.3: Physical and Chemical Properties
Classify the six underlined properties in the following paragraph as chemical or physical:
Fluorine is a pale yellow gas that reacts with most substances. The free element melts at −220 °C and boils at −188 °C. Finely divided metals burn in fluorine with a bright flame. Nineteen grams of fluorine will react with 1.0 gram of hydrogen.
Classify each of the following changes as physical or chemical:
1. condensation of steam
2. burning of gasoline
3. souring of milk
4. dissolving of sugar in water
5. melting of gold
(a) physical; (b) chemical; (c) chemical; (d) physical; (e) physical
Classify each of the following changes as physical or chemical:
1. coal burning
2. ice melting
3. mixing chocolate syrup with milk
4. explosion of a firecracker
5. magnetizing of a screwdriver
The volume of a sample of oxygen gas changed from 10 mL to 11 mL as the temperature changed. Is this a chemical or physical change?
physical
A 2.0-liter volume of hydrogen gas combined with 1.0 liter of oxygen gas to produce 2.0 liters of water vapor. Does oxygen undergo a chemical or physical change?
Explain the difference between extensive properties and intensive properties.
The value of an extensive property depends upon the amount of matter being considered, whereas the value of an intensive property is the same regardless of the amount of matter being considered.
Identify the following properties as either extensive or intensive.
1. volume
2. temperature
3. humidity
4. heat
5. boiling point
The density (d) of a substance is an intensive property that is defined as the ratio of its mass (m) to its volume (V).
$\mathrm{density=\dfrac{mass}{volume}}$ $\mathrm{d=\dfrac{m}{V}}$
Considering that mass and volume are both extensive properties, explain why their ratio, density, is intensive.
Being extensive properties, both mass and volume are directly proportional to the amount of substance under study. Dividing one extensive property by another will in effect “cancel” this dependence on amount, yielding a ratio that is independent of amount (an intensive property).
1.4: Measurements
Is one liter about an ounce, a pint, a quart, or a gallon?
Is a meter about an inch, a foot, a yard, or a mile?
about a yard
Indicate the SI base units or derived units that are appropriate for the following measurements:
1. (a) the length of a marathon race (26 miles 385 yards)
2. (b) the mass of an automobile
3. (c) the volume of a swimming pool
4. (d) the speed of an airplane
5. (e) the density of gold
6. (f) the area of a football field
7. (g) the maximum temperature at the South Pole on April 1, 1913
Indicate the SI base units or derived units that are appropriate for the following measurements:
1. (a) the mass of the moon
2. (b) the distance from Dallas to Oklahoma City
3. (c) the speed of sound
4. (d) the density of air
5. (e) the temperature at which alcohol boils
6. (f) the area of the state of Delaware
7. (g) the volume of a flu shot or a measles vaccination
(a) kilograms; (b) meters; (c) kilometers/second; (d) kilograms/cubic meter; (e) kelvin; (f) square meters; (g) cubic meters
Give the name and symbol of the prefixes used with SI units to indicate multiplication by the following exact quantities.
1. (a) 103
2. (b) 10−2
3. (c) 0.1
4. (d) 10−3
5. (e) 1,000,000
6. (f) 0.000001
Give the name of the prefix and the quantity indicated by the following symbols that are used with SI base units.
1. (a) c
2. (b) d
3. (c) G
4. (d) k
5. (e) m
6. (f) n
7. (g) p
8. (h) T
(a) centi-, $\times$ 10−2; (b) deci-, $\times$ 10−1; (c) Giga-, $\times$ 109; (d) kilo-, $\times$ 103; (e) milli-, $\times$ 10−3; (f) nano-, $\times$ 10−9; (g) pico-, $\times$ 10−12; (h) tera-, $\times$ 1012
A large piece of jewelry has a mass of 132.6 g. A graduated cylinder initially contains 48.6 mL water. When the jewelry is submerged in the graduated cylinder, the total volume increases to 61.2 mL.
1. (a) Determine the density of this piece of jewelry.
2. (b) Assuming that the jewelry is made from only one substance, what substance is it likely to be? Explain.
Visit this PhET density simulation and select the Same Volume Blocks.
1. (a) What are the mass, volume, and density of the yellow block?
2. (b) What are the mass, volume and density of the red block?
3. (c) List the block colors in order from smallest to largest mass.
4. (d) List the block colors in order from lowest to highest density.
5. (e) How are mass and density related for blocks of the same volume?
(a) 8.00 kg, 5.00 L, 1.60 kg/L; (b) 2.00 kg, 5.00 L, 0.400 kg/L; (c) red < green < blue < yellow; (d) If the volumes are the same, then the density is directly proportional to the mass.
Visit this PhET density simulation and select Custom Blocks and then My Block.
1. (a) Enter mass and volume values for the block such that the mass in kg is less than the volume in L. What does the block do? Why? Is this always the case when mass < volume?
2. (b) Enter mass and volume values for the block such that the mass in kg is more than the volume in L. What does the block do? Why? Is this always the case when mass > volume?
3. (c) How would (a) and (b) be different if the liquid in the tank were ethanol instead of water?
4. (d) How would (a) and (b) be different if the liquid in the tank were mercury instead of water?
Visit this PhET density simulation and select Mystery Blocks.
1. (a) Pick one of the Mystery Blocks and determine its mass, volume, density, and its likely identity.
2. (b) Pick a different Mystery Block and determine its mass, volume, density, and its likely identity.
3. (c) Order the Mystery Blocks from least dense to most dense. Explain.
(a) (b) Answer is one of the following. A/yellow: mass = 65.14 kg, volume = 3.38 L, density = 19.3 kg/L, likely identity = gold. B/blue: mass = 0.64 kg, volume = 1.00 L, density = 0.64 kg/L, likely identity = apple. C/green: mass = 4.08 kg, volume = 5.83 L, density = 0.700 kg/L, likely identity = gasoline. D/red: mass = 3.10 kg, volume = 3.38 L, density = 0.920 kg/L, likely identity = ice; and E/purple: mass = 3.53 kg, volume = 1.00 L, density = 3.53 kg/L, likely identity = diamond. (c) B/blue/apple (0.64 kg/L) < C/green/gasoline (0.700 kg/L) < D/red/ice (0.920 kg/L) < E/purple/diamond (3.53 kg/L) < A/yellow/gold (19.3 kg/L)
1.5: Measurement Uncertainty, Accuracy, and Precision
Express each of the following numbers in scientific notation with correct significant figures:
1. (a) 711.0
2. (b) 0.239
3. (c) 90743
4. (d) 134.2
5. (e) 0.05499
6. (f) 10000.0
7. (g) 0.000000738592
Express each of the following numbers in exponential notation with correct significant figures:
1. (a) 704
2. (b) 0.03344
3. (c) 547.9
4. (d) 22086
5. (e) 1000.00
6. (f) 0.0000000651
7. (g) 0.007157
(a) 7.04 $\times$ 102; (b) 3.344 $\times$ 10−2; (c) 5.479 $\times$ 102; (d) 2.2086 $\times$ 104; (e) 1.00000 $\times$ 103; (f) 6.51 $\times$ 10−8; (g) 7.157 $\times$ 10−3
Indicate whether each of the following can be determined exactly or must be measured with some degree of uncertainty:
1. (a) the number of eggs in a basket
2. (b) the mass of a dozen eggs
3. (c) the number of gallons of gasoline necessary to fill an automobile gas tank
4. (d) the number of cm in 2 m
5. (e) the mass of a textbook
6. (f) the time required to drive from San Francisco to Kansas City at an average speed of 53 mi/h
Indicate whether each of the following can be determined exactly or must be measured with some degree of uncertainty:
1. (a) the number of seconds in an hour
2. (b) the number of pages in this book
3. (c) the number of grams in your weight
4. (d) the number of grams in 3 kilograms
5. (e) the volume of water you drink in one day
6. (f) the distance from San Francisco to Kansas City
(a) exact; (b) exact; (c) uncertain; (d) exact; (e) uncertain; (f) uncertain
How many significant figures are contained in each of the following measurements?
1. (a) 38.7 g
2. (b) 2 $\times$ 1018 m
3. (c) 3,486,002 kg
4. (d) 9.74150 $\times$ 10−4 J
5. (e) 0.0613 cm3
6. (f) 17.0 kg
7. (g) 0.01400 g/mL
How many significant figures are contained in each of the following measurements?
1. (a) 53 cm
2. (b) 2.05 $\times$ 108 m
3. (c) 86,002 J
4. (d) 9.740 $\times$ 104 m/s
5. (e) 10.0613 m3
6. (f) 0.17 g/mL
7. (g) 0.88400 s
(a) two; (b) three; (c) five; (d) four; (e) six; (f) two; (g) five
The following quantities were reported on the labels of commercial products. Determine the number of significant figures in each.
1. (a) 0.0055 g active ingredients
2. (b) 12 tablets
3. (c) 3% hydrogen peroxide
4. (d) 5.5 ounces
5. (e) 473 mL
6. (f) 1.75% bismuth
7. (g) 0.001% phosphoric acid
8. (h) 99.80% inert ingredients
Round off each of the following numbers to two significant figures:
1. (a) 0.436
2. (b) 9.000
3. (c) 27.2
4. (d) 135
5. (e) 1.497 $\times$ 10−3
6. (f) 0.445
(a) 0.44; (b) 9.0; (c) 27; (d) 140; (e) 1.5 $\times$ 10−3; (f) 0.44
Round off each of the following numbers to two significant figures:
1. (a) 517
2. (b) 86.3
3. (c) 6.382 $\times$ 103
4. (d) 5.0008
5. (e) 22.497
6. (f) 0.885
Perform the following calculations and report each answer with the correct number of significant figures.
1. (a) 628 $\times$ 342
2. (b) (5.63 $\times$ 102) $\times$ (7.4 $\times$ 103)
3. (c)
4. $\dfrac{28.0}{13.483}$
5. (d) 8119 $\times$ 0.000023
6. (e) 14.98 + 27,340 + 84.7593
7. (f) 42.7 + 0.259
(a) 2.15 $\times$ 105; (b) 4.2 $\times$ 106; (c) 2.08; (d) 0.19; (e) 27,440; (f) 43.0
Perform the following calculations and report each answer with the correct number of significant figures.
1. (a) 62.8 $\times$ 34
2. (b) 0.147 + 0.0066 + 0.012
3. (c) 38 $\times$ 95 $\times$ 1.792
4. (d) 15 – 0.15 – 0.6155
5. (e) $8.78\times\left(\dfrac{0.0500}{0.478}\right)$
6. (f) 140 + 7.68 + 0.014
7. (g) 28.7 – 0.0483
8. (h) $\dfrac{(88.5−87.57)}{45.13}$
Consider the results of the archery contest shown in this figure.
1. (a) Which archer is most precise?
2. (b) Which archer is most accurate?
3. (c) Who is both least precise and least accurate?
(a) Archer X; (b) Archer W; (c) Archer Y
Classify the following sets of measurements as accurate, precise, both, or neither.
1. (a) Checking for consistency in the weight of chocolate chip cookies: 17.27 g, 13.05 g, 19.46 g, 16.92 g
2. (b) Testing the volume of a batch of 25-mL pipettes: 27.02 mL, 26.99 mL, 26.97 mL, 27.01 mL
3. (c) Determining the purity of gold: 99.9999%, 99.9998%, 99.9998%, 99.9999%
1.6: Mathematical Treatment of Measurement Results
Write conversion factors (as ratios) for the number of:
1. yards in 1 meter
2. liters in 1 liquid quart
3. pounds in 1 kilogram
(a) $\mathrm{\dfrac{1.0936\: yd}{1\: m}}$; (b) $\mathrm{\dfrac{0.94635\: L}{1\: qt}}$; (c) $\mathrm{\dfrac{2.2046\: lb}{1\: kg}}$
Write conversion factors (as ratios) for the number of:
1. (a) kilometers in 1 mile
2. (b) liters in 1 cubic foot
3. (c) grams in 1 ounce
The label on a soft drink bottle gives the volume in two units: 2.0 L and 67.6 fl oz. Use this information to derive a conversion factor between the English and metric units. How many significant figures can you justify in your conversion factor?
$\mathrm{\dfrac{2.0\: L}{67.6\: fl\: oz}=\dfrac{0.030\: L}{1\: fl\: oz}}$
Only two significant figures are justified.
The label on a box of cereal gives the mass of cereal in two units: 978 grams and 34.5 oz. Use this information to find a conversion factor between the English and metric units. How many significant figures can you justify in your conversion factor?
Soccer is played with a round ball having a circumference between 27 and 28 in. and a weight between 14 and 16 oz. What are these specifications in units of centimeters and grams?
68–71 cm; 400–450 g
A woman's basketball has a circumference between 28.5 and 29.0 inches and a maximum weight of 20 ounces (two significant figures). What are these specifications in units of centimeters and grams?
How many milliliters of a soft drink are contained in a 12.0-oz can?
355 mL
A barrel of oil is exactly 42 gal. How many liters of oil are in a barrel?
The diameter of a red blood cell is about 3 $\times$ 10−4 in. What is its diameter in centimeters?
8 $\times$ 10−4 cm
The distance between the centers of the two oxygen atoms in an oxygen molecule is 1.21 $\times$ 10−8 cm. What is this distance in inches?
Is a 197-lb weight lifter light enough to compete in a class limited to those weighing 90 kg or less?
yes; weight = 89.4 kg
A very good 197-lb weight lifter lifted 192 kg in a move called the clean and jerk. What was the mass of the weight lifted in pounds?
Many medical laboratory tests are run using 5.0 μL blood serum. What is this volume in milliliters?
5.0 $\times$ 10−3 mL
If an aspirin tablet contains 325 mg aspirin, how many grams of aspirin does it contain?
Use scientific (exponential) notation to express the following quantities in terms of the SI base units in [link]:
1. (a) 0.13 g
2. (b) 232 Gg
3. (c) 5.23 pm
4. (d) 86.3 mg
5. (e) 37.6 cm
6. (f) 54 μm
7. (g) 1 Ts
8. (h) 27 ps
9. (i) 0.15 mK
(a) 1.3 $\times$ 10−4 kg; (b) 2.32 $\times$ 108 kg; (c) 5.23 $\times$ 10−12 m; (d) 8.63 $\times$ 10−5 kg; (e) 3.76 $\times$ 10−1 m; (f) 5.4 $\times$ 10−5 m; (g) 1 $\times$ 1012 s; (h) 2.7 $\times$ 10−11 s; (i) 1.5 $\times$ 10−4 K
Complete the following conversions between SI units.
1. (a) 612 g = ________ mg
2. (b) 8.160 m = ________ cm
3. (c) 3779 μg = ________ g
4. (d) 781 mL = ________ L
5. (e) 4.18 kg = ________ g
6. (f) 27.8 m = ________ km
7. (g) 0.13 mL = ________ L
8. (h) 1738 km = ________ m
9. (i) 1.9 Gg = ________ g
Gasoline is sold by the liter in many countries. How many liters are required to fill a 12.0-gal gas tank?
45.4 L
Milk is sold by the liter in many countries. What is the volume of exactly 1/2 gal of milk in liters?
A long ton is defined as exactly 2240 lb. What is this mass in kilograms?
1.0160 $\times$ 103 kg
Make the conversion indicated in each of the following:
1. (a) the men’s world record long jump, 29 ft 4¼ in., to meters
2. (b) the greatest depth of the ocean, about 6.5 mi, to kilometers
3. (c) the area of the state of Oregon, 96,981 mi2, to square kilometers
4. (d) the volume of 1 gill (exactly 4 oz) to milliliters
5. (e) the estimated volume of the oceans, 330,000,000 mi3, to cubic kilometers.
6. (f) the mass of a 3525-lb car to kilograms
7. (g) the mass of a 2.3-oz egg to grams
Make the conversion indicated in each of the following:
1. (a) the length of a soccer field, 120 m (three significant figures), to feet
2. (b) the height of Mt. Kilimanjaro, at 19,565 ft the highest mountain in Africa, to kilometers
3. (c) the area of an 8.5 t 11-inch sheet of paper in cm2
4. (d) the displacement volume of an automobile engine, 161 in.3, to liters
5. (e) the estimated mass of the atmosphere, 5.6 t 1015 tons, to kilograms
6. (f) the mass of a bushel of rye, 32.0 lb, to kilograms
7. (g) the mass of a 5.00-grain aspirin tablet to milligrams (1 grain = 0.00229 oz)
1. (a) 394 ft
2. (b) 5.9634 km
3. (c) 6.0 $\times$ 102
4. (d) 2.64 L
5. (e) 5.1 $\times$ 1018 kg
6. (f) 14.5 kg
7. (g) 324 mg
Many chemistry conferences have held a 50-Trillion Angstrom Run (two significant figures). How long is this run in kilometers and in miles? (1 Å = 1 $\times$ 10−10 m)
A chemist’s 50-Trillion Angstrom Run (see Exercise) would be an archeologist’s 10,900 cubit run. How long is one cubit in meters and in feet? (1 Å = 1 $\times$ 10−8 cm)
0.46 m; 1.5 ft/cubit
The gas tank of a certain luxury automobile holds 22.3 gallons according to the owner’s manual. If the density of gasoline is 0.8206 g/mL, determine the mass in kilograms and pounds of the fuel in a full tank.
As an instructor is preparing for an experiment, he requires 225 g phosphoric acid. The only container readily available is a 150-mL Erlenmeyer flask. Is it large enough to contain the acid, whose density is 1.83 g/mL?
Yes, the acid's volume is 123 mL.
To prepare for a laboratory period, a student lab assistant needs 125 g of a compound. A bottle containing 1/4 lb is available. Did the student have enough of the compound?
A chemistry student is 159 cm tall and weighs 45.8 kg. What is her height in inches and weight in pounds?
62.6 in (about 5 ft 3 in.) and 101 lb
In a recent Grand Prix, the winner completed the race with an average speed of 229.8 km/h. What was his speed in miles per hour, meters per second, and feet per second?
Solve these problems about lumber dimensions.
(a) To describe to a European how houses are constructed in the US, the dimensions of “two-by-four” lumber must be converted into metric units. The thickness $\times$ width $\times$ length dimensions are 1.50 in. $\times$ 3.50 in. $\times$ 8.00 ft in the US. What are the dimensions in cm $\times$ cm $\times$ m?
(b) This lumber can be used as vertical studs, which are typically placed 16.0 in. apart. What is that distance in centimeters?
(a) 3.81 cm $\times$ 8.89 cm $\times$ 2.44 m; (b) 40.6 cm
The mercury content of a stream was believed to be above the minimum considered safe—1 part per billion (ppb) by weight. An analysis indicated that the concentration was 0.68 parts per billion. What quantity of mercury in grams was present in 15.0 L of the water, the density of which is 0.998 g/ml? $\mathrm{\left(1\: ppb\: Hg=\dfrac{1\: ng\: Hg}{1\: g\: water}\right)}$
Calculate the density of aluminum if 27.6 cm3 has a mass of 74.6 g.
2.70 g/cm3
Osmium is one of the densest elements known. What is its density if 2.72 g has a volume of 0.121 cm3?
Calculate these masses.
(a) What is the mass of 6.00 cm3 of mercury, density = 13.5939 g/cm3?
(b) What is the mass of 25.0 mL octane, density = 0.702 g/cm3?
(a) 81.6 g; (b) 17.6 g
Calculate these masses.
1. (a) What is the mass of 4.00 cm3 of sodium, density = 0.97 g/cm?
2. (b) What is the mass of 125 mL gaseous chlorine, density = 3.16 g/L?
Calculate these volumes.
1. (a) What is the volume of 25 g iodine, density = 4.93 g/cm3?
2. (b) What is the volume of 3.28 g gaseous hydrogen, density = 0.089 g/L?
(a) 5.1 mL; (b) 37 L
Calculate these volumes.
1. (a) What is the volume of 11.3 g graphite, density = 2.25 g/cm3?
2. (b) What is the volume of 39.657 g bromine, density = 2.928 g/cm3?
Convert the boiling temperature of gold, 2966 °C, into degrees Fahrenheit and kelvin.
5371 °F, 3239 K
Convert the temperature of scalding water, 54 °C, into degrees Fahrenheit and kelvin.
Convert the temperature of the coldest area in a freezer, −10 °F, to degrees Celsius and kelvin.
−23 °C, 250 K
Convert the temperature of dry ice, −77 °C, into degrees Fahrenheit and kelvin.
Convert the boiling temperature of liquid ammonia, −28.1 °F, into degrees Celsius and kelvin.
−33.4 °C, 239.8 K
The label on a pressurized can of spray disinfectant warns against heating the can above 130 °F. What are the corresponding temperatures on the Celsius and kelvin temperature scales?
The weather in Europe was unusually warm during the summer of 1995. The TV news reported temperatures as high as 45 °C. What was the temperature on the Fahrenheit scale?
113 °F | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/01%3A_Essential_Ideas_of_Chemistry/1.E%3A_Essential_Ideas_of_Chemistry_%28Exercises%29.txt |
This chapter will describe some of the fundamental chemical principles related to the composition of matter, including those central to the concept of molecular identity.
• 2.0: Prelude to Atoms
This chapter will describe some of the fundamental chemical principles related to the composition of matter, including those central to the concept of molecular identity.
• 2.1: Early Ideas in Atomic Theory
The ancient Greeks proposed that matter consists of extremely small particles called atoms. Dalton postulated that each element has a characteristic type of atom that differs in properties from atoms of all other elements, and that atoms of different elements can combine in fixed, small, whole-number ratios to form compounds. Samples of a particular compound all have the same elemental proportions by mass.
• 2.2: Evolution of Atomic Theory
Although no one has actually seen the inside of an atom, experiments have demonstrated much about atomic structure. Thomson’s cathode ray tube showed that atoms contain small, negatively charged particles called electrons. Millikan discovered that there is a fundamental electric charge—the charge of an electron. Rutherford’s gold foil experiment showed that atoms have a small, dense, positively charged nucleus; the positively charged particles within the nucleus are called protons.
• 2.3: Atomic Structure and Symbolism
An atom consists of a small, positively charged nucleus surrounded by electrons. The nucleus contains protons and neutrons; its diameter is about 100,000 times smaller than that of the atom. The mass of one atom is usually expressed in atomic mass units (amu), which is referred to as the atomic mass. An amu is defined as exactly $1/12$ of the mass of a carbon-12 atom and is equal to 1.6605 $\times$ 10−24 g.
• 2.4: Chemical Formulas
A molecular formula uses chemical symbols and subscripts to indicate the exact numbers of different atoms in a molecule or compound. An empirical formula gives the simplest, whole-number ratio of atoms in a compound. A structural formula indicates the bonding arrangement of the atoms in the molecule. Ball-and-stick and space-filling models show the geometric arrangement of atoms in a molecule. Isomers are compounds with the same molecular formula but different arrangements of atoms.
• 2.5: The Periodic Table
The discovery of the periodic recurrence of similar properties among the elements led to the formulation of the periodic table, in which the elements are arranged in order of increasing atomic number in rows known as periods and columns known as groups. Elements in the same group of the periodic table have similar chemical properties. Elements can be classified as metals, metalloids, and nonmetals, or as a main-group elements, transition metals, and inner transition metals.
• 2.6: Molecular and Ionic Compounds
Metals (particularly those in groups 1 and 2) tend to lose the number of electrons that would leave them with the same number of electrons as in the preceding noble gas in the periodic table. By this means, a positively charged ion is formed. Similarly, nonmetals (especially those in groups 16 and 17, and, to a lesser extent, those in Group 15) can gain the number of electrons needed to provide atoms with the same number of electrons as in the next noble gas in the periodic table.
• 2.7: Chemical Nomenclature
Chemists use nomenclature rules to clearly name compounds. Ionic and molecular compounds are named using somewhat-different methods. Binary ionic compounds typically consist of a metal and a nonmetal. The name of the metal is written first, followed by the name of the nonmetal with its ending changed to –ide. For example, K2O is called potassium oxide. If the metal can form ions with different charges, a Roman numeral in parentheses follows the name of the metal to specify its charge.
• 2.E: Atoms, Molecules, and Ions (Exercises)
These are homework exercises to accompany the Textmap created for "Chemistry" by OpenStax. Complementary General Chemistry question banks can be found for other Textmaps and can be accessed here. In addition to these publicly available questions, access to private problems bank for use in exams and homework is available to faculty only on an individual basis; please contact Delmar Larsen for an account with access permission.
02: Atoms Molecules and Ions
Your overall health and susceptibility to disease depends upon the complex interaction between your genetic makeup and environmental exposure, with the outcome difficult to predict. Early detection of biomarkers, substances that indicate an organism’s disease or physiological state, could allow diagnosis and treatment before a condition becomes serious or irreversible. Recent studies have shown that your exhaled breath can contain molecules that may be biomarkers for recent exposure to environmental contaminants or for pathological conditions ranging from asthma to lung cancer.
Scientists are working to develop biomarker “fingerprints” that could be used to diagnose a specific disease based on the amounts and identities of certain molecules in a patient’s exhaled breath. An essential concept underlying this goal is that of a molecule’s identity, which is determined by the numbers and types of atoms it contains, and how they are bonded together. This chapter will describe some of the fundamental chemical principles related to the composition of matter, including those central to the concept of molecular identity. | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/02%3A_Atoms_Molecules_and_Ions/2.0%3A_Prelude_to_Atoms.txt |
Skills to Develop
By the end of this section, you will be able to:
• State the postulates of Dalton’s atomic theory
• Use postulates of Dalton’s atomic theory to explain the laws of definite and multiple proportions
The language used in chemistry is seen and heard in many disciplines, ranging from medicine to engineering to forensics to art. The language of chemistry includes its own vocabulary as well as its own form of shorthand. Chemical symbols are used to represent atoms and elements. Chemical formulas depict molecules as well as the composition of compounds. Chemical equations provide information about the quality and quantity of the changes associated with chemical reactions.
This chapter will lay the foundation for our study of the language of chemistry. The concepts of this foundation include the atomic theory, the composition and mass of an atom, the variability of the composition of isotopes, ion formation, chemical bonds in ionic and covalent compounds, the types of chemical reactions, and the naming of compounds. We will also introduce one of the most powerful tools for organizing chemical knowledge: the periodic table.
Atomic Theory through the Nineteenth Century
The earliest recorded discussion of the basic structure of matter comes from ancient Greek philosophers, the scientists of their day. In the fifth century BC, Leucippus and Democritus argued that all matter was composed of small, finite particles that they called atomos, a term derived from the Greek word for “indivisible.” They thought of atoms as moving particles that differed in shape and size, and which could join together. Later, Aristotle and others came to the conclusion that matter consisted of various combinations of the four “elements”—fire, earth, air, and water—and could be infinitely divided. Interestingly, these philosophers thought about atoms and “elements” as philosophical concepts, but apparently never considered performing experiments to test their ideas.
The Aristotelian view of the composition of matter held sway for over two thousand years, until English schoolteacher John Dalton helped to revolutionize chemistry with his hypothesis that the behavior of matter could be explained using an atomic theory. First published in 1807, many of Dalton’s hypotheses about the microscopic features of matter are still valid in modern atomic theory. Here are the postulates of Dalton’s atomic theory.
1. Matter is composed of exceedingly small particles called atoms. An atom is the smallest unit of an element that can participate in a chemical change.
2. An element consists of only one type of atom, which has a mass that is characteristic of the element and is the same for all atoms of that element (Figure $1$). A macroscopic sample of an element contains an incredibly large number of atoms, all of which have identical chemical properties.
1. Atoms of one element differ in properties from atoms of all other elements.
2. A compound consists of atoms of two or more elements combined in a small, whole-number ratio. In a given compound, the numbers of atoms of each of its elements are always present in the same ratio (Figure $2$).
1. Atoms are neither created nor destroyed during a chemical change, but are instead rearranged to yield substances that are different from those present before the change (Figure $3$).
Dalton’s atomic theory provides a microscopic explanation of the many macroscopic properties of matter that you’ve learned about. For example, if an element such as copper consists of only one kind of atom, then it cannot be broken down into simpler substances, that is, into substances composed of fewer types of atoms. And if atoms are neither created nor destroyed during a chemical change, then the total mass of matter present when matter changes from one type to another will remain constant (the law of conservation of matter).
Example $1$: Testing Dalton’s Atomic Theory
In the following drawing, the green spheres represent atoms of a certain element. The purple spheres represent atoms of another element. If the spheres touch, they are part of a single unit of a compound. Does the following chemical change represented by these symbols violate any of the ideas of Dalton’s atomic theory? If so, which one?
Solution The starting materials consist of two green spheres and two purple spheres. The products consist of only one green sphere and one purple sphere. This violates Dalton’s postulate that atoms are neither created nor destroyed during a chemical change, but are merely redistributed. (In this case, atoms appear to have been destroyed.)
Exercise $1$
In the following drawing, the green spheres represent atoms of a certain element. The purple spheres represent atoms of another element. If the spheres touch, they are part of a single unit of a compound. Does the following chemical change represented by these symbols violate any of the ideas of Dalton’s atomic theory? If so, which one
Answer:
The starting materials consist of four green spheres and two purple spheres. The products consist of four green spheres and two purple spheres. This does not violate any of Dalton’s postulates: Atoms are neither created nor destroyed, but are redistributed in small, whole-number ratios.
Dalton knew of the experiments of French chemist Joseph Proust, who demonstrated that all samples of a pure compound contain the same elements in the same proportion by mass. This statement is known as the law of definite proportions or the law of constant composition. The suggestion that the numbers of atoms of the elements in a given compound always exist in the same ratio is consistent with these observations. For example, when different samples of isooctane (a component of gasoline and one of the standards used in the octane rating system) are analyzed, they are found to have a carbon-to-hydrogen mass ratio of 5.33:1, as shown in Table $1$.
Table $1$: Constant Composition of Isooctane
Sample Carbon Hydrogen Mass Ratio
A 14.82 g 2.78 g $\mathrm{\dfrac{14.82\: g\: carbon}{2.78\: g\: hydrogen}=\dfrac{5.33\: g\: carbon}{1.00\: g\: hydrogen}}$
B 22.33 g 4.19 g $\mathrm{\dfrac{22.33\: g\: carbon}{4.19\: g\: hydrogen}=\dfrac{5.33\: g\: carbon}{1.00\: g\: hydrogen}}$
C 19.40 g 3.64 g $\mathrm{\dfrac{19.40\: g\: carbon}{3.63\: g\: hydrogen}=\dfrac{5.33\: g\: carbon}{1.00\: g\: hydrogen}}$
It is worth noting that although all samples of a particular compound have the same mass ratio, the converse is not true in general. That is, samples that have the same mass ratio are not necessarily the same substance. For example, there are many compounds other than isooctane that also have a carbon-to-hydrogen mass ratio of 5.33:1.00.
Dalton also used data from Proust, as well as results from his own experiments, to formulate another interesting law. The law of multiple proportions states that when two elements react to form more than one compound, a fixed mass of one element will react with masses of the other element in a ratio of small, whole numbers. For example, copper and chlorine can form a green, crystalline solid with a mass ratio of 0.558 g chlorine to 1 g copper, as well as a brown crystalline solid with a mass ratio of 1.116 g chlorine to 1 g copper. These ratios by themselves may not seem particularly interesting or informative; however, if we take a ratio of these ratios, we obtain a useful and possibly surprising result: a small, whole-number ratio.
$\mathrm{\dfrac{\dfrac{1.116\: g\: Cl}{1\: g\: Cu}}{\dfrac{0.558\: g\: Cl}{1\: g\: Cu}}=\dfrac{2}{1}}$
This 2-to-1 ratio means that the brown compound has twice the amount of chlorine per amount of copper as the green compound.
This can be explained by atomic theory if the copper-to-chlorine ratio in the brown compound is 1 copper atom to 2 chlorine atoms, and the ratio in the green compound is 1 copper atom to 1 chlorine atom. The ratio of chlorine atoms (and thus the ratio of their masses) is therefore 2 to 1 (Figure $4$).
Example $2$: Laws of Definite and Multiple Proportions
A sample of compound A (a clear, colorless gas) is analyzed and found to contain 4.27 g carbon and 5.69 g oxygen. A sample of compound B (also a clear, colorless gas) is analyzed and found to contain 5.19 g carbon and 13.84 g oxygen. Are these data an example of the law of definite proportions, the law of multiple proportions, or neither? What do these data tell you about substances A and B?
Solution
In compound A, the mass ratio of carbon to oxygen is:
$\mathrm{\dfrac{1.33\: g\: O}{1\: g\: C}}$
In compound B, the mass ratio of carbon to oxygen is:
$\mathrm{\dfrac{2.67\: g\: O}{1\: g\: C}}$
The ratio of these ratios is:
$\mathrm{\dfrac{\dfrac{1.33\: g\: O}{1\: g\: C}}{\dfrac{2.67\: g\: O}{1\: g\: C}}=\dfrac{1}{2}}$
This supports the law of multiple proportions. This means that A and B are different compounds, with A having one-half as much carbon per amount of oxygen (or twice as much oxygen per amount of carbon) as B. A possible pair of compounds that would fit this relationship would be A = CO2 and B = CO.
Exercise $2$
A sample of compound X (a clear, colorless, combustible liquid with a noticeable odor) is analyzed and found to contain 14.13 g carbon and 2.96 g hydrogen. A sample of compound Y (a clear, colorless, combustible liquid with a noticeable odor that is slightly different from X’s odor) is analyzed and found to contain 19.91 g carbon and 3.34 g hydrogen. Are these data an example of the law of definite proportions, the law of multiple proportions, or neither? What do these data tell you about substances X and Y?
Answer:
In compound X, the mass ratio of carbon to hydrogen is $\mathrm{\dfrac{14.13\: g\: C}{2.96\: g\: H}}$. In compound Y, the mass ratio of carbon to oxygen is $\mathrm{\dfrac{19.91\: g\: C}{3.34\: g\: H}}$. The ratio of these ratios is $\mathrm{\dfrac{\dfrac{14.13\: g\: C}{2.96\: g\: H}}{\dfrac{19.91\: g\: C}{3.34\: g\: H}}=\dfrac{4.77\: g\: C/g\: H}{5.96\: g\: C/g\: H}=0.800=\dfrac{4}{5}}$. This small, whole-number ratio supports the law of multiple proportions. This means that X and Y are different compounds.
Summary
The ancient Greeks proposed that matter consists of extremely small particles called atoms. Dalton postulated that each element has a characteristic type of atom that differs in properties from atoms of all other elements, and that atoms of different elements can combine in fixed, small, whole-number ratios to form compounds. Samples of a particular compound all have the same elemental proportions by mass. When two elements form different compounds, a given mass of one element will combine with masses of the other element in a small, whole-number ratio. During any chemical change, atoms are neither created nor destroyed.
Glossary
Dalton’s atomic theory
set of postulates that established the fundamental properties of atoms
law of constant composition
(also, law of definite proportions) all samples of a pure compound contain the same elements in the same proportions by mass
law of definite proportions
(also, law of constant composition) all samples of a pure compound contain the same elements in the same proportions by mass
law of multiple proportions
when two elements react to form more than one compound, a fixed mass of one element will react with masses of the other element in a ratio of small whole numbers | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/02%3A_Atoms_Molecules_and_Ions/2.1%3A_Early_Ideas_in_Atomic_Theory.txt |
Learning Objectives
• Outline milestones in the development of modern atomic theory
• Summarize and interpret the results of the experiments of Thomson, Millikan, and Rutherford
• Describe the three subatomic particles that compose atoms
• Introduce the term isotopes
In the two centuries since Dalton developed his ideas, scientists have made significant progress in furthering our understanding of atomic theory. Much of this came from the results of several seminal experiments that revealed the details of the internal structure of atoms. Here, we will discuss some of those key developments, with an emphasis on application of the scientific method, as well as understanding how the experimental evidence was analyzed. While the historical persons and dates behind these experiments can be quite interesting, it is most important to understand the concepts resulting from their work.
Atomic Theory after the Nineteenth Century
If matter were composed of atoms, what were atoms composed of? Were they the smallest particles, or was there something smaller? In the late 1800s, a number of scientists interested in questions like these investigated the electrical discharges that could be produced in low-pressure gases, with the most significant discovery made by English physicist J. J. Thomson using a cathode ray tube. This apparatus consisted of a sealed glass tube from which almost all the air had been removed; the tube contained two metal electrodes. When high voltage was applied across the electrodes, a visible beam called a cathode ray appeared between them. This beam was deflected toward the positive charge and away from the negative charge, and was produced in the same way with identical properties when different metals were used for the electrodes. In similar experiments, the ray was simultaneously deflected by an applied magnetic field, and measurements of the extent of deflection and the magnetic field strength allowed Thomson to calculate the charge-to-mass ratio of the cathode ray particles. The results of these measurements indicated that these particles were much lighter than atoms (Figure $1$).
Based on his observations, here is what Thomson proposed and why: The particles are attracted by positive (+) charges and repelled by negative (−) charges, so they must be negatively charged (like charges repel and unlike charges attract); they are less massive than atoms and indistinguishable, regardless of the source material, so they must be fundamental, subatomic constituents of all atoms. Although controversial at the time, Thomson’s idea was gradually accepted, and his cathode ray particle is what we now call an electron, a negatively charged, subatomic particle with a mass more than one thousand-times less that of an atom. The term “electron” was coined in 1891 by Irish physicist George Stoney, from “electric ion.”
In 1909, more information about the electron was uncovered by American physicist Robert A. Millikan via his “oil drop” experiments. Millikan created microscopic oil droplets, which could be electrically charged by friction as they formed or by using X-rays. These droplets initially fell due to gravity, but their downward progress could be slowed or even reversed by an electric field lower in the apparatus. By adjusting the electric field strength and making careful measurements and appropriate calculations, Millikan was able to determine the charge on individual drops (Figure $2$).
Looking at the charge data that Millikan gathered, you may have recognized that the charge of an oil droplet is always a multiple of a specific charge, 1.6 $\times$ 10−19 C. Millikan concluded that this value must therefore be a fundamental charge—the charge of a single electron—with his measured charges due to an excess of one electron (1 times 1.6 $\times$ 10−19 C), two electrons (2 times 1.6 $\times$ 10−19 C), three electrons (3 times 1.6 $\times$ 10−19 C), and so on, on a given oil droplet. Since the charge of an electron was now known due to Millikan’s research, and the charge-to-mass ratio was already known due to Thomson’s research (1.759 $\times$ 1011 C/kg), it only required a simple calculation to determine the mass of the electron as well.
$\mathrm{Mass\: of\: electron=1.602\times 10^{-19}\:\cancel{C}\times \dfrac{1\: kg}{1.759\times 10^{11}\:\cancel{C}}=9.107\times 10^{-31}\:kg} \tag{2.3.1}$
Scientists had now established that the atom was not indivisible as Dalton had believed, and due to the work of Thomson, Millikan, and others, the charge and mass of the negative, subatomic particles—the electrons—were known. However, the positively charged part of an atom was not yet well understood. In 1904, Thomson proposed the “plum pudding” model of atoms, which described a positively charged mass with an equal amount of negative charge in the form of electrons embedded in it, since all atoms are electrically neutral. A competing model had been proposed in 1903 by Hantaro Nagaoka, who postulated a Saturn-like atom, consisting of a positively charged sphere surrounded by a halo of electrons (Figure $3$).
The next major development in understanding the atom came from Ernest Rutherford, a physicist from New Zealand who largely spent his scientific career in Canada and England. He performed a series of experiments using a beam of high-speed, positively charged alpha particles (α particles) that were produced by the radioactive decay of radium; α particles consist of two protons and two neutrons (you will learn more about radioactive decay in the chapter on nuclear chemistry). Rutherford and his colleagues Hans Geiger (later famous for the Geiger counter) and Ernest Marsden aimed a beam of α particles, the source of which was embedded in a lead block to absorb most of the radiation, at a very thin piece of gold foil and examined the resultant scattering of the α particles using a luminescent screen that glowed briefly where hit by an α particle.
What did they discover? Most particles passed right through the foil without being deflected at all. However, some were diverted slightly, and a very small number were deflected almost straight back toward the source (Figure $4$). Rutherford described finding these results: “It was quite the most incredible event that has ever happened to me in my life. It was almost as incredible as if you fired a 15-inch shell at a piece of tissue paper and it came back and hit you”1 (p. 68).
Here is what Rutherford deduced: Because most of the fast-moving α particles passed through the gold atoms undeflected, they must have traveled through essentially empty space inside the atom. Alpha particles are positively charged, so deflections arose when they encountered another positive charge (like charges repel each other). Since like charges repel one another, the few positively charged α particles that changed paths abruptly must have hit, or closely approached, another body that also had a highly concentrated, positive charge. Since the deflections occurred a small fraction of the time, this charge only occupied a small amount of the space in the gold foil. Analyzing a series of such experiments in detail, Rutherford drew two conclusions:
1. The volume occupied by an atom must consist of a large amount of empty space.
2. A small, relatively heavy, positively charged body, the nucleus, must be at the center of each atom.
This analysis led Rutherford to propose a model in which an atom consists of a very small, positively charged nucleus, in which most of the mass of the atom is concentrated, surrounded by the negatively charged electrons, so that the atom is electrically neutral (Figure $5$).
After many more experiments, Rutherford also discovered that the nuclei of other elements contain the hydrogen nucleus as a “building block,” and he named this more fundamental particle the proton, the positively charged, subatomic particle found in the nucleus. With one addition, which you will learn next, this nuclear model of the atom, proposed over a century ago, is still used today.
Another important finding was the discovery of isotopes. During the early 1900s, scientists identified several substances that appeared to be new elements, isolating them from radioactive ores. For example, a “new element” produced by the radioactive decay of thorium was initially given the name mesothorium. However, a more detailed analysis showed that mesothorium was chemically identical to radium (another decay product), despite having a different atomic mass. This result, along with similar findings for other elements, led the English chemist Frederick Soddy to realize that an element could have types of atoms with different masses that were chemically indistinguishable. These different types are called isotopes—atoms of the same element that differ in mass. Soddy was awarded the Nobel Prize in Chemistry in 1921 for this discovery.
One puzzle remained: The nucleus was known to contain almost all of the mass of an atom, with the number of protons only providing half, or less, of that mass. Different proposals were made to explain what constituted the remaining mass, including the existence of neutral particles in the nucleus. As you might expect, detecting uncharged particles is very challenging, and it was not until 1932 that James Chadwick found evidence of neutrons, uncharged, subatomic particles with a mass approximately the same as that of protons. The existence of the neutron also explained isotopes: They differ in mass because they have different numbers of neutrons, but they are chemically identical because they have the same number of protons. This will be explained in more detail later in this chapter.
Summary
Although no one has actually seen the inside of an atom, experiments have demonstrated much about atomic structure. Thomson’s cathode ray tube showed that atoms contain small, negatively charged particles called electrons. Millikan discovered that there is a fundamental electric charge—the charge of an electron. Rutherford’s gold foil experiment showed that atoms have a small, dense, positively charged nucleus; the positively charged particles within the nucleus are called protons. Chadwick discovered that the nucleus also contains neutral particles called neutrons. Soddy demonstrated that atoms of the same element can differ in mass; these are called isotopes.
Footnotes
1. Ernest Rutherford, “The Development of the Theory of Atomic Structure,” ed. J. A. Ratcliffe, in Background to Modern Science, eds. Joseph Needham and Walter Pagel, (Cambridge, UK: Cambridge University Press, 1938), 61–74. Accessed September 22, 2014, https://ia600508.us.archive.org/3/it...e032734mbp.pdf.
Glossary
alpha particle (α particle)
positively charged particle consisting of two protons and two neutrons
electron
negatively charged, subatomic particle of relatively low mass located outside the nucleus
isotopes
atoms that contain the same number of protons but different numbers of neutrons
neutron
uncharged, subatomic particle located in the nucleus
proton
positively charged, subatomic particle located in the nucleus
nucleus
massive, positively charged center of an atom made up of protons and neutrons | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/02%3A_Atoms_Molecules_and_Ions/2.2%3A_Evolution_of_Atomic_Theory.txt |
Learning Objectives
• Write and interpret symbols that depict the atomic number, mass number, and charge of an atom or ion
• Define the atomic mass unit and average atomic mass
• Calculate average atomic mass and isotopic abundance
The development of modern atomic theory revealed much about the inner structure of atoms. It was learned that an atom contains a very small nucleus composed of positively charged protons and uncharged neutrons, surrounded by a much larger volume of space containing negatively charged electrons. The nucleus contains the majority of an atom’s mass because protons and neutrons are much heavier than electrons, whereas electrons occupy almost all of an atom’s volume. The diameter of an atom is on the order of 10−10 m, whereas the diameter of the nucleus is roughly 10−15 m—about 100,000 times smaller. For a perspective about their relative sizes, consider this: If the nucleus were the size of a blueberry, the atom would be about the size of a football stadium (Figure $1$).
Atoms—and the protons, neutrons, and electrons that compose them—are extremely small. For example, a carbon atom weighs less than 2 $\times$ 10−23 g, and an electron has a charge of less than 2 $\times$ 10−19 C (coulomb). When describing the properties of tiny objects such as atoms, we use appropriately small units of measure, such as the atomic mass unit (amu) and the fundamental unit of charge (e). The amu was originally defined based on hydrogen, the lightest element, then later in terms of oxygen. Since 1961, it has been defined with regard to the most abundant isotope of carbon, atoms of which are assigned masses of exactly 12 amu. (This isotope is known as “carbon-12” as will be discussed later in this module.) Thus, one amu is exactly $1/12$ of the mass of one carbon-12 atom: 1 amu = 1.6605 $\times$ 10−24 g. (The Dalton (Da) and the unified atomic mass unit (u) are alternative units that are equivalent to the amu.) The fundamental unit of charge (also called the elementary charge) equals the magnitude of the charge of an electron (e) with e = 1.602 $\times$ 10−19 C.
A proton has a mass of 1.0073 amu and a charge of 1+. A neutron is a slightly heavier particle with a mass 1.0087 amu and a charge of zero; as its name suggests, it is neutral. The electron has a charge of 1− and is a much lighter particle with a mass of about 0.00055 amu (it would take about 1800 electrons to equal the mass of one proton. The properties of these fundamental particles are summarized in Table $1$. (An observant student might notice that the sum of an atom’s subatomic particles does not equal the atom’s actual mass: The total mass of six protons, six neutrons, and six electrons is 12.0993 amu, slightly larger than the 12.00 amu of an actual carbon-12 atom. This “missing” mass is known as the mass defect, and you will learn about it in the chapter on nuclear chemistry.)
Table $1$: Properties of Subatomic Particles
Name Location Charge (C) Unit Charge Mass (amu) Mass (g)
electron outside nucleus $−1.602 \times 10^{−19}$ 1− 0.00055 $0.00091 \times 10^{−24}$
proton nucleus $1.602 \times 10^{−19}$ 1+ 1.00727 $1.67262 \times 10^{−24}$
neutron nucleus 0 0 1.00866 $1.67493 \times10^{−24}$
The number of protons in the nucleus of an atom is its atomic number ($Z$). This is the defining trait of an element: Its value determines the identity of the atom. For example, any atom that contains six protons is the element carbon and has the atomic number 6, regardless of how many neutrons or electrons it may have. A neutral atom must contain the same number of positive and negative charges, so the number of protons equals the number of electrons. Therefore, the atomic number also indicates the number of electrons in an atom. The total number of protons and neutrons in an atom is called its mass number (A). The number of neutrons is therefore the difference between the mass number and the atomic number: A – Z = number of neutrons.
\begin{align*} \ce{atomic\: number\:(Z)\: &= \:number\: of\: protons\ mass\: number\:(A)\: &= \:number\: of\: protons + number\: of\: neutrons\ A-Z\: &= \:number\: of\: neutrons} \end{align*} \nonumber
Atoms are electrically neutral if they contain the same number of positively charged protons and negatively charged electrons. When the numbers of these subatomic particles are not equal, the atom is electrically charged and is called an ion. The charge of an atom is defined as follows:
Atomic charge = number of protons − number of electrons
As will be discussed in more detail later in this chapter, atoms (and molecules) typically acquire charge by gaining or losing electrons. An atom that gains one or more electrons will exhibit a negative charge and is called an anion. Positively charged atoms called cations are formed when an atom loses one or more electrons. For example, a neutral sodium atom (Z = 11) has 11 electrons. If this atom loses one electron, it will become a cation with a 1+ charge (11 − 10 = 1+). A neutral oxygen atom (Z = 8) has eight electrons, and if it gains two electrons it will become an anion with a 2− charge (8 − 10 = 2−).
Example $1$: Composition of an Atom
Iodine is an essential trace element in our diet; it is needed to produce thyroid hormone. Insufficient iodine in the diet can lead to the development of a goiter, an enlargement of the thyroid gland (Figure $2$).
The addition of small amounts of iodine to table salt (iodized salt) has essentially eliminated this health concern in the United States, but as much as 40% of the world’s population is still at risk of iodine deficiency. The iodine atoms are added as anions, and each has a 1− charge and a mass number of 127. Determine the numbers of protons, neutrons, and electrons in one of these iodine anions.
Solution
The atomic number of iodine (53) tells us that a neutral iodine atom contains 53 protons in its nucleus and 53 electrons outside its nucleus. Because the sum of the numbers of protons and neutrons equals the mass number, 127, the number of neutrons is 74 (127 − 53 = 74). Since the iodine is added as a 1− anion, the number of electrons is 54 [53 – (1–) = 54].
Exercise $1$
An ion of platinum has a mass number of 195 and contains 74 electrons. How many protons and neutrons does it contain, and what is its charge?
Answer
78 protons; 117 neutrons; charge is 4+
Chemical Symbols
A chemical symbol is an abbreviation that we use to indicate an element or an atom of an element. For example, the symbol for mercury is Hg (Figure $3$). We use the same symbol to indicate one atom of mercury (microscopic domain) or to label a container of many atoms of the element mercury (macroscopic domain).
The symbols for several common elements and their atoms are listed in Table $2$. Some symbols are derived from the common name of the element; others are abbreviations of the name in another language. Symbols have one or two letters, for example, H for hydrogen and Cl for chlorine. To avoid confusion with other notations, only the first letter of a symbol is capitalized. For example, Co is the symbol for the element cobalt, but CO is the notation for the compound carbon monoxide, which contains atoms of the elements carbon (C) and oxygen (O). All known elements and their symbols are in the periodic table.
Table $2$: Some Common Elements and Their Symbols
Element Symbol Element Symbol
aluminum Al iron Fe (from ferrum)
bromine Br lead Pb (from plumbum)
calcium Ca magnesium Mg
carbon C mercury Hg (from hydrargyrum)
chlorine Cl nitrogen N
chromium Cr oxygen O
cobalt Co potassium K (from kalium)
copper Cu (from cuprum) silicon Si
fluorine F silver Ag (from argentum)
gold Au (from aurum) sodium Na (from natrium)
helium He sulfur S
hydrogen H tin Sn (from stannum)
iodine I zinc Zn
Traditionally, the discoverer (or discoverers) of a new element names the element. However, until the name is recognized by the International Union of Pure and Applied Chemistry (IUPAC), the recommended name of the new element is based on the Latin word(s) for its atomic number. For example, element 106 was called unnilhexium (Unh), element 107 was called unnilseptium (Uns), and element 108 was called unniloctium (Uno) for several years. These elements are now named after scientists or locations; for example, element 106 is now known as seaborgium (Sg) in honor of Glenn Seaborg, a Nobel Prize winner who was active in the discovery of several heavy elements.
Isotopes
The symbol for a specific isotope of any element is written by placing the mass number as a superscript to the left of the element symbol (Figure $4$). The atomic number is sometimes written as a subscript preceding the symbol, but since this number defines the element’s identity, as does its symbol, it is often omitted. For example, magnesium exists as a mixture of three isotopes, each with an atomic number of 12 and with mass numbers of 24, 25, and 26, respectively. These isotopes can be identified as 24Mg, 25Mg, and 26Mg. These isotope symbols are read as “element, mass number” and can be symbolized consistent with this reading. For instance, 24Mg is read as “magnesium 24,” and can be written as “magnesium-24” or “Mg-24.” 25Mg is read as “magnesium 25,” and can be written as “magnesium-25” or “Mg-25.” All magnesium atoms have 12 protons in their nucleus. They differ only because a 24Mg atom has 12 neutrons in its nucleus, a 25Mg atom has 13 neutrons, and a 26Mg has 14 neutrons.
Information about the naturally occurring isotopes of elements with atomic numbers 1 through 10 is given in Table $2$. Note that in addition to standard names and symbols, the isotopes of hydrogen are often referred to using common names and accompanying symbols. Hydrogen-2, symbolized 2H, is also called deuterium and sometimes symbolized D. Hydrogen-3, symbolized 3H, is also called tritium and sometimes symbolized T.
Table $2$: Nuclear Compositions of Atoms of the Very Light Elements
Element Symbol Atomic Number Number of Protons Number of Neutrons Mass (amu) % Natural Abundance
hydrogen $\ce{^1_1H}$
(protium)
1 1 0 1.0078 99.989
$\ce{^2_1H}$
(deuterium)
1 1 1 2.0141 0.0115
$\ce{^3_1H}$
(tritium)
1 1 2 3.01605 — (trace)
helium $\ce{^3_2He}$ 2 2 1 3.01603 0.00013
$\ce{^4_2He}$ 2 2 2 4.0026 100
lithium $\ce{^6_3Li}$ 3 3 3 6.0151 7.59
$\ce{^7_3Li}$ 3 3 4 7.0160 92.41
beryllium $\ce{^9_4Be}$ 4 4 5 9.0122 100
boron $\ce{^{10}_5B}$ 5 5 5 10.0129 19.9
$\ce{^{11}_5B}$ 5 5 6 11.0093 80.1
carbon $\ce{^{12}_6C}$ 6 6 6 12.0000 98.89
$\ce{^{13}_6C}$ 6 6 7 13.0034 1.11
$\ce{^{14}_6C}$ 6 6 8 14.0032 — (trace)
nitrogen $\ce{^{14}_7N}$ 7 7 7 14.0031 99.63
$\ce{^{15}_7N}$ 7 7 8 15.0001 0.37
oxygen $\ce{^{16}_8O}$ 8 8 8 15.9949 99.757
$\ce{^{17}_8O}$ 8 8 9 16.9991 0.038
$\ce{^{18}_8O}$ 8 8 10 17.9992 0.205
fluorine $\ce{^{19}_9F}$ 9 9 10 18.9984 100
neon $\ce{^{20}_{10}Ne}$ 10 10 10 19.9924 90.48
$\ce{^{21}_{10}Ne}$ 10 10 11 20.9938 0.27
$\ce{^{22}_{10}Ne}$ 10 10 12 21.9914 9.25
Atomic Mass
Because each proton and each neutron contribute approximately one amu to the mass of an atom, and each electron contributes far less, the atomic mass of a single atom is approximately equal to its mass number (a whole number). However, the average masses of atoms of most elements are not whole numbers because most elements exist naturally as mixtures of two or more isotopes.
The mass of an element shown in a periodic table or listed in a table of atomic masses is a weighted, average mass of all the isotopes present in a naturally occurring sample of that element. This is equal to the sum of each individual isotope’s mass multiplied by its fractional abundance.
$\mathrm{average\: mass}=\sum_{i}(\mathrm{fractional\: abundance\times isotopic\: mass})_i \nonumber$
For example, the element boron is composed of two isotopes: About 19.9% of all boron atoms are 10B with a mass of 10.0129 amu, and the remaining 80.1% are 11B with a mass of 11.0093 amu. The average atomic mass for boron is calculated to be:
\begin{align*} \textrm{boron average mass} &=\mathrm{(0.199\times10.0129\: amu)+(0.801\times11.0093\: amu)}\ &=\mathrm{1.99\: amu+8.82\: amu}\ &=\mathrm{10.81\: amu} \end{align*} \nonumber
It is important to understand that no single boron atom weighs exactly 10.8 amu; 10.8 amu is the average mass of all boron atoms, and individual boron atoms weigh either approximately 10 amu or 11 amu.
Example $2$: Calculation of Average Atomic Mass
A meteorite found in central Indiana contains traces of the noble gas neon picked up from the solar wind during the meteorite’s trip through the solar system. Analysis of a sample of the gas showed that it consisted of 91.84% 20Ne (mass 19.9924 amu), 0.47% 21Ne (mass 20.9940 amu), and 7.69% 22Ne (mass 21.9914 amu). What is the average mass of the neon in the solar wind?
Solution
\begin{align*} \mathrm{average\: mass} &=\mathrm{(0.9184\times19.9924\: amu)+(0.0047\times20.9940\: amu)+(0.0769\times21.9914\: amu)}\ &=\mathrm{(18.36+0.099+1.69)\:amu}\ &=\mathrm{20.15\: amu} \end{align*} \nonumber
The average mass of a neon atom in the solar wind is 20.15 amu. (The average mass of a terrestrial neon atom is 20.1796 amu. This result demonstrates that we may find slight differences in the natural abundance of isotopes, depending on their origin.)
Exercise $2$
A sample of magnesium is found to contain 78.70% of 24Mg atoms (mass 23.98 amu), 10.13% of 25Mg atoms (mass 24.99 amu), and 11.17% of 26Mg atoms (mass 25.98 amu). Calculate the average mass of a Mg atom.
Answer
24.31 amu
We can also do variations of this type of calculation, as shown in the next example.
Example $3$: Calculation of Percent Abundance
Naturally occurring chlorine consists of 35Cl (mass 34.96885 amu) and 37Cl (mass 36.96590 amu), with an average mass of 35.453 amu. What is the percent composition of Cl in terms of these two isotopes?
Solution
The average mass of chlorine is the fraction that is 35Cl times the mass of 35Cl plus the fraction that is 37Cl times the mass of 37Cl.
$\mathrm{average\: mass=(fraction\: of\: ^{35}Cl\times mass\: of\: ^{35}Cl)+(fraction\: of\: ^{37}Cl\times mass\: of\: ^{37}Cl)} \nonumber$
If we let x represent the fraction that is 35Cl, then the fraction that is 37Cl is represented by 1.00 − x.
(The fraction that is 35Cl + the fraction that is 37Cl must add up to 1, so the fraction of 37Cl must equal 1.00 − the fraction of 35Cl.)
Substituting this into the average mass equation, we have:
\begin{align*} \mathrm{35.453\: amu} &=(x\times 34.96885\: \ce{amu})+[(1.00-x)\times 36.96590\: \ce{amu}]\ 35.453 &=34.96885x+36.96590-36.96590x\ 1.99705x &=1.513\ x&=\dfrac{1.513}{1.99705}=0.7576 \end{align*} \nonumber
So solving yields: x = 0.7576, which means that 1.00 − 0.7576 = 0.2424. Therefore, chlorine consists of 75.76% 35Cl and 24.24% 37Cl.
Exercise $3$
Naturally occurring copper consists of 63Cu (mass 62.9296 amu) and 65Cu (mass 64.9278 amu), with an average mass of 63.546 amu. What is the percent composition of Cu in terms of these two isotopes?
Answer
69.15% Cu-63 and 30.85% Cu-65
The occurrence and natural abundances of isotopes can be experimentally determined using an instrument called a mass spectrometer. Mass spectrometry (MS) is widely used in chemistry, forensics, medicine, environmental science, and many other fields to analyze and help identify the substances in a sample of material. In a typical mass spectrometer (Figure $5$), the sample is vaporized and exposed to a high-energy electron beam that causes the sample’s atoms (or molecules) to become electrically charged, typically by losing one or more electrons. These cations then pass through a (variable) electric or magnetic field that deflects each cation’s path to an extent that depends on both its mass and charge (similar to how the path of a large steel ball bearing rolling past a magnet is deflected to a lesser extent that that of a small steel BB). The ions are detected, and a plot of the relative number of ions generated versus their mass-to-charge ratios (a mass spectrum) is made. The height of each vertical feature or peak in a mass spectrum is proportional to the fraction of cations with the specified mass-to-charge ratio. Since its initial use during the development of modern atomic theory, MS has evolved to become a powerful tool for chemical analysis in a wide range of applications.
Video $1$: Watch this video from the Royal Society for Chemistry for a brief description of the rudiments of mass spectrometry.
Summary
An atom consists of a small, positively charged nucleus surrounded by electrons. The nucleus contains protons and neutrons; its diameter is about 100,000 times smaller than that of the atom. The mass of one atom is usually expressed in atomic mass units (amu), which is referred to as the atomic mass. An amu is defined as exactly $1/12$ of the mass of a carbon-12 atom and is equal to 1.6605 $\times$ 10−24 g.
Protons are relatively heavy particles with a charge of 1+ and a mass of 1.0073 amu. Neutrons are relatively heavy particles with no charge and a mass of 1.0087 amu. Electrons are light particles with a charge of 1− and a mass of 0.00055 amu. The number of protons in the nucleus is called the atomic number (Z) and is the property that defines an atom’s elemental identity. The sum of the numbers of protons and neutrons in the nucleus is called the mass number and, expressed in amu, is approximately equal to the mass of the atom. An atom is neutral when it contains equal numbers of electrons and protons.
Isotopes of an element are atoms with the same atomic number but different mass numbers; isotopes of an element, therefore, differ from each other only in the number of neutrons within the nucleus. When a naturally occurring element is composed of several isotopes, the atomic mass of the element represents the average of the masses of the isotopes involved. A chemical symbol identifies the atoms in a substance using symbols, which are one-, two-, or three-letter abbreviations for the atoms.
Key Equations
• $\mathrm{average\: mass}=\sum_{i}(\mathrm{fractional\: abundance \times isotopic\: mass})_i$
Glossary
anion
negatively charged atom or molecule (contains more electrons than protons)
atomic mass
average mass of atoms of an element, expressed in amu
atomic mass unit (amu)
(also, unified atomic mass unit, u, or Dalton, Da) unit of mass equal to $\dfrac{1}{12}$ of the mass of a 12C atom
atomic number (Z)
number of protons in the nucleus of an atom
cation
positively charged atom or molecule (contains fewer electrons than protons)
chemical symbol
one-, two-, or three-letter abbreviation used to represent an element or its atoms
Dalton (Da)
alternative unit equivalent to the atomic mass unit
fundamental unit of charge
(also called the elementary charge) equals the magnitude of the charge of an electron (e) with e = 1.602 $\times$ 10−19 C
ion
electrically charged atom or molecule (contains unequal numbers of protons and electrons)
mass number (A)
sum of the numbers of neutrons and protons in the nucleus of an atom
unified atomic mass unit (u)
alternative unit equivalent to the atomic mass unit | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/02%3A_Atoms_Molecules_and_Ions/2.3%3A_Atomic_Structure_and_Symbolism.txt |
Learning Objectives
• Symbolize the composition of molecules using molecular formulas and empirical formulas
• Represent the bonding arrangement of atoms within molecules using structural formulas
A molecular formula is a representation of a molecule that uses chemical symbols to indicate the types of atoms followed by subscripts to show the number of atoms of each type in the molecule. (A subscript is used only when more than one atom of a given type is present.) Molecular formulas are also used as abbreviations for the names of compounds.
The structural formula for a compound gives the same information as its molecular formula (the types and numbers of atoms in the molecule) but also shows how the atoms are connected in the molecule. The structural formula for methane contains symbols for one C atom and four H atoms, indicating the number of atoms in the molecule (Figure \(1\)). The lines represent bonds that hold the atoms together. (A chemical bond is an attraction between atoms or ions that holds them together in a molecule or a crystal.) We will discuss chemical bonds and see how to predict the arrangement of atoms in a molecule later. For now, simply know that the lines are an indication of how the atoms are connected in a molecule. A ball-and-stick model shows the geometric arrangement of the atoms with atomic sizes not to scale, and a space-filling model shows the relative sizes of the atoms.
Although many elements consist of discrete, individual atoms, some exist as molecules made up of two or more atoms of the element chemically bonded together. For example, most samples of the elements hydrogen, oxygen, and nitrogen are composed of molecules that contain two atoms each (called diatomic molecules) and thus have the molecular formulas H2, O2, and N2, respectively. Other elements commonly found as diatomic molecules are fluorine (F2), chlorine (Cl2), bromine (Br2), and iodine (I2). The most common form of the element sulfur is composed of molecules that consist of eight atoms of sulfur; its molecular formula is S8 (Figure \(2\)).
It is important to note that a subscript following a symbol and a number in front of a symbol do not represent the same thing; for example, H2 and 2H represent distinctly different species. H2 is a molecular formula; it represents a diatomic molecule of hydrogen, consisting of two atoms of the element that are chemically bonded together. The expression 2H, on the other hand, indicates two separate hydrogen atoms that are not combined as a unit. The expression 2H2 represents two molecules of diatomic hydrogen (Figure \(3\)).
Compounds are formed when two or more elements chemically combine, resulting in the formation of bonds. For example, hydrogen and oxygen can react to form water, and sodium and chlorine can react to form table salt. We sometimes describe the composition of these compounds with an empirical formula, which indicates the types of atoms present and the simplest whole-number ratio of the number of atoms (or ions) in the compound. For example, titanium dioxide (used as pigment in white paint and in the thick, white, blocking type of sunscreen) has an empirical formula of TiO2. This identifies the elements titanium (Ti) and oxygen (O) as the constituents of titanium dioxide, and indicates the presence of twice as many atoms of the element oxygen as atoms of the element titanium (Figure \(4\)).
As discussed previously, we can describe a compound with a molecular formula, in which the subscripts indicate the actual numbers of atoms of each element in a molecule of the compound. In many cases, the molecular formula of a substance is derived from experimental determination of both its empirical formula and its molecular mass (the sum of atomic masses for all atoms composing the molecule). For example, it can be determined experimentally that benzene contains two elements, carbon (C) and hydrogen (H), and that for every carbon atom in benzene, there is one hydrogen atom. Thus, the empirical formula is CH. An experimental determination of the molecular mass reveals that a molecule of benzene contains six carbon atoms and six hydrogen atoms, so the molecular formula for benzene is C6H6 (Figure \(5\)).
If we know a compound’s formula, we can easily determine the empirical formula. (This is somewhat of an academic exercise; the reverse chronology is generally followed in actual practice.) For example, the molecular formula for acetic acid, the component that gives vinegar its sharp taste, is C2H4O2. This formula indicates that a molecule of acetic acid (Figure \(6\)) contains two carbon atoms, four hydrogen atoms, and two oxygen atoms. The ratio of atoms is 2:4:2. Dividing by the lowest common denominator (2) gives the simplest, whole-number ratio of atoms, 1:2:1, so the empirical formula is CH2O. Note that a molecular formula is always a whole-number multiple of an empirical formula.
Example \(1\): Empirical and Molecular Formulas
Molecules of glucose (blood sugar) contain 6 carbon atoms, 12 hydrogen atoms, and 6 oxygen atoms. What are the molecular and empirical formulas of glucose?
Solution
The molecular formula is C6H12O6 because one molecule actually contains 6 C, 12 H, and 6 O atoms. The simplest whole-number ratio of C to H to O atoms in glucose is 1:2:1, so the empirical formula is CH2O.
Exercise \(1\)
A molecule of metaldehyde (a pesticide used for snails and slugs) contains 8 carbon atoms, 16 hydrogen atoms, and 4 oxygen atoms. What are the molecular and empirical formulas of metaldehyde?
Answer
Molecular formula, C8H16O4; empirical formula, C2H4O
It is important to be aware that it may be possible for the same atoms to be arranged in different ways: Compounds with the same molecular formula may have different atom-to-atom bonding and therefore different structures. For example, could there be another compound with the same formula as acetic acid, C2H4O2? And if so, what would be the structure of its molecules?
If you predict that another compound with the formula C2H4O2 could exist, then you demonstrated good chemical insight and are correct. Two C atoms, four H atoms, and two O atoms can also be arranged to form methyl formate, which is used in manufacturing, as an insecticide, and for quick-drying finishes. Methyl formate molecules have one of the oxygen atoms between the two carbon atoms, differing from the arrangement in acetic acid molecules. Acetic acid and methyl formate are examples of isomers—compounds with the same chemical formula but different molecular structures (Figure \(7\)). Note that this small difference in the arrangement of the atoms has a major effect on their respective chemical properties. You would certainly not want to use a solution of methyl formate as a substitute for a solution of acetic acid (vinegar) when you make salad dressing.
Many types of isomers exist (Figure \(8\)). Acetic acid and methyl formate are structural isomers, compounds in which the molecules differ in how the atoms are connected to each other. There are also various types of spatial isomers, in which the relative orientations of the atoms in space can be different. For example, the compound carvone (found in caraway seeds, spearmint, and mandarin orange peels) consists of two isomers that are mirror images of each other. S-(+)-carvone smells like caraway, and R-(−)-carvone smells like spearmint.
Summary
A molecular formula uses chemical symbols and subscripts to indicate the exact numbers of different atoms in a molecule or compound. An empirical formula gives the simplest, whole-number ratio of atoms in a compound. A structural formula indicates the bonding arrangement of the atoms in the molecule. Ball-and-stick and space-filling models show the geometric arrangement of atoms in a molecule. Isomers are compounds with the same molecular formula but different arrangements of atoms.
Glossary
empirical formula
formula showing the composition of a compound given as the simplest whole-number ratio of atoms
isomers
compounds with the same chemical formula but different structures
molecular formula
formula indicating the composition of a molecule of a compound and giving the actual number of atoms of each element in a molecule of the compound.
spatial isomers
compounds in which the relative orientations of the atoms in space differ
structural isomer
one of two substances that have the same molecular formula but different physical and chemical properties because their atoms are bonded differently
structural formula
shows the atoms in a molecule and how they are connected | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/02%3A_Atoms_Molecules_and_Ions/2.4%3A_Chemical_Formulas.txt |
Learning Objectives
• State the periodic law and explain the organization of elements in the periodic table
• Predict the general properties of elements based on their location within the periodic table
• Identify metals, nonmetals, and metalloids by their properties and/or location on the periodic table
As early chemists worked to purify ores and discovered more elements, they realized that various elements could be grouped together by their similar chemical behaviors. One such grouping includes lithium (Li), sodium (Na), and potassium (K): These elements all are shiny, conduct heat and electricity well, and have similar chemical properties. A second grouping includes calcium (Ca), strontium (Sr), and barium (Ba), which also are shiny, good conductors of heat and electricity, and have chemical properties in common. However, the specific properties of these two groupings are notably different from each other. For example: Li, Na, and K are much more reactive than are Ca, Sr, and Ba; Li, Na, and K form compounds with oxygen in a ratio of two of their atoms to one oxygen atom, whereas Ca, Sr, and Ba form compounds with one of their atoms to one oxygen atom. Fluorine (F), chlorine (Cl), bromine (Br), and iodine (I) also exhibit similar properties to each other, but these properties are drastically different from those of any of the elements above.
Dimitri Mendeleev in Russia (1869) and Lothar Meyer in Germany (1870) independently recognized that there was a periodic relationship among the properties of the elements known at that time. Both published tables with the elements arranged according to increasing atomic mass. But Mendeleev went one step further than Meyer: He used his table to predict the existence of elements that would have the properties similar to aluminum and silicon, but were yet unknown. The discoveries of gallium (1875) and germanium (1886) provided great support for Mendeleev’s work. Although Mendeleev and Meyer had a long dispute over priority, Mendeleev’s contributions to the development of the periodic table are now more widely recognized (Figure \(1\)).
By the twentieth century, it became apparent that the periodic relationship involved atomic numbers rather than atomic masses. The modern statement of this relationship, the periodic law, is as follows: the properties of the elements are periodic functions of their atomic numbers. A modern periodic table arranges the elements in increasing order of their atomic numbers and groups atoms with similar properties in the same vertical column (Figure \(2\)). Each box represents an element and contains its atomic number, symbol, average atomic mass, and (sometimes) name. The elements are arranged in seven horizontal rows, called periods or series, and 18 vertical columns, called groups. Groups are labeled at the top of each column. In the United States, the labels traditionally were numerals with capital letters. However, IUPAC recommends that the numbers 1 through 18 be used, and these labels are more common. For the table to fit on a single page, parts of two of the rows, a total of 14 columns, are usually written below the main body of the table.
Many elements differ dramatically in their chemical and physical properties, but some elements are similar in their behaviors. For example, many elements appear shiny, are malleable (able to be deformed without breaking) and ductile (can be drawn into wires), and conduct heat and electricity well. Other elements are not shiny, malleable, or ductile, and are poor conductors of heat and electricity. We can sort the elements into large classes with common properties: metals (elements that are shiny, malleable, good conductors of heat and electricity—shaded yellow); nonmetals (elements that appear dull, poor conductors of heat and electricity—shaded green); and metalloids (elements that conduct heat and electricity moderately well, and possess some properties of metals and some properties of nonmetals—shaded purple).
The elements can also be classified into the main-group elements (or representative elements) in the columns labeled 1, 2, and 13–18; the transition metals in the columns labeled 3–12; and inner transition metals in the two rows at the bottom of the table (the top-row elements are called lanthanides and the bottom-row elements are actinides; Figure \(3\)). The elements can be subdivided further by more specific properties, such as the composition of the compounds they form. For example, the elements in group 1 (the first column) form compounds that consist of one atom of the element and one atom of hydrogen. These elements (except hydrogen) are known as alkali metals, and they all have similar chemical properties. The elements in group 2 (the second column) form compounds consisting of one atom of the element and two atoms of hydrogen: These are called alkaline earth metals, with similar properties among members of that group. Other groups with specific names are the pnictogens (group 15), chalcogens (group 16), halogens (group 17), and the noble gases (group 18, also known as inert gases). The groups can also be referred to by the first element of the group: For example, the chalcogens can be called the oxygen group or oxygen family. Hydrogen is a unique, nonmetallic element with properties similar to both group 1 and group 17 elements. For that reason, hydrogen may be shown at the top of both groups, or by itself.
Example \(1\): Naming Groups of Elements
Atoms of each of the following elements are essential for life. Give the group name for the following elements:
1. chlorine
2. calcium
3. sodium
4. sulfur
Solution
The family names are as follows:
1. halogen
2. alkaline earth metal
3. alkali metal
4. chalcogen
Exercise \(1\)
Give the group name for each of the following elements:
1. krypton
2. selenium
3. barium
4. lithium
Answer a
noble gas
Answer b
chalcogen
Answer c
alkaline earth metal
Answer d
alkali metal
In studying the periodic table, you might have noticed something about the atomic masses of some of the elements. Element 43 (technetium), element 61 (promethium), and most of the elements with atomic number 84 (polonium) and higher have their atomic mass given in square brackets. This is done for elements that consist entirely of unstable, radioactive isotopes (you will learn more about radioactivity in the nuclear chemistry chapter). An average atomic weight cannot be determined for these elements because their radioisotopes may vary significantly in relative abundance, depending on the source, or may not even exist in nature. The number in square brackets is the atomic mass number (and approximate atomic mass) of the most stable isotope of that element.
Summary
The discovery of the periodic recurrence of similar properties among the elements led to the formulation of the periodic table, in which the elements are arranged in order of increasing atomic number in rows known as periods and columns known as groups. Elements in the same group of the periodic table have similar chemical properties. Elements can be classified as metals, metalloids, and nonmetals, or as a main-group elements, transition metals, and inner transition metals. Groups are numbered 1–18 from left to right. The elements in group 1 are known as the alkali metals; those in group 2 are the alkaline earth metals; those in 15 are the pnictogens; those in 16 are the chalcogens; those in 17 are the halogens; and those in 18 are the noble gases.
Glossary
actinide
inner transition metal in the bottom of the bottom two rows of the periodic table
alkali metal
element in group 1
alkaline earth metal
element in group 2
chalcogen
element in group 16
group
vertical column of the periodic table
halogen
element in group 17
inert gas
(also, noble gas) element in group 18
inner transition metal
(also, lanthanide or actinide) element in the bottom two rows; if in the first row, also called lanthanide, or if in the second row, also called actinide
lanthanide
inner transition metal in the top of the bottom two rows of the periodic table
main-group element
(also, representative element) element in columns 1, 2, and 12–18
metal
element that is shiny, malleable, good conductor of heat and electricity
metalloid
element that conducts heat and electricity moderately well, and possesses some properties of metals and some properties of nonmetals
noble gas
(also, inert gas) element in group 18
nonmetal
element that appears dull, poor conductor of heat and electricity
period
(also, series) horizontal row of the periodic table
periodic law
properties of the elements are periodic function of their atomic numbers.
periodic table
table of the elements that places elements with similar chemical properties close together
pnictogen
element in group 15
representative element
(also, main-group element) element in columns 1, 2, and 12–18
transition metal
element in columns 3–11
series
(also, period) horizontal row of the period table | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/02%3A_Atoms_Molecules_and_Ions/2.5%3A_The_Periodic_Table.txt |
Learning Objectives
• Define ionic and molecular (covalent) compounds
• Predict the type of compound formed from elements based on their location within the periodic table
• Determine formulas for simple ionic compounds
In ordinary chemical reactions, the nucleus of each atom (and thus the identity of the element) remains unchanged. Electrons, however, can be added to atoms by transfer from other atoms, lost by transfer to other atoms, or shared with other atoms. The transfer and sharing of electrons among atoms govern the chemistry of the elements. During the formation of some compounds, atoms gain or lose electrons, and form electrically charged particles called ions (Figure $1$).
You can use the periodic table to predict whether an atom will form an anion or a cation, and you can often predict the charge of the resulting ion. Atoms of many main-group metals lose enough electrons to leave them with the same number of electrons as an atom of the preceding noble gas. To illustrate, an atom of an alkali metal (group 1) loses one electron and forms a cation with a 1+ charge; an alkaline earth metal (group 2) loses two electrons and forms a cation with a 2+ charge, and so on. For example, a neutral calcium atom, with 20 protons and 20 electrons, readily loses two electrons. This results in a cation with 20 protons, 18 electrons, and a 2+ charge. It has the same number of electrons as atoms of the preceding noble gas, argon, and is symbolized Ca2+. The name of a metal ion is the same as the name of the metal atom from which it forms, so Ca2+ is called a calcium ion.
When atoms of nonmetal elements form ions, they generally gain enough electrons to give them the same number of electrons as an atom of the next noble gas in the periodic table. Atoms of group 17 gain one electron and form anions with a 1− charge; atoms of group 16 gain two electrons and form ions with a 2− charge, and so on. For example, the neutral bromine atom, with 35 protons and 35 electrons, can gain one electron to provide it with 36 electrons. This results in an anion with 35 protons, 36 electrons, and a 1− charge. It has the same number of electrons as atoms of the next noble gas, krypton, and is symbolized Br. (A discussion of the theory supporting the favored status of noble gas electron numbers reflected in these predictive rules for ion formation is provided in a later chapter of this text.)
Note the usefulness of the periodic table in predicting likely ion formation and charge (Figure $2$). Moving from the far left to the right on the periodic table, main-group elements tend to form cations with a charge equal to the group number. That is, group 1 elements form 1+ ions; group 2 elements form 2+ ions, and so on. Moving from the far right to the left on the periodic table, elements often form anions with a negative charge equal to the number of groups moved left from the noble gases. For example, group 17 elements (one group left of the noble gases) form 1− ions; group 16 elements (two groups left) form 2− ions, and so on. This trend can be used as a guide in many cases, but its predictive value decreases when moving toward the center of the periodic table. In fact, transition metals and some other metals often exhibit variable charges that are not predictable by their location in the table. For example, copper can form ions with a 1+ or 2+ charge, and iron can form ions with a 2+ or 3+ charge.
Example $1$: Composition of Ions
An ion found in some compounds used as antiperspirants contains 13 protons and 10 electrons. What is its symbol?
Solution
Because the number of protons remains unchanged when an atom forms an ion, the atomic number of the element must be 13. Knowing this lets us use the periodic table to identify the element as Al (aluminum). The Al atom has lost three electrons and thus has three more positive charges (13) than it has electrons (10). This is the aluminum cation, Al3+.
Exercise $1$
Give the symbol and name for the ion with 34 protons and 36 electrons.
Answer
Se2, the selenide ion
Example $2$: Formation of Ions
Magnesium and nitrogen react to form an ionic compound. Predict which forms an anion, which forms a cation, and the charges of each ion. Write the symbol for each ion and name them.
Solution
Magnesium’s position in the periodic table (group 2) tells us that it is a metal. Metals form positive ions (cations). A magnesium atom must lose two electrons to have the same number electrons as an atom of the previous noble gas, neon. Thus, a magnesium atom will form a cation with two fewer electrons than protons and a charge of 2+. The symbol for the ion is Mg2+, and it is called a magnesium ion.
Nitrogen’s position in the periodic table (group 15) reveals that it is a nonmetal. Nonmetals form negative ions (anions). A nitrogen atom must gain three electrons to have the same number of electrons as an atom of the following noble gas, neon. Thus, a nitrogen atom will form an anion with three more electrons than protons and a charge of 3−. The symbol for the ion is N3−, and it is called a nitride ion.
Exercise $2$
Aluminum and carbon react to form an ionic compound. Predict which forms an anion, which forms a cation, and the charges of each ion. Write the symbol for each ion and name them.
Answer
Al will form a cation with a charge of 3+: Al3+, an aluminum ion. Carbon will form an anion with a charge of 4−: C4−, a carbide ion.
The ions that we have discussed so far are called monatomic ions, that is, they are ions formed from only one atom. We also find many polyatomic ions. These ions, which act as discrete units, are electrically charged molecules (a group of bonded atoms with an overall charge). Some of the more important polyatomic ions are listed in Table $1$. Oxyanions are polyatomic ions that contain one or more oxygen atoms. At this point in your study of chemistry, you should memorize the names, formulas, and charges of the most common polyatomic ions. Because you will use them repeatedly, they will soon become familiar.
Table $1$: Common Polyatomic Ions
Name Formula Related Acid Formula
ammonium $\ce{NH4+}$
hydronium $\ce{H_3O^+}$
oxide $\ce{O^{2-}}$
peroxide $\ce{O_2^{2-}}$
hydroxide $\ce{OH^-}$
acetate $\ce{CH_3COO^-}$ acetic acid $\ce{CH_3COOH}$
cyanide $\ce{CN^-}$ hydrocyanic acid $\ce{HCN}$
azide $\ce{N_3^-}$ hydrazoic acid $\ce{HN_3}$
carbonate $\ce{CO_3^{2-}}$ carbonic acid $\ce{H_2CO_3}$
bicarbonate $\ce{HCO_3^-}$
nitrate $\ce{NO_3^-}$ nitric acid $\ce{HNO_3}$
nitrite $\ce{NO_2^-}$ nitrous acid $\ce{HNO_2}$
sulfate $\ce{SO_4^{2-}}$ sulfuric acid $\ce{H_2SO_4}$
hydrogen sulfate $\ce{HSO_4^-}$
sulfite $\ce{SO_3^{2-}}$ sulfurous acid $\ce{H_2SO_3}$
hydrogen sulfite $\ce{HSO_3^-}$
phosphate $\ce{PO_4^{3-}}$ phosphoric acid $\ce{H_3PO_4}$
hydrogen phosphate $\ce{HPO_4^{2-}}$
dihydrogen phosphate $\ce{H_2PO_4^-}$
perchlorate $\ce{ClO_4^-}$ perchloric acid $\ce{HClO_4}$
chlorate $\ce{ClO_3^-}$ chloric acid $\ce{HClO_3}$
chlorite $\ce{ClO_2^-}$ chlorous acid $\ce{HClO_2}$
hypochlorite $\ce{ClO^-}$ hypochlorous acid $\ce{HClO}$
chromate $\ce{CrO_4^{2-}}$ chromic acid $\ce{H_2CrO_4}$
dichromate $\ce{Cr_2O_7^{2-}}$ dichromic acid $\ce{H_2Cr_2O7}$
permanganate $\ce{MnO_4^-}$ permanganic acid $\ce{HMnO_4}$
Note that there is a system for naming some polyatomic ions; -ate and -ite are suffixes designating polyatomic ions containing more or fewer oxygen atoms. Per- (short for “hyper”) and hypo- (meaning “under”) are prefixes meaning more oxygen atoms than -ate and fewer oxygen atoms than -ite, respectively. For example, perchlorate is $\ce{ClO4-}$, chlorate is $\ce{ClO3-}$, chlorite is $\ce{ClO2-}$ and hypochlorite is ClO. Unfortunately, the number of oxygen atoms corresponding to a given suffix or prefix is not consistent; for example, nitrate is $\ce{NO3-}$ while sulfate is $\ce{SO4^{2-}}$. This will be covered in more detail in the next module on nomenclature.
The nature of the attractive forces that hold atoms or ions together within a compound is the basis for classifying chemical bonding. When electrons are transferred and ions form, ionic bonds result. Ionic bonds are electrostatic forces of attraction, that is, the attractive forces experienced between objects of opposite electrical charge (in this case, cations and anions). When electrons are “shared” and molecules form, covalent bonds result. Covalent bonds are the attractive forces between the positively charged nuclei of the bonded atoms and one or more pairs of electrons that are located between the atoms. Compounds are classified as ionic or molecular (covalent) on the basis of the bonds present in them.
Ionic Compounds
When an element composed of atoms that readily lose electrons (a metal) reacts with an element composed of atoms that readily gain electrons (a nonmetal), a transfer of electrons usually occurs, producing ions. The compound formed by this transfer is stabilized by the electrostatic attractions (ionic bonds) between the ions of opposite charge present in the compound. For example, when each sodium atom in a sample of sodium metal (group 1) gives up one electron to form a sodium cation, Na+, and each chlorine atom in a sample of chlorine gas (group 17) accepts one electron to form a chloride anion, Cl, the resulting compound, NaCl, is composed of sodium ions and chloride ions in the ratio of one Na+ ion for each Cl ion. Similarly, each calcium atom (group 2) can give up two electrons and transfer one to each of two chlorine atoms to form CaCl2, which is composed of Ca2+ and Cl ions in the ratio of one Ca2+ ion to two Cl ions.
A compound that contains ions and is held together by ionic bonds is called an ionic compound. The periodic table can help us recognize many of the compounds that are ionic: When a metal is combined with one or more nonmetals, the compound is usually ionic. This guideline works well for predicting ionic compound formation for most of the compounds typically encountered in an introductory chemistry course. However, it is not always true (for example, aluminum chloride, AlCl3, is not ionic).
You can often recognize ionic compounds because of their properties. Ionic compounds are solids that typically melt at high temperatures and boil at even higher temperatures. For example, sodium chloride melts at 801 °C and boils at 1413 °C. (As a comparison, the molecular compound water melts at 0 °C and boils at 100 °C.) In solid form, an ionic compound is not electrically conductive because its ions are unable to flow (“electricity” is the flow of charged particles). When molten, however, it can conduct electricity because its ions are able to move freely through the liquid (Figure $3$).
In every ionic compound, the total number of positive charges of the cations equals the total number of negative charges of the anions. Thus, ionic compounds are electrically neutral overall, even though they contain positive and negative ions. We can use this observation to help us write the formula of an ionic compound. The formula of an ionic compound must have a ratio of ions such that the numbers of positive and negative charges are equal.
Example $3$: Predicting the Formula of an Ionic Compound
The gemstone sapphire (Figure $4$) is mostly a compound of aluminum and oxygen that contains aluminum cations, Al3+, and oxygen anions, O2−. What is the formula of this compound?
Solution Because the ionic compound must be electrically neutral, it must have the same number of positive and negative charges. Two aluminum ions, each with a charge of 3+, would give us six positive charges, and three oxide ions, each with a charge of 2−, would give us six negative charges. The formula would be Al2O3.
Exercise $3$
Predict the formula of the ionic compound formed between the sodium cation, Na+, and the sulfide anion, S2−.
Answer
Na2S
Many ionic compounds contain polyatomic ions (Table $1$) as the cation, the anion, or both. As with simple ionic compounds, these compounds must also be electrically neutral, so their formulas can be predicted by treating the polyatomic ions as discrete units. We use parentheses in a formula to indicate a group of atoms that behave as a unit. For example, the formula for calcium phosphate, one of the minerals in our bones, is Ca3(PO4)2. This formula indicates that there are three calcium ions (Ca2+) for every two phosphate $\left(\ce{PO4^{3-}}\right)$ groups. The $\ce{PO4^{3-}}$ groups are discrete units, each consisting of one phosphorus atom and four oxygen atoms, and having an overall charge of 3−. The compound is electrically neutral, and its formula shows a total count of three Ca, two P, and eight O atoms.
Example $4$: Predicting the Formula of a Compound with a Polyatomic Anion
Baking powder contains calcium dihydrogen phosphate, an ionic compound composed of the ions Ca2+ and $\ce{H2PO4-}$. What is the formula of this compound?
Solution
The positive and negative charges must balance, and this ionic compound must be electrically neutral. Thus, we must have two negative charges to balance the 2+ charge of the calcium ion. This requires a ratio of one Ca2+ ion to two $\ce{H2PO4-}$ ions. We designate this by enclosing the formula for the dihydrogen phosphate ion in parentheses and adding a subscript 2. The formula is Ca(H2PO4)2.
Exercise $4$
Predict the formula of the ionic compound formed between the lithium ion and the peroxide ion, $\ce{O2^2-}$ (Hint: Use the periodic table to predict the sign and the charge on the lithium ion.)
Answer
Li2O2
Because an ionic compound is not made up of single, discrete molecules, it may not be properly symbolized using a molecular formula. Instead, ionic compounds must be symbolized by a formula indicating the relative numbers of its constituent ions. For compounds containing only monatomic ions (such as NaCl) and for many compounds containing polyatomic ions (such as CaSO4), these formulas are just the empirical formulas introduced earlier in this chapter. However, the formulas for some ionic compounds containing polyatomic ions are not empirical formulas. For example, the ionic compound sodium oxalate is comprised of Na+ and $\ce{C2O4^2-}$ ions combined in a 2:1 ratio, and its formula is written as Na2C2O4. The subscripts in this formula are not the smallest-possible whole numbers, as each can be divided by 2 to yield the empirical formula, NaCO2. This is not the accepted formula for sodium oxalate, however, as it does not accurately represent the compound’s polyatomic anion, $\ce{C2O4^2-}$.
Molecular Compounds
Many compounds do not contain ions but instead consist solely of discrete, neutral molecules. These molecular compounds (covalent compounds) result when atoms share, rather than transfer (gain or lose), electrons. Covalent bonding is an important and extensive concept in chemistry, and it will be treated in considerable detail in a later chapter of this text. We can often identify molecular compounds on the basis of their physical properties. Under normal conditions, molecular compounds often exist as gases, low-boiling liquids, and low-melting solids, although many important exceptions exist.
Whereas ionic compounds are usually formed when a metal and a nonmetal combine, covalent compounds are usually formed by a combination of nonmetals. Thus, the periodic table can help us recognize many of the compounds that are covalent. While we can use the positions of a compound’s elements in the periodic table to predict whether it is ionic or covalent at this point in our study of chemistry, you should be aware that this is a very simplistic approach that does not account for a number of interesting exceptions. Shades of gray exist between ionic and molecular compounds, and you’ll learn more about those later.
Example $5$: Predicting the Type of Bonding in Compounds
Predict whether the following compounds are ionic or molecular:
1. KI, the compound used as a source of iodine in table salt
2. H2O2, the bleach and disinfectant hydrogen peroxide
3. CHCl3, the anesthetic chloroform
4. Li2CO3, a source of lithium in antidepressants
Solution
1. Potassium (group 1) is a metal, and iodine (group 17) is a nonmetal; KI is predicted to be ionic.
2. Hydrogen (group 1) is a nonmetal, and oxygen (group 16) is a nonmetal; H2O2 is predicted to be molecular.
3. Carbon (group 14) is a nonmetal, hydrogen (group 1) is a nonmetal, and chlorine (group 17) is a nonmetal; CHCl3 is predicted to be molecular.
4. Lithium (group 1) is a metal, and carbonate is a polyatomic ion; Li2CO3 is predicted to be ionic.
Exercise $5$
Using the periodic table, predict whether the following compounds are ionic or covalent:
1. SO2
2. CaF2
3. N2H4
4. Al2(SO4)3
Answer a
molecular
Answer b
ionic
Answer c
molecular
Answer d
ionic
Summary
Metals (particularly those in groups 1 and 2) tend to lose the number of electrons that would leave them with the same number of electrons as in the preceding noble gas in the periodic table. By this means, a positively charged ion is formed. Similarly, nonmetals (especially those in groups 16 and 17, and, to a lesser extent, those in Group 15) can gain the number of electrons needed to provide atoms with the same number of electrons as in the next noble gas in the periodic table. Thus, nonmetals tend to form negative ions. Positively charged ions are called cations, and negatively charged ions are called anions. Ions can be either monatomic (containing only one atom) or polyatomic (containing more than one atom).
Compounds that contain ions are called ionic compounds. Ionic compounds generally form from metals and nonmetals. Compounds that do not contain ions, but instead consist of atoms bonded tightly together in molecules (uncharged groups of atoms that behave as a single unit), are called covalent compounds. Covalent compounds usually form from two or more nonmetals.
Glossary
covalent bond
attractive force between the nuclei of a molecule’s atoms and pairs of electrons between the atoms
covalent compound
(also, molecular compound) composed of molecules formed by atoms of two or more different elements
ionic bond
electrostatic forces of attraction between the oppositely charged ions of an ionic compound
ionic compound
compound composed of cations and anions combined in ratios, yielding an electrically neutral substance
molecular compound
(also, covalent compound) composed of molecules formed by atoms of two or more different elements
monatomic ion
ion composed of a single atom
polyatomic ion
ion composed of more than one atom
oxyanion
polyatomic anion composed of a central atom bonded to oxygen atoms | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/02%3A_Atoms_Molecules_and_Ions/2.6%3A_Molecular_and_Ionic_Compounds.txt |
Learning Objectives
• Derive names for common types of inorganic compounds using a systematic approach.
• Describe how to name binary covalent compounds including acids and oxyacids.
Nomenclature, a collection of rules for naming things, is important in science and in many other situations. This module describes an approach that is used to name simple ionic and molecular compounds, such as NaCl, CaCO3, and N2O4. The simplest of these are binary compounds, those containing only two elements, but we will also consider how to name ionic compounds containing polyatomic ions, and one specific, very important class of compounds known as acids (subsequent chapters in this text will focus on these compounds in great detail). We will limit our attention here to inorganic compounds, compounds that are composed principally of elements other than carbon, and will follow the nomenclature guidelines proposed by IUPAC. The rules for organic compounds, in which carbon is the principle element, will be treated in a later chapter on organic chemistry.
Ionic Compounds
To name an inorganic compound, we need to consider the answers to several questions. First, is the compound ionic or molecular? If the compound is ionic, does the metal form ions of only one type (fixed charge) or more than one type (variable charge)? Are the ions monatomic or polyatomic? If the compound is molecular, does it contain hydrogen? If so, does it also contain oxygen? From the answers we derive, we place the compound in an appropriate category and then name it accordingly. We will begin with the nomenclature rules for ionic compounds.
Compounds Containing Only Monatomic Ions
The name of a binary compound containing monatomic ions consists of the name of the cation (the name of the metal) followed by the name of the anion (the name of the nonmetallic element with its ending replaced by the suffix –ide). Some examples are given in Table \(2\).
Table \(1\): Names of Some Ionic Compounds
NaCl, sodium chloride Na2O, sodium oxide
KBr, potassium bromide CdS, cadmium sulfide
CaI2, calcium iodide Mg3N2, magnesium nitride
CsF, cesium fluoride Ca3P2, calcium phosphide
LiCl, lithium chloride Al4C3, aluminum carbide
Compounds Containing Polyatomic Ions
Compounds containing polyatomic ions are named similarly to those containing only monatomic ions, except there is no need to change to an –ide ending, since the suffix is already present in the name of the anion. Examples are shown in Table \(2\).
CL, ammonium chloride, C a S O subscript 4 calcium sulfate, and M g subscript 3 ( P O subscript 4 ) subscript 2 magnesium phosphate." data-quail-id="54" data-mt-width="1246">
Table \(2\): Names of Some Polyatomic Ionic Compounds
KC2H3O2, potassium acetate (NH4)Cl, ammonium chloride
NaHCO3, sodium bicarbonate CaSO4, calcium sulfate
Al2(CO3)3, aluminum carbonate Mg3(PO4)2, magnesium phosphate
Ionic Compounds in Your Cabinets
Every day you encounter and use a large number of ionic compounds. Some of these compounds, where they are found, and what they are used for are listed in Table \(3\). Look at the label or ingredients list on the various products that you use during the next few days, and see if you run into any of those in this table, or find other ionic compounds that you could now name or write as a formula.
Table \(3\): Everyday Ionic Compounds
Ionic Compound Name Use
NaCl sodium chloride ordinary table salt
KI potassium iodide added to “iodized” salt for thyroid health
NaF sodium fluoride ingredient in toothpaste
NaHCO3 sodium bicarbonate baking soda; used in cooking (and in antacids)
Na2CO3 sodium carbonate washing soda; used in cleaning agents
NaOCl sodium hypochlorite active ingredient in household bleach
CaCO3 calcium carbonate ingredient in antacids
Mg(OH)2 magnesium hydroxide ingredient in antacids
Al(OH)3 aluminum hydroxide ingredient in antacids
NaOH sodium hydroxide lye; used as drain cleaner
K3PO4 potassium phosphate food additive (many purposes)
MgSO4 magnesium sulfate added to purified water
Na2HPO4 sodium hydrogen phosphate anti-caking agent; used in powdered products
Na2SO3 sodium sulfite preservative
Compounds Containing a Metal Ion with a Variable Charge
Most of the transition metals can form two or more cations with different charges. Compounds of these metals with nonmetals are named with the same method as compounds in the first category, except the charge of the metal ion is specified by a Roman numeral in parentheses after the name of the metal. The charge of the metal ion is determined from the formula of the compound and the charge of the anion. For example, consider binary ionic compounds of iron and chlorine. Iron typically exhibits a charge of either 2+ or 3+, and the two corresponding compound formulas are FeCl2 and FeCl3. The simplest name, “iron chloride,” will, in this case, be ambiguous, as it does not distinguish between these two compounds. In cases like this, the charge of the metal ion is included as a Roman numeral in parentheses immediately following the metal name. These two compounds are then unambiguously named iron(II) chloride and iron(III) chloride, respectively. Other examples are provided in Table \(4\).
Table \(4\): Names of Some Transition Metal Ionic Compounds
Transition Metal Ionic Compound Name
FeCl3 iron(III) chloride
Hg2O mercury(I) oxide
HgO mercury(II) oxide
Cu3(PO4)2 copper(II) phosphate
Out-of-date nomenclature used the suffixes –ic and –ous to designate metals with higher and lower charges, respectively: Iron(III) chloride, FeCl3, was previously called ferric chloride, and iron(II) chloride, FeCl2, was known as ferrous chloride. Though this naming convention has been largely abandoned by the scientific community, it remains in use by some segments of industry. For example, you may see the words stannous fluoride on a tube of toothpaste. This represents the formula SnF2, which is more properly named tin(II) fluoride. The other fluoride of tin is SnF4, which was previously called stannic fluoride but is now named tin(IV) fluoride.
Example \(1\): Naming Ionic Compounds
Name the following ionic compounds, which contain a metal that can have more than one ionic charge:
1. Fe2S3
2. CuSe
3. GaN
4. CrCl3
5. Ti2(SO4)3
Solution
The anions in these compounds have a fixed negative charge (S2−, Se2, N3−, Cl, and \(\ce{SO4^2-}\)), and the compounds must be neutral. Because the total number of positive charges in each compound must equal the total number of negative charges, the positive ions must be Fe3+, Cu2+, Ga3+, Cr3+, and Ti3+. These charges are used in the names of the metal ions:
1. iron(III) sulfide
2. copper(II) selenide
3. gallium(III) nitride
4. chromium(III) chloride
5. titanium(III) sulfate
Exercise \(1\)
Write the formulas of the following ionic compounds:
1. chromium(III) phosphide
2. mercury(II) sulfide
3. manganese(II) phosphate
4. copper(I) oxide
5. chromium(VI) fluoride
Answer a
CrP
Answer b
HgS
Answer c
Mn3(PO4)2
Answer d
Cu2O
Answer e
CrF6
Erin Brokovich and Chromium Contamination
In the early 1990s, legal file clerk Erin Brockovich (Figure \(2\)) discovered a high rate of serious illnesses in the small town of Hinckley, California. Her investigation eventually linked the illnesses to groundwater contaminated by Cr(VI) used by Pacific Gas & Electric (PG&E) to fight corrosion in a nearby natural gas pipeline. As dramatized in the film Erin Brokovich (for which Julia Roberts won an Oscar), Erin and lawyer Edward Masry sued PG&E for contaminating the water near Hinckley in 1993. The settlement they won in 1996—\$333 million—was the largest amount ever awarded for a direct-action lawsuit in the US at that time.
Chromium compounds are widely used in industry, such as for chrome plating, in dye-making, as preservatives, and to prevent corrosion in cooling tower water, as occurred near Hinckley. In the environment, chromium exists primarily in either the Cr(III) or Cr(VI) forms. Cr(III), an ingredient of many vitamin and nutritional supplements, forms compounds that are not very soluble in water, and it has low toxicity. Cr(VI), on the other hand, is much more toxic and forms compounds that are reasonably soluble in water. Exposure to small amounts of Cr(VI) can lead to damage of the respiratory, gastrointestinal, and immune systems, as well as the kidneys, liver, blood, and skin.
Despite cleanup efforts, Cr(VI) groundwater contamination remains a problem in Hinckley and other locations across the globe. A 2010 study by the Environmental Working Group found that of 35 US cities tested, 31 had higher levels of Cr(VI) in their tap water than the public health goal of 0.02 parts per billion set by the California Environmental Protection Agency.
Molecular (Covalent) Compounds
The bonding characteristics of inorganic molecular compounds are different from ionic compounds, and they are named using a different system as well. The charges of cations and anions dictate their ratios in ionic compounds, so specifying the names of the ions provides sufficient information to determine chemical formulas. However, because covalent bonding allows for significant variation in the combination ratios of the atoms in a molecule, the names for molecular compounds must explicitly identify these ratios.
Compounds Composed of Two Elements
When two nonmetallic elements form a molecular compound, several combination ratios are often possible. For example, carbon and oxygen can form the compounds CO and CO2. Since these are different substances with different properties, they cannot both have the same name (they cannot both be called carbon oxide). To deal with this situation, we use a naming method that is somewhat similar to that used for ionic compounds, but with added prefixes to specify the numbers of atoms of each element. The name of the more metallic element (the one farther to the left and/or bottom of the periodic table) is first, followed by the name of the more nonmetallic element (the one farther to the right and/or top) with its ending changed to the suffix –ide. The numbers of atoms of each element are designated by the Greek prefixes shown in Table \(5\).
Table \(5\): Nomenclature Prefixes
Number Prefix Number Prefix
1 (sometimes omitted) mono- 6 hexa-
2 di- 7 hepta-
3 tri- 8 octa-
4 tetra- 9 nona-
5 penta- 10 deca-
When only one atom of the first element is present, the prefix mono- is usually deleted from that part. Thus, \(\ce{CO}\) is named carbon monoxide, and \(\ce{CO2}\) is called carbon dioxide. When two vowels are adjacent, the a in the Greek prefix is usually dropped. Some other examples are shown in Table \(6\).
Table \(6\): Names of Some Molecular Compounds Composed of Two Elements
Compound Name Compound Name
SO2 sulfur dioxide BCl3 boron trichloride
SO3 sulfur trioxide SF6 sulfur hexafluoride
NO2 nitrogen dioxide PF5 phosphorus pentafluoride
N2O4 dinitrogen tetroxide P4O10 tetraphosphorus decaoxide
N2O5 dinitrogen pentoxide IF7 iodine heptafluoride
There are a few common names that you will encounter as you continue your study of chemistry. For example, although NO is often called nitric oxide, its proper name is nitrogen monoxide. Similarly, N2O is known as nitrous oxide even though our rules would specify the name dinitrogen monoxide. (And H2O is usually called water, not dihydrogen monoxide.) You should commit to memory the common names of compounds as you encounter them.
Example \(2\): Naming Covalent Compounds
Name the following covalent compounds:
1. SF6
2. N2O3
3. Cl2O7
4. P4O6
Solution
Because these compounds consist solely of nonmetals, we use prefixes to designate the number of atoms of each element:
1. sulfur hexafluoride
2. dinitrogen trioxide
3. dichlorine heptoxide
4. tetraphosphorus hexoxide
Exercise \(2\)
Write the formulas for the following compounds:
1. phosphorus pentachloride
2. dinitrogen monoxide
3. iodine heptafluoride
4. carbon tetrachloride
Answer a
PCl5
Answer b
N2O
Answer c
IF7
Answer d
CCl4
Binary Acids
Some compounds containing hydrogen are members of an important class of substances known as acids. The chemistry of these compounds is explored in more detail in later chapters of this text, but for now, it will suffice to note that many acids release hydrogen ions, H+, when dissolved in water. To denote this distinct chemical property, a mixture of water with an acid is given a name derived from the compound’s name. If the compound is a binary acid (comprised of hydrogen and one other nonmetallic element):
1. The word “hydrogen” is changed to the prefix hydro-
2. The other nonmetallic element name is modified by adding the suffix -ic
3. The word “acid” is added as a second word
For example, when the gas \(\ce{HCl}\) (hydrogen chloride) is dissolved in water, the solution is called hydrochloric acid. Several other examples of this nomenclature are shown in Table \(7\).
Table \(7\): Names of Some Simple Acids
Name of Gas Name of Acid
HF(g), hydrogen fluoride HF(aq), hydrofluoric acid
HCl(g), hydrogen chloride HCl(aq), hydrochloric acid
HBr(g), hydrogen bromide HBr(aq), hydrobromic acid
HI(g), hydrogen iodide HI(aq), hydroiodic acid
H2S(g), hydrogen sulfide H2S(aq), hydrosulfuric acid
Oxyacids
Many compounds containing three or more elements (such as organic compounds or coordination compounds) are subject to specialized nomenclature rules that you will learn later. However, we will briefly discuss the important compounds known as oxyacids, compounds that contain hydrogen, oxygen, and at least one other element, and are bonded in such a way as to impart acidic properties to the compound (you will learn the details of this in a later chapter). Typical oxyacids consist of hydrogen combined with a polyatomic, oxygen-containing ion. To name oxyacids:
1. Omit “hydrogen”
2. Start with the root name of the anion
3. Replace –ate with –ic, or –ite with –ous
4. Add “acid”
For example, consider H2CO3 (which you might be tempted to call “hydrogen carbonate”). To name this correctly, “hydrogen” is omitted; the –ate of carbonate is replace with –ic; and acid is added—so its name is carbonic acid. Other examples are given in Table \(8\). There are some exceptions to the general naming method (e.g., H2SO4 is called sulfuric acid, not sulfic acid, and H2SO3 is sulfurous, not sulfous, acid).
Table \(8\): Names of Common Oxyacids
Formula Anion Name Acid Name
HC2H3O2 acetate acetic acid
HNO3 nitrate nitric acid
HNO2 nitrite nitrous acid
HClO4 perchlorate perchloric acid
H2CO3 carbonate carbonic acid
H2SO4 sulfate sulfuric acid
H2SO3 sulfite sulfurous acid
H3PO4 phosphate phosphoric acid
Summary
Chemists use nomenclature rules to clearly name compounds. Ionic and molecular compounds are named using somewhat-different methods. Binary ionic compounds typically consist of a metal and a nonmetal. The name of the metal is written first, followed by the name of the nonmetal with its ending changed to –ide. For example, K2O is called potassium oxide. If the metal can form ions with different charges, a Roman numeral in parentheses follows the name of the metal to specify its charge. Thus, FeCl2 is iron(II) chloride and FeCl3 is iron(III) chloride. Some compounds contain polyatomic ions; the names of common polyatomic ions should be memorized. Molecular compounds can form compounds with different ratios of their elements, so prefixes are used to specify the numbers of atoms of each element in a molecule of the compound. Examples include SF6, sulfur hexafluoride, and N2O4, dinitrogen tetroxide. Acids are an important class of compounds containing hydrogen and having special nomenclature rules. Binary acids are named using the prefix hydro-, changing the –ide suffix to –ic, and adding “acid;” HCl is hydrochloric acid. Oxyacids are named by changing the ending of the anion (-ate to –ic, and -ite to -ous) and adding “acid;” H2CO3 is carbonic acid.
Glossary
binary acid
compound that contains hydrogen and one other element, bonded in a way that imparts acidic properties to the compound (ability to release H+ ions when dissolved in water)
binary compound
compound containing two different elements.
oxyacid
compound that contains hydrogen, oxygen, and one other element, bonded in a way that imparts acidic properties to the compound (ability to release H+ ions when dissolved in water)
nomenclature
system of rules for naming objects of interest | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/02%3A_Atoms_Molecules_and_Ions/2.7%3A_Chemical_Nomenclature.txt |
2.1: Early Ideas in Atomic Theory
In the following drawing, the green spheres represent atoms of a certain element. The purple spheres represent atoms of another element. If the spheres of different elements touch, they are part of a single unit of a compound. The following chemical change represented by these spheres may violate one of the ideas of Dalton’s atomic theory. Which one?
The starting materials consist of one green sphere and two purple spheres. The products consist of two green spheres and two purple spheres. This violates Dalton’s postulate that that atoms are not created during a chemical change, but are merely redistributed.
Which postulate of Dalton’s theory is consistent with the following observation concerning the weights of reactants and products? When 100 grams of solid calcium carbonate is heated, 44 grams of carbon dioxide and 56 grams of calcium oxide are produced.
Identify the postulate of Dalton’s theory that is violated by the following observations: 59.95% of one sample of titanium dioxide is titanium; 60.10% of a different sample of titanium dioxide is titanium.
This statement violates Dalton’s fourth postulate: In a given compound, the numbers of atoms of each type (and thus also the percentage) always have the same ratio.
Samples of compound X, Y, and Z are analyzed, with results shown here.
Compound Description Mass of Carbon Mass of Hydrogen
X clear, colorless, liquid with strong odor 1.776 g 0.148 g
Y clear, colorless, liquid with strong odor 1.974 g 0.329 g
Z clear, colorless, liquid with strong odor 7.812 g 0.651 g
Do these data provide example(s) of the law of definite proportions, the law of multiple proportions, neither, or both? What do these data tell you about compounds X, Y, and Z?
Exercises
1. The existence of isotopes violates one of the original ideas of Dalton’s atomic theory. Which one?
2. How are electrons and protons similar? How are they different?
3. How are protons and neutrons similar? How are they different?
4. Predict and test the behavior of α particles fired at a “plum pudding” model atom.
1. Predict the paths taken by α particles that are fired at atoms with a Thomson’s plum pudding model structure. Explain why you expect the α particles to take these paths.
2. If α particles of higher energy than those in (a) are fired at plum pudding atoms, predict how their paths will differ from the lower-energy α particle paths. Explain your reasoning.
3. Now test your predictions from (a) and (b). Open the Rutherford Scattering simulation and select the “Plum Pudding Atom” tab. Set “Alpha Particles Energy” to “min,” and select “show traces.” Click on the gun to start firing α particles. Does this match your prediction from (a)? If not, explain why the actual path would be that shown in the simulation. Hit the pause button, or “Reset All.” Set “Alpha Particles Energy” to “max,” and start firing α particles. Does this match your prediction from (b)? If not, explain the effect of increased energy on the actual paths as shown in the simulation.
5. Predict and test the behavior of α particles fired at a Rutherford atom model.
1. (a) Predict the paths taken by α particles that are fired at atoms with a Rutherford atom model structure. Explain why you expect the α particles to take these paths.
2. (b) If α particles of higher energy than those in (a) are fired at Rutherford atoms, predict how their paths will differ from the lower-energy α particle paths. Explain your reasoning.
3. (c) Predict how the paths taken by the α particles will differ if they are fired at Rutherford atoms of elements other than gold. What factor do you expect to cause this difference in paths, and why?
4. (d) Now test your predictions from (a), (b), and (c). Open the Rutherford Scattering simulation and select the “Rutherford Atom” tab. Due to the scale of the simulation, it is best to start with a small nucleus, so select “20” for both protons and neutrons, “min” for energy, show traces, and then start firing α particles. Does this match your prediction from (a)? If not, explain why the actual path would be that shown in the simulation. Pause or reset, set energy to “max,” and start firing α particles. Does this match your prediction from (b)? If not, explain the effect of increased energy on the actual path as shown in the simulation. Pause or reset, select “40” for both protons and neutrons, “min” for energy, show traces, and fire away. Does this match your prediction from (c)? If not, explain why the actual path would be that shown in the simulation. Repeat this with larger numbers of protons and neutrons. What generalization can you make regarding the type of atom and effect on the path of α particles? Be clear and specific.
Solutions
1 Dalton originally thought that all atoms of a particular element had identical properties, including mass. Thus, the concept of isotopes, in which an element has different masses, was a violation of the original idea. To account for the existence of isotopes, the second postulate of his atomic theory was modified to state that atoms of the same element must have identical chemical properties.
2 Both are subatomic particles that reside in an atom’s nucleus. Both have approximately the same mass. Protons are positively charged, whereas neutrons are uncharged.
3 Both are subatomic particles that reside in an atom’s nucleus. Both have approximately the same mass. Protons are positively charged, whereas neutrons are uncharged.
4. (a) The plum pudding model indicates that the positive charge is spread uniformly throughout the atom, so we expect the α particles to (perhaps) be slowed somewhat by the positive-positive repulsion, but to follow straight-line paths (i.e., not to be deflected) as they pass through the atoms. (b) Higher-energy α particles will be traveling faster (and perhaps slowed less) and will also follow straight-line paths through the atoms. (c) The α particles followed straight-line paths through the plum pudding atom. There was no apparent slowing of the α particles as they passed through the atoms.
5. (a) The Rutherford atom has a small, positively charged nucleus, so most α particles will pass through empty space far from the nucleus and be undeflected. Those α particles that pass near the nucleus will be deflected from their paths due to positive-positive repulsion. The more directly toward the nucleus the α particles are headed, the larger the deflection angle will be. (b) Higher-energy α particles that pass near the nucleus will still undergo deflection, but the faster they travel, the less the expected angle of deflection. (c) If the nucleus is smaller, the positive charge is smaller and the expected deflections are smaller—both in terms of how closely the α particles pass by the nucleus undeflected and the angle of deflection. If the nucleus is larger, the positive charge is larger and the expected deflections are larger—more α particles will be deflected, and the deflection angles will be larger. (d) The paths followed by the α particles match the predictions from (a), (b), and (c).
2.3: Atomic Structure and Symbolism
In what way are isotopes of a given element always different? In what way(s) are they always the same?
Write the symbol for each of the following ions:
1. (a) the ion with a 1+ charge, atomic number 55, and mass number 133
2. (b) the ion with 54 electrons, 53 protons, and 74 neutrons
3. (c) the ion with atomic number 15, mass number 31, and a 3− charge
4. (d) the ion with 24 electrons, 30 neutrons, and a 3+ charge
(a) 133Cs+; (b) 127I; (c) 31P3−; (d) 57Co3+
Write the symbol for each of the following ions:
1. (a) the ion with a 3+ charge, 28 electrons, and a mass number of 71
2. (b) the ion with 36 electrons, 35 protons, and 45 neutrons
3. (c) the ion with 86 electrons, 142 neutrons, and a 4+ charge
4. (d) the ion with a 2+ charge, atomic number 38, and mass number 87
Open the Build an Atom simulation and click on the Atom icon.
1. (a) Pick any one of the first 10 elements that you would like to build and state its symbol.
2. (b) Drag protons, neutrons, and electrons onto the atom template to make an atom of your element. State the numbers of protons, neutrons, and electrons in your atom, as well as the net charge and mass number.
3. (c) Click on “Net Charge” and “Mass Number,” check your answers to (b), and correct, if needed.
4. (d) Predict whether your atom will be stable or unstable. State your reasoning.
5. (e) Check the “Stable/Unstable” box. Was your answer to (d) correct? If not, first predict what you can do to make a stable atom of your element, and then do it and see if it works. Explain your reasoning.
(a) Carbon-12, 12C; (b) This atom contains six protons and six neutrons. There are six electrons in a neutral 12C atom. The net charge of such a neutral atom is zero, and the mass number is 12. (c) The preceding answers are correct. (d) The atom will be stable since C-12 is a stable isotope of carbon. (e) The preceding answer is correct. Other answers for this exercise are possible if a different element of isotope is chosen.
Open the Build an Atom simulation
(a) Drag protons, neutrons, and electrons onto the atom template to make a neutral atom of Oxygen-16 and give the isotope symbol for this atom.
(b) Now add two more electrons to make an ion and give the symbol for the ion you have created.
Open the Build an Atom simulation
(a) Drag protons, neutrons, and electrons onto the atom template to make a neutral atom of Lithium-6 and give the isotope symbol for this atom.
(b) Now remove one electron to make an ion and give the symbol for the ion you have created.
(a) Lithium-6 contains three protons, three neutrons, and three electrons. The isotope symbol is 6Li or \(\ce{^6_3Li}\). (b) 6Li+ or \(\ce{^6_3Li+}\)
Determine the number of protons, neutrons, and electrons in the following isotopes that are used in medical diagnoses:
(a) atomic number 9, mass number 18, charge of 1−
(b) atomic number 43, mass number 99, charge of 7+
(c) atomic number 53, atomic mass number 131, charge of 1−
(d) atomic number 81, atomic mass number 201, charge of 1+
(e) Name the elements in parts (a), (b), (c), and (d).
The following are properties of isotopes of two elements that are essential in our diet. Determine the number of protons, neutrons and electrons in each and name them.
(a) atomic number 26, mass number 58, charge of 2+
(b) atomic number 53, mass number 127, charge of 1−
(a) Iron, 26 protons, 24 electrons, and 32 neutrons; (b) iodine, 53 protons, 54 electrons, and 74 neutrons
Give the number of protons, electrons, and neutrons in neutral atoms of each of the following isotopes:
(a) \(\ce{^{10}_5B}\)
(b) \(\ce{^{199}_{80}Hg}\)
(c) \(\ce{^{63}_{29}Cu}\)
(d) \(\ce{^{13}_6C}\)
(e) \(\ce{^{77}_{34}Se}\)
Give the number of protons, electrons, and neutrons in neutral atoms of each of the following isotopes:
(a) \(\ce{^7_3Li}\)
(b) \(\ce{^{125}_{52}Te}\)
(c) \(\ce{^{109}_{47}Ag}\)
(d) \(\ce{^{15}_7N}\)
(e) \(\ce{^{31}_{15}P}\)
(a) 3 protons, 3 electrons, 4 neutrons; (b) 52 protons, 52 electrons, 73 neutrons; (c) 47 protons, 47 electrons, 62 neutrons; (d) 7 protons, 7 electrons, 8 neutrons; (e) 15 protons, 15 electrons, 16 neutrons
Click on the site and select the “Mix Isotopes” tab, hide the “Percent Composition” and “Average Atomic Mass” boxes, and then select the element boron.
(a) Write the symbols of the isotopes of boron that are shown as naturally occurring in significant amounts.
(b) Predict the relative amounts (percentages) of these boron isotopes found in nature. Explain the reasoning behind your choice.
(c) Add isotopes to the black box to make a mixture that matches your prediction in (b). You may drag isotopes from their bins or click on “More” and then move the sliders to the appropriate amounts.
(d) Reveal the “Percent Composition” and “Average Atomic Mass” boxes. How well does your mixture match with your prediction? If necessary, adjust the isotope amounts to match your prediction.
(e) Select “Nature’s” mix of isotopes and compare it to your prediction. How well does your prediction compare with the naturally occurring mixture? Explain. If necessary, adjust your amounts to make them match “Nature’s” amounts as closely as possible.
Repeat Exercise using an element that has three naturally occurring isotopes.
Let us use neon as an example. Since there are three isotopes, there is no way to be sure to accurately predict the abundances to make the total of 20.18 amu average atomic mass. Let us guess that the abundances are 9% Ne-22, 91% Ne-20, and only a trace of Ne-21. The average mass would be 20.18 amu. Checking the nature’s mix of isotopes shows that the abundances are 90.48% Ne-20, 9.25% Ne-22, and 0.27% Ne-21, so our guessed amounts have to be slightly adjusted.
An element has the following natural abundances and isotopic masses: 90.92% abundance with 19.99 amu, 0.26% abundance with 20.99 amu, and 8.82% abundance with 21.99 amu. Calculate the average atomic mass of this element.
Average atomic masses listed by IUPAC are based on a study of experimental results. Bromine has two isotopes 79Br and 81Br, whose masses (78.9183 and 80.9163 amu) and abundances (50.69% and 49.31%) were determined in earlier experiments. Calculate the average atomic mass of bromine based on these experiments.
79.904 amu
Variations in average atomic mass may be observed for elements obtained from different sources. Lithium provides an example of this. The isotopic composition of lithium from naturally occurring minerals is 7.5% 6Li and 92.5% 7Li, which have masses of 6.01512 amu and 7.01600 amu, respectively. A commercial source of lithium, recycled from a military source, was 3.75% 6Li (and the rest 7Li). Calculate the average atomic mass values for each of these two sources.
The average atomic masses of some elements may vary, depending upon the sources of their ores. Naturally occurring boron consists of two isotopes with accurately known masses (10B, 10.0129 amu and 11B, 11.0931 amu). The actual atomic mass of boron can vary from 10.807 to 10.819, depending on whether the mineral source is from Turkey or the United States. Calculate the percent abundances leading to the two values of the average atomic masses of boron from these two countries.
Turkey source: 0.2649 (of 10.0129 amu isotope); US source: 0.2537 (of 10.0129 amu isotope)
The 18O:16O abundance ratio in some meteorites is greater than that used to calculate the average atomic mass of oxygen on earth. Is the average mass of an oxygen atom in these meteorites greater than, less than, or equal to that of a terrestrial oxygen atom?
2.4: Chemical Formulas
Explain why the symbol for an atom of the element oxygen and the formula for a molecule of oxygen differ.
The symbol for the element oxygen, O, represents both the element and one atom of oxygen. A molecule of oxygen, O2, contains two oxygen atoms; the subscript 2 in the formula must be used to distinguish the diatomic molecule from two single oxygen atoms.
Explain why the symbol for the element sulfur and the formula for a molecule of sulfur differ.
Write the molecular and empirical formulas of the following compounds:
(a)
(b)
(c)
(d)
(a) molecular CO2, empirical CO2; (b) molecular C2H2, empirical CH; (c) molecular C2H4, empirical CH2; (d) molecular H2SO4, empirical H2SO4
Write the molecular and empirical formulas of the following compounds:
(a)
(b)
(c)
(d)
Determine the empirical formulas for the following compounds:
1. (a) caffeine, C8H10N4O2
2. (b) fructose, C12H22O11
3. (c) hydrogen peroxide, H2O2
4. (d) glucose, C6H12O6
5. (e) ascorbic acid (vitamin C), C6H8O6
(a) C4H5N2O; (b) C12H22O11; (c) HO; (d) CH2O; (e) C3H4O3
Determine the empirical formulas for the following compounds:
1. (a) acetic acid, C2H4O2
2. (b) citric acid, C6H8O7
3. (c) hydrazine, N2H4
4. (d) nicotine, C10H14N2
5. (e) butane, C4H10
Write the empirical formulas for the following compounds:
(a)
(b)
(a) CH2O; (b) C2H4O
Open the Build a Molecule simulation and select the “Larger Molecules” tab. Select an appropriate atoms “Kit” to build a molecule with two carbon and six hydrogen atoms. Drag atoms into the space above the “Kit” to make a molecule. A name will appear when you have made an actual molecule that exists (even if it is not the one you want). You can use the scissors tool to separate atoms if you would like to change the connections. Click on “3D” to see the molecule, and look at both the space-filling and ball-and-stick possibilities.
1. (a) Draw the structural formula of this molecule and state its name.
2. (b) Can you arrange these atoms in any way to make a different compound?
Use the Build a Molecule simulation to repeat Exercise, but build a molecule with two carbons, six hydrogens, and one oxygen.
1. (a) Draw the structural formula of this molecule and state its name.
2. (b) Can you arrange these atoms to make a different molecule? If so, draw its structural formula and state its name.
3. (c) How are the molecules drawn in (a) and (b) the same? How do they differ? What are they called (the type of relationship between these molecules, not their names).
(a) ethanol
(b) methoxymethane, more commonly known as dimethyl ether
(c) These molecules have the same chemical composition (types and number of atoms) but different chemical structures. They are structural isomers.
Use the Build a Molecule simulation to repeat Exercise, but build a molecule with three carbons, seven hydrogens, and one chlorine.
1. Draw the structural formula of this molecule and state its name.
2. Can you arrange these atoms to make a different molecule? If so, draw its structural formula and state its name.
3. How are the molecules drawn in (a) and (b) the same? How do they differ? What are they called (the type of relationship between these molecules, not their names)?
2.5: The Periodic Table
Using the periodic table, classify each of the following elements as a metal or a nonmetal, and then further classify each as a main-group (representative) element, transition metal, or inner transition metal:
1. uranium
2. bromine
3. strontium
4. neon
5. gold
6. americium
7. rhodium
8. sulfur
9. carbon
10. potassium
(a) metal, inner transition metal; (b) nonmetal, representative element; (c) metal, representative element; (d) nonmetal, representative element; (e) metal, transition metal; (f) metal, inner transition metal; (g) metal, transition metal; (h) nonmetal, representative element; (i) nonmetal, representative element; (j) metal, representative element
Using the periodic table, classify each of the following elements as a metal or a nonmetal, and then further classify each as a main-group (representative) element, transition metal, or inner transition metal:
1. (a) cobalt
2. (b) europium
3. (c) iodine
4. (d) indium
5. (e) lithium
6. (f) oxygen
7. (g) cadmium
8. (h) terbium
9. (i) rhenium
Using the periodic table, identify the lightest member of each of the following groups:
1. (a) noble gases
2. (b) alkaline earth metals
3. (c) alkali metals
4. (d) chalcogens
(a) He; (b) Be; (c) Li; (d) O
Using the periodic table, identify the heaviest member of each of the following groups:
1. (a) alkali metals
2. (b) chalcogens
3. (c) noble gases
4. (d) alkaline earth metals
1. Use the periodic table to give the name and symbol for each of the following elements:
2. (a) the noble gas in the same period as germanium
3. (b) the alkaline earth metal in the same period as selenium
4. (c) the halogen in the same period as lithium
5. (d) the chalcogen in the same period as cadmium
(a) krypton, Kr; (b) calcium, Ca; (c) fluorine, F; (d) tellurium, Te
Use the periodic table to give the name and symbol for each of the following elements:
1. (a) the halogen in the same period as the alkali metal with 11 protons
2. (b) the alkaline earth metal in the same period with the neutral noble gas with 18 electrons
3. (c) the noble gas in the same row as an isotope with 30 neutrons and 25 protons
4. (d) the noble gas in the same period as gold
Write a symbol for each of the following neutral isotopes. Include the atomic number and mass number for each.
1. (a) the alkali metal with 11 protons and a mass number of 23
2. (b) the noble gas element with and 75 neutrons in its nucleus and 54 electrons in the neutral atom
3. (c) the isotope with 33 protons and 40 neutrons in its nucleus
4. (d) the alkaline earth metal with 88 electrons and 138 neutrons
(a) \(\ce{^{23}_{11}Na}\); (b) \(\ce{^{129}_{54}Xe}\); (c) \(\ce{^{73}_{33}As}\); (d) \(\ce{^{226}_{88}Ra}\)
Write a symbol for each of the following neutral isotopes. Include the atomic number and mass number for each.
1. (a) the chalcogen with a mass number of 125
2. (b) the halogen whose longest-lived isotope is radioactive
3. (c) the noble gas, used in lighting, with 10 electrons and 10 neutrons
4. (d) the lightest alkali metal with three neutrons
2.6: Molecular and Ionic Compounds
Using the periodic table, predict whether the following chlorides are ionic or covalent: KCl, NCl3, ICl, MgCl2, PCl5, and CCl4.
Ionic: KCl, MgCl2; Covalent: NCl3, ICl, PCl5, CCl4
Using the periodic table, predict whether the following chlorides are ionic or covalent: SiCl4, PCl3, CaCl2, CsCl, CuCl2, and CrCl3.
For each of the following compounds, state whether it is ionic or covalent. If it is ionic, write the symbols for the ions involved:
1. (a) NF3
2. (b) BaO,
3. (c) (NH4)2CO3
4. (d) Sr(H2PO4)2
5. (e) IBr
6. (f) Na2O
(a) covalent; (b) ionic, Ba2+, O2−; (c) ionic, \(\ce{NH4+}\), \(\ce{CO3^2-}\); (d) ionic, Sr2+, \(\ce{H2PO4-}\); (e) covalent; (f) ionic, Na+, O2−
For each of the following compounds, state whether it is ionic or covalent, and if it is ionic, write the symbols for the ions involved:
1. (a) KClO4
2. (b) MgC2H3O2
3. (c) H2S
4. (d) Ag2S
5. (e) N2Cl4
6. (f) Co(NO3)2
For each of the following pairs of ions, write the symbol for the formula of the compound they will form:
1. (a) Ca2+, S2−
2. (b) \(\ce{NH4+}\), \(\ce{SO4^2-}\)
3. (c) Al3+, Br
4. (d) Na+, \(\ce{HPO4^2-}\)
5. (e) Mg2+, \(\ce{PO4^3-}\)
(a) CaS; (b) (NH4)2CO3; (c) AlBr3; (d) Na2HPO4; (e) Mg3 (PO4)2
For each of the following pairs of ions, write the symbol for the formula of the compound they will form:
1. (a) K+, O2−
2. (b) \(\ce{NH4+}\), \(\ce{PO4^3-}\)
3. (c) Al3+, O2−
4. (d) Na+, \(\ce{CO3^2-}\)
5. (e) Ba2+, \(\ce{PO4^3-}\)
2.7: Chemical Nomenclature
Name the following compounds:
1. (a) CsCl
2. (b) BaO
3. (c) K2S
4. (d) BeCl2
5. (e) HBr
6. (f) AlF3
(a) cesium chloride; (b) barium oxide; (c) potassium sulfide; (d) beryllium chloride; (e) hydrogen bromide; (f) aluminum fluoride
Name the following compounds:
1. (a) NaF
2. (b) Rb2O
3. (c) BCl3
4. (d) H2Se
5. (e) P4O6
6. (f) ICl3
Write the formulas of the following compounds:
1. (a) rubidium bromide
2. (b) magnesium selenide
3. (c) sodium oxide
4. (d) calcium chloride
5. (e) hydrogen fluoride
6. (f) gallium phosphide
7. (g) aluminum bromide
8. (h) ammonium sulfate
(a) RbBr; (b) MgSe; (c) Na2O; (d) CaCl2; (e) HF; (f) GaP; (g) AlBr3; (h) (NH4)2SO4
Write the formulas of the following compounds:
1. (a) lithium carbonate
2. (b) sodium perchlorate
3. (c) barium hydroxide
4. (d) ammonium carbonate
5. (e) sulfuric acid
6. (f) calcium acetate
7. (g) magnesium phosphate
8. (h) sodium sulfite
Write the formulas of the following compounds:
1. (a) chlorine dioxide
2. (b) dinitrogen tetraoxide
3. (c) potassium phosphide
4. (d) silver(I) sulfide
5. (e) aluminum nitride
6. (f) silicon dioxide
(a) ClO2; (b) N2O4; (c) K3P; (d) Ag2S; (e) AlN; (f) SiO2
Write the formulas of the following compounds:
1. (a) barium chloride
2. (b) magnesium nitride
3. (c) sulfur dioxide
4. (d) nitrogen trichloride
5. (e) dinitrogen trioxide
6. (f) tin(IV) chloride
Each of the following compounds contains a metal that can exhibit more than one ionic charge. Name these compounds:
1. (a) Cr2O3
2. (b) FeCl2
3. (c) CrO3
4. (d) TiCl4
5. (e) CoO
6. (f) MoS2
(a) chromium(III) oxide; (b) iron(II) chloride; (c) chromium(VI) oxide; (d) titanium(IV) chloride; (e) cobalt(II) oxide; (f) molybdenum(IV) sulfide
Each of the following compounds contains a metal that can exhibit more than one ionic charge. Name these compounds:
1. (a) NiCO3
2. (b) MoO3
3. (c) Co(NO3)2
4. (d) V2O5
5. (e) MnO2
6. (f) Fe2O3
The following ionic compounds are found in common household products. Write the formulas for each compound:
1. (a) potassium phosphate
2. (b) copper(II) sulfate
3. (c) calcium chloride
4. (d) titanium dioxide
5. (e) ammonium nitrate
6. (f) sodium bisulfate (the common name for sodium hydrogen sulfate)
(a) K3PO4; (b) CuSO4; (c) CaCl2; (d) TiO2; (e) NH4NO3; (f) NaHSO4
The following ionic compounds are found in common household products. Name each of the compounds:
1. (a) Ca(H2PO4)2
2. (b) FeSO4
3. (c) CaCO3
4. (d) MgO
5. (e) NaNO2
6. (f) KI
What are the IUPAC names of the following compounds?
1. (a) manganese dioxide
2. (b) mercurous chloride (Hg2Cl2)
3. (c) ferric nitrate [Fe(NO3)3]
4. (d) titanium tetrachloride
5. (e) cupric bromide (CuBr2)
(a) manganese(IV) oxide; (b) mercury(I) chloride; (c) iron(III) nitrate; (d) titanium(IV) chloride; (e) copper(II) bromide | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/02%3A_Atoms_Molecules_and_Ions/2.E%3A_Atoms_Molecules_and_Ions_%28Exercises%29.txt |
The study of chemistry must at some point extend to the molecular level, for the physical and chemical properties of a substance are ultimately explained in terms of the structure and bonding of molecules. This module introduces some basic facts and principles that are needed for a discussion of organic molecules.
• 3.1: Electromagnetic Energy
Light and other forms of electromagnetic radiation move through a vacuum with a constant speed, c. This radiation shows wavelike behavior, which can be characterized by a frequency, ν, and a wavelength, λ, such that c = λν. Light is an example of a travelling wave. Other important wave phenomena include standing waves, periodic oscillations, and vibrations. Standing waves exhibit quantization, since their wavelengths are limited to discrete integer multiples of some characteristic lengths.
• 3.2: The Bohr Model
Bohr incorporated Planck’s and Einstein’s quantization ideas into a model of the hydrogen atom that resolved the paradox of atom stability and discrete spectra. The Bohr model of the hydrogen atom explains the connection between the quantization of photons and the quantized emission from atoms. Bohr described the hydrogen atom in terms of an electron moving in a circular orbit about a nucleus. He postulated that the electron was restricted to certain orbits characterized by discrete energies.
• 3.3: Development of Quantum Theory
Macroscopic objects act as particles. Microscopic objects (such as electrons) have properties of both a particle and a wave. but their exact trajectories cannot be determined. The quantum mechanical model of atoms describes the 3D position of the electron in a probabilistic manner according to a mathematical function called a wavefunction, often denoted as ψ. The squared magnitude of the wavefunction describes the distribution of the probability of finding the electron in a particular region in
• 3.4: Electronic Structure of Atoms (Electron Configurations)
The relative energy of the subshells determine the order in which atomic orbitals are filled. Electron configurations and orbital diagrams can be determined by applying the Pauli exclusion principle (no two electrons can have the same set of four quantum numbers) and Hund’s rule (whenever possible, electrons retain unpaired spins in degenerate orbitals). Electrons in the outermost orbitals, called valence electrons, are responsible for most of the chemical behavior of elements.
• 3.5: Periodic Variations in Element Properties
Electron configurations allow us to understand many periodic trends. Covalent radius increases as we move down a group because the n level (orbital size) increases. Covalent radius mostly decreases as we move left to right across a period because the effective nuclear charge experienced by the electrons increases, and the electrons are pulled in tighter to the nucleus. Anionic radii are larger than the parent atom, while cationic radii are smaller.
• 3.E: Electronic Structure and Periodic Properties (Exercises)
These are homework exercises to accompany the Textmap created for "Chemistry" by OpenStax. Complementary General Chemistry question banks can be found for other Textmaps and can be accessed here. In addition to these publicly available questions, access to private problems bank for use in exams and homework is available to faculty only on an individual basis; please contact Delmar Larsen for an account with access permission.
03: Electronic Structure and Periodic Properties
Learning Objectives
• Explain the basic behavior of waves, including traveling waves and standing waves
• Describe the wave nature of light
• Use appropriate equations to calculate related light-wave properties such as period, frequency, wavelength, and energy
• Distinguish between line and continuous emission spectra
• Describe the particle nature of light
The nature of light has been a subject of inquiry since antiquity. In the seventeenth century, Isaac Newton performed experiments with lenses and prisms and was able to demonstrate that white light consists of the individual colors of the rainbow combined together. Newton explained his optics findings in terms of a "corpuscular" view of light, in which light was composed of streams of extremely tiny particles travelling at high speeds according to Newton's laws of motion. Others in the seventeenth century, such as Christiaan Huygens, had shown that optical phenomena such as reflection and refraction could be equally well explained in terms of light as waves travelling at high speed through a medium called "luminiferous aether" that was thought to permeate all space. Early in the nineteenth century, Thomas Young demonstrated that light passing through narrow, closely spaced slits produced interference patterns that could not be explained in terms of Newtonian particles but could be easily explained in terms of waves. Later in the nineteenth century, after James Clerk Maxwell developed his theory of electromagnetic radiation and showed that light was the visible part of a vast spectrum of electromagnetic waves, the particle view of light became thoroughly discredited. By the end of the nineteenth century, scientists viewed the physical universe as roughly comprising two separate domains: matter composed of particles moving according to Newton's laws of motion, and electromagnetic radiation consisting of waves governed by Maxwell's equations. Today, these domains are referred to as classical mechanics and classical electrodynamics (or classical electromagnetism). Although there were a few physical phenomena that could not be explained within this framework, scientists at that time were so confident of the overall soundness of this framework that they viewed these aberrations as puzzling paradoxes that would ultimately be resolved somehow within this framework. As we shall see, these paradoxes led to a contemporary framework that intimately connects particles and waves at a fundamental level called wave-particle duality, which has superseded the classical view.
Visible light and other forms of electromagnetic radiation play important roles in chemistry, since they can be used to infer the energies of electrons within atoms and molecules. Much of modern technology is based on electromagnetic radiation. For example, radio waves from a mobile phone, X-rays used by dentists, the energy used to cook food in your microwave, the radiant heat from red-hot objects, and the light from your television screen are forms of electromagnetic radiation that all exhibit wavelike behavior.
Waves
A wave is an oscillation or periodic movement that can transport energy from one point in space to another. Common examples of waves are all around us. Shaking the end of a rope transfers energy from your hand to the other end of the rope, dropping a pebble into a pond causes waves to ripple outward along the water's surface, and the expansion of air that accompanies a lightning strike generates sound waves (thunder) that can travel outward for several miles. In each of these cases, kinetic energy is transferred through matter (the rope, water, or air) while the matter remains essentially in place. An insightful example of a wave occurs in sports stadiums when fans in a narrow region of seats rise simultaneously and stand with their arms raised up for a few seconds before sitting down again while the fans in neighboring sections likewise stand up and sit down in sequence. While this wave can quickly encircle a large stadium in a few seconds, none of the fans actually travel with the wave-they all stay in or above their seats.
Waves need not be restricted to travel through matter. As Maxwell showed, electromagnetic waves consist of an electric field oscillating in step with a perpendicular magnetic field, both of which are perpendicular to the direction of travel. These waves can travel through a vacuum at a constant speed of 2.998 × 108 m/s, the speed of light (denoted by c).
All waves, including forms of electromagnetic radiation, are characterized by, a wavelength (denoted by λ, the lowercase Greek letter lambda), a frequency (denoted by ν, the lowercase Greek letter nu), and an amplitude. As can be seen in Figure $1$, the wavelength is the distance between two consecutive peaks or troughs in a wave (measured in meters in the SI system). Electromagnetic waves have wavelengths that fall within an enormous range-wavelengths of kilometers (103 m) to picometers (10−12 m) have been observed. The frequency is the number of wave cycles that pass a specified point in space in a specified amount of time (in the SI system, this is measured in seconds). A cycle corresponds to one complete wavelength. The unit for frequency, expressed as cycles per second [s−1], is the hertz (Hz). Common multiples of this unit are megahertz, (1 MHz = 1 × 106 Hz) and gigahertz (1 GHz = 1 × 109 Hz). The amplitude corresponds to the magnitude of the wave's displacement and so, in Figure, this corresponds to one-half the height between the peaks and troughs. The amplitude is related to the intensity of the wave, which for light is the brightness, and for sound is the loudness.
The product of a wave's wavelength (λ) and its frequency (ν), λν, is the speed of the wave. Thus, for electromagnetic radiation in a vacuum:
$c=\mathrm{2.998×10^8\,ms^{−1}}=λν \label{6.2.1}$
Wavelength and frequency are inversely proportional: As the wavelength increases, the frequency decreases. The inverse proportionality is illustrated in Figure $2$. This figure also shows the electromagnetic spectrum, the range of all types of electromagnetic radiation. Each of the various colors of visible light has specific frequencies and wavelengths associated with them, and you can see that visible light makes up only a small portion of the electromagnetic spectrum. Because the technologies developed to work in various parts of the electromagnetic spectrum are different, for reasons of convenience and historical legacies, different units are typically used for different parts of the spectrum. For example, radio waves are usually specified as frequencies (typically in units of MHz), while the visible region is usually specified in wavelengths (typically in units of nm or angstroms).
Example $1$: Determining the Frequency and Wavelength of Radiation
A sodium streetlight gives off yellow light that has a wavelength of 589 nm (1 nm = 1 × 10−9 m). What is the frequency of this light?
Solution
We can rearrange the Equation \ref{6.2.1} to solve for the frequency:
$\nu=\dfrac{c}{λ} \nonumber$
Since c is expressed in meters per second, we must also convert 589 nm to meters.
$\nu=\mathrm{\left(\dfrac{2.998×10^8\:\cancel{m}s^{−1}}{589\cancel{nm}}\right)\left(\dfrac{1×10^9\cancel{nm}}{1\cancel{m}}\right)=5.09×10^{14}\,s^{−1}} \nonumber$
Exercise $1$
One of the frequencies used to transmit and receive cellular telephone signals in the United States is 850 MHz. What is the wavelength in meters of these radio waves?
Answer
0.353 m = 35.3 cm
Wireless Communication
Many valuable technologies operate in the radio (3 kHz-300 GHz) frequency region of the electromagnetic spectrum. At the low frequency (low energy, long wavelength) end of this region are AM (amplitude modulation) radio signals (540-2830 kHz) that can travel long distances. FM (frequency modulation) radio signals are used at higher frequencies (87.5-108.0 MHz). In AM radio, the information is transmitted by varying the amplitude of the wave (Figure $5$). In FM radio, by contrast, the amplitude is constant and the instantaneous frequency varies.
Other technologies also operate in the radio-wave portion of the electromagnetic spectrum. For example, 4G cellular telephone signals are approximately 880 MHz, while Global Positioning System (GPS) signals operate at 1.228 and 1.575 GHz, local area wireless technology (Wi-Fi) networks operate at 2.4 to 5 GHz, and highway toll sensors operate at 5.8 GHz. The frequencies associated with these applications are convenient because such waves tend not to be absorbed much by common building materials.
One particularly characteristic phenomenon of waves results when two or more waves come into contact: They interfere with each other. Figure $5$ shows the interference patterns that arise when light passes through narrow slits closely spaced about a wavelength apart. The fringe patterns produced depend on the wavelength, with the fringes being more closely spaced for shorter wavelength light passing through a given set of slits. When the light passes through the two slits, each slit effectively acts as a new source, resulting in two closely spaced waves coming into contact at the detector (the camera in this case). The dark regions in Figure $5$ correspond to regions where the peaks for the wave from one slit happen to coincide with the troughs for the wave from the other slit (destructive interference), while the brightest regions correspond to the regions where the peaks for the two waves (or their two troughs) happen to coincide (constructive interference). Likewise, when two stones are tossed close together into a pond, interference patterns are visible in the interactions between the waves produced by the stones. Such interference patterns cannot be explained by particles moving according to the laws of classical mechanics.
Dorothy Hodgkin
Because the wavelengths of X-rays (10-10,000 picometers [pm]) are comparable to the size of atoms, X-rays can be used to determine the structure of molecules. When a beam of X-rays is passed through molecules packed together in a crystal, the X-rays collide with the electrons and scatter. Constructive and destructive interference of these scattered X-rays creates a specific diffraction pattern. Calculating backward from this pattern, the positions of each of the atoms in the molecule can be determined very precisely. One of the pioneers who helped create this technology was Dorothy Crowfoot Hodgkin.
She was born in Cairo, Egypt, in 1910, where her British parents were studying archeology. Even as a young girl, she was fascinated with minerals and crystals. When she was a student at Oxford University, she began researching how X-ray crystallography could be used to determine the structure of biomolecules. She invented new techniques that allowed her and her students to determine the structures of vitamin B12, penicillin, and many other important molecules. Diabetes, a disease that affects 382 million people worldwide, involves the hormone insulin. Hodgkin began studying the structure of insulin in 1934, but it required several decades of advances in the field before she finally reported the structure in 1969. Understanding the structure has led to better understanding of the disease and treatment options.
Not all waves are travelling waves. Standing waves (also known as stationary waves) remain constrained within some region of space. As we shall see, standing waves play an important role in our understanding of the electronic structure of atoms and molecules. The simplest example of a standing wave is a one-dimensional wave associated with a vibrating string that is held fixed at its two end points. Figure $6$ shows the four lowest-energy standing waves (the fundamental wave and the lowest three harmonics) for a vibrating string at a particular amplitude. Although the string's motion lies mostly within a plane, the wave itself is considered to be one dimensional, since it lies along the length of the string. The motion of string segments in a direction perpendicular to the string length generates the waves and so the amplitude of the waves is visible as the maximum displacement of the curves seen in Figure $6$. The key observation from the figure is that only those waves having an integer number, n, of half-wavelengths between the end points can form. A system with fixed end points such as this restricts the number and type of the possible waveforms. This is an example of quantization, in which only discrete values from a more general set of continuous values of some property are observed. Another important observation is that the harmonic waves (those waves displaying more than one-half wavelength) all have one or more points between the two end points that are not in motion. These special points are nodes. The energies of the standing waves with a given amplitude in a vibrating string increase with the number of half-wavelengths n. Since the number of nodes is n – 1, the energy can also be said to depend on the number of nodes, generally increasing as the number of nodes increases.
An example of two-dimensional standing waves is shown in Figure $7$ which shows the vibrational patterns on a flat surface. Although the vibrational amplitudes cannot be seen like they could in the vibrating string, the nodes have been made visible by sprinkling the drum surface with a powder that collects on the areas of the surface that have minimal displacement. For one-dimensional standing waves, the nodes were points on the line, but for two-dimensional standing waves, the nodes are lines on the surface (for three-dimensional standing waves, the nodes are two-dimensional surfaces within the three-dimensional volume). Because of the circular symmetry of the drum surface, its boundary conditions (the drum surface being tightly constrained to the circumference of the drum) result in two types of nodes: radial nodes that sweep out all angles at constant radii and, thus, are seen as circles about the center, and angular nodes that sweep out all radii at constant angles and, thus, are seen as lines passing through the center. The upper left image in Figure $7$ shows two radial nodes, while the image in the lower right shows the vibrational pattern associated with three radial nodes and two angular nodes.
Blackbody Radiation and the Ultraviolet Catastrophe
The last few decades of the nineteenth century witnessed intense research activity in commercializing newly discovered electric lighting. This required obtaining a better understanding of the distributions of light emitted from various sources being considered. Artificial lighting is usually designed to mimic natural sunlight within the limitations of the underlying technology. Such lighting consists of a range of broadly distributed frequencies that form a continuous spectrum. Figure $8$ shows the wavelength distribution for sunlight. The most intense radiation is in the visible region, with the intensity dropping off rapidly for shorter wavelength ultraviolet (UV) light, and more slowly for longer wavelength infrared (IR) light.
In Figure $8$, the solar distribution is compared to a representative distribution, called a blackbody spectrum, that corresponds to a temperature of 5250 °C. The blackbody spectrum matches the solar spectrum quite well. A blackbody is a convenient, ideal emitter that approximates the behavior of many materials when heated. It is “ideal” in the same sense that an ideal gas is a convenient, simple representation of real gases that works well, provided that the pressure is not too high nor the temperature too low. A good approximation of a blackbody that can be used to observe blackbody radiation is a metal oven that can be heated to very high temperatures. The oven has a small hole allowing for the light being emitted within the oven to be observed with a spectrometer so that the wavelengths and their intensities can be measured. Figure $8$ shows the resulting curves for some representative temperatures. Each distribution depends only on a single parameter: the temperature. The maxima in the blackbody curves, λmax, shift to shorter wavelengths as the temperature increases, reflecting the observation that metals being heated to high temperatures begin to glow a darker red that becomes brighter as the temperature increases, eventually becoming white hot at very high temperatures as the intensities of all of the visible wavelengths become appreciable. This common observation was at the heart of the first paradox that showed the fundamental limitations of classical physics that we will examine.
Physicists derived mathematical expressions for the blackbody curves using well-accepted concepts from the theories of classical mechanics and classical electromagnetism. The theoretical expressions as functions of temperature fit the observed experimental blackbody curves well at longer wavelengths, but showed significant discrepancies at shorter wavelengths. Not only did the theoretical curves not show a peak, they absurdly showed the intensity becoming infinitely large as the wavelength became smaller, which would imply that everyday objects at room temperature should be emitting large amounts of UV light. This became known as the “ultraviolet catastrophe” because no one could find any problems with the theoretical treatment that could lead to such unrealistic short-wavelength behavior. Finally, around 1900, Max Planck derived a theoretical expression for blackbody radiation that fit the experimental observations exactly (within experimental error). Planck developed his theoretical treatment by extending the earlier work that had been based on the premise that the atoms composing the oven vibrated at increasing frequencies (or decreasing wavelengths) as the temperature increased, with these vibrations being the source of the emitted electromagnetic radiation. But where the earlier treatments had allowed the vibrating atoms to have any energy values obtained from a continuous set of energies (perfectly reasonable, according to classical physics), Planck found that by restricting the vibrational energies to discrete values for each frequency, he could derive an expression for blackbody radiation that correctly had the intensity dropping rapidly for the short wavelengths in the UV region.
$E=nhν,\:n=1,2,3,\:. . . \nonumber$
The quantity h is a constant now known as Planck's constant, in his honor. Although Planck was pleased he had resolved the blackbody radiation paradox, he was disturbed that to do so, he needed to assume the vibrating atoms required quantized energies, which he was unable to explain. The value of Planck's constant is very small, 6.626 × 10−34 joule seconds (J s), which helps explain why energy quantization had not been observed previously in macroscopic phenomena.
The Photoelectric Effect
The next paradox in the classical theory to be resolved concerned the photoelectric effect (Figure $10$). It had been observed that electrons could be ejected from the clean surface of a metal when light having a frequency greater than some threshold frequency was shone on it. Surprisingly, the kinetic energy of the ejected electrons did not depend on the brightness of the light, but increased with increasing frequency of the light. Since the electrons in the metal had a certain amount of binding energy keeping them there, the incident light needed to have more energy to free the electrons. According to classical wave theory, a wave's energy depends on its intensity (which depends on its amplitude), not its frequency. One part of these observations was that the number of electrons ejected within in a given time period was seen to increase as the brightness increased. In 1905, Albert Einstein was able to resolve the paradox by incorporating Planck's quantization findings into the discredited particle view of light (Einstein actually won his Nobel prize for this work, and not for his theories of relativity for which he is most famous).
Einstein argued that the quantized energies that Planck had postulated in his treatment of blackbody radiation could be applied to the light in the photoelectric effect so that the light striking the metal surface should not be viewed as a wave, but instead as a stream of particles (later called photons) whose energy depended on their frequency, according to Planck's formula, E = (or, in terms of wavelength using c = νλ, $E=\dfrac{hc}{λ}$). Electrons were ejected when hit by photons having sufficient energy (a frequency greater than the threshold). The greater the frequency, the greater the kinetic energy imparted to the escaping electrons by the collisions. Einstein also argued that the light intensity did not depend on the amplitude of the incoming wave, but instead corresponded to the number of photons striking the surface within a given time period. This explains why the number of ejected electrons increased with increasing brightness, since the greater the number of incoming photons, the greater the likelihood that they would collide with some of the electrons.
With Einstein's findings, the nature of light took on a new air of mystery. Although many light phenomena could be explained either in terms of waves or particles, certain phenomena, such as the interference patterns obtained when light passed through a double slit, were completely contrary to a particle view of light, while other phenomena, such as the photoelectric effect, were completely contrary to a wave view of light. Somehow, at a deep fundamental level still not fully understood, light is both wavelike and particle-like. This is known as wave-particle duality.
Example $2$: Calculating the Energy of Radiation
When we see light from a neon sign, we are observing radiation from excited neon atoms. If this radiation has a wavelength of 640 nm, what is the energy of the photon being emitted?
Solution
We use the part of Planck's equation that includes the wavelength, λ, and convert units of nanometers to meters so that the units of λ and c are the same.
\begin{align*} E&=\dfrac{hc}{λ} \[4pt] &=\mathrm{\dfrac{(6.626×10^{−34}\:J\cancel{s})(2.998×10^{8}\:m\cancel{s}^{−1})}{(640\cancel{nm})\left(\dfrac{1\:m}{10^9\cancel{nm}}\right)}}\[4pt] &=\mathrm{3.10×10^{−19}\:J} \end{align*} \nonumber
Exercise $2$
The microwaves in an oven are of a specific frequency that will heat the water molecules contained in food. (This is why most plastics and glass do not become hot in a microwave oven-they do not contain water molecules.) This frequency is about 3 × 109 Hz. What is the energy of one photon in these microwaves?
Answer
2 × 10−24 J
Example $3$: Photoelectric Effect
Identify which of the following statements are false and, where necessary, change the italicized word or phrase to make them true, consistent with Einstein's explanation of the photoelectric effect.
1. Increasing the brightness of incoming light increases the kinetic energy of the ejected electrons.
2. Increasing the wavelength of incoming light increases the kinetic energy of the ejected electrons.
3. Increasing the brightness of incoming light increases the number of ejected electrons.
4. Increasing the frequency of incoming light can increase the number of ejected electrons.
Solution
1. False. Increasing the brightness of incoming light has no effect on the kinetic energy of the ejected electrons. Only energy, not the number or amplitude, of the photons influences the kinetic energy of the electrons.
2. False. Increasing the frequency of incoming light increases the kinetic energy of the ejected electrons. Frequency is proportional to energy and inversely proportional to wavelength. Frequencies above the threshold value transfer the excess energy into the kinetic energy of the electrons.
3. True. Because the number of collisions with photons increases with brighter light, the number of ejected electrons increases.
4. True with regard to the threshold energy binding the electrons to the metal. Below this threshold, electrons are not emitted and above it they are. Once over the threshold value, further increasing the frequency does not increase the number of ejected electrons
Exercise $3$
Calculate the threshold energy in kJ/mol of electrons in aluminum, given that the lowest frequency photon for which the photoelectric effect is observed is $9.87 \times 10^{14}\; Hz$.
Answer
$3.94 \: kJ/mol$
Line Spectra
Another paradox within the classical electromagnetic theory that scientists in the late nineteenth century struggled with concerned the light emitted from atoms and molecules. When solids, liquids, or condensed gases are heated sufficiently, they radiate some of the excess energy as light. Photons produced in this manner have a range of energies, and thereby produce a continuous spectrum in which an unbroken series of wavelengths is present. Most of the light generated from stars (including our sun) is produced in this fashion. You can see all the visible wavelengths of light present in sunlight by using a prism to separate them. As can be seen in Figure $8$, sunlight also contains UV light (shorter wavelengths) and IR light (longer wavelengths) that can be detected using instruments but that are invisible to the human eye. Incandescent (glowing) solids such as tungsten filaments in incandescent lights also give off light that contains all wavelengths of visible light. These continuous spectra can often be approximated by blackbody radiation curves at some appropriate temperature, such as those shown in Figure $9$.
In contrast to continuous spectra, light can also occur as discrete or line spectra having very narrow line widths interspersed throughout the spectral regions such as those shown in Figure $5$. Exciting a gas at low partial pressure using an electrical current, or heating it, will produce line spectra. Fluorescent light bulbs and neon signs operate in this way (Figure $11$). Each element displays its own characteristic set of lines, as do molecules, although their spectra are generally much more complicated.
Each emission line consists of a single wavelength of light, which implies that the light emitted by a gas consists of a set of discrete energies. For example, when an electric discharge passes through a tube containing hydrogen gas at low pressure, the H2 molecules are broken apart into separate H atoms, and we see a blue-pink color. Passing the light through a prism produces a line spectrum, indicating that this light is composed of photons of four visible wavelengths, as shown in Figure $12$.
The origin of discrete spectra in atoms and molecules was extremely puzzling to scientists in the late nineteenth century, since according to classical electromagnetic theory, only continuous spectra should be observed. Even more puzzling, in 1885, Johann Balmer was able to derive an empirical equation that related the four visible wavelengths of light emitted by hydrogen atoms to whole integers. That equation is the following one, in which k is a constant:
$\dfrac{1}{λ}=k\left(\dfrac{1}{4}−\dfrac{1}{n^2}\right),\:n=3,\:4,\:5,\:6 \nonumber$
Other discrete lines for the hydrogen atom were found in the UV and IR regions. Johannes Rydberg generalized Balmer's work and developed an empirical formula that predicted all of hydrogen's emission lines, not just those restricted to the visible range, where, n1 and n2 are integers, n1 < n2, and $R_∞$ is the Rydberg constant (1.097 × 107 m−1).
$\dfrac{1}{λ}=R_∞\left(\dfrac{1}{n^2_1}−\dfrac{1}{n^2_2}\right) \nonumber$
Even in the late nineteenth century, spectroscopy was a very precise science, and so the wavelengths of hydrogen were measured to very high accuracy, which implied that the Rydberg constant could be determined very precisely as well. That such a simple formula as the Rydberg formula could account for such precise measurements seemed astounding at the time, but it was the eventual explanation for emission spectra by Neils Bohr in 1913 that ultimately convinced scientists to abandon classical physics and spurred the development of modern quantum mechanics.
Summary
Light and other forms of electromagnetic radiation move through a vacuum with a constant speed, c, of 2.998 × 108 m s−1. This radiation shows wavelike behavior, which can be characterized by a frequency, ν, and a wavelength, λ, such that c = λν. Light is an example of a travelling wave. Other important wave phenomena include standing waves, periodic oscillations, and vibrations. Standing waves exhibit quantization, since their wavelengths are limited to discrete integer multiples of some characteristic lengths. Electromagnetic radiation that passes through two closely spaced narrow slits having dimensions roughly similar to the wavelength will show an interference pattern that is a result of constructive and destructive interference of the waves. Electromagnetic radiation also demonstrates properties of particles called photons. The energy of a photon is related to the frequency (or alternatively, the wavelength) of the radiation as E = (or $E=\dfrac{hc}{λ}$), where h is Planck's constant. That light demonstrates both wavelike and particle-like behavior is known as wave-particle duality. All forms of electromagnetic radiation share these properties, although various forms including X-rays, visible light, microwaves, and radio waves interact differently with matter and have very different practical applications. Electromagnetic radiation can be generated by exciting matter to higher energies, such as by heating it. The emitted light can be either continuous (incandescent sources like the sun) or discrete (from specific types of excited atoms). Continuous spectra often have distributions that can be approximated as blackbody radiation at some appropriate temperature. The line spectrum of hydrogen can be obtained by passing the light from an electrified tube of hydrogen gas through a prism. This line spectrum was simple enough that an empirical formula called the Rydberg formula could be derived from the spectrum. Three historically important paradoxes from the late 19th and early 20th centuries that could not be explained within the existing framework of classical mechanics and classical electromagnetism were the blackbody problem, the photoelectric effect, and the discrete spectra of atoms. The resolution of these paradoxes ultimately led to quantum theories that superseded the classical theories.
Key Equations
• c = λν
• $E=hν=\dfrac{hc}{λ}$, where h = 6.626 × 10−34 J s
• $\dfrac{1}{λ}=R_∞\left(\dfrac{1}{n^2_1}−\dfrac{1}{n^2_2}\right)$
Glossary
amplitude
extent of the displacement caused by a wave (for sinusoidal waves, it is one-half the difference from the peak height to the trough depth, and the intensity is proportional to the square of the amplitude)
blackbody
idealized perfect absorber of all incident electromagnetic radiation; such bodies emit electromagnetic radiation in characteristic continuous spectra called blackbody radiation
continuous spectrum
electromagnetic radiation given off in an unbroken series of wavelengths (e.g., white light from the sun)
electromagnetic radiation
energy transmitted by waves that have an electric-field component and a magnetic-field component
electromagnetic spectrum
range of energies that electromagnetic radiation can comprise, including radio, microwaves, infrared, visible, ultraviolet, X-rays, and gamma rays; since electromagnetic radiation energy is proportional to the frequency and inversely proportional to the wavelength, the spectrum can also be specified by ranges of frequencies or wavelengths
frequency ($\nu$)
number of wave cycles (peaks or troughs) that pass a specified point in space per unit time
hertz (Hz)
the unit of frequency, which is the number of cycles per second, s−1
intensity
property of wave-propagated energy related to the amplitude of the wave, such as brightness of light or loudness of sound
interference pattern
pattern typically consisting of alternating bright and dark fringes; it results from constructive and destructive interference of waves
line spectrum
electromagnetic radiation emitted at discrete wavelengths by a specific atom (or atoms) in an excited state
node
any point of a standing wave with zero amplitude
photon
smallest possible packet of electromagnetic radiation, a particle of light
quantization
occurring only in specific discrete values, not continuous
standing wave
(also, stationary wave) localized wave phenomenon characterized by discrete wavelengths determined by the boundary conditions used to generate the waves; standing waves are inherently quantized
wave
oscillation that can transport energy from one point to another in space
wavelength (λ)
distance between two consecutive peaks or troughs in a wave
wave-particle duality
term used to describe the fact that elementary particles including matter exhibit properties of both particles (including localized position, momentum) and waves (including nonlocalization, wavelength, frequency) | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/03%3A_Electronic_Structure_and_Periodic_Properties/3.1%3A_Electromagnetic_Energy.txt |
Learning Objectives
• Describe the Bohr model of the hydrogen atom
• Use the Rydberg equation to calculate energies of light emitted or absorbed by hydrogen atoms
Following the work of Ernest Rutherford and his colleagues in the early twentieth century, the picture of atoms consisting of tiny dense nuclei surrounded by lighter and even tinier electrons continually moving about the nucleus was well established. This picture was called the planetary model, since it pictured the atom as a miniature “solar system” with the electrons orbiting the nucleus like planets orbiting the sun. The simplest atom is hydrogen, consisting of a single proton as the nucleus about which a single electron moves. The electrostatic force attracting the electron to the proton depends only on the distance between the two particles.
$F_{gravity} = G \dfrac{ m_1 m_2}{r^2} \nonumber$
with
• $G$ is a gravitational constant
• $m_1$ and $m_2$ are the masses of particle 1 and 2, respectively
• $r$ is the distance between the two particles
The electrostatic force has the same form as the gravitational force between two mass particles except that the electrostatic force depends on the magnitudes of the charges on the particles (+1 for the proton and −1 for the electron) instead of the magnitudes of the particle masses that govern the gravitational force.
$F_{electrostatic} = k \dfrac{ m_1 m_2}{r^2} \nonumber$
with
• $k$ is a constant
• $m_1$ and $m_2$ are the masses of particle 1 and 2, respectively
• $r$ is the distance between the two particles
Since forces can be derived from potentials, it is convenient to work with potentials instead, since they are forms of energy. The electrostatic potential is also called the Coulomb potential. Because the electrostatic potential has the same form as the gravitational potential, according to classical mechanics, the equations of motion should be similar, with the electron moving around the nucleus in circular or elliptical orbits (hence the label “planetary” model of the atom). Potentials of the form V(r) that depend only on the radial distance $r$ are known as central potentials. Central potentials have spherical symmetry, and so rather than specifying the position of the electron in the usual Cartesian coordinates (x, y, z), it is more convenient to use polar spherical coordinates centered at the nucleus, consisting of a linear coordinate r and two angular coordinates, usually specified by the Greek letters theta (θ) and phi (Φ). These coordinates are similar to the ones used in GPS devices and most smart phones that track positions on our (nearly) spherical earth, with the two angular coordinates specified by the latitude and longitude, and the linear coordinate specified by sea-level elevation. Because of the spherical symmetry of central potentials, the energy and angular momentum of the classical hydrogen atom are constants, and the orbits are constrained to lie in a plane like the planets orbiting the sun. This classical mechanics description of the atom is incomplete, however, since an electron moving in an elliptical orbit would be accelerating (by changing direction) and, according to classical electromagnetism, it should continuously emit electromagnetic radiation. This loss in orbital energy should result in the electron’s orbit getting continually smaller until it spirals into the nucleus, implying that atoms are inherently unstable.
In 1913, Niels Bohr attempted to resolve the atomic paradox by ignoring classical electromagnetism’s prediction that the orbiting electron in hydrogen would continuously emit light. Instead, he incorporated into the classical mechanics description of the atom Planck’s ideas of quantization and Einstein’s finding that light consists of photons whose energy is proportional to their frequency. Bohr assumed that the electron orbiting the nucleus would not normally emit any radiation (the stationary state hypothesis), but it would emit or absorb a photon if it moved to a different orbit. The energy absorbed or emitted would reflect differences in the orbital energies according to this equation:
$|ΔE|=|E_f−E_i|=h u=\dfrac{hc}{\lambda} \label{6.3.1}$
In this equation, h is Planck’s constant and Ei and Ef are the initial and final orbital energies, respectively. The absolute value of the energy difference is used, since frequencies and wavelengths are always positive. Instead of allowing for continuous values for the angular momentum, energy, and orbit radius, Bohr assumed that only discrete values for these could occur (actually, quantizing any one of these would imply that the other two are also quantized). Bohr’s expression for the quantized energies is:
$E_n=−\dfrac{k}{n^2} \label{6.3.2}$
with $n=1,2,3, ...$
In this expression, $k$ is a constant comprising fundamental constants such as the electron mass and charge and Planck’s constant. Inserting the expression for the orbit energies into the equation for $ΔE$ gives
$\color{red} ΔE=k \left(\dfrac{1}{n^2_1}−\dfrac{1}{n_2^2}\right)=\dfrac{hc}{\lambda} \label{6.3.3}$
or
$\dfrac{1}{\lambda}=\dfrac{k}{hc} \left(\dfrac{1}{n^2_1}−\dfrac{1}{n_2^2}\right) \label{6.3.4}$
The lowest few energy levels are shown in Figure $1$. One of the fundamental laws of physics is that matter is most stable with the lowest possible energy. Thus, the electron in a hydrogen atom usually moves in the $n = 1$ orbit, the orbit in which it has the lowest energy. When the electron is in this lowest energy orbit, the atom is said to be in its ground electronic state (or simply ground state). If the atom receives energy from an outside source, it is possible for the electron to move to an orbit with a higher $n$ value and the atom is now in an excited electronic state (or simply an excited state) with a higher energy. When an electron transitions from an excited state (higher energy orbit) to a less excited state, or ground state, the difference in energy is emitted as a photon. Similarly, if a photon is absorbed by an atom, the energy of the photon moves an electron from a lower energy orbit up to a more excited one.
We can relate the energy of electrons in atoms to what we learned previously about energy. The law of conservation of energy says that we can neither create nor destroy energy. Thus, if a certain amount of external energy is required to excite an electron from one energy level to another, that same amount of energy will be liberated when the electron returns to its initial state (Figure $2$). In effect, an atom can “store” energy by using it to promote an electron to a state with a higher energy and release it when the electron returns to a lower state. The energy can be released as one quantum of energy, as the electron returns to its ground state (say, from $n = 5$ to $n = 1$), or it can be released as two or more smaller quanta as the electron falls to an intermediate state, then to the ground state (say, from $n = 5$ to $n = 4$, emitting one quantum, then to $n = 1$, emitting a second quantum).
Since Bohr’s model involved only a single electron, it could also be applied to the single electron ions He+, Li2+, Be3+, and so forth, which differ from hydrogen only in their nuclear charges, and so one-electron atoms and ions are collectively referred to as hydrogen-like or hydrogenic atoms. The energy expression for hydrogen-like atoms is a generalization of the hydrogen atom energy, in which $Z$ is the nuclear charge (+1 for hydrogen, +2 for He, +3 for Li, and so on) and $k$ has a value of $2.179 \times 10^{–18}\; J$.
$\color{red} E_n=−\dfrac{kZ^2}{n^2} \label{6.3.5}$
The sizes of the circular orbits for hydrogen-like atoms are given in terms of their radii by the following expression, in which $a_o$ is a constant called the Bohr radius, with a value of $5.292 \times 10^{−11}\; m$:
$\color{red} r=\dfrac{n^2}{Z} a_0 \label{6.3.6}$
The equation also shows us that as the electron’s energy increases (as $n$ increases), the electron is found at greater distances from the nucleus. This is implied by the inverse dependence on $r$ in the Coulomb potential, since, as the electron moves away from the nucleus, the electrostatic attraction between it and the nucleus decreases, and it is held less tightly in the atom. Note that as $n$ gets larger and the orbits get larger, their energies get closer to zero, and so the limits $n⟶∞$ and $r⟶∞$ imply that $E = 0$ corresponds to the ionization limit where the electron is completely removed from the nucleus. Thus, for hydrogen in the ground state $n = 1$, the ionization energy would be:
$ΔE=E_{n⟶∞} −E_1=0+k=k \label{6.3.7}$
With three extremely puzzling paradoxes now solved (blackbody radiation, the photoelectric effect, and the hydrogen atom), and all involving Planck’s constant in a fundamental manner, it became clear to most physicists at that time that the classical theories that worked so well in the macroscopic world were fundamentally flawed and could not be extended down into the microscopic domain of atoms and molecules. Unfortunately, despite Bohr’s remarkable achievement in deriving a theoretical expression for the Rydberg constant, he was unable to extend his theory to the next simplest atom, He, which only has two electrons. Bohr’s model was severely flawed, since it was still based on the classical mechanics notion of precise orbits, a concept that was later found to be untenable in the microscopic domain, when a proper model of quantum mechanics was developed to supersede classical mechanics.
Example $1$: Calculating the Energy of an Electron in a Bohr Orbit
Early researchers were very excited when they were able to predict the energy of an electron at a particular distance from the nucleus in a hydrogen atom. If a spark promotes the electron in a hydrogen atom into an orbit with $n = 3$, what is the calculated energy, in joules, of the electron?
Solution
The energy of the electron is given by Equation $\ref{6.3.5}$:
$E=\dfrac{−kZ^2}{n^2} \nonumber$
The atomic number, $Z$, of hydrogen is 1; $k = 2.179 \times 10^{–18}\; J$; and the electron is characterized by an n value of $3$. Thus,
$E=\dfrac{−(2.179 \times 10^{−18}\;J)×(1)^2}{(3)^2}=−2.421 \times 10^{−19}\;J \nonumber$
Exercise $1$
The electron in Example $1$ in the $n=3$ state is promoted even further to an orbit with $n = 6$. What is its new energy?
Answer
TBD
Example $2$: Calculating Electron Transitions in a One–electron System
What is the energy (in joules) and the wavelength (in meters) of the line in the spectrum of hydrogen that represents the movement of an electron from Bohr orbit with n = 4 to the orbit with n = 6? In what part of the electromagnetic spectrum do we find this radiation?
Solution
In this case, the electron starts out with $n = 4$, so $n_1 = 4$. It comes to rest in the $n = 6$ orbit, so $n_2 = 6$. The difference in energy between the two states is given by this expression:
$ΔE=E_1−E_2=2.179 \times 10^{−18}\left(\dfrac{1}{n^2_1}−\dfrac{1}{n_2^2}\right) \nonumber$
$ΔE=2.179 \times 10^{−18} \left(\dfrac{1}{4^2}−\dfrac{1}{6^2}\right)\; J \nonumber$
$ΔE=2.179 \times 10^{−18} \left(\dfrac{1}{16}−\dfrac{1}{36}\right)\;J \nonumber$
$ΔE=7.566 \times 10^{−20}\;J \nonumber$
This energy difference is positive, indicating a photon enters the system (is absorbed) to excite the electron from the n = 4 orbit up to the $n = 6$ orbit. The wavelength of a photon with this energy is found by the expression $E=hc \lambda$. Rearrangement gives:
$\lambda=\dfrac{hc}{E} \nonumber$
From the figure of electromagnetic radiation, we can see that this wavelength is found in the infrared portion of the electromagnetic spectrum.
Exercise $2$
What is the energy in joules and the wavelength in meters of the photon produced when an electron falls from the $n = 5$ to the $n = 3$ level in a $He^+$ ion ($Z = 2$ for $He^+$)?
Answer
$6.198 \times 10^{–19}\; J$ and $3.205 \times 10^{−7}\; m$
Bohr’s model of the hydrogen atom provides insight into the behavior of matter at the microscopic level, but it is does not account for electron–electron interactions in atoms with more than one electron. It does introduce several important features of all models used to describe the distribution of electrons in an atom. These features include the following:
• The energies of electrons (energy levels) in an atom are quantized, described by quantum numbers: integer numbers having only specific allowed value and used to characterize the arrangement of electrons in an atom.
• An electron’s energy increases with increasing distance from the nucleus.
• The discrete energies (lines) in the spectra of the elements result from quantized electronic energies.
Of these features, the most important is the postulate of quantized energy levels for an electron in an atom. As a consequence, the model laid the foundation for the quantum mechanical model of the atom. Bohr won a Nobel Prize in Physics for his contributions to our understanding of the structure of atoms and how that is related to line spectra emissions.
Summary
Bohr incorporated Planck’s and Einstein’s quantization ideas into a model of the hydrogen atom that resolved the paradox of atom stability and discrete spectra. The Bohr model of the hydrogen atom explains the connection between the quantization of photons and the quantized emission from atoms. Bohr described the hydrogen atom in terms of an electron moving in a circular orbit about a nucleus. He postulated that the electron was restricted to certain orbits characterized by discrete energies. Transitions between these allowed orbits result in the absorption or emission of photons. When an electron moves from a higher-energy orbit to a more stable one, energy is emitted in the form of a photon. To move an electron from a stable orbit to a more excited one, a photon of energy must be absorbed. Using the Bohr model, we can calculate the energy of an electron and the radius of its orbit in any one-electron system.
Glossary
Bohr’s model of the hydrogen atom
structural model in which an electron moves around the nucleus only in circular orbits, each with a specific allowed radius; the orbiting electron does not normally emit electromagnetic radiation, but does so when changing from one orbit to another.
excited state
state having an energy greater than the ground-state energy
ground state
state in which the electrons in an atom, ion, or molecule have the lowest energy possible
quantum number
integer number having only specific allowed values and used to characterize the arrangement of electrons in an atom | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/03%3A_Electronic_Structure_and_Periodic_Properties/3.2%3A_The_Bohr_Model.txt |
Learning Objectives
• Extend the concept of wave–particle duality that was observed in electromagnetic radiation to matter as well
• Understand the general idea of the quantum mechanical description of electrons in an atom, and that it uses the notion of three-dimensional wave functions, or orbitals, that define the distribution of probability to find an electron in a particular part of space
• List and describe traits of the four quantum numbers that form the basis for completely specifying the state of an electron in an atom
Bohr’s model explained the experimental data for the hydrogen atom and was widely accepted, but it also raised many questions. Why did electrons orbit at only fixed distances defined by a single quantum number n = 1, 2, 3, and so on, but never in between? Why did the model work so well describing hydrogen and one-electron ions, but could not correctly predict the emission spectrum for helium or any larger atoms? To answer these questions, scientists needed to completely revise the way they thought about matter.
Behavior in the Microscopic World
We know how matter behaves in the macroscopic world—objects that are large enough to be seen by the naked eye follow the rules of classical physics. A billiard ball moving on a table will behave like a particle: It will continue in a straight line unless it collides with another ball or the table cushion, or is acted on by some other force (such as friction). The ball has a well-defined position and velocity (or a well-defined momentum, $p = mv$, defined by mass $m$ and velocity $v$) at any given moment. In other words, the ball is moving in a classical trajectory. This is the typical behavior of a classical object.
When waves interact with each other, they show interference patterns that are not displayed by macroscopic particles such as the billiard ball. For example, interacting waves on the surface of water can produce interference patters similar to those shown on Figure $1$. This is a case of wave behavior on the macroscopic scale, and it is clear that particles and waves are very different phenomena in the macroscopic realm.
As technological improvements allowed scientists to probe the microscopic world in greater detail, it became increasingly clear by the 1920s that very small pieces of matter follow a different set of rules from those we observe for large objects. The unquestionable separation of waves and particles was no longer the case for the microscopic world.
One of the first people to pay attention to the special behavior of the microscopic world was Louis de Broglie. He asked the question: If electromagnetic radiation can have particle-like character, can electrons and other submicroscopic particles exhibit wavelike character? In his 1925 doctoral dissertation, de Broglie extended the wave–particle duality of light that Einstein used to resolve the photoelectric-effect paradox to material particles. He predicted that a particle with mass m and velocity v (that is, with linear momentum p) should also exhibit the behavior of a wave with a wavelength value λ, given by this expression in which h is the familiar Planck’s constant
$\lambda=\dfrac{h}{mv}=\dfrac{h}{p} \label{6.4.1}$
This is called the de Broglie wavelength. Unlike the other values of λ discussed in this chapter, the de Broglie wavelength is a characteristic of particles and other bodies, not electromagnetic radiation (note that this equation involves velocity [v, m/s], not frequency [ν, Hz]. Although these two symbols are identical, they mean very different things). Where Bohr had postulated the electron as being a particle orbiting the nucleus in quantized orbits, de Broglie argued that Bohr’s assumption of quantization can be explained if the electron is considered not as a particle, but rather as a circular standing wave such that only an integer number of wavelengths could fit exactly within the orbit (Figure $2$).
For a circular orbit of radius r, the circumference is 2πr, and so de Broglie’s condition is:
$2πr=nλ \label{6.4.3}$
with $n=1,2,3,...$
Since the de Broglie expression relates the wavelength to the momentum and, hence, velocity, this implies:
$2πr=nλ=\dfrac{nh}{p}=\dfrac{nh}{mv}=\dfrac{nhr}{mvr}=\dfrac{nhr}{L} \label{6.4.3b}$
This expression can be rearranged to give Bohr’s formula for the quantization of the angular momentum:
$L=\dfrac{nh}{2π}=n \hbar \label{6.4.4}$
Classical angular momentum L for a circular motion is equal to the product of the radius of the circle and the momentum of the moving particle p.
$L=rp=rmv \;\;\; \text{(for a circular motion)} \label{6.4.5}$
Shortly after de Broglie proposed the wave nature of matter, two scientists at Bell Laboratories, C. J. Davisson and L. H. Germer, demonstrated experimentally that electrons can exhibit wavelike behavior by showing an interference pattern for electrons travelling through a regular atomic pattern in a crystal. The regularly spaced atomic layers served as slits, as used in other interference experiments. Since the spacing between the layers serving as slits needs to be similar in size to the wavelength of the tested wave for an interference pattern to form, Davisson and Germer used a crystalline nickel target for their “slits,” since the spacing of the atoms within the lattice was approximately the same as the de Broglie wavelengths of the electrons that they used. Figure $4$ shows an interference pattern.
The wave–particle duality of matter can be seen by observing what happens if electron collisions are recorded over a long period of time. Initially, when only a few electrons have been recorded, they show clear particle-like behavior, having arrived in small localized packets that appear to be random. As more and more electrons arrived and were recorded, a clear interference pattern that is the hallmark of wavelike behavior emerged. Thus, it appears that while electrons are small localized particles, their motion does not follow the equations of motion implied by classical mechanics, but instead it is governed by some type of a wave equation that governs a probability distribution even for a single electron’s motion. Thus the wave–particle duality first observed with photons is actually a fundamental behavior intrinsic to all quantum particles.
Example $1$: Calculating the Wavelength of a Particle
If an electron travels at a velocity of 1.000 × 107 m s–1 and has a mass of 9.109 × 10–28 g, what is its wavelength?
Solution
We can use de Broglie’s equation to solve this problem, but we first must do a unit conversion of Planck’s constant. You learned earlier that 1 J = 1 kg m2/s2. Thus, we can write h = 6.626 × 10–34 J s as 6.626 × 10–34 kg m2/s.
\begin{align*} λ&=\dfrac{h}{mv} \[4pt] &=\mathrm{\dfrac{6.626×10^{−34}\:kg\: m^2/s}{(9.109×10^{−31}\:kg)(1.000×10^7\:m/s)}}\[4pt] &= \mathrm{7.274×10^{−11}\:m} \end{align*} \nonumber
This is a small value, but it is significantly larger than the size of an electron in the classical (particle) view. This size is the same order of magnitude as the size of an atom. This means that electron wavelike behavior is going to be noticeable in an atom.
Exercise $1$
Calculate the wavelength of a softball with a mass of 100 g traveling at a velocity of 35 m s–1, assuming that it can be modeled as a single particle.
Answer
1.9 × 10–34 m.
We never think of a thrown softball having a wavelength, since this wavelength is so small it is impossible for our senses or any known instrument to detect (strictly speaking, the wavelength of a real baseball would correspond to the wavelengths of its constituent atoms and molecules, which, while much larger than this value, would still be microscopically tiny). The de Broglie wavelength is only appreciable for matter that has a very small mass and/or a very high velocity.
Werner Heisenberg considered the limits of how accurately we can measure properties of an electron or other microscopic particles. He determined that there is a fundamental limit to how accurately one can measure both a particle’s position and its momentum simultaneously. The more accurately we measure the momentum of a particle, the less accurately we can determine its position at that time, and vice versa. This is summed up in what we now call the Heisenberg uncertainty principle: It is fundamentally impossible to determine simultaneously and exactly both the momentum and the position of a particle. For a particle of mass m moving with velocity vx in the x direction (or equivalently with momentum px), the product of the uncertainty in the position, Δx, and the uncertainty in the momentum, Δpx , must be greater than or equal to $\dfrac{ℏ}{2}$ (recall that $ℏ=\dfrac{h}{2π}$, the value of Planck’s constant divided by 2π).
$Δx×Δp_x=(Δx)(mΔv)≥\dfrac{ℏ}{2} \nonumber$
This equation allows us to calculate the limit to how precisely we can know both the simultaneous position of an object and its momentum. For example, if we improve our measurement of an electron’s position so that the uncertainty in the position (Δx) has a value of, say, 1 pm (10–12 m, about 1% of the diameter of a hydrogen atom), then our determination of its momentum must have an uncertainty with a value of at least
$\left [Δp=mΔv=\dfrac{h}{(2Δx)} \right ]=\mathrm{\dfrac{(1.055×10^{−34}\:kg\: m^2/s)}{(2×1×10^{−12}\:m)}=5×10^{−23}\:kg\: m/s.} \nonumber$
The value of ħ is not large, so the uncertainty in the position or momentum of a macroscopic object like a baseball is too insignificant to observe. However, the mass of a microscopic object such as an electron is small enough that the uncertainty can be large and significant.
It should be noted that Heisenberg’s uncertainty principle is not just limited to uncertainties in position and momentum, but it also links other dynamical variables. For example, when an atom absorbs a photon and makes a transition from one energy state to another, the uncertainty in the energy and the uncertainty in the time required for the transition are similarly related, as ΔE Δt ≥ $\dfrac{ℏ}{2}$. As will be discussed later, even the vector components of angular momentum cannot all be specified exactly simultaneously.
Heisenberg’s principle imposes ultimate limits on what is knowable in science. The uncertainty principle can be shown to be a consequence of wave–particle duality, which lies at the heart of what distinguishes modern quantum theory from classical mechanics. Recall that the equations of motion obtained from classical mechanics are trajectories where, at any given instant in time, both the position and the momentum of a particle can be determined exactly. Heisenberg’s uncertainty principle implies that such a view is untenable in the microscopic domain and that there are fundamental limitations governing the motion of quantum particles. This does not mean that microscopic particles do not move in trajectories, it is just that measurements of trajectories are limited in their precision. In the realm of quantum mechanics, measurements introduce changes into the system that is being observed.
The Quantum–Mechanical Model of an Atom
Shortly after de Broglie published his ideas that the electron in a hydrogen atom could be better thought of as being a circular standing wave instead of a particle moving in quantized circular orbits, as Bohr had argued, Erwin Schrödinger extended de Broglie’s work by incorporating the de Broglie relation into a wave equation, deriving what is today known as the Schrödinger equation. When Schrödinger applied his equation to hydrogen-like atoms, he was able to reproduce Bohr’s expression for the energy and, thus, the Rydberg formula governing hydrogen spectra, and he did so without having to invoke Bohr’s assumptions of stationary states and quantized orbits, angular momenta, and energies; quantization in Schrödinger’s theory was a natural consequence of the underlying mathematics of the wave equation. Like de Broglie, Schrödinger initially viewed the electron in hydrogen as being a physical wave instead of a particle, but where de Broglie thought of the electron in terms of circular stationary waves, Schrödinger properly thought in terms of three-dimensional stationary waves, or wavefunctions, represented by the Greek letter psi, ψ. A few years later, Max Born proposed an interpretation of the wavefunction ψ that is still accepted today: Electrons are still particles, and so the waves represented by ψ are not physical waves but, instead, are complex probability amplitudes. The square of the magnitude of a wavefunction $∣ψ∣^2$ describes the probability of the quantum particle being present near a certain location in space. This means that wavefunctions can be used to determine the distribution of the electron’s density with respect to the nucleus in an atom. In the most general form, the Schrödinger equation can be written as:
$\hat{H}ψ=Eψ \nonumber$
$\hat{H}$ is the Hamiltonian operator, a set of mathematical operations representing the total energy of the quantum particle (such as an electron in an atom), ψ is the wavefunction of this particle that can be used to find the special distribution of the probability of finding the particle, and $E$ is the actual value of the total energy of the particle.
Schrödinger’s work, as well as that of Heisenberg and many other scientists following in their footsteps, is generally referred to as quantum mechanics.
Understanding Quantum Theory of Electrons in Atoms
The goal of this section is to understand the electron orbitals (location of electrons in atoms), their different energies, and other properties. The use of quantum theory provides the best understanding to these topics. This knowledge is a precursor to chemical bonding.
As was described previously, electrons in atoms can exist only on discrete energy levels but not between them. It is said that the energy of an electron in an atom is quantized, that is, it can be equal only to certain specific values and can jump from one energy level to another but not transition smoothly or stay between these levels.
The energy levels are labeled with an n value, where n = 1, 2, 3, …. Generally speaking, the energy of an electron in an atom is greater for greater values of n. This number, n, is referred to as the principal quantum number. The principal quantum number defines the location of the energy level. It is essentially the same concept as the n in the Bohr atom description. Another name for the principal quantum number is the shell number. The shells of an atom can be thought of concentric circles radiating out from the nucleus. The electrons that belong to a specific shell are most likely to be found within the corresponding circular area. The further we proceed from the nucleus, the higher the shell number, and so the higher the energy level (Figure $5$). The positively charged protons in the nucleus stabilize the electronic orbitals by electrostatic attraction between the positive charges of the protons and the negative charges of the electrons. So the further away the electron is from the nucleus, the greater the energy it has.
This quantum mechanical model for where electrons reside in an atom can be used to look at electronic transitions, the events when an electron moves from one energy level to another. If the transition is to a higher energy level, energy is absorbed, and the energy change has a positive value. To obtain the amount of energy necessary for the transition to a higher energy level, a photon is absorbed by the atom. A transition to a lower energy level involves a release of energy, and the energy change is negative. This process is accompanied by emission of a photon by the atom. The following equation summarizes these relationships and is based on the hydrogen atom:
\begin{align*} ΔE &=E_\ce{final}−E_\ce{initial} \[4pt] &=−2.18×10^{−18}\left(\dfrac{1}{n^2_\ce f}−\dfrac{1}{n^2_\ce i}\right)\:\ce J \end{align*} \nonumber
The values nf and ni are the final and initial energy states of the electron.
The principal quantum number is one of three quantum numbers used to characterize an orbital. An atomic orbital, which is distinct from an orbit, is a general region in an atom within which an electron is most probable to reside. The quantum mechanical model specifies the probability of finding an electron in the three-dimensional space around the nucleus and is based on solutions of the Schrödinger equation. In addition, the principal quantum number defines the energy of an electron in a hydrogen or hydrogen-like atom or an ion (an atom or an ion with only one electron) and the general region in which discrete energy levels of electrons in a multi-electron atoms and ions are located.
Another quantum number is l, the angular momentum quantum number. It is an integer that defines the shape of the orbital, and takes on the values, l = 0, 1, 2, …, n – 1. This means that an orbital with n = 1 can have only one value of l, l = 0, whereas n = 2 permits l = 0 and l = 1, and so on. The principal quantum number defines the general size and energy of the orbital. The l value specifies the shape of the orbital. Orbitals with the same value of l form a subshell. In addition, the greater the angular momentum quantum number, the greater is the angular momentum of an electron at this orbital.
Orbitals with l = 0 are called s orbitals (or the s subshells). The value l = 1 corresponds to the p orbitals. For a given n, p orbitals constitute a p subshell (e.g., 3p if n = 3). The orbitals with l = 2 are called the d orbitals, followed by the f-, g-, and h-orbitals for l = 3, 4, 5, and there are higher values we will not consider.
There are certain distances from the nucleus at which the probability density of finding an electron located at a particular orbital is zero. In other words, the value of the wavefunction ψ is zero at this distance for this orbital. Such a value of radius r is called a radial node. The number of radial nodes in an orbital is nl – 1.
Consider the examples in Figure $7$. The orbitals depicted are of the s type, thus l = 0 for all of them. It can be seen from the graphs of the probability densities that there are 1 – 0 – 1 = 0 places where the density is zero (nodes) for 1s (n = 1), 2 – 0 – 1 = 1 node for 2s, and 3 – 0 – 1 = 2 nodes for the 3s orbitals.
The s subshell electron density distribution is spherical and the p subshell has a dumbbell shape. The d and f orbitals are more complex. These shapes represent the three-dimensional regions within which the electron is likely to be found.
If an electron has an angular momentum (l ≠ 0), then this vector can point in different directions. In addition, the z component of the angular momentum can have more than one value. This means that if a magnetic field is applied in the z direction, orbitals with different values of the z component of the angular momentum will have different energies resulting from interacting with the field. The magnetic quantum number, called ml, specifies the z component of the angular momentum for a particular orbital. For example, for an s orbital, l = 0, and the only value of ml is zero. For p orbitals, l = 1, and ml can be equal to –1, 0, or +1. Generally speaking, ml can be equal to –l, –(l – 1), …, –1, 0, +1, …, (l – 1), l. The total number of possible orbitals with the same value of l (a subshell) is 2l + 1. Thus, there is one s-orbital for ml = 0, there are three p-orbitals for ml = 1, five d-orbitals for ml = 2, seven f-orbitals for ml = 3, and so forth. The principal quantum number defines the general value of the electronic energy. The angular momentum quantum number determines the shape of the orbital. And the magnetic quantum number specifies orientation of the orbital in space, as can be seen in Figure $7$.
Figure $8$ illustrates the energy levels for various orbitals. The number before the orbital name (such as 2s, 3p, and so forth) stands for the principal quantum number, n. The letter in the orbital name defines the subshell with a specific angular momentum quantum number l = 0 for s orbitals, 1 for p orbitals, 2 for d orbitals. Finally, there are more than one possible orbitals for l ≥ 1, each corresponding to a specific value of ml. In the case of a hydrogen atom or a one-electron ion (such as He+, Li2+, and so on), energies of all the orbitals with the same n are the same. This is called a degeneracy, and the energy levels for the same principal quantum number, n, are called degenerate energy levels. However, in atoms with more than one electron, this degeneracy is eliminated by the electron–electron interactions, and orbitals that belong to different subshells have different energies. Orbitals within the same subshell (for example ns, np, nd, nf, such as 2p, 3s) are still degenerate and have the same energy.
While the three quantum numbers discussed in the previous paragraphs work well for describing electron orbitals, some experiments showed that they were not sufficient to explain all observed results. It was demonstrated in the 1920s that when hydrogen-line spectra are examined at extremely high resolution, some lines are actually not single peaks but, rather, pairs of closely spaced lines. This is the so-called fine structure of the spectrum, and it implies that there are additional small differences in energies of electrons even when they are located in the same orbital. These observations led Samuel Goudsmit and George Uhlenbeck to propose that electrons have a fourth quantum number. They called this the spin quantum number, or ms.
The other three quantum numbers, n, l, and ml, are properties of specific atomic orbitals that also define in what part of the space an electron is most likely to be located. Orbitals are a result of solving the Schrödinger equation for electrons in atoms. The electron spin is a different kind of property. It is a completely quantum phenomenon with no analogues in the classical realm. In addition, it cannot be derived from solving the Schrödinger equation and is not related to the normal spatial coordinates (such as the Cartesian x, y, and z). Electron spin describes an intrinsic electron “rotation” or “spinning.” Each electron acts as a tiny magnet or a tiny rotating object with an angular momentum, even though this rotation cannot be observed in terms of the spatial coordinates.
The magnitude of the overall electron spin can only have one value, and an electron can only “spin” in one of two quantized states. One is termed the α state, with the z component of the spin being in the positive direction of the z axis. This corresponds to the spin quantum number $m_s=\dfrac{1}{2}$. The other is called the β state, with the z component of the spin being negative and $m_s=−\dfrac{1}{2}$. Any electron, regardless of the atomic orbital it is located in, can only have one of those two values of the spin quantum number. The energies of electrons having $m_s=−\dfrac{1}{2}$ and $m_s=\dfrac{1}{2}$ are different if an external magnetic field is applied.
Figure $9$ illustrates this phenomenon. An electron acts like a tiny magnet. Its moment is directed up (in the positive direction of the z axis) for the $\dfrac{1}{2}$ spin quantum number and down (in the negative z direction) for the spin quantum number of $−\ce{1/2}$. A magnet has a lower energy if its magnetic moment is aligned with the external magnetic field (the left electron) and a higher energy for the magnetic moment being opposite to the applied field. This is why an electron with $m_s=\dfrac{1}{2}$ has a slightly lower energy in an external field in the positive z direction, and an electron with $m_s=−\dfrac{1}{2}$ has a slightly higher energy in the same field. This is true even for an electron occupying the same orbital in an atom. A spectral line corresponding to a transition for electrons from the same orbital but with different spin quantum numbers has two possible values of energy; thus, the line in the spectrum will show a fine structure splitting.
The Pauli Exclusion Principle
An electron in an atom is completely described by four quantum numbers: n, l, ml, and ms. The first three quantum numbers define the orbital and the fourth quantum number describes the intrinsic electron property called spin. An Austrian physicist Wolfgang Pauli formulated a general principle that gives the last piece of information that we need to understand the general behavior of electrons in atoms. The Pauli exclusion principle can be formulated as follows: No two electrons in the same atom can have exactly the same set of all the four quantum numbers. What this means is that electrons can share the same orbital (the same set of the quantum numbers n, l, and ml), but only if their spin quantum numbers ms have different values. Since the spin quantum number can only have two values $\left(±\dfrac{1}{2}\right)$, no more than two electrons can occupy the same orbital (and if two electrons are located in the same orbital, they must have opposite spins). Therefore, any atomic orbital can be populated by only zero, one, or two electrons. The properties and meaning of the quantum numbers of electrons in atoms are briefly summarized in Table $1$.
Table $1$: Quantum Numbers, Their Properties, and Significance
Name Symbol Allowed values Physical meaning
principal quantum number n 1, 2, 3, 4, …. shell, the general region for the value of energy for an electron on the orbital
angular momentum or azimuthal quantum number l 0 ≤ ln – 1 subshell, the shape of the orbital
magnetic quantum number ml lmll orientation of the orbital
spin quantum number ms $\dfrac{1}{2},\:−\dfrac{1}{2}$ direction of the intrinsic quantum “spinning” of the electron
Example $2$: Working with Shells and Subshells
Indicate the number of subshells, the number of orbitals in each subshell, and the values of l and ml for the orbitals in the n = 4 shell of an atom.
Solution
For n = 4, l can have values of 0, 1, 2, and 3. Thus, s, p, d, and f subshells are found in the n = 4 shell of an atom. For l = 0 (the s subshell), ml can only be 0. Thus, there is only one 4s orbital. For l = 1 (p-type orbitals), m can have values of –1, 0, +1, so we find three 4p orbitals. For l = 2 (d-type orbitals), ml can have values of –2, –1, 0, +1, +2, so we have five 4d orbitals. When l = 3 (f-type orbitals), ml can have values of –3, –2, –1, 0, +1, +2, +3, and we can have seven 4f orbitals. Thus, we find a total of 16 orbitals in the n = 4 shell of an atom.
Exercise $2$
Identify the subshell in which electrons with the following quantum numbers are found:
1. n = 3, l = 1;
2. n = 5, l = 3;
3. n = 2, l = 0.
Answer a
3p
Answer b
5f
Answer c
2s
Example $3$: Maximum Number of Electrons
Calculate the maximum number of electrons that can occupy a shell with (a) n = 2, (b) n = 5, and (c) n as a variable. Note you are only looking at the orbitals with the specified n value, not those at lower energies.
Solution
(a) When n = 2, there are four orbitals (a single 2s orbital, and three orbitals labeled 2p). These four orbitals can contain eight electrons.
(b) When n = 5, there are five subshells of orbitals that we need to sum:
\begin{align*} &\phantom{+}\textrm{1 orbital labeled }5s\ &\phantom{+}\textrm{3 orbitals labeled }5p\ &\phantom{+}\textrm{5 orbitals labeled }5d\ &\phantom{+}\textrm{7 orbitals labeled }5f\ &\underline{+\textrm{9 orbitals labeled }5g}\ &\,\textrm{25 orbitals total} \end{align*}
Again, each orbital holds two electrons, so 50 electrons can fit in this shell.
(c) The number of orbitals in any shell n will equal n2. There can be up to two electrons in each orbital, so the maximum number of electrons will be 2 × n2
Exercise $3$
If a shell contains a maximum of 32 electrons, what is the principal quantum number, n?
Answer
n = 4
Example $4$: Working with Quantum Numbers
Complete the following table for atomic orbitals:
Table for Atomic Orbitals
Orbital n l ml degeneracy Radial nodes (no.)
4f
4 1
7 7 3
5d
Solution
The table can be completed using the following rules:
• The orbital designation is nl, where l = 0, 1, 2, 3, 4, 5, … is mapped to the letter sequence s, p, d, f, g, h, …,
• The ml degeneracy is the number of orbitals within an l subshell, and so is 2l + 1 (there is one s orbital, three p orbitals, five d orbitals, seven f orbitals, and so forth).
• The number of radial nodes is equal to n – l – 1.
Solution to Example 6.3.4
Orbital n l ml degeneracy Radial nodes (no.)
4f 4 3 7 0
4p 4 1 3 2
7f 7 3 7 3
5d 5 2 5 2
Exercise $4$
How many orbitals have l = 2 and n = 3?
Answer
The five degenerate 3d orbitals
Summary
Macroscopic objects act as particles. Microscopic objects (such as electrons) have properties of both a particle and a wave. Their exact trajectories cannot be determined. The quantum mechanical model of atoms describes the three-dimensional position of the electron in a probabilistic manner according to a mathematical function called a wavefunction, often denoted as ψ. Atomic wavefunctions are also called orbitals. The squared magnitude of the wavefunction describes the distribution of the probability of finding the electron in a particular region in space. Therefore, atomic orbitals describe the areas in an atom where electrons are most likely to be found.
An atomic orbital is characterized by three quantum numbers. The principal quantum number, n, can be any positive integer. The general region for value of energy of the orbital and the average distance of an electron from the nucleus are related to n. Orbitals having the same value of n are said to be in the same shell. The angular momentum quantum number, l, can have any integer value from 0 to n – 1. This quantum number describes the shape or type of the orbital. Orbitals with the same principal quantum number and the same l value belong to the same subshell. The magnetic quantum number, ml, with 2l + 1 values ranging from –l to +l, describes the orientation of the orbital in space. In addition, each electron has a spin quantum number, ms, that can be equal to $±\dfrac{1}{2}$. No two electrons in the same atom can have the same set of values for all the four quantum numbers.
Glossary
angular momentum quantum number (l)
quantum number distinguishing the different shapes of orbitals; it is also a measure of the orbital angular momentum
atomic orbital
mathematical function that describes the behavior of an electron in an atom (also called the wavefunction), it can be used to find the probability of locating an electron in a specific region around the nucleus, as well as other dynamical variables
d orbital
region of space with high electron density that is either four lobed or contains a dumbbell and torus shape; describes orbitals with l = 2. An electron in this orbital is called a d electron
electron density
a measure of the probability of locating an electron in a particular region of space, it is equal to the squared absolute value of the wave function ψ
f orbital
multilobed region of space with high electron density, describes orbitals with l = 3. An electron in this orbital is called an f electron
Heisenberg uncertainty principle
rule stating that it is impossible to exactly determine both certain conjugate dynamical properties such as the momentum and the position of a particle at the same time. The uncertainty principle is a consequence of quantum particles exhibiting wave–particle duality
magnetic quantum number (ml)
quantum number signifying the orientation of an atomic orbital around the nucleus; orbitals having different values of ml but the same subshell value of l have the same energy (are degenerate), but this degeneracy can be removed by application of an external magnetic field
p orbital
dumbbell-shaped region of space with high electron density, describes orbitals with l = 1. An electron in this orbital is called a p electron
Pauli exclusion principle
specifies that no two electrons in an atom can have the same value for all four quantum numbers
principal quantum number (n)
quantum number specifying the shell an electron occupies in an atom
quantum mechanics
field of study that includes quantization of energy, wave-particle duality, and the Heisenberg uncertainty principle to describe matter
s orbital
spherical region of space with high electron density, describes orbitals with l = 0. An electron in this orbital is called an s electron
shell
set of orbitals with the same principal quantum number, n
spin quantum number (ms)
number specifying the electron spin direction, either $+\dfrac{1}{2}$ or $−\dfrac{1}{2}$
subshell
set of orbitals in an atom with the same values of n and l
wavefunction (ψ)
mathematical description of an atomic orbital that describes the shape of the orbital; it can be used to calculate the probability of finding the electron at any given location in the orbital, as well as dynamical variables such as the energy and the angular momentum | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/03%3A_Electronic_Structure_and_Periodic_Properties/3.3%3A_Development_of_Quantum_Theory.txt |
Learning Objectives
• Derive the predicted ground-state electron configurations of atoms
• Identify and explain exceptions to predicted electron configurations for atoms and ions
• Relate electron configurations to element classifications in the periodic table
Having introduced the basics of atomic structure and quantum mechanics, we can use our understanding of quantum numbers to determine how atomic orbitals relate to one another. This allows us to determine which orbitals are occupied by electrons in each atom. The specific arrangement of electrons in orbitals of an atom determines many of the chemical properties of that atom.
Orbital Energies and Atomic Structure
The energy of atomic orbitals increases as the principal quantum number, $n$, increases. In any atom with two or more electrons, the repulsion between the electrons makes energies of subshells with different values of $l$ differ so that the energy of the orbitals increases within a shell in the order s < p < d < f. Figure $1$ depicts how these two trends in increasing energy relate. The 1s orbital at the bottom of the diagram is the orbital with electrons of lowest energy. The energy increases as we move up to the 2s and then 2p, 3s, and 3p orbitals, showing that the increasing n value has more influence on energy than the increasing l value for small atoms. However, this pattern does not hold for larger atoms. The 3d orbital is higher in energy than the 4s orbital. Such overlaps continue to occur frequently as we move up the chart.
Electrons in successive atoms on the periodic table tend to fill low-energy orbitals first. Thus, many students find it confusing that, for example, the 5p orbitals fill immediately after the 4d, and immediately before the 6s. The filling order is based on observed experimental results, and has been confirmed by theoretical calculations. As the principal quantum number, n, increases, the size of the orbital increases and the electrons spend more time farther from the nucleus. Thus, the attraction to the nucleus is weaker and the energy associated with the orbital is higher (less stabilized). But this is not the only effect we have to take into account. Within each shell, as the value of l increases, the electrons are less penetrating (meaning there is less electron density found close to the nucleus), in the order s > p > d > f. Electrons that are closer to the nucleus slightly repel electrons that are farther out, offsetting the more dominant electron–nucleus attractions slightly (recall that all electrons have −1 charges, but nuclei have +Z charges). This phenomenon is called shielding and will be discussed in more detail in the next section. Electrons in orbitals that experience more shielding are less stabilized and thus higher in energy. For small orbitals (1s through 3p), the increase in energy due to n is more significant than the increase due to l; however, for larger orbitals the two trends are comparable and cannot be simply predicted. We will discuss methods for remembering the observed order.
The arrangement of electrons in the orbitals of an atom is called the electron configuration of the atom. We describe an electron configuration with a symbol that contains three pieces of information ( Figure $2$):
1. The number of the principal quantum shell, n,
2. The letter that designates the orbital type (the subshell, l), and
3. A superscript number that designates the number of electrons in that particular subshell.
For example, the notation 2p4 (read "two–p–four") indicates four electrons in a p subshell (l = 1) with a principal quantum number (n) of 2. The notation 3d8 (read "three–d–eight") indicates eight electrons in the d subshell (i.e., l = 2) of the principal shell for which n = 3.
The Aufbau Principle
To determine the electron configuration for any particular atom, we can “build” the structures in the order of atomic numbers. Beginning with hydrogen, and continuing across the periods of the periodic table, we add one proton at a time to the nucleus and one electron to the proper subshell until we have described the electron configurations of all the elements. This procedure is called the Aufbau principle, from the German word Aufbau (“to build up”). Each added electron occupies the subshell of lowest energy available (in the order shown in Figure $3$), subject to the limitations imposed by the allowed quantum numbers according to the Pauli exclusion principle. Electrons enter higher-energy subshells only after lower-energy subshells have been filled to capacity. Figure $3$ illustrates the traditional way to remember the filling order for atomic orbitals.
Since the arrangement of the periodic table is based on the electron configurations, Figure $4$ provides an alternative method for determining the electron configuration. The filling order simply begins at hydrogen and includes each subshell as you proceed in increasing Z order. For example, after filling the 3p block up to Ar, we see the orbital will be 4s (K, Ca), followed by the 3d orbitals.
We will now construct the ground-state electron configuration and orbital diagram for a selection of atoms in the first and second periods of the periodic table. Orbital diagrams are pictorial representations of the electron configuration, showing the individual orbitals and the pairing arrangement of electrons. We start with a single hydrogen atom (atomic number 1), which consists of one proton and one electron. Referring to either Figure $3$ or $4$, we would expect to find the electron in the 1s orbital. By convention, the $m_s=+\dfrac{1}{2}$ value is usually filled first. The electron configuration and the orbital diagram are:
Following hydrogen is the noble gas helium, which has an atomic number of 2. The helium atom contains two protons and two electrons. The first electron has the same four quantum numbers as the hydrogen atom electron (n = 1, l = 0, ml = 0, $m_s=+\dfrac{1}{2}$). The second electron also goes into the 1s orbital and fills that orbital. The second electron has the same n, l, and ml quantum numbers, but must have the opposite spin quantum number, $m_s=−\dfrac{1}{2}$. This is in accord with the Pauli exclusion principle: No two electrons in the same atom can have the same set of four quantum numbers. For orbital diagrams, this means two arrows go in each box (representing two electrons in each orbital) and the arrows must point in opposite directions (representing paired spins). The electron configuration and orbital diagram of helium are:
The n = 1 shell is completely filled in a helium atom.
The next atom is the alkali metal lithium with an atomic number of 3. The first two electrons in lithium fill the 1s orbital and have the same sets of four quantum numbers as the two electrons in helium. The remaining electron must occupy the orbital of next lowest energy, the 2s orbital (Figure $3$ or $4$). Thus, the electron configuration and orbital diagram of lithium are:
An atom of the alkaline earth metal beryllium, with an atomic number of 4, contains four protons in the nucleus and four electrons surrounding the nucleus. The fourth electron fills the remaining space in the 2s orbital.
An atom of boron (atomic number 5) contains five electrons. The n = 1 shell is filled with two electrons and three electrons will occupy the n = 2 shell. Because any s subshell can contain only two electrons, the fifth electron must occupy the next energy level, which will be a 2p orbital. There are three degenerate 2p orbitals (ml = −1, 0, +1) and the electron can occupy any one of these p orbitals. When drawing orbital diagrams, we include empty boxes to depict any empty orbitals in the same subshell that we are filling.
Carbon (atomic number 6) has six electrons. Four of them fill the 1s and 2s orbitals. The remaining two electrons occupy the 2p subshell. We now have a choice of filling one of the 2p orbitals and pairing the electrons or of leaving the electrons unpaired in two different, but degenerate, p orbitals. The orbitals are filled as described by Hund’s rule: the lowest-energy configuration for an atom with electrons within a set of degenerate orbitals is that having the maximum number of unpaired electrons. Thus, the two electrons in the carbon 2p orbitals have identical n, l, and ms quantum numbers and differ in their ml quantum number (in accord with the Pauli exclusion principle). The electron configuration and orbital diagram for carbon are:
Nitrogen (atomic number 7) fills the 1s and 2s subshells and has one electron in each of the three 2p orbitals, in accordance with Hund’s rule. These three electrons have unpaired spins. Oxygen (atomic number 8) has a pair of electrons in any one of the 2p orbitals (the electrons have opposite spins) and a single electron in each of the other two. Fluorine (atomic number 9) has only one 2p orbital containing an unpaired electron. All of the electrons in the noble gas neon (atomic number 10) are paired, and all of the orbitals in the n = 1 and the n = 2 shells are filled. The electron configurations and orbital diagrams of these four elements are:
The alkali metal sodium (atomic number 11) has one more electron than the neon atom. This electron must go into the lowest-energy subshell available, the 3s orbital, giving a 1s22s22p63s1 configuration. The electrons occupying the outermost shell orbital(s) (highest value of n) are called valence electrons, and those occupying the inner shell orbitals are called core electrons ( Figure \PageIndex5\PageIndex5). Since the core electron shells correspond to noble gas electron configurations, we can abbreviate electron configurations by writing the noble gas that matches the core electron configuration, along with the valence electrons in a condensed format. For our sodium example, the symbol [Ne] represents core electrons, (1s22s22p6) and our abbreviated or condensed configuration is [Ne]3s1.
Similarly, the abbreviated configuration of lithium can be represented as [He]2s1, where [He] represents the configuration of the helium atom, which is identical to that of the filled inner shell of lithium. Writing the configurations in this way emphasizes the similarity of the configurations of lithium and sodium. Both atoms, which are in the alkali metal family, have only one electron in a valence s subshell outside a filled set of inner shells.
$\ce{Li:[He]}\,2s^1\ \ce{Na:[Ne]}\,3s^1 \nonumber$
The alkaline earth metal magnesium (atomic number 12), with its 12 electrons in a [Ne]3s2 configuration, is analogous to its family member beryllium, [He]2s2. Both atoms have a filled s subshell outside their filled inner shells. Aluminum (atomic number 13), with 13 electrons and the electron configuration [Ne]3s23p1, is analogous to its family member boron, [He]2s22p1.
The electron configurations of silicon (14 electrons), phosphorus (15 electrons), sulfur (16 electrons), chlorine (17 electrons), and argon (18 electrons) are analogous in the electron configurations of their outer shells to their corresponding family members carbon, nitrogen, oxygen, fluorine, and neon, respectively, except that the principal quantum number of the outer shell of the heavier elements has increased by one to n = 3. Figure $6$ shows the lowest energy, or ground-state, electron configuration for these elements as well as that for atoms of each of the known elements.
When we come to the next element in the periodic table, the alkali metal potassium (atomic number 19), we might expect that we would begin to add electrons to the 3d subshell. However, all available chemical and physical evidence indicates that potassium is like lithium and sodium, and that the next electron is not added to the 3d level but is, instead, added to the 4s level (Figure $3$ or $4$). As discussed previously, the 3d orbital with no radial nodes is higher in energy because it is less penetrating and more shielded from the nucleus than the 4s, which has three radial nodes. Thus, potassium has an electron configuration of [Ar]4s1. Hence, potassium corresponds to Li and Na in its valence shell configuration. The next electron is added to complete the 4s subshell and calcium has an electron configuration of [Ar]4s2. This gives calcium an outer-shell electron configuration corresponding to that of beryllium and magnesium.
Beginning with the transition metal scandium (atomic number 21), additional electrons are added successively to the 3d subshell. This subshell is filled to its capacity with 10 electrons (remember that for l = 2 [d orbitals], there are 2l + 1 = 5 values of ml, meaning that there are five d orbitals that have a combined capacity of 10 electrons). The 4p subshell fills next. Note that for three series of elements, scandium (Sc) through copper (Cu), yttrium (Y) through silver (Ag), and lutetium (Lu) through gold (Au), a total of 10 d electrons are successively added to the (n – 1) shell next to the n shell to bring that (n – 1) shell from 8 to 18 electrons. For two series, lanthanum (La) through lutetium (Lu) and actinium (Ac) through lawrencium (Lr), 14 f electrons (l = 3, 2l + 1 = 7 ml values; thus, seven orbitals with a combined capacity of 14 electrons) are successively added to the (n – 2) shell to bring that shell from 18 electrons to a total of 32 electrons.
Example $1$: Quantum Numbers and Electron Configurations
What is the electron configuration and orbital diagram for a phosphorus atom? What are the four quantum numbers for the last electron added?
Solution
The atomic number of phosphorus is 15. Thus, a phosphorus atom contains 15 electrons. The order of filling of the energy levels is 1s, 2s, 2p, 3s, 3p, 4s, . . . The 15 electrons of the phosphorus atom will fill up to the 3p orbital, which will contain three electrons:
The last electron added is a 3p electron. Therefore, n = 3 and, for a p-type orbital, l = 1. The ml value could be –1, 0, or +1. The three p orbitals are degenerate, so any of these ml values is correct. For unpaired electrons, convention assigns the value of $+\dfrac{1}{2}$ for the spin quantum number; thus, $m_s=+\dfrac{1}{2}$.
Exercise $1$
Identify the atoms from the electron configurations given:
1. [Ar]4s23d5
2. [Kr]5s24d105p6
Answer a
Mn
Answer b
Xe
The periodic table can be a powerful tool in predicting the electron configuration of an element. However, we do find exceptions to the order of filling of orbitals that are shown in Figure $3$ or $4$. For instance, the electron configurations of the transition metals chromium (Cr; atomic number 24) and copper (Cu; atomic number 29), among others, are not those we would expect. In general, such exceptions involve subshells with very similar energy, and small effects can lead to changes in the order of filling.
In the case of Cr and Cu, we find that half-filled and completely filled subshells apparently represent conditions of preferred stability. This stability is such that an electron shifts from the 4s into the 3d orbital to gain the extra stability of a half-filled 3d subshell (in Cr) or a filled 3d subshell (in Cu). Other exceptions also occur. For example, niobium (Nb, atomic number 41) is predicted to have the electron configuration [Kr]5s24d3. Experimentally, we observe that its ground-state electron configuration is actually [Kr]5s14d4. We can rationalize this observation by saying that the electron–electron repulsions experienced by pairing the electrons in the 5s orbital are larger than the gap in energy between the 5s and 4d orbitals. There is no simple method to predict the exceptions for atoms where the magnitude of the repulsions between electrons is greater than the small differences in energy between subshells.
Electron Configurations and the Periodic Table
As described earlier, the periodic table arranges atoms based on increasing atomic number so that elements with the same chemical properties recur periodically. When their electron configurations are added to the table (Figure $6$), we also see a periodic recurrence of similar electron configurations in the outer shells of these elements. Because they are in the outer shells of an atom, valence electrons play the most important role in chemical reactions. The outer electrons have the highest energy of the electrons in an atom and are more easily lost or shared than the core electrons. Valence electrons are also the determining factor in some physical properties of the elements.
Elements in any one group (or column) have the same number of valence electrons; the alkali metals lithium and sodium each have only one valence electron, the alkaline earth metals beryllium and magnesium each have two, and the halogens fluorine and chlorine each have seven valence electrons. The similarity in chemical properties among elements of the same group occurs because they have the same number of valence electrons. It is the loss, gain, or sharing of valence electrons that defines how elements react.
It is important to remember that the periodic table was developed on the basis of the chemical behavior of the elements, well before any idea of their atomic structure was available. Now we can understand why the periodic table has the arrangement it has—the arrangement puts elements whose atoms have the same number of valence electrons in the same group. This arrangement is emphasized in Figure $6$, which shows in periodic-table form the electron configuration of the last subshell to be filled by the Aufbau principle. The colored sections of Figure $6$ show the three categories of elements classified by the orbitals being filled: main group, transition, and inner transition elements. These classifications determine which orbitals are counted in the valence shell, or highest energy level orbitals of an atom.
1. Main group elements (sometimes called representative elements) are those in which the last electron added enters an s or a p orbital in the outermost shell, shown in blue and red in Figure $6$. This category includes all the nonmetallic elements, as well as many metals and the intermediate semimetallic elements. The valence electrons for main group elements are those with the highest n level. For example, gallium (Ga, atomic number 31) has the electron configuration [Ar]4s23d104p1, which contains three valence electrons (underlined). The completely filled d orbitals count as core, not valence, electrons.
2. Transition elements or transition metals. These are metallic elements in which the last electron added enters a d orbital. The valence electrons (those added after the last noble gas configuration) in these elements include the ns and (n – 1) d electrons. The official IUPAC definition of transition elements specifies those with partially filled d orbitals. Thus, the elements with completely filled orbitals (Zn, Cd, Hg, as well as Cu, Ag, and Au in Figure $6$) are not technically transition elements. However, the term is frequently used to refer to the entire d block (colored yellow in Figure $6$), and we will adopt this usage in this textbook.
3. Inner transition elements are metallic elements in which the last electron added occupies an f orbital. They are shown in green in Figure $6$. The valence shells of the inner transition elements consist of the (n – 2)f, the (n – 1)d, and the ns subshells. There are two inner transition series:
1. The lanthanide series: lanthanide (La) through lutetium (Lu)
2. The actinide series: actinide (Ac) through lawrencium (Lr)
Lanthanum and actinium, because of their similarities to the other members of the series, are included and used to name the series, even though they are transition metals with no f electrons.
Electron Configurations of Ions
We have seen that ions are formed when atoms gain or lose electrons. A cation (positively charged ion) forms when one or more electrons are removed from a parent atom. For main group elements, the electrons that were added last are the first electrons removed. For transition metals and inner transition metals, however, electrons in the s orbital are easier to remove than the d or f electrons, and so the highest ns electrons are lost, and then the (n – 1)d or (n – 2)f electrons are removed. An anion (negatively charged ion) forms when one or more electrons are added to a parent atom. The added electrons fill in the order predicted by the Aufbau principle.
Example $2$: Predicting Electron Configurations of Ions
What is the electron configuration and orbital diagram of:
1. Na+
2. P3–
3. Al2+
4. Fe2+
5. Sm3+
Solution
First, write out the electron configuration for each parent atom. We have chosen to show the full, unabbreviated configurations to provide more practice for students who want it, but listing the core-abbreviated electron configurations is also acceptable.
Next, determine whether an electron is gained or lost. Remember electrons are negatively charged, so ions with a positive charge have lost an electron. For main group elements, the last orbital gains or loses the electron. For transition metals, the last s orbital loses an electron before the d orbitals.
1. Na: 1s22s22p63s1. Sodium cation loses one electron, so Na+: 1s22s22p63s1 = Na+: 1s22s22p6.
2. P: 1s22s22p63s23p3. Phosphorus trianion gains three electrons, so P3−: 1s22s22p63s23p6.
3. Al: 1s22s22p63s23p1. Aluminum dication loses two electrons Al2+: 1s22s22p63s23p1 = Al2+: 1s22s22p63s1.
4. Fe: 1s22s22p63s23p64s23d6. Iron(II) loses two electrons and, since it is a transition metal, they are removed from the 4s orbital Fe2+: 1s22s22p63s23p64s23d6 = 1s22s22p63s23p63d6.
5. Sm: 1s22s22p63s23p64s23d104p65s24d105p66s24f6. Samarium trication loses three electrons. The first two will be lost from the 6s orbital, and the final one is removed from the 4f orbital. Sm3+: 1s22s22p63s23p64s23d104p65s24d105p66s24f6 = 1s22s22p63s23p64s23d104p65s24d105p64f5.
Exercise $2$
1. Which ion with a +2 charge has the electron configuration 1s22s22p63s23p63d104s24p64d5?
2. Which ion with a +3 charge has this configuration?
Answer a
Tc2+
Answer b
Ru3+
Summary
The relative energy of the subshells determine the order in which atomic orbitals are filled (1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, and so on). Electron configurations and orbital diagrams can be determined by applying the Pauli exclusion principle (no two electrons can have the same set of four quantum numbers) and Hund’s rule (whenever possible, electrons retain unpaired spins in degenerate orbitals).
Electrons in the outermost orbitals, called valence electrons, are responsible for most of the chemical behavior of elements. In the periodic table, elements with analogous valence electron configurations usually occur within the same group. There are some exceptions to the predicted filling order, particularly when half-filled or completely filled orbitals can be formed. The periodic table can be divided into three categories based on the orbital in which the last electron to be added is placed: main group elements (s and p orbitals), transition elements (d orbitals), and inner transition elements (f orbitals).
Glossary
Aufbau principle
procedure in which the electron configuration of the elements is determined by “building” them in order of atomic numbers, adding one proton to the nucleus and one electron to the proper subshell at a time
core electron
electron in an atom that occupies the orbitals of the inner shells
electron configuration
electronic structure of an atom in its ground state given as a listing of the orbitals occupied by the electrons
Hund’s rule
every orbital in a subshell is singly occupied with one electron before any one orbital is doubly occupied, and all electrons in singly occupied orbitals have the same spin
orbital diagram
pictorial representation of the electron configuration showing each orbital as a box and each electron as an arrow
valence electrons
electrons in the outermost or valence shell (highest value of n) of a ground-state atom; determine how an element reacts
valence shell
outermost shell of electrons in a ground-state atom; for main group elements, the orbitals with the highest n level (s and p subshells) are in the valence shell, while for transition metals, the highest energy s and d subshells make up the valence shell and for inner transition elements, the highest s, d, and f subshells are included | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/03%3A_Electronic_Structure_and_Periodic_Properties/3.4%3A_Electronic_Structure_of_Atoms_%28Electron_Configurations%29.txt |
Learning Objectives
• Describe and explain the observed trends in atomic size, ionization energy, and electron affinity of the elements
The elements in groups (vertical columns) of the periodic table exhibit similar chemical behavior. This similarity occurs because the members of a group have the same number and distribution of electrons in their valence shells. However, there are also other patterns in chemical properties on the periodic table. For example, as we move down a group, the metallic character of the atoms increases. Oxygen, at the top of Group 16 (6A), is a colorless gas; in the middle of the group, selenium is a semiconducting solid; and, toward the bottom, polonium is a silver-grey solid that conducts electricity.
As we go across a period from left to right, we add a proton to the nucleus and an electron to the valence shell with each successive element. As we go down the elements in a group, the number of electrons in the valence shell remains constant, but the principal quantum number increases by one each time. An understanding of the electronic structure of the elements allows us to examine some of the properties that govern their chemical behavior. These properties vary periodically as the electronic structure of the elements changes. They are (1) size (radius) of atoms and ions, (2) ionization energies, and (3) electron affinities.
Variation in Covalent Radius
The quantum mechanical picture makes it difficult to establish a definite size of an atom. However, there are several practical ways to define the radius of atoms and, thus, to determine their relative sizes that give roughly similar values. We will use the covalent radius (Figure $1$), which is defined as one-half the distance between the nuclei of two identical atoms when they are joined by a covalent bond (this measurement is possible because atoms within molecules still retain much of their atomic identity).
We know that as we scan down a group, the principal quantum number, n, increases by one for each element. Thus, the electrons are being added to a region of space that is increasingly distant from the nucleus. Consequently, the size of the atom (and its covalent radius) must increase as we increase the distance of the outermost electrons from the nucleus. This trend is illustrated for the covalent radii of the halogens in Table $1$ and Figure $1$. The trends for the entire periodic table can be seen in Figure $2$.
Table $1$: Covalent Radii of the Halogen Group Elements
Atom Covalent radius (pm) Nuclear charge
F 64 +9
Cl 99 +17
Br 114 +35
I 133 +53
At 148 +85
As shown in Figure $2$, as we move across a period from left to right, we generally find that each element has a smaller covalent radius than the element preceding it. This might seem counterintuitive because it implies that atoms with more electrons have a smaller atomic radius. This can be explained with the concept of effective nuclear charge, $Z_{eff}$. This is the pull exerted on a specific electron by the nucleus, taking into account any electron–electron repulsions. For hydrogen, there is only one electron and so the nuclear charge (Z) and the effective nuclear charge (Zeff) are equal. For all other atoms, the inner electrons partially shield the outer electrons from the pull of the nucleus, and thus:
$Z_\ce{eff}=Z−shielding \nonumber$
Shielding is determined by the probability of another electron being between the electron of interest and the nucleus, as well as by the electron–electron repulsions the electron of interest encounters. Core electrons are adept at shielding, while electrons in the same valence shell do not block the nuclear attraction experienced by each other as efficiently. Thus, each time we move from one element to the next across a period, Z increases by one, but the shielding increases only slightly. Thus, Zeff increases as we move from left to right across a period. The stronger pull (higher effective nuclear charge) experienced by electrons on the right side of the periodic table draws them closer to the nucleus, making the covalent radii smaller.
Thus, as we would expect, the outermost or valence electrons are easiest to remove because they have the highest energies, are shielded more, and are farthest from the nucleus. As a general rule, when the representative elements form cations, they do so by the loss of the ns or np electrons that were added last in the Aufbau process. The transition elements, on the other hand, lose the ns electrons before they begin to lose the (n – 1)d electrons, even though the ns electrons are added first, according to the Aufbau principle.
Example $1$: Sorting Atomic Radii
Predict the order of increasing covalent radius for Ge, Fl, Br, Kr.
Solution
Radius increases as we move down a group, so Ge < Fl (Note: Fl is the symbol for flerovium, element 114, NOT fluorine). Radius decreases as we move across a period, so Kr < Br < Ge. Putting the trends together, we obtain Kr < Br < Ge < Fl.
Exercise $1$
Give an example of an atom whose size is smaller than fluorine.
Answer
Ne or He
Variation in Ionic Radii
Ionic radius is the measure used to describe the size of an ion. A cation always has fewer electrons and the same number of protons as the parent atom; it is smaller than the atom from which it is derived (Figure $3$). For example, the covalent radius of an aluminum atom (1s22s22p63s23p1) is 118 pm, whereas the ionic radius of an Al3+ (1s22s22p6) is 68 pm. As electrons are removed from the outer valence shell, the remaining core electrons occupying smaller shells experience a greater effective nuclear charge Zeff (as discussed) and are drawn even closer to the nucleus.
Cations with larger charges are smaller than cations with smaller charges (e.g., V2+ has an ionic radius of 79 pm, while that of V3+ is 64 pm). Proceeding down the groups of the periodic table, we find that cations of successive elements with the same charge generally have larger radii, corresponding to an increase in the principal quantum number, n.
An anion (negative ion) is formed by the addition of one or more electrons to the valence shell of an atom. This results in a greater repulsion among the electrons and a decrease in $Z_{eff}$ per electron. Both effects (the increased number of electrons and the decreased Zeff) cause the radius of an anion to be larger than that of the parent atom ( Figure $3$). For example, a sulfur atom ([Ne]3s23p4) has a covalent radius of 104 pm, whereas the ionic radius of the sulfide anion ([Ne]3s23p6) is 170 pm. For consecutive elements proceeding down any group, anions have larger principal quantum numbers and, thus, larger radii.
Atoms and ions that have the same electron configuration are said to be isoelectronic. Examples of isoelectronic species are N3–, O2–, F, Ne, Na+, Mg2+, and Al3+ (1s22s22p6). Another isoelectronic series is P3–, S2–, Cl, Ar, K+, Ca2+, and Sc3+ ([Ne]3s23p6). For atoms or ions that are isoelectronic, the number of protons determines the size. The greater the nuclear charge, the smaller the radius in a series of isoelectronic ions and atoms.
Variation in Ionization Energies
The amount of energy required to remove the most loosely bound electron from a gaseous atom in its ground state is called its first ionization energy (IE1). The first ionization energy for an element, X, is the energy required to form a cation with +1 charge:
$\ce{X}(g)⟶\ce{X+}(g)+\ce{e-}\hspace{20px}\ce{IE_1} \nonumber$
The energy required to remove the second most loosely bound electron is called the second ionization energy (IE2).
$\ce{X+}(g)⟶\ce{X^2+}(g)+\ce{e-}\hspace{20px}\ce{IE_2} \nonumber$
The energy required to remove the third electron is the third ionization energy, and so on. Energy is always required to remove electrons from atoms or ions, so ionization processes are endothermic and IE values are always positive. For larger atoms, the most loosely bound electron is located farther from the nucleus and so is easier to remove. Thus, as size (atomic radius) increases, the ionization energy should decrease. Relating this logic to what we have just learned about radii, we would expect first ionization energies to decrease down a group and to increase across a period.
Figure $4$ graphs the relationship between the first ionization energy and the atomic number of several elements. Within a period, the values of first ionization energy for the elements (IE1) generally increases with increasing Z. Down a group, the IE1 value generally decreases with increasing Z. There are some systematic deviations from this trend, however. Note that the ionization energy of boron (atomic number 5) is less than that of beryllium (atomic number 4) even though the nuclear charge of boron is greater by one proton. This can be explained because the energy of the subshells increases as l increases, due to penetration and shielding (as discussed previously in this chapter). Within any one shell, the s electrons are lower in energy than the p electrons. This means that an s electron is harder to remove from an atom than a p electron in the same shell. The electron removed during the ionization of beryllium ([He]2s2) is an s electron, whereas the electron removed during the ionization of boron ([He]2s22p1) is a p electron; this results in a lower first ionization energy for boron, even though its nuclear charge is greater by one proton. Thus, we see a small deviation from the predicted trend occurring each time a new subshell begins.
Another deviation occurs as orbitals become more than one-half filled. The first ionization energy for oxygen is slightly less than that for nitrogen, despite the trend in increasing IE1 values across a period. Looking at the orbital diagram of oxygen, we can see that removing one electron will eliminate the electron–electron repulsion caused by pairing the electrons in the 2p orbital and will result in a half-filled orbital (which is energetically favorable). Analogous changes occur in succeeding periods (note the dip for sulfur after phosphorus in Figure $4$.
Removing an electron from a cation is more difficult than removing an electron from a neutral atom because of the greater electrostatic attraction to the cation. Likewise, removing an electron from a cation with a higher positive charge is more difficult than removing an electron from an ion with a lower charge. Thus, successive ionization energies for one element always increase. As seen in Table $2$, there is a large increase in the ionization energies (color change) for each element. This jump corresponds to removal of the core electrons, which are harder to remove than the valence electrons. For example, Sc and Ga both have three valence electrons, so the rapid increase in ionization energy occurs after the third ionization.
Table $2$: Successive Ionization Energies for Selected Elements (kJ/mol)
Element IE1 IE2 IE3 IE4 IE5 IE6 IE7
K 418.8 3051.8 4419.6 5876.9 7975.5 9590.6 11343
Ca 589.8 1145.4 4912.4 6490.6 8153.0 10495.7 12272.9
Sc 633.1 1235.0 2388.7 7090.6 8842.9 10679.0 13315.0
Ga 578.8 1979.4 2964.6 6180 8298.7 10873.9 13594.8
Ge 762.2 1537.5 3302.1 4410.6 9021.4 Not available Not available
As 944.5 1793.6 2735.5 4836.8 6042.9 12311.5 Not available
Example $2$: Ranking Ionization Energies
Predict the order of increasing energy for the following processes: IE1 for Al, IE1 for Tl, IE2 for Na, IE3 for Al.
Solution
Removing the 6p1 electron from Tl is easier than removing the 3p1 electron from Al because the higher n orbital is farther from the nucleus, so IE1(Tl) < IE1(Al). Ionizing the third electron from
$\ce{Al}\hspace{20px}\ce{(Al^2+⟶Al^3+ + e- )} \nonumber$
requires more energy because the cation Al2+ exerts a stronger pull on the electron than the neutral Al atom, so IE1(Al) < IE3(Al). The second ionization energy for sodium removes a core electron, which is a much higher energy process than removing valence electrons. Putting this all together, we obtain:
IE1(Tl) < IE1(Al) < IE3(Al) < IE2(Na).
Exercise $2$
Which has the lowest value for IE1: O, Po, Pb, or Ba?
Answer
Ba
Variation in Electron Affinities
The electron affinity [EA] is the energy change for the process of adding an electron to a gaseous atom to form an anion (negative ion).
$\ce{X}(g)+\ce{e-}⟶\ce{X-}(g)\hspace{20px}\ce{EA_1} \nonumber$
This process can be either endothermic or exothermic, depending on the element. The EA of some of the elements is given in Figure $6$. You can see that many of these elements have negative values of EA, which means that energy is released when the gaseous atom accepts an electron. However, for some elements, energy is required for the atom to become negatively charged and the value of their EA is positive. Just as with ionization energy, subsequent EA values are associated with forming ions with more charge. The second EA is the energy associated with adding an electron to an anion to form a –2 ion, and so on.
As we might predict, it becomes easier to add an electron across a series of atoms as the effective nuclear charge of the atoms increases. We find, as we go from left to right across a period, EAs tend to become more negative. The exceptions found among the elements of group 2 (2A), group 15 (5A), and group 18 (8A) can be understood based on the electronic structure of these groups. The noble gases, group 18 (8A), have a completely filled shell and the incoming electron must be added to a higher n level, which is more difficult to do. Group 2 (2A) has a filled ns subshell, and so the next electron added goes into the higher energy np, so, again, the observed EA value is not as the trend would predict. Finally, group 15 (5A) has a half-filled np subshell and the next electron must be paired with an existing np electron. In all of these cases, the initial relative stability of the electron configuration disrupts the trend in EA.
We also might expect the atom at the top of each group to have the largest EA; their first ionization potentials suggest that these atoms have the largest effective nuclear charges. However, as we move down a group, we see that the second element in the group most often has the greatest EA. The reduction of the EA of the first member can be attributed to the small size of the n = 2 shell and the resulting large electron–electron repulsions. For example, chlorine, with an EA value of –348 kJ/mol, has the highest value of any element in the periodic table. The EA of fluorine is –322 kJ/mol. When we add an electron to a fluorine atom to form a fluoride anion (F), we add an electron to the n = 2 shell. The electron is attracted to the nucleus, but there is also significant repulsion from the other electrons already present in this small valence shell. The chlorine atom has the same electron configuration in the valence shell, but because the entering electron is going into the n = 3 shell, it occupies a considerably larger region of space and the electron–electron repulsions are reduced. The entering electron does not experience as much repulsion and the chlorine atom accepts an additional electron more readily.
The properties discussed in this section (size of atoms and ions, effective nuclear charge, ionization energies, and electron affinities) are central to understanding chemical reactivity. For example, because fluorine has an energetically favorable EA and a large energy barrier to ionization (IE), it is much easier to form fluorine anions than cations. Metallic properties including conductivity and malleability (the ability to be formed into sheets) depend on having electrons that can be removed easily. Thus, metallic character increases as we move down a group and decreases across a period in the same trend observed for atomic size because it is easier to remove an electron that is farther away from the nucleus.
Summary
Electron configurations allow us to understand many periodic trends. Covalent radius increases as we move down a group because the n level (orbital size) increases. Covalent radius mostly decreases as we move left to right across a period because the effective nuclear charge experienced by the electrons increases, and the electrons are pulled in tighter to the nucleus. Anionic radii are larger than the parent atom, while cationic radii are smaller, because the number of valence electrons has changed while the nuclear charge has remained constant. Ionization energy (the energy associated with forming a cation) decreases down a group and mostly increases across a period because it is easier to remove an electron from a larger, higher energy orbital. Electron affinity (the energy associated with forming an anion) is more favorable (exothermic) when electrons are placed into lower energy orbitals, closer to the nucleus. Therefore, electron affinity becomes increasingly negative as we move left to right across the periodic table and decreases as we move down a group. For both IE and electron affinity data, there are exceptions to the trends when dealing with completely filled or half-filled subshells.
Glossary
covalent radius
one-half the distance between the nuclei of two identical atoms when they are joined by a covalent bond
effective nuclear charge
charge that leads to the Coulomb force exerted by the nucleus on an electron, calculated as the nuclear charge minus shielding
electron affinity
energy required to add an electron to a gaseous atom to form an anion
ionization energy
energy required to remove an electron from a gaseous atom or ion. The associated number (e.g., second ionization energy) corresponds to the charge of the ion produced (X2+)
isoelectronic
group of ions or atoms that have identical electron configurations | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/03%3A_Electronic_Structure_and_Periodic_Properties/3.5%3A_Periodic_Variations_in_Element_Properties.txt |
3.1: Formula Mass and the Mole Concept
What is the total mass (amu) of carbon in each of the following molecules?
1. (a) CH4
2. (b) CHCl3
3. (c) C12H10O6
4. (d) CH3CH2CH2CH2CH3
(a) 12.01 amu; (b) 12.01 amu; (c) 144.12 amu; (d) 60.05 amu
What is the total mass of hydrogen in each of the molecules?
1. (a) CH4
2. (b) CHCl3
3. (c) C12H10O6
4. (d) CH3CH2CH2CH2CH3
Calculate the molecular or formula mass of each of the following:
(a) P4
(b) H2O
(c) Ca(NO3)2
(d) CH3CO2H (acetic acid)
(e) C12H22O11 (sucrose, cane sugar).
(a) 123.896 amu; (b) 18.015 amu; (c) 164.086 amu; (d) 60.052 amu; (e) 342.297 amu
Determine the molecular mass of the following compounds:
(a)
(b)
(c)
(d)
Determine the molecular mass of the following compounds:
(a)
(b)
(c)
(d)
1. (a) 56.107 amu;
2. (b) 54.091 amu;
3. (c) 199.9976 amu;
4. (d) 97.9950 amu
Which molecule has a molecular mass of 28.05 amu?
(a)
(b)
(c)
Write a sentence that describes how to determine the number of moles of a compound in a known mass of the compound if we know its molecular formula.
Use the molecular formula to find the molar mass; to obtain the number of moles, divide the mass of compound by the molar mass of the compound expressed in grams.
Compare 1 mole of H2, 1 mole of O2, and 1 mole of F2.
1. (a) Which has the largest number of molecules? Explain why.
2. (b) Which has the greatest mass? Explain why.
Which contains the greatest mass of oxygen: 0.75 mol of ethanol (C2H5OH), 0.60 mol of formic acid (HCO2H), or 1.0 mol of water (H2O)? Explain why.
Formic acid. Its formula has twice as many oxygen atoms as the other two compounds (one each). Therefore, 0.60 mol of formic acid would be equivalent to 1.20 mol of a compound containing a single oxygen atom.
Which contains the greatest number of moles of oxygen atoms: 1 mol of ethanol (C2H5OH), 1 mol of formic acid (HCO2H), or 1 mol of water (H2O)? Explain why.
How are the molecular mass and the molar mass of a compound similar and how are they different?
The two masses have the same numerical value, but the units are different: The molecular mass is the mass of 1 molecule while the molar mass is the mass of 6.022 × 1023 molecules.
Calculate the molar mass of each of the following compounds:
1. (a) hydrogen fluoride, HF
2. (b) ammonia, NH3
3. (c) nitric acid, HNO3
4. (d) silver sulfate, Ag2SO4
5. (e) boric acid, B(OH)3
Calculate the molar mass of each of the following:
1. (a) S8
2. (b) C5H12
3. (c) Sc2(SO4)3
4. (d) CH3COCH3 (acetone)
5. (e) C6H12O6 (glucose)
(a) 256.528 g/mol; (b) 72.150 g mol−1; (c) 378.103 g mol−1; (d) 58.080 g mol−1; (e) 180.158 g mol−1
Calculate the empirical or molecular formula mass and the molar mass of each of the following minerals:
1. (a) limestone, CaCO3
2. (b) halite, NaCl
3. (c) beryl, Be3Al2Si6O18
4. (d) malachite, Cu2(OH)2CO3
5. (e) turquoise, CuAl6(PO4)4(OH)8(H2O)4
Calculate the molar mass of each of the following:
1. (a) the anesthetic halothane, C2HBrClF3
2. (b) the herbicide paraquat, C12H14N2Cl2
3. (c) caffeine, C8H10N4O2
4. (d) urea, CO(NH2)2
5. (e) a typical soap, C17H35CO2Na
(a) 197.382 g mol−1; (b) 257.163 g mol−1; (c) 194.193 g mol−1; (d) 60.056 g mol−1; (e) 306.464 g mol−1
Determine the number of moles of compound and the number of moles of each type of atom in each of the following:
1. (a) 25.0 g of propylene, C3H6
2. (b) 3.06 × 10−3 g of the amino acid glycine, C2H5NO2
3. (c) 25 lb of the herbicide Treflan, C13H16N2O4F (1 lb = 454 g)
4. (d) 0.125 kg of the insecticide Paris Green, Cu4(AsO3)2(CH3CO2)2
5. (e) 325 mg of aspirin, C6H4(CO2H)(CO2CH3)
Determine the mass of each of the following:
1. (a) 0.0146 mol KOH
2. (b) 10.2 mol ethane, C2H6
3. (c) 1.6 × 10−3 mol Na2 SO4
4. (d) 6.854 × 103 mol glucose, C6 H12 O6
5. (e) 2.86 mol Co(NH3)6Cl3
1. (a) 0.819 g;
2. (b) 307 g;
3. (c) 0.23 g;
4. (d) 1.235 × 106 g (1235 kg);
5. (e) 765 g
Determine the number of moles of the compound and determine the number of moles of each type of atom in each of the following:
1. (a) 2.12 g of potassium bromide, KBr
2. (b) 0.1488 g of phosphoric acid, H3PO4
3. (c) 23 kg of calcium carbonate, CaCO3
4. (d) 78.452 g of aluminum sulfate, Al2(SO4)3
5. (e) 0.1250 mg of caffeine, C8H10N4O2
Determine the mass of each of the following:
1. (a) 2.345 mol LiCl
2. (b) 0.0872 mol acetylene, C2H2
3. (c) 3.3 × 10−2 mol Na2 CO3
4. (d) 1.23 × 103 mol fructose, C6 H12 O6
5. (e) 0.5758 mol FeSO4(H2O)7
1. (a) 99.41 g;
2. (b) 2.27 g;
3. (c) 3.5 g;
4. (d) 222 kg;
5. (e) 160.1 g
The approximate minimum daily dietary requirement of the amino acid leucine, C6H13NO2, is 1.1 g. What is this requirement in moles?
Determine the mass in grams of each of the following:
1. (a) 0.600 mol of oxygen atoms
2. (b) 0.600 mol of oxygen molecules, O2
3. (c) 0.600 mol of ozone molecules, O3
(a) 9.60 g; (b) 19.2 g; (c) 28.8 g
A 55-kg woman has 7.5 × 10−3 mol of hemoglobin (molar mass = 64,456 g/mol) in her blood. How many hemoglobin molecules is this? What is this quantity in grams?
Determine the number of atoms and the mass of zirconium, silicon, and oxygen found in 0.3384 mol of zircon, ZrSiO4, a semiprecious stone.
zirconium: 2.038 × 1023 atoms; 30.87 g; silicon: 2.038 × 1023 atoms; 9.504 g; oxygen: 8.151 × 1023 atoms; 21.66 g
Determine which of the following contains the greatest mass of hydrogen: 1 mol of CH4, 0.6 mol of C6H6, or 0.4 mol of C3H8.
Determine which of the following contains the greatest mass of aluminum: 122 g of AlPO4, 266 g of Al2Cl6, or 225 g of Al2S3.
AlPO4: 1.000 mol
Al2Cl6: 1.994 mol
Al2S3: 3.00 mol
Diamond is one form of elemental carbon. An engagement ring contains a diamond weighing 1.25 carats (1 carat = 200 mg). How many atoms are present in the diamond?
The Cullinan diamond was the largest natural diamond ever found (January 25, 1905). It weighed 3104 carats (1 carat = 200 mg). How many carbon atoms were present in the stone?
3.113 × 1025 C atoms
One 55-gram serving of a particular cereal supplies 270 mg of sodium, 11% of the recommended daily allowance. How many moles and atoms of sodium are in the recommended daily allowance?
A certain nut crunch cereal contains 11.0 grams of sugar (sucrose, C12H22O11) per serving size of 60.0 grams. How many servings of this cereal must be eaten to consume 0.0278 moles of sugar?
0.865 servings, or about 1 serving.
A tube of toothpaste contains 0.76 g of sodium monofluorophosphate (Na2PO3F) in 100 mL.
1. What mass of fluorine atoms in mg was present?
2. How many fluorine atoms were present?
Which of the following represents the least number of molecules?
1. 20.0 g of H2O (18.02 g/mol)
2. 77.0 g of CH4 (16.06 g/mol)
3. 68.0 g of CaH2 (42.09 g/mol)
4. 100.0 g of N2O (44.02 g/mol)
5. 84.0 g of HF (20.01 g/mol)
20.0 g H2O represents the least number of molecules since it has the least number of moles.
3.2: Determining Empirical and Molecular Formulas
What information do we need to determine the molecular formula of a compound from the empirical formula?
Calculate the following to four significant figures:
1. (a) the percent composition of ammonia, NH3
2. (b) the percent composition of photographic “hypo,” Na2S2O3
3. (c) the percent of calcium ion in Ca3(PO4)2
(a) % N = 82.24%
% H = 17.76%;
(b) % Na = 29.08%
% S = 40.56%
% O = 30.36%;
(c) % Ca2+ = 38.76%
Determine the following to four significant figures:
1. the percent composition of hydrazoic acid, HN3
2. the percent composition of TNT, C6H2(CH3)(NO2)3
3. the percent of SO42– in Al2(SO4)3
Determine the percent ammonia, NH3, in Co(NH3)6Cl3, to three significant figures.
% NH3 = 38.2%
Determine the percent water in CuSO4∙5H2O to three significant figures.
Determine the empirical formulas for compounds with the following percent compositions:
(a) 15.8% carbon and 84.2% sulfur
(b) 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen
(a) CS2
(b) CH2O
Determine the empirical formulas for compounds with the following percent compositions:
(a) 43.6% phosphorus and 56.4% oxygen
(b) 28.7% K, 1.5% H, 22.8% P, and 47.0% O
A compound of carbon and hydrogen contains 92.3% C and has a molar mass of 78.1 g/mol. What is its molecular formula?
C6H6
Dichloroethane, a compound that is often used for dry cleaning, contains carbon, hydrogen, and chlorine. It has a molar mass of 99 g/mol. Analysis of a sample shows that it contains 24.3% carbon and 4.1% hydrogen. What is its molecular formula?
Determine the empirical and molecular formula for chrysotile asbestos. Chrysotile has the following percent composition: 28.03% Mg, 21.60% Si, 1.16% H, and 49.21% O. The molar mass for chrysotile is 520.8 g/mol.
Mg3Si2H3O8 (empirical formula), Mg6Si4H6O16 (molecular formula)
Polymers are large molecules composed of simple units repeated many times. Thus, they often have relatively simple empirical formulas. Calculate the empirical formulas of the following polymers:
1. Lucite (Plexiglas); 59.9% C, 8.06% H, 32.0% O
2. Saran; 24.8% C, 2.0% H, 73.1% Cl
3. polyethylene; 86% C, 14% H
4. polystyrene; 92.3% C, 7.7% H
5. Orlon; 67.9% C, 5.70% H, 26.4% N
A major textile dye manufacturer developed a new yellow dye. The dye has a percent composition of 75.95% C, 17.72% N, and 6.33% H by mass with a molar mass of about 240 g/mol. Determine the molecular formula of the dye.
C15H15N3
3.3: Molarity
Questions
Explain what changes and what stays the same when 1.00 L of a solution of NaCl is diluted to 1.80 L.
What information do we need to calculate the molarity of a sulfuric acid solution?
We need to know the number of moles of sulfuric acid dissolved in the solution and the volume of the solution.
What does it mean when we say that a 200-mL sample and a 400-mL sample of a solution of salt have the same molarity? In what ways are the two samples identical? In what ways are these two samples different?
Determine the molarity for each of the following solutions:
1. 0.444 mol of CoCl2 in 0.654 L of solution
2. 98.0 g of phosphoric acid, H3PO4, in 1.00 L of solution
3. 0.2074 g of calcium hydroxide, Ca(OH)2, in 40.00 mL of solution
4. 10.5 kg of Na2SO4·10H2O in 18.60 L of solution
5. 7.0 × 10−3 mol of I2 in 100.0 mL of solution
6. 1.8 × 104 mg of HCl in 0.075 L of solution
1. (a) 0.679 M;
2. (b) 1.00 M;
3. (c) 0.06998 M;
4. (d) 1.75 M;
5. (e) 0.070 M;
6. (f) 6.6 M
Determine the molarity of each of the following solutions:
1. 1.457 mol KCl in 1.500 L of solution
2. 0.515 g of H2SO4 in 1.00 L of solution
3. 20.54 g of Al(NO3)3 in 1575 mL of solution
4. 2.76 kg of CuSO4·5H2O in 1.45 L of solution
5. 0.005653 mol of Br2 in 10.00 mL of solution
6. 0.000889 g of glycine, C2H5NO2, in 1.05 mL of solution
Answers:
a.) 0.9713 M
b.) 5.25x10-3 M
c.) 6.122x10-2 M
d.) 7.62 M
e.) 0.5653 M
f.) 1.13x10-2 M
Consider this question: What is the mass of the solute in 0.500 L of 0.30 M glucose, C6H12O6, used for intravenous injection?
(a) Outline the steps necessary to answer the question.
(b) Answer the question.
(a) determine the number of moles of glucose in 0.500 L of solution; determine the molar mass of glucose; determine the mass of glucose from the number of moles and its molar mass; (b) 27 g
Consider this question: What is the mass of solute in 200.0 L of a 1.556-M solution of KBr?
1. (a) Outline the steps necessary to answer the question.
2. (b) Answer the question.
Answer:
(a)
1. Calculate to moles of KBr by multiplying the Molarity by the amount of solution (200.0 L)
2. Find the Molar Mass of KBr and convert moles of solute to grams
(b)
$\dfrac{1.556\:moles\:\ce{KBr}}{1\:\cancel{L}}\times 200.0\:\cancel{L}=311.2\:moles\:\ce{KBr}$
$311.2\:\cancel{moles}\:\ce{KBr}\times\dfrac{119.0\:g\:\ce{KBr}}{1\:\cancel{mole}\:\ce{KBr}}=37,030\:g$
37,030g; 37.03 kg
Calculate the number of moles and the mass of the solute in each of the following solutions:
1. (a) 2.00 L of 18.5 M H2SO4, concentrated sulfuric acid
2. (b) 100.0 mL of 3.8 × 10−5 M NaCN, the minimum lethal concentration of sodium cyanide in blood serum
3. (c) 5.50 L of 13.3 M H2CO, the formaldehyde used to “fix” tissue samples
4. (d) 325 mL of 1.8 × 10−6 M FeSO4, the minimum concentration of iron sulfate detectable by taste in drinking water
(a) 37.0 mol H2SO4;
3.63 × 103 g H2SO4;
(b) 3.8 × 10−6 mol NaCN;
1.9 × 10−4 g NaCN;
(c) 73.2 mol H2CO;
2.20 kg H2CO;
(d) 5.9 × 10−7 mol FeSO4;
8.9 × 10−5 g FeSO4
Calculate the number of moles and the mass of the solute in each of the following solutions:
1. 325 mL of 8.23 × 10−5 M KI, a source of iodine in the diet
2. 75.0 mL of 2.2 × 10−5 M H2SO4, a sample of acid rain
3. 0.2500 L of 0.1135 M K2CrO4, an analytical reagent used in iron assays
4. 10.5 L of 3.716 M (NH4)2SO4, a liquid fertilizer
Answers:
a. 2.67x10-5 moles KI; 4.44x10-3g KI
b. 1.7x10-6 moles H2SO4 ; 1.6x10-4 g H2SO4
c. 2.838x10-2 moles K2CrO4 ; 5.510g K2CrO4
d. 39.0 moles (NH4)2SO4 ; 5,160 g (NH4)2SO4
Consider this question: What is the molarity of KMnO4 in a solution of 0.0908 g of KMnO4 in 0.500 L of solution?
1. (a) Outline the steps necessary to answer the question.
2. (b) Answer the question.
(a) Determine the molar mass of KMnO4; determine the number of moles of KMnO4 in the solution; from the number of moles and the volume of solution, determine the molarity; (b) 1.15 × 10−3 M
Consider this question: What is the molarity of HCl if 35.23 mL of a solution of HCl contain 0.3366 g of HCl?
1. (a) Outline the steps necessary to answer the question.
2. (b) Answer the question.
Answer:
(a)
1. Convert g of HCl to moles of HCl and convert mL of solution to L of solution
2. Divide moles of HCl by L of solution
(b)
$0.3366\:\cancel{g}\:\ce{HCl}\times\dfrac{1\:mole\:\ce{HCl}}{36.46\:\cancel{g}\:\ce{HCl}}=9.232\times10^{-3}\:moles\:\ce{HCl}$
$35.23\:mL = 0.03523\:L$
$\dfrac{9.232\times10^{-3}\:moles\:\ce{HCl}}{0.03523\:L}=0.2621\:M\:\ce{HCl}$
0.2621 M ;
Calculate the molarity of each of the following solutions:
(a) 0.195 g of cholesterol, C27H46O, in 0.100 L of serum, the average concentration of cholesterol in human serum
(b) 4.25 g of NH3 in 0.500 L of solution, the concentration of NH3 in household ammonia
(c) 1.49 kg of isopropyl alcohol, C3H7OH, in 2.50 L of solution, the concentration of isopropyl alcohol in rubbing alcohol
(d) 0.029 g of I2 in 0.100 L of solution, the solubility of I2 in water at 20 °C
(a) 5.04 × 10−3 M;
(b) 0.499 M;
(c) 9.92 M;
(d) 1.1 × 10−3 M
Calculate the molarity of each of the following solutions:
1. 293 g HCl in 666 mL of solution, a concentrated HCl solution
2. 2.026 g FeCl3 in 0.1250 L of a solution used as an unknown in general chemistry laboratories
3. 0.001 mg Cd2+ in 0.100 L, the maximum permissible concentration of cadmium in drinking water
4. 0.0079 g C7H5SNO3 in one ounce (29.6 mL), the concentration of saccharin in a diet soft drink.
There is about 1.0 g of calcium, as Ca2+, in 1.0 L of milk. What is the molarity of Ca2+ in milk?
0.025 M
What volume of a 1.00-M Fe(NO3)3 solution can be diluted to prepare 1.00 L of a solution with a concentration of 0.250 M?
If 0.1718 L of a 0.3556-M C3H7OH solution is diluted to a concentration of 0.1222 M, what is the volume of the resulting solution?
0.5000 L
If 4.12 L of a 0.850 M-H3PO4 solution is be diluted to a volume of 10.00 L, what is the concentration the resulting solution?
What volume of a 0.33-M C12H22O11 solution can be diluted to prepare 25 mL of a solution with a concentration of 0.025 M?
1.9 mL
What is the concentration of the NaCl solution that results when 0.150 L of a 0.556-M solution is allowed to evaporate until the volume is reduced to 0.105 L?
What is the molarity of the diluted solution when each of the following solutions is diluted to the given final volume?
1. (a) 1.00 L of a 0.250-M solution of Fe(NO3)3 is diluted to a final volume of 2.00 L
2. (b) 0.5000 L of a 0.1222-M solution of C3H7OH is diluted to a final volume of 1.250 L
3. (c) 2.35 L of a 0.350-M solution of H3PO4 is diluted to a final volume of 4.00 L
4. (d) 22.50 mL of a 0.025-M solution of C12H22O11 is diluted to 100.0 mL
1. (a) 0.125 M;
2. (b) 0.04888 M;
3. (c) 0.206 M;
4. (e) 0.0056 M
What is the final concentration of the solution produced when 225.5 mL of a 0.09988-M solution of Na2CO3 is allowed to evaporate until the solution volume is reduced to 45.00 mL?
A 2.00-L bottle of a solution of concentrated HCl was purchased for the general chemistry laboratory. The solution contained 868.8 g of HCl. What is the molarity of the solution?
11.9 M
An experiment in a general chemistry laboratory calls for a 2.00-M solution of HCl. How many mL of 11.9 M HCl would be required to make 250 mL of 2.00 M HCl?
What volume of a 0.20-M K2SO4 solution contains 57 g of K2SO4?
1.6 L
The US Environmental Protection Agency (EPA) places limits on the quantities of toxic substances that may be discharged into the sewer system. Limits have been established for a variety of substances, including hexavalent chromium, which is limited to 0.50 mg/L. If an industry is discharging hexavalent chromium as potassium dichromate (K2Cr2O7), what is the maximum permissible molarity of that substance?
3.4: Other Units for Solution Concentrations
Questions
1. Consider this question: What mass of a concentrated solution of nitric acid (68.0% HNO3 by mass) is needed to prepare 400.0 g of a 10.0% solution of HNO3 by mass?
1. Outline the steps necessary to answer the question.
2. Answer the question.
2. What mass of a 4.00% NaOH solution by mass contains 15.0 g of NaOH?
3. What mass of solid NaOH (97.0% NaOH by mass) is required to prepare 1.00 L of a 10.0% solution of NaOH by mass? The density of the 10.0% solution is 1.109 g/mL.
4. What mass of HCl is contained in 45.0 mL of an aqueous HCl solution that has a density of 1.19 g cm–3 and contains 37.21% HCl by mass?
5. The hardness of water (hardness count) is usually expressed in parts per million (by mass) of $\ce{CaCO_3}$, which is equivalent to milligrams of $\ce{CaCO_3}$ per liter of water. What is the molar concentration of Ca2+ ions in a water sample with a hardness count of 175 mg CaCO3/L?
6. The level of mercury in a stream was suspected to be above the minimum considered safe (1 part per billion by weight). An analysis indicated that the concentration was 0.68 parts per billion. Assume a density of 1.0 g/mL and calculate the molarity of mercury in the stream.
7. In Canada and the United Kingdom, devices that measure blood glucose levels provide a reading in millimoles per liter. If a measurement of 5.3 mM is observed, what is the concentration of glucose (C6H12O6) in mg/dL?
8. A throat spray is 1.40% by mass phenol, $\ce{C_6H_5OH}$, in water. If the solution has a density of 0.9956 g/mL, calculate the molarity of the solution.
9. Copper(I) iodide (CuI) is often added to table salt as a dietary source of iodine. How many moles of CuI are contained in 1.00 lb (454 g) of table salt containing 0.0100% CuI by mass?
10. A cough syrup contains 5.0% ethyl alcohol, C2H5OH, by mass. If the density of the solution is 0.9928 g/mL, determine the molarity of the alcohol in the cough syrup.
11. D5W is a solution used as an intravenous fluid. It is a 5.0% by mass solution of dextrose ($\ce{C_6H_{12}O_6}$) in water. If the density of D5W is 1.029 g/mL, calculate the molarity of dextrose in the solution.
12. Find the molarity of a 40.0% by mass aqueous solution of sulfuric acid, $\ce{H_2SO_4}$, for which the density is 1.3057 g/mL.
Solutions
1
• (a) The dilution equation can be used, appropriately modified to accommodate mass-based concentration units: $\mathrm{\%\,mass_1 \times mass_1=\%\;mass_2 \times mass_2}$ This equation can be rearranged to isolate $\mathrm{mass_1}$ and the given quantities substituted into this equation.
• (b) 58.8 g
3. $\mathrm{114 \;g}$ 5. $1.75 \times 10^{−3} M$ 7 $\mathrm{95\: mg/dL}$ 9 $\mathrm{2.38 \times 10^{−4}\: mol}$ 11 $\mathrm{0.29 mol}$ | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/03%3A_Electronic_Structure_and_Periodic_Properties/3.E%3A_Electronic_Structure_and_Periodic_Properties_%28Exercises%29.txt |
A chemical bond is an attraction between atoms that allows the formation of chemical substances that contain two or more atoms. The bond is caused by the electrostatic force of attraction between opposite charges, either between electrons and nuclei, or as the result of a dipole attraction. All bonds can be explained by quantum theory, but, in practice, simplification rules allow chemists to predict the strength, directionality, and polarity of bonds. The octet rule and VSEPR theory are two examples. More sophisticated theories are valence bond theory which includes orbital hybridization and resonance, and the linear combination of atomic orbitals molecular orbital method. Electrostatics are used to describe bond polarities and the effects they have on chemical substances.
04: Chemical Bonding and Molecular Geometry
It has long been known that pure carbon occurs in different forms (allotropes) including graphite and diamonds. But it was not until 1985 that a new form of carbon was recognized: buckminsterfullerene, commonly known as a “buckyball.” This molecule was named after the architect and inventor R. Buckminster Fuller (1895–1983), whose signature architectural design was the geodesic dome, characterized by a lattice shell structure supporting a spherical surface. Experimental evidence revealed the formula, C60, and then scientists determined how 60 carbon atoms could form one symmetric, stable molecule. They were guided by bonding theory—the topic of this chapter—which explains how individual atoms connect to form more complex structures.
4.1: Ionic Bonding
Learning Objectives
• Explain the formation of cations, anions, and ionic compounds
• Predict the charge of common metallic and nonmetallic elements, and write their electron configurations
As you have learned, ions are atoms or molecules bearing an electrical charge. A cation (a positive ion) forms when a neutral atom loses one or more electrons from its valence shell, and an anion (a negative ion) forms when a neutral atom gains one or more electrons in its valence shell.
Compounds composed of ions are called ionic compounds (or salts), and their constituent ions are held together by ionic bonds: electrostatic forces of attraction between oppositely charged cations and anions. The properties of ionic compounds shed some light on the nature of ionic bonds. Ionic solids exhibit a crystalline structure and tend to be rigid and brittle; they also tend to have high melting and boiling points, which suggests that ionic bonds are very strong. Ionic solids are also poor conductors of electricity for the same reason—the strength of ionic bonds prevents ions from moving freely in the solid state. Most ionic solids, however, dissolve readily in water. Once dissolved or melted, ionic compounds are excellent conductors of electricity and heat because the ions can move about freely.
Neutral atoms and their associated ions have very different physical and chemical properties. Sodium atoms form sodium metal, a soft, silvery-white metal that burns vigorously in air and reacts explosively with water. Chlorine atoms form chlorine gas, Cl2, a yellow-green gas that is extremely corrosive to most metals and very poisonous to animals and plants. The vigorous reaction between the elements sodium and chlorine forms the white, crystalline compound sodium chloride, common table salt, which contains sodium cations and chloride anions (Figure Figure $1$). The compound composed of these ions exhibits properties entirely different from the properties of the elements sodium and chlorine. Chlorine is poisonous, but sodium chloride is essential to life; sodium atoms react vigorously with water, but sodium chloride simply dissolves in water.
The Formation of Ionic Compounds
Binary ionic compounds are composed of just two elements: a metal (which forms the cations) and a nonmetal (which forms the anions). For example, NaCl is a binary ionic compound. We can think about the formation of such compounds in terms of the periodic properties of the elements. Many metallic elements have relatively low ionization potentials and lose electrons easily. These elements lie to the left in a period or near the bottom of a group on the periodic table. Nonmetal atoms have relatively high electron affinities and thus readily gain electrons lost by metal atoms, thereby filling their valence shells. Nonmetallic elements are found in the upper-right corner of the periodic table.
As all substances must be electrically neutral, the total number of positive charges on the cations of an ionic compound must equal the total number of negative charges on its anions. The formula of an ionic compound represents the simplest ratio of the numbers of ions necessary to give identical numbers of positive and negative charges. For example, the formula for aluminum oxide, Al2O3, indicates that this ionic compound contains two aluminum cations, Al3+, for every three oxide anions, O2− [thus, (2 × +3) + (3 × –2) = 0].
It is important to note, however, that the formula for an ionic compound does not represent the physical arrangement of its ions. It is incorrect to refer to a sodium chloride (NaCl) “molecule” because there is not a single ionic bond, per se, between any specific pair of sodium and chloride ions. The attractive forces between ions are isotropic—the same in all directions—meaning that any particular ion is equally attracted to all of the nearby ions of opposite charge. This results in the ions arranging themselves into a tightly bound, three-dimensional lattice structure. Sodium chloride, for example, consists of a regular arrangement of equal numbers of Na+ cations and Cl anions (Figure Figure $2$).
The strong electrostatic attraction between Na+ and Cl ions holds them tightly together in solid NaCl. It requires 769 kJ of energy to dissociate one mole of solid NaCl into separate gaseous Na+ and Cl ions:
$\ce{NaCl}(s)⟶\ce{Na+}(g)+\ce{Cl-}(g)\hspace{20px}ΔH=\mathrm{769\:kJ} \nonumber$
Electronic Structures of Cations
When forming a cation, an atom of a main group element tends to lose all of its valence electrons, thus assuming the electronic structure of the noble gas that precedes it in the periodic table. For groups 1 (the alkali metals) and 2 (the alkaline earth metals), the group numbers are equal to the numbers of valence shell electrons and, consequently, to the charges of the cations formed from atoms of these elements when all valence shell electrons are removed. For example, calcium is a group 2 element whose neutral atoms have 20 electrons and a ground state electron configuration of 1s22s22p63s23p64s2. When a Ca atom loses both of its valence electrons, the result is a cation with 18 electrons, a 2+ charge, and an electron configuration of 1s22s22p63s23p6. The Ca2+ ion is therefore isoelectronic with the noble gas Ar.
For groups 12–17, the group numbers exceed the number of valence electrons by 10 (accounting for the possibility of full d subshells in atoms of elements in the fourth and greater periods). Thus, the charge of a cation formed by the loss of all valence electrons is equal to the group number minus 10. For example, aluminum (in group 13) forms 3+ ions (Al3+).
Exceptions to the expected behavior involve elements toward the bottom of the groups. In addition to the expected ions Tl3+, Sn4+, Pb4+, and Bi5+, a partial loss of these atoms’ valence shell electrons can also lead to the formation of Tl+, Sn2+, Pb2+, and Bi3+ ions. The formation of these 1+, 2+, and 3+ cations is ascribed to the inert pair effect, which reflects the relatively low energy of the valence s-electron pair for atoms of the heavy elements of groups 13, 14, and 15. Mercury (group 12) also exhibits an unexpected behavior: it forms a diatomic ion, $\ce{Hg_2^2+}$ (an ion formed from two mercury atoms, with an Hg-Hg bond), in addition to the expected monatomic ion Hg2+ (formed from only one mercury atom).
Transition and inner transition metal elements behave differently than main group elements. Most transition metal cations have 2+ or 3+ charges that result from the loss of their outermost s electron(s) first, sometimes followed by the loss of one or two d electrons from the next-to-outermost shell. For example, iron (1s22s22p63s23p63d64s2) forms the ion Fe2+ (1s22s22p63s23p63d6) by the loss of the 4s electrons and the ion Fe3+ (1s22s22p63s23p63d5) by the loss of the 4s electrons and one of the 3d electrons. Although the d orbitals of the transition elements are—according to the Aufbau principle—the last to fill when building up electron configurations, the outermost s electrons are the first to be lost when these atoms ionize. When the inner transition metals form ions, they usually have a 3+ charge, resulting from the loss of their outermost s electrons and a d or f electron.
Example $1$: Determining the Electronic Structures of Cations
There are at least 14 elements categorized as “essential trace elements” for the human body. They are called “essential” because they are required for healthy bodily functions, “trace” because they are required only in small amounts, and “elements” in spite of the fact that they are really ions. Two of these essential trace elements, chromium and zinc, are required as Cr3+ and Zn2+. Write the electron configurations of these cations.
Solution
First, write the electron configuration for the neutral atoms:
• Zn: [Ar]3d104s2
• Cr: [Ar]3d54s1
Next, remove electrons from the highest energy orbital. For the transition metals, electrons are removed from the s orbital first and then from the d orbital. For the p-block elements, electrons are removed from the p orbitals and then from the s orbital. Zinc is a member of group 12, so it should have a charge of 2+, and thus loses only the two electrons in its s orbital. Chromium is a transition element and should lose its s electrons and then its d electrons when forming a cation. Thus, we find the following electron configurations of the ions:
• Zn2+: [Ar]3d10
• Cr3+: [Ar]3d3
Exercise $1$
Potassium and magnesium are required in our diet. Write the electron configurations of the ions expected from these elements.
Answer
K+: [Ar], Mg2+: [Ne]
Electronic Structures of Anions
Most monatomic anions form when a neutral nonmetal atom gains enough electrons to completely fill its outer s and p orbitals, thereby reaching the electron configuration of the next noble gas. Thus, it is simple to determine the charge on such a negative ion: The charge is equal to the number of electrons that must be gained to fill the s and p orbitals of the parent atom. Oxygen, for example, has the electron configuration 1s22s22p4, whereas the oxygen anion has the electron configuration of the noble gas neon (Ne), 1s22s22p6. The two additional electrons required to fill the valence orbitals give the oxide ion the charge of 2– (O2–).
Example $2$: Determining the Electronic Structure of Anions
Selenium and iodine are two essential trace elements that form anions. Write the electron configurations of the anions.
Solution
Se2: [Ar]3d104s24p6
I: [Kr]4d105s25p6
Exercise $2$
Write the electron configurations of a phosphorus atom and its negative ion. Give the charge on the anion.
Answer
P: [Ne]3s23p3
P3–: [Ne]3s23p6
Summary
Atoms gain or lose electrons to form ions with particularly stable electron configurations. The charges of cations formed by the representative metals may be determined readily because, with few exceptions, the electronic structures of these ions have either a noble gas configuration or a completely filled electron shell. The charges of anions formed by the nonmetals may also be readily determined because these ions form when nonmetal atoms gain enough electrons to fill their valence shells.
Glossary
inert pair effect
tendency of heavy atoms to form ions in which their valence s electrons are not lost
ionic bond
strong electrostatic force of attraction between cations and anions in an ionic compound | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/04%3A_Chemical_Bonding_and_Molecular_Geometry/4.0%3A_Prelude_to_Chemical_Bonding_and_Molecular_Geometry.txt |
Learning Objectives
• Describe the formation of covalent bonds
• Define electronegativity and assess the polarity of covalent bonds
In ionic compounds, electrons are transferred between atoms of different elements to form ions. But this is not the only way that compounds can be formed. Atoms can also make chemical bonds by sharing electrons between each other. Such bonds are called covalent bonds. Covalent bonds are formed between two atoms when both have similar tendencies to attract electrons to themselves (i.e., when both atoms have identical or fairly similar ionization energies and electron affinities). For example, two hydrogen atoms bond covalently to form an H2 molecule; each hydrogen atom in the H2 molecule has two electrons stabilizing it, giving each atom the same number of valence electrons as the noble gas He.
Compounds that contain covalent bonds exhibit different physical properties than ionic compounds. Because the attraction between molecules, which are electrically neutral, is weaker than that between electrically charged ions, covalent compounds generally have much lower melting and boiling points than ionic compounds. In fact, many covalent compounds are liquids or gases at room temperature, and, in their solid states, they are typically much softer than ionic solids. Furthermore, whereas ionic compounds are good conductors of electricity when dissolved in water, most covalent compounds, being electrically neutral, are poor conductors of electricity in any state.
Formation of Covalent Bonds
Nonmetal atoms frequently form covalent bonds with other nonmetal atoms. For example, the hydrogen molecule, H2, contains a covalent bond between its two hydrogen atoms. Figure $1$ illustrates why this bond is formed. Starting on the far right, we have two separate hydrogen atoms with a particular potential energy, indicated by the red line. Along the x-axis is the distance between the two atoms. As the two atoms approach each other (moving left along the x-axis), their valence orbitals (1s) begin to overlap. The single electrons on each hydrogen atom then interact with both atomic nuclei, occupying the space around both atoms. The strong attraction of each shared electron to both nuclei stabilizes the system, and the potential energy decreases as the bond distance decreases. If the atoms continue to approach each other, the positive charges in the two nuclei begin to repel each other, and the potential energy increases. The bond length is determined by the distance at which the lowest potential energy is achieved.
It is essential to remember that energy must be added to break chemical bonds (an endothermic process), whereas forming chemical bonds releases energy (an exothermic process). In the case of H2, the covalent bond is very strong; a large amount of energy, 436 kJ, must be added to break the bonds in one mole of hydrogen molecules and cause the atoms to separate:
$\ce{H2}(g)⟶\ce{2H}(g)\hspace{20px}ΔH=\mathrm{436\:kJ} \nonumber$
Conversely, the same amount of energy is released when one mole of H2 molecules forms from two moles of H atoms:
$\ce{2H}(g)⟶\ce{H2}(g)\hspace{20px}ΔH=\mathrm{−436\:kJ} \nonumber$
Pure vs. Polar Covalent Bonds
If the atoms that form a covalent bond are identical, as in H2, Cl2, and other diatomic molecules, then the electrons in the bond must be shared equally. We refer to this as a pure covalent bond. Electrons shared in pure covalent bonds have an equal probability of being near each nucleus. In the case of Cl2, each atom starts off with seven valence electrons, and each Cl shares one electron with the other, forming one covalent bond:
$\ce{Cl + Cl⟶Cl2} \nonumber$
The total number of electrons around each individual atom consists of six nonbonding electrons and two shared (i.e., bonding) electrons for eight total electrons, matching the number of valence electrons in the noble gas argon. Since the bonding atoms are identical, Cl2 also features a pure covalent bond.
When the atoms linked by a covalent bond are different, the bonding electrons are shared, but no longer equally. Instead, the bonding electrons are more attracted to one atom than the other, giving rise to a shift of electron density toward that atom. This unequal distribution of electrons is known as a polar covalent bond, characterized by a partial positive charge on one atom and a partial negative charge on the other. The atom that attracts the electrons more strongly acquires the partial negative charge and vice versa. For example, the electrons in the H–Cl bond of a hydrogen chloride molecule spend more time near the chlorine atom than near the hydrogen atom. Thus, in an HCl molecule, the chlorine atom carries a partial negative charge and the hydrogen atom has a partial positive charge. Figure $2$ shows the distribution of electrons in the H–Cl bond. Note that the shaded area around Cl is much larger than it is around H. Compare this to Figure $1$, which shows the even distribution of electrons in the H2 nonpolar bond.
We sometimes designate the positive and negative atoms in a polar covalent bond using a lowercase Greek letter “delta,” δ, with a plus sign or minus sign to indicate whether the atom has a partial positive charge (δ+) or a partial negative charge (δ–). This symbolism is shown for the H–Cl molecule in Figure $\PageIndex{2b}$.
Electronegativity
Whether a bond is nonpolar or polar covalent is determined by a property of the bonding atoms called electronegativity. Electronegativity is a measure of the tendency of an atom to attract electrons (or electron density) towards itself. It determines how the shared electrons are distributed between the two atoms in a bond. The more strongly an atom attracts the electrons in its bonds, the larger its electronegativity. Electrons in a polar covalent bond are shifted toward the more electronegative atom; thus, the more electronegative atom is the one with the partial negative charge. The greater the difference in electronegativity, the more polarized the electron distribution and the larger the partial charges of the atoms.
Figure $3$ shows the electronegativity values of the elements as proposed by one of the most famous chemists of the twentieth century: Linus Pauling. In general, electronegativity increases from left to right across a period in the periodic table and decreases down a group. Thus, the nonmetals, which lie in the upper right, tend to have the highest electronegativities, with fluorine the most electronegative element of all (EN = 4.0). Metals tend to be less electronegative elements, and the group 1 metals have the lowest electronegativities. Note that noble gases are excluded from this figure because these atoms usually do not share electrons with others atoms since they have a full valence shell. (While noble gas compounds such as XeO2 do exist, they can only be formed under extreme conditions, and thus they do not fit neatly into the general model of electronegativity.)
Linus Pauling
Linus Pauling is the only person to have received two unshared (individual) Nobel Prizes: one for chemistry in 1954 for his work on the nature of chemical bonds and one for peace in 1962 for his opposition to weapons of mass destruction. He developed many of the theories and concepts that are foundational to our current understanding of chemistry, including electronegativity and resonance structures.
Pauling also contributed to many other fields besides chemistry. His research on sickle cell anemia revealed the cause of the disease—the presence of a genetically inherited abnormal protein in the blood—and paved the way for the field of molecular genetics. His work was also pivotal in curbing the testing of nuclear weapons; he proved that radioactive fallout from nuclear testing posed a public health risk.
Electronegativity versus Electron Affinity
We must be careful not to confuse electronegativity and electron affinity. The electron affinity of an element is a measurable physical quantity, namely, the energy released or absorbed when an isolated gas-phase atom acquires an electron, measured in kJ/mol. Electronegativity, on the other hand, describes how tightly an atom attracts electrons in a bond. It is a dimensionless quantity that is calculated, not measured. Pauling derived the first electronegativity values by comparing the amounts of energy required to break different types of bonds. He chose an arbitrary relative scale ranging from 0 to 4.
Electronegativity and Bond Type
The absolute value of the difference in electronegativity (ΔEN) of two bonded atoms provides a rough measure of the polarity to be expected in the bond and, thus, the bond type. When the difference is very small or zero, the bond is covalent and nonpolar. When it is large, the bond is polar covalent or ionic. The absolute values of the electronegativity differences between the atoms in the bonds H–H, H–Cl, and Na–Cl are 0 (nonpolar), 0.9 (polar covalent), and 2.1 (ionic), respectively. The degree to which electrons are shared between atoms varies from completely equal (pure covalent bonding) to not at all (ionic bonding). Figure $4$ shows the relationship between electronegativity difference and bond type.
A rough approximation of the electronegativity differences associated with covalent, polar covalent, and ionic bonds is shown in Figure $4$. This table is just a general guide, however, with many exceptions. For example, the H and F atoms in HF have an electronegativity difference of 1.9, and the N and H atoms in NH3 a difference of 0.9, yet both of these compounds form bonds that are considered polar covalent. Likewise, the Na and Cl atoms in NaCl have an electronegativity difference of 2.1, and the Mn and I atoms in MnI2 have a difference of 1.0, yet both of these substances form ionic compounds.
The best guide to the covalent or ionic character of a bond is to consider the types of atoms involved and their relative positions in the periodic table. Bonds between two nonmetals are generally covalent; bonding between a metal and a nonmetal is often ionic.
Some compounds contain both covalent and ionic bonds. The atoms in polyatomic ions, such as OH, $\ce{NO3-}$, and $\ce{NH4+}$, are held together by polar covalent bonds. However, these polyatomic ions form ionic compounds by combining with ions of opposite charge. For example, potassium nitrate, KNO3, contains the K+ cation and the polyatomic $\ce{NO3-}$ anion. Thus, bonding in potassium nitrate is ionic, resulting from the electrostatic attraction between the ions K+ and $\ce{NO3-}$, as well as covalent between the nitrogen and oxygen atoms in $\ce{NO3-}$.
Example $1$: Electronegativity and Bond Polarity
Bond polarities play an important role in determining the structure of proteins. Using the electronegativity values in Table A2, arrange the following covalent bonds—all commonly found in amino acids—in order of increasing polarity. Then designate the positive and negative atoms using the symbols δ+ and δ–:
C–H, C–N, C–O, N–H, O–H, S–H
Solution
The polarity of these bonds increases as the absolute value of the electronegativity difference increases. The atom with the δ– designation is the more electronegative of the two. Table $1$ shows these bonds in order of increasing polarity.
Table $1$: Bond Polarity and Electronegativity Difference
Bond ΔEN Polarity
C–H 0.4 $\overset{δ−}{\ce C}−\overset{δ+}{\ce H}$
S–H 0.4 $\overset{δ−}{\ce S}−\overset{δ+}{\ce H}$
C–N 0.5 $\overset{δ+}{\ce C}−\overset{δ−}{\ce N}$
N–H 0.9 $\overset{δ−}{\ce N}−\overset{δ+}{\ce H}$
C–O 1.0 $\overset{δ+}{\ce C}−\overset{δ−}{\ce O}$
O–H 1.4 $\overset{δ−}{\ce O}−\overset{δ+}{\ce H}$
Exercise $1$
Silicones are polymeric compounds containing, among others, the following types of covalent bonds: Si–O, Si–C, C–H, and C–C. Using the electronegativity values in Figure $3$, arrange the bonds in order of increasing polarity and designate the positive and negative atoms using the symbols δ+ and δ–.
Answer
Answer to Exercise 7.2.1
Bond Electronegativity Difference Polarity
C–C 0.0 nonpolar
C–H 0.4 $\overset{δ−}{\ce C}−\overset{δ+}{\ce H}$
Si–C 0.7 $\overset{δ+}{\ce{Si}}−\overset{δ−}{\ce C}$
Si–O 1.7 $\overset{δ+}{\ce{Si}}−\overset{δ−}{\ce O}$
Summary
Covalent bonds form when electrons are shared between atoms and are attracted by the nuclei of both atoms. In pure covalent bonds, the electrons are shared equally. In polar covalent bonds, the electrons are shared unequally, as one atom exerts a stronger force of attraction on the electrons than the other. The ability of an atom to attract a pair of electrons in a chemical bond is called its electronegativity. The difference in electronegativity between two atoms determines how polar a bond will be. In a diatomic molecule with two identical atoms, there is no difference in electronegativity, so the bond is nonpolar or pure covalent. When the electronegativity difference is very large, as is the case between metals and nonmetals, the bonding is characterized as ionic.
Glossary
bond length
distance between the nuclei of two bonded atoms at which the lowest potential energy is achieved
covalent bond
bond formed when electrons are shared between atoms
electronegativity
tendency of an atom to attract electrons in a bond to itself
polar covalent bond
covalent bond between atoms of different electronegativities; a covalent bond with a positive end and a negative end
pure covalent bond
(also, nonpolar covalent bond) covalent bond between atoms of identical electronegativities | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/04%3A_Chemical_Bonding_and_Molecular_Geometry/4.2%3A_Covalent_Bonding.txt |
Learning Objectives
• Write Lewis symbols for neutral atoms and ions
• Draw Lewis structures depicting the bonding in simple molecules
Thus far, we have discussed the various types of bonds that form between atoms and/or ions. In all cases, these bonds involve the sharing or transfer of valence shell electrons between atoms. In this section, we will explore the typical method for depicting valence shell electrons and chemical bonds, namely Lewis symbols and Lewis structures.
Lewis Symbols
We use Lewis symbols to describe valence electron configurations of atoms and monatomic ions. A Lewis symbol consists of an elemental symbol surrounded by one dot for each of its valence electrons:
Lewis symbols can be used to illustrate the formation of cations from atoms, as shown here for sodium and calcium:
Likewise, they can be used to show the formation of anions from atoms, as shown here for chlorine and sulfur:
Lewis Structures
We also use Lewis symbols to indicate the formation of covalent bonds, which are shown in Lewis structures, drawings that describe the bonding in molecules and polyatomic ions. For example, when two chlorine atoms form a chlorine molecule, they share one pair of electrons:
The Lewis structure indicates that each Cl atom has three pairs of electrons that are not used in bonding (called lone pairs) and one shared pair of electrons (written between the atoms). A dash (or line) is usually used to indicate a shared pair of electrons:
In the Lewis model, a single shared pair of electrons is a single bond. Each Cl atom interacts with eight valence electrons total: the six in the lone pairs and the two in the single bond.
The Octet Rule
The other halogen molecules (F2, Br2, I2, and At2) form bonds like those in the chlorine molecule: one single bond between atoms and three lone pairs of electrons per atom. This allows each halogen atom to have a noble gas electron configuration. The tendency of main group atoms to form enough bonds to obtain eight valence electrons is known as the octet rule.
The number of bonds that an atom can form can often be predicted from the number of electrons needed to reach an octet (eight valence electrons); this is especially true of the nonmetals of the second period of the periodic table (C, N, O, and F). For example, each atom of a group 14 element has four electrons in its outermost shell and therefore requires four more electrons to reach an octet. These four electrons can be gained by forming four covalent bonds, as illustrated here for carbon in CCl4 (carbon tetrachloride) and silicon in SiH4 (silane). Because hydrogen only needs two electrons to fill its valence shell, it is an exception to the octet rule and only needs to form one bond. The transition elements and inner transition elements also do not follow the octet rule since they have d and f electrons involved in their valence shells.
Group 15 elements such as nitrogen have five valence electrons in the atomic Lewis symbol: one lone pair and three unpaired electrons. To obtain an octet, these atoms form three covalent bonds, as in NH3 (ammonia). Oxygen and other atoms in group 16 obtain an octet by forming two covalent bonds:
Double and Triple Bonds
As previously mentioned, when a pair of atoms shares one pair of electrons, we call this a single bond. However, a pair of atoms may need to share more than one pair of electrons in order to achieve the requisite octet. A double bond forms when two pairs of electrons are shared between a pair of atoms, as between the carbon and oxygen atoms in CH2O (formaldehyde) and between the two carbon atoms in C2H4 (ethylene):
A triple bond forms when three electron pairs are shared by a pair of atoms, as in carbon monoxide (CO) and the cyanide ion (CN):
Writing Lewis Structures with the Octet Rule
For very simple molecules and molecular ions, we can write the Lewis structures by merely pairing up the unpaired electrons on the constituent atoms. See these examples:
For more complicated molecules and molecular ions, it is helpful to follow the step-by-step procedure outlined here:
1. Determine the total number of valence (outer shell) electrons among all the atoms. For cations, subtract one electron for each positive charge. For anions, add one electron for each negative charge.
2. Draw a skeleton structure of the molecule or ion, arranging the atoms around a central atom. (Generally, the least electronegative element should be placed in the center.) Connect each atom to the central atom with a single bond (one electron pair).
3. Distribute the remaining electrons as lone pairs on the terminal atoms (except hydrogen), completing an octet around each atom.
4. Place all remaining electrons on the central atom.
5. Rearrange the electrons of the outer atoms to make multiple bonds with the central atom in order to obtain octets wherever possible.
Let us determine the Lewis structures of SiH4, $\ce{CHO2-}$, NO+, and OF2 as examples in following this procedure:
1. Determine the total number of valence (outer shell) electrons in the molecule or ion.
• For a molecule, we add the number of valence electrons on each atom in the molecule:
\begin{align} &\phantom{+}\ce{SiH4}\ &\phantom{+}\textrm{Si: 4 valence electrons/atom × 1 atom = 4}\ &\underline{\textrm{+H: 1 valence electron/atom × 4 atoms = 4}}\ &\hspace{271px}\textrm{= 8 valence electrons} \end{align}
• For a negative ion, such as $\ce{CHO2-}$, we add the number of valence electrons on the atoms to the number of negative charges on the ion (one electron is gained for each single negative charge):
$\ce{CHO2-}\ \textrm{C: 4 valence electrons/atom × 1 atom} \hspace{6px}= \phantom{1}4\ \textrm{H: 1 valence electron/atom × 1 atom} \hspace{12px}= \phantom{1}1\ \textrm{O: 6 valence electrons/atom × 2 atoms = 12}\ \underline{+\hspace{100px}\textrm{1 additional electron} \hspace{9px}= \phantom{1}1}\ \hspace{264px}\textrm{= 18 valence electrons}$
• For a positive ion, such as NO+, we add the number of valence electrons on the atoms in the ion and then subtract the number of positive charges on the ion (one electron is lost for each single positive charge) from the total number of valence electrons:
$\ce{NO+}\ \textrm{N: 5 valence electrons/atom × 1 atom} = \phantom{−}5\ \textrm{O: 6 valence electron/atom × 1 atom}\hspace{5px} = \phantom{−}6\ \underline{\textrm{+ −1 electron (positive charge)} \hspace{44px}= −1}\ \hspace{260px}\textrm{= 10 valence electrons}$
• Since OF2 is a neutral molecule, we simply add the number of valence electrons:
$\phantom{+ }\ce{OF2}\ \phantom{+ }\textrm{O: 6 valence electrons/atom × 1 atom} \hspace{10px}= 6\ \underline{\textrm{+ F: 7 valence electrons/atom × 2 atoms} = 14}\ \hspace{280px}\textrm{= 20 valence electrons}$
2. Draw a skeleton structure of the molecule or ion, arranging the atoms around a central atom and connecting each atom to the central atom with a single (one electron pair) bond. (Note that we denote ions with brackets around the structure, indicating the charge outside the brackets:)
1. When several arrangements of atoms are possible, as for $\ce{CHO2-}$, we must use experimental evidence to choose the correct one. In general, the less electronegative elements are more likely to be central atoms. In $\ce{CHO2-}$, the less electronegative carbon atom occupies the central position with the oxygen and hydrogen atoms surrounding it. Other examples include P in POCl3, S in SO2, and Cl in $\ce{ClO4-}$. An exception is that hydrogen is almost never a central atom. As the most electronegative element, fluorine also cannot be a central atom.
2. Distribute the remaining electrons as lone pairs on the terminal atoms (except hydrogen) to complete their valence shells with an octet of electrons.
• There are no remaining electrons on SiH4, so it is unchanged:
1. Place all remaining electrons on the central atom.
• For SiH4, $\ce{CHO2-}$, and NO+, there are no remaining electrons; we already placed all of the electrons determined in Step 1.
• For OF2, we had 16 electrons remaining in Step 3, and we placed 12, leaving 4 to be placed on the central atom:
1. Rearrange the electrons of the outer atoms to make multiple bonds with the central atom in order to obtain octets wherever possible.
• SiH4: Si already has an octet, so nothing needs to be done.
• $\ce{CHO2-}$: We have distributed the valence electrons as lone pairs on the oxygen atoms, but the carbon atom lacks an octet:
• NO+: For this ion, we added eight outer electrons, but neither atom has an octet. We cannot add any more electrons since we have already used the total that we found in Step 1, so we must move electrons to form a multiple bond:
• This still does not produce an octet, so we must move another pair, forming a triple bond:
• In OF2, each atom has an octet as drawn, so nothing changes.
Example $1$: Writing Lewis Structures
NASA’s Cassini-Huygens mission detected a large cloud of toxic hydrogen cyanide (HCN) on Titan, one of Saturn’s moons. Titan also contains ethane (H3CCH3), acetylene (HCCH), and ammonia (NH3). What are the Lewis structures of these molecules?
Solution
Calculate the number of valence electrons.
• HCN: (1 × 1) + (4 × 1) + (5 × 1) = 10
• H3CCH3: (1 × 3) + (2 × 4) + (1 × 3) = 14
• HCCH: (1 × 1) + (2 × 4) + (1 × 1) = 10
• NH3: (5 × 1) + (3 × 1) = 8
Draw a skeleton and connect the atoms with single bonds. Remember that H is never a central atom:
Where needed, distribute electrons to the terminal atoms:
• HCN: six electrons placed on N
• H3CCH3: no electrons remain
• HCCH: no terminal atoms capable of accepting electrons
• NH3: no terminal atoms capable of accepting electrons
Where needed, place remaining electrons on the central atom:
• HCN: no electrons remain
• H3CCH3: no electrons remain
• HCCH: four electrons placed on carbon
• NH3: two electrons placed on nitrogen
Where needed, rearrange electrons to form multiple bonds in order to obtain an octet on each atom:
• HCN: form two more C–N bonds
• H3CCH3: all atoms have the correct number of electrons
• HCCH: form a triple bond between the two carbon atoms
• NH3: all atoms have the correct number of electrons
Exercise $1$
Both carbon monoxide, CO, and carbon dioxide, CO2, are products of the combustion of fossil fuels. Both of these gases also cause problems: CO is toxic and CO2 has been implicated in global climate change. What are the Lewis structures of these two molecules?
Answer
Fullerene Chemistry
Carbon soot has been known to man since prehistoric times, but it was not until fairly recently that the molecular structure of the main component of soot was discovered. In 1996, the Nobel Prize in Chemistry was awarded to Richard Smalley, Robert Curl, and Harold Kroto for their work in discovering a new form of carbon, the C60 buckminsterfullerene molecule. An entire class of compounds, including spheres and tubes of various shapes, were discovered based on C60. This type of molecule, called a fullerene, consists of a complex network of single- and double-bonded carbon atoms arranged in such a way that each carbon atom obtains a full octet of electrons. Because of their size and shape, fullerenes can encapsulate other molecules, so they have shown potential in various applications from hydrogen storage to targeted drug delivery systems. They also possess unique electronic and optical properties that have been put to good use in solar powered devices and chemical sensors.
Exceptions to the Octet Rule
Many covalent molecules have central atoms that do not have eight electrons in their Lewis structures. These molecules fall into three categories:
• Odd-electron molecules have an odd number of valence electrons, and therefore have an unpaired electron.
• Electron-deficient molecules have a central atom that has fewer electrons than needed for a noble gas configuration.
• Hypervalent molecules have a central atom that has more electrons than needed for a noble gas configuration.
Odd-electron Molecules
We call molecules that contain an odd number of electrons free radicals. Nitric oxide, NO, is an example of an odd-electron molecule; it is produced in internal combustion engines when oxygen and nitrogen react at high temperatures.
To draw the Lewis structure for an odd-electron molecule like NO, we follow the same five steps we would for other molecules, but with a few minor changes:
1. Determine the total number of valence (outer shell) electrons. The sum of the valence electrons is 5 (from N) + 6 (from O) = 11. The odd number immediately tells us that we have a free radical, so we know that not every atom can have eight electrons in its valence shell.
2. Draw a skeleton structure of the molecule. We can easily draw a skeleton with an N–O single bond: N–O
3. Distribute the remaining electrons as lone pairs on the terminal atoms. In this case, there is no central atom, so we distribute the electrons around both atoms. We give eight electrons to the more electronegative atom in these situations; thus oxygen has the filled valence shell:
1. Place all remaining electrons on the central atom. Since there are no remaining electrons, this step does not apply.
2. Rearrange the electrons to make multiple bonds with the central atom in order to obtain octets wherever possible. We know that an odd-electron molecule cannot have an octet for every atom, but we want to get each atom as close to an octet as possible. In this case, nitrogen has only five electrons around it. To move closer to an octet for nitrogen, we take one of the lone pairs from oxygen and use it to form a NO double bond. (We cannot take another lone pair of electrons on oxygen and form a triple bond because nitrogen would then have nine electrons:)
Electron-deficient Molecules
We will also encounter a few molecules that contain central atoms that do not have a filled valence shell. Generally, these are molecules with central atoms from groups 2 and 13 and outer atoms that are hydrogen or other atoms that do not form multiple bonds. For example, in the Lewis structures of beryllium dihydride, BeH2, and boron trifluoride, BF3, the beryllium and boron atoms each have only four and six electrons, respectively. It is possible to draw a structure with a double bond between a boron atom and a fluorine atom in BF3, satisfying the octet rule, but experimental evidence indicates the bond lengths are closer to that expected for B–F single bonds. This suggests the best Lewis structure has three B–F single bonds and an electron deficient boron. The reactivity of the compound is also consistent with an electron deficient boron. However, the B–F bonds are slightly shorter than what is actually expected for B–F single bonds, indicating that some double bond character is found in the actual molecule.
An atom like the boron atom in BF3, which does not have eight electrons, is very reactive. It readily combines with a molecule containing an atom with a lone pair of electrons. For example, NH3 reacts with BF3 because the lone pair on nitrogen can be shared with the boron atom:
Hypervalent Molecules
Elements in the second period of the periodic table (n = 2) can accommodate only eight electrons in their valence shell orbitals because they have only four valence orbitals (one 2s and three 2p orbitals). Elements in the third and higher periods (n ≥ 3) have more than four valence orbitals and can share more than four pairs of electrons with other atoms because they have empty d orbitals in the same shell. Molecules formed from these elements are sometimes called hypervalent molecules.Table $5$ shows the Lewis structures for two hypervalent molecules, PCl5 and SF6.
In some hypervalent molecules, such as IF5 and XeF4, some of the electrons in the outer shell of the central atom are lone pairs:
When we write the Lewis structures for these molecules, we find that we have electrons left over after filling the valence shells of the outer atoms with eight electrons. These additional electrons must be assigned to the central atom.
Example $2$: Octet Rule Violations
Xenon is a noble gas, but it forms a number of stable compounds. We examined $\ce{XeF4}$ earlier. What are the Lewis structures of $\ce{XeF2}$ and $\ce{XeF6}$?
Solution
We can draw the Lewis structure of any covalent molecule by following the six steps discussed earlier. In this case, we can condense the last few steps, since not all of them apply.
Step 1: Calculate the number of valence electrons:
$\ce{XeF2}$: 8 + (2 × 7) = 22
$\ce{XeF6}$: 8 + (6 × 7) = 50
Step 2: Draw a skeleton joining the atoms by single bonds. Xenon will be the central atom because fluorine cannot be a central atom:
Step 3: Distribute the remaining electrons.
XeF2: We place three lone pairs of electrons around each F atom, accounting for 12 electrons and giving each F atom 8 electrons. Thus, six electrons (three lone pairs) remain. These lone pairs must be placed on the Xe atom. This is acceptable because Xe atoms have empty valence shell d orbitals and can accommodate more than eight electrons. The Lewis structure of XeF2 shows two bonding pairs and three lone pairs of electrons around the Xe atom:
XeF6: We place three lone pairs of electrons around each F atom, accounting for 36 electrons. Two electrons remain, and this lone pair is placed on the Xe atom:
Exercise $2$: interhalogens
The halogens form a class of compounds called the interhalogens, in which halogen atoms covalently bond to each other. Write the Lewis structures for the interhalogens $\ce{BrCl3}$ and $\ce{ICl4-}$.
Answer
Summary
Valence electronic structures can be visualized by drawing Lewis symbols (for atoms and monatomic ions) and Lewis structures (for molecules and polyatomic ions). Lone pairs, unpaired electrons, and single, double, or triple bonds are used to indicate where the valence electrons are located around each atom in a Lewis structure. Most structures—especially those containing second row elements—obey the octet rule, in which every atom (except H) is surrounded by eight electrons. Exceptions to the octet rule occur for odd-electron molecules (free radicals), electron-deficient molecules, and hypervalent molecules.
Glossary
double bond
covalent bond in which two pairs of electrons are shared between two atoms
free radical
molecule that contains an odd number of electrons
hypervalent molecule
molecule containing at least one main group element that has more than eight electrons in its valence shell
Lewis structure
diagram showing lone pairs and bonding pairs of electrons in a molecule or an ion
Lewis symbol
symbol for an element or monatomic ion that uses a dot to represent each valence electron in the element or ion
lone pair
two (a pair of) valence electrons that are not used to form a covalent bond
octet rule
guideline that states main group atoms will form structures in which eight valence electrons interact with each nucleus, counting bonding electrons as interacting with both atoms connected by the bond
single bond
bond in which a single pair of electrons is shared between two atoms
triple bond
bond in which three pairs of electrons are shared between two atoms | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/04%3A_Chemical_Bonding_and_Molecular_Geometry/4.3%3A_Lewis_Symbols_and_Structures.txt |
Learning Objectives
• Compute formal charges for atoms in any Lewis structure
• Use formal charges to identify the most reasonable Lewis structure for a given molecule
• Explain the concept of resonance and draw Lewis structures representing resonance forms for a given molecule
Previously, we discussed how to write Lewis structures for molecules and polyatomic ions. In some cases, however, there is seemingly more than one valid structure for a molecule. We can use the concept of formal charges to help us predict the most appropriate Lewis structure when more than one is reasonable.
Calculating Formal Charge
The formal charge of an atom in a molecule is the hypothetical charge the atom would have if we could redistribute the electrons in the bonds evenly between the atoms. Another way of saying this is that formal charge results when we take the number of valence electrons of a neutral atom, subtract the nonbonding electrons, and then subtract the number of bonds connected to that atom in the Lewis structure.
Thus, we calculate formal charge as follows:
$\textrm{formal charge = # valence shell electrons (free atom) − # lone pair electrons − }\dfrac{1}{2}\textrm{ # bonding electrons} \nonumber$
We can double-check formal charge calculations by determining the sum of the formal charges for the whole structure. The sum of the formal charges of all atoms in a molecule must be zero; the sum of the formal charges in an ion should equal the charge of the ion.
We must remember that the formal charge calculated for an atom is not the actual charge of the atom in the molecule. Formal charge is only a useful bookkeeping procedure; it does not indicate the presence of actual charges.
Example $1$: Calculating Formal Charge from Lewis Structures
Assign formal charges to each atom in the interhalogen ion $\ce{ICl4-}$.
Solution
We divide the bonding electron pairs equally for all $\ce{I–Cl}$ bonds:
We assign lone pairs of electrons to their atoms. Each Cl atom now has seven electrons assigned to it, and the I atom has eight.
Subtract this number from the number of valence electrons for the neutral atom:
• I: 7 – 8 = –1
• Cl: 7 – 7 = 0
The sum of the formal charges of all the atoms equals –1, which is identical to the charge of the ion (–1).
Exercise $1$
Calculate the formal charge for each atom in the carbon monoxide molecule:
Answer
C −1, O +1
Example $2$: Calculating Formal Charge from Lewis Structures
Assign formal charges to each atom in the interhalogen molecule $\ce{BrCl3}$.
Solution
Assign one of the electrons in each Br–Cl bond to the Br atom and one to the Cl atom in that bond:
Assign the lone pairs to their atom. Now each Cl atom has seven electrons and the Br atom has seven electrons.
Subtract this number from the number of valence electrons for the neutral atom. This gives the formal charge:
• Br: 7 – 7 = 0
• Cl: 7 – 7 = 0
All atoms in $\ce{BrCl3}$ have a formal charge of zero, and the sum of the formal charges totals zero, as it must in a neutral molecule.
Exercise $2$
Determine the formal charge for each atom in $\ce{NCl3}$.
Answer
N: 0; all three Cl atoms: 0
Using Formal Charge to Predict Molecular Structure
The arrangement of atoms in a molecule or ion is called its molecular structure. In many cases, following the steps for writing Lewis structures may lead to more than one possible molecular structure—different multiple bond and lone-pair electron placements or different arrangements of atoms, for instance. A few guidelines involving formal charge can be helpful in deciding which of the possible structures is most likely for a particular molecule or ion.
Predicting Molecular Structure Guidelines
1. A molecular structure in which all formal charges are zero is preferable to one in which some formal charges are not zero.
2. If the Lewis structure must have nonzero formal charges, the arrangement with the smallest nonzero formal charges is preferable.
3. Lewis structures are preferable when adjacent formal charges are zero or of the opposite sign.
4. When we must choose among several Lewis structures with similar distributions of formal charges, the structure with the negative formal charges on the more electronegative atoms is preferable.
To see how these guidelines apply, let us consider some possible structures for carbon dioxide, $\ce{CO2}$. We know from our previous discussion that the less electronegative atom typically occupies the central position, but formal charges allow us to understand why this occurs. We can draw three possibilities for the structure: carbon in the center and double bonds, carbon in the center with a single and triple bond, and oxygen in the center with double bonds:
Comparing the three formal charges, we can definitively identify the structure on the left as preferable because it has only formal charges of zero (Guideline 1).
As another example, the thiocyanate ion, an ion formed from a carbon atom, a nitrogen atom, and a sulfur atom, could have three different molecular structures: $\ce{CNS^{–}}$, $\ce{NCS^{–}}$, or $\ce{CSN^{–}}$. The formal charges present in each of these molecular structures can help us pick the most likely arrangement of atoms. Possible Lewis structures and the formal charges for each of the three possible structures for the thiocyanate ion are shown here:
Note that the sum of the formal charges in each case is equal to the charge of the ion (–1). However, the first arrangement of atoms is preferred because it has the lowest number of atoms with nonzero formal charges (Guideline 2). Also, it places the least electronegative atom in the center, and the negative charge on the more electronegative element (Guideline 4).
Example $3$: Using Formal Charge to Determine Molecular Structure
Nitrous oxide, N2O, commonly known as laughing gas, is used as an anesthetic in minor surgeries, such as the routine extraction of wisdom teeth. Which is the likely structure for nitrous oxide?
Solution Determining formal charge yields the following:
The structure with a terminal oxygen atom best satisfies the criteria for the most stable distribution of formal charge:
The number of atoms with formal charges are minimized (Guideline 2), and there is no formal charge larger than one (Guideline 2). This is again consistent with the preference for having the less electronegative atom in the central position.
Exercise $3$
Which is the most likely molecular structure for the nitrite ($\ce{NO2-}$) ion?
Answer
$\ce{ONO^{–}}$
Resonance
You may have noticed that the nitrite anion in Example $3$ can have two possible structures with the atoms in the same positions. The electrons involved in the N–O double bond, however, are in different positions:
If nitrite ions do indeed contain a single and a double bond, we would expect for the two bond lengths to be different. A double bond between two atoms is shorter (and stronger) than a single bond between the same two atoms. Experiments show, however, that both N–O bonds in $\ce{NO2-}$ have the same strength and length, and are identical in all other properties.
It is not possible to write a single Lewis structure for $\ce{NO2-}$ in which nitrogen has an octet and both bonds are equivalent. Instead, we use the concept of resonance: if two or more Lewis structures with the same arrangement of atoms can be written for a molecule or ion, the actual distribution of electrons is an average of that shown by the various Lewis structures. The actual distribution of electrons in each of the nitrogen-oxygen bonds in $\ce{NO2-}$ is the average of a double bond and a single bond. We call the individual Lewis structures resonance forms. The actual electronic structure of the molecule (the average of the resonance forms) is called a resonance hybrid of the individual resonance forms. A double-headed arrow between Lewis structures indicates that they are resonance forms. Thus, the electronic structure of the $\ce{NO2-}$ ion is shown as:
We should remember that a molecule described as a resonance hybrid never possesses an electronic structure described by either resonance form. It does not fluctuate between resonance forms; rather, the actual electronic structure is always the average of that shown by all resonance forms. George Wheland, one of the pioneers of resonance theory, used a historical analogy to describe the relationship between resonance forms and resonance hybrids. A medieval traveler, having never before seen a rhinoceros, described it as a hybrid of a dragon and a unicorn because it had many properties in common with both. Just as a rhinoceros is neither a dragon sometimes nor a unicorn at other times, a resonance hybrid is neither of its resonance forms at any given time. Like a rhinoceros, it is a real entity that experimental evidence has shown to exist. It has some characteristics in common with its resonance forms, but the resonance forms themselves are convenient, imaginary images (like the unicorn and the dragon).
The carbonate anion, $\ce{CO3^2-}$, provides a second example of resonance:
One oxygen atom must have a double bond to carbon to complete the octet on the central atom. All oxygen atoms, however, are equivalent, and the double bond could form from any one of the three atoms. This gives rise to three resonance forms of the carbonate ion. Because we can write three identical resonance structures, we know that the actual arrangement of electrons in the carbonate ion is the average of the three structures. Again, experiments show that all three C–O bonds are exactly the same.
Summary
In a Lewis structure, formal charges can be assigned to each atom by treating each bond as if one-half of the electrons are assigned to each atom. These hypothetical formal charges are a guide to determining the most appropriate Lewis structure. A structure in which the formal charges are as close to zero as possible is preferred. Resonance occurs in cases where two or more Lewis structures with identical arrangements of atoms but different distributions of electrons can be written. The actual distribution of electrons (the resonance hybrid) is an average of the distribution indicated by the individual Lewis structures (the resonance forms).
Key Equations
• $\textrm{formal charge = # valence shell electrons (free atom) − # one pair electrons − }\dfrac{1}{2}\textrm{ # bonding electrons}$
Glossary
formal charge
charge that would result on an atom by taking the number of valence electrons on the neutral atom and subtracting the nonbonding electrons and the number of bonds (one-half of the bonding electrons)
molecular structure
arrangement of atoms in a molecule or ion
resonance
situation in which one Lewis structure is insufficient to describe the bonding in a molecule and the average of multiple structures is observed
resonance forms
two or more Lewis structures that have the same arrangement of atoms but different arrangements of electrons
resonance hybrid
average of the resonance forms shown by the individual Lewis structures | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/04%3A_Chemical_Bonding_and_Molecular_Geometry/4.4%3A_Formal_Charges_and_Resonance.txt |
Learning Objectives
• Describe the energetics of covalent and ionic bond formation and breakage
• Use the Born-Haber cycle to compute lattice energies for ionic compounds
• Use average covalent bond energies to estimate enthalpies of reaction
A bond’s strength describes how strongly each atom is joined to another atom, and therefore how much energy is required to break the bond between the two atoms. In this section, you will learn about the bond strength of covalent bonds, and then compare that to the strength of ionic bonds, which is related to the lattice energy of a compound.
Bond Strength: Covalent Bonds
Stable molecules exist because covalent bonds hold the atoms together. We measure the strength of a covalent bond by the energy required to break it, that is, the energy necessary to separate the bonded atoms. Separating any pair of bonded atoms requires energy; the stronger a bond, the greater the energy required to break it. The energy required to break a specific covalent bond in one mole of gaseous molecules is called the bond energy or the bond dissociation energy. The bond energy for a diatomic molecule, $D_{X–Y}$, is defined as the standard enthalpy change for the endothermic reaction:
$XY_{(g)}⟶X_{(g)}+Y_{(g)}\;\;\; D_{X−Y}=ΔH° \label{7.6.1}$
For example, the bond energy of the pure covalent H–H bond, $\Delta_{H–H}$, is 436 kJ per mole of H–H bonds broken:
$H_{2(g)}⟶2H_{(g)} \;\;\; D_{H−H}=ΔH°=436kJ \label{EQ2}$
Breaking a bond always require energy to be added to the molecule. Correspondingly, making a bond always releases energy.
Molecules with three or more atoms have two or more bonds. The sum of all bond energies in such a molecule is equal to the standard enthalpy change for the endothermic reaction that breaks all the bonds in the molecule. For example, the sum of the four C–H bond energies in CH4, 1660 kJ, is equal to the standard enthalpy change of the reaction:
The average C–H bond energy, $D_{C–H}$, is 1660/4 = 415 kJ/mol because there are four moles of C–H bonds broken per mole of the reaction. Although the four C–H bonds are equivalent in the original molecule, they do not each require the same energy to break; once the first bond is broken (which requires 439 kJ/mol), the remaining bonds are easier to break. The 415 kJ/mol value is the average, not the exact value required to break any one bond.
The strength of a bond between two atoms increases as the number of electron pairs in the bond increases. Generally, as the bond strength increases, the bond length decreases. Thus, we find that triple bonds are stronger and shorter than double bonds between the same two atoms; likewise, double bonds are stronger and shorter than single bonds between the same two atoms. Average bond energies for some common bonds appear in Table $2$, and a comparison of bond lengths and bond strengths for some common bonds appears in Table $2$. When one atom bonds to various atoms in a group, the bond strength typically decreases as we move down the group. For example, C–F is 439 kJ/mol, C–Cl is 330 kJ/mol, and C–Br is 275 kJ/mol.
Table $1$: Bond Energies (kJ/mol)
Bond Bond Energy Bond Bond Energy Bond Bond Energy
H–H 436 C–S 260 F–Cl 255
H–C 415 C–Cl 330 F–Br 235
H–N 390 C–Br 275 Si–Si 230
H–O 464 C–I 240 Si–P 215
H–F 569 N–N 160 Si–S 225
H–Si 395 $\mathrm{N=N}$ 418 Si–Cl 359
H–P 320 $\mathrm{N≡N}$ 946 Si–Br 290
H–S 340 N–O 200 Si–I 215
H–Cl 432 N–F 270 P–P 215
H–Br 370 N–P 210 P–S 230
H–I 295 N–Cl 200 P–Cl 330
C–C 345 N–Br 245 P–Br 270
$\mathrm{C=C}$ 611 O–O 140 P–I 215
$\mathrm{C≡C}$ 837 $\mathrm{O=O}$ 498 S–S 215
C–N 290 O–F 160 S–Cl 250
$\mathrm{C=N}$ 615 O–Si 370 S–Br 215
$\mathrm{C≡N}$ 891 O–P 350 Cl–Cl 243
C–O 350 O–Cl 205 Cl–Br 220
$\mathrm{C=O}$ 741 O–I 200 Cl–I 210
$\mathrm{C≡O}$ 1080 F–F 160 Br–Br 190
C–F 439 F–Si 540 Br–I 180
C–Si 360 F–P 489 I–I 150
C–P 265 F–S 285
Table $2$: Average Bond Lengths and Bond Energies for Some Common Bonds
Bond Bond Length (Å) Bond Energy (kJ/mol)
C–C 1.54 345
$\mathrm{C=C}$ 1.34 611
$\mathrm{C≡C}$ 1.20 837
C–N 1.43 290
$\mathrm{C=N}$ 1.38 615
$\mathrm{C≡N}$ 1.16 891
C–O 1.43 350
$\mathrm{C=O}$ 1.23 741
$\mathrm{C≡O}$ 1.13 1080
We can use bond energies to calculate approximate enthalpy changes for reactions where enthalpies of formation are not available. Calculations of this type will also tell us whether a reaction is exothermic or endothermic.
• An exothermic reactionH negative, heat produced) results when the bonds in the products are stronger than the bonds in the reactants.
• An endothermic reactionH positive, heat absorbed) results when the bonds in the products are weaker than those in the reactants.
The enthalpy change, ΔH, for a chemical reaction is approximately equal to the sum of the energy required to break all bonds in the reactants (energy “in”, positive sign) plus the energy released when all bonds are formed in the products (energy “out,” negative sign). This can be expressed mathematically in the following way:
$\Delta H=\sum D_{\text{bonds broken}}− \sum D_{\text{bonds formed}} \label{EQ3}$
In this expression, the symbol $\Sigma$ means “the sum of” and D represents the bond energy in kilojoules per mole, which is always a positive number. The bond energy is obtained from a table and will depend on whether the particular bond is a single, double, or triple bond. Thus, in calculating enthalpies in this manner, it is important that we consider the bonding in all reactants and products. Because D values are typically averages for one type of bond in many different molecules, this calculation provides a rough estimate, not an exact value, for the enthalpy of reaction.
Consider the following reaction:
$\ce{H_{2(g)} + Cl_{2(g)}⟶2HCl_{(g)}} \label{EQ4}$
or
$\ce{H–H_{(g)} + Cl–Cl_{(g)}⟶2H–Cl_{(g)}} \label{\EQ5}$
To form two moles of HCl, one mole of H–H bonds and one mole of Cl–Cl bonds must be broken. The energy required to break these bonds is the sum of the bond energy of the H–H bond (436 kJ/mol) and the Cl–Cl bond (243 kJ/mol). During the reaction, two moles of H–Cl bonds are formed (bond energy = 432 kJ/mol), releasing 2 × 432 kJ; or 864 kJ. Because the bonds in the products are stronger than those in the reactants, the reaction releases more energy than it consumes:
\begin {align*} ΔH&= \sum \mathrm{D_{bonds\: broken}}− \sum \mathrm{D_{bonds\: formed}}\[4pt] &=\mathrm{[D_{H−H}+D_{Cl−Cl}]−2D_{H−Cl}}\[4pt] &=\mathrm{[436+243]−2(432)=−185\:kJ} \end {align*} \nonumber
This excess energy is released as heat, so the reaction is exothermic. Table T2 gives a value for the standard molar enthalpy of formation of HCl(g), $ΔH^\circ_\ce f$, of –92.307 kJ/mol. Twice that value is –184.6 kJ, which agrees well with the answer obtained earlier for the formation of two moles of HCl.
Example $1$: Using Bond Energies to Approximate Enthalpy Changes
Methanol, CH3OH, may be an excellent alternative fuel. The high-temperature reaction of steam and carbon produces a mixture of the gases carbon monoxide, CO, and hydrogen, H2, from which methanol can be produced. Using the bond energies in Table $2$, calculate the approximate enthalpy change, ΔH, for the reaction here:
$CO_{(g)}+2H2_{(g)}⟶CH_3OH_{(g)} \nonumber$
Solution
First, we need to write the Lewis structures of the reactants and the products:
From this, we see that ΔH for this reaction involves the energy required to break a C–O triple bond and two H–H single bonds, as well as the energy produced by the formation of three C–H single bonds, a C–O single bond, and an O–H single bond. We can express this as follows (via Equation \ref{EQ3}):
\begin {align*} ΔH&= \sum D_{bonds\: broken}− \sum D_{bonds\: formed}\ ΔH&=\mathrm{[D_{C≡O}+2(D_{H−H})]−[3(D_{C−H})+D_{C−O}+D_{O−H}]} \end {align*} \nonumber
Using the bond energy values in Table $2$, we obtain:
\begin {align*} ΔH&=[1080+2(436)]−[3(415)+350+464]\ &=\ce{−107\:kJ} \end {align*} \nonumber
We can compare this value to the value calculated based on $ΔH^\circ_\ce f$ data from Appendix G:
\begin {align*} ΔH&=[ΔH^\circ_{\ce f}\ce{CH3OH}(g)]−[ΔH^\circ_{\ce f}\ce{CO}(g)+2×ΔH^\circ_{\ce f}\ce{H2}]\ &=[−201.0]−[−110.52+2×0]\ &=\mathrm{−90.5\:kJ} \end {align*} \nonumber
Note that there is a fairly significant gap between the values calculated using the two different methods. This occurs because D values are the average of different bond strengths; therefore, they often give only rough agreement with other data.
Exercise $1$
Ethyl alcohol, CH3CH2OH, was one of the first organic chemicals deliberately synthesized by humans. It has many uses in industry, and it is the alcohol contained in alcoholic beverages. It can be obtained by the fermentation of sugar or synthesized by the hydration of ethylene in the following reaction:
Using the bond energies in Table $2$, calculate an approximate enthalpy change, ΔH, for this reaction.
Answer
–35 kJ
Ionic Bond Strength and Lattice Energy
An ionic compound is stable because of the electrostatic attraction between its positive and negative ions. The lattice energy of a compound is a measure of the strength of this attraction. The lattice energy ($ΔH_{lattice}$) of an ionic compound is defined as the energy required to separate one mole of the solid into its component gaseous ions. For the ionic solid MX, the lattice energy is the enthalpy change of the process:
$MX_{(s)}⟶Mn^+_{(g)}+X^{n−}_{(g)} \;\;\;\;\; ΔH_{lattice} \label{EQ6}$
Note that we are using the convention where the ionic solid is separated into ions, so our lattice energies will be endothermic (positive values). Some texts use the equivalent but opposite convention, defining lattice energy as the energy released when separate ions combine to form a lattice and giving negative (exothermic) values. Thus, if you are looking up lattice energies in another reference, be certain to check which definition is being used. In both cases, a larger magnitude for lattice energy indicates a more stable ionic compound. For sodium chloride, ΔHlattice = 769 kJ. Thus, it requires 769 kJ to separate one mole of solid NaCl into gaseous Na+ and Cl ions. When one mole each of gaseous Na+ and Cl ions form solid NaCl, 769 kJ of heat is released.
The lattice energy $ΔH_{lattice}$ of an ionic crystal can be expressed by the following equation (derived from Coulomb’s law, governing the forces between electric charges):
$ΔH_{lattice}=\dfrac{C(Z^+)(Z^−)}{R_o} \label{EQ7}$
in which
• $\ce{C}$ is a constant that depends on the type of crystal structure;
• $Z^+$ and $Z^–$ are the charges on the ions; and
• $R_o$ is the interionic distance (the sum of the radii of the positive and negative ions).
Thus, the lattice energy of an ionic crystal increases rapidly as the charges of the ions increase and the sizes of the ions decrease. When all other parameters are kept constant, doubling the charge of both the cation and anion quadruples the lattice energy. For example, the lattice energy of LiF (Z+ and Z = 1) is 1023 kJ/mol, whereas that of MgO (Z+ and Z = 2) is 3900 kJ/mol (Ro is nearly the same—about 200 pm for both compounds).
Different interatomic distances produce different lattice energies. For example, we can compare the lattice energy of MgF2 (2957 kJ/mol) to that of MgI2 (2327 kJ/mol) to observe the effect on lattice energy of the smaller ionic size of F as compared to I.
Example $2$: Lattice Energy Comparisons
The precious gem ruby is aluminum oxide, Al2O3, containing traces of Cr3+. The compound Al2Se3 is used in the fabrication of some semiconductor devices. Which has the larger lattice energy, Al2O3 or Al2Se3?
Solution
In these two ionic compounds, the charges Z+ and Z are the same, so the difference in lattice energy will mainly depend upon Ro. The O2– ion is smaller than the Se2 ion. Thus, Al2O3 would have a shorter interionic distance than Al2Se3, and Al2O3 would have the larger lattice energy.
Exercise $2$
Zinc oxide, ZnO, is a very effective sunscreen. How would the lattice energy of ZnO compare to that of NaCl?
Answer
ZnO would have the larger lattice energy because the Z values of both the cation and the anion in ZnO are greater, and the interionic distance of ZnO is smaller than that of NaCl.
The Born-Haber Cycle
It is not possible to measure lattice energies directly. However, the lattice energy can be calculated using the equation given in the previous section or by using a thermochemical cycle. The Born-Haber cycle is an application of Hess’s law that breaks down the formation of an ionic solid into a series of individual steps:
• $ΔH^\circ_\ce f$, the standard enthalpy of formation of the compound
• IE, the ionization energy of the metal
• EA, the electron affinity of the nonmetal
• $ΔH^\circ_s$, the enthalpy of sublimation of the metal
• D, the bond dissociation energy of the nonmetal
• ΔHlattice, the lattice energy of the compound
Figure $1$ diagrams the Born-Haber cycle for the formation of solid cesium fluoride.
We begin with the elements in their most common states, Cs(s) and F2(g). The $ΔH^\circ_\ce s$ represents the conversion of solid cesium into a gas, and then the ionization energy converts the gaseous cesium atoms into cations. In the next step, we account for the energy required to break the F–F bond to produce fluorine atoms. Converting one mole of fluorine atoms into fluoride ions is an exothermic process, so this step gives off energy (the electron affinity) and is shown as decreasing along the y-axis. We now have one mole of Cs cations and one mole of F anions. These ions combine to produce solid cesium fluoride. The enthalpy change in this step is the negative of the lattice energy, so it is also an exothermic quantity. The total energy involved in this conversion is equal to the experimentally determined enthalpy of formation, $ΔH^\circ_\ce f$, of the compound from its elements. In this case, the overall change is exothermic.
Hess’s law can also be used to show the relationship between the enthalpies of the individual steps and the enthalpy of formation. Table $3$ shows this for cesium fluoride, CsF.
Table $3$: Enthalpies of Select Transitions
Enthalpy of sublimation of Cs(s) $\ce{Cs}(s)⟶\ce{Cs}(g)\hspace{20px}ΔH=ΔH^\circ_s=\mathrm{77\:kJ/mol}$
One-half of the bond energy of F2 $\dfrac{1}{2}\ce{F2}(g)⟶\ce{F}(g)\hspace{20px}ΔH=\dfrac{1}{2}D=\mathrm{79\:kJ/mol}$
Ionization energy of Cs(g) $\ce{Cs}(g)⟶\ce{Cs+}(g)+\ce{e-}\hspace{20px}ΔH=IE=\ce{376\:kJ/mol}$
Negative of the electron affinity of F $\ce{F}(g)+\ce{e-}⟶\ce{F-}(g)\hspace{20px}ΔH=−EA=\ce{-328\:kJ/mol}$
Negative of the lattice energy of CsF(s) $\ce{Cs+}(g)+\ce{F-}(g)⟶\ce{CsF}(s)\hspace{20px}ΔH=−ΔH_\ce{lattice}=\:?$
Enthalpy of formation of CsF(s), add steps 1–5
$ΔH=ΔH^\circ_f=ΔH^\circ_s+\dfrac{1}{2}D+IE+(−EA)+(−ΔH_\ce{lattice})$
$\ce{Cs}(s)+\dfrac{1}{2}\ce{F2}(g)⟶\ce{CsF}(s)=\ce{-554\:kJ/mol}$
Thus, the lattice energy can be calculated from other values. For cesium chloride, using this data, the lattice energy is:
$ΔH_\ce{lattice}=\mathrm{(411+109+122+496+368)\:kJ=770\:kJ} \nonumber$
The Born-Haber cycle may also be used to calculate any one of the other quantities in the equation for lattice energy, provided that the remainder is known. For example, if the relevant enthalpy of sublimation $ΔH^\circ_s$, ionization energy (IE), bond dissociation enthalpy (D), lattice energy ΔHlattice, and standard enthalpy of formation $ΔH^\circ_\ce f$ are known, the Born-Haber cycle can be used to determine the electron affinity of an atom.
Lattice energies calculated for ionic compounds are typically much larger than bond dissociation energies measured for covalent bonds. Whereas lattice energies typically fall in the range of 600–4000 kJ/mol (some even higher), covalent bond dissociation energies are typically between 150–400 kJ/mol for single bonds. Keep in mind, however, that these are not directly comparable values. For ionic compounds, lattice energies are associated with many interactions, as cations and anions pack together in an extended lattice. For covalent bonds, the bond dissociation energy is associated with the interaction of just two atoms.
Summary
The strength of a covalent bond is measured by its bond dissociation energy, that is, the amount of energy required to break that particular bond in a mole of molecules. Multiple bonds are stronger than single bonds between the same atoms. The enthalpy of a reaction can be estimated based on the energy input required to break bonds and the energy released when new bonds are formed. For ionic bonds, the lattice energy is the energy required to separate one mole of a compound into its gas phase ions. Lattice energy increases for ions with higher charges and shorter distances between ions. Lattice energies are often calculated using the Born-Haber cycle, a thermochemical cycle including all of the energetic steps involved in converting elements into an ionic compound.
Key Equations
• Bond energy for a diatomic molecule: $\ce{XY}(g)⟶\ce{X}(g)+\ce{Y}(g)\hspace{20px}\ce{D_{X–Y}}=ΔH°$
• Enthalpy change: ΔH = ƩDbonds broken – ƩDbonds formed
• Lattice energy for a solid MX: $\ce{MX}(s)⟶\ce M^{n+}(g)+\ce X^{n−}(g)\hspace{20px}ΔH_\ce{lattice}$
• Lattice energy for an ionic crystal: $ΔH_\ce{lattice}=\mathrm{\dfrac{C(Z^+)(Z^-)}{R_o}}$
Footnotes
1. This question is taken from the Chemistry Advanced Placement Examination and is used with the permission of the Educational Testing Service.
Glossary
bond energy
(also, bond dissociation energy) energy required to break a covalent bond in a gaseous substance
Born-Haber cycle
thermochemical cycle relating the various energetic steps involved in the formation of an ionic solid from the relevant elements
lattice energy (ΔHlattice)
energy required to separate one mole of an ionic solid into its component gaseous ions | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/04%3A_Chemical_Bonding_and_Molecular_Geometry/4.5%3A_Strengths_of_Ionic_and_Covalent_Bonds.txt |
Learning Objectives
• Predict the structures of small molecules using valence shell electron pair repulsion (VSEPR) theory
• Explain the concepts of polar covalent bonds and molecular polarity
• Assess the polarity of a molecule based on its bonding and structure
Thus far, we have used two-dimensional Lewis structures to represent molecules. However, molecular structure is actually three-dimensional, and it is important to be able to describe molecular bonds in terms of their distances, angles, and relative arrangements in space (Figure $1$). A bond angle is the angle between any two bonds that include a common atom, usually measured in degrees. A bond distance (or bond length) is the distance between the nuclei of two bonded atoms along the straight line joining the nuclei. Bond distances are measured in Ångstroms (1 Å = 10–10 m) or picometers (1 pm = 10–12 m, 100 pm = 1 Å).
VSEPR Theory
Valence shell electron-pair repulsion theory (VSEPR theory) enables us to predict the molecular structure, including approximate bond angles around a central atom, of a molecule from an examination of the number of bonds and lone electron pairs in its Lewis structure. The VSEPR model assumes that electron pairs in the valence shell of a central atom will adopt an arrangement that minimizes repulsions between these electron pairs by maximizing the distance between them. The electrons in the valence shell of a central atom form either bonding pairs of electrons, located primarily between bonded atoms, or lone pairs. The electrostatic repulsion of these electrons is reduced when the various regions of high electron density assume positions as far from each other as possible.
VSEPR theory predicts the arrangement of electron pairs around each central atom and, usually, the correct arrangement of atoms in a molecule. We should understand, however, that the theory only considers electron-pair repulsions. Other interactions, such as nuclear-nuclear repulsions and nuclear-electron attractions, are also involved in the final arrangement that atoms adopt in a particular molecular structure.
As a simple example of VSEPR theory, let us predict the structure of a gaseous BeF2 molecule. The Lewis structure of BeF2 (Figure $2$) shows only two electron pairs around the central beryllium atom. With two bonds and no lone pairs of electrons on the central atom, the bonds are as far apart as possible, and the electrostatic repulsion between these regions of high electron density is reduced to a minimum when they are on opposite sides of the central atom. The bond angle is 180° (Figure $2$).
Figure $3$ illustrates this and other electron-pair geometries that minimize the repulsions among regions of high electron density (bonds and/or lone pairs). Two regions of electron density around a central atom in a molecule form a linear geometry; three regions form a trigonal planar geometry; four regions form a tetrahedral geometry; five regions form a trigonal bipyramidal geometry; and six regions form an octahedral geometry.
Electron-pair Geometry versus Molecular Structure
It is important to note that electron-pair geometry around a central atom is not the same thing as its molecular structure. The electron-pair geometries shown in Figure $3$ describe all regions where electrons are located, bonds as well as lone pairs. Molecular structure describes the location of the atoms, not the electrons.
We differentiate between these two situations by naming the geometry that includes all electron pairs the electron-pair geometry. The structure that includes only the placement of the atoms in the molecule is called the molecular structure. The electron-pair geometries will be the same as the molecular structures when there are no lone electron pairs around the central atom, but they will be different when there are lone pairs present on the central atom.
For example, the methane molecule, CH4, which is the major component of natural gas, has four bonding pairs of electrons around the central carbon atom; the electron-pair geometry is tetrahedral, as is the molecular structure (Figure $4$). On the other hand, the ammonia molecule, NH3, also has four electron pairs associated with the nitrogen atom, and thus has a tetrahedral electron-pair geometry. One of these regions, however, is a lone pair, which is not included in the molecular structure, and this lone pair influences the shape of the molecule (Figure $5$).
Small distortions from the ideal angles in Figure $5$ can result from differences in repulsion between various regions of electron density. VSEPR theory predicts these distortions by establishing an order of repulsions and an order of the amount of space occupied by different kinds of electron pairs. The order of electron-pair repulsions from greatest to least repulsion is:
lone pair-lone pair > lone pair-bonding pair > bonding pair-bonding pair
This order of repulsions determines the amount of space occupied by different regions of electrons. A lone pair of electrons occupies a larger region of space than the electrons in a triple bond; in turn, electrons in a triple bond occupy more space than those in a double bond, and so on. The order of sizes from largest to smallest is:
lone pair > triple bond > double bond > single bond
Consider formaldehyde, H2CO, which is used as a preservative for biological and anatomical specimens. This molecule has regions of high electron density that consist of two single bonds and one double bond. The basic geometry is trigonal planar with 120° bond angles, but we see that the double bond causes slightly larger angles (121°), and the angle between the single bonds is slightly smaller (118°).
In the ammonia molecule, the three hydrogen atoms attached to the central nitrogen are not arranged in a flat, trigonal planar molecular structure, but rather in a three-dimensional trigonal pyramid (Figure $6$) with the nitrogen atom at the apex and the three hydrogen atoms forming the base. The ideal bond angles in a trigonal pyramid are based on the tetrahedral electron pair geometry. Again, there are slight deviations from the ideal because lone pairs occupy larger regions of space than do bonding electrons. The H–N–H bond angles in NH3 are slightly smaller than the 109.5° angle in a regular tetrahedron (Figure $6$) because the lone pair-bonding pair repulsion is greater than the bonding pair-bonding pair repulsion. The ideal molecular structures are predicted based on the electron-pair geometries for various combinations of lone pairs and bonding pairs.
According to VSEPR theory, the terminal atom locations (Xs in Figure $7$) are equivalent within the linear, trigonal planar, and tetrahedral electron-pair geometries (the first three rows of the table). It does not matter which X is replaced with a lone pair because the molecules can be rotated to convert positions. For trigonal bipyramidal electron-pair geometries, however, there are two distinct X positions (Figure $7$a): an axial position (if we hold a model of a trigonal bipyramid by the two axial positions, we have an axis around which we can rotate the model) and an equatorial position (three positions form an equator around the middle of the molecule). The axial position is surrounded by bond angles of 90°, whereas the equatorial position has more space available because of the 120° bond angles. In a trigonal bipyramidal electron-pair geometry, lone pairs always occupy equatorial positions because these more spacious positions can more easily accommodate the larger lone pairs.
Theoretically, we can come up with three possible arrangements for the three bonds and two lone pairs for the ClF3 molecule (Figure $7$). The stable structure is the one that puts the lone pairs in equatorial locations, giving a T-shaped molecular structure.
When a central atom has two lone electron pairs and four bonding regions, we have an octahedral electron-pair geometry. The two lone pairs are on opposite sides of the octahedron (180° apart), giving a square planar molecular structure that minimizes lone pair-lone pair repulsions.
Predicting Electron Pair Geometry and Molecular Structure
The following procedure uses VSEPR theory to determine the electron pair geometries and the molecular structures:
1. Write the Lewis structure of the molecule or polyatomic ion.
2. Count the number of regions of electron density (lone pairs and bonds) around the central atom. A single, double, or triple bond counts as one region of electron density.
3. Identify the electron-pair geometry based on the number of regions of electron density: linear, trigonal planar, tetrahedral, trigonal bipyramidal, or octahedral (Figure $7$, first column).
4. Use the number of lone pairs to determine the molecular structure (Figure $7$ ). If more than one arrangement of lone pairs and chemical bonds is possible, choose the one that will minimize repulsions, remembering that lone pairs occupy more space than multiple bonds, which occupy more space than single bonds. In trigonal bipyramidal arrangements, repulsion is minimized when every lone pair is in an equatorial position. In an octahedral arrangement with two lone pairs, repulsion is minimized when the lone pairs are on opposite sides of the central atom.
The following examples illustrate the use of VSEPR theory to predict the molecular structure of molecules or ions that have no lone pairs of electrons. In this case, the molecular structure is identical to the electron pair geometry.
Example $1$: Predicting Electron-pair Geometry and Molecular Structure
Predict the electron-pair geometry and molecular structure for each of the following:
1. carbon dioxide, CO2, a molecule produced by the combustion of fossil fuels
2. boron trichloride, BCl3, an important industrial chemical
Solution
(a) We write the Lewis structure of CO2 as:
This shows us two regions of high electron density around the carbon atom—each double bond counts as one region, and there are no lone pairs on the carbon atom. Using VSEPR theory, we predict that the two regions of electron density arrange themselves on opposite sides of the central atom with a bond angle of 180°. The electron-pair geometry and molecular structure are identical, and CO2 molecules are linear.
(b) We write the Lewis structure of BCl3 as:
Thus we see that BCl3 contains three bonds, and there are no lone pairs of electrons on boron. The arrangement of three regions of high electron density gives a trigonal planar electron-pair geometry. The B–Cl bonds lie in a plane with 120° angles between them. BCl3 also has a trigonal planar molecular structure.
The electron-pair geometry and molecular structure of BCl3 are both trigonal planar. Note that the VSEPR geometry indicates the correct bond angles (120°), unlike the Lewis structure shown above.
Exercise $1$
Carbonate, $\ce{CO3^2-}$, is a common polyatomic ion found in various materials from eggshells to antacids. What are the electron-pair geometry and molecular structure of this polyatomic ion?
Answer
The electron-pair geometry is trigonal planar and the molecular structure is trigonal planar. Due to resonance, all three C–O bonds are identical. Whether they are single, double, or an average of the two, each bond counts as one region of electron density.
Example $2$: Predicting Electron-pair Geometry and Molecular Structure
Two of the top 50 chemicals produced in the United States, ammonium nitrate and ammonium sulfate, both used as fertilizers, contain the ammonium ion. Predict the electron-pair geometry and molecular structure of the $\ce{NH4+}$ cation.
Solution
We write the Lewis structure of $\ce{NH4+}$ as:
Exercise $2$
Identify a molecule with trigonal bipyramidal molecular structure.
Answer
Any molecule with five electron pairs around the central atoms including no lone pairs will be trigonal bipyramidal. $\ce{PF5}$ is a common example
The next several examples illustrate the effect of lone pairs of electrons on molecular structure.
Example $3$: Lone Pairs on the Central Atom
Predict the electron-pair geometry and molecular structure of a water molecule.
Solution
The Lewis structure of H2O indicates that there are four regions of high electron density around the oxygen atom: two lone pairs and two chemical bonds:
Exercise $3$
The hydronium ion, H3O+, forms when acids are dissolved in water. Predict the electron-pair geometry and molecular structure of this cation.
Answer
electron pair geometry: tetrahedral; molecular structure: trigonal pyramidal
Example $4$: SF4 Sulfur tetrafluoride,
Predicting Electron-pair Geometry and Molecular Structure: SF4, is extremely valuable for the preparation of fluorine-containing compounds used as herbicides (i.e., SF4 is used as a fluorinating agent). Predict the electron-pair geometry and molecular structure of a SF4 molecule.
Solution
The Lewis structure of SF4 indicates five regions of electron density around the sulfur atom: one lone pair and four bonding pairs:
Exercise $4$
Predict the electron pair geometry and molecular structure for molecules of XeF2.
Answer
The electron-pair geometry is trigonal bipyramidal. The molecular structure is linear.
Example $4$: XeF4
Of all the noble gases, xenon is the most reactive, frequently reacting with elements such as oxygen and fluorine. Predict the electron-pair geometry and molecular structure of the XeF4 molecule.
Solution
The Lewis structure of XeF4 indicates six regions of high electron density around the xenon atom: two lone pairs and four bonds:
Exercise $4$
In a certain molecule, the central atom has three lone pairs and two bonds. What will the electron pair geometry and molecular structure be?
Answer
electron pair geometry: trigonal bipyramidal; molecular structure: linear
Molecular Structure for Multicenter Molecules
When a molecule or polyatomic ion has only one central atom, the molecular structure completely describes the shape of the molecule. Larger molecules do not have a single central atom, but are connected by a chain of interior atoms that each possess a “local” geometry. The way these local structures are oriented with respect to each other also influences the molecular shape, but such considerations are largely beyond the scope of this introductory discussion. For our purposes, we will only focus on determining the local structures.
Example $5$: Predicting Structure in Multicenter Molecules
The Lewis structure for the simplest amino acid, glycine, H2NCH2CO2H, is shown here. Predict the local geometry for the nitrogen atom, the two carbon atoms, and the oxygen atom with a hydrogen atom attached:
Solution
Consider each central atom independently. The electron-pair geometries:
• nitrogen––four regions of electron density; tetrahedral
• carbon (CH2)––four regions of electron density; tetrahedral
• carbon (CO2)—three regions of electron density; trigonal planar
• oxygen (OH)—four regions of electron density; tetrahedral
The local structures:
• nitrogen––three bonds, one lone pair; trigonal pyramidal
• carbon (CH2)—four bonds, no lone pairs; tetrahedral
• carbon (CO2)—three bonds (double bond counts as one bond), no lone pairs; trigonal planar
• oxygen (OH)—two bonds, two lone pairs; bent (109°)
Exercise $5$
Another amino acid is alanine, which has the Lewis structure shown here. Predict the electron-pair geometry and local structure of the nitrogen atom, the three carbon atoms, and the oxygen atom with hydrogen attached:
Answer
electron-pair geometries: nitrogen––tetrahedral; carbon (CH)—tetrahedral; carbon (CH3)—tetrahedral; carbon (CO2)—trigonal planar; oxygen (OH)—tetrahedral; local structures: nitrogen—trigonal pyramidal; carbon (CH)—tetrahedral; carbon (CH3)—tetrahedral; carbon (CO2)—trigonal planar; oxygen (OH)—bent (109°)
Example $6$: Molecular Simulation
Using this molecular shape simulator allows us to control whether bond angles and/or lone pairs are displayed by checking or unchecking the boxes under “Options” on the right. We can also use the “Name” checkboxes at bottom-left to display or hide the electron pair geometry (called “electron geometry” in the simulator) and/or molecular structure (called “molecular shape” in the simulator).
Build the molecule HCN in the simulator based on the following Lewis structure:
$\mathrm{H–C≡N}$
Click on each bond type or lone pair at right to add that group to the central atom. Once you have the complete molecule, rotate it to examine the predicted molecular structure. What molecular structure is this?
Solution
The molecular structure is linear.
Exercise $6$
Build a more complex molecule in the simulator. Identify the electron-group geometry, molecular structure, and bond angles. Then try to find a chemical formula that would match the structure you have drawn.
Answer
Answers will vary. For example, an atom with four single bonds, a double bond, and a lone pair has an octahedral electron-group geometry and a square pyramidal molecular structure. XeOF4 is a molecule that adopts this structure.
Molecular Polarity and Dipole Moment
As discussed previously, polar covalent bonds connect two atoms with differing electronegativities, leaving one atom with a partial positive charge (δ+) and the other atom with a partial negative charge (δ–), as the electrons are pulled toward the more electronegative atom. This separation of charge gives rise to a bond dipole moment. The magnitude of a bond dipole moment is represented by the Greek letter mu (µ) and is given by
$μ=Qr \label{7.6.X}$
where
• $Q$ is the magnitude of the partial charges (determined by the electronegativity difference) and
• $r$ is the distance between the charges:
This bond moment can be represented as a vector, a quantity having both direction and magnitude (Figure $12$). Dipole vectors are shown as arrows pointing along the bond from the less electronegative atom toward the more electronegative atom. A small plus sign is drawn on the less electronegative end to indicate the partially positive end of the bond. The length of the arrow is proportional to the magnitude of the electronegativity difference between the two atoms.
A whole molecule may also have a separation of charge, depending on its molecular structure and the polarity of each of its bonds. If such a charge separation exists, the molecule is said to be a polar molecule (or dipole); otherwise the molecule is said to be nonpolar. The dipole moment measures the extent of net charge separation in the molecule as a whole. We determine the dipole moment by adding the bond moments in three-dimensional space, taking into account the molecular structure.
For diatomic molecules, there is only one bond, so its bond dipole moment determines the molecular polarity. Homonuclear diatomic molecules such as Br2 and N2 have no difference in electronegativity, so their dipole moment is zero. For heteronuclear molecules such as CO, there is a small dipole moment. For HF, there is a larger dipole moment because there is a larger difference in electronegativity.
When a molecule contains more than one bond, the geometry must be taken into account. If the bonds in a molecule are arranged such that their bond moments cancel (vector sum equals zero), then the molecule is nonpolar. This is the situation in CO2 (Figure $\PageIndex{13A}$). Each of the bonds is polar, but the molecule as a whole is nonpolar. From the Lewis structure, and using VSEPR theory, we determine that the CO2 molecule is linear with polar C=O bonds on opposite sides of the carbon atom. The bond moments cancel because they are pointed in opposite directions. In the case of the water molecule (Figure $\PageIndex{13B}$), the Lewis structure again shows that there are two bonds to a central atom, and the electronegativity difference again shows that each of these bonds has a nonzero bond moment. In this case, however, the molecular structure is bent because of the lone pairs on O, and the two bond moments do not cancel. Therefore, water does have a net dipole moment and is a polar molecule (dipole).
The OCS molecule has a structure similar to CO2, but a sulfur atom has replaced one of the oxygen atoms. To determine if this molecule is polar, we draw the molecular structure. VSEPR theory predicts a linear molecule:
The C–O bond is considerably polar. Although C and S have very similar electronegativity values, S is slightly more electronegative than C, and so the C-S bond is just slightly polar. Because oxygen is more electronegative than sulfur, the oxygen end of the molecule is the negative end.
Chloromethane, CH3Cl, is another example of a polar molecule. Although the polar C–Cl and C–H bonds are arranged in a tetrahedral geometry, the C–Cl bonds have a larger bond moment than the C–H bond, and the bond moments do not completely cancel each other. All of the dipoles have a upward component in the orientation shown, since carbon is more electronegative than hydrogen and less electronegative than chlorine:
When we examine the highly symmetrical molecules BF3 (trigonal planar), CH4 (tetrahedral), PF5 (trigonal bipyramidal), and SF6 (octahedral), in which all the polar bonds are identical, the molecules are nonpolar. The bonds in these molecules are arranged such that their dipoles cancel. However, just because a molecule contains identical bonds does not mean that the dipoles will always cancel. Many molecules that have identical bonds and lone pairs on the central atoms have bond dipoles that do not cancel. Examples include H2S and NH3. A hydrogen atom is at the positive end and a nitrogen or sulfur atom is at the negative end of the polar bonds in these molecules:
To summarize, to be polar, a molecule must:
1. Contain at least one polar covalent bond.
2. Have a molecular structure such that the sum of the vectors of each bond dipole moment does not cancel.
Properties of Polar Molecules
Polar molecules tend to align when placed in an electric field with the positive end of the molecule oriented toward the negative plate and the negative end toward the positive plate (Figure $14$). We can use an electrically charged object to attract polar molecules, but nonpolar molecules are not attracted. Also, polar solvents are better at dissolving polar substances, and nonpolar solvents are better at dissolving nonpolar substances.
Example $7$: Polarity Simulations
Open the molecule polarity simulation and select the “Three Atoms” tab at the top. This should display a molecule ABC with three electronegativity adjustors. You can display or hide the bond moments, molecular dipoles, and partial charges at the right. Turning on the Electric Field will show whether the molecule moves when exposed to a field, similar to Figure $14$.
Use the electronegativity controls to determine how the molecular dipole will look for the starting bent molecule if:
1. A and C are very electronegative and B is in the middle of the range.
2. A is very electronegative, and B and C are not.
Solution
1. Molecular dipole moment points immediately between A and C.
2. Molecular dipole moment points along the A–B bond, toward A.
Exercise $7$
Determine the partial charges that will give the largest possible bond dipoles.
Answer
The largest bond moments will occur with the largest partial charges. The two solutions above represent how unevenly the electrons are shared in the bond. The bond moments will be maximized when the electronegativity difference is greatest. The controls for A and C should be set to one extreme, and B should be set to the opposite extreme. Although the magnitude of the bond moment will not change based on whether B is the most electronegative or the least, the direction of the bond moment will.
Summary
VSEPR theory predicts the three-dimensional arrangement of atoms in a molecule. It states that valence electrons will assume an electron-pair geometry that minimizes repulsions between areas of high electron density (bonds and/or lone pairs). Molecular structure, which refers only to the placement of atoms in a molecule and not the electrons, is equivalent to electron-pair geometry only when there are no lone electron pairs around the central atom. A dipole moment measures a separation of charge. For one bond, the bond dipole moment is determined by the difference in electronegativity between the two atoms. For a molecule, the overall dipole moment is determined by both the individual bond moments and how these dipoles are arranged in the molecular structure. Polar molecules (those with an appreciable dipole moment) interact with electric fields, whereas nonpolar molecules do not.
Glossary
axial position
location in a trigonal bipyramidal geometry in which there is another atom at a 180° angle and the equatorial positions are at a 90° angle
bond angle
angle between any two covalent bonds that share a common atom
bond distance
(also, bond length) distance between the nuclei of two bonded atoms
bond dipole moment
separation of charge in a bond that depends on the difference in electronegativity and the bond distance represented by partial charges or a vector
dipole moment
property of a molecule that describes the separation of charge determined by the sum of the individual bond moments based on the molecular structure
electron-pair geometry
arrangement around a central atom of all regions of electron density (bonds, lone pairs, or unpaired electrons)
equatorial position
one of the three positions in a trigonal bipyramidal geometry with 120° angles between them; the axial positions are located at a 90° angle
linear
shape in which two outside groups are placed on opposite sides of a central atom
molecular structure
structure that includes only the placement of the atoms in the molecule
octahedral
shape in which six outside groups are placed around a central atom such that a three-dimensional shape is generated with four groups forming a square and the other two forming the apex of two pyramids, one above and one below the square plane
polar molecule
(also, dipole) molecule with an overall dipole moment
tetrahedral
shape in which four outside groups are placed around a central atom such that a three-dimensional shape is generated with four corners and 109.5° angles between each pair and the central atom
trigonal bipyramidal
shape in which five outside groups are placed around a central atom such that three form a flat triangle with 120° angles between each pair and the central atom, and the other two form the apex of two pyramids, one above and one below the triangular plane
trigonal planar
shape in which three outside groups are placed in a flat triangle around a central atom with 120° angles between each pair and the central atom
valence shell electron-pair repulsion theory (VSEPR)
theory used to predict the bond angles in a molecule based on positioning regions of high electron density as far apart as possible to minimize electrostatic repulsion
vector
quantity having magnitude and direction | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/04%3A_Chemical_Bonding_and_Molecular_Geometry/4.6%3A_Molecular_Structure_and_Polarity.txt |
7.1: Ionic Bonding
Q7.1.1
Does a cation gain protons to form a positive charge or does it lose electrons?
S7.1.1
The protons in the nucleus do not change during normal chemical reactions. Only the outer electrons move. Positive charges form when electrons are lost.
Q7.1.2
Iron(III) sulfate [Fe2(SO4)3] is composed of Fe3+ and $\ce{SO4^2-}$ ions. Explain why a sample of iron(III) sulfate is uncharged.
Q7.1.3
Which of the following atoms would be expected to form negative ions in binary ionic compounds and which would be expected to form positive ions: P, I, Mg, Cl, In, Cs, O, Pb, Co?
S7.1.3
P, I, Cl, and O would form anions because they are nonmetals. Mg, In, Cs, Pb, and Co would form cations because they are metals.
Q7.1.4
Which of the following atoms would be expected to form negative ions in binary ionic compounds and which would be expected to form positive ions: Br, Ca, Na, N, F, Al, Sn, S, Cd?
Q7.1.5
Predict the charge on the monatomic ions formed from the following atoms in binary ionic compounds:
1. P
2. Mg
3. Al
4. O
5. Cl
6. Cs
S7.1.5
P3–; Mg2+; Al3+; O2–; Cl; Cs+
Q7.1.6
Predict the charge on the monatomic ions formed from the following atoms in binary ionic compounds:
1. I
2. Sr
3. K
4. N
5. S
6. In
1. I-
2. Sr2+
3. K+
4. N3-
5. S2-
6. In3+
Q7.1.7
Write the electron configuration for each of the following ions:
1. As3–
2. I
3. Be2+
4. Cd2+
5. O2–
6. Ga3+
7. Li+
8. (h) N3–
9. (i) Sn2+
10. (j) Co2+
11. (k) Fe2+
12. (l) As3+
S7.1.7
[Ar]4s23d104p6; [Kr]4d105s25p6 1s2 [Kr]4d10; [He]2s22p6; [Ar]3d10; 1s2 (h) [He]2s22p6 (i) [Kr]4d105s2 (j) [Ar]3d7 (k) [Ar]3d6, (l) [Ar]3d104s2
Q7.1.8
Write the electron configuration for the monatomic ions formed from the following elements (which form the greatest concentration of monatomic ions in seawater):
1. Cl
2. Na
3. Mg
4. Ca
5. K
6. Br
7. Sr
8. (h) F
Q7.1.9
Write out the full electron configuration for each of the following atoms and for the monatomic ion found in binary ionic compounds containing the element:
1. Al
2. Br
3. Sr
4. Li
5. As
6. S
S7.1.9
1s22s22p63s23p1; Al3+: 1s22s22p6; 1s22s22p63s23p63d104s24p5; 1s22s22p63s23p63d104s24p6; 1s22s22p63s23p63d104s24p65s2;
Sr2+: 1s22s22p63s23p63d104s24p6; 1s22s1;
Li+: 1s2; 1s22s22p63s23p63d104s24p3; 1s22s22p63s23p63d104s24p6; 1s22s22p63s23p4; 1s22s22p63s23p6
Q7.1.10
From the labels of several commercial products, prepare a list of six ionic compounds in the products. For each compound, write the formula. (You may need to look up some formulas in a suitable reference.)
7.3: Covalent Bonding
Why is it incorrect to speak of a molecule of solid NaCl?
NaCl consists of discrete ions arranged in a crystal lattice, not covalently bonded molecules.
What information can you use to predict whether a bond between two atoms is covalent or ionic?
Predict which of the following compounds are ionic and which are covalent, based on the location of their constituent atoms in the periodic table:
1. Cl2CO
2. MnO
3. NCl3
4. CoBr2
5. K2S
6. CO
7. CaF2
8. (h) HI
9. (i) CaO
10. (j) IBr
11. (k) CO2
ionic: (b), (d), (e), (g), and (i); covalent: (a), (c), (f), (h), (j), and (k)
Explain the difference between a nonpolar covalent bond, a polar covalent bond, and an ionic bond.
From its position in the periodic table, determine which atom in each pair is more electronegative:
1. Br or Cl
2. N or O
3. S or O
4. P or S
5. Si or N
6. Ba or P
7. N or K
Cl; O; O; S; N; P; N
From its position in the periodic table, determine which atom in each pair is more electronegative:
1. N or P
2. N or Ge
3. S or F
4. Cl or S
5. H or C
6. Se or P
7. C or Si
From their positions in the periodic table, arrange the atoms in each of the following series in order of increasing electronegativity:
1. C, F, H, N, O
2. Br, Cl, F, H, I
3. F, H, O, P, S
4. Al, H, Na, O, P
5. Ba, H, N, O, As
H, C, N, O, F; H, I, Br, Cl, F; H, P, S, O, F; Na, Al, H, P, O; Ba, H, As, N, O
From their positions in the periodic table, arrange the atoms in each of the following series in order of increasing electronegativity:
1. As, H, N, P, Sb
2. Cl, H, P, S, Si
3. Br, Cl, Ge, H, Sr
4. Ca, H, K, N, Si
5. Cl, Cs, Ge, H, Sr
Which atoms can bond to sulfur so as to produce a positive partial charge on the sulfur atom?
N, O, F, and Cl
Which is the most polar bond?
1. C–C
2. C–H
3. N–H
4. O–H
5. Se–H
Identify the more polar bond in each of the following pairs of bonds:
1. HF or HCl
2. NO or CO
3. SH or OH
4. PCl or SCl
5. CH or NH
6. SO or PO
7. CN or NN
HF; CO; OH; PCl; NH; PO; CN
Which of the following molecules or ions contain polar bonds?
1. O3
2. S8
3. $\ce{O2^2-}$
4. $\ce{NO3-}$
5. CO2
6. H2S
7. $\ce{BH4-}$
7.4: Lewis Symbols and Structures
Q7.4.1
Write the Lewis symbols for each of the following ions:
1. As3–
2. I
3. Be2+
4. O2–
5. Ga3+
6. Li+
7. N3–
eight electrons:
eight electrons:
no electrons
Be2+;
eight electrons:
no electrons
Ga3+;
no electrons
Li+;
eight electrons:
Q7.4.2
Many monatomic ions are found in seawater, including the ions formed from the following list of elements. Write the Lewis symbols for the monatomic ions formed from the following elements:
1. Cl
2. Na
3. Mg
4. Ca
5. K
6. Br
7. Sr
8. F
Q7.4.3
Write the Lewis symbols of the ions in each of the following ionic compounds and the Lewis symbols of the atom from which they are formed:
1. MgS
2. Al2O3
3. GaCl3
4. K2O
5. Li3N
6. KF
(a)
;
(b)
;
(c)
;
(d)
;
(e)
;
(f)
In the Lewis structures listed here, M and X represent various elements in the third period of the periodic table. Write the formula of each compound using the chemical symbols of each element:
(a)
(b)
(c)
(d)
Write the Lewis structure for the diatomic molecule P2, an unstable form of phosphorus found in high-temperature phosphorus vapor.
Write Lewis structures for the following:
1. H2
2. HBr
3. PCl3
4. SF2
5. H2CCH2
6. HNNH
7. H2CNH
8. (h) NO
9. (i) N2
10. (j) CO
11. (k) CN
Write Lewis structures for the following:
1. O2
2. H2CO
3. AsF3
4. ClNO
5. SiCl4
6. H3O+
7. $\ce{NH4+}$
8. (h) $\ce{BF4-}$
9. (i) HCCH
10. (j) ClCN
11. (k) $\ce{C2^2+}$
(a)
In this case, the Lewis structure is inadequate to depict the fact that experimental studies have shown two unpaired electrons in each oxygen molecule.
(b)
;
(c)
;
(d)
;
(e)
;
(f)
;
(g)
;
(h)
;
(i)
;
(j)
;
(k)
Write Lewis structures for the following:
1. ClF3
2. PCl5
3. BF3
4. $\ce{PF6-}$
Write Lewis structures for the following:
1. SeF6
2. XeF4
3. $\ce{SeCl3+}$
4. Cl2BBCl2 (contains a B–B bond)
SeF6:
;
XeF4:
;
$\ce{SeCl3+}$:
;
Cl2BBCl2:
Write Lewis structures for:
1. $\ce{PO4^3-}$
2. $\ce{ICl4-}$
3. $\ce{SO3^2-}$
4. HONO
Correct the following statement: “The bonds in solid PbCl2 are ionic; the bond in a HCl molecule is covalent. Thus, all of the valence electrons in PbCl2 are located on the Cl ions, and all of the valence electrons in a HCl molecule are shared between the H and Cl atoms.”
Two valence electrons per Pb atom are transferred to Cl atoms; the resulting Pb2+ ion has a 6s2 valence shell configuration. Two of the valence electrons in the HCl molecule are shared, and the other six are located on the Cl atom as lone pairs of electrons.
Write Lewis structures for the following molecules or ions:
1. SbH3
2. XeF2
3. Se8 (a cyclic molecule with a ring of eight Se atoms)
Methanol, H3COH, is used as the fuel in some race cars. Ethanol, C2H5OH, is used extensively as motor fuel in Brazil. Both methanol and ethanol produce CO2 and H2O when they burn. Write the chemical equations for these combustion reactions using Lewis structures instead of chemical formulas.
Many planets in our solar system contain organic chemicals including methane (CH4) and traces of ethylene (C2H4), ethane (C2H6), propyne (H3CCCH), and diacetylene (HCCCCH). Write the Lewis structures for each of these molecules.
Carbon tetrachloride was formerly used in fire extinguishers for electrical fires. It is no longer used for this purpose because of the formation of the toxic gas phosgene, Cl2CO. Write the Lewis structures for carbon tetrachloride and phosgene.
Identify the atoms that correspond to each of the following electron configurations. Then, write the Lewis symbol for the common ion formed from each atom:
1. 1s22s22p5
2. 1s22s22p63s2
3. 1s22s22p63s23p64s23d10
4. 1s22s22p63s23p64s23d104p4
5. 1s22s22p63s23p64s23d104p1
The arrangement of atoms in several biologically important molecules is given here. Complete the Lewis structures of these molecules by adding multiple bonds and lone pairs. Do not add any more atoms.
the amino acid serine:
urea:
pyruvic acid:
uracil:
carbonic acid:
(a)
;
(b)
;
(c)
;
(d)
;
(e)
A compound with a molar mass of about 28 g/mol contains 85.7% carbon and 14.3% hydrogen by mass. Write the Lewis structure for a molecule of the compound.
A compound with a molar mass of about 42 g/mol contains 85.7% carbon and 14.3% hydrogen by mass. Write the Lewis structure for a molecule of the compound.
Two arrangements of atoms are possible for a compound with a molar mass of about 45 g/mol that contains 52.2% C, 13.1% H, and 34.7% O by mass. Write the Lewis structures for the two molecules.
How are single, double, and triple bonds similar? How do they differ?
Each bond includes a sharing of electrons between atoms. Two electrons are shared in a single bond; four electrons are shared in a double bond; and six electrons are shared in a triple bond.
7.5: Formal Charges and Resonance
Write resonance forms that describe the distribution of electrons in each of these molecules or ions.
1. selenium dioxide, OSeO
2. nitrate ion, $\ce{NO3-}$
3. nitric acid, HNO3 (N is bonded to an OH group and two O atoms)
4. benzene, C6H6:
the formate ion:
Write resonance forms that describe the distribution of electrons in each of these molecules or ions.
1. sulfur dioxide, SO2
2. carbonate ion, $\ce{CO3^2-}$
3. hydrogen carbonate ion, $\ce{HCO3-}$ (C is bonded to an OH group and two O atoms)
4. pyridine:
the allyl ion:
(a)
;
(b)
;
(c)
;
(d)
;
(e)
Write the resonance forms of ozone, O3, the component of the upper atmosphere that protects the Earth from ultraviolet radiation.
Sodium nitrite, which has been used to preserve bacon and other meats, is an ionic compound. Write the resonance forms of the nitrite ion, $\ce{NO2-}$.
In terms of the bonds present, explain why acetic acid, CH3CO2H, contains two distinct types of carbon-oxygen bonds, whereas the acetate ion, formed by loss of a hydrogen ion from acetic acid, only contains one type of carbon-oxygen bond. The skeleton structures of these species are shown:
Write the Lewis structures for the following, and include resonance structures where appropriate. Indicate which has the strongest carbon-oxygen bond.
1. CO2
2. CO
(a)
(b)
CO has the strongest carbon-oxygen bond because there is a triple bond joining C and O. CO2 has double bonds.
Toothpastes containing sodium hydrogen carbonate (sodium bicarbonate) and hydrogen peroxide are widely used. Write Lewis structures for the hydrogen carbonate ion and hydrogen peroxide molecule, with resonance forms where appropriate.
Determine the formal charge of each element in the following:
1. HCl
2. CF4
3. PCl3
4. PF5
H: 0, Cl: 0; C: 0, F: 0; P: 0, Cl 0; P: 0, F: 0
Determine the formal charge of each element in the following:
1. H3O+
2. $\ce{SO4^2-}$
3. NH3
4. $\ce{O2^2-}$
5. H2O2
Calculate the formal charge of chlorine in the molecules Cl2, BeCl2, and ClF5.
Cl in Cl2: 0; Cl in BeCl2: 0; Cl in ClF5: 0
Calculate the formal charge of each element in the following compounds and ions:
1. F2CO
2. NO
3. $\ce{BF4-}$
4. $\ce{SnCl3-}$
5. H2CCH2
6. ClF3
7. SeF6
8. (h) $\ce{PO4^3-}$
Draw all possible resonance structures for each of these compounds. Determine the formal charge on each atom in each of the resonance structures:
1. O3
2. SO2
3. $\ce{NO2-}$
4. $\ce{NO3-}$
;
(b)
;
(c)
;
(d)
Based on formal charge considerations, which of the following would likely be the correct arrangement of atoms in nitrosyl chloride: ClNO or ClON?
Based on formal charge considerations, which of the following would likely be the correct arrangement of atoms in hypochlorous acid: HOCl or OClH?
HOCl
Based on formal charge considerations, which of the following would likely be the correct arrangement of atoms in sulfur dioxide: OSO or SOO?
Draw the structure of hydroxylamine, H3NO, and assign formal charges; look up the structure. Is the actual structure consistent with the formal charges?
The structure that gives zero formal charges is consistent with the actual structure:
Iodine forms a series of fluorides (listed here). Write Lewis structures for each of the four compounds and determine the formal charge of the iodine atom in each molecule:
1. IF
2. IF3
3. IF5
4. IF7
Write the Lewis structure and chemical formula of the compound with a molar mass of about 70 g/mol that contains 19.7% nitrogen and 80.3% fluorine by mass, and determine the formal charge of the atoms in this compound.
NF3;
Which of the following structures would we expect for nitrous acid? Determine the formal charges:
Sulfuric acid is the industrial chemical produced in greatest quantity worldwide. About 90 billion pounds are produced each year in the United States alone. Write the Lewis structure for sulfuric acid, H2SO4, which has two oxygen atoms and two OH groups bonded to the sulfur.
7.6: Strengths of Ionic and Covalent Bonds
Which bond in each of the following pairs of bonds is the strongest?
1. C–C or $\mathrm{C=C}$
2. C–N or $\mathrm{C≡N}$
3. $\mathrm{C≡O}$ or $\mathrm{C=O}$
4. H–F or H–Cl
5. C–H or O–H
6. C–N or C–O
Using the bond energies in Table, determine the approximate enthalpy change for each of the following reactions:
1. $\ce{H2}(g)+\ce{Br2}(g)⟶\ce{2HBr}(g)$
2. $\ce{CH4}(g)+\ce{I2}(g)⟶\ce{CH3I}(g)+\ce{HI}(g)$
3. (c) $\ce{C2H4}(g)+\ce{3O2}(g)⟶\ce{2CO2}(g)+\ce{2H2O}(g)$
1. −114 kJ;
2. 30 kJ;
3. (c) −1055 kJ
Using the bond energies in Table, determine the approximate enthalpy change for each of the following reactions:
1. $\ce{Cl2}(g)+\ce{3F2}(g)⟶\ce{2ClF3}(g)$
2. $\mathrm{H_2C=CH_2}(g)+\ce{H2}(g)⟶\ce{H3CCH3}(g)$
3. (c) $\ce{2C2H6}(g)+\ce{7O2}(g)⟶\ce{4CO2}(g)+\ce{6H2O}(g)$
When a molecule can form two different structures, the structure with the stronger bonds is usually the more stable form. Use bond energies to predict the correct structure of the hydroxylamine molecule:
The greater bond energy is in the figure on the left. It is the more stable form.
How does the bond energy of HCldiffer from the standard enthalpy of formation of HCl(g)?
Using the standard enthalpy of formation data in Appendix G, show how the standard enthalpy of formation of HCl(g) can be used to determine the bond energy.
$\ce{HCl}(g)⟶\dfrac{1}{2}\ce{H2}(g)+\dfrac{1}{2}\ce{Cl2}(g)\hspace{20px}ΔH^\circ_1=−ΔH^\circ_{\ce f[\ce{HCl}(g)]}\ \dfrac{1}{2}\ce{H2}(g)⟶\ce{H}(g)\hspace{105px}ΔH^\circ_2=ΔH^\circ_{\ce f[\ce H(g)]}\ \underline{\dfrac{1}{2}\ce{Cl2}(g)⟶\ce{Cl}(g)\hspace{99px}ΔH^\circ_3=ΔH^\circ_{\ce f[\ce{Cl}(g)]}}\ \ce{HCl}(g)⟶\ce{H}(g)+\ce{Cl}(g)\hspace{58px}ΔH^\circ_{298}=ΔH^\circ_1+ΔH^\circ_2+ΔH^\circ_3$
\begin{align} D_\ce{HCl}=ΔH^\circ_{298}&=ΔH^\circ_{\ce f[\ce{HCl}(g)]}+ΔH^\circ_{\ce f[\ce H(g)]}+ΔH^\circ_{\ce f[\ce{Cl}(g)]}\ &=\mathrm{−(−92.307\:kJ)+217.97\:kJ+121.3\:kJ}\ &=\mathrm{431.6\:kJ} \end{align}
Using the standard enthalpy of formation data in Appendix G, calculate the bond energy of the carbon-sulfur double bond in CS2.
Using the standard enthalpy of formation data in Appendix G, determine which bond is stronger: the S–F bond in SF4(g) or in SF6(g)?
The S–F bond in SF4 is stronger.
Using the standard enthalpy of formation data in Appendix G, determine which bond is stronger: the P–Cl bond in PCl3(g) or in PCl5(g)?
Complete the following Lewis structure by adding bonds (not atoms), and then indicate the longest bond:
The C–C single bonds are longest.
Use the bond energy to calculate an approximate value of ΔH for the following reaction. Which is the more stable form of FNO2?
Use principles of atomic structure to answer each of the following:1
1. The radius of the Ca atom is 197 pm; the radius of the Ca2+ ion is 99 pm. Account for the difference.
2. The lattice energy of CaO(s) is –3460 kJ/mol; the lattice energy of K2O is –2240 kJ/mol. Account for the difference.
3. (c) Given these ionization values, explain the difference between Ca and K with regard to their first and second ionization energies.
Element First Ionization Energy (kJ/mol) Second Ionization Energy (kJ/mol)
K 419 3050
Ca 590 1140
The first ionization energy of Mg is 738 kJ/mol and that of Al is 578 kJ/mol. Account for this difference.
When two electrons are removed from the valence shell, the Ca radius loses the outermost energy level and reverts to the lower n = 3 level, which is much smaller in radius. The +2 charge on calcium pulls the oxygen much closer compared with K, thereby increasing the lattice energy relative to a less charged ion. (c) Removal of the 4s electron in Ca requires more energy than removal of the 4s electron in K because of the stronger attraction of the nucleus and the extra energy required to break the pairing of the electrons. The second ionization energy for K requires that an electron be removed from a lower energy level, where the attraction is much stronger from the nucleus for the electron. In addition, energy is required to unpair two electrons in a full orbital. For Ca, the second ionization potential requires removing only a lone electron in the exposed outer energy level. In Al, the removed electron is relatively unprotected and unpaired in a p orbital. The higher energy for Mg mainly reflects the unpairing of the 2s electron.
The lattice energy of LiF is 1023 kJ/mol, and the Li–F distance is 200.8 pm. NaF crystallizes in the same structure as LiF but with a Na–F distance of 231 pm. Which of the following values most closely approximates the lattice energy of NaF: 510, 890, 1023, 1175, or 4090 kJ/mol? Explain your choice.
For which of the following substances is the least energy required to convert one mole of the solid into separate ions?
1. MgO
2. SrO
3. (c) KF
4. CsF
5. MgF2
(d)
The reaction of a metal, M, with a halogen, X2, proceeds by an exothermic reaction as indicated by this equation: $\ce{M}(s)+\ce{X2}(g)⟶\ce{MX2}(s)$. For each of the following, indicate which option will make the reaction more exothermic. Explain your answers.
1. a large radius vs. a small radius for M+2
2. a high ionization energy vs. a low ionization energy for M
3. (c) an increasing bond energy for the halogen
4. a decreasing electron affinity for the halogen
5. an increasing size of the anion formed by the halogen
The lattice energy of LiF is 1023 kJ/mol, and the Li–F distance is 201 pm. MgO crystallizes in the same structure as LiF but with a Mg–O distance of 205 pm. Which of the following values most closely approximates the lattice energy of MgO: 256 kJ/mol, 512 kJ/mol, 1023 kJ/mol, 2046 kJ/mol, or 4008 kJ/mol? Explain your choice.
4008 kJ/mol; both ions in MgO have twice the charge of the ions in LiF; the bond length is very similar and both have the same structure; a quadrupling of the energy is expected based on the equation for lattice energy
Which compound in each of the following pairs has the larger lattice energy? Note: Mg2+ and Li+ have similar radii; O2– and F have similar radii. Explain your choices.
1. MgO or MgSe
2. LiF or MgO
3. (c) Li2O or LiCl
4. Li2Se or MgO
Which compound in each of the following pairs has the larger lattice energy? Note: Ba2+ and
K+ have similar radii; S2– and Cl have similar radii. Explain your choices.
1. K2O or Na2O
2. K2S or BaS
3. (c) KCl or BaS
4. BaS or BaCl2
Na2O; Na+ has a smaller radius than K+; BaS; Ba has a larger charge than K; (c) BaS; Ba and S have larger charges; BaS; S has a larger charge
Which of the following compounds requires the most energy to convert one mole of the solid into separate ions?
1. MgO
2. SrO
3. (c) KF
4. CsF
5. MgF2
Which of the following compounds requires the most energy to convert one mole of the solid into separate ions?
1. K2S
2. K2O
3. (c) CaS
4. Cs2S
5. CaO
(e)
The lattice energy of KF is 794 kJ/mol, and the interionic distance is 269 pm. The Na–F
distance in NaF, which has the same structure as KF, is 231 pm. Which of the following values is the closest approximation of the lattice energy of NaF: 682 kJ/mol, 794 kJ/mol, 924 kJ/mol, 1588 kJ/mol, or 3175 kJ/mol? Explain your answer.
7.7: Molecular Structure and Polarity
Explain why the HOH molecule is bent, whereas the HBeH molecule is linear.
The placement of the two sets of unpaired electrons in water forces the bonds to assume a tetrahedral arrangement, and the resulting HOH molecule is bent. The HBeH molecule (in which Be has only two electrons to bond with the two electrons from the hydrogens) must have the electron pairs as far from one another as possible and is therefore linear.
What feature of a Lewis structure can be used to tell if a molecule’s (or ion’s) electron-pair geometry and molecular structure will be identical?
Explain the difference between electron-pair geometry and molecular structure.
Space must be provided for each pair of electrons whether they are in a bond or are present as lone pairs. Electron-pair geometry considers the placement of all electrons. Molecular structure considers only the bonding-pair geometry.
Why is the H–N–H angle in NH3 smaller than the H–C–H bond angle in CH4? Why is the H–N–H angle in $\ce{NH4+}$ identical to the H–C–H bond angle in CH4?
Explain how a molecule that contains polar bonds can be nonpolar.
As long as the polar bonds are compensated (for example. two identical atoms are found directly across the central atom from one another), the molecule can be nonpolar.
As a general rule, MXn molecules (where M represents a central atom and X represents terminal atoms; n = 2 – 5) are polar if there is one or more lone pairs of electrons on M. NH3 (M = N, X = H, n = 3) is an example. There are two molecular structures with lone pairs that are exceptions to this rule. What are they?
Predict the electron pair geometry and the molecular structure of each of the following molecules or ions:
1. SF6
2. PCl5
3. (c) BeH2
4. $\ce{CH3+}$
1. Both the electron geometry and the molecular structure are octahedral.
2. Both the electron geometry and the molecular structure are trigonal bipyramid.
3. (c) Both the electron geometry and the molecular structure are linear.
4. Both the electron geometry and the molecular structure are trigonal planar.
Identify the electron pair geometry and the molecular structure of each of the following molecules or ions:
1. $\ce{IF6+}$
2. CF4
3. (c) BF3
4. $\ce{SiF5-}$
5. BeCl2
What are the electron-pair geometry and the molecular structure of each of the following molecules or ions?
1. ClF5
2. $\ce{ClO2-}$
3. (c) $\ce{TeCl4^2-}$
4. PCl3
5. SeF4
6. $\ce{PH2-}$
electron-pair geometry: octahedral, molecular structure: square pyramidal; electron-pair geometry: tetrahedral, molecular structure: bent; (c) electron-pair geometry: octahedral, molecular structure: square planar; electron-pair geometry: tetrahedral, molecular structure: trigonal pyramidal; electron-pair geometry: trigonal bypyramidal, molecular structure: seesaw; electron-pair geometry: tetrahedral, molecular structure: bent (109°)
Predict the electron pair geometry and the molecular structure of each of the following ions:
1. H3O+
2. $\ce{PCl4-}$
3. (c) $\ce{SnCl3-}$
4. $\ce{BrCl4-}$
5. ICl3
6. XeF4
7. (g) SF2
Identify the electron pair geometry and the molecular structure of each of the following molecules:
1. ClNO (N is the central atom)
2. CS2
3. (c) Cl2CO (C is the central atom)
4. Cl2SO (S is the central atom)
5. SO2F2 (S is the central atom)
6. XeO2F2 (Xe is the central atom)
7. (g) $\ce{ClOF2+}$ (Cl is the central atom)
electron-pair geometry: trigonal planar, molecular structure: bent (120°); electron-pair geometry: linear, molecular structure: linear; (c) electron-pair geometry: trigonal planar, molecular structure: trigonal planar; electron-pair geometry: tetrahedral, molecular structure: trigonal pyramidal; electron-pair geometry: tetrahedral, molecular structure: tetrahedral; electron-pair geometry: trigonal bipyramidal, molecular structure: seesaw; (g) electron-pair geometry: tetrahedral, molecular structure: trigonal pyramidal
Predict the electron pair geometry and the molecular structure of each of the following:
1. IOF5 (I is the central atom)
2. POCl3 (P is the central atom)
3. (c) Cl2SeO (Se is the central atom)
4. ClSO+ (S is the central atom)
5. F2SO (S is the central atom)
6. $\ce{NO2-}$
7. (g) $\ce{SiO4^4-}$
Which of the following molecules and ions contain polar bonds? Which of these molecules and ions have dipole moments?
1. ClF5
2. $\ce{ClO2-}$
3. (c) $\ce{TeCl4^2-}$
4. PCl3
5. SeF4
6. $\ce{PH2-}$
7. (g) XeF2
All of these molecules and ions contain polar bonds. Only ClF5, $\ce{ClO2-}$, PCl3, SeF4, and $\ce{PH2-}$ have dipole moments.
Which of the molecules and ions in Exercise contain polar bonds? Which of these molecules and ions have dipole moments?
1. H3O+
2. $\ce{PCl4-}$
3. (c) $\ce{SnCl3-}$
4. $\ce{BrCl4-}$
5. ICl3
6. XeF4
7. (g) SF2
Which of the following molecules have dipole moments?
1. CS2
2. SeS2
3. (c) CCl2F2
4. PCl3 (P is the central atom)
5. ClNO (N is the central atom)
SeS2, CCl2F2, PCl3, and ClNO all have dipole moments.
Identify the molecules with a dipole moment:
1. SF4
2. CF4
3. (c) Cl2CCBr2
4. CH3Cl
5. H2CO
The molecule XF3 has a dipole moment. Is X boron or phosphorus?
P
The molecule XCl2 has a dipole moment. Is X beryllium or sulfur?
Is the Cl2BBCl2 molecule polar or nonpolar?
nonpolar
There are three possible structures for PCl2F3 with phosphorus as the central atom. Draw them and discuss how measurements of dipole moments could help distinguish among them.
Describe the molecular structure around the indicated atom or atoms:
1. the sulfur atom in sulfuric acid, H2SO4 [(HO)2SO2]
2. the chlorine atom in chloric acid, HClO3 [HOClO2]
3. (c) the oxygen atom in hydrogen peroxide, HOOH
4. the nitrogen atom in nitric acid, HNO3 [HONO2]
5. the oxygen atom in the OH group in nitric acid, HNO3 [HONO2]
6. the central oxygen atom in the ozone molecule, O3
7. (g) each of the carbon atoms in propyne, CH3CCH
8. (h) the carbon atom in Freon, CCl2F2
9. (i) each of the carbon atoms in allene, H2CCCH2
tetrahedral; trigonal pyramidal; (c) bent (109°); trigonal planar; bent (109°); bent (109°); (g) CH3CCH tetrahedral, CH3CCH linear; (h) tetrahedral; (i) H2CCCH2 linear; H2CCCH2 trigonal planar
Draw the Lewis structures and predict the shape of each compound or ion:
1. CO2
2. $\ce{NO2-}$
3. (c) SO3
4. $\ce{SO3^2-}$
A molecule with the formula AB2, in which A and B represent different atoms, could have one of three different shapes. Sketch and name the three different shapes that this molecule might have. Give an example of a molecule or ion for each shape.
A molecule with the formula AB3, in which A and B represent different atoms, could have one of three different shapes. Sketch and name the three different shapes that this molecule might have. Give an example of a molecule or ion that has each shape.
Draw the Lewis electron dot structures for these molecules, including resonance structures where appropriate:
1. $\ce{CS3^2-}$
2. CS2
3. (c) CS
predict the molecular shapes for $\ce{CS3^2-}$ and CS2 and explain how you arrived at your predictions
(a)
;
(b)
;
(c)
;
$\ce{CS3^2-}$ includes three regions of electron density (all are bonds with no lone pairs); the shape is trigonal planar; CS2 has only two regions of electron density (all bonds with no lone pairs); the shape is linear
What is the molecular structure of the stable form of FNO2? (N is the central atom.)
A compound with a molar mass of about 42 g/mol contains 85.7% carbon and 14.3% hydrogen. What is its molecular structure?
The Lewis structure is made from three units, but the atoms must be rearranged:
Use the simulation to perform the following exercises for a two-atom molecule:
1. Adjust the electronegativity value so the bond dipole is pointing toward B. Then determine what the electronegativity values must be to switch the dipole so that it points toward A.
2. With a partial positive charge on A, turn on the electric field and describe what happens.
3. (c) With a small partial negative charge on A, turn on the electric field and describe what happens.
4. Reset all, and then with a large partial negative charge on A, turn on the electric field and describe what happens.
Use the simulation to perform the following exercises for a real molecule. You may need to rotate the molecules in three dimensions to see certain dipoles.
1. Sketch the bond dipoles and molecular dipole (if any) for O3. Explain your observations.
2. Look at the bond dipoles for NH3. Use these dipoles to predict whether N or H is more electronegative.
3. (c) Predict whether there should be a molecular dipole for NH3 and, if so, in which direction it will point. Check the molecular dipole box to test your hypothesis.
The molecular dipole points away from the hydrogen atoms.
Use the Molecule Shape simulator to build a molecule. Starting with the central atom, click on the double bond to add one double bond. Then add one single bond and one lone pair. Rotate the molecule to observe the complete geometry. Name the electron group geometry and molecular structure and predict the bond angle. Then click the check boxes at the bottom and right of the simulator to check your answers.
Use the Molecule Shape simulator to explore real molecules. On the Real Molecules tab, select H2O. Switch between the “real” and “model” modes. Explain the difference observed.
The structures are very similar. In the model mode, each electron group occupies the same amount of space, so the bond angle is shown as 109.5°. In the “real” mode, the lone pairs are larger, causing the hydrogens to be compressed. This leads to the smaller angle of 104.5°.
Use the Molecule Shape simulator to explore real molecules. On the Real Molecules tab, select “model” mode and S2O. What is the model bond angle? Explain whether the “real” bond angle should be larger or smaller than the ideal model angle. | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/04%3A_Chemical_Bonding_and_Molecular_Geometry/4.E%3A_Chemical_Bonding_and_Molecular_Geometry_%28Exercises%29.txt |
We have examined the basic ideas of bonding, showing that atoms share electrons to form molecules with stable Lewis structures and that we can predict the shapes of those molecules by valence shell electron pair repulsion (VSEPR) theory. These ideas provide an important starting point for understanding chemical bonding. But these models sometimes fall short in their abilities to predict the behavior of real substances. How can we reconcile the geometries of s, p, and d atomic orbitals with molecular shapes that show angles like 120° and 109.5°? Furthermore, we know that electrons and magnetic behavior are related through electromagnetic fields.
05: Advanced Theories of Covalent Bonding
We have examined the basic ideas of bonding, showing that atoms share electrons to form molecules with stable Lewis structures and that we can predict the shapes of those molecules by valence shell electron pair repulsion (VSEPR) theory. These ideas provide an important starting point for understanding chemical bonding. But these models sometimes fall short in their abilities to predict the behavior of real substances. How can we reconcile the geometries of s, p, and d atomic orbitals with molecular shapes that show angles like 120° and 109.5°? Furthermore, we know that electrons and magnetic behavior are related through electromagnetic fields. Both N2 and O2 have fairly similar Lewis structures that contain lone pairs of electrons.
Yet oxygen demonstrates very different magnetic behavior than nitrogen. We can pour liquid nitrogen through a magnetic field with no visible interactions, while liquid oxygen is attracted to the magnet and floats in the magnetic field. We need to understand the additional concepts of valence bond theory, orbital hybridization, and molecular orbital theory to understand these observations.
5.1: Valence Bond Theory
Learning Objectives
• Describe the formation of covalent bonds in terms of atomic orbital overlap
• Define and give examples of σ and π bonds
As we know, a scientific theory is a strongly supported explanation for observed natural laws or large bodies of experimental data. For a theory to be accepted, it must explain experimental data and be able to predict behavior. For example, VSEPR theory has gained widespread acceptance because it predicts three-dimensional molecular shapes that are consistent with experimental data collected for thousands of different molecules. However, VSEPR theory does not provide an explanation of chemical bonding.
There are successful theories that describe the electronic structure of atoms. We can use quantum mechanics to predict the specific regions around an atom where electrons are likely to be located: A spherical shape for an s orbital, a dumbbell shape for a p orbital, and so forth. However, these predictions only describe the orbitals around free atoms. When atoms bond to form molecules, atomic orbitals are not sufficient to describe the regions where electrons will be located in the molecule. A more complete understanding of electron distributions requires a model that can account for the electronic structure of molecules. One popular theory holds that a covalent bond forms when a pair of electrons is shared by two atoms and is simultaneously attracted by the nuclei of both atoms. In the following sections, we will discuss how such bonds are described by valence bond theory and hybridization.
Valence bond theory describes a covalent bond as the overlap of half-filled atomic orbitals (each containing a single electron) that yield a pair of electrons shared between the two bonded atoms. We say that orbitals on two different atoms overlap when a portion of one orbital and a portion of a second orbital occupy the same region of space. According to valence bond theory, a covalent bond results when two conditions are met:
1. an orbital on one atom overlaps an orbital on a second atom and
2. the single electrons in each orbital combine to form an electron pair.
The mutual attraction between this negatively charged electron pair and the two atoms’ positively charged nuclei serves to physically link the two atoms through a force we define as a covalent bond. The strength of a covalent bond depends on the extent of overlap of the orbitals involved. Orbitals that overlap extensively form bonds that are stronger than those that have less overlap.
The energy of the system depends on how much the orbitals overlap. Figure $1$ illustrates how the sum of the energies of two hydrogen atoms (the colored curve) changes as they approach each other. When the atoms are far apart there is no overlap, and by convention we set the sum of the energies at zero. As the atoms move together, their orbitals begin to overlap. Each electron begins to feel the attraction of the nucleus in the other atom. In addition, the electrons begin to repel each other, as do the nuclei. While the atoms are still widely separated, the attractions are slightly stronger than the repulsions, and the energy of the system decreases. (A bond begins to form.) As the atoms move closer together, the overlap increases, so the attraction of the nuclei for the electrons continues to increase (as do the repulsions among electrons and between the nuclei). At some specific distance between the atoms, which varies depending on the atoms involved, the energy reaches its lowest (most stable) value. This optimum distance between the two bonded nuclei is the bond distance between the two atoms. The bond is stable because at this point, the attractive and repulsive forces combine to create the lowest possible energy configuration. If the distance between the nuclei were to decrease further, the repulsions between nuclei and the repulsions as electrons are confined in closer proximity to each other would become stronger than the attractive forces. The energy of the system would then rise (making the system destabilized), as shown at the far left of Figure $1$.
The bond energy is the difference between the energy minimum (which occurs at the bond distance) and the energy of the two separated atoms. This is the quantity of energy released when the bond is formed. Conversely, the same amount of energy is required to break the bond. For the $H_2$ molecule shown in Figure $1$, at the bond distance of 74 pm the system is $7.24 \times 10^{−19}\, J$ lower in energy than the two separated hydrogen atoms. This may seem like a small number. However, we know from our earlier description of thermochemistry that bond energies are often discussed on a per-mole basis. For example, it requires $7.24 \times 10^{−19}\; J$ to break one H–H bond, but it takes $4.36 \times 10^5\; J$ to break 1 mole of H–H bonds. A comparison of some bond lengths and energies is shown in Table $1$. We can find many of these bonds in a variety of molecules, and this table provides average values. For example, breaking the first C–H bond in CH4 requires 439.3 kJ/mol, while breaking the first C–H bond in $\ce{H–CH2C6H5}$ (a common paint thinner) requires 375.5 kJ/mol.
Table $1$: Representative Bond Energies and Lengths
Bond Length (pm) Energy (kJ/mol) Bond Length (pm) Energy (kJ/mol)
H–H 74 436 C–O 140.1 358
H–C 106.8 413 $\mathrm{C=O}$ 119.7 745
H–N 101.5 391 $\mathrm{C≡O}$ 113.7 1072
H–O 97.5 467 H–Cl 127.5 431
C–C 150.6 347 H–Br 141.4 366
$\mathrm{C=C}$ 133.5 614 H–I 160.9 298
$\mathrm{C≡C}$ 120.8 839 O–O 148 146
C–N 142.1 305 $\mathrm{O=O}$ 120.8 498
$\mathrm{C=N}$ 130.0 615 F–F 141.2 159
$\mathrm{C≡N}$ 116.1 891 Cl–Cl 198.8 243
In addition to the distance between two orbitals, the orientation of orbitals also affects their overlap (other than for two s orbitals, which are spherically symmetric). Greater overlap is possible when orbitals are oriented such that they overlap on a direct line between the two nuclei. Figure $2$ illustrates this for two p orbitals from different atoms; the overlap is greater when the orbitals overlap end to end rather than at an angle.
The overlap of two s orbitals (as in H2), the overlap of an s orbital and a p orbital (as in HCl), and the end-to-end overlap of two p orbitals (as in Cl2) all produce sigma bonds (σ bonds), as illustrated in Figure $3$. A σ bond is a covalent bond in which the electron density is concentrated in the region along the internuclear axis; that is, a line between the nuclei would pass through the center of the overlap region. Single bonds in Lewis structures are described as σ bonds in valence bond theory.
A pi bond (π bond) is a type of covalent bond that results from the side-by-side overlap of two p orbitals, as illustrated in Figure $4$. In a π bond, the regions of orbital overlap lie on opposite sides of the internuclear axis. Along the axis itself, there is a node, that is, a plane with no probability of finding an electron.
While all single bonds are σ bonds, multiple bonds consist of both σ and π bonds. As the Lewis structures suggest, O2 contains a double bond, and N2 contains a triple bond. The double bond consists of one σ bond and one π bond, and the triple bond consists of one σ bond and two π bonds. Between any two atoms, the first bond formed will always be a σ bond, but there can only be one σ bond in any one location. In any multiple bond, there will be one σ bond, and the remaining one or two bonds will be π bonds. These bonds are described in more detail later in this chapter.
As seen in Table $1$, an average carbon-carbon single bond is 347 kJ/mol, while in a carbon-carbon double bond, the π bond increases the bond strength by 267 kJ/mol. Adding an additional π bond causes a further increase of 225 kJ/mol. We can see a similar pattern when we compare other σ and π bonds. Thus, each individual π bond is generally weaker than a corresponding σ bond between the same two atoms. In a σ bond, there is a greater degree of orbital overlap than in a π bond.
Example $1$: Counting σ and π Bonds
Butadiene, C4H6, is used to make synthetic rubber. Identify the number of σ and π bonds contained in this molecule.
Solution
There are six σ C–H bonds and one σ C–C bond, for a total of seven from the single bonds. There are two double bonds that each have a π bond in addition to the σ bond. This gives a total nine σ and two π bonds overall.
Exercise $1$
Identify each illustration as depicting a σ or π bond:
1. side-by-side overlap of a 4p and a 2p orbital
2. end-to-end overlap of a 4p and 4p orbital
3. end-to-end overlap of a 4p and a 2p orbital
Answer
(a) is a π bond with a node along the axis connecting the nuclei while (b) and (c) are σ bonds that overlap along the axis.
Summary
Valence bond theory describes bonding as a consequence of the overlap of two separate atomic orbitals on different atoms that creates a region with one pair of electrons shared between the two atoms. When the orbitals overlap along an axis containing the nuclei, they form a σ bond. When they overlap in a fashion that creates a node along this axis, they form a π bond.
Glossary
overlap
coexistence of orbitals from two different atoms sharing the same region of space, leading to the formation of a covalent bond
node
plane separating different lobes of orbitals, where the probability of finding an electron is zero
pi bond (π bond)
covalent bond formed by side-by-side overlap of atomic orbitals; the electron density is found on opposite sides of the internuclear axis
sigma bond (σ bond)
covalent bond formed by overlap of atomic orbitals along the internuclear axis
valence bond theory
description of bonding that involves atomic orbitals overlapping to form σ or π bonds, within which pairs of electrons are shared | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/05%3A_Advanced_Theories_of_Covalent_Bonding/5.0%3A_Prelude_to_Covalent_Bonding.txt |
Learning Objectives
• Explain the concept of atomic orbital hybridization
• Determine the hybrid orbitals associated with various molecular geometries
Thinking in terms of overlapping atomic orbitals is one way for us to explain how chemical bonds form in diatomic molecules. However, to understand how molecules with more than two atoms form stable bonds, we require a more detailed model. As an example, let us consider the water molecule, in which we have one oxygen atom bonding to two hydrogen atoms. Oxygen has the electron configuration 1s22s22p4, with two unpaired electrons (one in each of the two 2p orbitals). Valence bond theory would predict that the two O–H bonds form from the overlap of these two 2p orbitals with the 1s orbitals of the hydrogen atoms. If this were the case, the bond angle would be 90°, as shown in Figure \(1\), because p orbitals are perpendicular to each other. Experimental evidence shows that the bond angle is 104.5°, not 90°. The prediction of the valence bond theory model does not match the real-world observations of a water molecule; a different model is needed.
Quantum-mechanical calculations suggest why the observed bond angles in H2O differ from those predicted by the overlap of the 1s orbital of the hydrogen atoms with the 2p orbitals of the oxygen atom. The mathematical expression known as the wave function, ψ, contains information about each orbital and the wavelike properties of electrons in an isolated atom. When atoms are bound together in a molecule, the wave functions combine to produce new mathematical descriptions that have different shapes. This process of combining the wave functions for atomic orbitals is called hybridization and is mathematically accomplished by the linear combination of atomic orbitals, LCAO, (a technique that we will encounter again later). The new orbitals that result are called hybrid orbitals. The valence orbitals in an isolated oxygen atom are a 2s orbital and three 2p orbitals. The valence orbitals in an oxygen atom in a water molecule differ; they consist of four equivalent hybrid orbitals that point approximately toward the corners of a tetrahedron (Figure \(2\)). Consequently, the overlap of the O and H orbitals should result in a tetrahedral bond angle (109.5°). The observed angle of 104.5° is experimental evidence for which quantum-mechanical calculations give a useful explanation: Valence bond theory must include a hybridization component to give accurate predictions.
The following ideas are important in understanding hybridization:
1. Hybrid orbitals do not exist in isolated atoms. They are formed only in covalently bonded atoms.
2. Hybrid orbitals have shapes and orientations that are very different from those of the atomic orbitals in isolated atoms.
3. A set of hybrid orbitals is generated by combining atomic orbitals. The number of hybrid orbitals in a set is equal to the number of atomic orbitals that were combined to produce the set.
4. All orbitals in a set of hybrid orbitals are equivalent in shape and energy.
5. The type of hybrid orbitals formed in a bonded atom depends on its electron-pair geometry as predicted by the VSEPR theory.
6. Hybrid orbitals overlap to form σ bonds. Unhybridized orbitals overlap to form π bonds.
In the following sections, we shall discuss the common types of hybrid orbitals.
sp Hybridization
The beryllium atom in a gaseous BeCl2 molecule is an example of a central atom with no lone pairs of electrons in a linear arrangement of three atoms. There are two regions of valence electron density in the BeCl2 molecule that correspond to the two covalent Be–Cl bonds. To accommodate these two electron domains, two of the Be atom’s four valence orbitals will mix to yield two hybrid orbitals. This hybridization process involves mixing of the valence s orbital with one of the valence p orbitals to yield two equivalent sp hybrid orbitals that are oriented in a linear geometry (Figure \(3\)). In this figure, the set of sp orbitals appears similar in shape to the original p orbital, but there is an important difference. The number of atomic orbitals combined always equals the number of hybrid orbitals formed. The p orbital is one orbital that can hold up to two electrons. The sp set is two equivalent orbitals that point 180° from each other. The two electrons that were originally in the s orbital are now distributed to the two sp orbitals, which are half filled. In gaseous BeCl2, these half-filled hybrid orbitals will overlap with orbitals from the chlorine atoms to form two identical σ bonds.
We illustrate the electronic differences in an isolated Be atom and in the bonded Be atom in the orbital energy-level diagram in Figure \(4\). These diagrams represent each orbital by a horizontal line (indicating its energy) and each electron by an arrow. Energy increases toward the top of the diagram. We use one upward arrow to indicate one electron in an orbital and two arrows (up and down) to indicate two electrons of opposite spin.
When atomic orbitals hybridize, the valence electrons occupy the newly created orbitals. The Be atom had two valence electrons, so each of the sp orbitals gets one of these electrons. Each of these electrons pairs up with the unpaired electron on a chlorine atom when a hybrid orbital and a chlorine orbital overlap during the formation of the Be–Cl bonds. Any central atom surrounded by just two regions of valence electron density in a molecule will exhibit sp hybridization. Other examples include the mercury atom in the linear HgCl2 molecule, the zinc atom in Zn(CH3)2, which contains a linear C–Zn–C arrangement, and the carbon atoms in HCCH and CO2.
sp2 Hybridization
The valence orbitals of a central atom surrounded by three regions of electron density consist of a set of three sp2 hybrid orbitals and one unhybridized p orbital. This arrangement results from sp2 hybridization, the mixing of one s orbital and two p orbitals to produce three identical hybrid orbitals oriented in a trigonal planar geometry (Figure \(5\)).
Although quantum mechanics yields the “plump” orbital lobes as depicted in Figure \(5\), sometimes for clarity these orbitals are drawn thinner and without the minor lobes, as in Figure \(6\), to avoid obscuring other features of a given illustration. We will use these “thinner” representations whenever the true view is too crowded to easily visualize.
The observed structure of the borane molecule, BH3, suggests sp2 hybridization for boron in this compound. The molecule is trigonal planar, and the boron atom is involved in three bonds to hydrogen atoms ( Figure \(7\)).
We can illustrate the comparison of orbitals and electron distribution in an isolated boron atom and in the bonded atom in BH3 as shown in the orbital energy level diagram in Figure \(8\). We redistribute the three valence electrons of the boron atom in the three sp2 hybrid orbitals, and each boron electron pairs with a hydrogen electron when B–H bonds form.
Any central atom surrounded by three regions of electron density will exhibit sp2 hybridization. This includes molecules with a lone pair on the central atom, such as ClNO (Figure \(9\)), or molecules with two single bonds and a double bond connected to the central atom, as in formaldehyde, CH2O, and ethene, H2CCH2.
sp3 Hybridization
The valence orbitals of an atom surrounded by a tetrahedral arrangement of bonding pairs and lone pairs consist of a set of four sp3 hybrid orbitals. The hybrids result from the mixing of one s orbital and all three p orbitals that produces four identical sp3 hybrid orbitals (Figure \(10\)). Each of these hybrid orbitals points toward a different corner of a tetrahedron.
A molecule of methane, CH4, consists of a carbon atom surrounded by four hydrogen atoms at the corners of a tetrahedron. The carbon atom in methane exhibits sp3 hybridization. We illustrate the orbitals and electron distribution in an isolated carbon atom and in the bonded atom in CH4 in Figure \(11\). The four valence electrons of the carbon atom are distributed equally in the hybrid orbitals, and each carbon electron pairs with a hydrogen electron when the C–H bonds form.
In a methane molecule, the 1s orbital of each of the four hydrogen atoms overlaps with one of the four sp3 orbitals of the carbon atom to form a sigma (σ) bond. This results in the formation of four strong, equivalent covalent bonds between the carbon atom and each of the hydrogen atoms to produce the methane molecule, CH4.
The structure of ethane, C2H6, is similar to that of methane in that each carbon in ethane has four neighboring atoms arranged at the corners of a tetrahedron—three hydrogen atoms and one carbon atom (Figure \(10\)). However, in ethane an sp3 orbital of one carbon atom overlaps end to end with an sp3 orbital of a second carbon atom to form a σ bond between the two carbon atoms. Each of the remaining sp3 hybrid orbitals overlaps with an s orbital of a hydrogen atom to form carbon–hydrogen σ bonds. The structure and overall outline of the bonding orbitals of ethane are shown in Figure \(12\). The orientation of the two CH3 groups is not fixed relative to each other. Experimental evidence shows that rotation around σ bonds occurs easily.
An sp3 hybrid orbital can also hold a lone pair of electrons. For example, the nitrogen atom in ammonia is surrounded by three bonding pairs and a lone pair of electrons directed to the four corners of a tetrahedron. The nitrogen atom is sp3 hybridized with one hybrid orbital occupied by the lone pair.
The molecular structure of water is consistent with a tetrahedral arrangement of two lone pairs and two bonding pairs of electrons. Thus we say that the oxygen atom is sp3 hybridized, with two of the hybrid orbitals occupied by lone pairs and two by bonding pairs. Since lone pairs occupy more space than bonding pairs, structures that contain lone pairs have bond angles slightly distorted from the ideal. Perfect tetrahedra have angles of 109.5°, but the observed angles in ammonia (107.3°) and water (104.5°) are slightly smaller. Other examples of sp3 hybridization include CCl4, PCl3, and NCl3.
sp3d and sp3d2 Hybridization
To describe the five bonding orbitals in a trigonal bipyramidal arrangement, we must use five of the valence shell atomic orbitals (the s orbital, the three p orbitals, and one of the d orbitals), which gives five sp3d hybrid orbitals. With an octahedral arrangement of six hybrid orbitals, we must use six valence shell atomic orbitals (the s orbital, the three p orbitals, and two of the d orbitals in its valence shell), which gives six sp3d2 hybrid orbitals. These hybridizations are only possible for atoms that have d orbitals in their valence subshells (that is, not those in the first or second period).
In a molecule of phosphorus pentachloride, PCl5, there are five P–Cl bonds (thus five pairs of valence electrons around the phosphorus atom) directed toward the corners of a trigonal bipyramid. We use the 3s orbital, the three 3p orbitals, and one of the 3d orbitals to form the set of five sp3d hybrid orbitals (Figure \(13\)) that are involved in the P–Cl bonds. Other atoms that exhibit sp3d hybridization include the sulfur atom in SF4 and the chlorine atoms in ClF3 and in \(\ce{ClF4+}\). (The electrons on fluorine atoms are omitted for clarity.)
The sulfur atom in sulfur hexafluoride, SF6, exhibits sp3d2 hybridization. A molecule of sulfur hexafluoride has six bonding pairs of electrons connecting six fluorine atoms to a single sulfur atom. There are no lone pairs of electrons on the central atom. To bond six fluorine atoms, the 3s orbital, the three 3p orbitals, and two of the 3d orbitals form six equivalent sp3d2 hybrid orbitals, each directed toward a different corner of an octahedron. Other atoms that exhibit sp3d2 hybridization include the phosphorus atom in \(\ce{PCl6-}\), the iodine atom in the interhalogens \(\ce{IF6+}\), IF5, \(\ce{ICl4-}\), \(\ce{IF4-}\) and the xenon atom in XeF4.
Assignment of Hybrid Orbitals to Central Atoms
The hybridization of an atom is determined based on the number of regions of electron density that surround it. The geometrical arrangements characteristic of the various sets of hybrid orbitals are shown in Figure \(16\). These arrangements are identical to those of the electron-pair geometries predicted by VSEPR theory. VSEPR theory predicts the shapes of molecules, and hybrid orbital theory provides an explanation for how those shapes are formed. To find the hybridization of a central atom, we can use the following guidelines:
1. Determine the Lewis structure of the molecule.
2. Determine the number of regions of electron density around an atom using VSEPR theory, in which single bonds, multiple bonds, radicals, and lone pairs each count as one region.
3. Assign the set of hybridized orbitals from Figure \(16\) that corresponds to this geometry.
It is important to remember that hybridization was devised to rationalize experimentally observed molecular geometries, not the other way around.
The model works well for molecules containing small central atoms, in which the valence electron pairs are close together in space. However, for larger central atoms, the valence-shell electron pairs are farther from the nucleus, and there are fewer repulsions. Their compounds exhibit structures that are often not consistent with VSEPR theory, and hybridized orbitals are not necessary to explain the observed data.
For example, we have discussed the H–O–H bond angle in H2O, 104.5°, which is more consistent with sp3 hybrid orbitals (109.5°) on the central atom than with 2p orbitals (90°). Sulfur is in the same group as oxygen, and H2S has a similar Lewis structure. However, it has a much smaller bond angle (92.1°), which indicates much less hybridization on sulfur than oxygen. Continuing down the group, tellurium is even larger than sulfur, and for H2Te, the observed bond angle (90°) is consistent with overlap of the 5p orbitals, without invoking hybridization. We invoke hybridization where it is necessary to explain the observed structures.
Example \(1\): Assigning Hybridization
Ammonium sulfate is important as a fertilizer. What is the hybridization of the sulfur atom in the sulfate ion, \(\ce{SO4^2-}\)?
Solution
The Lewis structure of sulfate shows there are four regions of electron density.
Exercise \(1\)
What is the hybridization of the selenium atom in SeF4?
Answer
The selenium atom is sp3d hybridized.
Example \(2\): Assigning Hybridization
Urea, NH2C(O)NH2, is sometimes used as a source of nitrogen in fertilizers. What is the hybridization of each nitrogen and carbon atom in urea?
Solution
The Lewis structure of urea is
The carbon atom is surrounded by three regions of electron density, positioned in a trigonal planar arrangement. The hybridization in a trigonal planar electron pair geometry is sp2 (Figure \(16\)), which is the hybridization of the carbon atom in urea.
Exercise \(1\)
Acetic acid, H3CC(O)OH, is the molecule that gives vinegar its odor and sour taste. What is the hybridization of the two carbon atoms in acetic acid?
Answer
H3C, sp3; C(O)OH, sp2
Summary
We can use hybrid orbitals, which are mathematical combinations of some or all of the valence atomic orbitals, to describe the electron density around covalently bonded atoms. These hybrid orbitals either form sigma (σ) bonds directed toward other atoms of the molecule or contain lone pairs of electrons. We can determine the type of hybridization around a central atom from the geometry of the regions of electron density about it. Two such regions imply sp hybridization; three, sp2 hybridization; four, sp3 hybridization; five, sp3d hybridization; and six, sp3d2 hybridization. Pi (π) bonds are formed from unhybridized atomic orbitals (p or d orbitals).
Footnotes
1. Note that orbitals may sometimes be drawn in an elongated “balloon” shape rather than in a more realistic “plump” shape in order to make the geometry easier to visualize.
Glossary
hybrid orbital
orbital created by combining atomic orbitals on a central atom
hybridization
model that describes the changes in the atomic orbitals of an atom when it forms a covalent compound
sp hybrid orbital
one of a set of two orbitals with a linear arrangement that results from combining one s and one p orbital
sp2 hybrid orbital
one of a set of three orbitals with a trigonal planar arrangement that results from combining one s and two p orbitals
sp3 hybrid orbital
one of a set of four orbitals with a tetrahedral arrangement that results from combining one s and three p orbitals
sp3d hybrid orbital
one of a set of five orbitals with a trigonal bipyramidal arrangement that results from combining one s, three p, and one d orbital
sp3d2 hybrid orbital
one of a set of six orbitals with an octahedral arrangement that results from combining one s, three p, and two d orbitals | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/05%3A_Advanced_Theories_of_Covalent_Bonding/5.2%3A_Hybrid_Atomic_Orbitals.txt |
Learning Objectives
• Describe multiple covalent bonding in terms of atomic orbital overlap
• Relate the concept of resonance to π-bonding and electron delocalization
The hybrid orbital model appears to account well for the geometry of molecules involving single covalent bonds. Is it also capable of describing molecules containing double and triple bonds? We have already discussed that multiple bonds consist of σ and π bonds. Next we can consider how we visualize these components and how they relate to hybrid orbitals. The Lewis structure of ethene, C2H4, shows us that each carbon atom is surrounded by one other carbon atom and two hydrogen atoms.
The π bond in the C=C double bond results from the overlap of the third (remaining) 2p orbital on each carbon atom that is not involved in hybridization. This unhybridized p orbital (lobes shown in red and blue in Figure \(2\)) is perpendicular to the plane of the sp2 hybrid orbitals. Thus the unhybridized 2p orbitals overlap in a side-by-side fashion, above and below the internuclear axis and form a π bond.
In an ethene molecule, the four hydrogen atoms and the two carbon atoms are all in the same plane. If the two planes of sp2 hybrid orbitals tilted relative to each other, the p orbitals would not be oriented to overlap efficiently to create the π bond. The planar configuration for the ethene molecule occurs because it is the most stable bonding arrangement. This is a significant difference between σ and π bonds; rotation around single (σ) bonds occurs easily because the end-to-end orbital overlap does not depend on the relative orientation of the orbitals on each atom in the bond. In other words, rotation around the internuclear axis does not change the extent to which the σ bonding orbitals overlap because the bonding electron density is symmetric about the axis. Rotation about the internuclear axis is much more difficult for multiple bonds; however, this would drastically alter the off-axis overlap of the π bonding orbitals, essentially breaking the π bond.
In molecules with sp hybrid orbitals, two unhybridized p orbitals remain on the atom (Figure \(3\)). We find this situation in acetylene, H−C≡C−H, which is a linear molecule. The sp hybrid orbitals of the two carbon atoms overlap end to end to form a σ bond between the carbon atoms (Figure \(4\)). The remaining sp orbitals form σ bonds with hydrogen atoms. The two unhybridized p orbitals per carbon are positioned such that they overlap side by side and, hence, form two π bonds. The two carbon atoms of acetylene are thus bound together by one σ bond and two π bonds, giving a triple bond.
Hybridization involves only σ bonds, lone pairs of electrons, and single unpaired electrons (radicals). Structures that account for these features describe the correct hybridization of the atoms. However, many structures also include resonance forms. Remember that resonance forms occur when various arrangements of π bonds are possible. Since the arrangement of π bonds involves only the unhybridized orbitals, resonance does not influence the assignment of hybridization.
For example, molecule benzene has two resonance forms (Figure \(5\)). We can use either of these forms to determine that each of the carbon atoms is bonded to three other atoms with no lone pairs, so the correct hybridization is sp2. The electrons in the unhybridized p orbitals form π bonds. Neither resonance structure completely describes the electrons in the π bonds. They are not located in one position or the other, but in reality are delocalized throughout the ring. Valence bond theory does not easily address delocalization. Bonding in molecules with resonance forms is better described by molecular orbital theory.
Example \(1\): Assignment of Hybridization Involving Resonance
Some acid rain results from the reaction of sulfur dioxide with atmospheric water vapor, followed by the formation of sulfuric acid. Sulfur dioxide, \(\ce{SO2}\), is a major component of volcanic gases as well as a product of the combustion of sulfur-containing coal. What is the hybridization of the \(S\) atom in \(\ce{SO2}\)?
Solution
The resonance structures of \(\ce{SO2}\) are
The sulfur atom is surrounded by two bonds and one lone pair of electrons in either resonance structure. Therefore, the electron-pair geometry is trigonal planar, and the hybridization of the sulfur atom is sp2.
Exercise \(1\)
Another acid in acid rain is nitric acid, HNO3, which is produced by the reaction of nitrogen dioxide, NO2, with atmospheric water vapor. What is the hybridization of the nitrogen atom in NO2? (Note: the lone electron on nitrogen occupies a hybridized orbital just as a lone pair would.)
Answer
sp2
Summary
Multiple bonds consist of a σ bond located along the axis between two atoms and one or two π bonds. The σ bonds are usually formed by the overlap of hybridized atomic orbitals, while the π bonds are formed by the side-by-side overlap of unhybridized orbitals. Resonance occurs when there are multiple unhybridized orbitals with the appropriate alignment to overlap, so the placement of π bonds can vary. | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/05%3A_Advanced_Theories_of_Covalent_Bonding/5.3%3A_Multiple_Bonds.txt |
Learning Objectives
• Outline the basic quantum-mechanical approach to deriving molecular orbitals from atomic orbitals
• Describe traits of bonding and antibonding molecular orbitals
• Calculate bond orders based on molecular electron configurations
• Write molecular electron configurations for first- and second-row diatomic molecules
• Relate these electron configurations to the molecules’ stabilities and magnetic properties
For almost every covalent molecule that exists, we can now draw the Lewis structure, predict the electron-pair geometry, predict the molecular geometry, and come close to predicting bond angles. However, one of the most important molecules we know, the oxygen molecule O2, presents a problem with respect to its Lewis structure. We would write the following Lewis structure for O2:
This electronic structure adheres to all the rules governing Lewis theory. There is an O=O double bond, and each oxygen atom has eight electrons around it. However, this picture is at odds with the magnetic behavior of oxygen. By itself, O2 is not magnetic, but it is attracted to magnetic fields. Thus, when we pour liquid oxygen past a strong magnet, it collects between the poles of the magnet and defies gravity. Such attraction to a magnetic field is called paramagnetism, and it arises in molecules that have unpaired electrons. And yet, the Lewis structure of O2 indicates that all electrons are paired. How do we account for this discrepancy?
Magnetic susceptibility measures the force experienced by a substance in a magnetic field. When we compare the weight of a sample to the weight measured in a magnetic field (Figure $1$), paramagnetic samples that are attracted to the magnet will appear heavier because of the force exerted by the magnetic field. We can calculate the number of unpaired electrons based on the increase in weight.
Experiments show that each O2 molecule has two unpaired electrons. The Lewis-structure model does not predict the presence of these two unpaired electrons. Unlike oxygen, the apparent weight of most molecules decreases slightly in the presence of an inhomogeneous magnetic field. Materials in which all of the electrons are paired are diamagnetic and weakly repel a magnetic field. Paramagnetic and diamagnetic materials do not act as permanent magnets. Only in the presence of an applied magnetic field do they demonstrate attraction or repulsion.
Molecular orbital theory (MO theory) provides an explanation of chemical bonding that accounts for the paramagnetism of the oxygen molecule. It also explains the bonding in a number of other molecules, such as violations of the octet rule and more molecules with more complicated bonding (beyond the scope of this text) that are difficult to describe with Lewis structures. Additionally, it provides a model for describing the energies of electrons in a molecule and the probable location of these electrons. Unlike valence bond theory, which uses hybrid orbitals that are assigned to one specific atom, MO theory uses the combination of atomic orbitals to yield molecular orbitals that are delocalized over the entire molecule rather than being localized on its constituent atoms. MO theory also helps us understand why some substances are electrical conductors, others are semiconductors, and still others are insulators. Table $1$ summarizes the main points of the two complementary bonding theories. Both theories provide different, useful ways of describing molecular structure.
Table $1$: Comparison of Bonding Theories
Valence Bond Theory Molecular Orbital Theory
considers bonds as localized between one pair of atoms considers electrons delocalized throughout the entire molecule
creates bonds from overlap of atomic orbitals (s, p, d…) and hybrid orbitals (sp, sp2, sp3…) combines atomic orbitals to form molecular orbitals (σ, σ*, π, π*)
forms σ or π bonds creates bonding and antibonding interactions based on which orbitals are filled
predicts molecular shape based on the number of regions of electron density predicts the arrangement of electrons in molecules
needs multiple structures to describe resonance
Molecular orbital theory describes the distribution of electrons in molecules in much the same way that the distribution of electrons in atoms is described using atomic orbitals. Using quantum mechanics, the behavior of an electron in a molecule is still described by a wave function, Ψ, analogous to the behavior in an atom. Just like electrons around isolated atoms, electrons around atoms in molecules are limited to discrete (quantized) energies. The region of space in which a valence electron in a molecule is likely to be found is called a molecular orbital (Ψ2). Like an atomic orbital, a molecular orbital is full when it contains two electrons with opposite spin.
We will consider the molecular orbitals in molecules composed of two identical atoms (H2 or Cl2, for example). Such molecules are called homonuclear diatomic molecules. In these diatomic molecules, several types of molecular orbitals occur.
The mathematical process of combining atomic orbitals to generate molecular orbitals is called the linear combination of atomic orbitals (LCAO). The wave function describes the wavelike properties of an electron. Molecular orbitals are combinations of atomic orbital wave functions. Combining waves can lead to constructive interference, in which peaks line up with peaks, or destructive interference, in which peaks line up with troughs (Figure $2$). In orbitals, the waves are three dimensional, and they combine with in-phase waves producing regions with a higher probability of electron density and out-of-phase waves producing nodes, or regions of no electron density.
There are two types of molecular orbitals that can form from the overlap of two atomic s orbitals on adjacent atoms. The two types are illustrated in Figure $3$. The in-phase combination produces a lower energy σs molecular orbital (read as "sigma-s") in which most of the electron density is directly between the nuclei. The out-of-phase addition (which can also be thought of as subtracting the wave functions) produces a higher energy $σ^∗_s$ molecular orbital (read as "sigma-s-star") molecular orbital in which there is a node between the nuclei. The asterisk signifies that the orbital is an antibonding orbital. Electrons in a σs orbital are attracted by both nuclei at the same time and are more stable (of lower energy) than they would be in the isolated atoms. Adding electrons to these orbitals creates a force that holds the two nuclei together, so we call these orbitals bonding orbitals. Electrons in the $σ^∗_s$ orbitals are located well away from the region between the two nuclei. The attractive force between the nuclei and these electrons pulls the two nuclei apart. Hence, these orbitals are called antibonding orbitals. Electrons fill the lower-energy bonding orbital before the higher-energy antibonding orbital, just as they fill lower-energy atomic orbitals before they fill higher-energy atomic orbitals.
In p orbitals, the wave function gives rise to two lobes with opposite phases, analogous to how a two-dimensional wave has both parts above and below the average. We indicate the phases by shading the orbital lobes different colors. When orbital lobes of the same phase overlap, constructive wave interference increases the electron density. When regions of opposite phase overlap, the destructive wave interference decreases electron density and creates nodes. When p orbitals overlap end to end, they create σ and σ* orbitals (Figure $4$). If two atoms are located along the x-axis in a Cartesian coordinate system, the two px orbitals overlap end to end and form σpx (bonding) and $σ^∗_{px}$ (antibonding) (read as "sigma-p-x" and "sigma-p-x star," respectively). Just as with s-orbital overlap, the asterisk indicates the orbital with a node between the nuclei, which is a higher-energy, antibonding orbital.
The side-by-side overlap of two p orbitals gives rise to a pi ($π$) bonding molecular orbital and a $π^*$ antibonding molecular orbital, as shown in Figure $5$. In valence bond theory, we describe π bonds as containing a nodal plane containing the internuclear axis and perpendicular to the lobes of the p orbitals, with electron density on either side of the node. In molecular orbital theory, we describe the π orbital by this same shape, and a π bond exists when this orbital contains electrons. Electrons in this orbital interact with both nuclei and help hold the two atoms together, making it a bonding orbital. For the out-of-phase combination, there are two nodal planes created, one along the internuclear axis and a perpendicular one between the nuclei.
In the molecular orbitals of diatomic molecules, each atom also has two sets of p orbitals oriented side by side (py and pz), so these four atomic orbitals combine pairwise to create two π orbitals and two $π^*$ orbitals. The $π_{py}$ and $π^∗_{py}$ orbitals are oriented at right angles to the $π_{pz}$ and $π^∗_{pz}$ orbitals. Except for their orientation, the πpy and πpz orbitals are identical and have the same energy; they are degenerate orbitals. The $π^∗_{py}$ and $π^∗_{pz}$ antibonding orbitals are also degenerate and identical except for their orientation. A total of six molecular orbitals results from the combination of the six atomic p orbitals in two atoms: $σ_{px}$ and $σ^∗_{px}$, $π_{py}$ and $π^∗_{py}$, $π_{pz}$ and $π^∗_{pz}$.
Example $1$: Molecular Orbitals
Predict what type (if any) of molecular orbital would result from adding the wave functions so each pair of orbitals shown overlap. The orbitals are all similar in energy.
Solution
1. This is an in-phase combination, resulting in a σ3p orbital
2. This will not result in a new orbital because the in-phase component (bottom) and out-of-phase component (top) cancel out. Only orbitals with the correct alignment can combine.
3. This is an out-of-phase combination, resulting in a $π^∗_{3p}$ orbital.
Exercise $1$
Label the molecular orbital shown as $σ$ or $π$, bonding or antibonding and indicate where the node occurs.
Answer
The orbital is located along the internuclear axis, so it is a $σ$ orbital. There is a node bisecting the internuclear axis, so it is an antibonding orbital.
Application: Computational Chemistry in Drug Design
While the descriptions of bonding described in this chapter involve many theoretical concepts, they also have many practical, real-world applications. For example, drug design is an important field that uses our understanding of chemical bonding to develop pharmaceuticals. This interdisciplinary area of study uses biology (understanding diseases and how they operate) to identify specific targets, such as a binding site that is involved in a disease pathway. By modeling the structures of the binding site and potential drugs, computational chemists can predict which structures can fit together and how effectively they will bind (Figure $6$). Thousands of potential candidates can be narrowed down to a few of the most promising candidates. These candidate molecules are then carefully tested to determine side effects, how effectively they can be transported through the body, and other factors. Dozens of important new pharmaceuticals have been discovered with the aid of computational chemistry, and new research projects are underway.
Molecular Orbital Energy Diagrams
The relative energy levels of atomic and molecular orbitals are typically shown in a molecular orbital diagram (Figure $7$). For a diatomic molecule, the atomic orbitals of one atom are shown on the left, and those of the other atom are shown on the right. Each horizontal line represents one orbital that can hold two electrons. The molecular orbitals formed by the combination of the atomic orbitals are shown in the center. Dashed lines show which of the atomic orbitals combine to form the molecular orbitals. For each pair of atomic orbitals that combine, one lower-energy (bonding) molecular orbital and one higher-energy (antibonding) orbital result. Thus we can see that combining the six 2p atomic orbitals results in three bonding orbitals (one σ and two π) and three antibonding orbitals (one σ* and two π*).
We predict the distribution of electrons in these molecular orbitals by filling the orbitals in the same way that we fill atomic orbitals, by the Aufbau principle. Lower-energy orbitals fill first, electrons spread out among degenerate orbitals before pairing, and each orbital can hold a maximum of two electrons with opposite spins (Figure $7$). Just as we write electron configurations for atoms, we can write the molecular electronic configuration by listing the orbitals with superscripts indicating the number of electrons present. For clarity, we place parentheses around molecular orbitals with the same energy. In this case, each orbital is at a different energy, so parentheses separate each orbital. Thus we would expect a diatomic molecule or ion containing seven electrons (such as $\ce{Be2+}$) would have the molecular electron configuration $(σ_{1s})^2(σ^∗_{1s})^2(σ_{2s})^2(σ^∗_{2s})^1$. It is common to omit the core electrons from molecular orbital diagrams and configurations and include only the valence electrons.
Bond Order
The filled molecular orbital diagram shows the number of electrons in both bonding and antibonding molecular orbitals. The net contribution of the electrons to the bond strength of a molecule is identified by determining the bond order that results from the filling of the molecular orbitals by electrons.
When using Lewis structures to describe the distribution of electrons in molecules, we define bond order as the number of bonding pairs of electrons between two atoms. Thus a single bond has a bond order of 1, a double bond has a bond order of 2, and a triple bond has a bond order of 3. We define bond order differently when we use the molecular orbital description of the distribution of electrons, but the resulting bond order is usually the same. The MO technique is more accurate and can handle cases when the Lewis structure method fails, but both methods describe the same phenomenon.
In the molecular orbital model, an electron contributes to a bonding interaction if it occupies a bonding orbital and it contributes to an antibonding interaction if it occupies an antibonding orbital. The bond order is calculated by subtracting the destabilizing (antibonding) electrons from the stabilizing (bonding) electrons. Since a bond consists of two electrons, we divide by two to get the bond order. We can determine bond order with the following equation:
$\textrm{bond order}=\dfrac{(\textrm{number of bonding electrons})−(\textrm{number of antibonding electrons})}{2} \nonumber$
The order of a covalent bond is a guide to its strength; a bond between two given atoms becomes stronger as the bond order increases. If the distribution of electrons in the molecular orbitals between two atoms is such that the resulting bond would have a bond order of zero, a stable bond does not form. We next look at some specific examples of MO diagrams and bond orders.
Bonding in Diatomic Molecules
A dihydrogen molecule (H2) forms from two hydrogen atoms. When the atomic orbitals of the two atoms combine, the electrons occupy the molecular orbital of lowest energy, the σ1s bonding orbital. A dihydrogen molecule, H2, readily forms because the energy of a H2 molecule is lower than that of two H atoms. The σ1s orbital that contains both electrons is lower in energy than either of the two 1s atomic orbitals.
A molecular orbital can hold two electrons, so both electrons in the H2 molecule are in the σ1s bonding orbital; the electron configuration is $(σ_{1s})^2$. We represent this configuration by a molecular orbital energy diagram (Figure $8$) in which a single upward arrow indicates one electron in an orbital, and two (upward and downward) arrows indicate two electrons of opposite spin.
A dihydrogen molecule contains two bonding electrons and no antibonding electrons so we have
$\ce{bond\: order\: in\: H2}=\dfrac{(2−0)}{2}=1 \nonumber$
Because the bond order for the H–H bond is equal to 1, the bond is a single bond.
A helium atom has two electrons, both of which are in its 1s orbital. Two helium atoms do not combine to form a dihelium molecule, He2, with four electrons, because the stabilizing effect of the two electrons in the lower-energy bonding orbital would be offset by the destabilizing effect of the two electrons in the higher-energy antibonding molecular orbital. We would write the hypothetical electron configuration of He2 as $(σ_{1s})^2(σ^∗_{1s})^2$ as in Figure $9$. The net energy change would be zero, so there is no driving force for helium atoms to form the diatomic molecule. In fact, helium exists as discrete atoms rather than as diatomic molecules. The bond order in a hypothetical dihelium molecule would be zero.
$\ce{bond\: order\: in\: He2}=\dfrac{(2−2)}{2}=0 \nonumber$
A bond order of zero indicates that no bond is formed between two atoms.
The Diatomic Molecules of the Second Period
Eight possible homonuclear diatomic molecules might be formed by the atoms of the second period of the periodic table: Li2, Be2, B2, C2, N2, O2, F2, and Ne2. However, we can predict that the Be2 molecule and the Ne2 molecule would not be stable. We can see this by a consideration of the molecular electron configurations (Table $1$).
Table $1$: Electron Configuration and Bond Order for Molecular Orbitals in Homonuclear Diatomic Molecules of Period Two Elements
Molecule Electron Configuration Bond Order
Li2 $(σ_{2s})^2$ 1
Be2 (unstable) $(σ_{2s})^2(σ^∗_{2s})^2$ 0
B2 $(σ_{2s})^2(σ^∗_{2s})^2(π_{2py},\:π_{2pz})^2$ 1
C2 $(σ_{2s})^2(σ^∗_{2s})^2(π_{2py},\:π_{2pz})^4$ 2
N2 $(σ_{2s})^2(σ^∗_{2s})^2(π_{2py},\:π_{2pz})^4(σ_{2px})^2$ 3
O2 $(σ_{2s})^2(σ^∗_{2s})^2(σ_{2px})^2(π_{2py},\:π_{2pz})^4(π^∗_{2py},π^∗_{2pz})^2$ 2
F2 $(σ_{2s})^2(σ^∗_{2s})^2(σ_{2px})^2(π_{2py},\:π_{2pz})^4(π^∗_{2py},\:π^∗_{2pz})^4$ 1
Ne2 (unstable) $(σ_{2s})^2(σ^∗_{2s})^2(σ_{2px})^2(π_{2py},\:π_{2pz})^4(π^∗_{2py},π^∗_{2pz})^4(σ^∗_{2px})^2$ 0
We predict valence molecular orbital electron configurations just as we predict electron configurations of atoms. Valence electrons are assigned to valence molecular orbitals with the lowest possible energies. Consistent with Hund’s rule, whenever there are two or more degenerate molecular orbitals, electrons fill each orbital of that type singly before any pairing of electrons takes place.
As we saw in valence bond theory, σ bonds are generally more stable than π bonds formed from degenerate atomic orbitals. Similarly, in molecular orbital theory, σ orbitals are usually more stable than π orbitals. However, this is not always the case. The MOs for the valence orbitals of the second period are shown in Figure $10$. Looking at Ne2 molecular orbitals, we see that the order is consistent with the generic diagram shown in the previous section. However, for atoms with three or fewer electrons in the p orbitals (Li through N) we observe a different pattern, in which the σp orbital is higher in energy than the πp set. Obtain the molecular orbital diagram for a homonuclear diatomic ion by adding or subtracting electrons from the diagram for the neutral molecule.
This switch in orbital ordering occurs because of a phenomenon called s-p mixing. s-p mixing does not create new orbitals; it merely influences the energies of the existing molecular orbitals. The σs wavefunction mathematically combines with the σp wavefunction, with the result that the σs orbital becomes more stable, and the σp orbital becomes less stable (Figure $11$). Similarly, the antibonding orbitals also undergo s-p mixing, with the σs* becoming more stable and the σp* becoming less stable.
s-p mixing occurs when the s and p orbitals have similar energies. The energy difference between 2s and 2p orbitals in O, F, and Ne is greater than that in Li, Be, B, C, and N. Because of this, O2, F2, and Ne exhibit negligible s-p mixing (not sufficient to change the energy ordering), and their MO diagrams follow the normal pattern, as shown in Figure $7$. All of the other period 2 diatomic molecules do have s-p mixing, which leads to the pattern where the σp orbital is raised above the πp set.
Using the MO diagrams shown in Figure $11$, we can add in the electrons and determine the molecular electron configuration and bond order for each of the diatomic molecules. As shown in Table $1$, Be2 and Ne2 molecules would have a bond order of 0, and these molecules do not exist.
The combination of two lithium atoms to form a lithium molecule, Li2, is analogous to the formation of H2, but the atomic orbitals involved are the valence 2s orbitals. Each of the two lithium atoms has one valence electron. Hence, we have two valence electrons available for the σ2s bonding molecular orbital. Because both valence electrons would be in a bonding orbital, we would predict the Li2 molecule to be stable. The molecule is, in fact, present in an appreciable concentration in lithium vapor at temperatures near the boiling point of the element. All of the other molecules in Table $1$ with a bond order greater than zero are also known.
The O2 molecule has enough electrons to half fill the $(π^∗_{2py},\:π^∗_{2pz})$ level. We expect the two electrons that occupy these two degenerate orbitals to be unpaired, and this molecular electronic configuration for O2 is in accord with the fact that the oxygen molecule has two unpaired electrons ( Figure $10$). The presence of two unpaired electrons has proved to be difficult to explain using Lewis structures, but the molecular orbital theory explains it quite well. In fact, the unpaired electrons of the oxygen molecule provide a strong piece of support for the molecular orbital theory.
Application: Band Theory in Extended Systems
When two identical atomic orbitals on different atoms combine, two molecular orbitals result (e.g., $H_2$ in Figure $8$). The bonding orbital is lower in energy than the original atomic orbitals because the atomic orbitals are in-phase in the molecular orbital. The antibonding orbital is higher in energy than the original atomic orbitals because the atomic orbitals are out-of-phase.
In a solid, similar things happen, but on a much larger scale. Remember that even in a small sample there are a huge number of atoms (typically > 1023 atoms), and therefore a huge number of atomic orbitals that may be combined into molecular orbitals. When N valence atomic orbitals, all of the same energy and each containing one (1) electron, are combined, N/2 (filled) bonding orbitals and N/2 (empty) antibonding orbitals will result. Each bonding orbital will show an energy lowering as the atomic orbitals are mostly in-phase, but each of the bonding orbitals will be a little different and have slightly different energies. The antibonding orbitals will show an increase in energy as the atomic orbitals are mostly out-of-phase, but each of the antibonding orbitals will also be a little different and have slightly different energies. The allowed energy levels for all the bonding orbitals are so close together that they form a band, called the valence band. Likewise, all the antibonding orbitals are very close together and form a band, called the conduction band. Figure $12$) shows the bands for three important classes of materials: insulators, semiconductors, and conductors.
In order to conduct electricity, electrons must move from the filled valence band to the empty conduction band where they can move throughout the solid. The size of the band gap, or the energy difference between the top of the valence band and the bottom of the conduction band, determines how easy it is to move electrons between the bands. Only a small amount of energy is required in a conductor because the band gap is very small. This small energy difference is “easy” to overcome, so they are good conductors of electricity. In an insulator, the band gap is so “large” that very few electrons move into the conduction band; as a result, insulators are poor conductors of electricity. Semiconductors conduct electricity when “moderate” amounts of energy are provided to move electrons out of the valence band and into the conduction band. Semiconductors, such as silicon, are found in many electronics.
Semiconductors are used in devices such as computers, smartphones, and solar cells. Solar cells produce electricity when light provides the energy to move electrons out of the valence band. The electricity that is generated may then be used to power a light or tool, or it can be stored for later use by charging a battery. As of December 2014, up to 46% of the energy in sunlight could be converted into electricity using solar cells.
Example $2$: Molecular Orbital Diagrams, Bond Order, and Number of Unpaired Electrons
Draw the molecular orbital diagram for the oxygen molecule, O2. From this diagram, calculate the bond order for O2. How does this diagram account for the paramagnetism of O2?
Solution
We draw a molecular orbital energy diagram similar to that shown in Figure $7$. Each oxygen atom contributes six electrons, so the diagram appears as shown in Figure $7$.
We calculate the bond order as
$\ce{O2}=\dfrac{(8−4)}{2}=2 \nonumber$
Oxygen's paramagnetism is explained by the presence of two unpaired electrons in the (π2py, π2pz)* molecular orbitals.
Exercise $2$
The main component of air is N2. From the molecular orbital diagram of N2, predict its bond order and whether it is diamagnetic or paramagnetic.
Answer
N2 has a bond order of 3 and is diamagnetic.
Example $3$: Ion Predictions with MO Diagrams
Give the molecular orbital configuration for the valence electrons in $\ce{C2^2-}$. Will this ion be stable?
Solution
Looking at the appropriate MO diagram, we see that the π orbitals are lower in energy than the σp orbital. The valence electron configuration for C2 is
$(σ_{2s})^2(σ^∗_{2s})^2(π_{2py},\:π_{2pz})^4$.
Adding two more electrons to generate the $\ce{C2^2-}$ anion will give a valence electron configuration of
$(σ_{2s})^2(σ^∗_{2s})^2(π_{2py},\:π_{2pz})^4(σ_{2px})^2$
Since this has six more bonding electrons than antibonding, the bond order will be 3, and the ion should be stable.
Exercise $3$
How many unpaired electrons would be present on a $\ce{Be2^2-}$ ion? Would it be paramagnetic or diamagnetic?
Answer
two, paramagnetic
Key Concepts and Summary
Molecular orbital (MO) theory describes the behavior of electrons in a molecule in terms of combinations of the atomic wave functions. The resulting molecular orbitals may extend over all the atoms in the molecule. Bonding molecular orbitals are formed by in-phase combinations of atomic wave functions, and electrons in these orbitals stabilize a molecule. Antibonding molecular orbitals result from out-of-phase combinations of atomic wave functions and electrons in these orbitals make a molecule less stable. Molecular orbitals located along an internuclear axis are called σ MOs. They can be formed from s orbitals or from p orbitals oriented in an end-to-end fashion. Molecular orbitals formed from p orbitals oriented in a side-by-side fashion have electron density on opposite sides of the internuclear axis and are called π orbitals.
We can describe the electronic structure of diatomic molecules by applying molecular orbital theory to the valence electrons of the atoms. Electrons fill molecular orbitals following the same rules that apply to filling atomic orbitals; Hund’s rule and the Aufbau principle tell us that lower-energy orbitals will fill first, electrons will spread out before they pair up, and each orbital can hold a maximum of two electrons with opposite spins. Materials with unpaired electrons are paramagnetic and attracted to a magnetic field, while those with all-paired electrons are diamagnetic and repelled by a magnetic field. Correctly predicting the magnetic properties of molecules is in advantage of molecular orbital theory over Lewis structures and valence bond theory. | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/05%3A_Advanced_Theories_of_Covalent_Bonding/5.4%3A_Molecular_Orbital_Theory.txt |
Chapter Exercises
1. Explain how σ and π bonds are similar and how they are different.
2. Draw a curve that describes the energy of a system with H and Cl atoms at varying distances. Then, find the minimum energy of this curve two ways.
1. Use the bond energy found in Table 8.2.1 to calculate the energy for one single HCl bond (Hint: How many bonds are in a mole?)
2. Use the enthalpy of reaction and the bond energies for $H_2$ and $Cl_2$ to solve for the energy of one mole of HCl bonds. $H_{2(g)}+Cl_{2(g)} \rightleftharpoons 2HCl_{(g)} \;\;\; ΔH^∘_{rxn}=−184.7\; kJ/mol$
3. Explain why bonds occur at specific average bond distances instead of the atoms approaching each other infinitely close.
4. Use valence bond theory to explain the bonding in F2, HF, and ClBr. Sketch the overlap of the atomic orbitals involved in the bonds.
5. Use valence bond theory to explain the bonding in O2. Sketch the overlap of the atomic orbitals involved in the bonds in O2.
6. How many σ and π bonds are present in the molecule HCN?
7. A friend tells you N2 has three π bonds due to overlap of the three p-orbitals on each N atom. Do you agree?
8. Draw the Lewis structures for CO2 and CO, and predict the number of σ and π bonds for each molecule.
1. CO2
2. CO
Solutions
1. Similarities: Both types of bonds result from overlap of atomic orbitals on adjacent atoms and contain a maximum of two electrons. Differences: σ bonds are stronger and result from end-to-end overlap and all single bonds are σ bonds; π bonds between the same two atoms are weaker because they result from side-by-side overlap, and multiple bonds contain one or more π bonds (in addition to a σ bond).
2
When H and Cl are separate (the x axis) the energy is at a particular value. As they approach, it decreases to a minimum at 127 pm (the bond distance), and then it increases sharply as you get closer.
1. (a) H–Cl431 kJ/mol 427kJmol×mol6.022×1023bonds×1000 JkJ=7.09×10−19
2. (b) We know Hess’s law related to bond energies: ΔH°=ƩΔH∘BDE(broken)−ƩΔH∘BDE(formed) We are given the enthalpy of reaction
$−184.7 kJ/mol=(ΔH∘BDE(H–H)+ΔH∘BDE(Cl–Cl))−(2ΔH∘BDE(H–Cl))$
$H–H is 436 kJ/mol and Cl–Cl is 243$
$–184.7 kJ/mol = (436 + 243) – 2x = 679 – 2x$
$2x = 863.7 kJ/mol$
$x = 432\; kJ/mol$
This is very close to the value from part (a).
3. The specific average bond distance is the distance with the lowest energy. At distances less than the bond distance, the positive charges on the two nuclei repel each other, and the overall energy increases.
4. The single bond present in each molecule results from overlap of the relevant orbitals: F 2p orbitals in F2, the H 1s and F 2p orbitals in HF, and the Cl 3p orbital and Br 4p orbital in ClBr.
5. Bonding: One σ bond and one π bond. The s orbitals are filled and do not overlap. The p orbitals overlap along the axis to form a σ bond and side by side to form the π bond.
6. $\ce{H–C≡N}$ has two σ (H–C and C–N) and two π (making the CN triple bond).
7. No, two of the p orbitals (one on each N) will be oriented end-to-end and will form a σ bond.
8. (a) 2 σ 2 π;
(b) 1 σ 2 π;
Chemistry End of Chapter Exercises
Why is the concept of hybridization required in valence bond theory?
Hybridization is introduced to explain the geometry of bonding orbitals in valance bond theory.
Give the shape that describes each hybrid orbital set:
(a) sp2
(b) sp3d
(c) sp
(d) sp3d2
Explain why a carbon atom cannot form five bonds using sp3d hybrid orbitals.
There are no d orbitals in the valence shell of carbon.
What is the hybridization of the central atom in each of the following?
(a) BeH2
(b) SF6
(c) $\ce{PO4^3-}$
(d) PCl5
A molecule with the formula AB3 could have one of four different shapes. Give the shape and the hybridization of the central A atom for each.
trigonal planar, sp2; trigonal pyramidal (one lone pair on A) sp3; T-shaped (two lone pairs on A sp3d, or (three lone pairs on A) sp3d2
Methionine, CH3SCH2CH2CH(NH2)CO2H, is an amino acid found in proteins. Draw a Lewis structure of this compound. What is the hybridization type of each carbon, oxygen, the nitrogen, and the sulfur?
Sulfuric acid is manufactured by a series of reactions represented by the following equations:
$\ce{S8}(s)+\ce{8O2}(g)⟶\ce{8SO2}(g)$
$\ce{2SO2}(g)+\ce{O2}(g)⟶\ce{2SO3}(g)$
$\ce{SO3}(g)+\ce{H2O}(l)⟶\ce{H2SO4}(l)$
Draw a Lewis structure, predict the molecular geometry by VSEPR, and determine the hybridization of sulfur for the following:
(a) circular S8 molecule
(b) SO2 molecule
(c) SO3 molecule
(d) H2SO4 molecule (the hydrogen atoms are bonded to oxygen atoms)
(a) Each S has a bent (109°) geometry, sp3
(b) Bent (120°), sp2
(c) Trigonal planar, sp2
(d) Tetrahedral, sp3
Two important industrial chemicals, ethene, C2H4, and propene, C3H6, are produced by the steam (or thermal) cracking process:
$\ce{2C3H8}(g)⟶\ce{C2H4}(g)+\ce{C3H6}(g)+\ce{CH4}(g)+\ce{H2}(g)$
For each of the four carbon compounds, do the following:
(a) Draw a Lewis structure.
(b) Predict the geometry about the carbon atom.
(c) Determine the hybridization of each type of carbon atom.
For many years after they were discovered, it was believed that the noble gases could not form compounds. Now we know that belief to be incorrect. A mixture of xenon and fluorine gases, confined in a quartz bulb and placed on a windowsill, is found to slowly produce a white solid. Analysis of the compound indicates that it contains 77.55% Xe and 22.45% F by mass.
(a) What is the formula of the compound?
(b) Write a Lewis structure for the compound.
(c) Predict the shape of the molecules of the compound.
(d) What hybridization is consistent with the shape you predicted?
(a) XeF2
(b)
(c) linear (d) sp3d
Consider nitrous acid, HNO2 (HONO).
(a) Write a Lewis structure.
(b) What are the electron pair and molecular geometries of the internal oxygen and nitrogen atoms in the HNO2 molecule?
(c) What is the hybridization on the internal oxygen and nitrogen atoms in HNO2?
Strike-anywhere matches contain a layer of KClO3 and a layer of P4S3. The heat produced by the friction of striking the match causes these two compounds to react vigorously, which sets fire to the wooden stem of the match. KClO3 contains the $\ce{ClO3-}$ ion. P4S3 is an unusual molecule with the skeletal structure.
(a) Write Lewis structures for P4S3 and the $\ce{ClO3-}$ ion.
(b) Describe the geometry about the P atoms, the S atom, and the Cl atom in these species.
(c) Assign a hybridization to the P atoms, the S atom, and the Cl atom in these species.
(d) Determine the oxidation states and formal charge of the atoms in P4S3 and the $\ce{ClO3-}$ ion.
(a)
(b) P atoms, trigonal pyramidal; S atoms, bent, with two lone pairs; Cl atoms, trigonal pyramidal; (c) Hybridization about P, S, and Cl is, in all cases, sp3; (d) Oxidation states P +1, S $−1\dfrac{1}{3}$, Cl +5, O –2. Formal charges: P 0; S 0; Cl +2: O –1
Identify the hybridization of each carbon atom in the following molecule. (The arrangement of atoms is given; you need to determine how many bonds connect each pair of atoms.)
Write Lewis structures for NF3 and PF5. On the basis of hybrid orbitals, explain the fact that NF3, PF3, and PF5 are stable molecules, but NF5 does not exist.
Phosphorus and nitrogen can form sp3 hybrids to form three bonds and hold one lone pair in PF3 and NF3, respectively. However, nitrogen has no valence d orbitals, so it cannot form a set of sp3d hybrid orbitals to bind five fluorine atoms in NF5. Phosphorus has d orbitals and can bind five fluorine atoms with sp3d hybrid orbitals in PF5.
In addition to NF3, two other fluoro derivatives of nitrogen are known: N2F4 and N2F2. What shapes do you predict for these two molecules? What is the hybridization for the nitrogen in each molecule?
Chemistry End of Chapter Exercises
The bond energy of a C–C single bond averages 347 kJ mol−1; that of a C≡C triple bond averages 839 kJ mol−1. Explain why the triple bond is not three times as strong as a single bond.
A triple bond consists of one σ bond and two π bonds. A σ bond is stronger than a π bond due to greater overlap.
For the carbonate ion, $\ce{CO3^2-}$, draw all of the resonance structures. Identify which orbitals overlap to create each bond.
A useful solvent that will dissolve salts as well as organic compounds is the compound acetonitrile, H3CCN. It is present in paint strippers.
(a) Write the Lewis structure for acetonitrile, and indicate the direction of the dipole moment in the molecule.
(b) Identify the hybrid orbitals used by the carbon atoms in the molecule to form σ bonds.
(c) Describe the atomic orbitals that form the π bonds in the molecule. Note that it is not necessary to hybridize the nitrogen atom.
(a)
(b) The terminal carbon atom uses sp3 hybrid orbitals, while the central carbon atom is sp hybridized. (c) Each of the two π bonds is formed by overlap of a 2p orbital on carbon and a nitrogen 2p orbital.
For the molecule allene, $\mathrm{H_2C=C=CH_2}$, give the hybridization of each carbon atom. Will the hydrogen atoms be in the same plane or perpendicular planes?
Identify the hybridization of the central atom in each of the following molecules and ions that contain multiple bonds:
(a) ClNO (N is the central atom)
(b) CS2
(c) Cl2CO (C is the central atom)
(d) Cl2SO (S is the central atom)
(e) SO2F2 (S is the central atom)
(f) XeO2F2 (Xe is the central atom)
(g) $\ce{ClOF2+}$ (Cl is the central atom)
(a) sp2; (b) sp; (c) sp2; (d) sp3; (e) sp3; (f) sp3d; (g) sp3
Describe the molecular geometry and hybridization of the N, P, or S atoms in each of the following compounds.
(a) H3PO4, phosphoric acid, used in cola soft drinks
(b) NH4NO3, ammonium nitrate, a fertilizer and explosive
(c) S2Cl2, disulfur dichloride, used in vulcanizing rubber
(d) K4[O3POPO3], potassium pyrophosphate, an ingredient in some toothpastes
For each of the following molecules, indicate the hybridization requested and whether or not the electrons will be delocalized:
(a) ozone (O3) central O hybridization
(b) carbon dioxide (CO2) central C hybridization
(c) nitrogen dioxide (NO2) central N hybridization
(d) phosphate ion ($\ce{PO4^3-}$) central P hybridization
(a) sp2, delocalized; (b) sp, localized; (c) sp2, delocalized; (d) sp3, delocalized
For each of the following structures, determine the hybridization requested and whether the electrons will be delocalized:
(a) Hybridization of each carbon
(b) Hybridization of sulfur
(c) All atoms
Draw the orbital diagram for carbon in CO2 showing how many carbon atom electrons are in each orbital.
Each of the four electrons is in a separate orbital and overlaps with an electron on an oxygen atom.
Chemistry End of Chapter Exercises
Sketch the distribution of electron density in the bonding and antibonding molecular orbitals formed from two s orbitals and from two p orbitals.
How are the following similar, and how do they differ?
(a) σ molecular orbitals and π molecular orbitals
(b) ψ for an atomic orbital and ψ for a molecular orbital
(c) bonding orbitals and antibonding orbitals
(a) Similarities: Both are bonding orbitals that can contain a maximum of two electrons. Differences: σ orbitals are end-to-end combinations of atomic orbitals, whereas π orbitals are formed by side-by-side overlap of orbitals. (b) Similarities: Both are quantum-mechanical constructs that represent the probability of finding the electron about the atom or the molecule. Differences: ψ for an atomic orbital describes the behavior of only one electron at a time based on the atom. For a molecule, ψ represents a mathematical combination of atomic orbitals. (c) Similarities: Both are orbitals that can contain two electrons. Differences: Bonding orbitals result in holding two or more atoms together. Antibonding orbitals have the effect of destabilizing any bonding that has occurred.
If molecular orbitals are created by combining five atomic orbitals from atom A and five atomic orbitals from atom B combine, how many molecular orbitals will result?
Can a molecule with an odd number of electrons ever be diamagnetic? Explain why or why not.
An odd number of electrons can never be paired, regardless of the arrangement of the molecular orbitals. It will always be paramagnetic.
Can a molecule with an even number of electrons ever be paramagnetic? Explain why or why not.
Why are bonding molecular orbitals lower in energy than the parent atomic orbitals?
Bonding orbitals have electron density in close proximity to more than one nucleus. The interaction between the bonding positively charged nuclei and negatively charged electrons stabilizes the system.
Calculate the bond order for an ion with this configuration:
$(σ_{2s})^2(σ^∗_{2s})^2(σ_{2px})^2(π_{2py},\:π_{2pz})^4(π^∗_{2py},\:π^∗_{2pz})^3$
Explain why an electron in the bonding molecular orbital in the H2 molecule has a lower energy than an electron in the 1s atomic orbital of either of the separated hydrogen atoms.
The pairing of the two bonding electrons lowers the energy of the system relative to the energy of the nonbonded electrons.
Predict the valence electron molecular orbital configurations for the following, and state whether they will be stable or unstable ions.
(a) $\ce{Na2^2+}$
(b) $\ce{Mg2^2+}$
(c) $\ce{Al2^2+}$
(d) $\ce{Si2^2+}$
(e) $\ce{P2^2+}$
(f) $\ce{S2^2+}$
(g) $\ce{F2^2+}$
(h) $\ce{Ar2^2+}$
Determine the bond order of each member of the following groups, and determine which member of each group is predicted by the molecular orbital model to have the strongest bond.
(a) H2, $\ce{H2+}$, $\ce{H2-}$
(b) O2, $\ce{O2^2+}$, $\ce{O2^2-}$
(c) Li2, $\ce{Be2+}$, Be2
(d) F2, $\ce{F2+}$, $\ce{F2-}$
(e) N2, $\ce{N2+}$, $\ce{N2-}$
(a) H2 bond order = 1, $\ce{H2+}$ bond order = 0.5, $\ce{H2-}$ bond order = 0.5, strongest bond is H2; (b) O2 bond order = 2, $\ce{O2^2+}$ bond order = 3; $\ce{O2^2-}$ bond order = 1, strongest bond is $\ce{O2^2+}$; (c) Li2 bond order = 1, $\ce{Be2+}$ bond order = 0.5, Be2 bond order = 0, strongest bond is $\ce{Li2}$;(d) F2 bond order = 1, $\ce{F2+}$ bond order = 1.5, $\ce{F2-}$ bond order = 0.5, strongest bond is $\ce{F2+}$; (e) N2 bond order = 3, $\ce{N2+}$ bond order = 2.5, $\ce{N2-}$ bond order = 2.5, strongest bond is N2
For the first ionization energy for an N2 molecule, what molecular orbital is the electron removed from?
Compare the atomic and molecular orbital diagrams to identify the member of each of the following pairs that has the highest first ionization energy (the most tightly bound electron) in the gas phase:
(a) H and H2
(b) N and N2
(c) O and O2
(d) C and C2
(e) B and B2
(a) H2; (b) N2; (c) O; (d) C2; (e) B2
Which of the period 2 homonuclear diatomic molecules are predicted to be paramagnetic?
A friend tells you that the 2s orbital for fluorine starts off at a much lower energy than the 2s orbital for lithium, so the resulting σ2s molecular orbital in F2 is more stable than in Li2. Do you agree?
Yes, fluorine is a smaller atom than Li, so atoms in the 2s orbital are closer to the nucleus and more stable.
True or false: Boron contains 2s22p1 valence electrons, so only one p orbital is needed to form molecular orbitals.
What charge would be needed on F2 to generate an ion with a bond order of 2?
2+
Predict whether the MO diagram for S2 would show s-p mixing or not.
Explain why $\ce{N2^2+}$ is diamagnetic, while $\ce{O2^4+}$, which has the same number of valence electrons, is paramagnetic.
N2 has s-p mixing, so the π orbitals are the last filled in $\ce{N2^2+}$. O2 does not have s-p mixing, so the σp orbital fills before the π orbitals.
Using the MO diagrams, predict the bond order for the stronger bond in each pair:
(a) B2 or $\ce{B2+}$
(b) F2 or $\ce{F2+}$
(c) O2 or $\ce{O2^2+}$
(d) $\ce{C2+}$ or $\ce{C2-}$ | textbooks/chem/General_Chemistry/Chemistry_-_Atoms_First_1e_(OpenSTAX)/05%3A_Advanced_Theories_of_Covalent_Bonding/5.E%3A_Advanced_Theories_of_Covalent_Bonding_%28Exercises%29.txt |
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