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Our basis for understanding chemical bonding and the structures of molecules is the electron orbital description of the structure and valence of atoms, as provided by quantum mechanics. We assume an understanding of the periodicity of the elements based on the nuclear structure of the atom and our deductions concerning valence based on electron orbitals.
Section 2: Goals
Our model of valence describes a chemical bond as resulting from the sharing of a pair of electrons in the valence shell of the bonded atoms. This sharing allows each atom to complete an octet of electrons in its valence shell, at least in the sense that we count the shared electrons as belonging to both atoms. However, it is not clear that this electron counting picture has any basis in physical reality. What is meant, more precisely, by the sharing of the electron pair in a bond, and why does this result in the bonding of two atoms together? Indeed, what does it mean to say that two atoms are bound together? Furthermore, what is the significance of sharing a pair of electrons? Why aren’t chemical bonds formed by sharing one or three electrons, for example?
We seek to understand how the details of chemical bonding are related to the properties of the molecules formed, particularly in terms of the strengths of the bonds formed.
Section 3: Observation 1: Bonding with a Single Electron
We began our analysis of the energies and motions of the electrons in atoms by observing the properties of the simplest atom, hydrogen, with a single electron. Similarly, to understand the energies and motions of electrons which lead to chemical bonding, we begin our observations with the simplest particle with a chemical bond, which is the H+2 molecular ion. Each hydrogen nucleus has a charge of +1. An H+2 molecular ion therefore has a single electron. It seems inconsistent with our notions of valence that a single electron, rather than an electron pair, can generate a chemical bond. However, these concepts have been based on observations on molecules, not molecular ions like H+2. And it is indeed found that H+2
is a stable bound molecular ion.
What forces and motions hold the two hydrogen nuclei close together in the H+2
ion? It is worth keeping in mind that the two nuclei must repel one another, since they are both positively charged. In the absence of the electron, the two nuclei would accelerate away from one another, rather than remaining in close proximity. What is the role of the electron? Clearly, the electron is attracted to both nuclei at the same time, and, in turn, each nucleus is attracted to the electron. The effect of this is illustrated in Fig. 1. In Fig. 1a, the electron is “outside” of the two nuclei. In this position, the electron is primarily attracted to the nucleus on the left, to which it is closer. More importantly, the nucleus on the right feels a greater repulsion from the other nucleus than attraction to the electron, which is farther away. As a result, the nucleus on the right experiences a strong force driving it away from the hydrogen atom on the left. This arrangement does not generate chemical bonding, therefore. By contrast, in Fig. 1b, the electron is between the two nuclei. In this position, the electron is roughly equally attracted to the two nuclei, and very importantly, each nucleus feels an attractive force to the electron which is greater than the repulsive force generated by the other nucleus. Focusing on the electron’s energy, the proximity of the two nuclei provides it a doubly attractive environment with a very low potential energy. If we tried to pull one of the nuclei away, this would raise the potential energy of the electron, since it would lose attraction to that nucleus. Hence, to pull one nucleus away requires us to add energy to the molecular ion. This is what is meant by a chemical bond: the energy of the electrons is lower when the atoms are in close proximity than when the atoms are far part. This “holds” the nuclei close together, since we must do work (add energy) to take the nuclei apart.
Note that the chemical bond in Fig. 1b results from the electron’s position between the nuclei. On first thought, this appears to answer our question of what we mean by “sharing an electron pair” to form a chemical bond. An electron positioned between two nuclei is “shared” to the extent that its potential energy is lowered due to attraction to both nuclei simultaneously.
On second thought, though, this description must be inaccurate. We have learned our study of Energy Levels in Atoms that an electron must obey the uncertainty principle and that, as a consequence, the electron does not have a definite position, between the nuclei or otherwise. We can only hope to specify a probability for observing an electron in a particular location. This probability is, from quantum mechanics, provided by the wave function. What does this probability distribution look like for the
molecular ion?
To answer this question, we begin by experimenting with a distribution that we know: the 1s electron orbital in a hydrogen atom. This we recall has the symmetry of a sphere, with equal probability in all directions away from the nucleus. To create an
molecular ion from a hydrogen atom, we must add a bare second hydrogen nucleus (an ion). Imagine bringing this nucleus closer to the hydrogen atom from a very great distance (see Fig. 2a). As the ion approaches the neutral atom, both the hydrogen atom’s nucleus and electron respond to the electric potential generated by the positive charge. The electron is attracted and the hydrogen atom nucleus is repelled. As a result, the distribution of probability for the electron about the nucleus must become distorted, so that the electron has a greater probability of being near the ion and the nucleus has a greater probability of being farther from the ion. This distortion, illustrated in Fig. 2b, is called “polarization”: the hydrogen atom has become like a “dipole”, with greater negative charge to one side and greater positive charge to the other.
This polarization must increase as the ion approaches the hydrogen atom until, eventually, the electron orbital must be sufficiently distorted that there is equal probability for observing the electron in proximity to either hydrogen nucleus (see Fig. 2c). The electron probability distribution in Fig. 2c now describes the motion of the electron, not in a hydrogen atom, but in an
molecular ion. As such, we refer to this distribution as a “molecular orbital.”
We note that the molecular orbital in Fig. 2c is more delocalized than the atomic orbital in Fig. 2a, and this is also important in producing the chemical bond. We recall from the discussion of Atomic Energy Levels that the energy of an electron in an orbital is determined, in part, by the compactness of the orbital. The more the orbital confines the motion of the electron, the higher is the kinetic energy of the electron, an effect we referred to as the “confinement energy.” Applying this concept to the orbitals in Fig. 2, we can conclude that the confinement energy is lowered when the electron is delocalized over two nuclei in a molecular orbital. This effect contributes significantly to the lowering of the energy of an electron resulting from sharing by two nuclei.
Recall that the electron orbitals in the hydrogen atom are described by a set of quantum numbers. One of these quantum numbers is related to the symmetry or shape of the atomic orbital and is generally depicted by a letter. Recall that an “s” orbital is spherical in shape, and a “p” orbital has two lobes aligned along one axis. Similarly, the molecular orbitals for the
molecular ion are described by a set of numbers which give the symmetry (or shape) of the orbital. For our purposes, we need only one of these descriptors, based on the symmetry of the orbital along the bond: if the molecular orbital has the symmetry of a cylinder, we refer to it as a “
orbital.” The orbital in Fig. 2c satisfies this condition.
We conclude that chemical bonding results from an electron in a molecular orbital which has substantial probability for the electron to be between two nuclei. However, this example illustrates chemical bonding with a single electron. Our rules of valence indicate that bonding typically occurs with a pair of electrons, rather than a single electron. Furthermore, this model of bonding does not tell us how to handle molecules with many electrons (say,
) where most of the electrons do not participate in the bonding at all. | textbooks/chem/General_Chemistry/Concept_Development_Studies_in_Chemistry_(Hutchinson)/9%3A_Chemical_Bonding_and_Molecular_Energy_Levels/Section_1%3A_Foundation.txt |
We now consider molecules with more than one electron. These are illustrated most easily by diatomic molecules (molecules with only two atoms) formed by like atoms, beginning with the hydrogen molecule, . The most direct experimental observation of a chemical bond is the amount of energy required to break it. This is called the bond energy, or somewhat less precisely, the bond strength. Experimentally, it is observed that the bond energy of the hydrogen molecule is 458 kJ/mol. By contrast, the bond energy of the molecular ion is 269 kJ/mol. Therefore, the bond in is stronger than the bond in . Thus, the pair of shared electrons in generates a stronger attractive force than does the single electron in
.
Before deducing an explanation of this in terms of electron orbitals, we first recall the valence shell electron pair description of the bonding in
. Each hydrogen atom has a single electron. By sharing these two electrons, each hydrogen atom can fill its valence shell, attaining the electron configuration of helium.
How does this translate into the electron orbital picture of electron sharing that we have just described for the
molecular ion? There are two ways to deduce the answer to this question, and, since they are both useful, we will work through them both. The first way is to imagine that we form an molecule by starting with an molecular ion and adding an electron to it. As a simple approximation, we might imagine that the first electron’s probability distribution (its orbital) is not affected by the addition of the second electron. The second electron must have a probability distribution describing its location in the molecule as well. We recall that, in atoms, it is possible to put two electrons into a single electron orbital, provided that the two electrons have opposite values of the spin quantum number, ms. Therefore, we expect this to be true for molecules as well, and we place the added second electron in into the same orbital as the first. This results in two electrons in the region between the two nuclei, thus adding to the force of attraction of the two nuclei into the bond. This explains our observation that the bond energy of is almost (although not quite) twice the bond energy of
.
The second way to understand the electron orbital picture of
is to imagine that we form the molecule by starting with two separated hydrogen atoms. Each of these atoms has a single electron in a 1s orbital. As the two atoms approach one another, each electron orbital is polarized in the direction of the other atom. Once the atoms are close enough together, these two orbitals become superimposed. Now we must recall that these orbitals describe the wave-like motion of the electron, so that, when these two wave functions overlap, they must interfere, either constructively or destructively. In Fig. 3, we see the consequences of constructive and destructive interference. We can deduce that, in the electron orbitals from the atoms must constructively interfere, because that would increase the electron probability in the region between the nuclei, resulting in bonding as before. Therefore, the molecular orbital describing the two electrons in
can be understood as resulting from the constructive overlap of two atomic 1s electron orbitals.
We now add to our observations of diatomic molecules by noting that, of the diatomic molecules formed from like atoms of the first ten elements,
, , , , , , and are stable molecules with chemical bonds, whereas , , and are not bound. In examining the electron configurations of the atoms of these elements, we discover a correspondence with which diatomic molecules are bound and which ones are not.
all have odd numbers of electrons, so that at least one electron in each atom is unpaired. By contrast, He, Be, and Ne all have even numbers of electrons, none of which are unpaired. The other atoms, C and O both have an even number of electrons. However, as deduced in our understanding of the electron configurations in atoms, electrons will, when possible, distribute themselves into different orbitals of the same energy so as to reduce the effect of their mutual repulsion. Thus, in C and O, there are three 2p orbitals into which 2 and 4 electrons are placed, respectively. Therefore, in both atoms, there are two unpaired electrons. We conclude that bonds will form between atoms if and only if there are unpaired electrons in these atoms.
In
, the unpaired electrons from the separated atoms become paired in a molecular orbital formed from the overlap of the 1s atomic electron orbitals. In the case of a hydrogen atom, then, there are of course no paired electrons in the atom to worry about. In all other atoms, there certainly are paired electrons, regardless of whether there are or are not unpaired electrons. For example, in a lithium atom, there are two paired electrons in a 1s orbital and an unpaired electron in the 2s orbital. To form , the unpaired electron from each atom can be placed into a molecular orbital formed from the overlap of the 2s atomic electron orbitals. However, what becomes of the two electrons paired in the 1s orbital in a Li atom during the bonding of
?
To answer this question, we examine
, in which each atom begins with only the two 1s electrons. As we bring the two He atoms together from a large distance, these 1s orbitals should become polarized, as in the hydrogen atom. When the polarized 1s orbitals overlap, constructive interference will again result in a molecular orbital, just as in . Yet, we observe that is not a stable bound molecule. The problem which prevents bonding for arises from the Pauli Exclusion Principle: only two of the four electrons in can be placed into this bonding molecular orbital. The other two must go into a different orbital with a different probability distribution. To deduce the form of this new orbital, we recall that the bonding orbital discussed so far arises from the constructive interference of the atomic orbitals, as shown in Fig. 3. We could, instead, have assumed destructive interference of these orbitals. Destructive interference of two waves eliminates amplitude in the region of overlap of the waves, also shown in Fig. 3. In the case of the atomic orbitals, this means that the molecular orbital formed from destructive interference decreases probability for the electron to be between in the nuclei. Therefore, it increases probability for the electron to be outside the nuclei, as in Fig. 1a. As discussed there, this arrangement for the electron does not result in bonding; instead, the nuclei repel each other and the atoms are forced apart. This orbital is thus called an anti-bonding orbital. This orbital also has the symmetry of a cylinder along the bond axis, so it is also a orbital; to indicate that it is an anti-bonding orbital, we designate it with an asterisk,
In
, both the bonding and the anti-bonding orbitals must be used in order to accommodate four electrons. The two electrons in the bonding orbital lower the energy of the molecule, but the two electrons in the anti-bonding orbital raise it. Since two He atoms will not bind together, then the net effect must be that the anti-bonding orbital more than offsets the bonding orbital.
We have now deduced an explanation for why the paired electrons in an atom do not contribute to bonding. Both bonding and anti-bonding orbitals are always formed when two atomic orbitals overlap. When the electrons are already paired in the atomic orbitals, then there are too many electrons for the bonding molecular orbital. The extra electrons must go into the anti-bonding orbital, which raises the energy of the molecule, preventing the bond from forming.
Returning to the
example discussed above, we can develop a simple picture of the bonding. The two 1s electrons from each atom do not participate in the bonding, since the anti-bonding more than offsets the bonding. Thus, the paired “core” electrons remain in their atomic orbitals, unshared, and we can ignore them in describing the bond. The bond is formed due to overlap of the 2s orbitals and sharing of these electrons only. This is also consistent with our earlier view that the core electrons are closer to the nucleus, and thus unlikely to be shared by two atoms.
The model we have constructed seems to describe fairly well the bonding in the bound diatomic molecules listed above. For example, in a fluorine atom, the only unpaired electron is in a 2p orbital. Recall that a 2p orbital has two lobes, directed along one axis. If these lobes are assumed to lie along the axis between the two nuclei in
, then we can overlap them to form a bonding orbital. Placing the two unpaired electrons into this orbital then results in a single shared pair of electrons and a stable molecular bond. | textbooks/chem/General_Chemistry/Concept_Development_Studies_in_Chemistry_(Hutchinson)/9%3A_Chemical_Bonding_and_Molecular_Energy_Levels/Section_4%3A_Observation_2%3A_Bonding_and_Non-Bonding_in_Diatomic_Molecules.txt |
The energies of electrons in molecular orbitals can be observed directly by measuring the ionization energy. This is the energy required to remove an electron, in this case, from a molecule:
H2(g)→H+2(g)+e−(g)
The measured ionization energy of H2
is 1488 kJ/mol. This number is primarily important in comparison to the ionization energy of a hydrogen atom, which is 1312 kJ/mol. Therefore, it requires more energy to remove an electron from the hydrogen molecule than from the hydrogen atom, so we can conclude that the electron has a lower energy in the molecule. If we attempt to pull the atoms apart, we must raise the energy of the electron. Hence, energy is required to break the bond, so the molecule is bound.
We conclude that a bond is formed when the energy of the electrons in the molecule is lower than the energy of the electrons in the separated atoms. This conclusion seems consistent with our previous view of shared electrons in bonding molecular orbitals.
As a second example, we consider the nitrogen molecule, N2
. We find that the ionization energy of molecular nitrogen is 1503 kJ/mol, and that of atomic nitrogen is 1402 kJ/mol. Once again, we conclude that the energy of the electrons in molecular nitrogen is lower than that of the electrons in the separated atoms, so the molecule is bound.
As a third example, we consider fluorine, F2
. In this case, we find that the ionization energy of molecular fluorine is 1515 kJ/mol, which is smaller than the ionization energy of a fluorine atom, 1681 kJ/mol. This seems inconsistent with the bonding orbital concept we have developed above, which states that the electrons in the bond have a lower energy than in the separated atoms. If the electron being ionized has a higher energy in F2 than in F, why is F2
a stable molecule? Apparently, we need a more complete description of the molecular orbital concept of chemical bonding.
To proceed further, we compare bond energies in several molecules. Recall that the bond energy (or bond strength) is the energy required to separate the bonded atoms. We observe that the bond energy of N2
is 956 kJ/mol. This is very much larger than the bond energy of H2, 458 kJ/mol, and of F2, which is 160 kJ/mol. We can account for the unusually strong bond in nitrogen using both our valence shell electron pair sharing model and our electron orbital descriptions. A nitrogen atom has three unpaired electrons in its valence shell, because the three 2p electrons distribute themselves over the three 2p orbitals, each oriented along a different axis. Each of these unpaired electrons is available for sharing with a second nitrogen atom. The result, from valence shell electron pair sharing concepts, is that three pairs of electrons are shared between two nitrogen atoms, and we call the bond in N2 a “triple bond.” It is somewhat intuitive that the triple bond in N2 should be much stronger than the single bond in H2 or in F2
.
Now consider the molecular orbital description of bonding in N2
. Each of the three 2p atomic orbitals in each nitrogen atom must overlap to form a bonding molecular orbital, if we are to accommodate three electron pairs. Each 2p orbital is oriented along a single axis. One 2p orbital from each atom is oriented in the direction of the other atom, that is, along the bond axis. When these two atomic orbitals overlap, they form a molecular orbital which has the symmetry of a cylinder and which is therefore a σ orbital. Of course, they also form a σ*orbital. The two electrons are then paired in the bonding orbital.
The other two 2p orbitals on each nitrogen atom are perpendicular to the bond axis. The constructive overlap between these orbitals from different atoms must therefore result in a molecular orbital somewhat different that what we have discussed before. As shown in Fig. 4, the molecular orbital which results now does not have the symmetry of a cylinder, and in fact, looks something more like a cylinder cut into two pieces. This we call a orbital. There are two such orbitals since there are two sets of p orbitals perpendicular to the bond axis. Figure 4 also shows that an anti-bonding orbital is formed from the destructive overlap of 2p orbitals, and this is called a orbital. There are also two orbitals formed from destructive overlap of 2p orbitals. In , the three shared electron pairs are thus in a single orbital and in two
orbitals. Each of these orbitals is a bonding orbital, therefore all six electrons have their energy lowered in comparison to the separated atoms.
This is depicted in Fig. 5 in what is called a “molecular orbital energy diagram.” Each pair of atomic orbitals, one from each atom, is overlapped to form a bonding and an anti-bonding orbital. The three 2p orbitals from each atom form one
and pair and two and pairs. The lowering of the energies of the electrons in the and orbitals is apparent. The ten n=2 electrons from the nitrogen atoms are then placed pairwise, in order of increasing energy, into these molecular orbitals. Note that, in agreement with the Pauli Exclusion Principle, each pair in a single orbital consists of one spin up and one spin down electron.
Recall now that we began the discussion of bonding in because of the curious result that the ionization energy of an electron in is less than that of an electron in an F atom. By comparing the molecular orbital energy level diagrams for and we are now prepared to answer this puzzle. There are five p electrons in each fluorine atom. These ten electrons must be distributed over the molecular orbitals whose energies are shown in Fig. 6. (Note that the ordering of the bonding 2p orbitals differ between and .) We place two electrons in the orbital, four more in the two orbitals, and four more in the two orbitals. Overall, there are six electrons in bonding orbitals and four in anti-bonding orbitals. Since is a stable molecule, we must conclude that the lowering of energy for the electrons in the bonding orbitals is greater than the raising of energy for the electrons in the antibonding orbitals. Overall, this distribution of electrons is, net, equivalent to having two electrons paired in a single bonding orbital.
This also explains why the ionization energy of is less than that of an F atom. The electron with the highest energy requires the least energy to remove from the molecule or atom. The molecular orbital energy diagram in Fig. 6 clearly shows that the highest energy electrons in
are in anti-bonding orbitals. Therefore, one of these electrons is easier to remove than an electron in an atomic 2p orbital, because the energy of an anti-bonding orbital is higher than that of the atomic orbitals. (Recall that this is why an anti-bonding orbital is, indeed, anti-bonding.) Therefore, the ionization energy of molecular fluorine is less than that of atomic fluorine. This clearly demonstrates the physical reality and importance of the anti-bonding orbitals.
A particularly interesting case is the oxygen molecule,
. In completing the molecular orbital energy level diagram for oxygen, we discover that we must decide whether to pair the last two electrons in the same orbital, or whether they should be separated into different
orbitals. To determine which, we note that oxygen molecules are paramagnetic, meaning that they are strongly attracted to a magnetic field. To account for this paramagnetism, we recall that electron spin is a magnetic property. In most molecules, all electrons are paired, so for each “spin up” electron there is a “spin down” electron and their magnetic fields cancel out. When all electrons are paired, the molecule is diamagnetic meaning that it responds only weakly to a magnetic field.
If the electrons are not paired, they can adopt the same spin in the presence of a magnetic field. This accounts for the attraction of the paramagnetic molecule to the magnetic field. Therefore, for a molecule to be paramagnetic, it must have unpaired electrons. The correct molecular orbital energy level diagram for an
molecule is shown in Fig. 7.
In comparing these three diatomic molecules, we recall that N2 has the strongest bond, followed by O2 and F2. We have previously accounted for this comparison with Lewis structures, showing that N2 is a triple bond, O2 is a double bond, and F2 is a single bond. The molecular orbital energy level diagrams in Figs. 5 to 7 cast a new light on this analysis. Note that, in each case, the number of bonding electrons in these molecules is eight. The difference in bonding is entirely due to the number of antibonding electrons: 2 for N2 , 4 for O2 , and six for F2
. Thus, the strength of a bond must be related to the relative numbers of bonding and antibonding electrons in the molecule. Therefore, we now define the bond order as
BondOrder=12(# bonding electrons−# antibonding electrons)
Note that, defined this way, the bond order for N2
is 3, for O2 is 2, and for F2 is 1, which agrees with our conclusions from Lewis structures. We conclude that we can predict the relative strengths of bonds by comparing bond orders.
Section 6: Review and Discussion Questions
Why does an electron shared by two nuclei have a lower potential energy than an electron on a single atom? Why does an electron shared by two nuclei have a lower kinetic energy than an electron on a single atom? How does this sharing result in a stable molecule? How can this affect be measured experimentally?
Explain why the bond in an H2
molecule is almost twice as strong as the bond in the H+2 ion. Explain why the H2 bond is less than twice as strong as the H+2
bond.
Be2
is not a stable molecule. What information can we determine from this observation about the energies of molecular orbitals?
Less energy is required to remove an electron from an F2
molecule than to remove an electron from an F atom. Therefore, the energy of that electron is higher in the molecule than in the atom. Explain why, nevertheless, F2 is a stable molecule, i.e., the energy of an F2
molecule is less than the energy of two F atoms.
Why do the orbitals of an atom "hybridize" when forming a bond?
Calculate the bond orders of the following molecules and predict which molecule in each pair has the stronger bond:
C2
or C+2
B2
or B+2
F2
or F−2
O2
or O+2
Which of the following diatomic molecules are paramagnetic: CO, Cl2
, NO, N2
?
B2
is observed to be paramagnetic. Using this information, draw an appropriate molecular orbital energy level diagram for B2. | textbooks/chem/General_Chemistry/Concept_Development_Studies_in_Chemistry_(Hutchinson)/9%3A_Chemical_Bonding_and_Molecular_Energy_Levels/Section_5%3A_Observation_3%3A_Ionization_energies_of_diatomic_molecule.txt |
Skills to Develop
• Describe neutralization using chemical equations
Acids are familiar to you from everyday life, because acids are sour things, like vinegar and lemons. Bases might be less familiar to you. Common bases in regular life are baking soda and antacids that people take if they are having stomach trouble. Also, many soaps are basic, although this is just because of how they are made, not necessary to make them soapy. Bases make things feel slippery, and they taste bitter. They can be used to clean greasy things. Common acids and bases will react with each other and when they react completely, the products are usually a salt solution that isn't sour or bitter. This is called neutralization. Acid-base neutralization reactions are what make most cakes fluffy, because sometimes these reactions generate a gas that makes holes in the cake.
It is important to know that there are many different definitions of acid and base. This page describes the simplest and most specific definition. Practicing chemists use broader definitions that stretch the general concept to a variety of situations that are important in more advanced chemistry.
How does Neutralization work?
This description describes acid-base reactions in water. (It works a little bit differently in other solvents, but you don't need to think about that too much until you study more advanced chemistry). An acid is an electrolyte (strong or weak) that produces H+ ions when it dissolves in water. Hydrogen ions are also called protons, because a hydrogen nucleus is just a proton (unless it is a heavier isotope, but this is rare). Acids are sometimes called "proton donors" meaning they give away protons, but this is not a very good word, because the protons are pulled away by the solvent, not dropped by the solute. Examples of strong acids include HCl and H2SO4, which are called hydrochloric acid and sulfuric acid. But sulfuric acid has 2 protons bound to sulfate ion, and only one comes off completely (just like the weak electrolytes discussed here), so a solution of sulfuric acid will have hydrogen ions, bisulfate ions, and sulfate ions in solution. Just like in the case of precipitation reactions, if a base is added, both protons might come off completely and react with the base. Acids are called monoprotic, diprotic, etc. depending on how many acidic protons they have. HCl, acetic acid (vinegar, CH3COOH) and nitric acid (HNO3) are monoprotic acids. (Acetic acid has other protons, but only the last one is acidic.) Sulfuric acid and many others are diprotic acids.
A base is an electrolyte (strong or weak) that produces hydroxide ions when dissolved in water solution. This could be because it is a hydroxide salt, like NaOH, or because it takes hydrogen ions from water, leaving hydroxide behind. A good example of this is ammonia, NH3, which is sometimes used in house cleaning products. Ammonia reacts with water to make ammonium hydroxide (but only a little bit, ~1% of the ammonia reacts):
$NH_{3}(aq) + H_{2}O(l) \rightarrow NH_{4}^{+}(aq) + OH^{–}(aq)$
In general, bases react with hydrogen ions. This is how neutralization happens. The acid produces hydrogen ions, and the base produces hydroxide ions. These react together to make water. The anion that came from the acid and the cation are left, so if you evaporate the water, you would get a salt. The general reaction looks like this:
$X^{–}(aq) + H^{+}(aq) + M^{+}(aq) + OH^{–}(aq) \rightarrow H_{2}O(l) + X^{–}(aq) + M^{+}(aq)$
Overall, the reaction is:
$H^{+}(aq) + OH^{–}(aq) \rightarrow H_{2}O(l)$
Thus, the hydrogen ions, which makes acids acidic, are consumed, and the hydroxide which makes bases basic is also consumed, and if the moles of acid and base are equal, only neutral water and a salt is left. (Actually, it is a little bit more complicated than this if the acid or base is weak. The solution will only really become neutral when the moles are equal if both are strong.)
Need to Know Acids and Bases
Name Formula Comment
Ammonia NH3 weak base
Alkali Metal Hydroxides LiOH, NaOH, KOH, RbOH, CsOH soluble strong bases
Alkaline Earth Metal Hydroxides Ca(OH)2, Sr(OH)2, Ba(OH)2 soluble strong bases
Acetic Acid HC2H3O2 vinegar, weak acid
Perchloric Acid HClO4 strong acid
Chloric Acid HClO3 strong acid
Nitric Acid HNO3 strong acid
Carbonic Acid H2CO3 a weak acid, in sodas
Bicarbonate HCO3 a weak base (baking soda)
Sulfuric acid H2SO4 diprotic strong acid
Bisulfate HSO4 weak acid
Hydrohalic Acids HCl, HBr, HI strong acids
Hydrofluoric Acid HF weak acid, but very dangerous
Example
Q: How do acid-base reactions make cake fluffy?
A: Usually cakes include an acidic ingredient (this varies) and sodium bicarbonate, a base. When they react, the proton from the acid is transferred to the bicarbonate, making the weak acid carbonic acid. Carbonic acid is the product of an acid anhydride reaction between carbon dioxide and water. This reaction can be reversed, or carbonic acid can decompose into water and carbon dioxide. Especially at the high temperatures inside a baking cake, this decomposition will happen, and produce carbon dioxide gas. The pressure of the hot gas will form bubbles inside the cake, making it fluffy.
In the previous section (Precipitation), instead of having hydroxide react with hydrogen ions to form water, the acid base reaction made carbonic acid from protons and bicarbonate. In general, a base is something that will bind tightly to a proton. Bicarbonate and carbonate ions are bases, and so are sulfide ions. Both of these reactions can produce a gas, either carbon dioxide or hydrogen sulfide. In the lab, sodium bicarbonate is usually used to neutralize acid spills. When it reacts with acid it produces bubbles, so it's easy to see when the reaction finishes.
Most chemists agree that acid-base reactions are combination reactions without redox. This is a much more general definition than described here, but after you read about redox, go over these examples and convince yourself that they all fit that definition. What other examples can you think of that also fit this definition?
Summary
Acids are strong/weak electrolytes that produce H+ ions when dissolved in water. A notable property of acids is that they have a sour taste. Bases are strong/weak electrolytes that produce hydroxide ions (OH-) when dissolved in water. They are also known to have a bitter taste as well as a slippery or soapy texture. Neutralization is a type of reaction in which acids react with bases to form a salt solution (water and a salt), usually.
Outside Links
(18 min, skip the last 3 min, optional: skip the first 4 min) | textbooks/chem/General_Chemistry/General_Chemistry_Supplement_(Eames)/Chemical_Reactions_and_Interactions/Acid-Base_Reactions.txt |
Skills to Develop
• Describe the various types of combination reactions using chemical equations
Combination reactions describe a reaction like this:
$A + B \rightarrow C$
in which two or more reactants become one product (are combined). The problem with this term is that it doesn't give you much chemical insight because there are many different types of reactions that follow this pattern. So we'll break it into groups that reflect what's actually happening a little better.
Combination of Elements to Make Ionic Substances
In this category, an elemental metal and an elemental non-metal react to make an ionic substance that is neutral and has each ion in its correct charge state or valence. For instance,
$2Na(s) + Cl_{2}(g) → 2NaCl(s)$
$2Mg(s) + O_{2}(g) → 2MgO(s)$
$2Al(s) + 3O_{2}(g) → Al_{2}O_{3}(s)$
If the metal is a transition metal, it will be much harder to predict the correct charge on the metal ion in the ionic compound. You can check the element info in the nomenclature section or the links from the periodic table section. As you practice, you'll start to get a sense for what common charges are, but even then it is often good to check, because it might not be what you expect! For example, what's the charge on iron in Fe3O4 (magnetite)?
Under what circumstances do these reactions happen? Often, an elemental metal and non-metal "want" to make an ionic compound, because this is a more stable state (think about a heavy ball on a table: it can easily roll to the ground, where it has less potential energy, so the table isn't a stable state; if the heavy ball is in a small hole in the ground, it can't easily move, and if it did, it would have more potential energy, so the hole is a stable, low energy state). However, that doesn't necessarily mean the reaction will just happen on its own. That depends on how easily the reaction can happen (think about a place you want to go, but don't go because traveling there is very inconvenient).
For instance, the alkali metals and the halogens react pretty easily, so they will often react even without a "push." Oxygen is very reactive, which is why things burn, but you have to light them on fire to get them started. This is good, because otherwise we would burn in air at room temperature! Many of these elemental combination reactions might need a high temperature to get started, even if they want to happen. It won't be hard to remember that alkalis, alkaline earth metals and halogens react easily, because this is why they are very hard to find in elemental form! Oxygen and nitrogen are very abundant in elemental form because it is hard for them to react even if they want to. Nitrogen in particular reacts only with lithium metal and a few complicated compounds at room temperature, although it will react with many other elements at high temperatures. Most metals aren't found in elemental form in nature (except for ones that don't want to react, like gold), but if you find them in elemental form in your house, then probably they don't react easily.
Combination of Elemental Non-metals into Covalent Compounds
These reactions involve elemental forms of elements like H, C, N, O, Cl, S, P, etc. It will often be hard to predict the product because these elements can often combine in different ratios (this is where the law of multiple proportions comes from!). You can always expect that H will have a valence of 1, and O will usually have a valence of 2. Many of these reactions will happen quickly if you get them started with a little heat, especially if oxygen or a halogen is involved. Otherwise, they might happen very slowly or not at all except under special circumstances that we will talk more about later. Some examples:
$C(s) + O_{2}(g) → CO_{2}(g)\; (fast,\; once\; lit)$
$N_{2}(g) + 3H_{2}(g) → 2NH_{3}(g)\; (usually\; very\; slow)$
Basic Anhydrides
Basic anhydrides are compounds that turn into a base (a hydroxide salt) when you add water. They are metal oxides. Here's an example:
$CaO(s) + H_{2}O(l) \rightarrow Ca(OH)_{2}(aq)$
If the metal is an alkali or alkaline earth, the reaction probably happens quickly and produces a lot of heat. If the metal is a transition metal, the reaction might not happen so easily or at all.
Acid Anhydrides
Acid anhydrides are compounds that turn into an acid when you add water. They are non-metal oxides. These are a little more complicated than basic anhydrides, so don't worry too much about them right now. Here's an example:
$SO_{3}(g) + H_{2}O(l) \rightarrow H_{2}SO_{4}(aq)$
Other Combination Reactions
There are many other circumstances in which a combination reaction could happen. The types listed here are the simple ones that are good to know in the beginning. | textbooks/chem/General_Chemistry/General_Chemistry_Supplement_(Eames)/Chemical_Reactions_and_Interactions/Combination_Reactions.txt |
Skills to Develop
• Identify a general chemical equation for combustion reactions
• Describe some of the characteristics of combustion reactions
Combustion reactions are common and very important. Combustion means burning, usually in oxygen but sometimes with other oxidants such as fluorine. A combustion reaction happens quickly, producing heat, and usually light and fire. Combustion describes how the reaction happens, not the reactants and products. Chemists as early as Lavoisier suggested that people get their energy from combustion-like reactions, but even though the products and reactants are the same when you burn food in a fire and in your body, the way it happens is different. In a combustion reaction, the thing that burns (the reactant that isn't O2 or F2) is called the fuel. Combustion reactions are a type of redox reaction.
The classic chemistry class combustion reaction involves a compound of C and H reacting with O2 to form CO2 and H2O. Sometimes the reactant has some other elements, like O, S or N in it. The O will form water, the S will form usually SO2and the N will often produce N2, but sometimes a little bit of a nitrogen oxide. For class purposes, you can usually write equations in which carbon dioxide is produced. In real life, often some or a lot of CO is produced, depending on how much oxygen is present and other factors. In general, most elements in a compound that is combusted will form oxides, but you won't be able to say for sure how much of each oxide will be produced (CO or CO2, SO2 or SO3, etc). Here are some example equations. When you balance combustion reactions, usually start with the C, and the fuel, and do the oxygen last.
$C_{3}H_{8}(g) + 5O_{2}(g) \rightarrow 3CO_{2}(g) + 4H_{2}O(g)$
$C_{6}H_{12}O_{6}(s) + 12O_{2}(g) \rightarrow 6CO_{2}(g) + 6H_{2}O(g)$
Reaction 4.2 is sugar burning, which may also represent an animal or plant using stored energy.
Decomposition Reactions
Skills to Develop
• Describe decomposition reactions with chemical equations
Decomposition reactions are very common, and this word is used so much that many chemists just say "decomp". When people use this word in the lab, they might just mean that something didn't work, or that a chemical reacted in an unwanted way, especially while it was sitting in a bottle for a long time. The official meaning of decomposition is a little bit more specific, and means a reaction in which one chemical splits into two or more chemicals, like this:
$A \rightarrow B + C$
Decomposition reactions are often undesirable, but not always. For instance, many explosions are decompositions, and explosives are very important for many purposes other than weapons. Decomposition reactions might be hard to predict at first.
Some Simple Decomposition Patterns
The decomposition reactions in intro chemistry classes often result from heating a substance. For instance, when heated or struck, a salt of a complex anion (chlorate, carbonate, azide) may lose a gas (oxygen, carbon dioxide or nitrogen) leaving behind a simpler salt or metal. This could happen explosively, depending on the compound. Or, when heated, a metal hydroxide loses water to form the metal oxide (the reverse of the basic anhydride combination reaction). Here are some examples:
$CaCO_{3}(s) \rightarrow CaO(s) + CO_{2}(g)$
$2NaN_{3}(s) \rightarrow 2Na(s) + 3N_{2}g)$
$2KClO_{3}(s) \rightarrow 2KCl(s) + 3O_{2}(g)$
$Cu(OH)_{2}(s) \rightarrow CuO(s) + H_{2}O(g\; or\; l)$
Note
Students often get confused into thinking that combination reactions, the opposite of decomposition reactions, are called composition reactions. Actually, composition is not a type of reaction, rather, it has a different meaning. Composition means the ratio of elements in a compound, such as 75% C and 25% H. | textbooks/chem/General_Chemistry/General_Chemistry_Supplement_(Eames)/Chemical_Reactions_and_Interactions/Combustion_Reactions.txt |
Skills to Develop
• Define the 3 types of electrolytes
• Describe the relationship between solubility and electrolyte strength
Electrolyte means a solute that allows a water solution to conduct electricity. Electrolytes produce ions when they dissolve in solution. Salts are usually electrolytes, while molecular substances usually aren't, unless they are acids or bases.
Strong and Weak Electrolytes
The situation gets a little bit more complicated, though, because of the distinction between strong electrolytes and weak electrolytes. A strong electrolyte, like NaCl, splits up completely into sodium and chloride ions in solution. Likewise, a strong acid like HCl splits up completely into hydrogen and chloride ions in solution. Salts are often strong electrolytes, and strong acids are always strong electrolytes. Weak acids are weak electrolytes, and most other molecular compounds are non-electrolytes.
Many textbooks incorrectly state that all salts or ionic compounds are strong electrolytes. However, many ionic compounds or salts of transition metals or alkaline earth metals are not strong electrolytes. When they dissolve, some of the ions separate, but some stay together. Here are some examples:
$K_{2}SO_{4}(s) + water \rightarrow K^{+}(aq)\; (70\%) + KSO_{4}^{–}(aq)\; (30\%) + SO_{4}^{2–}(aq)\; (for\; a\; 0.36\; M\; solution)$
$CdI_{2}(s) + water \rightarrow Cd^{2+}(aq)\; (2\%) + CdI^{+}(aq)\; (22\%) + CdI_{2}(aq)\; (76\%) + I^{–}(aq)\; (for\; a\; 0.5\; M\; solution)$
As you can see, these salts are electrolytes (they do produce ions) but if you do calculations assuming that they separate completely into sulfate, potassium, cadmium(II) and iodide ions, you might get very wrong answers! They are not strong electrolytes. In general, the lower the concentration and the lower the charges on the ions, the "stronger" the electrolytes will be. Alkali metals other than lithium are usually strong electrolytes especially when the anion also has a small charge, and in dilute solutions (<0.1M). Alkaline earth metal compounds are weaker electrolytes, and other metals are even weaker still. We'll talk more about why this is later.
Solubility and Electrolyte Strength
Solubility can cause some confusion here. For instance, it's possible that a compound is a strong electrolyte, but just not very soluble. For this reason, it would not be able to produce a solution with lots of ions, because it isn't soluble, not because the ions are still attached to each other in the solution. In general, it makes sense to guess that if the ions in a compound are very strongly attracted to each other, the compound will be less soluble, and also it might be a weaker electrolyte because even in solution the ions will be bonded to each other. However, electrolytes also look stronger at lower concentrations, because if the ions split up, they are less likely to find each other again.
Overview
Skills to Develop
• Categorize the various types of reactions analyzed as acid-base or redox
So far we've seen a variety of types of reaction. This section is intended to help you fit all the pieces together. You've seen combination reactions and decomposition reactions. Sometimes these are the same reactions, just going in opposite directions. Likewise, dissolution and precipitation are opposite processes.
Many chemists think of reactions as falling into 2 main categories: acid-base type reactions and redox reactions. In an acid-base type reaction, an under-populated nucleus makes a bond with an over-populated nucleus, but the electrons don't change their primary loyalty. (The electrons from the over-populated nucleus do appreciate the better benefits they get from the under-populated nucleus, which has more pension money than it can spend on its own population.) In the classic acid-base reaction, the electrons on water really like oxygen as a home, but they are feeling a little crowded and poor; alliance with a hydrogen ion provides lots of money to make them happier, and a nice convenient vacation destination. In contrast, a redox reaction is any reaction in which electrons change their primary loyalties. Bonds between nuclei may change or not, but oxidation numbers do change. Try going through all the examples and deciding which category they fit and why.
Example of the classic acid-base reaction on the left and a redox reaction on the right. | textbooks/chem/General_Chemistry/General_Chemistry_Supplement_(Eames)/Chemical_Reactions_and_Interactions/Electrolyte_Strength.txt |
Skills to Develop
• Describe what occurs in a redox reaction
• Identify redox patterns in the periodic table
Redox reactions are reactions in which electrons shift allegiance. Allegiance means loyalty or commitment to a group, like your allegiance to your family or your country. If you decide to leave your home and become a citizen of a new country, you have shifted allegiance. Of course, you might be able to share your allegiance between your old country and your new country by being a citizen of both countries at the same time. This is similar to what electrons do. They have allegiance to a nucleus, and sometimes during a reaction they will shift allegiance to a different nucleus. They might shift allegiance completely, if the new nucleus is much better for them than the old one, or they might shift partially, and share their loyalty between the two. (Also, electrons, like people, can spend some time near nuclei they don't belong to, just like you can visit countries where you don't live, and then go back near their main nucleus, although electrons' trips are much shorter than ours!)
Redox Patterns in the Periodic Table
All neutral atoms have a population of electrons equal to their number of protons. Atoms, just like countries, always have elite citizens that they treat really well (power, money...) and these elite citizens are way too comfortable to shift allegiance unless they get kidnapped or something. Unlike countries, atoms always have only a small number of non-elite citizens, between 0 and 8, roughly, but usually no more than 3 or 4. These non-elite electrons will easily shift allegiance if they get a better offer.
What are the good parts of the periodic table? Fluorine is like heaven for electrons, they will basically never leave. If fluorine is accepting immigrants, electrons will leave anywhere else to move to F. Oxygen is second best. (Fluorine seems nasty to us because we don't want our citizen electrons to leave us for fluorine. We have jobs for them to do!) The noble gases are aloof nations, their populations of electrons hardly ever change. Their citizens don't want to leave, and they don't want any troublesome immigrants either. In general, the upper right corner of the periodic table are good comfortable nations to live in. The bottom left is such a bad place to live that electrons will destroy property in their rush to leave. (Watch this video for an example) This shows "standard of living" across the periodic table. Red elements will tend to keep their own electrons and attract electrons from other nuclei; yellow elements will tend to lose their non-elite electrons.
Ionic substances are like alliances between nations with really different standards of living. The non-elite electrons of the metal will try to shift allegiance to the non-metal. In molecular substances, also called covalent, the electrons will have dual citizenship, although they might have slightly greater loyalty to one nuclei than another, because these are alliances between similarly comfortable non-metal nuclei.
Oxidation States: Redox from the Nuclei's Perspective
With all these dual citizen electrons, with partial loyalty to several nuclei, it can get hard to count the population of electrons at each nucleus. Oxidation number is one procedure for counting. In this procedure, electrons are counted as belonging completely to the nuclei they feel the greatest loyalty to. The number of electrons that feel primary loyalty to each nucleus is counted, and subtracted the number of protons in that nucleus. But there are shortcuts to make this faster using valence rules. For example, in CO2, oxygen has a normal valence of 2. This means that it can accept 2 extra citizen electrons per nucleus, so it has an oxidation number of -2. Therefore, because there are 2 O nuclei, carbon loses 4 citizen electrons, who transfer their primary loyalty to O. Thus, carbon has an oxidation number of +4, because it has 4 fewer truly loyal electrons than protons.
How to count oxidation states:
• In elemental form, all nuclei have oxidations number 0
• In single-nucleus ions, the oxidation number is the charge
• Alkali nuclei always have an oxidation number of +1, unless in metallic form
• Alkaline earth nuclei always have an oxidation number of +2, unless in metallic form
• Oxygen nearly always has an oxidation number of -2, unless it is in an O-O unit like peroxide
• Non-metals have negative oxidation numbers when bound to metals and positive oxidation numbers when bound to O or F.
• Halogens have oxidation number of -1 unless bound to O or F. F is always -1 unless in elemental form.
• The sum of all the oxidation numbers of a molecule or ion is its overall charge (0 if neutral, +2 if a dipositive cation, etc)
Types of Redox Reactions
Oxidation refers to a process in which something loses electrons, and has its oxidation number increase. This usually happens to compounds that react with oxygen gas, which is why it is called oxidation. Reduction refers to a process in which something gains electrons, and its oxidation number is reduced. Actually this is not quite where the word came from. When a metal oxide is reduced to the elemental metal and an elemental non-metal or non-metal compound, the mass of solid decreases because the non-metal usually leaves as a gas. This reduction in quantity is where the word comes from. When you think about it, for one thing to be oxidized, another thing must be reduced (because electrons can't appear out of nowhere), which is why we often use the combination word redox.
Many of the reaction types we've already seen involve redox. For instance, combination of elements is a redox reaction. Decomposition reactions are often redox reactions. Combustion reactions are redox reactions. (Dissolution/precipitation and acid-base are not redox reactions.) The examples of redox you've already seen involve electrons shifting loyalty at the same time that new bonds are made between nuclei. It's also possible for electrons to abandon one nucleus for another separate nucleus that isn't bonded to the first one. For example, if an elemental metal is placed in a solution of salt or acid, the metal can lose some electrons to the cation in the solution. The metal forms cations, and the former cation usually leaves the solution, forming hydrogen gas or elemental metal. Here are some examples:
$Zn(s) + 2HBr(aq) \rightarrow ZnBr_{2}(aq) + H_{2}(g)$
$Fe(s) + Pb(NO_{3})_{2}(aq) \rightarrow Fe(NO_{3})_{2}(aq) + Pb(s)$
All that happened in these reactions is that the disloyal electrons left a less-comfortable home country for a more comfortable one. These are called displacement reactions because one cation is displaced (replaced) by another. To predict what reactions will occur on this pattern, you can use the activity series, which is like a ranking of "worst places to live". It's pretty good for making predictions, because in general electrons will move to places lower on the list, but it doesn't take into account all the specific circumstances, like current population pressures. You'll learn more reliable ways to make predictions later. The activity series is:
1. Worst place to live/easiest to oxidize: the alkali and alkaline earth metals
2. Bad: the active metals, aluminum, zinc, iron, and others from the middle of the transition metals
3. Better: tin, lead, hydrogen from acid, copper
4. Best place to live, hardest to oxidize: the noble metals, silver, mercury, platinum, gold
Copper metal doesn't react with acid, because copper comes after hydrogen in the list. However, some acids have other oxidants in them that are stronger than hydrogen ions. For instance, nitrate ions are oxidizing, and Cu will react with nitric acid. Even gold reacts with a concentrated acid mixture, aqua regia, made of nitric and hydrochloric acids. | textbooks/chem/General_Chemistry/General_Chemistry_Supplement_(Eames)/Chemical_Reactions_and_Interactions/Redox_Reactions.txt |
Skills to Develop
• Describe what occurs in a precipitation reaction
Precipitation is the process of a compound coming out of solution. It is the opposite of dissolution or solvation. In dissolution, the solute particles separate from each other and are surrounded by solvent molecules. In precipitation, the solute particles find each other and form a solid together. This solid is called the precipitate or sometimes abbreviated "ppt".
Solubility Equilibria
Precipitation and dissolution are a great example of a dynamic equilibrium (also described here). Any time there is a solution with a little bit of solid solute in it, both processes will be happening at once. Some molecules or ions will leave the solid and become solvated, and some solvated solute particles will bump into the solid and get stuck there. The rates of the 2 processes determine the overall effect: if precipitation happens faster, then a lot of solid can come out of the solution very quickly. If dissolution happens faster, than the solid will dissolve. As the solution becomes more concentrated, the rate of precipitation will increase and the rate of dissolution will decrease, so that eventually the concentration will stop changing, and this is equilibrium. When equilibrium is reached, the solution is saturated, and that concentration defines the solubility of the solute. Solubility is the maximum possible concentration, and it is given in M, g/L, or other units. Solubility changes with temperature, so if you look up solubility data it will specify the temperature.
Precipitation Reactions
Precipitation can happen for various reasons, such as that you cooled a solution, or removed some solvent by evaporation, or both. (This is often used as a way to purify a compound.) You can also have a precipitation reaction, when you mix two solutions together and a new combination of ions is super-saturated in the combined solution. For example, maybe you mixed a solution of silver(I) nitrate and sodium chloride. Silver(I) chloride is very insoluble, so it will precipitate, leaving soluble sodium nitrate in solution. Precipitation reactions can be a good way to prepare a salt you want from some other salts with the right anion and cation. Precipitation reactions can also be used to detect the presence of particular ions in solution. For instance, you might test for chloride, iodide and bromide in an unknown solution by adding silver(I) ions and looking for precipitation.
Predicting Precipitation Reactions
Beginning chemistry students usually memorize a list of solubility rules. Here it is (these rules will be a little bit different in different textbooks, because people might not have exactly the same definition of soluble or insoluble):
• Most nitrate and acetate salts are soluble
• Most alkali cation and ammonium salts are soluble
• Most chloride, bromide and iodide salts are soluble, except those of Ag(I), Pb(II) and Hg(I)
• Most sulfate salts are soluble, except those of barium, calcium and Pb(II)
• Most hydroxide salts are only slightly soluble, except those of sodium and potassium
• Most sulfide, carbonate and phosphate salts are only slightly soluble
You can use this list to predict when precipitation reactions will occur. For this purpose, you don't usually have to worry about whether the compounds are strong or weak electrolytes, you can think of the ions as being separate. The reason is that usually some of the ions will be separate, and once those precipitate with a new partner, more of the original compound ions will separate from each other, and the process will continue.
Writing Equations for Precipitation Reactions
Chemists may write equations in different ways to emphasize the important parts. For instance, we might write an equation like this, which describes mixing 2 solutions of different soluble salts and getting a precipitate:
$AgNO_{3}(aq) + NaCl(aq) \rightarrow AgCl(s) + NaNO_{3}(aq)$
Alternately, we might write the same reaction just focusing on the part that forms the precipitate, and leaving out the spectator ions that don't really do anything, just stay in solution:
$Ag^{+}(aq) + Cl^{–}(aq) \rightarrow AgCl(s)$
Chemistry students are sometimes asked to prove their understanding of dissociation by writing out all the ions separately, like this:
$Ag^{+}(aq) + NO_{3}^{–}(aq) + Cl^{–}(aq) + Na^{+} \rightarrow AgCl(s) + Na^{+} + NO_{3}^{–}(aq)$
No real chemist would be likely to do this because it is a nuisance. (It's also a little funny because many salts aren't strong electrolytes, so teachers might be telling their students to write an equation that doesn't show what's really happening.) However, it does help show what it means to be a spectator ion, since they are the same on both sides when you write it like this.
What Determines Solubility?
Solubility depends on the relative stability of the solid and solvated states for a particular compound. For instance, if it has very strong interactions between molecules or ions in the solid state, then it won't be very soluble unless the solvation interations are also very strong. (Ionic salts are a good example: usually they have strong interactions in the solid and solvated states.) If the interactions in the solid are weak, the compound can still be insoluble in polar solvents if the interactions with the solvent are weaker than the Coulomb interactions of the solvent molecules with other solvent molecules. (This is why wax is insoluble in water: it is non-polar, so the wax-wax interactions are weak, but the wax-water interactions are weaker than the water-water interactions.) We can't explain what makes these interactions strong or weak well until after we study chemical bonding, but in general ionic compounds with larger charges on the ions and smaller ions are less soluble, because they can have stronger Coulomb interactions in the solid. | textbooks/chem/General_Chemistry/General_Chemistry_Supplement_(Eames)/Chemical_Reactions_and_Interactions/Solubility_and_Precipitation.txt |
Skills to Develop
• Describe factors affecting dissolution
Dissolution means the process of dissolving or forming a solution. When dissolution happens, the solute separates into ions or molecules, and each ion or molecule is surrounded by molecules of solvent. The interactions between the solute particles and the solvent molecules is called solvation. A solvated ion or molecule is surrounded by solvent.
A sodium ion solvated by water, from Wikimedia Commons.
Technically a solvent can mean anything that is the more abundant component of a homogeneous mixture, but usually it means a volatile liquid that things can easily dissolve in. (Volatile means that it can easily evaporate, like water or alcohol.) The most common solvent is water. When you scuba dive in the ocean, you will need to rinse your gear with water afterwards to remove the salt. The salt dissolves in the water, gets washed away, and then the water evaporates, leaving the gear clean. This is the typical action of a solvent.
Solvents are either polar or non-polar. A polar solvent has partial negative and positive charges. For instance, water has a partial negative charge on O and a partial positive charge on H. The symbol δ means a partial charge, less than the charge on one proton or electron, such as δ+ or δ–. This helps the solvent interact with (solvate) ions and polar molecules through Coulomb interactions. A non-polar solvent is one that is electrically neutral all over, or almost so. Oil, or the gas in your car, are examples of non-polar liquids that could be used as solvents. Non-polar solvents are only good for dissolving non-polar solutes, which is why water, salt and sugar don't mix into oil.
The Wild Ionists
Skills to Develop
• Discuss the significance of electrolytes and osmosis
Faraday proved that pure water doesn't conduct electricity, but conductivity increases when some types of solutes are added. These solutes are called electrolytes. For a long time, people thought that the ions that let water conduct electricity (to conduct, charged particles must move, such as electrons through a metal wire or ions through solution) were formed by the electricity.
Svante Arrhenius was the son of a minor university employee in Sweden. For his doctoral research, he wanted to study whether molecular weights of compounds could be measured using the conductivity of their solutions (a subject his adviser told him not to work on―he often ignored his advisers' advice). To do this, he needed to understand how the conductivity of electrolyte solutions depends on concentration, but he found that his data were unexpectedly confusing. Wisely, he realized that explaining the conductivities would actually be more interesting and useful than his original plan to measure molecular weights. He found that the electrolytes fell into 2 types: strong and weak. Strong electrolytes (like salt and HCl) easily conduct electricity, and their conductivity is proportional to the concentration. Weak electrolytes (like vinegar and ammonia) conduct electricity much less, but their conductivity is not directly proportional to concentration. As concentration decreases, conductivity does too, but conductivity/mass of solute increases. He proposed that electrolytes had an active state (that conducts) and an inactive state (that doesn't conduct electricity); strong electrolytes were entirely in the active state, while weak electrolytes would have some molecules in the active state and some in the inactive state. As a weak electrolyte solution was diluted, the % of molecules in the active state would increase toward 100%. But this was vague, and his advisers didn't like it and gave him a very low grade. Because the grade was so low, he had to travel and do research without pay in other labs for many years to establish himself, instead of becoming a professor himself.
Arrhenius travelled and met other scientists. A professor named Ostwald helped introduce him. He became friends with Nernst (who would later do some important work on thermodynamics and electrochemistry). Nernst was a wild young man who thought about becoming an actor, had scars from fighting, and often got very drunk. He also began to correspond (write letters) with van't Hoff, whom we mentioned earlier. Van't Hoff had moved on from carbon chemistry and was studying osmosis. Osmosis occurs when water moves across a membrane (like the membrane of a biological cell) to go from a dilute solution on one side to a concentrated solution on the other. This can create a pressure difference. This is how some antibiotics kill bacteria: the bacteria explode from the pressure because the solution outside the cell is more dilute than the solution inside the cell. Van't Hoff found that the osmotic pressure depends on the concentration of the solutions, but for electrolytes, there's an extra "fudge factor"; the concentration seems higher than expected. It appeared that the electrolytes were breaking into pieces, and these pieces increases the apparent concentration of the solution when determining osmotic pressure. Arrhenius noticed that the "fudge factor" in Van't Hoff's data was related to the fractions in Arrhenius' active states.
At this time, electrons hadn't been discovered yet. People thought that ions form in solution only when electricity is passed through. However, Arrhenius realized that electrolytes must be splitting into charged pieces (ions) when they dissolved, even without electricity, because these charged pieces affected van't Hoff's osmotic pressure measurements. The idea that salt, which forms so eagerly from the elements, would split up completely in water, seemed crazy to most scientists. (Now, we know that splitting into ions vs splitting into atoms is very different!) But Ostwald, Nernst and van't Hoff agreed with Arrhenius, and they were called "the wild army of the Ionists". They convinced other chemists pretty quickly that this new theory of ionization was correct.
Acknowledgement
This discussion was inspired by first chapter of the book Cathedrals of Science by Patrick Coffey. | textbooks/chem/General_Chemistry/General_Chemistry_Supplement_(Eames)/Chemical_Reactions_and_Interactions/Solutions_Solvation_and_Dissociation.txt |
Skills to Develop
• Distinguish physical properties from chemical properties
• Understand the difference between homogeneous and heterogeneous mixtures
If everything is made of the same 4 elements in different ratios, perhaps you can adjust the ratios of elements through various processes and change one material into another. This was called transmutation. In particular, people wanted to change inexpensive metals into gold.
There were known examples of one material turning into another. For instance, there's a reddish mineral called cinnabar, and if you heat it, you get silvery liquid mercury. Oddly, if you heat it again, you get another red solid. We'll explain that in chemical terms soon (it's a chemical change!) but it looked like a proof of principle for transmutation, with one substance becoming another.
Also, some people claimed to be able to make gold. The problem here was that they weren't distinguishing between a pure substance and a mixture. A pure substance is composed of a single type of molecule. Gold (theoretically) is a pure substance: not just a pure substance, but a pure element. However, you can make compounds that look a little like gold, that are yellow and shiny, by mixing different metals, such as copper and tin. We call that a mixture (or more specifically an alloy, which is a mixture of metals).
A compound is a substance that contains multiple elements. Water is a compound of hydrogen and oxygen. However, water is a pure substance, because each molecule of water is the same. Air is a mixture, not a pure substance, because it contains different types of molecules, some of which are compounds, like carbon dioxide, which is a compound of carbon and oxygen. The difference between a compound and a mixture is that in a compound you always have the same ratio of the elements: in carbon dioxide, the ratio is there in the name: one carbon atom, two (di) oxygen atoms. Carbon dioxide is CO2. In a mixture, the ratio can vary. Air contains nitrogen (N2) and oxygen (O2) molecules and many other components, and a sample from a lecture hall and a sample from a forest would probably have slightly different ratios.
Alchemists did not always distinguish between mixtures and pure compounds or elements, for instance gold vs bronze (an alloy of copper and tin). We can distinguish different materials by using the properties of the materials. Properties are things like what temperature it melts at, whether it dissolves in acid, and so on. We can distinguish physical properties and chemical properties. Melting point is a physical property, and solubility in acid is a chemical property.
Alchemists developed many of the techniques of chemistry that we still use. For instance, a heterogeneous mixture of a solid and a liquid, such as sugar in oil, could be separated by filtration. If you have a solution, like sugar in water, that is homogeneous, because all the sugar has dissolved, you can't filter it. Instead, you could let the water evaporate slowly until big crystals grow. (We call that rock candy in English.) This process is called recrystallization, and is used to purify solids. To purify liquids, you can use distillation, which is based on different boiling points: heat at a temperature where one component boils and the other doesn't, and collect the vapor. Sublimation is similar: some solids will vaporize with heat, and can then be recollected from the vapor on a cold surface, where they solidify.
Overall, alchemists were sometimes excellent experimentalists, and they definitely spent time "in the lab." However, their explanations and reasons for beliefs may seem strange from a modern perspective. For instance, if a theory had parallels to Christian religious events, that might be considered evidence that it was correct. The other "unscientific" thing about their practice was that they often reported results in a way that was intended to confuse the reader, if they reported the results at all, and they often looked to ancient texts as an authority, even when there was no evidence that the authors had accomplished anything. Finally, the goals they chose were very ambitious, so instead of trying to look at the simplest questions first, to get clear answers, they used many complicated procedures and got results that are hard to explain even now.
Summary
Transmutation involves changing one material into another by adjusting the ratios of elements through various processes. A pure substance has a chemical composition of only one type of molecule .A compound is a substance that contains 2 or more types of atoms chemically bonded together. Mixtures have two or more substances physically combined, meaning that each component retains its chemical composition and properties. The components of a heterogeneous mixture are not uniformly combined while the components of a homogeneous mixture are. A substance's physical properties can be observed without changing its chemical composition (i.e. color, volume, melting point) while its chemical properties are observed through chemical changes (i.e. burning, rusting). | textbooks/chem/General_Chemistry/General_Chemistry_Supplement_(Eames)/Chemistry_Basics/Alchemy.txt |
Skills to Develop
• Distinguish physical changes from chemical changes
• Identify the three states of matter and their respective properties
• Understand some of the history behind identifying the three states of matter
• Define matter and its components (i.e. mass, weight, volume)
Humanity's first chemical knowledge was mostly technology, like metal working, ceramics, cooking, etc. Early civilizations learned to control fire, to cast metals and make alloys, to make glass and ceramics, and so forth. The first chemical thinking, as opposed to chemical applications, asked: What is matter? Matter is stuff. It's what we are made of, what the earth and the air are made of. Matter is anything that has mass... what is mass? It's the amount of stuff. Not how much space it occupies (that's volume) but how much stuff is there. We measure mass using weight, which is how strongly the stuff in question is attracted to the earth by gravity.
What is matter made of? Well, one philosopher of ancient Greece proposed that all matter is made of water. He observed that water can "become air" by evaporation, or become solid by freezing into ice. He reasoned therefore that water can convert into everything, and matter is made of water. Now, we call those changes physical changes. The water is still water when it boils and turns into steam. The water is still water when it freezes into ice. We changed its temperature, not its nature.
Another Greek philosopher said that everything was made of air: when air becomes less compressed, it becomes fire, and when more compressed, it turns into water, stones, and so forth. He offered the proof that when you breathe through open lips, the air is warm, and when you compress it by breathing through puckered lips, it's cold, and condenses into liquid or solid. Air turning into stone would be a chemical change, in modern terms. In terms of Dalton's Atomic Theory, a chemical change means that the atoms form new combinations, like one atom of A combining with an atom of B. Before A and B were separate, but now they are attached. That's a chemical change.
Others proposed 5 elements, with distinct shapes: octahedra, tetrahedra, cubes, etc. For a long while, the four element model (earth, air, fire, water) was popular. (In Greece it was proposed by a man who was asked to become king of his city, but created a democracy instead. Then he declared himself a god and jumped into a volcano. It's said that the volcano tossed back his sandals to prove he wasn't a god.) This model, which Plato and Aristotle also used, suggested that all matter was composed of these four elements in different ratios. For instance, when wood is burned, you get smoke (air), ash (earth), pitch (a viscous liquid, here identified with water), and fire, so wood is made of all these things. Wood burning is a great example of a chemical change.
From a modern perspective, in all these theories an important element was lacking: experimentation. The Greeks preferred thinking to trying things, and you might say that it shows in their theories. For them, the fundamental difference was what we now call the state of matter or the phase of matter. Solid, liquid and gas are the three main states of matter.
• Ice, wood and stone are all solids
• Water is a liquid, like oil
• Air is a gas, like steam, and like the gas that you use in a stove
The three states are fundamentally different in nature. Gases take up as much space as they can, in whatever shape they can: the molecules are far apart, and try to spread out. Liquids change shape but have a constant volume. Solids have a constant shape and volume. Now we know that you can change the state of matter just by changing the temperature and pressure, and the molecules stay the same. So the wrong theories of the Greeks were based on not recognizing the difference between physical changes, which are changes in the state of matter, and chemical changes, changes in the combination of atoms.
Summary
Matter is anything that has volume and mass. Mass is the amount of matter in a substance. Physical changes involve processes that change the form of a substance but not its chemical composition. Chemical changes involves processes in which a new substances are formed with new chemical compositions. Weight is the force resulting from a mass being pulled towards a particular location as a result of gravity. The three states of matter are: solid, liquid, and gas. A solid is an object with a fixed shape and volume. Liquids are fluids have a constant volume but are capable of changing their shape. Gases are fluids that do not have constant shape or volume so they take up as much space as they can within a given container. | textbooks/chem/General_Chemistry/General_Chemistry_Supplement_(Eames)/Chemistry_Basics/Ancient_History.txt |
Skills to Develop
• Define the sub-atomic particles making up an atom
• Distinguish isotopes from typical atoms of an element
• Identify the atomic mass and mass number of an element
• Understand how to utilize units such as coulombs, atomic mass units, and angstroms
Atoms of an element are not all identical and indivisible as Dalton said. They are made of three mains types of particles.
Particle Mass (amu) Charge
Electrons 5.5 x 10–4 Negative (1–)
Protons 1.0073 Positive (1+)
Neutrons 1.0087 Neutral
The protons and neutrons together form a very small dense center of the atom, called a nucleus. The nucleus contains all the mass of the atom except the small mass of electrons. The electrons move around the nucleus, and occupy a much bigger space than the nucleus, so that most of the atom is empty space.
An element is defined by the number of protons, or atomic number, which is equal to the number of electrons in the neutral element. Atoms of an element can have different numbers of neutrons, resulting in different masses. Isotopes are atoms that have the same number of protons, but different numbers of neutrons. Chemical properties depend mostly on the atomic number, so isotopes are nearly the same chemically. The atomic mass is the average mass of the atom, including all the different isotopes that are likely to be present. If you want to show what isotope of an element you can use special notation, like this: 12C. This is read as carbon-12. The 12 is the mass number, or the number of protons + neutrons. Since carbon always has 6 protons, 12C must have 6 neutrons also.
We use some very small units to describe atoms. For instance, the charge of a proton or electron (which we will use as the unit of charge) is 1.602 x 10–19 coulombs (C). The atomic mass unit, or amu, is 1.661 x 10–24 g. Sizes of atoms are usually measured in angstroms or Å, which is 1 x 10–10 m. The diameter of most atoms is 1-5 Å. The diameter of nuclei is roughly 10–4 Å. Electrons are smaller than nuclei, so most of the atom is empty.
Atomic Theory
Skills to Develop
• Describe the law of conservation of mass, the law of definite proportions, and the law of multiple proportions
• Describe Dalton's atomic theory
The law of conservation of mass was established by Lavoisier, although others had used it before. It said that in any chemical reaction, the total mass of products is the same as the total mass of reactants. No matter is created or destroyed during the reaction.
There was a debate over whether elements always combine in exactly the same ratio, which is called composition. It was well known that the elemental ratios in many materials are approximately constant. Water is made by burning hydrogen in oxygen. The composition is about 11.1% hydrogen and 88.9% oxygen, by mass. The oxides of metals also generally have consistent compositions. For instance, mercuric oxide is 92.6% mercury, and mercurous oxide is 96.2% mercury; no bigger or smaller ratios were known. Berthollet, a respected scientist, argued that the consistent ratios observed arise from the conditions of the experiment; for instance, the most insoluble or volatile composition will be preferentially produced because it removes itself from the reaction. In the absence of such influences, he believed the ratios could vary continuously.
Another scientist, Proust, said that the proportions were always the same, and eventually persuaded Berthollet of the law of definite proportions. Proust was more careful to study only pure compounds, and knew how precise his measurements were, so that even though the exact numbers didn't always come out the same, once he rounded to the correct number of significant figures, the ratios were the same, no matter how the material was prepared. Review the difference between accuracy and precision on the previous page (Lavoisier), and how to use the correct number of significant figures. (Some types of materials, like minerals or alloys, can have variable proportions; Berthollet was also right that in solution compositions might be possible that were not observed in the isolated substances. Berthollet's arguments on composition were inspired by his correct understanding of chemical equilibrium, but because he was wrong about definite proportions for pure compounds, this contribution to scientific knowledge was not recognized for a long time.)
Studying the data of people like Proust and Lavoisier, Dalton noticed a remarkable pattern. Carbon and nitrogen both combine with oxygen in several different definite ratios to form several different products. Likewise sulfur and phosphorus, which is why there are sulfates and sulfites. For example, carbon combines with oxygen in the ratios of 3:4 and 3:8 (3g carbon for each 4g of oxygen). Of course, 2 x 4 = 8. One compound had exactly twice as much oxygen as the other. Nitrogen combines with oxygen in ratios of 7:4, 7:8, and 7:16. In each case, the amount of oxygen doubles. This is called the law of multiple proportions.
These laws all made good sense if each element has atoms that combine as whole atoms. An atom is a very small, distinct thing, like a ball. Now we know that atoms can be divided into smaller parts, but an atom is the smallest amount of an element you can have, because if you divide it into smaller parts, the properties will change.
The Atomic Theory
John Dalton's atomic theory:
• Matter is composed of atoms
• Atoms come in different types, called elements
• Atoms of each element have a distinct mass
• Each atom of a given element is identical to every other atom of that element
• Atoms are not created, destroyed or changed when chemical changes occur
In more modern terminology, we say that atoms of different elements are combined to make molecules. Chemical reactions change how the atoms are combined, but the number of each type of atom doesn't change. | textbooks/chem/General_Chemistry/General_Chemistry_Supplement_(Eames)/Chemistry_Basics/Atomic_Structure.txt |
Skills to Develop
• Describe the three types of radioactivity
• Understand the difference between atoms. ions, and isotopes
• Define the subatomic particles that make up atoms
At the end of the last section (Valence), we mentioned the problem of the masses of iodine and tellurium, which were "out of order" in the periodic table. This was only one of several problems facing chemists at this point. Another was the rare earth elements (the lanthanoides and actinoides, at the bottom of the periodic table). The rare earth elements were extremely similar to each other, so it was very hard to separate them, determine the masses, and fit them into the periodic table.
Another problem was ions. Ions were charged chemical entities that moved through solutions carrying electric current when electricity was applied. Chemists knew they existed, but could not explain how they formed. Cations were positively charged, while anions were negatively charged. Hydrogen was known to form positive ions with a particular charge. But where did the charges come from? Berzelius had thought that the atoms themselves were charged, but this didn't make sense after Avogadro's hypothesis was accepted, since in that case the two O atoms in O2 would repel each other, not form a molecule. Of course, this was mostly a problem to physical chemists, because physicists mostly still didn't believe in atoms.
Experiments by Davy, Faraday and others showed that solids usually conduct electricity without changing themselves; liquids are decomposed by electrical current, and gases conduct badly and only at high voltage, but become better conductors at low pressure. Glass tubes of low pressure gas glow; examples include neon lights in store windows, fluorescent light bulbs, and yellow sodium lamps sometimes used at street lights. When glass tubes full of very low pressure gas were made by expert glassblower Geissler after 1850, a glowing spot would appear somewhere on the glass when the tube was connected to a high voltage. This spot could be moved by a magnet, suggesting that it was negatively charged. A beam or ray was coming out of the cathode (the electrode to which cations move), and this beam caused the glass to glow. (This is exactly how old TVs work, from before there were flat screens.)
But what was the beam that came out of the cathode? Some thought it was negatively-charged particles, while others (especially physicists) thought it was a wave. Thomson convinced himself that the beam is a negatively charged particle, which can be moved using electric or magnetic fields. He estimated that the particle has a charge/mass ratio 1000x greater than the hydrogen ion, but with the opposite charge. The ray that comes out of the cathode is the same whatever the material, and he found that you can generate the same particle by heating metal filaments very hot or exposing a metal surface to UV light (the light that gives you sunburns).
What are cathode rays? Electrons! When Thomson suggested in 1897 that either they had a much higher charge or a much smaller mass than hydrogen ions, many people were skeptical. However, careful measurements by Millikan (starting 1908, continuing for almost 10 years!) of falling droplets of mist in an electric or magnetic field revealed that electrons have the same charge as hydrogen ions, but are almost ~1/2000 less massive, so in fact Thomson guessed electrons were twice as big as they really are.
In summary, electrons have an equal but opposite charge to hydrogen ions, but hydrogen ions weigh 2000x more than electrons. Electrons are part of every kind of atom.
Other than electrons, what are atoms made of? X-rays were discovered in 1895 when Roentgen noticed a ray that passed through the glass of his cathode ray tube and developed photographic film placed in its path, even if it was wrapped in dark paper. Inspired by this discovery, Becquerel studied the fluorescence (this means glowing after being exposed to light) of minerals and whether this could develop film through dark paper. He used sunlight to make the minerals fluoresce, but discovered that some minerals fluoresce even without sunlight. Minerals that contain the elements uranium and thorium have this property, which is called radioactivity. Three types of radioactivity were discovered, α (positive charge), β (electrons, same as cathode rays) and γ (light, similar to x-rays). This radiation is a property of the element, and is not changed by chemical combinations. It was also discovered that radioactive elements decay into products that closely resemble known elements, except that the masses are different. These decay products have exactly the same chemical properties as the known elements, and once mixed together, they can't be separated by chemical means. But the masses were different. The evidence against Dalton's "identical and indivisible atoms of each element" was getting strong.
Rutherford studied the α-radiation, and found that it had a positive charge, and an e/m (charge/mass) ratio 1/2 that of a hydrogen ion. Later he proved that it is a charged form of the element helium (He).
Finally, using an early version of the mass spectrometer, which measures e/m by accelerating charged particles in an electric or magnetic field, Aston determined that neon (Ne) atoms have masses of 20 or 22, not 20.2, then the accepted value. Here's another case where knowing the precision of your experiment is important: if he couldn't tell the difference between 20.0 and 20.2, this wouldn't have been useful! The two versions of Ne are called isotopes. Isotopes are atoms of the same element with have the same chemical properties, but have different masses.
What is the structure of an atom? At this point, it was clear that Dalton had not been entirely correct, because atoms of one element could exist in different forms, or isotopes. It was also becoming clear that atoms could be divided into smaller parts, like electrons. But how were the electrons and the positive parts (like the part of He that is α-radiation) arranged, and what was the positive particle? It seemed like there was a lot of empty space in atoms, because cathode rays could pass through aluminum windows. Many theories were considered, but ultimately two of Rutherford's students, Geiger and Marsden, found an answer. They observed that α-rays directed at a thin gold (Au) foil usually passed through the foil (though not usually quite straight), but sometimes the foil knocked them back in the direction they came from. These α-particles must have bumped into (or been repelled by a large positive charge in) something very heavy, to have so large a change in direction. In 1911, Rutherford proposed that atoms consist of a small dense particle, the nucleus, which contains most of the atom's mass and all of its positive charge, with electrons moving through a large empty space around the nucleus. But they were a little confused about the mass and charge of the nucleus.
In 1913, a very young chemist named Henry Moseley measured the fluorescence wavelength produced when x-rays shine on pure elemental samples. He discovered that if you graph the square root of the fluorescence frequency vs the element's location in the periodic table (such as H, 1; C, 6; Hg, 80) you get a straight line. The location in the periodic table is called atomic number. This confirmed the ordering of Te/I in the periodic table and a few other similar problems. He was also able to determine where undiscovered elements were missing in the table, and clarify identification of rare earth elements, which was very difficult because they were so similar. He was 26 years old, and he died a year later in WWI. Moseley's work showed that the ordering of the periodic table is based on the atomic number, which is the number of positive charges in the nucleus and also the number of electrons. The unit of positive charge is a hydrogen nucleus, or proton.
The mass of the nucleus continued to be confusing even after the amount of positive charge had been determined by Moseley's atomic numbers. The atomic weights were about double what they should be if the nucleus held only protons. A proposal to explain this was that there were some electrons in the nucleus, canceling out the charge of half the protons. Later radiochemistry experiments showed the existence of the neutron, which has about the mass of a proton and no charge. This was understood to be the extra mass in nuclei. Now, though, we know that neutrons can decay into protons and electrons (and some other less familiar particles), so the earlier idea wasn't far off.
Summary
Atoms are composed of protons, neutrons, and electrons. Protons and neutrons make most of the atom's mass and all of its positive charge in the nucleus. Protons are subatomic particles that contain a positive charge. Essentially, they are hydrogen nuclei. Neutrons are subatomic particles that have the same mass as a proton but not charge. Electrons are subatomic particles with a negative charge that are found outside of the atom's nucleus. An element's atomic number provides its location on the periodic table as well as describe how many protons are in an atoms of that particular element. Ions are atoms that are either positively or negatively charged due to the number of protons or electrons in an atom not being equal to each other. Isotopes are atoms that have a different number of neutrons in the nucleus than normal. There are three main types of radioactivity: α , β, and γ. α (Alpha) radiation involves the emission of a helium nucleus, which is two protons and two neutrons. β (Beta) radiation involves the emission of an electron from the atom's nucleus. γ (Gamma) radiation involves the emission of high-energy electromagnetic radiation. | textbooks/chem/General_Chemistry/General_Chemistry_Supplement_(Eames)/Chemistry_Basics/Discovering_Sub-atomic_Particles.txt |
Skills to Develop
• Distinguish between static and current electricity
• Describe the electrostatic force and Coulomb's Law
• Define ionic compounds, an electrolyte, and the components of an electrolyte
• Describe electrolysis and Berzelius' dualistic theory
The earliest studies of electricity focused on electrostatics. Static electricity can be produced when certain materials are rubbed together, like silk or hair on some metals or plastics (this is called the triboelectric effect). This leads to a separation of charge, with positive charge on the silk or hair and negative charge on the metal or plastic. You've probably noticed how this can lead to things sticking to each other if they have opposite charges, and you can also observe that the same charges will repel each other. Benjamin Franklin (one of the few early American scientists) proposed that the positive and negative charges resulted from having either too much or too little of the same "electrical fluid".
The electrostatic force had been studied by many scientists, some of whom suggested that it followed a law similar to Newton's law for gravity, but Coulomb gets credit for this, because he did many experiments that improved understanding. In 1785, he published the law in its current form. The equation is:
$F=\frac{kQq}{r^{2}}$
where F is the force, k is a constant, Q and q are two charges, and r is the distance between Q and Q. Like charges (+/+ or -/-) repel and opposite charges (+/-) attract.
Current electricity, like the electricity used in any electrical device today, was discovered a little later. Galvani was studying physiology, and noticed that the legs of dead frogs twitched when in contact with two different metals. He attributed this to "animal electricity." Volta did many more experiments and discovered that animals were unnecessary for the effect. He put a piece of paper soaked in saltwater (the electrolyte, because it conducts electricity) between disks of two different metals, then connected the metals with wire and noticed that electrical current flowed. When he stacked these on top of each other, alternating the metals, the effect increased. This device for generating current came to be called a "voltaic pile."
Volta published his pile reports in 1800. That same year, two English scientists published their results of splitting water into hydrogen and oxygen using a Voltaic pile. This is an example of electrolysis, which means using electricity to break chemical compounds.
Humphry Davy was a scientist and popular lecturer who had discovered "laughing gas" or nitrous oxide, which is still used by doctors and dentists. He did many electrochemical studies after 1800. He noticed that during electrolysis of various compounds, hydrogen, metals, or bases appear around the negative pole, and oxygen or acid appears around the positive pole. Based on this, he guessed he could use electricity to break chemical bonds. The strong bases KOH and NaOH (potassium hydroxide and sodium hydroxide) had been known for a long time. (A base neutralizes an acid.) Sodium and potassium were believed to be elements but had not been isolated. Davy was able to isolate them by melting and electrolyzing the solid sodium hydroxide and potassium hydroxide. The solid bases didn't conduct electricity, but once melted they did conduct, and then they separated into shiny metal and a gas.
K and Na are alkali metals. When Davy isolated them, he found that they were light, soft metals that react vigorously and spontaneously with water and air. So they have to be stored under oil. Davy also isolated magnesium, calcium, strontium, and barium (Mg, Ca, Sr, and Ba) using a slightly modified procedure involving mercury. These are the alkaline earth metals. They are also soft, light metals, but they are not quite as reactive as the alkali metals.
Berzelius was a very influential chemist who worked very hard and very carefully to determine atomic weights, without using Gay-Lussac's law or Avogadro's hypothesis, and whom we mentioned earlier because he introduced the idea of isomerism. He also did many electrochemical studies and took Davy's hypothesis, that chemical and electrical attraction are the same force, much farther. His dualistic theory said: elements have positive or negative polarity, and chemical reactions partially neutralize that polarity. For example:
Cu (positive) + O (negative) → CuO (approximately neutral)
He used the terms electropositive and electronegative, which are still used today with a somewhat different meaning. He thought oxygen was the most electronegative element, while metals were generally electropositive. He also considered the element polarity to be on a spectrum, so that sulfur, for instance, was positive with respect to oxygen and negative with respect to metals. Thus it could combine with both; this is very different from the modern understanding of Coulomb attraction because he did not have a clear sense of neutrality.
Ionic compounds are usually composed of a metal and a non-metal. For instance, salt, and many rocks, are ionic compounds. They are different from molecular compounds because they do not have distinct molecules that vaporize as a unit. They are held together by charges, not quite as Berzelius thought, but close enough that his theory worked all right for them.
However, Berzelius' theory did not work well for molecular compounds! He said Avogadro must be wrong because two oxygen (O) atoms would have the same polarity and thus repel each other. Now we know that chemical bonding is more complicated than Berzelius realized. O atoms don't repel each other, and do form a diatomic molecule.
Berzelius' theory delayed the progress of science for about 50 years because he didn't believe Avogadro's hypothesis, which was needed for the next big breakthrough in chemistry.
Faraday was born to a poor family, and discovered chemistry when working in a book-making shop, where we worked on Jane Marcet's Conversations on Chemistry (one of the most popular textbooks at the time, it is surprising that it was written by a woman). He became Davy's assistant, and became a great experimentalist. He helped invent the words used for electrochemistry, such as ion, anion, cation, electrolyte and electrolysis. At that time, anion and cation meant the parts of the electrolyte (the salt used to make water conductive) that appeared at each electrode. Faraday studied the amount of current necessary to produce an amount of an electrolysis product. For instance, the current that produced 1g of hydrogen produced 8g of oxygen, 36g of chlorine, 125g of iodine, 104g of lead or 58g of tin. Faraday called the O, Cl and I anions, meaning that they were produced at the positive pole, and Sn and Pb cations, meaning that they were produced at the negative pole. These numbers could have solved the atomic weights problem, just like Avogadro's hypothesis, but Faraday wasn't interested in atomic weights or theories, and Berzelius, who was, didn't believe Faraday's electrical laws, just like he didn't believe Avogadro's hypothesis. | textbooks/chem/General_Chemistry/General_Chemistry_Supplement_(Eames)/Chemistry_Basics/Electricity_and_Electrochemistry.txt |
Skills to Develop
• Describe Gay-Lussac's Law and Avogadro's hypothesis
Dalton's atomic theory led to a new question: if each element has atoms with a characteristic mass, what are those masses? If water is 88.9% O, 11.1% H, what is the atomic mass of O in terms of H (if you assume H is 1.00)? To answer, you need the formula: H2O (2 H atoms for 1 O atom) (11.1)/2 = 5.55 → (88.9)/5.55 = 16.0 (this is the atomic weight of O, assuming H is ~1)
But early chemists didn't know the formulas, because they didn't know the atomic weights! Dalton assumed that the simplest formula (example: HO) was right, but this was usually wrong! How could they figure out the formulas?
In 1808, Gay-Lussac published results that showed what volumes of gases combined with each other in chemical reactions. For instance (O is oxygen, H is hydrogen, N is nitrogen, L is liters):
• 2 L H + 1 L O → 2 L water (steam)
• 3 L H + 1 L N → 2 L ammonia (NH3)
• 2 L "carbonic oxide" + 1 L O → 2 L "carbonic acid" (using the old names for the compounds)
• 1 L N + 1 L O → 2 L nitric oxide (NO)
Based on Gay-Lussac's Law, we can guess the following: equal volumes of gas at the same temperature (T) and pressure (P) have the same number of "particles".
What are these "particles?" Many of them are molecules, strongly-bonded collection of atoms. Molecules usually remain intact when vaporized into the gas phase. Molecular compounds (compounds containing molecules, as compared to ionic compounds discussed later) are usually made of non-metal elements such as C, O, S, P, H, Cl, etc.
How was Gay-Lussac's law received? Dalton didn't believe in it because the densities of the gases seem wrong. Oxygen gas is denser than steam (water gas), even though water is oxygen plus hydrogen. Also, Gay-Lussac didn't push the conclusions of his law as far as he could have because Berthollet (who believed that combining ratios of elements could vary) was his mentor.
In 1811, Avogadro explained the problems with Gay-Lussac's law by saying that equal volumes of gases (at the same T and P) have the same number of molecules. The elemental gases were present not as single atoms but as diatomic molecules (such as H2, O2, N2). Now the equations from Gay-Lussac's law are:
• 2 L H2 + 1 L O2 → 2 L H2O
• 3 L H2 + 1 L N2 → 2 L NH3
• 2 L CO + 1 L O2 → 2 L CO2
• 1 L N2 + 1 L O2 → 2 L NO
This resolves the issue of gas density (O2 is denser than H2O because H2 weighs less than O). Avogadro's hypothesis could have cleared up all the confusion about formulas, and allowed good atomic weight calculations. But people either ignored him or said it was impossible.
Introduction
Skills to Develop
• Understand some of the historically fundamental aspects of chemistry
Chemistry is different from physics. The chemical theories that we use to understand and predict aren't written in math. They also aren't based on anything we can see directly, like a ball falling. Instead, they are based in images, patterns, symmetry and imagination. How do we know what metaphors we can use to imagine, and predict accurately, things that we can't see? When you think about it, it's amazing how much chemists were able to figure out.
John Dalton proposed his atomic theory around 1805. He said:
1. All matter is composed of atoms
2. Atoms come in different types, called elements
3. Atoms of each element have a distinct mass
4. Each atom of a given element is identical to every other atom of that element
5. Atoms are not created, destroyed or changed when chemical changes occur
Scientists scoffed, laughed at him, or at least remained skeptical, for at least 50 years after that, because how could he know? But he was almost completely right.
A little later, in 1830, Jons Jacob Berzelius proposed that even if you have the same composition (the same number of atoms of each element) you can have two different molecules (a combination of atoms), because the atoms are arranged differently. The first example discovered was silver fulminate (AgCNO, very explosive) and silver cyanate (AgOCN, a non-explosive grayish powder).
And then in 1874 Jacobus Henricus van't Hoff proposed that when carbon atoms form bonds to four other atoms, those other atoms around them have a particular arrangement in space: a tetrahedron. A critic said:
"A Dr. J. H. van't Hoff who is employed at the Veterinary School in Utrecht appears to find exact chemical research unsuited to his tastes. He finds it more suitable to mount Pegasus (obviously loaned from the Veterinary School) and to proclaim ... how, during his flight to the top of the chemical Parnassus, the atoms appeared to be arranged in the universe." (In ancient Greek mythology, Pegasus is a horse with wings, and Parnassus is a mountain associated with art and knowledge because the Muses were said to live there.)
You can tell from that how weird it seemed, at the time, to claim a particular spatial arrangement of tiny particles that nobody had ever seen. But van't Hoff, like Dalton and Berzelius, was right. How did they manage this? That will be one of the questions we answer in this class.
In fact, before the physicists had admitted that atoms exist, before they had been proven directly, chemists had already published many chemical structures showing how atoms were arranged in molecules.
Summary
Chemistry is the study of the composition and structure of chemical substances and the changes that they undergo. An element is one type of atom. An atom is the smallest particle of an element that retains all of the element's properties. A molecule is a combination of atoms of the same element. A substance's chemical composition involves having a piece of the substance being made of the same number of atoms as well as the the same type(s) of atoms. | textbooks/chem/General_Chemistry/General_Chemistry_Supplement_(Eames)/Chemistry_Basics/Gay-Lussac%27s_Law.txt |
Skills to Develop
• Understand why Lavoisier is sometimes called "the Father of Modern Chemistry"
• Distinguish accuracy from precision
So what happened to turn alchemy, which was like magical potion-brewing in Harry Potter, into the science of chemistry? It was measurement. Careful, careful measurement of quantities, such as masses, volumes, densities, temperatures, pressures.
An early hero of measurement was Antoine Lavoisier. He was one of the first true chemical scientists. He conducted careful experiments, and tried to draw no conclusions except those required by his data. He said fact, idea, and word should be as closely connected as possible: that you can't improve your language without improving your thinking, and you can't improve your thinking without improving your language. So he pioneered a systematic chemical nomenclature that is essentially what we use today. Remarkably, if you read his text, written in 1789, intended to introduce chemistry to beginners, much of it is still perfectly understandable and even correct by modern standards.
Lavoisier first describes the states of matter: gases, liquids and solids. He points out when a solid material is heated, it tends to expand, becoming first a liquid, which takes up a constant volume, but can be poured, unlike a solid. More heating, and it becomes a gas, which he describes as elastic because it will expand or compress to different volumes depending on the pressure. Unlike the Greek philosophers, he understood that this is a physical change, not a chemical change, and he has a good submicroscopic-scale intuition of what's happening: the particles of the material don't change, they just get further apart.
He recognized the following as elements: oxygen, nitrogen, hydrogen, sulfur, phosphorus, chlorine and fluorine (although he did not know their elemental forms), carbon, iron, copper, silver, gold, mercury, lead, tin, antimony, arsenic, bismuth, cobalt, manganese, molybdenum, nickel, platinum, tungsten, and zinc.
He burned sulfur and phosphorus and charcoal (carbon) and made careful observations, often using the bell jar over a bucket of mercury as shown in the drawing from his book, Figure 1. This is an example of a chemical change or chemical reaction, in which reactant chemicals turn into different product chemicals. If you light the sulfur in the dish labeled D under the bell jar of air, it burns until it goes out leaving some extra sulfur. The air remaining in the jar is no longer good for breathing. If you put a mouse in the jar, it will die, just as the flame did. This demonstrates the concept of limiting reactant. The reaction or burning stopped when it ran out of oxygen, leaving primarily nitrogen (and a few trace other gases) in the jar. Priestley, another scientist, showed him how to prepare pure oxygen gas, and he used this to do many burning experiments as well.
Lavoisier was obsessed with measurement. He developed elaborate apparatus for measuring everything. He would burn phosphorus, as shown in Figure 1, and observe the formation of a white flaky product. The phosphorus (the reactant in this case) wasn't water soluble, but the product was, so he collected the product very carefully, separating it from the unreacted phosphorus by washing with water. After drying, he could measure how much phosphorus had burned, how much oxygen had been consumed (because he knew the density of oxygen gas), and how much product had formed. He found that the mass of product was the sum of the masses of reactant consumed, in every experiment. This is the law of conservation of mass (which, actually, some earlier alchemists and chemists had also used). He also observed that the phosphorus has no taste, but the product, which he called phosphoric acid, is sour. He knew from these experiments that in many cases elements combine in only certain proportions, and also that oxygen can combine with sulfur, phosphorus, etc in two different ratios. He gave us the terminology we still use today: sulfuric acid is composed of sulfur and more oxygen, sulfurous acid is composed of sulfur and less oxygen. -ous means less oxygen; -ic means more oxygen. See the nomenclature page for details.
Lavoisier paid close attention to accuracy and precision. For instance, in the experiment we just described, he measured the volume of gas in the bell jar, before and after the reaction, but noted that after the reaction, you must wait until the temperature returns to what it was when you measured originally. If the gas is hot when you measure its volume after the reaction, it will have expanded, and your standard density will not apply. This would introduce a systematic error into the measurements: each time you perform the experiment, you will think that there is more gas leftover than there actually is, and your measurement won't be accurate. If the average result of your experiment is near the correct value, it is accurate. However, if your experiment gives very different numbers each time, even if the average is correct and the experiment is accurate, it is not precise. Precision is the difference between meeting "around 2 o'clock" and meeting "at 3 minutes and 27 seconds before 2 pm." Precision is how specific you are, how much detail you use. Lavoisier also helped develop the system of units (kg, L, m) that are currently in use in Korea and many other countries.
Overall, while he didn't do very many original experiments that nobody else had done before, he did his experiments very carefully, so they were as accurate and precise as possible, and then he thought about them clearly and created words to describe the chemicals and ideas that helped make everything clearer. If you read a chemistry textbook written before Lavoiser, you will be very confused because the names for chemicals would be based on history (and would sound like they came from Harry Potter), rather than being based on what the chemicals are. If you read a chemistry text written after Lavoiser, you will recognize the language as similar to what we use today.
Summary
Accuracy describes how close a measured value is to the actual value. Precision describes how well a group of measured values agree with each other. The law of conservation of mass states that matter can neither be created nor destroyed by a chemical or physical process. This results in the sum of the masses of reactant consumed in any experiment is equal to the mass of product. Chemical changes involve changing a substance's chemical identity such that new substances are formed. Physical changes involve altering a substance without changing its chemical identity. Combustion and rusting are two examples of chemical processes while boiling and melting are examples of physical processes. Chemical reactions involve turning reactants, chemicals that get consumed in the process of chemical change, into products, chemicals produced through the process of chemical change that have a different composition from the reactants. A limiting reactant determines, or limits, the amount of product that can be produced from a chemical reaction.
Meet the Periodic Table
Skills to Develop
• Associate family name to description (especially for groups 1 - 2 and 16 - 18)
The Periodic Table is an essential tool for chemists. I have provided a simple version here. It shows the symbols, atomic numbers, and average atomic masses for each element. If you point at a symbol, it will show the element name. You can search for an element by name or symbol and it will be highlighted so it's easy to find.
How do you use the periodic table? It can help you predict many important properties of elements. To make these predictions, you will need to know a little about the different families or groups, which are the columns of the table. The term for the rows is period. Here is some info about the important groups.
Group Number Family Name Description
1 Alkali Metals soft, extremely reactive metals, valence 1, almost always M+ ions
2 Alkaline Earth Metals soft, less reactive metals, valence 2, almost always M2+ ions
13 Boron Group non-metals and metals, valence 3
14 Carbon Group non-metals and metals, valence 4
15 Pnictogens non-metals, valence 3
16 Chalcogens non-metals, valence 2, often X2– ions
17 Halogens non-metals, very reactive, volatile elements, valence 1, usually X ions
18 Noble Gases very unreactive monatomic gases, valence 0
3-12 Transition Metals metals with multiple valences and ionic forms, initially hard to fit into periodic table, many exist as M2+
*, ** Rare Earth Metals (lanthanoides, actinoides) similar to each other, most have M3+ ionic form
To learn your way around the table, try going to this much fancier periodic table. Notice how metals are on the left and bottom of the periodic table, while non-metals are on the right and top. Metals are shiny, and they conduct heat and electricity. Non-metals don't conduct, and are often softer or easier to break than metals. Some elements are called metalloids because they are in between metals and non-metals, and you can see that the metalloids are also in between the metals and non-metals in the periodic table. The most reactive elements are on the edges of the table (groups 1 and 17), and the most reactive non-metals are O and F, in the top right corner. | textbooks/chem/General_Chemistry/General_Chemistry_Supplement_(Eames)/Chemistry_Basics/Lavoisier.txt |
Chemical nomenclature is the names we use for chemicals. For instance, H2O is called "water", and CH4 (the gas you burn in a stove) is called "methane." You should learn the chemical nomenclature here on this page now, so that you will be able to understand when it is used.
The Basics
Here is some important info about how we write chemicals.
1. Elements have symbols of one or two letters. The first is a capital letter (ABC). If there is a second letter, it is a lower-case letter (abc). For instance, "m" is one unit, and "M" is a different unit. "K" is the symbol for one type of constant, and "k" is the symbol for a different type. You need to remember that capital letter symbols are usually different from lower-case symbols. For instance, Co is cobalt, a metal element next to iron, and CO is carbon monoxide, a poisonous gas made of one carbon atom and one oxygen atom.
1. We write the charge of a chemical using a superscript, which looks like this: H+or H. If we write just H, that means an H atom, which is one proton and one electron. H+ means 1 H atom – 1 electron, so it means just one proton, also called hydrogen ion. If we write H this means one hydrogen atom + one electron, so a proton and 2 electrons, also called hydride ion. If there's an number in the superscript, that says how many electrons are added or removed. For instance, Ca2+ is a calcium atom – 2 electrons, or calcium ion. S2– is sulfide, or sulfur + 2 electrons.
1. We indicate the number of atoms of a particular type using a subscript, like this: CO2. This means one carbon atom and 2 oxygen atoms. If we write O2 that means the oxygen molecule, which is two atoms of oxygen connected together. Sometimes people might write O2 to mean the same thing. If the number comes first, though, it has a different meaning. 2 O means 2 atoms of oxygen that aren't connected to anything.
1. If we want to show how many protons and neutrons are present in an atom, we can use the mass number, as a superscript before the element symbol, such as13C. This means carbon with (protons + neutrons) = 13. You can tell that this is different from the charge, because the charge will always include + or – and come after the symbol.
1. Most molecules or ions that are stable have an even number of electrons. If they have an odd number of electrons, this is called a radical. For instance, H is a radical, because it has one electron. Because this is unusual, it might be indicated with a dot, like this: H. For instance, water is H2O, and if you remove hydrogen ion, you are left with hydroxide ion, OH. If you remove Hfrom water, you are left with OH molecule, which is neutral. This is also called hydroxyl radical, written OH.
1. The phase of a substance is often indicated by a letter in () after the symbol. For instance, He is almost always a gas, written He(g). If it's a liquid (4.2K or below, less than -269° C) that is written He(l). You'll probably never hear about He(s), since it would be very hard to make it a solid. You might also see something written with (aq), which means "dissolved in water." For instance, NaCl(aq) means salt dissolved in water so there is no solid left. Or you might just see K+(aq), meaning potassium ions dissolved in water.
Elements
There are lots of elements and you don't need to memorize them all. Here are a few that you should learn right now, though, because they are common or important, so that you won't be confused when they are mentioned later. They are organized by their type.
1. Non-metals
• Light elements: the elements with smallest mass
• Hydrogen (H): exists as H2 or in combination with other elements, such as in water
• Helium (He): named after the sun, because it was discovered in the sun before being discovered on Earth (we'll explain how later); it doesn't react with anything
• Major gases in air
• Oxygen (O): we get most of our energy from reactions with oxygen, when we breathe or when we burn fuel; O2 is 21% of air
• Nitrogen (N): often the limiting factor for agriculture or population growth, even though N2 is 78% of air, because it only reacts under special circumstances.
• Halogens: reactive elements that make salts; common negative ions
• Fluorine (F): the lightest halogen and most reactive element in the periodic table, people say that it killed the first two chemists who tried to isolate F2
• Chlorine (Cl): part of normal salt, NaCl, it is common in the ocean and in your body
• Bromine (Br): one of only two elements that are liquid at room temperature, bromine is also found in salts and minerals
• Iodine (I): a soft, shiny silver solid that easily evaporates to a purple gas, iodine can be used to disinfect cuts and is essential for human brains, suggesting that humans may have evolved to live near the ocean, which provides sources of iodine in fish and seaweed
• Main group solid non-metals: non-conductive and usually soft materials
• Carbon (C): the element on which biology is based, also found in diamond, graphite, coal, and charcoal
• Silicon (Si): the basis of the electronics industry; also a main component of sand, glass, and most rocks
• Sulfur (S): a smelly yellow solid, used to make strong acid in industry, also common in minerals and essential to life
• Phosphorus (P): first isolated from urine, although common in minerals; essential for life, often glows
1. Metals: soft or hard, light or heavy, usually solid electric conductors
• Alkali metals: soft, light, common soluble positive ions
• Lithium (Li): the lightest alkali, used in batteries and anti-depressants
• Sodium (Na, you may know it as natrium): common in the ocean and salt
• Potassium (K, you may know it as kalium): also common, high concentration inside cells
• Alkaline Earth metals: like alkalis, but less so, less reactive, less soluble, positive ions more common in rocks, but also abundant in ocean
• Magnesium (Mg): common in rocks, essential for life, especially photosynthesis
• Calcium (Ca): common in biomaterials such as bone, tooth, and shells, also essential for muscles
• Main group metals
• Aluminum (Al, also called aluminium): requires lots of energy to produce the metal from the mineral sources, but very common and useful metal
• Tin (Sn, from Latin stannum): used since ancient times, especially in alloys such as bronze; still used in solder and many other applications
• Lead (Pb, from Latin plumbum): very heavy, soft, sweet-tasting toxic metal, commonly used since ancient times, now used to shield radiation and in bullets, among many other uses
• Transition Metals: a widely varied group, often characterized by complex chemical properties
• Iron (Fe, from Latin ferrum): most abundant element on earth, essential in steel, with complex reaction properties essential to life
• Copper (Cu, from Latin cuprum): less reactive metal, with characteristic colors, commonly used in coins and electronics
• Silver (Ag, from Latin argentum): used in jewelry, coins and other ornaments and utensils since ancient times, it didn't tarnish until after the industrial revolution, and now also used in electronics
• Gold (Au, from Latin aurum): used since ancient times in coins and jewelry, to color stained glass, also in dentistry and other applications
• Mercury (Hg, from Latin hydrargyrum): also called quicksilver, because it is a silver liquid, it is toxic but very important in the history of science; it may be familiar from thermometers
Common Positive Ions (Cations)
"[element name](charge in Roman numerals if needed) ion"
Cation is another word for positive ion. The common positive ions are the ions of the alkali and alkaline earth metals and ammonium, NH4+. The alkali metals form +1 cations, such as Na+ and K+. The alkaline earth metals form +2 cations, such as Ca2+ and Mg2+. The hydrogen ion, H+ is a very common cation. For these cations, you can call them "[element name] ion", such as sodium ion or calcium ion.
You'll also see transition metal cations or main group metal cations, but it is harder to predict what charge they will have, especially because some of them can have different charges, like iron, which is commonly Fe2+ or Fe3+. The charge on a transition metal cation can also be indicated using Roman numerals in parentheses, which looks like Fe(II) or Fe(III). The Roman numerals you will need to know for chemistry are:
1 2 3 3 5 6 7 8 9 10
I II III IV V VI VII VIII IX X
For cations that have uncertain charge, you should call them "[element name](charge in Roman numerals) ion." For instance, iron(II) ion or sometimes just Fe(II).
Sometimes people use special names for these ions, in which the higher charge ion is called "[name]-ic ion" and the lower charge ion is called "[name]-ous ion," such as ferrous for Fe(II) and ferric for Fe(III), or cuprous ion for Cu(I) and cupric ion for Cu(II). I think this is most common for Fe, and I've never heard anyone call nickel(II) nickelous ion because that sounds ridiculous.
Here's a list of common cations with less predictable charges:
• Ag+
• Cu+
• Cu2+
• Fe2+
• Fe3+
• Hg22+
• Hg2+
• Pb2+
• Sn2+
• Al3+
Elements not on the list above, that you may see soon anyway: zinc(II): Zn2+, cadmium(II): Cd2+, cobalt(II): Co2+, manganese(II): Mn2+, nickel(II): Ni2+, chromium(III): Cr3+.
Common Negative Ions (Anions)
"[base name] + (-ide,-ate, or -ite)"
Anion is another word for negative ion. Common negative ions are the halide ions, formed from the halogen elements: fluoride, F; chloride, Cl; bromide, Br; and iodide, I. As you may have noticed, the names of anions have "-ide" at the end when they are formed from elements. Other examples include oxide, O2–, sulfide, S2–, and nitride, N3–.
There are also many important polyatomic anions, which means anions that include more than one atom. These include toxic cyanide ion, CN, common hydroxide ion, OH, and peroxide ion, O22. Other important anions include acetate ion (C2H3O2), which is in vinegar, the chlorate ion (ClO3), the perchlorate ion (ClO4) which is often explosive, the nitrate ion (NO3), the carbonate ion (CO32) found in shells, the sulfate ion (SO42), and the phosphate ion (PO43). All of these end in "-ate", which means that they have more oxygen. Also, notice that "per-___-ate" means more oxygen than just "-ate", as in perchlorate.
Less common but still important are some "-ite" anions, which have less oxygen, such as nitrite (NO2), sulfite (SO32), chlorite (ClO2) and hypochlorite (ClO). Notice that "hypo-___-ite" means less oxygen than just "-ite" as in hypochlorite. Sulfite and nitrite are used to preserve foods. Sulfite salts are used in wine, dried fruit and preserved radish (mu). Nitrite salts are used in preserved meats.
One more rule says that if you take an anion like carbonate or sulfate and add one hydrogen ion, then you call that "bicarbonate" (HCO3) or "bisulfate" (HSO4). Or you might see it called "hydrogen carbonate" or "hydrogen sulfate." Note that because we added a hydrogen ion, the charge on the bicarbonate ion is one less than the charge on the carbonate ion. Also, note that "disulfate" is S2O72, quite different from bisulfate.
Chemical Nomenclature for Ionic Compounds
"[cation name] + [anion name]"
Ionic compounds are compounds that include at least two components, a positive ion and a negative ion. Often the positive ion is a metal element ion and the negative ion is a non-metal ion. To name an ionic compound, you usually just give the cation followed by the anion, such as "sodium chloride" or "ammonium nitrate." If the cation is the type that could have different charges, than you should say what the charge is, such as "mercury(I) iodide" or "cupric sulfate."
Chemical Nomenclature for Acids
"(hydro if -ide)[anion base name] + (-ic if -ide, -ate; -ous if -ite) + acid"
Acid usually means an anion combined with the hydrogen ion as the cation. For instance, HCl is a common acid, which is the hydrogen ion and the chloride anion. If the anion ends in "-ide" then usually the acid is called "hydro-___-ic acid" such as "hydrochloric acid" for HCl. You'll see this for all the "hydrohalic acids" which are H + a halogen, such as "hydrofluoric acid" or "hydroiodic acid." You might also see "hydrocyanic acid" for HCN. If the anion ends in "-ate" than you call the acid "___-ic acid," such as "sulfuric acid," which is H2SO4, or "nitric acid." HNO3. If the anion ends in "-ite" than the acid name is "___-ous acid." such as "hypochlorous acid" for HClO. Notice that earlier "-ic" and "-ous" meant more and less charge for cations, such as ferric and ferrous ions of iron. Now it also means more and less oxygen in acids.
Chemical Nomenclature for Non-metal Compounds
"(prefix, not mono)[less anion-like atom name] + (prefix)[more anion-like atom name]-ide"
Non-metal compounds are often called covalent compounds. They are named following a different rule from ionic compounds. You will need these "prefixes" which indicate how many of each type of atom are present:
1 2 3 3 5 6 7 8 9 10
mono di tri tetra penta hexa hepta octa nona deca
The prefixes come from Greek. You will put the element that is more left on the periodic table first, unless it is oxygen, which is always last unless it is in a compound with fluorine. This follows the same pattern as ionic compounds. In ionic compounds, the cation is written first, and you will notice that it is usually more to the left in the periodic table than the anion, which is written last. When you name covalent compounds, the atom that's more like an anion is written last. Fluorine is always most "anionic," and oxygen is next most "anionic," so they will always be last. (Fluorine is actually most electronegative, but we will study this concept much later, which is why right now I'm calling it "anion-like.") If both atoms are in the same group (same column of the periodic table) then the lower one is named first. Notice that the two most "anion-like," F and O, are in the upper right of the periodic table. The atom written second, that's more "anion-like" is named like an anion, with the "-ide" ending. For example, CO: carbon is on the left, so we can write "monocarbon monoxide." Actually people usually just call it "carbon monoxide." You can skip "mono" for the first element. For instance, SO3 is called "sulfur trioxide" and N2O4 is called "dinitrogen tetroxide." XeO2 is xenon dioxide, even though xenon is more to the right than oxygen, because oxygen is more like an anion than anything except fluorine. If the compound involves hydrogen, then you can leave out the prefixes, such as "hydrogen chloride" for HCl or "hydrogen sulfide" for H2S, because the numbers of each atom can be predicted as if it were an ionic substance. But actually many compounds of hydrogen have special names, such as "ammonia" for NH3, "methane" for CH4, "borane" for BH3, "silane" for SiH4 and "phosphine" for PH3. You should learn the first two of these now.
Summary
Key Info for Common Elements
Element name Symbol Atomic number Commonly found as...
Hydrogen H 1 H2(g), water (H2O), acid (H+(aq))
Helium He 2 He(g), He(l) if you want to make things very cold
Lithium Li 3 Li+ always, (aq) or in solids with anions, lithium metal Li(s) only in chem class
Carbon C 6 Covalent compounds, making 4 bonds
Nitrogen N 7 N2(g) in air, in ammonia (NH3(g or l)), in basic covalent compounds, in proteins
Oxygen O 8 O2(g) in air, in water, in rock and glass, usually combined with Si
Fluorine F 9 F(aq) or with cations in rock, in covalent compounds with carbon (non-stick pans)
Sodium (Natrium) Na 11 Na+(aq) or with anions in salts, sodium metal (Na(s)) only in chem class
Magnesium Mg 12 Mg2+(aq) or with anions in salts and rocks
Aluminum (Aluminium) Al 13 Al3+ with anions in rocks and salts, industrially made Al(s) metal
Silicon Si 14 Industrially made Si(s) in computer chips, Si(IV) oxides in sand, glass, most rocks
Phosphorus P 15 Phosphates: PO43, P2O74, etc. in rock, DNA
Sulfur S 16 S8(s), S2– or sulfate (SO42) in salts or rocks
Chlorine Cl 17 Cl(aq) or with cations in salts, Cl2(g) or ClO(aq) in disinfectants
Potassium (kalium) K 19 K+(aq) or with anions in salts, potassium metal (K(s)) only in chem class
Calcium Ca 20 Ca2+(aq) or with anions in salts and rocks
Iron Fe 26 Fe(s) metal industrially made, Fe(II) or Fe(III) oxides or sulfides and other minerals
Copper Cu 29 Natural Cu(s) metal, Cu(I) or Cu(II) salts or minerals, usually blue or green
Bromine Br 35 Br(aq) or with cations in salts
Silver Ag 47 Natural Ag(s) metal, Ag(I) or Ag(II) in salts or sulfide minerals
Tin Sn 50 Industrially made Sn(s) in alloys, Sn(II) or Sn(IV) salts and oxide or sulfide minerals
Iodine I 53 I(aq) or with cations in salts
Gold Au 79 Natural Au(s) metal, rarely Au(I) or Au(III) salts
Mercury Hg 80 Natural (but rare) metal Hg(l), Hg(II) sulfides and halides, Hg(I) exists as Hg22+
Lead Pb 82 Pb(s) in alloys, Pb(II) sulfide, carbonate and sulfate minerals, sometimes Pb(IV) salts or minerals
Common Polyatomic Ions
Name Formula Where you find it
Ammonium NH4+ Soluble salts, fertilizer
Cyanide CN Toxic, in some plant products and dyes
Hydroxide OH In bases and some minerals
Peroxide O22 In bleaches and disinfectants
Acetate C2H3O2 In vinegar
Perchlorate ClO4 Soluble salts and explosives, a strong acid
Chlorate ClO3 Similar to perchlorates but less stable
Chlorite ClO2 Disinfectants and bleaches, some explosive salts
Hypochlorite ClO Disinfectants and bleaches, most salts unstable
Nitrate NO3 Soluble salts, fertilizer, explosives, a strong acid
Nitrite NO2 Preservatives for food
Carbonate CO32 In rock, seashells, cement; a base
Bicarbonate HCO3 A base (baking soda), in soda, in blood
Sulfate SO42 In salts, plaster, detergent, a strong acid
Bisulfate HSO4 Food additives
Sulfite SO32 Salts, food preservatives
Phosphate PO43 Salts, rock, fertilizers, a strong acid, in ATP and DNA | textbooks/chem/General_Chemistry/General_Chemistry_Supplement_(Eames)/Chemistry_Basics/Nomenclature.txt |
Skills to Develop
• Understand what is the scientific method and how to apply it
You know what science is. A method is a way to do something. The scientific method is how scientific knowledge is produced. You have probably learned science before as a set of knowledge that is just given to you, and you have to memorize facts and procedures to follow to get answers. This way of learning that you have probably experienced is completely different from how science professionals think of science and do science. Now it is time to learn science like a scientist. Because scientists create new knowledge, they have to be able to think independently. Even if this seems very hard or unfamiliar, you can learn to make new scientific conclusions yourself.
The basic steps of the scientific method involve collecting observations about what actually happens, thinking about what the observations mean, making guesses about what will be observed in the future, and then making observations to see if the guesses are right. The observations are evidence or data, the results of experiments. The guesses are usually called hypotheses. If the hypotheses fit a lot of data and seem to work well, then they are called theories. In science, a law describes a pattern of consistent results. For instance, you know that when you drop things, they fall. That could be called the law of gravity. If we had a good explanation of why that happens, we would call that a theory of gravity. Gravity is easy to describe (in Newton's law, which describes how things move on Earth and how planets and moons and stars moves in space) but very hard to explain: how can objects that are very far apart feel a force towards each other?
The real test of whether something is science is whether it works reliably. An experiment is good and useful if you can repeat it (if someone different can repeat it) and get the same result. There will be some small differences in the data because measurements are always a little imprecise and there might be some small differences in the way the experiment was done, but the data should be the same within the "error range." Likewise, a law is good if observations always follow it within the context it describes. And a theory is good if it explains all the available evidence. For theories, it's also important that it be possible to prove the theory wrong, that it makes predictions that can be tested. Otherwise, it might be true but it isn't useful.
A useful skill for learning science is building "mental models" of how things work. A model is a bigger or smaller version of something, depending on the size of the original object, that shows all its parts and how they are related. For instance, an architect builds a small 3-D model of a new building before people start constructing the building. The architect also has complex computer models of the building showing how it stays up, how the doors don't bump into each other, etc. These models help the architect predict whether their design will be successful or not.
Summary
The scientific method is a structured approach to gathering and analyzing data, and drawing conclusions about various subjects or phenomena. Observations consist of evidence or data that may verify or falsify ideas about what actually happens in a given event. These ideas are usually hypotheses, educated guesses about how or why the particular event occurred which could lead to experiments to test the guesses. If a hypothesis seems reasonable and fits with a lot of data through rigorous testing, the hypothesis evolves into a theory. Good experiments have replicable results, meaning that other scientists should be able to come up with the same results or conclusions after repeating the experiment. A scientific law describes what will be observed in the future given a pattern of consistent results on the phenomenon. A model is a visualization of something to show how all of its parts are related.
Valence and the Periodic Table
Skills to Develop
• Understand and describe how the periodic table was first organized
Many chemists were interested in knowing the "chemical equivalents" of different substances. For instance, what mass of acid A neutralizes base B? These numbers were easy to measure and practically useful. But what about elements? For instance, what mass of element A reacts with 1 g of element B? Knowing these numbers could be very useful, but it was hard to relate the equivalent masses to the actual masses of the atoms because they didn't know the formulas. There were two good ways available to figure this out. One was Avogadro's hypothesis based on Gay-Lussac's law, which allowed chemists to relate the equivalent masses to equivalent volumes. The other was Faraday's law, which measured masses of elements produced by a set amount of current. However, Berzelius, who was working hardest on this problem, didn't believe either Avogadro or Faraday.
Berzelius' beliefs held up the progress of science for about 50 years. Eventually, Cannizzaro revived Avogadro's hypothesis at a big meeting of chemists in 1860. His paper explaining how to calculate molecular weights was distributed to everyone, including Julius Lothar Meyer, who wrote that when he read it "doubts disappeared and a feeling of quiet certainty took their place". Avogadro's hypothesis let chemists figure out atomic weights and formulas together. Once Cannizzaro convinced most chemists to accept it, chemists were able to study the actual atomic weights and formulas.
Scientists soon observed patterns in the valence of the different elements. Valence is the number of connections an atom tends to form. H is defined to have a valence of 1. For instance:
• methane, CH4: carbon atoms have a valence of 4
• water, H2O: oxygen has a valence of 2
• lithium oxide, Li2O: lithium has a valence of 1
• hydrogen sulfide, H2S: sulfur has a valence of 2
• aluminum oxide, Al2O3: aluminum has a valence of 3
By the 1860s, ~60 elements were known. Using Cannizzaro's atomic weights, Mendeleev and Lothar Meyer made a great discovery, the periodic law: If you arrange the elements by their atomic weights, there is a periodic repetition in properties such as valence. The modern version of this periodic arrangement is the Periodic Table.
Also, within a group (sharing a valence) properties like density, boiling point, heat capacity, etc follow a simple progression. Mendeleev used this to predict the properties of undiscovered elements.
Here's a smaller version of the periodic table that leaves out the elements Mendeleev and Meyer found most problematic (transition metals, rare earths) and the group that hadn't been discovered yet (noble gases). Notice how the the valences repeat every 7 elements when they are arranged according to atomic mass.
But there were some problems with the table. For instance, tellurium (Te) was clearly a chalcogen, in the oxygen family, and iodine (I) was clearly a halogen, based on their properties, but the weights were wrong. (Check the table!) Mendeleev said that the atomic weights must not have been determined correctly, but they were correct. | textbooks/chem/General_Chemistry/General_Chemistry_Supplement_(Eames)/Chemistry_Basics/Scientific_Method.txt |
Skills to Develop
• Write and interpret chemical equations
Chemical equations are a way to show what happens in a chemical reaction. A chemical equation looks something like this:
$A \rightarrow B$
In this case, A represents a reactant or reagent, and B represents a product. Usually one element doesn't turn into another element, so A and B might represent different molecules that are too complicated to just write the formulas. There might also be multiple reactants and products, like this:
$A + B \rightarrow C + D$
In this class, we will mostly study reactions of simple molecules, so we will use their formulas in equations, like this:
$Cl_{2} + Mg \rightarrow MgCl_{2}$
Often, we might want to show the state of the reactants and products, so we can use (g, l, s, or aq) to show if it is a gas, liquid, solid, or in a water solution (solutions in water are called aqueous solutions). For the example above, this becomes, assuming we do the reaction dry
$Cl_{2}(g) + Mg(s) \rightarrow MgCl_{2}(s)$
Notice that the numbers showing how many atoms of each element are in the formula go after the element, as a subscript like this. When we write chemical equations, usually we want them to be balanced equations, which means that they have the same number of each kind of particle on each side of the equation. Here's an example of an unbalanced equation:
$Br_{2}(l) + Na(s) \rightarrow NaBr(s)$
In this example, there are 2 bromine atoms on the left, and only 1 on the right. We always have to balance the number of each type of nucleus on each side, like this:
$Br_{2}(l) + 2Na(s) \rightarrow 2NaBr(s)OR like this:\frac{1}{2}Br_{2}(l) + Na(s) \rightarrow NaBr(s)$
Notice that we have put the number of sodium atoms needed for the reaction in front of the symbol for sodium, and not in a subscript after the symbol. This is called a coefficient, and it is different because it tells us how many of a molecule we need, not how many atoms are in the molecule, like this:
$2H_{2}(g) + O_{2}(g) \rightarrow 2H_{2}O(l)$
We also have to balance the number of electrons on each side. The easiest way to do this is usually to make sure the charges on both sides add up to the same number. For example, here's an equation that isn't balanced for electrons, even though it is balanced for nuclei:
$Mg(s) + Ag^{+}(aq) \rightarrow Mg^{2+}(aq) + Ag(s)$
To balance this equation we have to make both the charges and the nuclei balanced, like this:
$Mg(s) + 2Ag^{+}(aq) \rightarrow Mg^{2+}(aq) + 2Ag(s)$
Equations Need to Represent What's Actually Happening in a Reaction
This means that the number of each type of particle must be the same on both sides of the equation, because particles (nuclei or electrons) can't appear or disappear (except under special circumstances, which we call nuclear chemistry, so don't worry about that right now). This is why equations need to be balanced.
The second important thing is that the formulas in the equation need to match the actual molecules that are used or produced in the reaction. So if you are given the formulas, and you change them instead of changing the coefficients when you balance the equations, the equation and the reaction it represents has changed! You can't do this. On the other hand, if you are trying to write a chemical equation but you aren't sure what the formulas are, you can definitely use balancing to help you decide, and in this case you could change the formulas and the coefficients, as long as the formulas you use match all the information you have.
Balancing Equations to Avoid Mistakes
Here's an example of how balancing equations can help you avoid mistakes, based on some wrong answers my students gave on an exam last year. The question was, what reaction happens between calcium ions and carbonic acid in the ocean (a water solution)? To answer this, we have to translate it into formulas. Calcium ions: this is an alkaline earth metal, so Ca2+. Carbonic acid is H2CO3. Carbonic acid will dissociate a little bit, making some hydrogen ion (H+), some bicarbonate ion (HCO3), and some carbonate ion (CO32). Calcium carbonate is insoluble, so it will form a solid, ionic material. You could write the reaction like this:
$Ca^{2+}(aq) + H_{2}CO_{3}(aq) \rightarrow CaCO_{3}(s) + 2H^{+}(aq)$
This equation is balanced. What if you wrote it like this?
$Ca^{2+}(aq) +H_{2}CO_{3}(aq) \rightarrow CaCO_{3}(s)+ H_{2}(aq)$
Now it isn't balanced, because the charge on the left side is 2+ and the charge on the right side is 0. This means that there are 2 more electrons on the right than on the left, which is a problem! They can't come from nowhere, so this isn't a complete equation: we are missing the source of the electrons. The mistake may have been confusing 2H+ with H2 (the difference between 2 before H and 2 after H is important!), or it might have been confusing Ca2+ with Ca metal. For instance, this reaction is fine:
$Ca(s) + H_{2}CO_{3}(aq) \rightarrow CaCO_{3}(s)+ H_{2}(aq)$
However, in the context of the question, it would be very surprising to have Ca(s) in the ocean, because alkaline earth metals, like alkali metals, react with water. So we have lots of Ca2+, Mg2+, and Na+ ions in the ocean, but no atomic Ca, Mg, or Na in the ocean.
Outside Links
• Khan Academy: Balancing Chemical Equations (14 min)
• CrashCourse Chemistry: Stoichiometry (13 min) | textbooks/chem/General_Chemistry/General_Chemistry_Supplement_(Eames)/Chemistry_Calculations/Chemical_Equations.txt |
Skills to Develop
• Use dimensional analysis to calculate solutions with the correct units
A dimension is any measurable extent, such as length, time, and mass. Units help describe the measurement according to certain standards. In the metric system for example, a one-dimensional (1-D) length is measured in meters (m) a two-dimensional (2-D) area is measured in meters squared (m2), and a three-dimensional (3-D) volume is measured in meters cubed (m3). Other types of quantities (time, mass, temperature) are measured using different units because they have different dimensions. Analysis means to think about something, often focusing on one part at a time. Putting it all together, dimensional analysis means thinking about units piece by piece. Dimensional analysis can by to correctly go between different types of units, to catch mistakes in one's calculations, and to make many useful calculations in real life.
Essentially, dimensional analysis means multiplying by one. You collect a set of "conversion factors" or ratios that equal one, and then multiply a quantity that you are interested in by those "ones." For example, if you want to know how many seconds it would take to get from New York to Philadelphia, you'd do it like this:
First, using the express train it takes 2.5 hours to get to Philadelphia from a station in New York. Then, we know that 1 hour = 60 minutes and 1 minute = 60 seconds, so (1h / 60 min) = 1, and (1 min / 60 s) = 1. Now, all we have to do it multiply our starting number (2.5 h) by "one" twice, making sure that the units cancel correctly so that we have only seconds at the end.
$(2.5\; \cancel{ h}) \left(\dfrac{60\; \cancel{min}}{1\; \cancel{ h}}\right)\left(\dfrac{60 s}{1 \; \cancel{min}}\right) = 9.0 \times 10^3 s$
If each part is not put in the right place, the units will come out wrong. For example:
$\left(\dfrac{1}{2.5\;\cancel{ h}}\right)\left(\dfrac{1 \,\cancel{h}}{60\,m \cancel{ min}}\right)\left(\dfrac{1 \, \cancel{min}}{60 \,s}\right) = 1.1 \times 10^{-4} s^{-1}$
In this case, we put the starting quantity on the bottom, so we got s-1 when the units are canceled out. Here is an example of not being able to cancel out the units correctly:
$(2.5 h)\left(\dfrac{1 h}{60 min}\right)\left(\dfrac{60 s}{1 min}\right) = 2.5 s\cdot h^2\cdot min^{-2} The important part is that if you check the units to make sure that they come out right, you can be pretty sure you set the calculation up right! Here is an example of how dimensional analysis can help. A student was calculating initial velocity (v0) from this equation: \[d = (v_0)t + \dfrac{at^2}{2} \nonumber$
But the student had derived the equation incorrectly, and used this equation instead:
$v_0 = \dfrac{d}{t} - \dfrac{at^2}{2} \nonumber$
So the student had the wrong answer, but didn't know that because he just put the numbers for d, t, and a into his calculator using the wrong equation. If he had checked the units, he would have seen that (d/t) has units of meters per second (m/s) while (at2)/2 has units of meters (m).
Dimensional analysis is often useful when you want to estimate some quantity in the real world. For instance, maybe you want to know how much money you spend on coffee each month. If you spend \$5 per cup and have 2 cups per day, and there are approximately 30 days in a month, than you can set up a calculation just like those above to calculate dollars per month spent on coffee. This works for many important, less obvious situations, for instance in business, to get an approximate idea of some quantity. | textbooks/chem/General_Chemistry/General_Chemistry_Supplement_(Eames)/Chemistry_Calculations/Dimensional_Analysis.txt |
Skills to Develop
• Distinguish the parts of a solution
• Define the units for concentration
Concentration means how much of something there is in a given volume, kind of like density, except that it describes solutions. A solution is some compound, called the solute that is dissolved in another, more abundant compound, called the solvent. To be a solution, the molecules or ions of the solute must be separated from each other and surrounded by solvent molecules or ions. If very small bits of one compound are mixed into another compound, but not actually dissolved into molecules or ions, that is called a suspension (a solid in a liquid) or emulsion (two liquids). In a solution, both the solute and solvent can be any phase, solid, liquid or gas. The concentration of a solution is the amount of solute divided by the total amount of solution, usually. However, there are many different units used for concentration, and some of them assume that there is so much more solvent than solute that you can use amount of solvent instead of amount of solution.
Units for Concentration
There are many, many units used for concentration. Some are mostly used to describe concentrated solutions (that have a lot of solute) and others are mostly used to describe dilute solutions (that have very little solute). The most common unit in chemistry is molarity (abbreviated M), which is moles of solute divided by liters of solution.
$Molarity=\frac{(moles\; of\; solute)}{(liters\; of\; solution)}$
For example, if you dissolve 1 mol of NaCl in 1L of water, that is a 1 M (read "1 molar") NaCl solution.
Another common unit is weight %, which means
$Weight\; (Mass)\; \%=\frac{(mass\; of\; solute)}{(mass\; of\; solution)}\; \times 100\%$
This is often used for very concentrated solutions. For very dilute solutions, you'll see ppm (parts per million) or ppb (parts per billion):
$ppm=\frac{(mass\; of\; solute)}{(mass\; of\; solution)}\; \times 10^{6}$
$ppb=\frac{(mass\; of\; solute)}{(mass\; of\; solution)}\; \times 10^{9}$
Many compounds are important even at these very low concentrations. Some chemicals used in agriculture or industry are called "endocrine disrupters" and studies suggest that they can be dangerous for living things at the ppb level. (Example abstract of a scientific paper: note that ppm is described as "high concentration.") In semiconductors, which are what computer chips and LEDs are made of, ppm-level solid solutions enable the essential properties.
Example
HCl (hydrogen chloride) is a really nasty, dangerous gas, but dissolved in water it makes a convenient acid (hydrochloric acid) for many applications in the lab or in industry. When working with an HCl solutions in the lab, we often want to measure the mass of volume of solution used and know how many moles of HCl we added. Unfortunately, when you buy HCl, it usually comes as concentrated HCl, and the bottle will say something like "32% by weight, density 1.1593." Convert this to molarity.
To solve this, we need to think of it like a unit conversion. A good trick for dealing with % quantities is just to translate that into g/g, like this: 32 weight % = (32 g HCl)/(100 g solution). Now we do our usual unit conversion:
$\left(\dfrac{32\; \cancel{g\; HCl}}{100\; \cancel{g\; solution}}\right) \left(\dfrac{1\; mol\; HCl}{36.46\; \cancel{g\; HCl}}\right) \left(\dfrac{1.1593\; \cancel{g}}{1\; \cancel{mL\; solution}}\right) \left(\dfrac{\cancel{1000\; mL}}{1\; L}\right)=10\; M\; HCl$
Here we started with units of g HCl/g solution, and converted the g HCl to moles, and the g solution to liters using the density, which is (g/ml). Notice that the number of significant figures is fairly low. This is because the gas HCl can evaporate out of the solution, just like the water, so the concentration might change a little over time. If you want to know the concentration more precisely, see the next section on titrations to find out how to measure it.
Now suppose you want to make 2.5 M HCl using the 10 M HCl. You will need to dilute it, which means adding solvent to decrease the concentration. If you want to make 1 L of 2.5 M HCl, how much 10 M HCl do you dilute?
To solve this, we can still think about it like a unit conversion. We want 1 L of 2.5 M HCl solution. In the first step, we will convert to the number of moles of HCl needed to make that solution. In the second step, we will find how many ml of 10 M solution have this number of moles HCl.
$(1\; \cancel{L\; of\; 2.5\; M\; HCl}) \left(\dfrac{2.5\; \cancel{mol\; HCl}}{1\; \cancel{L\; of\; 2.5\; M\; HCl}}\right) \left(\dfrac{1000\; mL\; of\; 10\; M\; HCl}{10\; \cancel{mol\; HCl}}\right)=250\; mL\; of\; 10\; M\; HCl$
To make the solution, we will take 250 mL of 10 M HCl, and we will put it in a 1 L volumetric flask, and then add water slowly, with mixing, until the volume reaches the 1 L mark. The reason we do it this way, instead of just adding 750 mL of water, is that the density can change because of the solute. If we use the volumetric flask, we can be sure that we have exactly 2.5 mol HCl in exactly 1 L of solution. | textbooks/chem/General_Chemistry/General_Chemistry_Supplement_(Eames)/Chemistry_Calculations/Molarity.txt |
Skills to Develop
• Define a mole
• Distinguish theoretical yield from actual yield
Moles are a convenient unit used in chemistry to convert between amounts of a substance in grams and numbers of atoms or molecules. This is useful because we usually measure how much of a molecule is used or produced in a reaction by massing it, but as a chemical equation shows, the reaction will happen between atoms or molecules. For example, suppose we combine 1.0 g of calcium oxide (CaO) with 1.0 g of water (H2O). The product we get is Ca(OH)2. Here's the equation:
$CaO(s) + H_{2}O(l) \rightarrow Ca(OH)_{2}(s\; or\; aq)$
This is balanced. Thus every molecule of water reacts with one CaO formula unit (it's not called a molecule because it's an ionic solid, and each Ca2+ ion is surrounded with oxide ions that it interacts with equally). How much calcium hydroxide is produced by this reaction? Once all of one reactant has been used, whatever is left of the other will stop reacting, because of the law of definite proportions: we won't change the ratio of O:H:Ca in the product. So will we get solid calcium hydroxide with calcium oxide left over, or will we have water left over, and thus get Ca(OH)2(aq)? To answer this question, we can convert both masses (1 g of each) to the number of molecules or formula weights, but this would be inconvenient because the number would be very very big!
Instead, we use moles. A mole (abbreviation: mol) is like a pair, which means 2 of something. You can have a pair of people, a pair of apples, whatever. A mole is 6.022 x 1023 of something. This is a convenient quantity because it converts amu (atomic mass units) to grams. The atomic weight of carbon is (on average) 12.011 amu/atom. It is also 12.011g/mol. In other words, 1g = 6.02 x 1023 amu. Usually, a mol of a substance is a useful, practical amount, somewhere between a few grams and a few kg. The number of things in a mole, 6.022 x 1023, is called Avogadro's number, and abbreviated as NA. It is named after Avogadro, the scientist who proposed that a liter of any gas at the same temperature and pressure has the same number of molecules in it. To summarize:
1 mole of [thing] = NA things = 6.022 x 1023 things
So the way to answer the question above is to convert both quantities to moles. The maximum amount of product that can be formed is the smaller number of moles. The formula weight is just the sum of the atomic weights.
$(1.0\; \cancel{ g\; CaO}) \left(\dfrac{1\; \cancel{mol\; CaO}}{56.08\; \cancel{ g\; CaO}}\right) \left(\dfrac{10^{3}\; mmol\; CaO}{1\; \cancel{mol\; CaO}}\right) = 17.8\; mmol\; CaO$
$(1.0\; \cancel{ g\; H_{2}O}) \left(\dfrac{1\; \cancel{mol\; H_{2}O}}{18.01\; \cancel{ g\; H_{2}O}}\right) \left(\dfrac{10^{3}\; mmol\; H_{2}O}{1\; \cancel{mol\; H_{2}O}}\right)= 55.5\; mmol\; H_{2}O$
After we make 17.8 mmol (milimoles) of Ca(OH)2, we will use up all the CaO, so the reaction won't continue. The maximum amount of Ca(OH)2 possible to make is 17.8 mmol. If we wanted to know the theoretical yield (maximum mass of product) of Ca(OH)2, we could do it in a one-step calculation like this:
$(1.0\; \cancel{ g\; CaO}) \left(\dfrac{1\; \cancel{mol\; CaO}}{56.08\; \cancel{ g\; CaO}}\right)\left(\dfrac{1 \; \cancel{mol\; Ca(OH)_{2}}}{1 \; \cancel{mol\; CaO}}\right) \left(\dfrac{74.09\; g\; Ca(OH)_{2}}{1\; \cancel{ mol\; Ca(OH)_{2}}}\right) = 1.3\; g\; Ca(OH)_{2}$
Here, we knew that the limiting reactant (or limiting reagent), which is the reactant that will run out first, is CaO because the masses are the same, the coefficients in the equation are the same, and the formula weight of CaO is bigger than the molecular weight of water. So we start with the limiting reactant mass, convert it to moles (using 1 mol = 56.08 g), then "convert" between moles of CaO and moles of Ca(OH)2 using the coefficients from the balanced equation (1 mol of CaO produces 1 mol of Ca(OH)2), then we convert to g of Ca(OH)2 (using 1 mol = 74.09 g). This is just an example of using dimensional analysis to convert units. We check to make sure we have always multiplied by 1 (because 1 mol CaO = 56g CaO, so (1 mol CaO/56 g CaO)=1), and that the units cancel out to leave the correct final units (g Ca(OH)2), and we can be pretty sure that we got it right.
Here's a slightly more complicated example. This time, we add 2.0 g of water to 2.5 g of Li2O. This will produce LiOH as the major product. What is the most LiOH (in g) that could be produced, also called the theoretical yield?
To answer, first we need to write and balance the chemical equation. It's going to look pretty similar to the previous one, because this is a similar reaction.
$Li_{2}O(s) + H_{2}O(l) \rightarrow 2LiOH(s\; or\; aq)$
Which is the limiting reactant? The formula weights are 18.01 g and 29.88 g. Water is still in excess, which means it will be left over. Here's the unit conversion:
$(2.5\; \cancel{ g\; Li_{2}O}) \left(\dfrac{1\; \cancel{mol\; Li_{2}O}}{29.88\; \cancel{ g\; Li_{2}O}}\right)\left(\dfrac{2 \; \cancel{mol\; LiOH}}{1 \; \cancel{mol\; Li_{2}O}}\right) \left(\dfrac{23.95\; g\; LiOH}{1\; \cancel{ mol\; LiOH}}\right) = 4.0\; g\; LiOH$
Note
This time we used 1 mol Li2O producing 2 mol LiOH, using the coefficients from the balanced equation.
Outside Links
• Khan Academy: The Mole and Avogadro's Number (10 min)
• Khan Academy: Molecular and Empirical Formulas (15 min)
• CrashCourse Chemistry: Stoichiometry (13 min) | textbooks/chem/General_Chemistry/General_Chemistry_Supplement_(Eames)/Chemistry_Calculations/Moles.txt |
Skills to Develop
• Be able to understand how to interpret and report significant figures
Suppose you were measuring the diameter of the box below and you needed to report its circumference. You used a ruler and found the diameter to be 31 mm. Then you do the calculation on your calculator and get 97.389372261284 mm. So what do you say the circumference of the circle was? You certainly don't know the circumference more precisely than you knew the diameter, which was between 30 and 32 mm. Going through the calculations with these values would tell you that the circumference was between 94.25 mm and 100.53 mm. This is where significant figures come in handy.
Significant figures, often called sig figs, are the number of digits in a given value, or number. For instance, 18 has 2 sig figs, and 3.456 has 4 sig figs. However, both 10 and 1000 have only 1 sig fig. The reason is because the zeros have to be there to show what the number is, so they don't count as significant digits. What about 1001? It has 4 sig figs. We could have "rounded" it to 1000, showing that the last digit wasn't significant, but we didn't. This shows that the 1 on the right is significant, and so if the smallest digit (representing 1s) is significant, then the bigger ones (representing 10s and 100s) must be also. However, if the zeros are significant, then a period or decimal point would be added to the end. For example, if given a problem in which 20. mL are used, then there are 2 sig figs in the number 20.
You may forget to include the decimal point, particularly in your lab notebook when working in the lab. But you can assume that you used the standard measuring tools in the lab and use the significant figures based on the tools' accuracy. For example, a graduated cylinder could be accurate to 2 mL. So recording 20. mL would be like saying the measurement was between 18 - 22 mL. This means that recording the data with 2 sig figs would be correct. Generally, including an extra sig fig, especially in the middle of calculations is reasonable.
When measuring a quantity, the significant figures describe how precise the measurement was by listing the digits in a measured value which are known with certainty. For example, suppose you measure the length of a box with a normal ruler with increments, or markings, for millimeters (mm). You can be sure that your measurement is no more than 1 mm different from the real length of the box if you measured carefully. So, for instance, you could report the length as 31 mm or 3.1 cm. You wouldn't round to 3 cm or 30 mm since you were able to measure the box more precisely than that. But since you were only able to accurately measure to the nearest millimeter, the certainty of your measurement is within a millimeter of the reading. So you would read your measurement of 3.1 cm as 31 mm ± 1 mm.
Now suppose you wanted to know the length of the box much more precisely. To do that you will need a better tool. For instance, you could use a dial caliper to measure to the nearest 0.02 mm. Now you could report your length as, say, 31.14 mm, which means that your are certain your measurement was between 31.12 mm and 31.16 mm. If you measured with a ruler but wrote 31.1 mm, or 31.12 mm, people going through your numbers would probably think that you used a better tool than you actually did, so that would be almost dishonest
.
A dial caliper.
Going back to the original problem, one rule for using sig figs when doing multiplication with a measured number is to report the answer with the same number of sig figs and the number you started with: 2 sig figs, so 97 mm. The uncertainty is a little bigger than it was before, 97 ± 3 mm. But you shouldn't write 100 mm (1 sig fig) because you don't mean 0 - 200 mm, but you also don't mean 90 - 110 mm (sig fig change!).
If you wanted to say 100 mm with 2 sig figs, you would have to write it as 10. cm or use scientific notation and write it as 1.0 x 102 mm.
Some numbers are counted or defined, meaning that they are not measured. These are exact numbers. For instance, there are exactly 1000 grams (g) in 1 kilogram (kg), because that's the definition. Or if you use a volumetric pipette to add 1.00 mL of liquid twice, then the total amount added was 2 x 1.00 mL = 2.00 mL. You used the pipette exactly twice, so the 2 is exact, and you don't have to round to 1 sig fig (2 mL) for the total volume. | textbooks/chem/General_Chemistry/General_Chemistry_Supplement_(Eames)/Chemistry_Calculations/Significant_Figures.txt |
Skills to Develop
• Perform stoichiometry calculations
• Distinguish dynamic equilibrium from equilibrium
• Describe why actual yield may be less than theoretical yield
Stoichiometry is a general term for relationships between amounts of substances in chemical reactions. It also describes calculations done to determine how much of a substance will be used in a reaction, left over after a reaction, produced by a reaction, etc.
How do you use it?
The calculations of theoretical yield in the previous section are simple stoichiometry calculations. To do stoichiometry calculations, you'll need:
1. A balanced chemical equation: you may need to write the equation and balance it yourself, based on what you know
2. Formula weights or molecular weights (abbreviated FW or MW) for relevant compounds: you will often need to calculate these for yourself using the periodic table and the formulas; just add up the atomic weights according to the formula
3. Usually, you'll need to figure out how many moles react; to do this, you'll need to decide what is the limiting reactant, by comparing the number of moles of each compound that are present and the coefficients in the equation
4. Finally, once you have all the parts you can do a dimensional analysis-style unit conversion to find the answer. Make sure you understand what each step does and that each step is multiplying by 1. You might also need to do some other addition or subtraction, depending on what the question asks for, such as subtracting the amount of a compound that reacts from the initial amount to find the part left over
Stoichiometry and Reaction Concepts
Doing stoichiometry calculations isn't just a procedure to solve a problem in a book. It is a way to describe things that actually happen, and it is related to concepts and understanding of chemistry.
Stoichiometry calculations are based on the conservation of mass (see the Lavoisier page) and the idea that particles like nuclei and electrons aren't created or destroyed during reactions, just rearranged. These understandings allow us to make these calculations.
Stoichiometry calculations are also related to the concept of chemical equilibrium. Equilibrium means a stable state in which opposing forces are balanced. For instance, when you stand on one foot, to maintain your balance and not fall over, if you move a little to one side, you will need to correct that and move back a little in the other direction so you don't lose your balance. In chemistry, equilibrium means a state in which 2 opposite processes are occuring, but at the same rate. For instance, the reaction may go in the forward direction (to the right), and at the same time some molecules of product are turning into reactants, going in the reverse direction (to the left). In many chemical reactions, both directions are possible; when they are happening at the same rate, that is called dynamic equilibrium, which means "moving equilibrium", because the individual molecules are moving back and forth between "reactant" and "product" (which are really just defined by how you write the equation), but the total amounts of reactant and product aren't changing.
When we do stoichiometry calculations, we assume that the reaction will be complete, meaning that the limiting reactant will react completely, so none is left, forming as much product as possible. However, this is only sometimes what happens. There are several reason why this might not happen.
Possible Reasons why Actual Yield is Less than Theoretical Yield
• Rate describes how fast a chemical reaction happens. Some chemical reactions are very slow, like the ones that dissolve rock and change the shape of mountains; others are very fast, or in between. In general, the rate of a reaction will depend on the conditions, such as temperature. Right now, we will mostly talk about reactions that happen pretty quickly, but if a reaction doesn't finish before you measure it, it could be because it's too slow. This is discussed more when you study chemical kinetics.
• The position of the equilibrium is another possible reason a reaction might not produce as much product as you calculate. For all reactions, equilibrium is reached before all the reactant is used; it might use almost all the reactant, so that you don't notice the tiny bit left over, or it might reach equilibrium with more than half the reactant left. If equilibrium is reached (if the reaction is fast enough), the predicted theoretical yield will only be reached if the equilibrium is almost all product. Later, you'll learn ways to predict where the equilibrium is. For now, just know that reactions can go both directions, and that the equilibrium isn't always all product.
The last possible complication is side or competing reactions. This means that some other reactions you haven't thought of or learned about might be happening, that use up some of the reactant or product. Because chemistry is really complicated and very few materials are actually truly pure, almost every real system or reaction will have some side reactions. Sometimes they are very important and prevent the thing you want to happen from happening; other times they don't matter at all.
For your calculations, you don't usually need to really worry about rate, equilibrium or side reactions, because we haven't learned strategies for dealing with them yet. However, to keep your mental models of chemical reactions matching real chemical systems, it's important to know that real reactions are more complicated than we make them seem in the first few weeks of chem class. | textbooks/chem/General_Chemistry/General_Chemistry_Supplement_(Eames)/Chemistry_Calculations/Stoichiometry.txt |
Skills to Develop
• Perform and interpret titration calculations
A titration is a laboratory technique used to precisely measure molar concentration of an unknown solution using a known solution. The basic process involves adding a standard solution of one reagent to a known amount of the unknown solution of a different reagent. For instance, you might add a standard base solution to an mystery acid solution. As the addition takes place, the two reagents in the solutions, in this the acid and base, react. You also add an indicator, which is a molecule that changes color when the original reagent (the acid in the mystery solution, say) is completely consumed by reaction with the standard solution reagent. If you know exactly how much standard was added before the color change, you can calculate how many moles of the unknown were present at the beginning, and thus the concentration of the unknown.
Many of the standard reagents you might use in the lab, especially HCl and NaOH, which are very common and important, are hard to prepare at precise concentration without titration. The reason is that HCl is purchased as a concentrated solution, which can vary a little in concentration because both the HCl and the water can evaporate. NaOH can be purchased as a solid, but it is hygroscopic which means that it absorbs water from the air. It can absorb so much water that it actually dissolves. For this reason, even if you buy it dry, once you open the bottle, it might start to absorb water, and it would be difficult to know when you measure it what % water it is. Thus, if you work in a biochemistry lab, for instance, you might want to control the pH of your solutions by adding a little bit of dilute HCl or NaOH, because chloride and sodium ions are very common and probably are already included in the solution, but you might want to know how concentrated your solutions are. To determine this, you would use a standard solution made of some easier-to-mass acid or base to titrate the solution you actually want to use. Once titrated, you could dilute it precisely to the concentration you want. Some other reagents you might want standard solutions of react with air; these you might also titrate if they have been waiting a long time so you know what the current concentration is.
Titrations might seem a little old-fashioned. Actually, the number of automated titration machines available (try a google search!) suggest that titrations are still very important in industry. One reason might be that titrations can be good for studying newly discovered molecules, for instance to measure the molecular weight and other properties that we will study more later.
Traditionally, you take a known mass or volume of the unknown solution and put it in a flask with the indicator. Then you add the standard solution in a buret, which is a special tube for adding solution slowly and measuring the volume added at the end. These days, it might be easier to use a plastic squeeze bottle instead of a buret. You put the standard solution in the squeeze bottle, get the mass of the bottle, do the titration, and then mass the bottle again. Now you know exactly how much standard was added!
Example $1$
We have a solution of HCl whose concentration is known imprecisely (~2.5 M). (We made this solution in the previous section on molarity.) We want to determine the concentration more precisely. We have a solution of NaOH that is known to be 5.1079 M. We place 100.00 ml of the HCl solution in a flask with a drop of an indicator that will change color when the solution is no longer acidic. Then we add NaOH slowly until the indicator color changes. At this point, we have added 46.67 ml NaOH. Calculate the precise concentration of the HCl.
Solution
To answer, we need to know that the reaction is
$HCl + NaOH \rightarrow NaCl + H_{2}O$
So the ratio is 1 HCl:1 NaOH. We calculate the number of moles of NaOH added:
$(5.1079\; mol/L)(46.67\; mL)= 238.4\; mmol$
This is also the number of moles of HCl in the original 100.00 mL of solution, because the reaction ratio is 1:1. To calculate the concentration of the HCl solution, we just divide the number of moles of HCl by the volume.
$(238.4\; mmol)/(100.00\; mL) = 2.384\; M$
We could do this in one step using dimensional analysis:
$(46.67\; \cancel{mL\; NaOH}) \left(\dfrac{5.1079\; \cancel{mol\; NaOH}}{1000\; \cancel{mL}}\right) \left(\dfrac{1\; mol\; HCl}{1\; \cancel{mol\; NaOH}}\right) \left(\dfrac{1}{100.00\; mL}\right)=2.384\; M$
Now, we diluted 250 mL of the original stock solution to 1.00 L to make this solution. What is the concentration of the 10 M HCl, precisely? First we calculate the moles of HCl in the whole "2.5 M" solution, which is equal to the number of moles of HCl in quantity of stock solution ("10 M") that was used to make it (250ml).
$(1.00\; \cancel{L\; of\; diluted\; solution}) \left(\dfrac{2.384\; mol\; HCl}{1\; \cancel{L\; of\; diluted\; solution}}\right) \left(\dfrac{1}{0.250\; L\; of\; conc.\; solution}\right)=9.536\; M$
Why so low? HCl is a gas, and can evaporate out of solution. The stock solution must have been pretty old. | textbooks/chem/General_Chemistry/General_Chemistry_Supplement_(Eames)/Chemistry_Calculations/Titrations.txt |
Skills to Develop
• Describe the interactions involved with effusion and diffusion
In this section we discuss movements of gases. This is actually a very complicated field (called fluid dynamics) and we will not go very deep.
Effusion
Effusion is the movement of a gas through a tiny hole into a vacuum. We want to know the rate of effusion, which is how much gas moves through the hole per unit time. We assume that the gas particles don't bump into each other while they move through the hole (this means it's a narrow hole in a thin wall). So the rate of effusion just depends on how often the particles bump the hole. This depends on their density and speed. Because at a given temperature, all gases have the same kinetic energy, their speed is inversely proportional to m1/2, the square root of the mass. Thus, the relative effusion rates for different gases at the same temperature is
$\frac{Effusion\; rate\; for\; gas\; 1}{Effusion\; rate\; for\; gas\; 2} = \frac{M_{2}^{\frac{1}{2}}}{M_{1}^{\frac{1}{2}}}$
where M1 and M2 are the molecular weights of gas 1 and gas 2.
Diffusion
Diffusion is a more complicated process. It means the movement of gases through each other or the spreading of one gas through another. Because there are many collisions, the gases move much slower than we might expect from the average speeds near 400-700 m/s. (This is why it will take a moment to smell perfume when someone walks into a room.) Technically, many processes that sound like this are not exactly simple diffusion. We have to be careful about whether there are pressure differences or flows of gases (like wind). If there are then the process isn't simple diffusion and it won't follow the equations for simple diffusion perfectly. (The equation for simple diffusion is the same as for effusion, but for different reasons, see below.) For our purposes, when you want to predict relative rates of movement of gases, you can start with the effusion/diffusion equation. It will be exactly right in a few situations, and close enough in some others. The other situations you can study in a more advanced class if you are interested.
Diffusion Equation Explanation
Why is the diffusion equation the same as the effusion equation, but for a different reason? In simple diffusion, 2 gases move in opposite directions through a medium with the same pressure everywhere. If the pressure in the medium is constant, then the collisions of one gas with the medium are balanced by the collisions of the other gas. The momentum given to the medium by one gas in an average collision is mV, where m is the mass and V is the diffusion velocity (which is different from the average speed of the particles: it's the overall rate of movement of the gas). The number of collisions is proportional to nv, where n is the number of particles and v is their average speed. Since there is no pressure difference,
$\left(m_{1}V_{1}\right) \left(n_{1}v_{1}\right) = \left(m_{2}V_{2}\right) \left(n_{2}v_{2}\right)$
When we rearrange, the relative diffusion flux (nV, amount of particles moving times speed of diffusion) of the gases 1 and 2 is
$\frac{n_{1}V_{1}}{n_{2}V_{2}} = \frac{m_{2}v_{2}}{m_{1}v_{1}}$
Because v is proportional to m-1/2, this gives us the same result as the effusion equation. However, the reason is different. | textbooks/chem/General_Chemistry/General_Chemistry_Supplement_(Eames)/Gases/Diffusion_and_Effusion.txt |
Skills to Develop
• Explain the following laws within the Ideal Gas Law
Boyle's Law
Boyle was an Irish nobleman who is often described as one of the first modern chemists, as opposed to the old alchemists. However, many of his ideas and experiments came from earlier chemist/alchemists.
Boyle observed that for a particular sample of gas at a constant temperature, if the pressure or volume is changed, the initial and final pressure and volume are related:
\[P_{0}V_{0} = P_{1}V_{1}\]
This is now called Boyle's Law. Later it was shown that Boyle's law is an approximate law, not an exact law. Most gases follow it pretty closely at normal pressures, but less closely at large pressures.
Charles' Law and Absolute Temperature
Charles made the first solo flight in a hot-air balloon filled with hydrogen. He also observed that at constant pressure, the volume of a gas increases linearly with temperature. This can be written
\[V = kT\]
Notably, although he couldn't measure volumes at very low temperatures, all the lines for different gases pointed to the same temperature point. The temperature at which each gas was predicted to have V = 0 was the same, although the slopes were different. This temperature is now called 0 K = -273 °C, or absolute zero. Since gases can't have negative volume, this temperature seems to be special: the lowest possible temperature. Although in fact gases won't have zero volume at absolute zero (they'll be solids, and solids have volume), modern theory does still consider absolute zero special. In fact, we have to use temperature in Kelvin for any gas law problem.
Avogadro's Hypothesis
We discussed this earlier. Although often called Avogadro's Law, it was actually a hypothesis. The hypothesis is that at the same temperature and pressure, all gases have the same number of particles (molecules). Avogadro guessed that this was true based on Gay-Lussac's law, but he had no way to measure it directly, so it couldn't really be called a law. However, now we can be pretty sure that it is approximately true.
Ideal Gas Law
The Ideal Gas Law combines Boyle, Charles and Avogadro's laws. The Ideal Gas Law says that
\[PV = nRT\]
where P is pressure, V is volume, T is temperature, n is the number of moles, and R is the molar gas constant. We can express it another way too:
\[PV = nk_{B}T\]
where everything is the same except n is now the number of particles, and kB is the Boltzmann constant. As you can see, the gas constant R is just the Boltzmann constant multiplied by Avogadro's number (the number of particles in a mole). The Boltzman constant essentially provides a conversion factor between temperature in K and energy in J. Because temperature and energy are closely connected, kB appears in many important equations.
You can use the Ideal Gas Law to make predictions about how gases will react when you change pressure, volume or temperature. It gives you a good intuition for what gases do. The predictions it makes aren't always very accurate: they're pretty good at normal temperature and pressure, but actually for most engineering work they aren't good enough, so people use other equations or data tables instead. The Ideal Gas Law is a scientific law: it describes mathematically what happens under certain conditions, in this case low pressure. In the next section we'll describe the theory that explains the behavior of gases, which will also tell us when we should expect the Ideal Gas Law to be inaccurate.
You can use the ideal gas law to make various calculations, including with density and molar mass. We need to be careful with the units, because there are so many pressure units. Check that your value of R has the right unit for pressure, and isn't using an energy unit (because sometimes it's convenient to use energy units for R, but not in the ideal gas equation). Also make sure you use temperature in K. For example of a calculation, the ideal gas law says that the molar volume of any gas should be almost the same, and under standard conditions (1 atm and 0 °C) it should be close to 22.4 L.
Intro to Gases
Skills to Develop
• Describe some of the history behind defining "gas"
We've described gases before, so you already know that gases are a phase of matter, like liquids and solids. Like liquids, gases can flow and change shape to fit their container; unlike liquids, gases can expand or be compressed quite a lot. They will always increase in volume to fill their container, and if you squeeze them, you can force them into a small space.
The most familiar gases are probably air and steam. People have been thinking about these gases for thousands of years. In the early study of chemistry, studying gases was very important. People didn't know at first that there are different gases made of different atoms and molecules (all gases seemed like "air"). Some very early work suggested different types of gases, while other work (like Boyle's Law) suggested that all gases were the same, since they seemed to follow the same law. It was also hard to study gases at first, because they often broke the containers scientists tried to collect them in. Lavoisier calls gases "airs"; Van Helmont created the word "gas" using the Greek word for "chaos" because when he tried to collect gases as products of reactions, the containers often shattered.
Hales, a biologist, found a good way to collect gases. He filled a bottle with water, then set it upside-down in a dish of water. He put a sample of wood, sugar, seashell, etc at the closed end of a bent metal tube, and the open end into the upside-down bottle. Then he heated the closed end, so whatever gases formed went into the bottle. But Hales thought all these gases were "air", so he just measured the amount, instead of studying their properties.
A little after Hales' work, soon Black, Priestley, Cavendish and others were discovering all the different gases: carbon dioxide, nitrogen, oxygen, hydrogen, ammonia, nitrous oxide... these studies, and especially Gay-Lussac's law, were important for discovering the basic facts of chemistry. | textbooks/chem/General_Chemistry/General_Chemistry_Supplement_(Eames)/Gases/Gas_Laws.txt |
Skills to Develop
• Define the kinetic-molecular theory and its relationship to the ideal gas equation
Boyle's Law was published around 1660. In 1718, a mathematician named Bernoulli proposed an explanation for Boyle's Law. Although this was almost a hundred years before Dalton's Atomic Theory, atomistic theories (also sometimes called "corpuscular theories") had been around a long time. Boyle himself had made some arguments similar to Dalton's that he borrowed from Sennert. Bernoulli assumed that the gas was made of many small particles moving quickly. They move straight until they bump into another particle or a wall, then they bounce off according to conservation of momentum. Pressure comes from the impact when these particles bump into the walls.
How does the pressure depend on the volume? Bernoulli gives this explanation. If we compress a gas, the particles will bump the walls more often. This happens for 2 reasons: first, there are more particles in the layer next to the wall, where they can bump it. Second, particles moving away from the wall are more likely to bump into another particle, change direction, and bump the wall again. Imagine we have a cubic container, with each side length s. If volume is decreased from 1 to s3, then the number of particles in the layer next to the wall increases by s2/1. Also, the number of collisions between the wall-layer particles and the wall increases by s/1. Combining these, the number of collisions increases by s3 when the volume decreases by 1/s3. The pressure is the number of impacts multiplied by the momentum of the particles, mv, where m is the mass of a particle and v is the average velocity. When you calculate the average momentum change from each collision and the average number of collisions per area of wall, the result is P = nmv2/3V, where n is the number of particles and V is the volume. You can see that this matches Boyle's Law: PV = nmv2/3 = constant. Later, mv2, the kinetic energy, was shown to be proportional to the temperature: kT = mv2. This is the ideal gas equation.
How fast do gas particles move? Because the average kinetic energy is proportional to the temperature, heavier gases move slower than light gases at the same temperature. There will be a big range of speeds for different molecules, because they change speed as they bump off each other. For N2 at 0 °C, the range might be 0-1300 m/s, with the average speed about 500 m/s. As the temperature gets bigger, the range gets bigger and so does the average speed. So the particles are usually moving very fast! How far do they go between collisions? This depends on the conditions, but the average distance between collisions might be 10-7 m, so not far!
Thus, we imagine a gas as many small particles that bump off each other perfectly elastically (with conservation of momentum). They behave like hard little balls, and don't attract each other at all. Their kinetic energy depends on the temperature. In the derivation of the ideal gas law, we assume that there are no attractive forces between the particles and that the particles don't take up any space. These two assumptions are obviously incorrect: if there are no attractions between particles, there would be no liquids or solids. Likewise, the particles do take up a little space. Since we know that attractive forces become important at low temperatures, and that the volume of the particles will be important we the volume is relatively low (meaning pressure is high) we can predict that the ideal gas equation works best a high temperatures and low pressures.
Partial Pressures
Skills to Develop
• Define partial pressure using math equations
What if we have a mixture of different gases? We might want to know how much of each there is. We can define the mole fraction of a particular gas A as the number of moles of A divided by the total number of moles of gas:
$x_{A} = \frac{n_{A}}{n_{total}}$
Then we can define the partial pressure of A as:
$p_{A} = x_{A}P$
This works for real or ideal gases. Partial pressures are often used to describe concentrations of gases. If the gas is ideal, then the partial pressure is the same pressure that the gas would make if all the other components of the mixture weren't there, because:
$p_{A} = \frac{n_{A}RT}{V}$
In a real gas, this might not be true because the different gases might interact a little bit differently. | textbooks/chem/General_Chemistry/General_Chemistry_Supplement_(Eames)/Gases/Kinetic-Molecular_Theory.txt |
Skills to Develop
• Describe the significance (and applications) of measuring pressure
What is Pressure?
Pressure is defined as force/area. For instance, the pressure from snow on a roof would be the weight of the snow divided by the area of the roof. In chemistry, usually pressure comes from gases. When you blow up a balloon, you put gas inside. The gas molecules bump into each other and into the walls of the balloon. Although the force from each molecule bumping the balloon is very small, when you put enough air in, all the collisions add up and make the balloon stretch and get bigger.
The absence of pressure is called vacuum. For hundreds of years, people thought that vacuums were impossible and unnatural: "nature abhors a vacuum." This isn't actually true. Miners had noticed that they could only pump water about 10 m up a pipe; the vacuum at the top from the pump wasn't strong enough to lift the water any higher. In 1641, Berti tried an experiment. He built a giant pipe about 13 m tall next to his house and filled it with water. The top of the tube was sealed. The bottom was in a big bucket. The he made a hole at the bottom so water could flow out of the tube into the bucket. Water flowed down until the water column was about 10.3 m high. Then it stopped.
Torricelli, later a companion of Galileo, experimented further. He noticed that a column of mercury (similar to the column of water in Berti's experiment) will be 760 mm high. Water has a density of 1 g/mL, while mercury has a density of 13.5 g/mL. 10.3 m/13.5 = 763 mm. From this he concluded that the column of liquid is held up not by pull from the vacuum above but by push from the weight of air (the atmosphere) on the open surface of the bucket. The weight of the mercury column, water column and air column were the same. Traditionally, pressure was measured using a barometer, which originally was just a tube with a column of liquid. The greater the height of the liquid, the higher the external pressure. Barometers can be used to measure pressure, which can be used to predict weather. Often low pressure means rain or storm is coming.
How do we Measure Pressure?
We can use a column of liquid, or other methods. One involves a small flexible container with a vacuum inside it, prevented from collapsing by springs. It expands or shrinks depending on the pressure, and this can be measured. Now there are also even smaller, simpler electrical barometers.
What Units should we use?
There are a crazy number of units for pressure! A traditional unit is the torr or mmHg. This just means the height of a column of mercury. Atmospheric pressure is about 760 torr or mmHg, as discussed above. You might also see mmH2O, which uses the same concept, but in this case because water is less dense than mercury, atmospheric pressure is about 10.3 mH2O.
There are some more modern units also. In SI, we use Pascals: 1 Pa = N/m2. Often it's more convenient to use bar: 1 bar = 105 Pa. Another unit you may have used before is the atmosphere: 1 atm = 1.01325 bar = 760 torr. The atm is close to the average atmospheric pressure.
When you use pressure units, because there are so many of them, you should be extra careful to check your units and make sure they cancel properly (see Dimensional Analysis for more). If you are using SI units in the rest of your calculation (like forces in N, mass in kg, etc) your pressure will probably come out in Pa.
How do we Control Pressure?
Often we want to control pressures, making them either higher or lower than atmospheric pressure. For instance, in chemical industry, many reactions are run at high pressures. In the lab, we might use low pressure to pull a liquid through a filter or evaporate a solvent. We also can use vacuum techniques to do "air-free" chemistry, if we want to study molecules that react with water or air, by removing all the air from our containers before adding the chemicals. Many important instruments used in physics and chemistry, like electron microscopes, only work under vacuum.
To make a vacuum, we usually use a pump to remove air. We can't make a perfect vacuum that doesn't have any gas molecules, but we can reduce the number of molecules to quite low levels. To get ultrahigh vacuum, or even just high vacuum, we might use 2 or more different types of pumps. Some pumps work by repeatedly expanding a volume, so that the gas expands into the bigger space, then gets separated from the area that is being evacuated. (See some images on Wikipedia.) Or we could absorb the gas molecules onto a surface to remove them the space being evacuated. Normal vacuum cleaners used in homes might be at 0.2 atm, while ultrahigh vacuum in a lab might be 100 nPa.
To make very high pressures, scientists sometimes use diamond anvils. For instance, geochemists who study how rocks form might put a bit of water and mineral powder between the tips of two small, pointy diamonds (just like you might see in an engagement ring). Then they push the diamonds together, and the force gets concentrated onto just the tiny tips of the diamonds, so the pressure is huge, like 3 million atm. And because the diamonds are clear, the scientists can watch what happens right through the diamond!
Figure 3: Cross section of a Diamond Anvil Cell. The following things are included: The two diamonds in between which the pressure is created. The sample. A Ruby which is usually used as a pressure indicator. The Gasket which seals the sample chamber The casing with the screws. Tightening of the screws moves the casings and the diamond closer together and builds pressure. The backing plate which holds the diamond in place. Electromagnetic rays which pass through the sample chamber to allow measurements. (CC-SA-BY-3.0; Tobias1984)
Real Gases
Skills to Develop
• Describe how real gases differ from ideal gases
• Derive the Van der Waals equation from the ideal gas equation
By the kinetic-molecular theory of gases, we imagine a gas as many small particles that bump off each other perfectly elastically (with conservation of momentum). They behave like hard little balls, and don't attract each other at all. Their kinetic energy depends on the temperature. In the derivation of the ideal gas law, we assume that there are no attractive forces between the particles and that the particles don't take up any space. These two assumptions are obviously incorrect: if there are no attractions between particles, there would be no liquids or solids. Also, the particles do take up a little space. Since we know that attractive forces become important at low temperatures, and that the volume of the particles will be important when the volume is relatively low (meaning pressure is high) we can predict that the ideal gas equation works best at high temperatures and low pressures.
If we want to make another equation that is closer to the real behavior of gases, we can make a few changes in the Ideal Gas Equation. First, we will assume that the particles have some volume. Instead of V we will use (V - nb) where n is the number of molecules or moles, and b is a constant for each different gas that means roughly how big it is.
Second, we need to include the effect of attractions between particles. If the particles attract each other, they will stay closer together and pump the walls a little less, so the observed pressure will be lower than we would expect. The higher the concentration of the gas (bigger n/V), the more important the attractive forces are. Actually the attractive forces depend on (n/V)2, because this tells us how many other particles each particle can interact with. So we replace P with (P + a(n/V)2), where a is a constant that depends on the gas, and tells approximately how big the attractive forces are. This makes sense because bigger attractive forces happen when the particles are very close together (V is small), and they cause the pressure to seem lower.
Putting this together, we have the Van der Waals equation:
$\left(P + a \left(\dfrac{n}{V}\right)^{2} \right)(V - nb) = nRT$
This equation describes real gases pretty well, although there are other equations used also. The constants a and b are found by fitting the real data for each gas to this equation. You can look up values of a and b in tables. | textbooks/chem/General_Chemistry/General_Chemistry_Supplement_(Eames)/Gases/Pressure.txt |
Skills to Develop
• Establish a general procedure for drawing Lewis structures
• Describe the interactions between atoms using Lewis structures (what happens to the valence electrons)
Everyone who has studied chemistry should be able to draw Lewis structures. Although there are many complicated situations, and some people try to stretch Lewis structures to be an accurate description of molecules even when they don't work well, the basic idea is simple. Here's how I draw Lewis structures.
1. You count the valence electrons. Count the valence electrons for each atom, add them up, and add or remove electrons if there is an overall charge.
2. You figure out what the connections between atoms are. Sometimes you might look this up. Other times, you have to guess. If the molecule is linear (like HCN) usually it is written in the correct order. If it is a polyatomic ion, like sulfate or nitrate, usually you put the heavy atom, or the atom to the left in the periodic table, in the center. You should probably not put all the atoms in a line if there are more than 4 (single-bonded chains are usually very unstable, except for carbon). Elements like N, C, S, P, Cl and the heavier elements in these groups can easily connect to 4 other atoms, so often they go in the middle. O should not connect to more than 2 atoms, and often only connects to one. If O connects to 2 atoms, usually at least one is C or H. H and F will almost always make just one bond. (Hydrogen bonds, which you may have heard of, are much weaker than the covalent bonds shown by Lewis structures.)
3. Once you have chosen an arrangement of atoms, add the right number of electrons. Try to make sure every element gets the right number of electrons, using lone pairs of electrons which are not shared, or shared pairs (which are bonds). You can draw single, double, or triple bonds. Make sure that H has 2 electrons (never more) and C, N, O, F have 8 electrons (never more, and not less unless the molecule has an odd number of electrons). The heavy elements under C-F should have at least eight electrons, and they can also connect to 6 or even 7 other atoms. B often has 6 electrons, and Be often has 4. Move the electrons around until it works. Make sure your final structure has the right total number of electrons, and that none of the atoms have too many or too few.
4. Unpaired electrons are called radicals, and you should avoid them. When you draw the Lewis structure, make all the electrons paired unless there is an odd number of electrons. All electrons should be in lone pairs or bonding pairs. (There are molecules, like O2, which have unpaired electrons even though they could all be paired, but you can't predict that with Lewis structures, so assume they are all paired.)
Table \(1\): Acceptable Numbers of Electrons and Connected Atoms for Common Elements
Element Number of Electrons
(including shared)
Number of Connected Atoms Exceptions
H 2 1 Sometimes connects to 2 atoms, in H-bonding or with boron
Be 0, 4, 6, 8 2 - 4 Can be shown as ionic (0 electrons) or covalent (4, 6, 8)
B 6, 8 3, 4 Sometimes has less than 8 electrons
C 8 1 - 4 Less electrons if compound is a radical
N 8 1 - 4 Less electrons if compound is a radical
O 8 1 - 2 Sometimes connects to 3 atoms, such as in H-bonding
F 8 1 Sometimes connects to 2 atoms, such as in H-bonding
P and below 8 or more 3 - 6
S and below 8 or more 2 - 6
Cl and below 8 or more 1 - 6 Can connect to 7 atoms
Xe 8 or more 0 - 6 Xe compounds with O and F are known
Example \(1\)
First, let's do hydrogen cyanide, the poison that might have killed Lewis. The formula is HCN. As usual, this is the correct order of the atoms. The number of valence electrons in the molecule is (1 + 4 + 5) = 10. When I'm putting the electrons in, I usually start by putting each atom's valence electrons around it, then I connect the dots into lines. (These steps are shown in the picture).
For a second example, let's do the tetrafluoroborate ion, BF4. In this case, we have to put B in the middle, because F shouldn't make more than 1 bond. We count electrons: (3 + 7 x 4 + 1) = 32. Remember to count +1 for the negative charge on the ion. Because B needs to make 4 bonds, we'll give it the extra electron. Then we'll connect the electrons into bonds. In this case, you know that F pulls on electrons much harder than B, so the "shared pairs" will probably be closer to F, even though the picture doesn't show that. | textbooks/chem/General_Chemistry/General_Chemistry_Supplement_(Eames)/Lewis_Bonding_Theory/Drawing_Lewis_Structures.txt |
Skills to Develop
• Define Lewis acids and Lewis bases
Previously we said that an acid produces H+ when dissolved in water, and a base produces OH when dissolved in water. Then the acid and base (meaning H+ and OH) can react (without redox) to make water. This is a pretty good definition, but it is kind of small.
Lewis explained that in many reactions that form new bonds, both electrons in the new bond come from 1 atom (or 1 reactant) only, instead of 1 electron coming from each. He called all these reactions acid-base reactions. The picture shows water forming from the elements, in a redox process, and water forming from hydrogen ion and hydroxide ion, in an acid base process. It also shows how the tetrafluoroborate ion, BF4 can form from boron trifluoride and fluoride ion.
In general, if we can draw a good Lewis structure by making two molecules or ions share an electron pair, it's likely that the reaction can happen. For instance, BH3 can react with NH3, because N has an extra lone pair, and B only has 6 electrons and 3 connected atoms. So a Lewis acid is something than can fit 2 more electrons from a different molecule. It can share another molecule's lone pair. A Lewis base is any molecule or ion with a lone pair to share. It's easy to see what can be a Lewis base just by drawing a Lewis structure. Lewis acids are usually cations, like H+or Al3+. Boron is a famous Lewis base because it often makes electron-deficient compounds, like BH3, in which it only has 6 electrons. Try drawing Lewis structures for an acid-forming anhydride combination reaction. Is the anhydride a Lewis acid or base?
Lewis and Lagmuir
Skills to Develop
• List some of the achievements from Gilbert Newton Lewis and Irving Langmuir
• Describe the relationship between Lewis and Langmuir
You have probably already heard of Gilbert Newton Lewis' ideas. He was one of the great chemists of the 1900s, and developed the basic concepts of chemical bonding still taught to all chemistry students today. He also introduced an important acid-base theory, and did a lot of work in physical chemistry, making the energy concepts (like enthalpy and equilibrium relationships) truly useful to chemists. Although I think it is generally agreed now that he deserved at least one Nobel Prize, he did not get it at all because he did not advertise his work, and made himself a little unpopular.
You probably haven't heard of Irving Langmuir, another great chemist from the same time. He worked on surface chemistry, and did most of his research in an industrial lab, where he invented a better light bulb and hydrogen arc welding, which is much hotter than previous welding techniques. When he and Lewis first met, they were friendly and impressed with each other. Lewis vigorously defended Langmuir when another chemist tried to steal credit for one of Langmuir's ideas. Later, though, Lewis wrote a few brilliant papers on chemical bonding, then served in World War I. The papers were not noticed, until Langmuir read them and was instantly converted. He published several more papers on the same topic, applying Lewis' ideas, and then traveled widely speaking about the theory. It became well-known and accepted because of Langmuir's efforts. Although Langmuir credited Lewis for the original idea, perhaps he didn't give Lewis as much credit as he should have. And other people associated the theory with Langmuir, because he was the one who popularized it. Lewis began to feel that Langmuir had stolen his ideas.
The story has a sad end. When Lewis was 70, he still had not won the Nobel Prize, while Langmuir had won it for his work on surface chemistry. One day Langmuir visited University of California at Berkeley, where Lewis had built the chemistry department and run it for years. Langmuir gave a talk, and a few days later was given an award by the Berkeley chemistry department. That day, Lewis had planned an experiment using hydrogen cyanide, a deadly poison, as a solvent. About an hour after Lewis started work, he was found dead. It's not clear exactly how he died: perhaps it was a heart attack, or perhaps it was a suicide using the hydrogen cyanide. Whichever it was, it may not be coincidence that Lewis had lunch with Langmuir that day. | textbooks/chem/General_Chemistry/General_Chemistry_Supplement_(Eames)/Lewis_Bonding_Theory/Lewis_Acid-Base_Theory.txt |
Skills to Develop
• Describe 2 types of dipoles
• Distinguish paramagnetism from diamagnetism
In 1923, Lewis wrote that the study of spectra and magnetism are the 2 best ways to learn about chemical bonding. Some types of spectra are discussed here. In this section, we will describe the magnetic evidence Lewis used. You are familiar with normal magnets, like the ones used to stick restaurant menus to fridges. In this section, however, we are more interested in the magnetic properties of molecular materials, rather than metallic or semi-ionic solids. Most materials are non-magnetic. The molecular materials that are magnetic still wouldn't stick to your fridge, because they only really act magnetic in a magnetic field. Normally, each molecular magnetic has a random direction because the interactions between the molecules are weak. (Just like how molecular materials melt at lower temperature than metals or rocks). If you put them in a magnetic field, if they are cold enough, they will start to line up with the field. If they are too hot, they will continue to move randomly. This is called paramagnetism.
Dipoles
Let's start by reviewing electric dipoles. An electric dipole is something that had separated electrical charges. For instance, an HF molecule or a water molecule have an electric dipole moment defined as
$\mu = q \times d$
where q is the partial charge on each end of the molecule and d is the distance between the charges.
Generally, solvents with large dipoles have high dielectric constants. In Coulomb's law, the dielectric constant D reduces the force between charges.
$F= \frac{kQq}{Dr^{2}}$
Bigger D means less force. This explains how ionic substances dissolve in solvents with large dipole moments and large D. When the ions separate and have water between them, they no longer attract each other strongly.
Although you can have a point charge, magnets are always dipoles (have a North and South pole). Just like solvents and other materials have dielectric constants, they also have magnetic permeabilities.
Magnetic Measurements
How do you measure magnetic moments or magnetic dipoles in molecules? The old way is to put the molecular material between the poles of a big magnet and see what forces are present. For instance, if you put a paramagnetic material (that is cold enough) between the N and S magnetic poles, it will line up with the field and be attracted by the field. Thus if you weigh a paramagnetic material in a magnetic field, it will be heavier than without the field. Most materials are diamagnetic, and get lighter in magnetic fields, because they are repelled by the field. So the easiest way to see if a material is magnetic is to weigh it with and without a magnetic field. (Now there is a fancier method also, called SQUID, which is too complicated to explain here.)
How is Magnetism Related to Chemical Bonding?
Lewis correctly realized that molecules with an odd number of electrons (such as NO, with 7+8 =15, but not CO, with 6+8 =14) have some unusual properties. They are paramagnetic, usually strongly colored, and quite reactive. The huge majority of molecules have even numbers of electrons, and are diamagnetic. Thus, paramagnetism seems to be a result of unpaired electrons, while diamagnetism is a result of paired electrons. Transition metal compounds are often colored and magnetic; most compounds of C,H,O,N etc (such as in your body) have little color, are diamagnetic, and not so reactive. Basically, almost all molecules have all their electrons in pairs. | textbooks/chem/General_Chemistry/General_Chemistry_Supplement_(Eames)/Lewis_Bonding_Theory/Magnetism.txt |
Skills to Develop
• Distinguish polarity from polarizability
• Define polarizability
Polarity means separation, in this case of electrical charge. If a bonding pair of electrons are pulled more toward one atom and away from the other, this will cause the first atom to be partially negatively charged, and the second to be partially positively charged. This will create an electric dipole moment, such as the the dipole moment in water that makes water so good as a solvent.
Although covalent and ionic substances might seem really different, with Lewis theory we can think of them as being basically similar. At one extreme, we have a complete transfer of an electron, such as in CsF, making a positive and negative ion that are then attracted to each other. At the other extreme, we have a completely equally shared pair, such as in F2. In between, we have bonds with unequal sharing. Either way, usually all atoms will have a noble gas electron configuration, either by sharing or by losing electrons.
In addition to average polarity of a bond, we can have temporary polarity. The electrons move around, and sometimes it will happen that both of them move toward one atom. In general, the more electrons an atom has, the looser they are held. For example, iodine has 53 electrons, which is a lot! They can move around pretty easily. Even though I2 has a non-polar bond on average, because the atoms are the same, it can easily become polar because the electrons are held loosely. In solution, I2 can split a little bit into I+ and I.
To keep these ideas separate, polarity means the permanent average separation of charge, and polarizability means the ability to become polarized temporarily. In a non-polar polarizable molecule like I2, the average polarity is 0, but if we take many precise measurements of the instantaneous (very short time) polarity, many of them will be far from 0. In a polar, non-polarizable molecule like HF, all the instantaneous measurements will be very similar, but the average will not be 0. AgI is both polar and polarizable. Both of these have important effects on properties of materials, such as reactivity, solubility, and boiling point, that we will talk about later.
The Octet and Other Stable Groups
Skills to Develop
• Explain the stable groups described by Lewis
Lewis considered the "group of two" to be of greatest importance for understanding molecules and chemistry. However, he had also described the "group of eight", which just said that many atoms gain, lose or share electrons until they have 8 valence electrons. But he later said that the "group of eight" was less fundamental than the "group of two". This may be in part because Langmuir invented the term "octet" to replace "group of eight" and tried to force many compounds to fit it, even if they didn't seem to. Lewis recognized that not all stable compounds follow the octet rule. Although it usually works pretty well for elements in the p-block, the transition metals usually follow the "18-electron rule". You can see that the rule matches the periodic table: hydrogen and helium want 2 electrons, p-block elements want 8, d-block elements 18, etc.
The essence of Lewis' theory is that stable compounds can be predicted and understood using what are now called "Lewis dot structures". These show the arrangement of valence electrons in a molecule. For stable molecules, it is usually possible to draw a structure in which electrons are shared so that every atom has its octet (8 electrons), either by adding, losing, or sharing electrons.
The Two-Electron Bond
Skills to Develop
• Describe Lewis' theory for bonds between atoms
The facts described in the previous section, that almost all molecules have all their electrons paired, lead Lewis to the conclusion that electron pairs are of central importance in chemistry. He proposed that in atoms, and especially in molecules, electrons are usually paired. He proposed this theory a little after Bohr's theory of quantum mechanics, which proposed orbits for electrons, and thought that 2 electrons could fit in each orbit. Lewis imagined that when 2 H atoms form a molecule, the 2 electrons would share an orbit "between" the 2 atoms. He didn't talk about what the orbits would really look like, but he did propose that the electrons could be imagined as sitting on the corners of tetrahedra. When atoms made bonds, they could share electrons on the points of the tetrahedra, as shown. Thus they could make single, double or triple bonds (such as in F2, O2 and N2).
In Lewis' theory, there is no fundamental difference between covalent and ionic bonding. (Actually, he invented the idea of covalent bonding, while Langmuir invented the word "covalent," but it had been known for a while that molecules made of only non-metals had different properties from those made of metals and non-metals together). Electrons can be shared between atoms, and the sharing is more equal if the atoms are similar or the same, and more unequal if the atoms are different. Two shared electrons make one chemical bond. | textbooks/chem/General_Chemistry/General_Chemistry_Supplement_(Eames)/Lewis_Bonding_Theory/Polarity_of_Chemical_Bonds.txt |
Molecular Orbital Theory is another theory to explain chemical bonding using orbitals. It was developed about the same time Valence Bond Theory was developed, primarily by Mulliken and Hund. (Mulliken was mentioned earlier because he proposed a definition of electronegativity, and Hund because of Hund's Rule.) It is a little harder to learn than Valence Bond Theory, but very useful.
Molecular Orbital Theory
Skills to Develop
• Distinguish and describe the significance of frontier MOs
Frontier means a border area, between two things (often, between 2 countries). In this case, we are interested in the MOs at the border between occupied and empty. The frontier MOs are called the HOMO and the LUMO. HOMO is Highest Occupied MO, the highest-energy MO that has electrons in it. LUMO is Lowest Unoccupied MO, the lowest-energy MO that doesn't have any electrons in it.
Frontier MOs are very important for reactivity. Earlier, we said that most reactions can be called either Lewis acid/base or redox. In a Lewis acid/base reaction, an electron pair from the base is shared with the acid. What this really means is that the base has a HOMO that is pretty high-energy (a lone pair), and the acid has a LUMO that is pretty low energy. (A lone pair in MO theory is an electron pair in a non-bonding MO. A low LUMO usually means an empty valence orbital, like on B or on a cation.) We can make a bonding and anti-bonding combination of the base HOMO and acid LUMO, and that will stabilize the electrons from the base's HOMO, lowering the total energy. This creates a bond between the acid and base.
In a redox reaction, the oxidant has a low LUMO, and the reductant has a high HOMO, but this time the oxidant LUMO is lower than the reductant HOMO, so that the electrons in the reductant HOMO move completely to the oxidant LUMO. Often the energy match is bad, so that no covalent bond forms, just the electron moves. Sometimes a covalent bond forms also. This depends on the AO energies, which depends on the electronegativity, just like you would expect. You can still predict covalent/ionic bonding based on electronegativity.
Molecules with low HOMO and high LUMO, a big HOMO-LUMO gap, are not very reactive! Hydrocarbons are a good example (like oil, etc). They do burn easily, but you have to get them hot first. At room temperature, they don't react. This is why if you want to store something reactive like K metal, you probably keep it in a bottle of oil.
Example \(1\): CO Toxicity
You probably know that CO is toxic (which is why you shouldn't stay in a garage with a car running, because the CO from incomplete combustion can kill you). The reason CO is toxic is because it binds metal ions really tightly. You use Fe(II) ions in your blood to bind O2 and carry it to your cells. CO binds to the Fe(II) more tightly than O2, so if you breathe too much CO, your cells won't get any oxygen, because all the Fe(II) in your blood bound CO instead. We can understand how and why CO binds Fe(II) using MO theory. Go back and look at the MO diagram for CO. The HOMO is a slightly-bonding orbital that is mostly on carbon. It is pretty high-energy. The LUMO is a π* orbital that is also mostly on carbon, and it is kind of low-energy because the splitting of π MOs is smaller than σ MOs (because of less overlap). CO has a small gap between the HOMO and LUMO. Fe(II) also has a small gap between HOMO and LUMO, because it has 6 electrons in 3d orbitals. The HOMO is high, because 3d is not so stable, and the LUMO is low, because it is also 3d and not much higher than the HOMO (there is a gap because the other atoms around the Fe(II) in hemoglobin make the d orbitals different energies). So what can happen is that the HOMO on CO makes a σ bond with the LUMO on Fe(II), and the HOMO on Fe(II) makes a π bond with the LUMO on CO. This "multiple bond" between CO and Fe(II) makes CO toxic. And because the HOMO and LUMO of CO are big on carbon, you won't be surprised that the bond is Fe-C=O, not Fe-O=C.
MO interactions between Fe and CO frontier MOs. Top: σ interaction. Bottom: π interactions (2 CO LUMOs, 2 Fe HOMOs, related by 90° rotation)
Intro to MO Theory
Skills to Develop
• Compare and contrast MO Theory and Valence Bond Theory
Molecular Orbital (MO) Theory is another theory to explain chemical bonding using orbitals. It was developed about the same time Valence Bond Theory was developed, primarily by Mulliken and Hund. (Mulliken was mentioned earlier because he proposed a definition of electronegativity, and Hund because of Hund's Rule.) It is a little harder to learn than Valence Bond Theory, but very useful.
Why do we need another theory after learning Valence Bond Theory? Although Valence Bond Theory works well to explain some properties of certain types of molecules, like shape and bond strengths in organic molecules, there are many situations when it doesn't work very well. For example, it isn't very good for predicting the magnetic properties of molecules. From Valence Bond Theory, you would not expect O2 to have 2 unpaired electrons, but it does. It also isn't good for predicting the spectroscopic properties of molecules, including what color they are. For example, you can do Photoelectron Spectroscopy on molecules, in which you knock electrons off molecules using high-energy photons. By knowing the energy of the photons and the kinetic energy of the photoelectrons, you can find the binding energies of the electrons in the molecule. For water, you would expect from the Lewis structure that there are 2 different binding energies for valence electrons (because they are either lone pairs or bonding pairs) and another higher energy for the O 1s electrons. However, the data shows 4 different energies, meaning that the 4 electron pairs in the Lewis structure all have different energies. MO Theory can explain this. MO theory is also good for predicting how strong bonds are, for predicting stability of weird molecules (like C2), and for describing bonding in molecules that have resonance structures. | textbooks/chem/General_Chemistry/General_Chemistry_Supplement_(Eames)/Molecular_Orbital_Theory/Frontier_MOs%3A_An_Acid-Base_Theory.txt |
Skills to Develop
• Construct MO diagrams for simple diatomic molecules
Overview
In this section, we will compare MO diagrams for diatomic molecules X-X, from Li2 to Ne2. We will predict their bond order and see how the energies of the different orbitals change. We will also compare our predictions to experimental evidence. First, though, we need to talk about a new effect, s-p mixing.
s-p Mixing
Let's think about the orbitals we use to make MO diagrams for the first row elements, Li-Ne. We don't have to worry about 1s orbitals, because their interactions don't have much effect on the properties of the molecule. The core orbitals are completely full, so there can't be any net bonding between them. They are also smaller, so they have worse overlap with core orbitals on other atoms, and lower energy, so they have bad energy match with valence orbitals on other atoms.
We will use 2s and 2p orbitals. The 2p orbitals that make π combinations don't do anything new. But we have to think more about the σ orbitals. Both 2s and 2pz can make σ bonds. Do they do this separately? It turns out that sometimes they do and sometimes they don't. In molecules like F2, they mostly interact separately, because the energy match between them is very bad. F 2s is much lower energy than F 2p. So you get an MO diagram as shown in the figure. You can see that including the 2s orbitals does not change the bond order, because both the bonding and anti-bonding combination are filled.
However, in elements with less difference between the 2s and 2p energies, like carbon, the 2s and 2pz orbitals can mix. This means that we can add some p-character to the bonding s combination, so the overlap is better. And we can add some p-character to the s anti-bonding combination so that the overlap is worse, and it is less anti-bonding. (Basically they form sp hybrids, like you have seen before. However, they are unequal hybrids, with more s in one and more p in the other.) We can picture sp hybrids making 4 combinations. We get a strongly bonding combination, a weakly bonding combination, a weakly anti-bonding combination, and a strongly anti-bonding combination. Or you can think of it as making a bonding, 2 non-bonding, and an anti-bonding combination. The combinations are shown in the figure.
What effect does sp-mixing have on the MO diagram? The 2s bonding orbital becomes more bonding, and gets lower in energy. The 2p anti-bonding orbital becomes even more anti-bonding, and gets higher in energy. The two in the middle might be harder to predict (especially because the molecule will use whatever ratios of s and p in the hybrids is best for lowering the total energy) but probably the lower one gets lower and the higher one gets higher. We would have to use complicated computer calculations to know exactly, but we can make pretty useful predictions just by thinking about it.
MO Diagrams
Now we're ready to look at MO diagrams for the first row series. Note that the molecules that aren't common and stable might still be more stable in the gas phase than single atoms. For instance, in a gas of Li metal, diatomic molecules will form.
The figure shows a summary of the energy levels, so we can see how they change. The number of electrons increases as we move from right to left. You can also see that the gap between the s and p-based orbitals gets bigger from left to right. (Also see that here.) The result is that sp-mixing decreases from left to right, because orbitals only mix well if they have similar energies. Between N and O the σ(2p) crosses below the π MOs. We can match up the MO diagrams with the experimental data, such as bond length and bond strength, which suggests the bond order, and the magnetic properties, which tell us how many unpaired electrons the molecule has.
Let's go through the list. Also, you can check the table below for some data.
• Li2: Using Lewis structures, we predict 1 bond. MO theory also predicts 1 bond. This is a longer and weaker bond than in H2 because we are using 2s orbitals instead of 1s orbitals. The Coulomb attractions between the nuclei and bonding electrons are weaker because they are farther apart, and there is electron-electron repulsion. However, in the gas phase the molecule is stable.
• Be2: Using Lewis structures, we predict 2 bonds. MO theory predicts <1 bond, however, because 2 electrons are in a bonding orbital and 2 are in a weakly anti-bonding orbital. The molecule is not normally stable, but has been detected at very low temperatures in the gas phase.
• B2: Using Lewis structures, we predict 3 bonds and no unpaired electrons. Experimentally, B2 is paramagnetic, with 2 unpaired electrons. MO theory can explain this once we include sp-mixing, which makes the 2p σ bonding orbital mostly non-bonding, moving it above the π bonding orbitals. Following Hund's Rule, we put electrons in the π orbitals unpaired. What is the bond order? It depends how much sp-mixing there is, which we can't know without computer calculations, but it should be between 1 and 2 bonds. This is a shorter and stronger bond than in Li2 so probably it is close to a double bond.
• C2: Using Lewis structures, we probably aren't sure how to draw it. MO theory suggests that it has 2 π bonds, and a partial σ bond (we don't know how anti-bonding the 2s σ* orbital is because we aren't sure about sp-mixing). The bond length and strength match a bond order of 2-3.
• N2: Using Lewis structures, we predict a triple bond. MO theory also predicts 3 bonds, which match the experimental very short strong bond. Notice that in MO theory, the "lone pairs" (the 2 mostly non-bonding σ MOs) are shared over both atoms, and have different energies.
• O2: Using Lewis structures, we predict 2 bonds and no unpaired electrons. MO theory also predicts 2 bonds, but correctly predicts 2 unpaired electrons. This is a longer and weaker bond than in F2 as we would expect. If we reduce O2 by adding electrons, they go into the π* orbitals, and make the bond weaker; this matches experimentsm showing that peroxide has a longer, weaker bond.
• F2: Using Lewis structures, we predict 1 bond. MO theory also predicts 1 bond. This matches experiments showing a long, weak bond.
• Ne2: Using Lewis structures, we predict no bond. MO theory also predicts no bond. Ne2 has never been detected experimentally.
Bond Enthalpies (kcal/mol) and Bond Lengths (Å) for the Homonuclear Diatomic Molecules
Molecule D (kcal/mol) r (Å)
H2 103 0.74
Li2 26 2.67
B2 66 1.59
C2 144 1.24
N2 225 1.10
O2 118 1.21
F2 37 1.42 | textbooks/chem/General_Chemistry/General_Chemistry_Supplement_(Eames)/Molecular_Orbital_Theory/MO_Diagrams_for_First_Row_Diatomic_Molecules.txt |
Skills to Develop
• Construct MO diagrams for simple diatomic compounds
What's Different if we have 2 Elements?
If the elements are different, the main thing is that the AOs won't have the same initial energy. If they don't have the same energy, then the splitting will be smaller. Also, the bonding MO will have a higher % of the lower energy AO and the anti-bonding MO will have a higher % of the higher energy MO. This is exactly the same as saying that it will make a polar covalent bond. The electrons will usually be in the bonding MO, not the anti-bonding MO. The bonding MO has more of the lower energy AO, so the electrons will spend more time next to the atom with lower AOs, which is the same as the more electronegative atom. The bigger the difference in electronegativity/AO energy, the smaller the splitting and the more the bonding MO looks like the AO of the electronegative atom. When the difference is really big, the bond becomes completely ionic, and the "bonding MO" basically is the lower energy AO.
Example \(1\): CO
This diagram is based on calculations and comparison to experiments. (But it is not drawn exactly, just approximately.) It would be hard to guess all the details, especially about the sp-mixing and the shapes and sizes of MOs. Notice that the bonding orbitals are bigger on the oxygen and the antibonding orbitals are bigger on the carbon. We will talk more about the consequences of this later.
MO Diagrams for Linear Triatomic Molecules
Skills to Develop
• Construct MO diagrams for simple linear triatomic molecules and/or compounds
What's Different if we have 3 Atoms?
We won't go into the details of MO theory for complex molecules, because that's a topic for more advanced classes, like Inorganic Chemistry. However, we'll show a couple examples of simple molecules so you can get the idea. We'll focus on molecules with 1 central atom and some others around it. In this case, first we combine the AOs of the outer atoms into sensible combinations, then we see how these combinations interact with the AOs on the central atom. Don't forget about net overlap, which is important for deciding how the orbitals interact. If they don't have net overlap, they can't interact.
2
BeH2 might be hard to make, but it's very nice and simple as an example! Using the electron domain model we predict that it will be linear. Be has 2s and 2p orbitals, and it is in the middle. H has 1s orbitals; there are 2 H atoms on the outside. We need to make combinations of the H AOs, and we'll use the same combinations we used to make H2, except now they aren't touching. These combinations will match the 2s and 2pz on Be, as shown in the figure.
Forming MOs for BeH2.
Then we can put the MO diagram together just the way we usually do, starting with the outside, drawing in bonding, non-bonding and anti-bonding MOs, and filling the electrons. The bond order is 2.
2
This is a little more complicated example. Now we have to make combinations of 4 different AOs from oxygen. We can combine 2s orbitals, 2px orbitals, 2py orbitals and 2pz orbitals. Each pair will make the same add/subtract combinations we've seen before. However, for a basic diagram, we will include only the O 2p orbitals, because the O 2s orbitals are much lower in energy; they have a bad energy match so they won't interact very much. The O 2p combinations that we will use are shown in the figure, and labeled with which carbon AO they match.
Then we combine these oxygen AO combinations with the AOs on C. We still make bonding and anti-bonding combinations just like before. And there are some non-bonding orbitals too. Just as we expect from Lewis structures, there are 2 σ bonds and 2 π bonds. However, each of these is delocalized over the whole molecule. The bonds aren't just between 2 atoms at a time, they connect all 3 atoms.
MO Diagrams for Water and Nitrate Ion
Skills to Develop
• Construct MO diagrams for simple non-linear molecules and/or compounds
Non-linear Molecules
If you take an Inorganic Chemistry class, you'll learn ways to make MO diagrams for more complicated, non-linear molecules. Here we won't really explain these methods, just show some results. First we'll look at an MO diagram for water, because it's pretty simple and water is very important. Then we'll look at the π MOs for the nitrate ion, so we can see the difference between MO theory and valence bond theory.
Example \(1\): Water
This diagram is based on calculations and comparison to experiments. (But it is not drawn exactly, just approximately.) It would be hard to guess all the details, especially about the sp-mixing and the shapes and sizes of MOs. Notice that the bonding orbitals are bigger on the oxygen and the antibonding orbitals are bigger on the hydrogen. This produces the polarity that makes water a good solvent, because there is more electron density on the O and less electron density on the H.
We start by making the same H AO combinations we used for H2 and also for BeH2. Then we do sp-mixing on O, making one orbital pointed toward the H atoms and one pointed away (which will be mostly non-bonding, like a lone pair). Then we make bonding and anti-bonding combinations of H and O orbitals that match.
Example \(2\): Nitrate ion π MOs
In the section on multiple bonds using Valence Bond Theory, we talked about nitrate ion (NO3), which has 1 π bond shared over 4 atoms (3 different resonance structures). That might have seemed kind of weird, so let's look at it again using MO theory, which treats resonance much more naturally. Let's just think about the 4 2pz orbitals, on 1 N and 3 O atoms. These orbitals point out of the plane of the molecule and will be used to make π bonds. First we will make combinations of the oxygen AOs. The combinations are shown in the figure. Don't worry about where these combinations came from too much, because that's a little too advanced for us (you'll see, if you take an Inorganic Chemistry class).
You can see that 2 of the combinations don't match the AOs on N, so they will be non-bonding MOs (like 2 lone pairs shared over 3 O atoms). One combination matches the N orbital, so it will make bonding and anti-bonding combinations. There are 6 π electrons in nitrate, so the bonding and non-bonding MOs will be filled. This means there is one π bond shared over the 4 atoms, which is just what valence bond theory says also. | textbooks/chem/General_Chemistry/General_Chemistry_Supplement_(Eames)/Molecular_Orbital_Theory/MO_Diagrams_for_Heterodiatomic_Molecules.txt |
Skills to Develop
• Construct MO diagrams for H2 and He2
• Define bond order in theory and calculation
MO Theory for H2
Let's find the MO description of H2. We are going to use one electron on each H atom, each in a 1s orbital. When we bring the 2 atoms next to each other, we can make 2 MOs out of the 2 1s AOs. As always, we will do a + combination and a — combination. These combinations are illustrated in the figure below.
Notice that the + combination produces an MO with more electron density between the nuclei, because the waves interfere constructively. The — combination produces an MO with a node between the nuclei, and not much electron density there, because the waves interfere destructively. If you think about the Coulomb's law forces, and imagine putting 2 electrons in the + MO, they will usually be between the nuclei, and the attractions between the electrons and nuclei will hold the nuclei together, making the molecule. For this reason, the + MO is called a bonding MO. On the other hand, if we imagine putting 2 electrons in the — MO, they will usually be on the outside of the nuclei, so the repulsion between the nuclei will push them apart, and no molecule will form. For this reason, the — MO is called an anti-bonding MO.
You've seen before that more nodes means higher energy. (For instance, if you try to swing a jump rope so it has a standing wave with 2 nodes, that is much harder than making a standing wave with no nodes.) So it makes sense that the bonding MO (no nodes) is lower in energy than the anti-bonding MO (1 node). Thus, in H2, both electrons will go in the bonding MO, and the molecule is stable. In fact, the bonding orbital will be lower in energy than the AOs it was made from, because of the increased Coulomb attractions. The anti-bonding orbital will be higher in energy than the AOs because of the increased Coulomb repulsions. We can represent this with an MO diagram, shown in the figure.
MO Theory for He2
Now let's think about He2. We still have a combination of 2 AOs, both 1s. The bonding and anti-bonding orbitals will look very similar. But now we have 4 electrons, so we will have to put 2 electrons in each MO. Because the molecule He2 does not exist, we can conclude that the anti-bonding orbital increases in energy more than the bonding orbital decreases in energy, so that He2 is higher energy than 2 He.
Bond Order in MO Theory
One great thing about MO theory is that it makes it really simple to think about partial bonds and weird molecules, like radicals. The table shows some data for a few examples.
Molecule Bond Length
(Å)
Bond energy
(kcal/mol)
He2 * *
H2+ 1.06 61
He2+ 1.08 55
H2 0.74 103
Now try drawing the MO diagram for each molecule. What do you notice? If we calculate
net bonding electrons = (number of bonding electrons) — (number of anti-bonding electrons)
we get:
Molecule Bonding
Electrons
Anti-Bonding
Electrons
Net Bonding
Electrons
Bond Length
(Å)
Bond energy
(kcal/mol)
He2 2 2 0 * *
H2+ 1 0 1 1.06 61
He2+ 2 1 1 1.08 55
H2 2 0 2 0.74 103
The number of net bonding electrons predicts the length and strength of the bond! Since we normally think of a chemical bond as a 2-electron bond, we can define the bond order like this:
$Bond\; Order = \frac{\left(bonding\; e^{-}\right) - \left(anti \mbox{-} bonding\; e^{-}\right)}{2}$
If bond order = 1, there is a single bond (H2, for instance). If bond order = 0, we expect no bond. If bond order = 0.5, we have a 1-electron bond or a half bond. It's approximately half as strong as a 2-electron bond.
Contributors
Mixing Orbitals to Make MOs
Skills to Develop
• Describe and illustrate MOs in a molecule
Molecular Orbitals or MOs are orbitals in a molecule. Remember that atomic orbitals (AOs) are "electron probability clouds." We can't know exactly where an electron is or what path it follows, because of the Uncertainty Principle, but we can find a mathematical function that tells us how likely we are to find the electron at each position around an atom. If we think of electrons behaving like waves, which they do, the wavefunction Ψ tells us the amplitude of the "electron wave" at each point around the atom. Ψ2 tells us the probability of finding the electron there. When we combine atoms to make molecules, there's wave interference between the electron waves on different atoms. The electrons in a molecule occupy molecular orbitals, or MOs, that are combinations of the atomic orbitals made using the basic rules of wave interference.
MOs have the same basic properties that atomic orbitals have. Each MO can hold 2 electrons with opposite spins. Ψ2 still represents the probability of finding the electron. Each MO has a specific energy, and removing an electron from that MO requires that much energy. However, MOs can have very different shapes compared to MOs, and they can be spread out over many atoms in a molecule.
We make MOs in a way similar to how we made hybrid orbitals. If we combine 2 AOs (call them φA and φB) on different atoms, we will make 2 new MOs (call them ψ1 and ψ2).
$\Psi_{1} = C \left(\varphi_{A} + \varphi_{B}\right)$
$\Psi_{2} = C \left(\varphi_{A} - \varphi_{B}\right)$
We will make one MO by adding the AOs and the other by subtracting the AOs. C is a constant that "normalizes" the wavefunction so that the total probability of finding the electron over the whole atom is 1. | textbooks/chem/General_Chemistry/General_Chemistry_Supplement_(Eames)/Molecular_Orbital_Theory/MO_Theory%3A_Simplest_Examples.txt |
Skills to Develop
• Construct MO diagrams for simple diatomic molecules and/or compounds
We saw two simple MO diagrams in the section on H2. Now let's think about how to make some slightly more complicated MO diagrams. First, we need to know a little about how big the energy splitting between the bonding and anti-bonding MOs is. Splitting is the energy difference between the bonding and anti-bonding orbitals. Usually the bonding orbital goes down almost as much as the anti-bonding orbital goes up, so the average energy stays almost the same. The size of the splitting depends on the energy match and the overlap.
Energy match means how close the orbital energies are. The reason we can consider only valence orbitals is that the core orbitals have much lower energy, and the higher empty orbitals have much higher energy, than the valence electrons. Interactions between completely completely empty orbitals don't matter because there are no electrons. Interactions between completely filled orbitals are usually repulsive, which is why the noble gases don't usually make bonds. Interactions between partially-filled valence orbitals and either core orbitals or higher shell orbitals aren't important because the energy match is bad, so the splitting is almost zero.
We talked about overlap a little in the previous section. Overlap means how much the orbitals touch. For example, usually σ combinations have bigger overlap than π combinations, because the orbitals are pointed right toward each other. You can see this in the pictures on the previous page. For this reason, usually σ MOs have bigger splitting than π MOs.
MO Diagram for HF
In the last section, we talked about the bonding, non-bonding, and anti-bonding MOs in HF. Now let's put these ideas together to make an MO diagram for HF.
1. We need to know what orbitals we are using. We are only going to consider valence orbitals. H has a 1s orbital. F has a 2s orbital and 3 2p orbitals (x,y,z).
2. We want to know the energies of the orbitals. We can use photoelectron spectroscopy data, which tells us the energy of the different orbitals. Here is some data you can use. We see that H 1s orbital has energy -13.6 eV, F 2s has energy -40 eV and F 2p has energy -18.7 eV. Because there is a big energy difference, more than 12eV, between H 1s and F 2s (bad energy match), we can put just H 1s and F 2p in our diagram.
3. We draw the AOs on the outside of the diagram and include the right number of electrons. H has 1 valence electron, and F has 7 valence electrons. We are only including the F 2p orbitals, which have 5 electrons; the F 2s orbital holds 2 electrons and isn't in the diagram (it's like a lone pair).
1. We remember which orbitals interact. Before, we saw that bonding and anti-bonding combinations only form between H 1s and the F 2p orbital that points straight toward it. The other 2p orbitals are non-bonding, so we can draw them in the middle at the same starting energy.
1. The H 1s and F 2pz make a bonding and anti-bonding combination, so we draw these new MO energy levels. We don't know exactly how big the splitting is, but that's ok, don't worry about it.
1. We put the same number of valence electrons we had in the AOs on the outside into the MOs at the center, starting at the bottom. In this case, we have 6 electrons. They will go into the bonding MO and the 2 non-bonding MOs. The anti-bonding MO is empty because it is too high in energy. We're done!
MO Diagram for F2
Let's do another example. This time we'll do F2, which is a little more complicated. This time we'll use σ and π bonds.
1. What orbitals we are using? F has a 2s orbital and 3 2p orbitals (x,y,z). For now, let's just consider the 2p orbitals. We'll see what happens with 2s orbital in the next section.
2. We don't need to worry about the energies this time, because they all start the same.
3. We draw the AOs on the outside of the diagram and include the right number of electrons. F has 7 valence electrons. We are only including the F 2p orbitals, which have 5 electrons; the F 2s orbital holds 2 electrons and isn't in the diagram yet.
1. We remember which orbitals interact. Before, we saw that we make σ bond and anti-bonding combinations using the 2p orbitals that point toward each other. Let's draw those in.
1. We also make π bonding and anti-bonding combinations using the other 2p orbitals. Because these have less overlap than the σ combinations, the splitting will be smaller, so we'll draw these in between the σ bonding and anti-bonding orbitals. There are 2 of each (π bonding and anti-bonding), so we draw 2 lines for each. They are the same energy because the are the same except rotated 90°.
1. We put the same number of valence electrons we had in the AOs on the outside into the MOs at the center. In this case, we have 10 electrons. They will go into the σ bonding MO and all 4 π MOs. Only the σ* MO is left empty. We're done!
Reading MO Diagrams
When you look at an MO diagram, you can see what AOs are included by checking the outside of the diagram. The middle shows you how they combine and approximately what the energies of the combinations are. You can tell which orbitals are bonding because they have lower energy than the AOs. Non-bonding orbitals have about the same energy as the AOs, and anti-bonding orbitals have higher energy than the AOs. You can use the number of electrons in each type of MO to find the bond order. You can also tell how many unpaired electrons there are, which tells you about the magnetic properties. You can make some guesses about colors, because these usually come from having a small gap between full and empty orbitals. Finally, you can tell which parts of the molecule are most reactive: they will have a low energy empty MO or a high energy full MO. We'll see some more examples of this later. | textbooks/chem/General_Chemistry/General_Chemistry_Supplement_(Eames)/Molecular_Orbital_Theory/Reading_and_Writing_MO_Diagrams.txt |
Skills to Develop
• Label the parts of an MO diagram
Bonding, Anti-bonding, and Non-bonding MOs
In the previous section, we introduced bonding and anti-bonding MOs. Bonding MOs have more electron density between the nuclei, and lower energy than the atomic orbitals they were made from. Putting electrons in bonding orbitals tends to make a bond between the nuclei, because when the electrons spend time between the nuclei, both nuclei are attracted to the negative charge between them.
Anti-bonding orbitals have less electron density between the nuclei, because they have a node there. They have higher energy than the AOs they were made of. Putting electrons in anti-bonding orbitals tends to break bonds, because with the electrons on the outside, the nuclei repel each other and all the electrons repel each other.
There is a third type of MO: non-bonding MOs. In order to make a bond, orbitals have to have net overlap, which means that if you multiply them together and take the integral over the whole molecule, the integral isn't 0. Consider the HF molecule. The H 1s orbital can make a bond with the F 2s or F 2pz orbital. However, there is no net overlap between the H 1s and the F 2px and 2py. This is shown in the figure. When we multiply Ψ1stimes Ψ2p, the top half is + * + = +. The bottom half is + * — = —. Except for the sign, the top half and bottom half are symmetrical, so when we add up the values of Ψ1sΨ2p everywhere, the top half and bottom half cancel out, and ∫ Ψ1sΨ2p dV = 0. (Note that multiplying the wavefunctions is different from adding them, which would give us the wave interference patterns we saw in hybrids and MOs.) For this reason, the F 2px and 2py orbitals in HF are called non-bonding orbitals. In contrast, the 2pz orbital does have net overlap with 1s, because │Ψ1sΨ2p│ on the red side in the figure is bigger than │Ψ1sΨ2p│ on the blue side.
The figure below shows a summary illustration of bonding, non-bonding, and anti-bonding orbitals in HF.
σ and π MOs
The bonding and anti-bonding MOs shown above are both σ-type MOs. Recall from the section on multiple bonds that we can classify bonds as σ or π bonds. σ-bonds are symmetrical around the bond (if you rotate them around the bond, they don't change). π-bonds change sign when you rotate them 180° around the bond.
If we think about making MOs for F2, we can imagine making a bonding and anti-bonding combination of the 2pz orbitals, which point toward each other. This will make a σ-bonding MO and an σ-anti-bonding MO. We can also make σ-type combinations of the 2s orbitals. When we combine the 2px and 2py orbitals, these are perpendicular to each other, so they will make π-type bonding and anti-bonding combinations. These are shown below. Note that for both σ combinations and π combinations electron density increases between the nuclei in bonding MOs and decreases between nuclei in anti-bonding MOs.
Naming MOs
The easiest way to name and label MOs is using σ and π. Anti-bonding character is shown using a *, such as π* which means a π-type anti-bonding orbital. Each MO in the figure above is labeled in this way. There are other more complicated ways to name MOs, but you won't learn them unless you take Inorganic Chemistry. | textbooks/chem/General_Chemistry/General_Chemistry_Supplement_(Eames)/Molecular_Orbital_Theory/Types_of_MOs.txt |
Periodic trends are specific patterns that are present in the periodic table that illustrate different aspects of a certain element, including its size and its electronic properties. Major periodic trends include: electronegativity, ionization energy, electron affinity, atomic radius, melting point, and metallic character. Periodic trends, arising from the arrangement of the periodic table, provide chemists with an invaluable tool to quickly predict an element's properties. These trends exist because of the similar atomic structure of the elements within their respective group families or periods, and because of the periodic nature of the elements.
Periodic Trends
Skills to Develop
• List and explain the properties of metals
Remember that metals are on the left and bottom of the periodic table. Based on the periodic trends in the last 3 sections, this means that they are usually bigger, more likely to lose electrons, and less likely to gain electrons, than the non-metals.
Elemental Properties
In the elemental form, metals are usually shiny, can be bent or stretched, and conduct heat and electricity. This is because metals hold their valence electrons kind of loosely, because of the low IE. It's a general pattern that the closer an atom is to the noble gas electron configuration, the fewer bonds it makes. Metals are far from the noble gas configuration, so they usually make bonds to many neighbors. Thus, they have have structures in which each atom touches many neighbors (sometimes 6, usually 8 or 12). Because they don't have very many electrons, the valence electrons are shared by many atoms in a "delocalized ocean" of electrons that aren't really attached to particular atoms. These "free floating" electrons allow the metal to conduct heat and electricity. Also, because the bonds in a metal aren't pointed directly from atom to atom, they don't break easily. This means the metals can be bent and stretched. The technical words are malleable (can be pounded into foil) and ductile (can be stretched into wire). The sea of valence electrons allows the nuclei to be pushed around without separating. I don't think there's a simple explanation for why metals are shiny, though.
Reaction Patterns
Metals can mix with each other to form alloys that are similar to elemental metals. When they react with non-metals, they usually lose electrons to form cations. The cations are then attracted to the anions, so the result are ionic or sort of ionic compounds. The easier it is for a metal to lose electrons, the more reactive it is. If you haven't already, watch this video about the reactivity of the alkali metals. The metals on the right that have bigger ionization energies form more covalent compounds with non-metals, with more electron sharing and less pure electrostatic attraction. These metals are also more flexible about what ionic charge they have. See why by going to Ptable and watching the purple move right as you go from IE1 to IE4 or higher. The transition metals stay green: they don't take a sudden jump to much higher IE.
Characteristics of Nonmetals
Skills to Develop
• List some characteristics of nonmetals
• Contrast metals and nonmetals
Remember that non-metals are on the right and top of the periodic table. Based on the periodic trends in the last 4 sections, this means that they are usually smaller, more likely to gain electrons, and less likely to lose electrons, than the metals.
Elemental Properties
In the elemental form, non-metals can be gas, liquid or solid. They aren't shiny (lustrous) and they don't conduct heat or electricity well. Usually their melting points are lower than for metals, although there are exceptions. The solids usually break easily, and can't bend like metals. It's a general pattern that the closer an atom is to the noble gas electron configuration, the fewer bonds it makes. Non-metals are close to the noble gas configuration, so they usually make a few bonds to a few neighbors. The noble gases make no bonds, and are monatomic (single atoms); halogens make 1 bond to 1 other atom, etc. This means that they don't usually form extended structures (except diamond and graphite). Instead, they form separate molecules. These molecules aren't held together tightly, so solids can easily melt or break. The electrons are held tightly by just 1 - 2 atoms, so they can't conduct electricity.
Reaction Patterns
Non-metals can react with each other to form compounds in which electrons are shared. These compounds have some of the same characteristics as the elementals forms: usually they melt or boil at relatively low temperature and don't conduct heat or electricity. When non-metals react with metals, they usually gain electrons to form anions. The cations are then attracted to the anions, so the result are ionic or sort of ionic compounds. The more a non-metal wants to gain electrons, the more reactive it is. Thus, the halogens are all reactive, but iodine is pretty safe, while bromine, chlorine and especially fluorine are really nasty and dangerous! Oxygen only seems safe and friendly to us because we are adapted to it. When oxygen first appeared in the atmosphere due to photosynthesis, most of the early life forms probably died from it; we descended from the survivors. | textbooks/chem/General_Chemistry/General_Chemistry_Supplement_(Eames)/Periodic_Trends/Characteristics_of_Metals.txt |
Skills to Develop
• Explain the difference between nuclear charge and effective nuclear charge
The reason electrons are attached to atoms is the Coulomb's law attraction between the positively charged nucleus and the negatively charged electrons. Without the nuclear charge holding on to the electrons, they would have no reason to stay in orbitals near nuclei. So it makes sense that energy of the orbitals and their size depend on the nuclear charge. For instance, equivalent orbitals get lower in energy and smaller (more density closer to the nucleus) when nuclear charge increases. (If you want to see where nuclear charge is in the orbital equations, follow this link and click on an orbital; then scroll down to see the equation. Zeff is the effective nuclear charge.)
There are 2 reasons an electron might not spend that much time actually right next to the nucleus, even though there is an attraction. One is angular momentum. Remember from physics that angular momentum is basically the momentum in the "around the center" direction multiplied by r, the distance from the center. Thus, if the electron is "in" the nucleus, r = 0, and angular momentum is 0. Remember also the second quantum number is the angular momentum quantum number ℓ, which corresponds to s, p, d and f orbitals. For s orbitals, ℓ = 0, and angular momentum is zero, so the electron can hang out at the nucleus. For the other shapes (p,d,f) angular momentum is not zero, and there is a node at the nucleus. So only s orbitals can hang out right at the nucleus.
The other reason an electron might not be able to get that close to the nucleus is because of the other electrons. Electrons repel each other, and because of the exclusion principle there can only be 2 electrons in each orbital. So think about lithium, element 3. Its first 2 electrons are in 1s. The third electron is in 2s. Because the principal quantum number is bigger, the 2s electron is usually farther from the nucleus than the 1s electrons. That means that often the 1s electrons are between the 2s electron and the nucleus. This means that the total Coulomb attraction felt by the 2s electron is smaller than you would calculate using just nuclear charge and distance, because there is also the electron-electron repulsion. This is the basis of the effective nuclear charge. Because of the electron-electron repulsions, the outer electrons in a many-electron atom feel less attraction toward the nucleus, so they feel like the nuclear charge is smaller than it actually is. So effective nuclear charge is always smaller than actual nuclear charge.
Effective nuclear charge depends on the type of electron. Electrons in s orbitals, even 4s or 5s, still spend some time right at the nucleus, and when they are there, they feel the full nuclear charge, so on average the s electrons feel a nuclear charge closer to the actual nuclear charge. Electrons in d or f orbitals really don't get very close to the nucleus, so they really get blocked, or shielded by inner electrons. They feel a smaller effective nuclear charge than s electrons. And electrons in p orbitals are in between.
Usually if you do any calculation of orbitals for many-electron atoms, you will use effective nuclear charge instead of actual nuclear charge. The size and energy of the orbitals will depend on effective nuclear charge, not on actual nuclear charge. Size and energy of orbitals determines some very important chemical properties, including the size of the element (as an atom, ion, or in a molecule) and how easily it loses or gains electrons. Electrons take up most of the space in an atom, so orbital size tells you size. Losing electrons requires the energy by which they are bound, which is the roughly same as the orbital energy. Adding electrons only works if the orbital where they will go is lower energy than where they came from. | textbooks/chem/General_Chemistry/General_Chemistry_Supplement_(Eames)/Periodic_Trends/Effective_Nuclear_Charge.txt |
Skills to Develop
• Define electron affinity
• Describe the periodic trend of electron affinity
Electron affinity is a measure of how much an atom wants to gain an electron, becoming an anion. Unfortunately 2 different definitions are used: intro textbooks use 1 definition and everyone else uses the other! I think you should use the standard advanced definition, according to which electron affinity EA = IE0, the energy of this reaction:
$A^{−}(g) \rightarrow A(g) + e^{−}(g)$
The other definition just has the opposite sign, because it's the energy of this process:
$A(g) + e^{−}(g) \rightarrow A^{−}(g)$
Why does Electron Affinity matter?
Like ionization energy, electron affinity tells us how likely an atom is to steal electrons from other atoms, or just convince them to transfer their loyalty partially.
Predicting Relative Electron Affinities
It pretty much follows the same pattern you would expect based on ionization energy. If the new electron goes into an orbital that feels high effective nuclear charge, gaining an electron is good. If it goes into an orbital that is in a higher shell or subshell, or that feels low effective nuclear charge, gaining an electron is bad. The only other thing is that it's easier to add electrons to bigger atoms, because then they can spread out and not have so much electron-electron repulsion. For this reason, chlorine has more favorable EA than fluorine (in the gas phase, fluorine is still much more reactive).
Look at EA for yourself!
Go to Ptable's electron affinity page. See the general trend (bigger middle and right) with big effects at the end and middle of each block (noble gases, mercury family, alkaline earths, nitrogen, manganese).
Ionization Energy
Skills to Develop
• Describe the significance and periodic trend of ionization energy
Ionization energy is the energy needed to remove an electron from an atom or ion. Unlike atomic radii, we can and do measure ionization energies in the gas phase, when the atom or ion is not interacting with anything else. The first ionization energy, IE1, is the energy of this reaction
$A(g) \rightarrow A^{+}(g)+e^{-}(g)$
The second ionization energy, IE2, is the energy of
$A^{+}(g) \rightarrow A^{2+}(g) + e^{-}(g)$
You can also measure third, fourth, etc. All of these assume that the highest energy electron is knocked off. If you are removing core electrons, you would write them like this: IE1s, to show what orbital the electron comes from.
How do you Measure Ionization Energy?
Generally you do this by shining high energy photons on the material you want to study. UV will ionize valence electrons and X-ray can ionize core electrons. You can vary the wavelength of the photons, and also measure the kinetic energy of the electrons that come off, to see what the binding energy (orbital energy) for each electron is. You can also measure the ionization energy of molecules, which is very interesting because it can tell you how the orbital energies change because of bonding.
Why does it matter?
Ionization energy tells us how likely an atom is to form a cation, and if so, what charge. In general, it tells us how tightly the electron is bound, how stable it is. It can tell us the energies of real orbitals, the effects electrons have on each other, and help us predict reactivity and properties of molecules. (We'll talk more about this when we get to bonding!)
Predicting Relative Ionization Energies
Ionization energy depends on orbital energy, which depends on the type of orbital and the effective nuclear charge. Thus, it follows predictable patterns in the periodic table. As you go down, n increases, and the energy of the orbital increases. That means the orbital is less stable, so it's easier to pull off the electron, so ionization energy decreases. (Stable, bound electrons have negative energies; electrons that aren't in an atom have 0 energy.) As you go across the periodic table, usually the type of orbital is the same, and the effective nuclear charge increases, making the orbital more stable, so ionization energy increases. But when you change subshells, the ionization energy might increase less, because the new subshell is less stable. Also, remember Hund's rule: electrons are more stable when they don't share the same orbital. So if you have to put a new electron into an occupied orbital, that also makes the ionization energy increase less.
You can predict any relative IE just by thinking about how big the Coulomb forces are. Bigger n, means bigger distance, weaker force. Removing an electron from a neutral atom is easier than removing an electron from a cation, because of the charge. Also consider effective nuclear charge and electron-electron repulsions (especially in the same orbital). In summary, mostly IE increases up and to the right, because of low shells and high effective nuclear charge.
Look at IE for yourself!
Go to Ptable's ionization energy page. You can look at first, second, third, etc. See the general trend (bigger up and right), and also notice some exceptions as you change blocks (s-block or d-block to p-block). | textbooks/chem/General_Chemistry/General_Chemistry_Supplement_(Eames)/Periodic_Trends/Electron_Affinity.txt |
Skills to Develop
• Describe the significance of and periodic trend for atomic radii
What are Atomic Radii?
Atomic radii are the radii of atoms, a measure of how big the atoms are. That seems kind of simple. It's a useful idea because if we know the radii, we can predict how big molecules are, whether different parts will touch each other, etc. This is good for designing molecules for particular purposes, or interpreting data, and other things.
How do you Measure Atomic Radii?
This is where the trouble starts! Atoms are more like clouds than metal balls. It's easy to measure the diameter of a metal ball, because it doesn't change; the ball is hard. But how do you measure the size of a single cloud that is touching many other clouds? Remember from the nuclear model of the atom that the nucleus is very tiny. Almost all the space in an atom is taken by the electrons in their orbitals, and the orbitals, like clouds, don't have obvious edges; also, they can change size and shape when other atoms are nearby because of Coulomb forces from other nuclei and electrons. We usually measure the size of atoms using X-ray crystallography. Basically we shine X-rays on a crystal (like a salt crystal or a diamond, a crystal is an orderly arrangement of atoms with sharp, straight edges). The X-rays interact with the electrons in the crystal, and the picture we get from the X-rays can tell us the electron density at each point in the crystal. Most atoms have lots of electron density because they have 6 or more electrons. We can see how far apart these electron density peaks are, and that tells us where the atoms are. Then we guess their size based on how far away they are from each other. I say "guess" because the distance between atoms depends on whether they are bonding or not bonding, etc. So the size depends on the type of environment we look at. We can't look at all types of atoms in the same environment because they have different bonding properties, but by looking at lots of different situations we can see some general trends.
Predicting Relative Sizes
Because electrons are what take up space in atoms, the result is that the size of the biggest filled orbital determines the size of the atom or ion. Sizes of orbitals depend on the quantum numbers (n = 1, n = 3, etc.) and also on the effective nuclear charge. An orbital of any type will get smaller as the effective nuclear charge gets bigger, because the attraction to the nucleus is greater. As you go down the periodic table, usually atoms get bigger because n gets bigger (there are electrons in higher shells). Effective nuclear charge does get bigger too going down the periodic table, but this effect is smaller than the change in shell. As you go left across the periodic table, the effective nuclear charge increases. The number of electrons also increases, but they are usually in the same shell or subshell, so the effective nuclear charge increase is more important, and the atoms or ions get smaller going left. There are many exceptions, but for now bigger to the right and down is what you should remember.
The sizes of ions follow a simple pattern. When you remove electrons, making cations, there are less electrons and less electron-electron repulsions, so the cation is smaller than the atom. The more electrons you take off, the smaller it gets. Anions are the opposite. When you add electrons, they repel each other, and there are more of them, so anions are bigger than atoms, and get bigger as you add electrons. You can combine these patterns with the general pattern for atoms to predict relative size of ions.
In general, to predict relative sizes, ask: How many electrons? What shell? What effective nuclear charge? Find the differences between the particles you are comparing, and see what effect the differences should have.
Check out the Atomic Radii for yourself!
Go to Ptable's radius page, and look at the radii. Try selecting different types of radii, like calculated, empirical, covalent, and Van der Waals. Notice that some types of radii only have data for some elements, and that the patterns change a little depending on how you measure. | textbooks/chem/General_Chemistry/General_Chemistry_Supplement_(Eames)/Periodic_Trends/Sizes_of_Atoms_and_Ions.txt |
Skills to Develop
• Describe how to calculate the shielding constant using Slater's rules
• Identify a periodic trend for effective nuclear charge
• Calculate effective nuclear charge
Effective nuclear charge is really important, because it determines the size and energy of orbitals, which determine most properties of atoms. So it's useful to be able to predict effective nuclear charge! Slater's rules give a simple approximation of effective nuclear charge that works pretty well.
Based on the last section, we can expect that effective nuclear charge will depend on the number of electrons that might get between, so it depends on the electron we are looking at. For any electron, to find the effective nuclear charge it feels, we need to know how many other electrons might get in the way, and how much time it spends near the nucleus. Based on these, we will calculate a shielding constant, S. Then,
\[Z_{eff}=Z-S\]
where Z is the actual nuclear charge (which is the same as the atomic number) and Zeff is the effective nuclear charge.
To calculate S, we will write out all the electrons in atom until we get to the group of the electron we want, like this:
(1s)(2s,2p)(3s,3p)(3d)(4s,4p)(4d)(4f)(5s,5p) etc.
1. Each other electron (not counting the one we have picked) in the same group () as the chosen electron, contributes 0.35 to S. (This means electrons in the same group shield each other 35%.
2. If the chosen electron is d or f, every electron in groups to the left contributes 1.00 to S. (This means that d and f electrons are shielded 100% by electrons with lower n or same n and lower ℓ.)
3. If the chosen electron is s or p, all the electrons in the next lower shell (n - 1) contribute 0.85 to S. (This means that s and p electrons are shielded 85% by the electrons one shell lower.) And all the electrons in even lower shells contribute 1.00 to S. (All electrons in shells n - 2 or lower shield 100%.)
The outcome of this is that Zeff changes suddenly when going from one period to another. As you go from Li to Be, Zeff (for the new electron) increases, because you add one proton (Z + 1) and it is only shielded 35% (S + 0.35). When you get to B, you added one proton, and it still shields 35%. So Zeff increases until you go from Ne to Na. Now, suddenly, the (1s) electrons shield 100% instead of 85%, and the (2s,2p) shield 85% instead of 35%! So Zeff goes down suddenly. From Na to Ar, Zeff increases slowly again. From Ar to K, it drops again.
For an example, let's calculate Zeff for a d electron in Zn, atomic number 30. Notice that although 4s is filled, we don't include it because it comes to the right of the d electrons we are looking at.
(1s)(2s,2p)(3s,3p)(3d)
\[S=18(1)+9(0.35)=21.15\]
\[Z_{eff}=31-26=5\]
You can see that just like changing periods (going to a new shell), going from the d-block to the p-block also gives a drop in Zeff (partly because you actually are going to a new shell, as well as subshell). | textbooks/chem/General_Chemistry/General_Chemistry_Supplement_(Eames)/Periodic_Trends/Slater%27s_Rules_for_Effective_Nuclear_Charge.txt |
Skills to Develop
• Define and illustrate dipole-dipole forces
Dipole-dipole forces are probably the simplest to understand. You probably already know that in an ionic solid like NaCl, the solid is held together by Coulomb attractions between the oppositely-charges ions. The Na+ and Cl- ions alternate so the Coulomb forces are attractive. Dipole-dipole forces work the same way, except that the charges are smaller. A good example is HF (this is also an example of a special type of dipole-dipole force called a hydrogen bonding). In HF, the bond is a very polar covalent bond. That means there is a partial negative (δ-) charge on F and partial positive (δ+) charge on H, and the molecule has a permanent dipole (the electrons always spend more time on F). In the liquid or solid HF, the molecules arrange themselves so that the δ- and δ+ are close together. These partial charges attract each other, and this attraction is what we call dipole-dipole forces. Any molecule with a permanent dipole has dipole-dipole forces that hold the molecules next to each other as a solid or liquid.
Hydrogen Bonding
Skills to Develop
• Define and illustrate hydrogen bonds
It's a general rule that "hard" things like to bond with other hard things, and "soft" things like to bond with other soft things. (See the previous section for an explanation of hard and soft.) The reason for this is because hard-hard combinations are very favorable. Remember that hard means "hard to polarize", which usually means small and highly charged. Hard-hard combinations are based on strong Coulomb forces, between relatively big charges relatively close together.
Hydrogen bonds are a special case of hard-hard interaction that occurs in covalent molecules. Hydrogen that is bonded to very electronegative elements (N, O and F) will have a big δ+ (partial positive charge). N, O or F bonded to C or H will have a big δ- (partial negative charge). If the N, O or F has lone pairs, these can make interactions with a hydrogen on another molecule. The partial charges are pretty big, because of the big difference in electronegativity. Also, they can get very close together because these elements (especially H) are very small.
Hydrogen bonds are also an example of a dipole-dipole force, but they are extra big dipole-dipole forces because the charges are big (for partial charges) and the distances are short. They are also an example of Lewis acid-base interactions (because the electrons in the interaction come from a lone pair on N, O or F). And they are also an example of a HOMO/LUMO interaction. The H which is bound to a very electronegative atom has a low LUMO, because the energy match is bad between very different electronegativity atoms, leading to low splitting. The lone pair on the N, O or F is a high HOMO, because it is non-bonding, not bonding. These can then make a new bond.
Hydrogen bonds are very important. Because they are very strong, water is a liquid over a much wider temperature range than we would expect otherwise. Hydrogen bonds are also extremely important in biochemistry. They help hold proteins in their correct shape, help DNA store genetic information, and help enzymes make reactions go quickly. | textbooks/chem/General_Chemistry/General_Chemistry_Supplement_(Eames)/Phases_and_Intermolecular_Forces/Dipole-dipole_Forces.txt |
Skills to Develop
• Describe phases and phase changes on a molecular level
Phases of matter mean the state of a material, like solid, liquid or gas. How can we predict whether a material will be a solid, liquid or gas under certain conditions? We have to know about the forces that hold the material together.
In solids, the separate molecules or ions are held tightly in their positions by some type of force. They might vibrate a little bit in place, but they don't move around. For this reason, solids don't easily change shape. You might be able to bend or break them, but usually they don't change shape by themselves.
In liquids, the separate molecules or ions are held close together by some type of force, but they can move around while staying close together. For this reason, a liquid can change shape to fit whatever is holding it. But like solids, liquids have approximately constant volume. Even if the liquid flows into a new shape, the distance between the molecules doesn't change, so the total volume stays the same.
In a gas, usually the molecules bounce around as though there are no forces between them. (See Kinetic-Molecular Theory of Gases.) At very high pressure, when they are forced to be close together, we might start to notice that there are some forces between the molecules (because the pressure is less than we expect from the Ideal Gas Law) but usually they move around separately. Because they aren't really attracted to each other and have a lot of kinetic energy (at least at normal temperatures), they fill the whole space they have. So gases can take any shape, and also can change volume a lot.
The higher the temperature, the more kinetic energy the molecules or ions have. With more kinetic energy, it's harder for them to stay in their place in a solid, or not to bounce right out of a liquid and become a gas. So as we increase the temperature, we might see phase transitions from solid to liquid to gas. We can think about these transitions using equilibrium, like when we think about reaction and solubility equilibria. Think about a liquid in a closed container, like a bottle half full of water. Some of the molecules have bigger kinetic energy and some have smaller kinetic energy. If a molecule with big kinetic energy is on the surface of the liquid, it might fly off and enter the gas state. At the same time, other molecules in the gas state might bump the surface of the liquid, and if they don't have very much kinetic energy, they might stay there and join the liquid. This is a dynamic equilibrium: the molecules go back and forth between the 2 states. If we increase the temperature, the average kinetic energy increases, and that means the molecules are more likely to have enough kinetic energy to go into or stay in the gas phase. Liquid molecules will become gas molecules more often, and gas molecules will become liquid molecules less often. Then the equilibrium will move, so a bigger % of the total molecules are gas.
In many cases, there is a specific temperature above which all of a material goes from solid to liquid (a melting point) or from liquid to gas (a boiling point). What temperature that is depends on the strength and type of forces between the molecules and ions. If the forces are strong, then more kinetic energy is needed to make the molecules move around or separate and become gas, which means the melt point and boiling point are higher. If the forces between molecules are very weak, then the material may be a gas, and it may be hard to cool it enough to make a liquid.
In the next sections, we will talk about the forces between molecules that determine boiling points and melting points and other important properties. (Thus, they are called intermolecular forces, to separate them from the forces inside molecules that hold the molecules together.) These forces are often called Van der Waals forces after Johannes van der Waals, who wrote the equation for real gases. Van der Waals figured out that the reason gas pressures are often lower than we expect (at high pressure) is because there are attractions between the molecules. Actually Van der Waals was the son of a carpenter who wasn't allowed to enter university because he didn't have the right expensive primary education. But he took classes anyway, became a teacher, and eventually they changed the rules for university admission, so he got a doctorate, became famous, became a professor, and won the Nobel Prize. | textbooks/chem/General_Chemistry/General_Chemistry_Supplement_(Eames)/Phases_and_Intermolecular_Forces/Intro_to_Phases_and_Intermolecular_Forces.txt |
Skills to Develop
• Explain some of the properties of liquids
• Distinguish cohesive and adhesive forces
In general, liquids are harder to describe than gases (in which interactions between particles are simple collisions) and solids, in which particles stay mostly still in an organized arrangement. So we will only describe some properties here. Certain properties of liquids also depend on the intermolecular forces, like the viscosity and surface tension. These roughly describe the shapes liquids take when poured, or as droplets, etc.
Viscosity
Viscosity means how thick or sticky a liquid is. For instance, water pours easily and quickly, so it is pretty low viscosity. Honey is a thick, sticky liquid that pours slowly, so it has higher viscosity. Viscosity depends on how easily the molecules can flow past each other. The smaller they are, and the weaker the forces between them, the easier they flow. If the molecules are big and flexible, they might be able to get a bit tangled together, and that could make them flow more slowly.
Surface Tension
Surface tension means how much the liquid wants to minimize its surface area. If the intermolecular forces are big, then molecules would rather be inside the liquid where they have favorable intermolecular interactions instead of being on the surface. This could make the liquid pull itself into rounded shapes to make the surface area smaller. You've probably seen water do this, like on a non-stick pan, because water has strong hydrogen-bonds. In this case, they are called cohesive forces, which means forces that pull the material together. (A cohesive team is very close and works well together.) On the other hand, water can have good interactions with glass surfaces, so it doesn't mind so much spreading out on glass. In fact, the forces that make water stick to glass, called adhesive forces are bigger than the cohesive forces in water, which is why the water in a glass tube is higher around the edges.
In contrast, mercury doesn't spread out over glass, and in a glass tube, it is higher in the middle. This means that mercury has bigger cohesive forces than adhesive forces toward glass.
London Dispersion Forces
Skills to Develop
• Define London dispersion forces
It's not too hard to see why dipole-dipole forces hold molecules like HF or H2O together in the solid or liquid phase. However, let's think about the halogens. F2 and Cl2 are gases, Br2 is a liquid, and I2 is a solid at room temperature. But I2 has no dipole moment to make attractions between the molecules. But actually, although I2 has no permanent dipole moment, it can have a temporary dipole moment. We mentioned this before, when we talked about polarizability. Go back and read that section.
London dispersion forces can explain how liquids and solids form in molecules with no permanent dipole moment. "Dispersion" means the way things are distributed or spread out. Because the electrons move around a lot, sometimes they may move in a way that creates a temporary dipole moment. The more electrons an atom has, the more easily this can happen, because the electrons are held more loosely, far from the nucleus. (Basically, the energy gaps between orbitals become smaller as we move to higher shells, allowing the electrons to more easily move into excited states, occupying orbitals higher than they need to. This gives them more flexibility to move around and create temporary dipole moments.) The technical word for an element that is polarizable, or able to have temporary dipoles, is "soft". In other words, it can squish and change shape. Elements that can't polarize easily (which usually means low atomic number) are called "hard".
Fluorine is really really hard. In F2, both F atoms are holding all the electrons really tightly, trying to grab them and not share. In contrast, iodine is really soft. It's electrons are far away from the nucleus, and they can move around easily. If they all happen to move one direction, creating a temporary dipole, the other molecules nearby can adjust, making more dipoles to attract the first one. These are called induced dipoles, because they appear in response to the original accidental dipole. Lots of induced dipoles can create attraction between molecules, called London dispersion forces.
London dispersion forces are always present, but they vary widely in strength. In light atoms, they are very small, because there aren't many electrons and they are held tightly. In large atoms, they can be very big, because the atoms are very soft and easy to polarize. Generally, London dispersion forces depend on the atomic or molecular weight of the material. Heavier atoms or molecules have more electrons, and stronger London forces. This means that they are harder to melt or boil. This explains the states of the halogen molecules at room temperature. | textbooks/chem/General_Chemistry/General_Chemistry_Supplement_(Eames)/Phases_and_Intermolecular_Forces/Liquids.txt |
Skills to Develop
• Describe the relationship between heat (energy), bonding forces, and phase changes
Most phase changes occur at specific temperature-pressure combinations. For instance, at atmospheric pressure, water melts at 0 °C and boils at 100 °C. In this section, we will talk about when and how they happen. The names of the different phase changes are shown below:
Predicting Phase-Change Temperatures
We can predict the relative temperature at which phase changes will happen using intermolecular forces. If the intermolecular forces are strong, then the melting point and boiling point will be high. If the intermolecular forces are weak, the melting and boiling point will be low.
London forces vary widely in strength based on the number of electrons present. The number of electrons is related to the molecular or atomic weight. Heavy elements or molecules, like iodine or wax, are solids at room temperature because they have relatively strong London forces, which correlate with big molecular weights. London forces are always present, but in small molecules or atoms, like helium, they are quite weak.
Dipole-dipole forces are present in molecules with a permanent dipole. We can predict this by drawing a Lewis structure, identifying polar bonds using electronegativity, predicting the shape of the molecule, and seeing if the bond dipoles on different molecules can touch. If they can, there will be dipole-dipole forces. The bigger the dipoles (bigger electronegativity difference, etc.) and the closer together they can get, the bigger the dipole-dipole forces are.
Hydrogen bonds occur only when there are H atoms bonded to N, O, or F and lone pairs on N, O, or F. Look for both of these in molecules to see if they can hydrogen bond.
If you can find which types of intermolecular forces are present in a molecule, you can make some guesses about which molecules have higher or lower melting or boiling points. For instance, let's compare methane (CH4), silane (SiH4), hydrogen sulfide (H2S) and water (H2O). Methane and silane are non-polar, because of the tetrahedral shape and also the small electronegativity differences. Because these don't have dipole-dipole forces, the boiling point will depend on how strong the London forces are. Silane is heavier, so it has bigger London forces and a higher boiling point. Between water and hydrogen sulfide, both are polar, and have dipole-dipole forces, so they have higher boiling points than methane or silane. But water has hydrogen bonds, which are extra-strong dipole-dipole forces. Water boils much hotter than hydrogen sulfide.
Energy and Phase Changes
We can't really explain phase changes in terms of energy without entropy, which we haven't talked about yet. For now, we can just say that as we add energy to a substance, it usually gets hotter and the particles have more kinetic energy. This will make it easier for them go from solid to liquid, or liquid to gas. Gases have more energy than liquids, which have more energy than solids. As we increase the temperature, the stable form of the substance goes from solid to liquid to gas. The transition temperatures (melting point, boiling point) are the temperatures at which both phases are stable and in equilibrium. Actually, there will be some gas in equilibrium with solid and liquid all the time, because a few molecules can always escape the solid/liquid, but the solid or liquid won't be present above certain temperatures.
For instance, imagine heating a solid. The molecules start moving more, and the temperature increases as predicted by the heat capacity. At some point, they have so much energy that it's hard for them to stay in the orderly solid, so the solid starts to melt. As we add more heat, the temperature doesn't change, because all the heat we add goes into melting the solid. The solid can't get any hotter than it is, and the liquid can't increase it's temperature because its kinetic energy is absorbed to melt the remaining solid. The amount of energy needed to melt the solid is the enthalpy of fusion. When all the solid is melted, if we keep adding heat, the temperature will rise again. As the temperature rises, the vapor pressure increases, because more molecules have enough kinetic energy to escape. Still, most of the molecules are in the liquid form, because the total pressure pushes on the liquid and keeps it from expanding into a gas. When the temperature increases to the boiling point, then the vapor pressure will be equal to the outside pressure. Now, because the vapor pressure is equal to the atmospheric pressure, bubbles form in the liquid. It can expand into a gas, because it's pressure is the same as the atmospheric pressure. The temperature will stay constant again as all the liquid become gas, while you add the enthalpy of vaporization. Then if you keep heating the temperature of the gas will increase. This is shown in the diagram below: | textbooks/chem/General_Chemistry/General_Chemistry_Supplement_(Eames)/Phases_and_Intermolecular_Forces/Phase_Changes.txt |
Skills to Develop
• Identify and describe the parts of a phase diagram
You know that phase changes usually depend on temperature, which determines the kinetic energy of atoms and molecules. We mentioned before that they also depend on pressure. In the section on phase changes we said that the boiling point is the place where vapor pressure is the same at the external pressure, so clearly boiling point depends on pressure! Melting temperature also depends on pressure (usually the density of solid and liquid are different, so it makes sense) but not nearly as much as boiling point, since the volume changes are smaller. We use phase diagrams to show how the transition temperatures depend on temperature and pressure both.
Look at the diagram. Notice that the gas phase is on the bottom, where the pressure is low. Solid is on the left, where the temperature is low. Liquid is in between. The red line shows the sublimation point: along this line, a low pressure, solid turns directly into gas without going through liquid. The point where liquid become stable is called the triple point, where all three phases (solid, liquid and gas) are all in equilibrium. The blue line is the boiling point. Notice that the boiling temperature changes a lot with a change in pressure. The solid green line shows the melting point of most liquids. Notice that the melting point doesn't depend on pressure nearly as much as the boiling point (which makes sense, because the change in volume from solid to liquid is small). Most liquids are less dense than the solid phase, so higher pressure increase the melting point. The dotted green line shows the melting point for water. Water is denser as a liquid, so higher pressures decrease the melting temperature.
The second red point in the diagram is the critical point. The dotted black lines show the area where a supercritical fluid exists. This is the high-temperature, high-pressure part of the diagram. Because the temperature is high, the molecules have lots of kinetic energy, so a liquid form isn't really stable because the intermolecular forces aren't strong enough to hold such energetic molecules together. However, the pressure is so high that the molecules can't really get away from each other either, so they bump into each other a lot, and feel some attractions, and don't really act like a normal gas (certainly not an ideal gas!). Past the critical point, there's no distinct liquid or gas, just a supercritical fluid with some special properties.
Supercritical fluids can make good solvents. For instance, supercritical CO2 is commonly used because it is a safe, inert, inexpensive non-polar solvent. Most non-polar solvents are not very safe (toxic and flammable), and disposing of them is expensive; supercritical CO2 avoids these problems.
Vapor Pressure
Skills to Develop
• Define vapor pressure
Whenever we have a liquid with some space above it, some of the molecules on the surface of the liquid might escape into the space and become gaseous if they have enough kinetic energy. If the liquid is in a closed bottle, then the molecules in the gas phase can't get away completely. There will be the liquid, and above it, a gas of the same molecule. The gas particles will sometimes bump the liquid, and if they have small enough kinetic energy, they might stay in it. Thus, molecules can go back and forth between the liquid and the gas. As the amount of gas increases, it is more likely to bump the surface and get stuck, so the rates of becoming gas and becoming liquid become the same. An equilibrium will be established. The partial pressure of that gas above the liquid at equilibrium will depend on the type of molecule and the temperature. If the temperature of the liquid increases, more molecules will have enough kinetic energy to escape the liquid and be gas. If the molecule has strong intermolecular forces, it will take more kinetic energy to escape the liquid.
In an open container, a liquid like water will completely evaporate eventually, even at low temperatures (even ice will disappear eventually, because solids also have vapor pressure). This happens because when the molecules become gas, they can diffuse away instead of staying near the surface, and maybe bumping it and getting stuck again.
The vapor pressure of a liquid doesn't depend on the pressure in general (at least not much). The presence of other gas molecules won't really affect the rates of the liquid molecules leaving the surface or returning to it, because their collisions with the liquid don't matter. When we collect gases over a liquid, like in Hales' method, we should include a correction for the vapor pressure of the liquid. We can find this in a table if we know the temperature.
Some liquids have a high vapor pressure and others have low vapor pressure. This depends on the intermolecular forces, like London dispersion forces, dipole-dipole forces, and hydrogen bonds. If the intermolecular forces are strong, the vapor pressure will be low. If they are weak, it will be high. Volatile liquids have high vapor pressure. | textbooks/chem/General_Chemistry/General_Chemistry_Supplement_(Eames)/Phases_and_Intermolecular_Forces/Phase_Diagrams.txt |
Skills to Develop
• Illustrate the general shape of atomic orbitals
• Identify the relationship between quantum numbers
The "old quantum mechanics" introduced the idea of quantization, and it was good for describing the position of spectroscopic lines of single electron atoms. But it couldn't predict how strong the lines were. The "new quantum mechanics" of Schrodinger and Heisenberg took the wave-particle duality to its logical conclusion. While Bohr's model was still "classical" in that there was a defined orbit or trajectory (a path that could be calculated) for the electron, Schrodinger and Heisenberg changed that. Basically, where is a wave? If a particle behaves as a wave, you can't point to the exact spot where it is. Also, if light energy is quantized, it turns out that you can't measure the path taken by an electron without changing the path. In microscopy, which is using microscopes to look at small things, you can't separate things that are closer together than the wavelength used. This is why we use X-ray diffraction and electron microscopes with very short wavelengths to look at atoms and molecules. So if you want to measure the electron's path with light, and measure it precisely on an atomic scale, you have to use a short wavelength of light. But if you use a short wavelength then it has a lot of energy (E = hν), enough to change the electron's direction. This is the basis of the Uncertainty Principle.
The result of this (and also of Schrodinger's 3-D, space-filling wavefunction) is that we no longer describe electrons using orbits, or defined paths. Instead we talk about orbitals, which are defined by wavefunctions Ψ(x,y,z) like the one calculated in the previous section. It turns out (although I think Schrodinger didn't exactly intend this at first) that Ψ(x,y,z)2 is the probability of finding the "particle" at position (x,y,z).
Schrodinger found the standing waves or Ψ(x,y,z) for the electron in a hydrogen atom. These are much more complicated that the wavefunction we found because they are in 3-D and also have a potential energy to worry about. (But they aren't that complicated: you can look at them here.) There are an infinite number of functions that solve the equation, just like in the simple example. In the 1-D example, we used 1 quantum number n to determine the energies. As n increased, so did the energy, and so did the number of nodes (places where the amplitude is zero). In 3-D, the standing waves are specified by 3 quantum numbers. (Actually, Bohr and others were using 3 quantum numbers before Schrodinger published this, because it makes sense in 3-D.) The principal quantum number n corresponds roughly to the radius of the orbit in the Bohr model, or to the "most probable distance from the nucleus" in the orbital model. This also corresponds to the energy: the electron is lower energy when it is close to the nucleus, because of the Coulomb attraction. The second quantum number ℓ gives the shape. In the Bohr model this meant ellipse or circle. In Schrodinger's model the shapes are 3-D. You can look at them in detail on Wikipedia (scroll down to the table of images) or with an animated applet like this. This is also a good time to check out the "orbitals" tab on Ptable. Pick an element, then move over the arrows to see a picture of each orbital.
Finally, the third quantum number tells the orientation. In the Bohr model this meant what plane the orbit was in. In the figure, you can see that the last 3 orbitals have the same shape but point along different directions (x, y and z).
Notice that standard pictures of orbitals show most of them with 2 colors. This represents the phase of the orbital (whether Ψ is + or − at that point). This doesn't effect the energy or the probability of finding the particle there, but it is very important when atoms interact with each other, like forming bonds, which is what chemistry is all about! When waves interact with each other, it matters a lot whether the peak (high point) overlaps with the trough (low point) or not. This is true of water waves, sound waves, light waves and electron waves.
There are lots of ways of showing orbitals, but usually they just draw a surface so that you have 50% or 90% or whatever chance of finding the electron inside the surface. (Always, the probability of finding the electron becomes 0 as you get far enough from the nucleus.) The second orbital in the picture is shown as a slice, so you can see that the inside near the nucleus has the opposite phase as the outside. Wherever the color changes, there must be a node (Ψ passes through 0 as it goes from + or −). Thus the first orbital has no node, which makes sense because more nodes mean higher energy. The second has 1 spherical node. The next three have a planar (flat) node. If you look at more orbitals, you'll notice that the nodes keep increasing.
Here's the pattern of orbital shapes and quantum numbers.
• The principal quantum number n tells you how big the orbital is and what the energy is. It also tells you how many spherical nodes the orbital has. The values go from n = 1, 2, 3, 4, 5, 6, 7... they could keep going but we haven't found elements that need that many orbitals yet! The number of spherical nodes in the orbital is n − ℓ − 1.
• The angular momentum quantum number ℓ tells you the shape. If ℓ = 0, the orbital has spherical symmetry (it's round), and we say it is an "s orbital". If ℓ = 1, the orbital has one planar node, and it's called a "p orbital". See the figure below for d and f orbitals, with 2 and 3 planar nodes each. The possible values of ℓ are 0, 1, ... n − 1.
• The magnetic quantum number m tells the orientation of the orbital. The possible values are -ℓ, -ℓ + 1, ... 0, 1, ℓ − 1, ℓ. For instance, for the p orbitals, it can be -1, 0, 1. You can remember the number of orientations using the table below.
A geometric pattern for remembering the number of each type of orbitals and the possible values of the quantum numbers. | textbooks/chem/General_Chemistry/General_Chemistry_Supplement_(Eames)/Quantum_Chemistry/Atomic_Orbitals.txt |
Skills to Develop
• Describe the basic wave properties and resonance
• Distinguish the 2 types of wave interference
• Explain the significance of standing waves
Understanding waves is essential to understand the rest of this section, because we will be talking about many different types of waves. Light has wave properties, and standing electron waves explain many chemical properties.
Basic Properties of Waves
You have probably studied waves a little bit in a previous class. Waves include sound waves, ocean waves, and light waves. You have probably also studied sine and cosine waves in math class. Here will will review some of these topics. A wave is defined as "a periodic disturbance that moves through space." The disturbance could mean the high and low parts of an ocean wave, or the high and low pressure parts of a sound wave, etc. Periodic means that the disturbance repeats. As an example, consider a graph of y = sin(x). This is a wave. In this case, the amplitude is 1, because this is the distance between the average value of y and the maximum value. The wavelength is 2π. In non-mathematical waves, wavelength (abbreviated λ) is given in meters. A cycle is when the wave goes through one wavelength, so it ends at the same part of the wave where it started. The frequency of the wave is the number of cycles/second. So if you look at one position, you will see the wave go up and down at that point. The time it takes for the wave to go through one cycle at that point (from the top to the bottom to the top again) is the period. Frequency is 1/period, (we'll abbreviate it ν) and is usually reported in Hertz, which is just s-1. You can think of it as cycles/second, but sometimes people use radians/second instead (you can convert between these using 1 cycle = 2π radians). Because wavelength is meters/cycle and frequency is cycles/second, wavelength x frequency is meters/second or velocity.
Wave Interference
This describes what happens when waves interact with each other, or overlap. The best way to see what this means is to watch: try watching the video on the Double Slit experiment to see waves interacting. Basically, to find the total wave made of several different waves, you just add the amplitude at each point. For instance, if the high part (amplitude 1, say) of one wave overlaps the low part of another (-2, say, this wave is bigger), the result is a smaller wave (-1, here), called destructive interference. If the 2 peaks (high parts) overlap, then the wave get bigger, called constructive interference. In the graph below, the sum wave has big peaks in regions of constructive interference, and small peaks in regions of destructive interference.
Standing Waves
An important type of wave for us is a standing wave. Standing waves are present in musical instruments, and are how the instrument produces sound of the correct frequency (or pitch, the term used in music). Standing waves don't travel, because they are confined in the instrument. For instance, in a stringed instrument like a guitar or a violin, the whole string vibrates, but the wave doesn't travel because it is trapped in the string, and the string has end points (where it is attached to the instrument, or held down by the musician's finger). Because the ends are fixed in place, the amplitude at these points has to stay zero, but the amplitude in the middle of the string can change. This limits the wavelengths possible to the string. The properties of the string, like its mass/length and tension, will determine the velocity, and the frequency is determined from the combination of wavelength and velocity.
What does velocity mean? The standing wave comes from the wave traveling down the string, reflecting off the fixed ends, reversing direction, reflecting again... when the frequency/wavelength/velocity all match up, the wave reflections reinforce each other, or interfere constructively, which is called resonance. Resonance generally describes a vibration that is reinforced by the particular properties of the surroundings because it matches their natural frequencies. When waves with the wrong wavelength bounce back they will cause destructive interference with themselves. The result is that the string can easily vibrate at its natural, resonant frequencies, but can't really vibrate at other frequencies because those waves cancel themselves out. For instance, a harp (an instrument with many strings) will laugh with you, because some of the strings will match some of the frequencies in your laugh and start to resonate, creating sound waves in the air. Here's a 1 min video of using resonance to break a glass. Here's an amazing video showing the location of 2-D standing waves using metal plates and the bow of a violin. | textbooks/chem/General_Chemistry/General_Chemistry_Supplement_(Eames)/Quantum_Chemistry/Describing_Waves.txt |
Skills to Develop
• Explain the significance of spectroscopy, wave-particle duality
• Describe the contributions of Bohr and Planck to "old quantum mechanics"
The history of quantum mechanics is often divided into 2 parts: the "Old Quantum Mechanics" and the "New Quantum Mechanics." Before either quantum mechanics, physicists described the world using "Classical Mechanics," which is like what you've probably studied before in physics class: Newtonian mechanics (forces, accelerations, etc), electricity and magnetism using Maxwell's equations. . . all these approaches work well for big things that aren't moving too fast. There are two qualities of classical mechanics that quantum mechanics altered. First, in classical mechanics energy and velocity and such quantities can have any value. If you drop a ball, it accelerates smoothly from 0 to a final velocity, rather than moving jerkily from step to step. The "Old Quantum Mechanics" got rid of the assumptions that energy and velocity should be "continuous." The other aspect of classical mechanics was the idea of a trajectory: the path taken by a moving object. If you launch a missile or a space shuttle, you can calculate almost exactly the path it will follow; if you have perfect knowledge of the forces acting on it, you can calculate its path perfectly. The "New Quantum Mechanics" which is described more in the later section on orbitals essentially said that there is no trajectory for small particles, so you can only describe them using statistics, not actually know where they are or what path they follow.
Spectroscopy
The discovery of spectroscopy in 1859 presented a problem for chemists. Spectroscopy essentially means looking at the wavelengths of light that are absorbed or emitted from a sample. Any time you look at colors, you are doing a sort of spectroscopy, because color comes from particular wavelengths of light. Paint absorbs certain wavelengths, so what we see is the color of the other wavelengths. Computer screens can emit certain wavelengths of light so we see the colors that correspond to those wavelengths. But you can look more carefully, by splitting light into its component wavelengths, using a prism or a diffraction grating. If you do this with sunlight, as Newton did, you see a rainbow. If you do this with the light emitted by a very hot sample of an element, you will see just a few separate lines: the sample emits only specific wavelengths, instead of a smooth rainbow with a little of each. But chemists had trouble explaining why only these particular wavelengths were emitted. The wavelengths are called lines, and the collection of lines is the spectrum.
In 1885 (25 years after the introduction of the spectrometer), Balmer, a teacher, studied the 4 lines emitted by very hot hydrogen atoms, and noticed a pattern. The wavelengths could be calculated from
$\lambda = C\frac{n^{2}}{n^{2}-2^{2}}$
where C is a constant and n is a whole number (3, 4, 5, or 6 for the 4 known lines). He suggested that there might be other lines corresponding to replacing the 22 in his formula with 32, 42, etc. Rydberg later rewrote Balmer's formula as follows, using the wavenumber (1/λ) which is still commonly used in labs today with the unit cm−1
$\frac{1}{\lambda}=R \left(\dfrac{1}{n_{f}^{2}}-\dfrac{1}{n_{i}^{2}}\right)$
where R is called the Rydberg constant, and nf and ni are integers, and ni is larger than nf. Until 1908, the only known lines had nf = 2, but then finally lines given by nf = 3 were found, and later lines with nf = 4, 5 were found also, proving Balmer correct.
The "Old Quantum Mechanics"
This is a good time to review the previous section Discovery of Sub-Atomic Particles. Recall the use of gas discharge tubes in discovering X-rays. This started a long debate about whether X-rays were waves (like light) or particles, like electrons. They couldn't be charged particles, because they weren't affected by magnets, but they didn't seem to reflect or diffract either. (Diffract means waves changing direction after passing through a gap about the size of the wavelength.) Eventually Stokes realized that X-rays were light with a very short wavelength, so it wouldn't diffract without very tiny gaps. Von Laue proposed that salt crystals could be used to diffract X-rays, because the gaps between ions in crystals are very small. This is the basis of X-ray diffraction, which is now used commonly to find structures of molecules. Also, remember the young chemist Moseley who figured out the fundamental order of the periodic table? He was essentially doing X-ray spectroscopy, using an electron beam to make elemental samples emit X-rays. He measured the wavelengths of these X-rays using diffraction by salt crystals. Like Balmer, he found a fascinating pattern in his line spectra.
Review as well Rutherford's nuclear model of the atom in the same section, proposed in 1911 (around the time that Balmer's hypothesis was proved correct). After various proposals, his experiments showed clearly that there was a dense, central part of the atom with most of the weight and the positive charge, which is called the nucleus. Around the nucleus, electrons move, occupying a much larger volume than the nucleus. But the physicists had some problems with this model. If the electrons weren't moving, they should "fall" into the nucleus. If they were moving around the nucleus, when they changed direction (to stay near the nucleus) they should emit light. The oscillating charged particles would be like a miniature cell phone tower, emitting radiation at high frequency. Emitting light would cause them to lose kinetic energy, and very quickly they should fall into the nucleus. To solve this problem, Bohr employed quantum theory, which had been introduced by Planck in 1900.
Planck used quantum theory to explain blackbody or thermal radiation. Recall the problem with Rutherford's model, that acceleration (including direction change) of charged particles produces light (not necessarily visible light, but electromagnetic radiation of some wavelength). The atoms and ions in normal objects are always moving a little, and the hotter they are, the more they move. These movements mean that they bump into each other and change direction; these collisions result in emission of radiation if charged particles are involved. The light emitted by normal objects at normal temperatures is usually IR or lower energy (animals emit IR thermal radiation, which is how night-vision goggles work); incandescent (normal old-fashioned) light bulbs emit visible light and IR because the filament is much hotter than room temperature. In English, we talk about "red-hot" or "white-hot" to describe temperatures: glass or metal that is being shaped is often red-hot, hot enough to emit red light. White-hot is even hotter, hot enough to emit white light (all the colors). But it was very hard to explain the spectrum of thermal radiation produced by objects at different temperatures (what wavelengths of light were produced and how much of each). Planck had to introduce quantization to get a good model. He proposed that light isn't just a wave, but it comes small separate packages, called quanta. The energy of a quantum of light is
$E=h u$
where h is Planck's constant (6.626 x 10-34 Js) and ν (Greek letter read as "nu") is the frequency of the light in Hz. Most scientists didn't like this idea (it seemed very strange!) but it worked.
Wave-Particle Duality
The Wave-Particle Duality means that something can behave like a wave and also behave like a particle. This was first applied to light. Before quantization was proposed, physicists knew that light behaves as a wave (as described in the previous section. When Planck proposed quantization, he thought it was a property of atoms, not light. However, Einstein applied Planck's theory to explain the photoelectric effect, and clearly showed that light is a particle. The photoelectric effect means that when light (such as UV light) shines on a metal surface, sometimes it knocks electrons off the surface, creating a "photocurrent". It turns out that photocurrent is only produced when light has a sufficient frequency, and that increased intensity of light only produces more photocurrent, not photoelectrons with higher kinetic energy. And higher frequencies of light produce photoelectrons with more kinetic energy, but not more electrons. Einstein explained this by saying that light comes in particles with energy proportional to frequency (E = hν, which is the same formula Planck used in which ν is the frequency and h is a constant). To create a photoelectron, the light particle or photon must have more energy than the energy holding the electron on the surface. Any extra energy in the photon turns into kinetic energy for the electron.
Bohr's Model
Bohr used Planck's quantum concept to try to explain the Rutherford model of the atom. He focused on the hydrogen atom, with just one electron around the nucleus. Surprisingly, it was dimensional analysis that lead him to Planck's theory. He liked the "solar system" model, in which electrons orbit the nucleus like planets orbit the sun. He realized that an orbit would have a characteristic radius (distance between nucleus and electron) and also that h2/m•e2 would have units of length (check this yourself, but note that you'll need to include another quantity, the permitivity of free space). So he used classical physics, including Coulomb's law and
$a = \frac{v^{2}}{r}$
to look for an equation for stable orbits (he just assumed that in a stable orbit, the acceleration wouldn't cause radiation). He got part of the way (calculating the total energy of the orbits including kinetic and potential energy), then got stuck. Luckily, a spectroscopist then introduced him to Balmer's formula. When he saw that, the answer suddenly became clear to him. He assumed that the lines in the spectrum come from electrons moving from one stable orbit to another, and the wavelength of the light emitted equals the energy difference of the orbits. He amazed everyone by deriving the Rydberg constant (which was experimentally known to be 109677 cm−1) in terms of fundamental constants:
$R=\frac{e^{4}m}{8\epsilon_{0}^{2}h^{3}c}$
and it worked out correct! (Check for yourself; ε0 is again the permitivity of free space.) So Bohr's model worked well for explaining the hydrogen spectra: in stable orbits, which had only certain allowed energies, there was no radiation; light (a single photon) was absorbed or emitted when changing orbits, and its wavelength matched the energy difference of the orbits. This theory also worked to explain some lines in the sun's spectrum, which came from He+; to calculate these you have to use the atomic number Z to account for the larger charge in the nucleus. It turned out that his formula works for all single-electron atoms.
Essentially, his model predicted stable orbits with energies determined by a quantum number, corresponding to the integers in Rydberg's formula. Orbits with one value of the quantum number are called a shell. He also figured out that orbits didn't have to be circles, and didn't have to all be in the same flat plane. He added 2 more quantum numbers to indicate the shape (circle or ellipse) and the orientation of the orbits.
However, his theory didn't work for multi-electron atoms very well. Bohr worked on extending it for more electrons, and he found that this was easiest for alkali atoms. Bohr found that he could do ok by treating the alkali atoms as having only one electron, and including all the other electrons as part of a bigger "nucleus" with a +1 charge. Bohr used spectroscopic data to arrange the elements in roughly the same pattern as a modern periodic table, with periods of 2, 8, 8, 18, 18, and 32 elements, and some divisions between the electrons in each period, that he did not get exactly right. Each row consisted of filling one shell. Recall that early periodic tables didn't have the modern shape; it was at this time that the modern shape started to emerge. Other chemists we'll study later (Lewis and Langmuir) were arranging the elements by chemical properties, with similar results. The mathematical basis of this pattern didn't become clear until the "new quantum mechanics" was introduced. | textbooks/chem/General_Chemistry/General_Chemistry_Supplement_(Eames)/Quantum_Chemistry/Discovery_of_Quantization.txt |
Skills to Develop
• Predict and illustrate electron configurations
The wavefunctions in the previous section were derived for single-electron atoms only. It's really hard if not totally impossible to solve the Schrodinger equation perfectly when there are 2 or more electrons, because of electron-electron repulsions which make the Hamiltonian incredibly complicated. However, it turns out that we can approximate the orbitals in other atoms with small modifications of the hydrogen wavefunctions, like decreasing the nuclear charge to account for electron repulsion.
Electron Configurations
Bohr figured out the number of electrons in each shell, where a shell is all the electrons with the same principal quantum number. The pattern he used, which you can verify with the periodic table, was 2, 8, 8, 18, 18, 32, 32. However, he was not as clear about the arrangement into subshells, which are groups of electrons with the same principle and angular momentum quantum numbers, such as 1s or 2p. This became much clearer after Schrodinger's wavefunctions for the hydrogen atom were introduced, because the solution wavefunctions clearly had only certain allowable ℓ values and m values. Although all the orbitals in a shell have the same energy in a single electron atom, when there are more electrons the subshells have different energies. This is because the larger ℓ is within a shell, the farther the electron usually is from the nucleus. In s orbitals, the electrons are often close to the nucleus, so other electrons don't block the nuclear charge much. In d and f orbitals, the electrons are far from the nucleus and do get blocked. Thus, orbitals tend to fill in the order of lowest n and lowest ℓ first.
Orbitals with the same energy are called degenerate. (In other contexts, degenerate means immoral, or other bad things.) In general, the more complications you add, like more electrons, neighboring atoms, magnetic fields, etc, the fewer orbitals are degenerate. So for a hydrogen atom by itself, all the orbitals in each shell are degenerate. When you move to a lonely helium atom, the orbitals in the subshells are degenerate. When you make chemical bonds, the orbitals in subshells are no longer degenerate. When you apply a magnetic field, the electrons in the same orbital are not degenerate.
Simple Rules for Predicting Electron Configurations of Atoms
How to predict electron configurations for elements? Electron configuration just means how many electrons in each orbital. You will need to know the number of electrons you are using. You can get this from the periodic table (the atomic number − charge if it's an ion). Or you might just be given the atomic number. If you have a periodic table, it's easy, because of the shape of the table. If you go to the periodic table on this site, you'll see it has a blue block, a red block, a yellow block and a green block. The code is blue = s; yellow = p; red = d; and green = f. The alkali metals and alkaline earth metals have their highest energy electrons in an s orbital. The highest energy electrons, the ones in the highest shell, are called valence electrons. The main group (yellow) including halogens and noble gases, have their highest electrons in p orbitals. The transition metals have theirs in d orbitals, and the lanthanoides and actinoids in f orbitals. Notice that each section is 2, 6, 10 or 14 elements across, because there are 1,3,5,7 orbitals in the s, p, d, and f subshells, and each orbital holds 2 electrons. So you can write the electron configuration just by looking at the periodic table. Follow your way along, writing how many electrons are in each subshell, until you have the right number of electrons total. For instance, for Al, you would have 1s22s22p63s23p1. Notice that we indicate the number of electrons in each subshell using superscripts we we write electron configurations.
Another trick you can use to write electron configurations is to follow the arrows on this diagram. The filling pattern is named "aufbau" which means "building up" in German, or so I've heard. Notice that you don't actually fill all the subshells in one shell before moving on to the next. This can be a little confusing (espcially because the s orbitals are only slightly lower than the previous d orbitals, and in ions the energy order switches back!). Either use this diagram or a periodic table, and you'll get the right answers as long as you are careful to count from the top of the periodic table. Notice how the first row of d comes after the 4th row of s? But we can have 3d orbitals, so the first row of d is 3d, not 4d.
Often we don't want to write out all the electrons for a heavy atoms, because the inner ones are just the same as all the other elements. So we might show only the valence electrons, which means in this case the ones added on this row of the periodic table, or after the last noble gas. So we might write Ti as [Ar]4s23d2, which just means that it has the same configuration as argon, + 4s23d2.
The next question comes when not all the orbitals in a subshell are filled. Do you pair electrons or put them in separate orbitals? For instance, consider carbon. You can represent its electron configuration like this, with the arrows up and down representing the electron spin.
But which of the 3 possibilities is it? Remember that electron spin showed up in spectroscopy with magnetic fields. The electrons are like tiny magnets, and when they are paired (one up, one down) they cancel each other out and don't seem magnetic unless you look really hard. But you know that there are lots of very magnetic materials out there (maybe you use them on your fridge; you also use them in headphones and other devices). The reason they are magnetic is because the bottom arrangment is the lowest energy. Electrons prefer to have the same spin in different orbitals, where they repel each other less (among other more complicated reasons). This is called Hund's Rule. The reason the best magnets are transition metals or lanthanoides is that d and f orbitals have space to spread out 5 or 7 electrons with the same spin.
The only other thing you need to know about electron configurations right now is that there are some exceptions to the rules about what order they fill in (mostly in the d and f blocks). These aren't very important. The other thing you might want to know is whether the electron configuration in isolated atoms is important to chemists. Mostly only the valence electrons are affected by bonding, so the elemental configuration will probably still apply to the core electrons. And the atomic electron configurations match the reactivity patterns that the periodic table was originally based on, so that's useful. The other reason is that these rules apply almost the same to real chemical situations. Even if the orbitals are more complicated molecular orbitals (we'll study these later) instead of atomic orbitals, you still fill them from lowest to highest and use Hund's rule. | textbooks/chem/General_Chemistry/General_Chemistry_Supplement_(Eames)/Quantum_Chemistry/Electron_Configurations.txt |
Skills to Develop
• Describe the Pauli Exclusion Principle
Many-electron Atoms
If there are multiple electrons, which orbits do they occupy? In atoms larger than hydrogen, the larger nuclear charge means that the orbits should be closer to the nucleus. If all the electrons are in the lowest energy orbit, even if repulsion expands the orbit a little, then the size of elements would get smaller as they get heavier. Also, it would get harder and harder to pull electrons off heavier elements, because they would be bound much more tightly to the more charged nuclei. Both of these are the opposite of the facts: sizes of elements increase a little as you go down the periodic table, and it is hardest to remove electrons from the lightest elements. Thus, it was clear that only some electrons could be in the lowest orbit.
Pauli was a theoretician "so good that experimental apparatus broke when he walked through the lab door." (It's a joke that theorists, who usually just think, can't handle lab equipment.) He suggested that theory and data would fit well if each orbit (defined by 3 quantum numbers) could hold exactly 2 electrons. Thus as soon as each orbit was full, with 2, the next electron would have to go in the next higher orbit. Using this rule, Bohr and others determined the electron configurations (which electrons were in which state or orbit) of most of the elements, and found that the order matched the periodic table nicely! The movement to the next principal quantum number started a new period, so the theory fit all the data on valence and other properties.
Electron Spin
Pauli didn't know at the time why each orbital holds 2 electrons, but later it was found that a fourth quantum number was needed, because strong magnetic fields split the energy of the 2 electrons in one orbit into 2 different levels. (This is the basis of many important experimental techniques, especially NMR, used by chemists, and MRI, used in medical imaging.) A proposed explanation for this is that the electrons can "spin" in either direction, just like the earth spinning to give us day and night. This isn't exactly what happens, but the word spin is still used to describe this effect. The spin quantum number has values of +1/2 or −1/2. Thus, Pauli's idea is now called the Pauli Exclusion Principle, and it says that no 2 electrons in one atom can have exactly the same quantum numbers. Two electrons can occupy one orbit, but they must have opposite spins.
Electron Configurations
Bohr figured out the number of electrons in each shell, where a shell is all the electrons with the same principal quantum number. The pattern he used, which you can verify with the periodic table, was 2, 8, 8, 18, 18, 32, 32.
Based on Bohr's model, you can find the number of valence electrons or electrons in the highest shell by looking at the periodic table. (For now, let's not worry about transition metals, and lanthanoides.) Alkalis have 1 valence electron, because we have just started a new shell. Alkaline earths have 2, B has 3, C has 4, N has 5. Chalcogens have 6, halogens have 7, noble gases have 8. (You can see that this is kind of close to the valence we mentioned earlier.
Light as a Wave
Skills to Develop
• Relate the wavelength and frequency of light using a mathematical equation
• Define electromagnetic radiation
Light is a little different from sound waves, water waves and string waves, because it can move through a vacuum. In general, the velocity of light is constant, c = 3.00 x 108 m/s. So for light, wavelength λ and frequency ν are related by
$c = \lambda ν$
Light is called electromagnetic radiation, which basically means that it radiates (travels) and the wave part is oscillating electric and magnetic fields. You don't need to worry about the details of this now. The properties of different types of light depend on their wavelength. X-rays used in medicine have very short wavelengths and high frequencies; UV (ultraviolet) light gives you sunburns and helps you make vitamin D, visible light is normal light, IR (infrared) is used for night-vision goggles, microwave is used in cooking, and radio is used in radios. | textbooks/chem/General_Chemistry/General_Chemistry_Supplement_(Eames)/Quantum_Chemistry/Electron_Configurations_According_to_Bohr_and_Pauli.txt |
Skills to Develop
• Derive Schrodinger's Equation (optional)
• Verbalize the solution of Schrodinger's Equation (optional)
Many introductory chemistry textbooks introduce the Schrodinger Equation, but students don't understand what it means. This section is optional; if you want to know where orbitals come from, it can help you understand. It will be easier to follow this section if you know a little calculus (basically, what a derivative is).
The Schrodinger Equation is the starting point for describing the motions of electrons as waves. De Broglie suggested that their stable "orbits" in the Bohr model were standing waves analogous to those in a guitar string. Schrodinger extended this theory using the wave equation and wavefunction. Instead of circular orbits, Schrodinger's waves were 3D and took up the whole space of the atom, more like vibration of air in a spherical flute than the vibration of a circular string. The wavefunction Ψ(x, y, z, t) describes the amplitude of the electron vibration at each point in space and time. Oddly, Schrodinger seems to have proposed the wavefunction without fully understanding what it means, but it worked! Here we will describe the time-independent Schrodinger equation for simplicity, which describes the standing waves. We will also consider only a 1-dimensional system, such as a particle that only moves linearly, also for simplicity. Thus, we will find Ψ(x) for a very simple situation.
Schrodinger proposed that a standing wave is described by the wavefunction Ψ when the it fits the following differential equation
$H\Psi = E\Psi$
where H is the Hamiltonian operator, which finds the total energy of the system E. (This approach uses the linear algebra concept of an eigenfunction and eigenstate, but don't worry if you don't know what these are.) Kinetic energy KE is given by
$KE = \frac{p^{2}}{2m}$
where p is the momentum (p = mv). For a particle moving in 1D (along x) Schrodinger assumed that a permissible general form of Ψ is
$\Psi (x,\; t) = Ae^{\frac{i(px\; -\; Et)}{\hbar}}$
where A is a constant and i is the imaginary number (i2 = -1). (This comes from the equations E = hν and the de Broglie relationship λ = h/p. These equations connect energy to time and distance to momentum through Planck's constant. These are also the quantities that are mutually limited by the Uncertainty Principle.) If this is true, the derivative of the wavefunction with respect to x is
$\frac{d\Psi}{dx}=\frac{ip}{\hbar}\Psi$
Notice that this is kind of like the equation HΨ = EΨ in that we get the original wavefunction multiplied by some important quantity, like energy or momentum. So the momentum operator p (like the Hamiltonian operator, which gives the energy) gives the momentum p, and can be written like this:
$\textbf{p}\Psi(x,\; t)=-i\hbar \frac{d\Psi}{dx}$
We can write the Schrodinger equation using this Hamiltonian (which gives total energy, KE + PE)
$H\Psi (x)=-\frac{\hbar^{2}}{2m} \frac{d^{2}\Psi}{dx^{2}}+V(x)\Psi (x)=E\Psi (x)$
The potential energy is given by V(x), which just depends on the position. The kinetic energy is calculated using the equation above, using the square of the momentum operator (thus, the first derivative in the momentum operator becomes a second derivative when the operator is squared). Now, if we choose a function V(x) we can find the wavefunctions that fit! We will use a simple example: a particle in a box (in 1-D). The potential is 0 inside the box and infinite outside the box. So we will just know that the particle has to be inside the box, but use V = 0. Then our Schrodinger Equation looks like this
$H\Psi (x)=-\frac{\hbar^{2}}{2m} \frac{d^{2}\Psi}{dx^{2}}=E\Psi (x)$
or basically the second derivative of Ψ is a constant times Ψ. There are different forms of the solution, but we'll just choose a simple one.
$\frac{d}{dx}\; sin(ax)=a\; cos(ax)$
$\frac{d}{dx}\; cos(ax)=-a\; sin(ax)$
Thus,
$\frac{d^{2}}{dx^{2}}\; sin(ax)=-a^{2}\; sin(ax)$
So we can pick sin(ax) or cos(ax) or a sum of these for the wavefunction:
$\Psi (x) = sin(ax) + cos(bx)$
So far, there is no quantization! The coefficient a can have any value. But just like a string on a guitar, the amplitude of Ψ has to be 0 at the edges of the box. If we just use Ψ(x) = sin(ax), then if the box is from x = 0 to x = L, we need to have an integer number of half-wavelengths in the box. So
$a=\frac{n\pi}{L}$
so that
$\Psi (0) = \Psi (L)=0$
To summarize,
$\Psi = sin \left(\dfrac{n\pi x}{L}\right)$
is a solution of the Schrodinger Equation for the 1-D particle-in-a-box system. Try putting this in and see what the energy is! You should get:
$E_{n}=\frac{n^{2}\hbar ^{2}\pi ^{2}}{2mL^{2}}$
There are an infinite number of solutions, or wavefunctions that satisfy the Schrodinger Equation, corresponding to n = 1, 2, 3... and any sum of these wavefunctions is also a solution. What do they mean? The amplitude of the particle wave is given by Ψ. The next section explains the meaning of the wavefunction in more detail, now that you have been introduced to the math. | textbooks/chem/General_Chemistry/General_Chemistry_Supplement_(Eames)/Quantum_Chemistry/Particle_in_a_Box.txt |
Skills to Develop
• Describe some of the contributions of de Broglie and Heisenberg to "new quantum mechanics"
The "New Quantum Mechanics" was introduced a little after the "Old Quantum Mechanics" which was developed by Planck and Bohr. This is when quantum mechanics became very confusing to people. In the old quantum theory, we could think about particles moving in normal ways. The electrons orbit the nucleus just like planets orbit the sun. But this was wrong. The new quantum theory says that tiny particles behave in ways that are totally different from normal objects that we can see, like basketballs or planets. In fact, it is impossible to know exactly where they are or what they do.
Einstein had convinced physicists that light was a particle (now called photon) in 1905. Later in 1925, de Broglie proposed that particles like electrons could also be waves. De Broglie was a French prince who was initially interested in history, but during World War I he joined the army as a radio operator and became interested in waves. He was also a lover of music. These qualities lead him to interpret Bohr's atoms as "musical instruments." Bohr assumed that his energy levels were quantized, but didn't explain why. De Broglie proposed that quantization arose from the same effects as the quantized frequencies (fundamental and overtone) present in a guitar string, that make it work as an instrument. (Review standing waves in this section if needed.) De Broglie thought that electrons in "orbits" like in the Bohr model were like strings on instruments. A string has to have nodes at each end; an electron in an orbit also has to be a standing wave, with a whole number of wavelengths around its path. In other words,
$n\lambda = 2\pi r$
where n is an integer. Thus, the wave interferes constructively with itself when it makes a full circle.
According to his theory,
$\lambda = \frac{h}{mv}$
He derived this equation using the radii Bohr calculated for his orbits. You can see that the wavelength effect is only important for very small particles, like electrons (check this out for yourself!). He thought that the "wave-ness" of electrons would be present outside of atomic orbits as well, and this proved true. A beam of electrons accelerated by high voltage is predicted to have similar wavelength to X-rays, and like X-rays, can be diffracted by a crystal. The diffraction pattern let researchers measure the wavelength of electrons (whose velocity was known, because they were accelerated by a known voltage) just as Moseley had measured the wavelength of emitted X-rays to determine atomic numbers. Later the wavelengths of atoms were measured as well. The experimental wavelengths matched de Broglie's prediction, and the wave-particle duality for electrons was accepted, just like the wave-particle duality for light.
Heisenberg developed a way to work with the allowed energies of atoms using matrices. He believed that we must work with the quantities we can observe experimentally, like energies (via spectroscopy). This led him to the Uncertainty Principle which says that certain pairs of quantities (position and momentum, energy and time) can't be measured precisely at the same time. For instance, if you want to know exactly where a particle is, you can't also know exactly what its velocity is. This was part of the new "statistical" quantum mechanics. In classical mechanics, you can know everything about particles (like balls): exactly where they are, exactly where they are going, and predict exactly the results of collisions or other events. In the new quantum mechanics, it was argued that this perfect knowledge of particles was simply impossible. For instance, exactly where is a wave?
Although Heisenberg's matrix quantum mechanics is probably used more often because it works well with computer calculations, an equivalent description developed by Schrodinger gives more intuition, so we will describe it in detail. The methods give the same answers but use very different math. Schrödinger extended de Broglie's theory by applying well-known physics of waves (for instance, in instruments) to describe electrons. There's a detailed example in the next section. | textbooks/chem/General_Chemistry/General_Chemistry_Supplement_(Eames)/Quantum_Chemistry/The_New_Quantum_Mechanics.txt |
Skills to Develop
• Explain the properties of some covalent-network solids
• Describe semiconductors and some of their properties
You can read a quick introduction to covalent-network solids in the intro page. The basic idea is that to make a network of covalent bonds, each atom (or many of the atoms) have to make 3 or 4 bonds to other atoms. This means that covalent-network solids usually include carbon, silicon, and their neighbors in the periodic table. Here, we'll focus on simple, orderly structures like diamond, graphite, and pure silicon. There are also covalent-network solid oxides, like the silicates, in which oxygen atoms connect 2 silicon atoms, and each silicon atom connects to 4 oxygens. Partially covalent oxides are what most rocks and ceramics are made of.
Graphite
Carbon has 2 allotropes, or pure elemental forms. The more stable form is graphite, a dark, slippery material used in pencils and lubricants. Remember that carbon typically makes 4 bonds. The structure of graphite is flat hexagonal sheets; a single sheet is called graphene. Each carbon atom makes 3 σ bonds and the leftover p orbitals form a delocalized π-bond network over the whole sheet, very similar to the π-bonding in benzene. The π-bond system actually forms bands, like in a metal, allowing graphite to conduct electricity along the sheets. Weak interactions, like London dispersion forces, (called π-stacking in this case) hold the sheets loosely together. Because they can slide past each other (especially when impurity atoms are trapped in between) graphite is a good lubricant.
Diamond Structure
The other bulk allotrope of carbon is diamond. In diamond, each carbon makes 4 bonds in tetrahedral directions to other carbon atoms. The structure is like the zinc blende ionic structure, except that all the atoms are the same. The properties of diamond (insulator, hard) come from the strong covalent bonds. Remember that C-C bonds are some of the strongest covalent bonds. It's easiest to think about diamond as forming with sp3 hybrid orbitals on each atom. It's hard to imagine the MO interactions, but the main thing we can know is that because there is good energy match and good overlap between the atoms, the splitting (energy difference) between bonding and anti-bonding MOs will be big (and there won't be any non-bonding MOs, because each atom has 4 orbitals and makes 4 bonds). There will still be bands like in metals, but now there are 2 bands with a big energy gap in the middle. The low-energy, bonding band is called the valence band, and there are exactly enough electrons to fill it. The high-energy, anti-bonding band is called the conduction band, and it is empty.
If it was hard to understand the previous paragraph, let's just imagine making 1 bond between 2 atoms. We use one sp3 hybrid on each atom. These point right at each other, have exactly the same energy, and have good overlap. We can imagine drawing an MO diagram just like for H2, with a big energy gap between the bonding and anti-bonding orbitals. Each atom has 4 electrons total, and makes 4 bonds just like this, so each bond gets 1 electron from each atom: just enough to fill up the bonding MO. When we multiply this over all the bonds in the diamond, we get the full valence band and the empty conduction band. The energy gap between them is called the band gap. In diamond, the band gap is big, and so diamond is an insulator.
Semiconductors
Silicon has a structure just like diamond. (It doesn't make graphite, because mostly π-bonds are weak for the second-row elements.) However, silicon has a smaller band gap than diamond. Germanium, which is below silicon in the periodic table, has the same structure, and an even lower band gap. Generally, as we go down the periodic table, covalent bond strengths get smaller, which is the same as saying that the splitting between bonding and antibonding orbitals gets smaller. It's hard to say why, but if you assume that heavier atoms have worse orbital overlap, you will usually make good predictions.
Semiconductors are materials that conduct electricity just a little bit. They are the basis for all computing and electronics. Semiconductivity comes from having a not-quite-full valence band, a not-quite-empty conduction band, or both. If there are a few electrons in the conduction band, they can conduct electricity just like in a metal (except less, because there aren't very many of them!). If there are a few electrons missing from the valence band, the empty spots are called holes and they can move around, also conducting electricity. If the band gap of a material is not too big compared to the thermal energy, then a few electrons can be in the conduction band even though there is room for them in the valence band (because of thermal energy). For this reason, semiconductors conduct better at higher temperature, because more electrons will be in the conduction band.
Doping Semiconductors
The other way to make a semiconductor, with a material whose band gap is too big at room temperature, is by adding some impurity atoms with different numbers of electrons, which is called doping. Imagine that you have silicon with just a few nitrogen atoms replacing silicon atoms. Each N has an extra electron. Now there will be a few electrons in the conduction band, so it's a semiconductor! And we can control the conductivity by controlling how much N we add. We can do the same thing if we add a little bit of boron, which has 3 electrons. In this case, there would be a few holes in the valence band. These are called p-type (positive, less electrons, like with B) or n-type (negative, extra electrons, like with N) semiconductors. Lots of important devices, like LEDs, solar cells and transistors (the basis of computer chips) are made of layers of p-type and n-type semiconductors.
Compound Semiconductors
We can also make semiconductors that are compounds, like gallium arsenide (formula GaAs). These have an AB formula, and there are always 8 valence electrons in the formula. For example, Ga has 3 and As has 5, so GaAs has 8. Another example is ZnSe, (Zn has 2 valence electrons in 4s, Se has 6 valence electrons just like O). Thus, they have the same number of valence electrons as C or Si. They also have the same structure (zinc blende, just like diamond except that that atoms alternate types). By combining many different elements, we can change various properties, like band gap, to be exactly what we want. Generally, the band gap increases as the 2 elements are farther apart (the bonding becomes more ionic, and ionic solids aren't conductive). We can even mix 3 or more elements, like in CdZnTe (the actual formula would probably be written (Cd,Zn)Te, meaning that the number of Cd + Zn = Te). To make semiconductors with the right properties, we can use different combinations of elements and also dope with many different elements.
Crystalline Solid Structures
Skills to Develop
• Define a unit cell and its parts
Many solids are crystalline, which means that they have atoms or ions or molecules arranged in an ordered pattern. For instance, think about NaCl. This ionic solid has an alternating arrangement of Na+ and Cl- ions, as shown in the image below.
In a crystal structure, we can think about the unit cells, which are the smallest repeating unit of the structure. Basically, a unit cell is a box, and if we stack many unit cells, we get the total structure. Here's the unit cell for the NaCl lattice above.
Not all salt structures are the same. For instance, here is the cesium chloride structure, which is a little different from the NaCl structure.
Here's another view of CsCl. You can see 2 different (equally valid) unit cells (the shaded cubes), each of which has an ion at each corner of the unit cell box, and the other ion in the center of the box. If we repeat either unit cell, we'll get the crystal structure.
We can describe the unit cell using lattice vectors which are usually called a, b and c. These show the lengths of each side of the unit cell box. The angles between the lattice vectors are called α, β and γ. Here's a general unit cell, showing the lattice vectors and angles. The lengths a, b and c can be the same or different, and the angles can be 90° or not. There are different names for unit cells with particular shapes, but you don't need to learn them now, except for cubic, which means that the sides are the same and all the angles are 90°.
In simple salts, the atoms might be on the lattice points (the corners of the unit cells) but they don't have to be. Inside each unit cell, there will be some arrangement of atoms or molecules, called a motif. We can make to whole crystal by repeating the unit cells that contain the motifs. | textbooks/chem/General_Chemistry/General_Chemistry_Supplement_(Eames)/Solids/Covalent-Network_Solids%3A_Semiconductors_and_Insulators.txt |
Skills to Develop
• Relate the different types of solids to the 3 main types of bonding
• Describe some of the properties of solids
Solids are one of the most interesting and important topics in general chemistry because they are important for so many things. We use solids as building materials, to make tools of all kinds, to make computer chips, energy storage devices, solar cells, catalysts... Nearly all technology depends on the properties of solids.
Types of Solids
We can categorize solids in many ways. The most common way is by the type of bonding. There are three main types of bonding. So far we have focused on covalent bonding. There are two types of covalent solids: molecular and covalent-network. Molecular solids are made of covalent molecules, and the solids are held together by Van der Waals forces between the molecules. These molecular solids usually have low melting points and are easy to break, because the forces between molecules are weak. Covalent-network solids are different because there are no separate molecules: the whole solid is held together by covalent bonds between atoms. For example, in diamond, each C atom makes 4 covalent bonds to 4 other C atoms in a tetrahedral arrangement. Covalent-network solids usually have high melting points and are relatively hard because they are held together by very strong bonds. Diamond is the hardest material known, because C-C bonds are among the strongest single bonds known. (Could you make a covalent-network solid using the even-stronger bonds?)
We have also talked a little bit about ionic bonding. Basically, you probably know that most ionic materials are solids at room temperature, and they are made of ions, such as a metal cation and a non-metal anion. Ionic solids are held together by strong electrostatic (Coulomb's law) forces between the ions, so they are usually hard and have high boiling points.
There are also many solids that are on the border between covalent and ionic: they are made of metals and non-metals, but both covalent bonding and ionic bonding are important. This includes most rocks, minerals and ceramics. For instance, sapphire, another of the hardest materials, is aluminum oxide.
Finally, there is metallic bonding. Metallic bonding is sort of like covalent bonding, because it involves sharing electrons. The simplest model of metallic bonding is the "sea of electrons" model, which imagines that the atoms sit in a sea of valence electrons that are delocalized (spread out) over all the atoms. Because there aren't specific bonds between individual atoms, metals are more flexible than covalent-network solids. The atoms can move around and the electron sea will keep holding them together. Some metals are very hard and have very high melting points, while others are soft and have low melting points. This depends roughly on the number of valence electrons that form the sea.
There are actually some other types of solids also. One way we can separate solids into categories is crystalline or amorphous. Crystalline solids have ordered arrangements of atoms or ions. Amorphous (which means "no-shape") solids have disorganized arrangements. Metals, ionic solids, covalent-network and molecular solids are usually at least sort of crystalline and orderly, although they are never completely perfect. Crystalline solids often have faces or facets (flat surfaces at particular angles), like most of the gem-stones used to make jewelry. An important type of amorphous materials are polymers, which include most plastics. Polymers are usually made of really long, skinny covalent molecules that are tangled together. They are often flexible. There are also amorphous stones, like opal (which is usually polished to a rounded shape, because it doesn't make nice facets).
We can also classify solids by their size. Nanomaterials are solids that have very small sizes. Nanomaterials can be made from metallic or covalent-network materials, but because of their small size, they have different properties.
Properties of Solids
As you read about solids, think about how the structure relates to the important properties of the solids that determine how we can use it. Here we will describe some of these properties.
For structural materials (materials that we use to build, support, protect) the strength and hardness can be very important. Hardness means how easy it is to change the shape of the material, such as by stretching or denting it, or how elastic it is in a collision. Strength of materials means how well they resist applied forces, such as compression or stretching. This is important if you want to build a bridge or a building, for example. We will describe solids as brittle if they break into pieces, such as ionic or covalent solids. Metals don't usually break into pieces, but they can stretch into wires (ductility) and be pounded into sheets (malleability).
Conductivity is another important type of property. Do the materials conduct electricity? If so, how much? Do they conduct sound or heat, and how well? We might also want to know about the reactivity of solids. Are they inert (non-reactive) at high temperature in acid? Do they catalyze (speed up) certain reactions on their surfaces? Do they absorb or emit light, and if so, at what energy? Are they magnetic? Are they porous (full of tiny holes)? All these properties can be important for different applications.
Valence and Solids
As you read about solids, it can be helpful to remember the ideas about valence we have discussed earlier. Different atoms have different numbers of valence electrons, which determines how many electrons they usually lose or gain, and how many bonds they usually make. Another way to think about this is with coordination number, which is how many other atoms one atom interacts with. In molecules, coordination numbers are usually pretty low, like 1-4. In ionic solids, each ion might have 2-8 neighbors. In metals, each atom might have 12 neighbors. The stable coordination number of an atom or ion depends on the situation, and how many electrons it needs. To form a covalent-network solid, at least some of the atoms in the formula have to make many bonds (like carbon or silicon, which usually makes 4 bonds). You can't make a good network with only atoms that form 2 bonds, because you can't have branches. In molecular solids, all the bonds are inside the molecule. In metals, in order to gain the electrons they want, because they start with fewer valence electrons, they usually have high coordination numbers, which requires that they have certain structures.
Defects
A really important topic in solids is defects or places where the arrangement of atoms isn't perfect. All examples of a particular molecule are exactly the same, because if one of the atoms is different or the arrangement is different, it would be a different molecule. But solids can be mostly one formula and structure, with some impurities or places where the structure isn't exactly orderly. In fact, solids always have defects. Many of the most important properties of solids depend on the defects, like having a few atoms of different types, or an atom missing, etc. So remember that sometimes the bulk description (how most of the solid is) isn't actually the most important part. The most important part might be the rare spot where a few atoms aren't in their places. This can make solids very confusing to study! | textbooks/chem/General_Chemistry/General_Chemistry_Supplement_(Eames)/Solids/Intro_to_Solids.txt |
Skills to Develop
• Describe the coordination number and some characteristics of a unit cell
• Relate lattice energy and some of the properties of ionic solids
Ionic solids are made of lattices of alternating ions. They can't be close-packed structures, because then like ions would have to touch. We've already seen 2 common ionic unit cells: the CsCl structure and the NaCl structure, on the lattices page. We'll show a few more below, but first, let's talk about interpreting unit cells a little more.
Interpreting Unit Cells
When you look at a unit cell, it's good to be able to figure out what the coordination number of each atom is, and what the overall formula is. When you do this, you need to remember that the unit cell repeats. So to count the number of atoms of each type in the unit cell, we have to count the atoms at the corners as 1/8, the atoms on edges as 1/4, and the atoms on faces as 1/2. The rest of these border atoms are in other cells. When you count coordination number, you also have to remember that atoms might have neighbors in other cells. Here's a formula that can help you:
$\frac{number\; of\; cations\; per\; unit\; cell}{number\; of\; anions\; per\; unit\; cell} = \frac{anion\; coordination\; number}{cation\; coordination\; number}$
The key thing is that the number of cations and anions in the unit cell is the same as the overall formula, and, for example, if the formula is AB2, then the coordination number of A is double the coordination number of B.
Unit Cells for Ionic Crystals
In addition to the ones we saw before, a common structure for the formula AB is called the zinc blende or ZnS structure. Let's analyze this unit cell as shown below. We can see that the grey ions have tetrahedral coordination, and there are 4 of them inside the unit cell. What is the coordination number of the yellow ions, and how many of them are there? It might be a little hard to answer the first question. Let's count them first: each corner has a yellow ion, but since they are on corners, 8 x 1/8 = 1. Then there is a yellow ion on each face of the unit cell, and there are 6 faces. 6 x 1/2 = 3. So there are 4 yellow ions inside each unit cell. This is the same as the number of gray ions, so the formula is AB, as we said. Since the formula is AB, the coordination number of the yellow ions must also be 4. Try to convince yourself that this is true. If you think about each of the corner positions as the same (each corner atom has the interactions of all the corners added together) you can see that they also have tetrahedral coordination.
Let's look at an example of an AB2 structure. Below is the rutile structure, which is quite common, and occurs in many materials including the very common TiO2, which is used in everything from sunscreen to paper to paint. Let's analyze this one also. In the rutile structure, we have a red ion on each corner and one in the middle, so that's 2 red per unit cell. Then we have 2 gray in the middle and 4 on faces (it's a little hard to be sure they're on the faces from the picture, but they are), so that's 4 gray ions total. The formula is AB2. What are the coordination numbers? The gray in the middle is clearly octahedral (6-coordinate). The gray on the corners is also octahedral. If we consider all the interactions between the corners and the inside of the cell, we have 4 bonds to reds that are fully inside, and 4 bonds to reds on faces. We have to count the reds on faces as 1/2, so the coordination number is still 6 for the corners. We can count the red coordination number as 3 easily for the middle, and as 3 for the reds on faces also if we are careful. This fits the equation.
There are many, many other structures. Some are very regular and symmetric, and some are distorted. Some structures are more complex to fit more complex formulas. We don't need to worry about the others now. If you are curious why some materials have one structure and some have another, there are 2 main effects to consider. First, it depends on the relative sizes of the ions. If the ions are similar sizes, the CsCl structure is good, because it has the highest coordination number (more favorable interactions between opposite ions). If the ions aren't the same size, then in the CsCl structure the big ions might touch each other, so the NaCl structure (with coordination number 6) becomes more stable, so the big ions can be farther apart. For even less equal sizes, the zincblende structure might be necessary. The other effect is how covalent the bonds are. If the bonds are partly covalent, even if the sizes aren't so different, the zincblende structure might be better because it allows better orbital overlap (just like in diamond, which has the same structure but with all the atoms the same).
Properties of Ionic Solids
Ionic solids are usually insulators, because electrons are held tightly to the ions. They are hard and have high melting points, because ionic bonds are strong. However, they are brittle because if the crystal gets knocked so that the wrong ions are touching, it will shatter.
A key concept for ionic bonding is the lattice energy. If you compare the electron affinity for a non-metal and the ionization energy for a metal, you'll notice that forming ions is endothermic. However, remember that electron affinities and ionization energies are defined for gases. Making ionic solids from the elements is very exothermic, because bringing the ions together into the crystal lattice is releases a lot of heat. The lattice energy is the energy released when the solid forms from ions in the gas phase. The lattice energy includes all the ion-ion interactions in an infinite lattice, so it is a little complicated to calculate. The main thing to know is that the smaller the ions and the larger the charge, the bigger the lattice energy is, and the more stable the solid. | textbooks/chem/General_Chemistry/General_Chemistry_Supplement_(Eames)/Solids/Ionic_Solids.txt |
Skills to Develop
• Describe the different types of unit cells and alloys
Most metals want to have high coordination numbers (lots of neighbors for each atom). The highest coordination number possible with spheres is 12. There are 2 regular arrangements that give 12 neighbors, and they are called close-packing. These are the densest possible ways to arrange spheres. Both are based on hexagonal planes of atoms that are stacked. Each plane's atoms sit in the holes of the lower plane. Look at the figure.
The gray A spheres represent the bottom plane. The holes in this plane are labelled B and C. The next layer will go over either the B holes or the C holes. It doesn't matter which. The 2 types of close-packing come from what happens to the third layer. If the layers are ABAB (which is equivalent to ACAC or BCBC), that forms one type of close packing, called hexagonal close-packed, or hcp. The other type of close packing has layers ABCABC and is called cubic close-packed or ccp. Cubic close-packed is also called face-centered cubic or fcc because it is a cubic structure with an atom on each face and corner of the cube. Each atom in a close-packed structure has 12 neighbors (6 in its plane, and 3 in the planes above and below).
Not all metals make close-packed structures. A few have a simple cubic structure (just one atom on each corner of the cubic unit cell), so in this case each atom has a coordination number of 6. Another more common structure is called body-centered cubic or bcc, in which there is an atom on each corner and in the center of the cubic unit cell, so that each atom has 8 neighbors. Here's a different view of the bcc unit cell, using a "space-filling" picture of the atoms. Notice that each corner of the unit cell actually has 1/8 of an atom. (If there were a whole atom on each corner, then we'd get the wrong structure when we stack the unit cells.)
Alloys
Alloys are metal solutions or compounds. Some alloys have different metal atoms arranged in a regular structure. These alloys have a precise formula, such as Ni3Al. Others have a random replacement of one type of metal atom with another, like gold/silver alloys. Both gold and silver have the fcc/ccp structure and the atoms are similar sizes, so they can make mixtures with almost any % Au and Ag. This is called a substitutional alloy because one type of atom substitutes for another. A third type have a smaller atom (like carbon in steel) that occupies some of the holes between the majority metal atom. This is called an interstitial alloy. Alloys are important for many applications because they have different properties compared to the pure metals.
Metallic Bonding
Skills to Develop
• Describe metallic bonding using MO theory
We previously discussed the "electron-sea model" of metallic bonding in the intro section and metal characteristics sections. This is a good time to review those sections, before we describe a better model. Why do we need a different model for metallic bonding than the electron-sea model? Although it explains some general properties of metals, like malleability and conductivity, it doesn't explain the relative properties of metals, like their hardness and melt points. These properties depend on how strong the bonding in the metals is. By the electron-sea model, we might think that more electrons makes the bonds stronger, so hardness and melt point would increase across the periodic table.
We can explain these properties using MO theory. In this case, we imagine combining many atomic orbitals (1 or more for each atom) to make and equal number of MOs that extend over the whole solid. Some MOs will have fewer nodes and be lower energy, while others will have more nodes and be higher energy. Each MO can hold no more than 2 electrons. Filling the lower energy MOs (bonding MOs) makes the bonds stronger, which is why alkali metals have low melt points and are soft (not many bonding MOs filled). Filling the higher energy MOs (antibonding MOs) makes the bonds weaker, which is why Cu, Ag, Au and Zn are soft and melt at low temperatures (Hg is a liquid at RT!).
When we have a really big number of MOs, some interesting things happen. One mole of Fe is about 7 mL or 7 cc (a pretty small amount). In each atom of Fe, we have 1 4s orbital, 5 3d orbitals, and 3 4p orbitals that can be involved in bonding. In a mole of Fe, we have 9 times Avogadro's number of AOs that can be involved in bonding. That's about 1024 AOs, which means the same number of MOs. There just isn't space to have much difference in energy between all those MOs. The MOs have to have energies not too different from the AOs (they are definitely higher and lower, but not by too much), so they are limited to a relatively narrow range of energies. This means that the energies form bands rather than separate energy levels.
We can start to see how this happens by imagining a line of Li atoms. As we increase the number of Li atoms, the orbitals get closer together. When the number of Li atoms reaches infinity, the MOs become infinitely close together.
Band theory explains the conductivity of metals: for electrons to move, electrons have to be able to change MOs without gaining much energy. If a band is partly-filled with electrons, they can easily change states because there are empty states almost the same energy as the full states. We can tell that the bands from s, p and d orbitals must overlap in metals, because the alkaline earth metals are conductors. If there were an energy gap between the s band and the p or d band, they would not conduct because the s band would be full.
Other Solids: Polymers Nanomaterials Foams etc.
Skills to Develop
• Describe some of the properties and/or characteristics of polymers and nanomaterials
There are many other types of solids, or other important qualities that distinguish solids, like catalysis, porosity, etc. We haven't described minerals or ceramics very well, or superconductors, magnetic materials, etc. For now, we'll just describe two more types: polymers and nanomaterials. Polymers are familiar and important from life, because we use them for almost everything. Nanomaterials are less familiar, but are starting to be used in commercial products although we don't always understand them too well.
Polymers
Polymers are also called macromolecules, which means big molecules. Polymers are usually long chains made of small molecules covalently bonded together. The important properties of the polymer come from the "intermolecular" or non-covalent forces that determine how the polymer chains fold up or tangle together. Most important biomolecules (like DNA, proteins and starch) are polymers. We also make a wide variety of polymers, usually starting with hydrocarbon molecules from oil, that are used in clothing, plastics, construction materials, etc. Many different molecules and reactions can be used to make polymers. The properties of the polymer will often depend on the structure, which is usually amorphous or disorganized. If the macromolecules are more orderly and rigid, the structure may be more crystalline and less flexible, with a higher melting point (because the molecules can pack together well with relatively strong intermolecular forces). If the macromolecules are very different sizes and have irregular shapes, then the polymer might be more flexible. Sometimes covalent bonds are added between the molecules, which can also make the material tougher. (For example, rubber is made by stronger by heating with sulfur, which can connect the chains together.)
Nanomaterials
Nanomaterials are very small solids, like extremely fine powders only smaller. They have dimensions measured in nanometers, or 10-9 meters. At this scale, they are too small to have all the normal properties of solids, because they have so much edge. For example, nanomaterials have bigger band gaps than the same material in bigger chunks. The smaller the particles, the bigger the band gap. (Molecules have bigger HOMO/LUMO gaps than solid materials; nanoparticles are in between molecules and normal solids.) People are pretty excited about nanoparticles (they're popular!) but it's good to remember that we don't really know much about their health effects yet, except to say that they are complicated and need more study. I don't recommend running out and buying the new nano-shampoo or nano-silver handwipes or whatever you might encounter unless it's noticeably superior to the non-nano alternative.
Nanostructures and nanoparticles can be natural or manufactured. For example, most birds that look blue don't have blue pigment, they have nanostructures on their feathers that scatter light. These days, most electronic devices are based on nanostructures also: each individual transistor in a computer chip is nanosized, and it may not be possible to make them much smaller than they are now without too many size-based complications. | textbooks/chem/General_Chemistry/General_Chemistry_Supplement_(Eames)/Solids/Metal_Crystal_Structures.txt |
Skills to Develop
• Understand the fundamental concepts of thermodynamics
Thermodynamics
Thermodynamics is the study of heat, energy, and work and how they move. This is important because we have to move or generate heat to stay comfortable in winter and summer; we need to do work by moving things for many different purposes; we need to generate chemical energy to live and grow our bodies, etc.
System and Surroundings
In thermodynamics, we often separate the universe (that is, everything that exists) into 2 parts: the system, which is the small part we are interested in, and the surroundings which is everything outside the system. This will help us think about how heat, energy and work move between parts of the universe.
Open, Closed, and Isolated Systems
Open systems allow energy and matter (stuff) to enter and leave the system. A pan on the stove is an open system because water can evaporate or be poured in, and heat can enter the pan if the stove is turned on, and leave the pan also. A closed system does not allow matter to enter or leave, but does allow energy to enter or leave. A covered pot on the stove is approximately a closed system. An isolated system does not allow either matter or energy to enter or leave. A thermos or cooler is approximately an isolated system. There are no truly isolated systems.
State Functions
State functions are quantities that don't depend on path. Your bank balance is a good example. It doesn't matter how the money entered your bank account, the total amount there at any given time is what it is and you can measure it easily. It doesn't matter if you put it in all at once, or a little bit every month, or put a lot in then spent it slowly... any time you want to know how much is there, you just check. Most of the quantities you know are state functions, like pressure, volume, temperature, location, etc. But some quantities that are important in thermodynamics, like heat and work, are only defined by a process, so they aren't state functions.
Extensive vs. Intensive
Extensive refers to properties that depend on how much stuff there is. For instance, the volume or pressure created by a sample of gas depend on how much gas is in the sample, so they are extensive. Intensive quantities don't depend on how much there is. For instance, temperature, density, etc. If you divide a sample in 2, it does not change temperature. Density is a ratio of 2 extensive properties, so it is intensive.
Calorimetry and Reaction Enthalpy
Skills to Develop
• Perform and describe calorimetry calculations
How do you measure ΔH?
If you want to measure ΔH, you usually use calorimetry, which just means measuring heat. The usual way this is done is by measuring how much the temperature of a system increases when the process occurs. For instance, perhaps we have 2 solutions (like an acid and a base solution) and we mix them in a thermos. We measure the temperature of the solutions before mixing and also after the reaction. Because we run the reaction in a thermos, we expect that almost all the heat from the reaction will stay in the thermos. Also, we don't close the thermos all the way, so the pressure is always atmospheric pressure. The reaction enthalpy is related to the temperature change, but how, exactly?
Heat Capacity
Heat capacity tells us how much heat is needed to increase the temperature of an object or substance by a certain amount. The unit calorie is the energy needed to increase the temperature of 1 g of water by 1 degree (either Celsius or Kelvin). 1 calorie = 4.18 J. Generally, the heat capacity is
$C=\frac{q}{\Delta T}$
Depending on what is convenient, heat capacity can be defined in different ways. Sometimes it is defined using heat transferred at constant pressure, and other times heat transferred at constant volume. Specific heat is the heat capacity per unit of mass:
$C=\dfrac{q}{\Delta T\; \times m}$
Molar heat capacity is the heat capacity per mole of substance. If you know the heat capacity of the system, you can calculate ΔH using the temperature change data from a calorimetry experiment. In the example of the acid-base reaction in the thermos, you would want the heat capacity of the thermos and you would add the heat capacity of the solution, which you could calculate using the specific or molar heat of water.
Doing Calorimetry Calculations
First, you need to be a little careful about whether the experiment was done at constant pressure or constant volume. This will determine whether you calculate enthalpy or internal energy of reaction. Second, make sure you figure out the heat capacity of the system correctly. You might have to add up heat capacities of different parts of the system using (mass x specific heat) or (moles x molar heat capacity) of each part. Once you've figured that out, you can usually think of it as a "unit conversion" and use dimensional analysis to combine all the quantities you know to find the quantity you want. | textbooks/chem/General_Chemistry/General_Chemistry_Supplement_(Eames)/Thermochemistry/Basic_Definitions.txt |
Skills to Develop
• Perform enthalpy calculations and describe the process and results
What is Fuel?
One of the most important applications of chemistry is the study of fuels. Fuel basically means a chemical that can provide energy. We use fuel to do nearly everything. Food provides fuel to our bodies. Gasoline provides fuel for our cars. Many different chemicals, from coal and gas to uranium, provide fuel for the power plants that make electric power. When we use a fuel, it becomes a reactant in some type of reaction. Most often, fuels are burned in air to provide heat, and the heat is converted to work. For this reason, we often want to know how much heat we can get out of a chemical reaction.
Enthalpy
When our systems includes chemicals that participate in a reaction, the system may gain or lose heat because of the reaction, and the system may also do work, or work may be done on the system. How would the system do work? Suppose you burn some propane (C3H8). Write a balanced equation for this reaction, and convince yourself that there are more moles of gas present in the products than the reactants. Also, you know that burning propane raises the temperature. If you burn some propane, the system will increase in volume, because there is more gas, and it is hotter. (If you burn the propane in a very strong container, then the volume will stay the same and the pressure will increase.) When the volume of the system increases in an open container, it will push on the surroundings, like the atmosphere, and do pV work. We might be able to use this work (in an internal combustion engine like in a car, for instance) but only if we control the volume. If we do the reaction open to the atmosphere, at constant pressure (atmospheric pressure), then we won't be able to use the work done by the reaction.
In chemistry, because reactions are often done at atmospheric pressure, we often want to know how much heat is available from a reaction that occurs at constant pressure. To make this convenient, we can define a new quantity, related to internal energy, which is called enthalpy, abbreviated H:
\[H=E+pV\]
Here's why this definition is useful. The work done by a reaction at constant pressure is
\[w=-p \Delta V\]
where p is the pressure, and ΔV is the change in volume of the reaction system, and the sign of w is negative because the system is doing work. The change in enthalpy for a reaction is
\[\Delta H=\Delta (E+pV) = \Delta E + p\Delta V (at\; constant\; p)\]
Since ΔE = q + w,
\[\Delta H = (q + w) + p\Delta V = (q + w) − w = q\]
Thus, the enthalpy change in a reaction tells us exactly how much heat the reaction can provide if it runs at constant pressure. We don't need to worry about calculating the work done by the reaction pushing back the atmosphere because it is already removed from the definition of enthalpy. Enthalpy is a state function because E, p and V are all state functions. Enthalpy doesn't have a molecular meaning like internal energy, but usually pV is small, so enthalpy is similar to internal energy.
When ΔH is positive, then heat has entered the system, and the process is called endothermic. Endothermic reactions feel cold to the touch because they pull heat from your hand into the reaction system. Evaporation is endothermic, which is why sweat helps us cool off. When ΔH is negative, heat leaves the system, and the process is called exothermic. Exothermic processes feel hot. This is why flames will burn you. | textbooks/chem/General_Chemistry/General_Chemistry_Supplement_(Eames)/Thermochemistry/Fuels_and_Enthalpy.txt |
Skills to Develop
• Calculate enthalpies of reactions using Hess' Law
• Define "standard state"
What is Hess' Law?
Hess' Law is an early statement of the law of conservation of energy (1840). It says that the heat liberated by a process doesn't depend on how the process happens (only on the starting and ending states: in other words, it's a state function). Now we know we should really use enthalpy for this, not heat, because enthalpy is a state function, so this is true, while heat is a process. Hess' Law lets us break a reaction or process into a series of small, easily measured steps, and then we can add up the ΔH of the steps to find the change in enthalpy of the whole thing. Hess' law is a great way to think about chemical processes and make predictions. We'll see lots of applications of Hess' law, but right now let's start with finding reaction enthalpies using standard enthalpies of formation.
Standard Enthalpies of Formation
Standard enthalpies of formation help us predict reaction enthalpies for many reactions if the products and reactants are well-studied, even if the specific reaction is new. To do this, we imagine that we take the reactants and separate them into their pure elements in a standard state. The standard state is the element in its most stable form at room temperature and atmospheric pressure. Then we take the elements and recombine them to make the products. The reaction enthalpy is equal to the difference in the enthalpies of these processes.
Example
Let's look at a specific example. Here are some enthalpies of formation (in kJ/mol of reaction):
$C(s) + O_{2}(g) \rightarrow CO_{2}(g)\; \Delta H_{f} = -394\; kJ C(s) + 2H_{2}(g) \rightarrow CH_{4}(g)\; \Delta H_{f} = -75\; kJ 2H_{2}(g) + O_{2}(g) \rightarrow 2H_{2}O(l)\; \Delta H_{f} = -572\; kJ$
Let's use these enthalpies of formation to calculate the enthalpy of combustion for 1 mol of methane. The reaction we want is
$CH_{4}(g) + 2O_{2}(g) \rightarrow CO_{2}(g) + 2H_{2}O(g)$
If we reverse a reaction, we change the sign on ΔH, and if we multiply the reaction by a constant coefficient, we multiply ΔH by the same coefficient. Let's combine the formation constant equations so they add up to the reaction we want:
CH4 (g) C(s) + 2H2(g) ΔHf = 75 kJ
C(s) + O2(g) CO2(g) ΔHf = -394 kJ
2H2(g) + O2(g) 2H2O(l) ΔHf = -572 kJ
That's almost right but we're missing the state of the water:
$H_{2}O(l)\rightarrow H_{2}O(g)\; \Delta H=44\; kJ$
The full set is
CH4(g) C(s) + 2H2(g) ΔHf = 75 kJ
C(s) + O2(g) CO2(g) ΔHf = -394 kJ
2H2(g) + O2(g) 2H2O(l) ΔHf = -572 kJ
2H2O(l) 2H2O(g) ΔHf = 88 kJ
CH4(g) + 2O2(g) CO2(g) + 2H2O(g) ΔH = ?
Note that everything but the desired reaction cancels. Now we just add up the enthalpies of each step, and we find that the enthalpy of combustion of 1 mole of methane is -803 kJ.
General Procedure for Hess' Law Calculations
Determine the equation for the desired process (the process for which you want to know the enthalpy change). Break it into steps for which you can look up the enthalpy changes. This probably means steps like formation from elements, and changes of state. (Later, we'll include other processes like ionization, etc) Arrange the steps so that everything cancels out leaving just the desired reaction. Make sure the coefficients on equations are correct (multiply the equation and ΔH by a constant if needed) and that all the components are in the correct state (like the example above, we had to convert from liquid water to gaseous water). Then just add it up! | textbooks/chem/General_Chemistry/General_Chemistry_Supplement_(Eames)/Thermochemistry/Hess%27_Law_and_Enthalpy_of_Formation.txt |
Skills to Develop
• Describe the 2 theories for heat
• Explain Mayer's argument
For a long time, physicists and chemists debated whether heat was a fluid (like a mysterious liquid) or came from the motion of particles. Many early scientists, like Newton, had thought that heat might be caused by small movement of particles, and greater heat meant greater velocities or kinetic energies. Lavoisier, however, thought that heat was a massless fluid that he called "caloric."
Count Rumford observed that the process of boring cannon (drilling the hole in the middle of the brass cannon) produced a lot of heat, especially when the drill was dull or blunt. He showed that the heat produced was related to the amount of mechanical work done by the drill. Davy (mentioned earlier here) showed that even at 0°C, two ice cubes would melt when rubbed together. This frictional heating is also a way that people sometimes start fires in the wilderness.
Other scientists liked the fluid theory. Lavoisier thought heat was a fluid that caused the atoms it surrounded to separate (which is why, he said, density usually decreases as you heat a substance). One important contribution he made was to show that the heat generated by human or animal metabolism (oxidizing food with oxygen from breathing) produces the same amount of energy as combusting the food (which is often if not always true). Carnot, who will be very important later (in the development of the second law of thermodynamics), also thought that heat was a liquid, because like liquids it "flows downhill" from hot objects to cold objects. He thought that like power generation from a waterfall, the amount of heat that moves and the distance it falls (change in temperature) determine the available power. However, later he realized that some of the heat is lost when it is converted to mechanical energy (work), which means it can't be a fluid like water (water isn't lost when falling water is used to drive a motor).
Mayer used data other people collected on heat capacities of air at constant pressure or constant volume to calculate the relationship between the energy defined as force x distance (like the modern unit joule) and energy defined by change in temperature of a substance (like the modern unit calorie, the energy needed to raise the temperature of 1g of water 1°C). Imagine heating a sample of air in a fixed volume container or in a chamber with a piston, so that it is always at atmospheric pressure. One sample does work when heated (by expanding against atmospheric pressure) and the other does not. The difference in heat required to get the same temperature change in the 2 containers must be equivalent to the work done by the system with the piston. Mayer argued that heat, work, and chemical energy are all interconvertible, meaning they are all energy in different forms.
Joule (for whom the unit joule is named) was an English beer-brewer who did the studies mentioned earlier that lead to the first law. You might have learned Joule's law in a physics class, that heat produced by electricity, Q, is
$Q\; \alpha\; I^{2}R$
where I is current and R is resistance. He compared heat produced by electricity and heat produced by mechanical work (heating water using a paddlewheel powered by a falling weight) and thus showed the equivalence of mechanical work and heat. In other words, Joule and Carnot had showed that heat can be used to generate work (like in a power plant) and work can be used to generate heat (like with the cannon drills or paddlewheel). Kelvin combined these ideas and used them to propose the Kelvin temperature scale (but the details of that can wait until we study Carnot's discoveries in more detail). | textbooks/chem/General_Chemistry/General_Chemistry_Supplement_(Eames)/Thermochemistry/History_of_Thermodynamics.txt |
Learning Objectives
• Define energy and its two main types listed below
• Describe the conservation of energy
You are probably familiar with these types of energy from a physics class. Energy is the capacity (ability, sort of) to do work. You have a sense of what work is from regular life, it's things that require effort. Energy and work have the same units. Kinetic energy is the energy that comes from motion. The equation for kinetic energy is
$KE=\frac{1}{2}mv^{2}$
where KE is kinetic energy, m is mass, and v is velocity. This definition should make sense: big things moving fast have the most energy, the most ability to shove other things or knock them over, etc. Potential energy is energy that comes from position and a force. For instance, gravitational potential energy is the energy that things have if they are high up. If they fall, their potential energy will turn into kinetic energy because they are accelerated by gravity. The equation for potential energy from gravity is
$PE=mgh$
where PE is potential energy, m is mass, g is the acceleration of gravity, and h is the height. This makes the units of energy very clear: mass x distance x acceleration, or force x distance, which comes to kg•m2s-2. In chemistry, the force that leads to potential energy is almost always the Coulomb force, not gravity. In this case, the potential energy from 2 charges near each other is
$PE=\frac{kQq}{d}$
where q and Q are the 2 charges, d is the distance between them, and k is a constant, 8.99 x 109 J•m•C-2. (Joules, J, are the SI unit of energy, and coulombs, C, are the SI unit of charge.) When the charges have the same sign, they repel and will accelerate away from each other if allowed to move; the potential energy has a positive sign. When the charges have the opposite sign, they attract each other and have negative potential energy. If they are allowed to get closer together, the potential energy will get more negative. If they are separated, d gets bigger and the potential energy approaches zero.
Conservation of Energy
You probably learned about conservation of energy already in a physics class. For instance, if you have a pendulum as shown, at Position 1 the weight has some potential energy, but no kinetic energy. When you release the weight, the weight falls, moving through Position 2. At Position 2, some of the potential energy has been converted to kinetic energy. Finally, at Position 3, all the potential energy has been converted to kinetic energy. As it passes 3, the process is reversed, and kinetic energy is converted to potential energy. When the weight reaches Position 4, all the kinetic energy has been converted back to the same amount of potential energy it started with at 1. This is just one example of conservation of energy. It is a general observation that the amount of energy in the universe doesn't change, and the amount of energy in a particular system doesn't change unless there is a flow of energy in or out. | textbooks/chem/General_Chemistry/General_Chemistry_Supplement_(Eames)/Thermochemistry/Kinetic_and_Potential_Energy.txt |
Skills to Develop
• Develop applications of the First Law of Thermodynamics
• Calculate internal energy and elaborate on the results
The First Law of Thermodynamics is equivalent to the law of conservation of energy, which was described previously here. However, instead of describing a system in which energy changes form (KE to PE and back) but the total amount doesn't change, now we will describe a system in which energy can move in and out. The 2 ways it can do this are work and heat.
Imagine we have a system that is approximately isolated, like a thermos of water, and we do work on it in various ways. First we use the energy from a falling weight to mix it very vigorously, and we see what happens to the temperature of the system. Then we run some electric current through a resistor dipped in the system, and see what happens to the temperature of the system. And we observe that the same amount of work always raises the temperature the same amount. If we do the same experiments on a cup of water that isn't isolated like the thermos (but is otherwise the same), we find that the temperature doesn't increase as much when we do the same amount of work without the insulating thermos. As the system got warmer, some energy moved from the system to the surroundings. Without the thermos, heat leaves the system when it gets warmer than the surroundings, so we have to do more work to get the same increase in temperature.
Energy is the capacity to do work. Any system can do work. Perhaps it can heat a gas, causing it to expand against a force, or fall, compressing a spring, etc. But an isolated system's capacity to do work won't change. For instance, if we use a system to do work, then wait a long time, the system won't regain its original ability to do work. More concretely, if we use a falling weight to do some other work, like lifting another weight or driving a motor, the weight won't be able to do more work until we lift it to its original height. It won't float back up by itself, ready to fall again. If we use a very hot block of metal to boil water and drive motors, the block will get cooler. And if we leave it isolated and wait, it won't heat back up again by itself so we can boil more water. The capacity of a system to do work is called its internal energy. The internal energy of a system can change if work is done on or by the system, or if heat enters or leaves the system. If no work is done and no heat flows, then the internal energy of the system can't change. We can write this as an equation:
\[\Delta E=q+w\]
where ΔE is the change in internal energy (some people use U instead), q is heat, and w is work. Heat q is positive when it flows into the system, and negative when it flows out. Work w is positive when it is done to the system and negative when it is done by the system. (Actually, some people use the opposite signs for work, in which case the equation is E=q-w) This should make sense. The internal energy of the system increases when we put energy in; it decreases when we take energy out.
Molecularly, internal energy means all the kinetic and potential energy of each particle in the system. Internal energy is a state function, like temperature and pressure. Work and heat are not state functions: they depend on processes. You can't look at an object and determine how much work or heat it has because that doesn't make sense, but you can measure what its temperature is, what its volume is, and what its internal energy is. | textbooks/chem/General_Chemistry/General_Chemistry_Supplement_(Eames)/Thermochemistry/The_First_Law_of_Thermodynamics.txt |
Skills to Develop
• Define the Zeroth Law of Thermodynamics
Equilibria Again
We discussed equilibrium reactions before, especially in the stoichiometry section. We mentioned chemical equilibria, between reactants and products. We also mentioned solubility equilibria in the precipitation section. Now let's introduce some other types of equilibrium. Suppose we put a partially-filled balloon on each end of a tube with a piston. Initially the piston is clamped in place. If we remove the clamp so the piston can move, it will move until the pressure on each side is equal. At this point, the balloons will probably also be the same size (unless they are different size balloons), because pressure determines the size of the balloons. The changes the volume on each side until the pressures are equal, and then the system reaches mechanical equilibrium, because the piston won't move any more. Pressure is the property that tells us whether systems will be at mechanical equilibrium; if we clamp the piston and put on a new balloon, if it's at the same pressure as the other balloons then the piston won't move when the clamp is removed.
Top: initial position. Bottom: after reaching mechanical equilibrium.
Now we can introduce energy can flow in and out; maybe they have metal walls, or some other wall that conducts heat). We set them in contact, so they are touching each other. Maybe we can observe some sort of change in the systems, like that the pressure inside one increases and the other decreases. This might mean that one was hotter and the other colder, so the pressure inside changed as they approached the same temperature once they were touching. If no change happens when they touch, we could say that they were in thermal equilibrium. (If we leave them in contact long enough, they will reach thermal equilibrium, but that is different from being in thermal equilibrium when they first touch.) If we have a system A, and we find that it is in thermal equilibrium with another system B, and also with another system C, then we know without doing the experiment that B and C are also in thermal equilibrium with each other. That statement is the Zeroth Law of Thermodynamics. The property that tells us if 2 systems are already at thermal equilibrium is the temperature. This is a familiar word, but technically this concept of thermal equilibrium is how it is defined: the Zeroth Law introduces the concept and property of temperature.
We measure temperature using a thermometer. In science we use 2 scales to measure temperature, the Celsius or centigrade scale, on which water freezes at 0 °C and boils at 100 °C; and the Kelvin scale, on which water freezes at 273.15 K and boils at 373.15 K. The Celsius scale is probably what you are used to from regular life (although I am used to the Fahrenheit scale), and it is convenient because it focuses on water, which is very important for our lives. The Kelvin scale is good for science and thermodynamics especially, because 0 K is the lowest possible temperature.
Work and Heat
Skills to Develop
• Compare and contrast work and heat
In physics, work is defined as motion over some distance against an opposing force. (A force is an acceleration multiplied by a mass.) For instance, lifting an object against the force of gravity. The amount of work is the opposing force multiplied by the distance the object moved. The amount of work done in lifting an object is the force of gravity (the mass of the object times the acceleration of gravity) times the height you lifted it. There are lots of other ways systems can do work: a stretched spring can expand and lift a weight, a battery or burning gas can drive a motor and move a car up hill or forward against friction and wind.
In chemistry, work is often defined in terms of a change in volume against a pressure. (Pressure is force divided by area, so convince yourself that P x V has the same units as F x d.) For instance, atmospheric pressure is constant at ~1 atm. If you have a sample of gas at higher pressure, and you let it come to mechanical equilibrium with the atmosphere, it will expand to some new volume. The work it does expanding against atmospheric pressure is (1 atm)(ΔV), where ΔV is (final volume - initial volume). In general, the work done by gases expanding is called pV work, and is
$w=\int{pdV}$
For our purposes, we won't be dealing with integrals, so we won't calculate work this way unless the pressure is constant, in which case we can use the simpler equation
$w=-P\Delta V$
where P is the constant pressure and ΔV is the change in volume of the system. The negative sign accounts for the fact that if the system is compressed, ΔV is negative, and work was done to the system, so w is positive.
There are several different definitions of heat. For instance, we can heat water to make tea, or enjoy the heat when we enter a warm room in the winter. In thermodynamics the meaning of heat is more precise: it is a process, a way energy can move. Heat is energy that moves from a hot object to a cold object. When heat leaves the system, it has a negative sign, and when it enters the system, it has a positive sign.
Heat and work are the ways that energy can move between objects. When you think about the molecules, the difference between work and heat is very simple. Work involves an orderly motion of molecules, like all the molecules in an object moving the same direction. Heat involves disorderly or random motions of molecules. | textbooks/chem/General_Chemistry/General_Chemistry_Supplement_(Eames)/Thermochemistry/The_Zeroth_Law_of_Thermodynamics.txt |
Valence bond theory is one of two basic theories, along with molecular orbital theory, that were developed to use the methods of quantum mechanics to explain chemical bonding. It focuses on how the atomic orbitals of the dissociated atoms combine to give individual chemical bonds when a molecule is formed. In contrast, molecular orbital theory has orbitals that cover the whole molecule.
Valence Bond Theory
Skills to Develop
• List the common 3-D structures and their characteristics
Why do 3-D Structures Matter?
3D structures are important because they can have big effects on the properties of molecules. For instance, water has a dipole moment because it is bent, not linear. If it were linear, it would not be a great solvent for polar compounds, and all life would be completely different. In general, shapes of molecules will influence how they react, because they determine polarity (which can help pull molecules together to react) and fit (whether the reactive parts can get close to each other). Shapes of molecules can have other effects as well. Consider the following:
Formula NaF MgF2 AlF3 SiF4 PF5 SF6
Melting Point (°C) 980 1400 1040 -77 -83 -55
Notice that there is an abrupt change in the melt point between AlF3and SiF4. You might take this as the change from an ionic compound to a covalent compound, but the difference in electronegativity between Al and Si isn't that large, and both are much less electronegative than F. We can explain the change better using molecular structure than bond type. In the first 3, the structure is typical of ionic compounds. Instead of clearly-defined molecules, there is an alternating lattice of positive and negative ions. Each cation is surrounded by 6 fluoride ions, and each fluoride ion is surrounded by 6, 3, or 2 cations (depending on the formula). All the ion-ion attractions must be loosened to melt the solid, which requires high temperature. In SiF4, however, each Si is surrounded by 4 "fluoride ions" (because radius decreases across the periodic table) which naturally makes a tetrahedron, and thus clearly separated molecules. When the molecules pack in the solid, the F atoms touch other F atoms, not Si atoms, and are not attracted much at all. This is only more true for PF5 and SF6, in which the F atoms surround the central atom even more completely. Thus, these 3 compounds are molecular gases.
How do we Measure 3-D Structures?
We can measure 3D structures with several different techniques, but X-ray crystallography is probably the most common. For small molecules, it can usually tell you the exact positions of all the atoms as long as you can grow a good crystal. This data includes the bond lengths and bond angles.
Describing 3-D Structures
You'll need to learn the names of the common geometries, which describe the shape of bonds around each atom. They are organized in the following diagram based on how many bonds the atom makes. If the molecule only has one central atom, the geometry of the molecule is the geometry of the central atom. If there are several atoms that have bonds to 2 or more atoms, we can describe the geometry at each.
Predicting Molecular Shapes (Bond Angles)
Lewis structures are a great way to predict the shapes of molecules. The basic idea is that while all electrons repel each other, electrons with the same spin repel each other even more. The 2 paired electrons with opposite spin that make up a bond can be in the same general area between the bonding nuclei (although they will still try to avoid each other within this area), but electrons with the same spin really have to give each other space. The result is that the bonding and non-bonding electron pairs each take up their own area and try to stay as far away from each other as they can. The area occupied by a lone pair or a bond is called a domain. (We only worry about the valence electrons. We can think of the core electrons having their own area closer to the nucleus, although it's actually a little more complicated than this.) For the same reason, the different parts of a molecule usually spread out also, so the different parts don't get too close and bump their electron pairs.
You can picture each pair of electrons as being a big soft balloon, as in the diagram. Bonding pairs are a little narrower because they are attracted to 2 nuclei and try to stay between them. Lone pairs are a little flatter and take up more space, because there's no other nucleus pulling them away in a specific direction. So we can expect bond angles are a little smaller than we might think, because lone pairs take up more space and push the bonds closer together. Multiple bonds also take up more space than single bonds, but still act like a single domain, because they are held by the same two atoms.
Bond Lengths
Bond lengths depend on the size of the atoms and the strength of the bonds. In general, bond length decreases going from single bond to double bond to triple bond.
Some Average Bond Lengths, in Å
Bond r (Å) Bond r (Å) Bond r (Å)
C—C 1.54 C=C 1.34 C≡C 1.20
C—N 1.47 C=N 1.38 C≡N 1.16
N—N 1.45 N=N 1.25 N≡N 1.10
C—O 1.43 C=O 1.20 C≡O 1.13 | textbooks/chem/General_Chemistry/General_Chemistry_Supplement_(Eames)/Valence_Bond_Theory/3-D_Structures_of_Molecules.txt |
Skills to Develop
• Describe the significance of dipole moments
Dipole moments are a measure of how much how much charge separation exists in a bond or a molecule. We have talked about similar ideas before, because molecular dipole moments are important for solvation. We defined dipoles here when introducing magnetism, and discussed polarity also. To review, the electric dipole moment is defined as
$\mu = q \times d$
where q is the partial charge on each end and d is the distance between the charges. "Each end" could mean each end of a bond (each atom), or each end of a molecule, like water. The total dipole moment of a molecule is the vector sum of all the bond dipoles.
Why do Dipole Moments Matter?
Dipole moments tell us, on average, where the electrons in a molecule are. They can also tell us the shape of molecules. For instance, if H2O were linear, the 2 O-H bond dipoles would cancel each other out, and the whole molecule would be non-polar. Since we know water is polar, it has to have a bent shape.
Also, molecular dipole moments are very important for many properties, such as ability to dissolve solutes, melting and boiling points, and reactivity in general. Dipole moments (actually, change in dipole moments due to molecular vibration) are also involved in whether molecules absorb certain energies of light, which is important for the greenhouse effect. CO2 and methane cause climate change because they can absorb IR light. Although these molecules are too symmetrical to have permanent dipole moments, vibrations can produce small temporary dipole moments that allow them to absorb light.
How do we Measure Dipole Moments?
We can measure molecular dipole moments by measuring the dielectric constant of a material (how much the material weakens the Coulomb forces when it is between the charges). Generally, materials with large dipoles have high dielectric constants. In Coulomb's law, the dielectric constant D reduces the force between charges.
$F= \frac{kQq}{Dr^{2}}$
Bigger D means less force. To measure bond dipole moments, we can measure the dielectric constants of diatomic molecules (in the gas phase, if the diatomic molecule doesn't exist as a solid or liquid).
Electronegativity
Skills to Develop
• Describe and explain the periodic trend for electronegativity
• Discuss the significance of electronegativity
Electronegativity is a measure of how much an atom attracts electrons. For instance, a more electronegative atom will be easily reduced, while a less electronegative atom will be easily oxidized. In covalent bonds, more electronegative atoms "pull harder" on the bonding electrons, so the shared electrons may spend more than half their time with the more electronegative atom, giving it a partial negative charge. (For example, O in water has a partial negative charge because it is more electronegative than H.) This will also influence the bond dipole moment.
How do we Define Electronegativity?
One way we can define electronegativity is by saying it is proportional to the sum of ionization energy and electron affinity. If it's hard to take electrons away, and easy to add electrons, the electronegativity is big.
$EN = constant \times (IE + EA)$
We choose the constant so that F has EN=4, and then use the same constant for all other elements.
However, electron affinities are not known exactly for most elements, so the first definition (the Mulliken definition, we'll talk more about Mulliken in the next section) is limited. Pauling proposed a different definition, which gives similar results but uses easier measurements.
Pauling's electronegativity scale is based on ionic resonance energies. Consider 2 diatomic molecules A2 and B2. We can measure the bond energy of each. We can also describe the bonding using resonance between a covalent structure, A—A and ionic structures [A+][A]. If we make the molecule AB, what is its bond energy? In this case, because A isn't the same as B, one of the ionic resonance structures will be more important, and contribute to increased resonance stabilization. We can model the bond in AB using resonance between the covalent and ionic structures, A—B, [A+][B], and [A][B+]. We can guess that the pure covalent A—B bond strength is the average of the A—A and B—B covalent bonds, because this depends on factors like distance between the nuclei and repulsion between the electrons. The extra resonance energy from increased stability of one of the ionic structures should make the A—B bond energy bigger than the average of the A—A and B—B bonds, and this is observed. (Convince yourself using the data in the previous section.) The bigger the difference between the elements, the more stable the ionic structure becomes, and the greater the resonance energy is. The difference between the average of the A—A and B—B bonds and the experimental A—B bond energy is used as the basis of Pauling's electronegativity scale.
Predicting Relative Electronegativities
It pretty much follows the same pattern you would expect based on ionization energy and electron affinity. Thus, in general, electronegativity is big in the upper right of the periodic table and decreases down and to the left. It's good to know that H's electronegativity is between B and C. (Even though H is written on the far left of the periodic table, it is sort of in between an alkali metal and a halogen.)
Look at EN for yourself!
Go to Ptable's electron affinity page. See the general trend (bigger up and right) with some relatively high electronegativities among the heavy transition metals, like gold, as well.
Using Electronegativity
We can use electronegativity as a convenient way to predict the polarization of covalent bonds (in other words, how ionic they are). At the extreme, we can use it to predict whether compounds are covalent or ionic, which suggests also that it correlates roughly with metallic or non-metallic character. We can also use it to predict good Lewis structures, which usually have the negative formal charges on the more electronegative atoms. (If you follow this rule consistently, you'll predict much more ionic bonding than you might expect, which is actually consistent with more sophisticated models.)
The Pauling definition of electronegativity leads us to the conclusion that "combination of elements" reactions should always or nearly always be exothermic if only single bonds are present in the products and reactants. For N2 and O2 the multiple bond energy is greater than 2 times or 3 times the single bond energy. For this reason, compounds between N and O and elements of similar electronegativity like Cl might have positive heats of formation, meaning that they can exothermically decompose to the elements. Many reactive compounds that decompose to form elemental gases, such as explosives and bleaches, are compounds of Cl, N, and O. | textbooks/chem/General_Chemistry/General_Chemistry_Supplement_(Eames)/Valence_Bond_Theory/Dipole_Moments.txt |
Skills to Develop
• Determine and illustrate formal charges for Lewis structures
When you draw Lewis structures, sometimes the electrons are shared in a way which seems "unfair." For instance, in (CH3)3NO, to give N 8 electrons (and not more, since N can't have more than 8), you have to draw a single bond to oxygen. If you imagine a reaction between (CH3)3N and an oxygen atom, both electrons that form the bond to O come from N (the former lone pair). (This is a rare example of a reaction that is both a Lewis acid-base reaction and a redox reaction.) However, once the bond is made, these electrons are shared. If you imagine that they are shared equally, then there is a single positive charge on N and negative charge on O. These are called formal charges
How are Formal Charges Different from Oxidation Numbers?
You might remember oxidation numbers from the discussion of redox chemistry earlier. Oxidation numbers are found by assuming that the bonding electrons are entirely owned by the atom that pulls hardest. Formal charges, in contrast, are calculated by assuming that the bonding electrons are shared evenly between the two atoms. The truth is usually somewhere in between. Both formal charges and oxidation numbers are used for "bookkeeping" or counting purposes. They don't tell you much about the real position of the electrons in the bond.
How Should You Find Formal Charges?
To find formal charges in a Lewis structure, for each atom, you should count how many electrons it "owns". Count all of its lone pair electrons, and half of its bonding electrons. The difference between the atom's number of valence electrons and the number it owns is the formal charge. For example, in NH3, N has 1 lone pair (2 electrons) and 3 bonds (6 electrons total, so count 6/2 =3), so it owns 5 electrons, which is the same as the number of valence electrons. The formal charge is 0. For each H atom, it has 1 bond and thus 1 electron, so its formal charge is also 0. This is good, because all the formal charges of each atom must add up to the total charge on the molecule or ion. For the ammonium ion, NH4+, each H is still 0. Now N has 4 bonds and no lone pairs, so it owns 4 electrons. 5 - 4 = +1, so N has a +1 charge. This matches the +1 charge of the whole ion.
What Do You Need to Do with Formal Charges?
When you write Lewis structures, you should include formal charges next to each atom with a formal charge that isn't 0. (Usually, you circle the charge so it's clear.) This can also help you tell which Lewis structures are good. Usually negative formal charges should be on atoms that pull electrons strongly (like O or F, elements from the top right of the periodic table that have high ionization energies and high electron affinities). Positive formal charges should be on elements that pull electrons less. Big formal charges (more than 2) are usually bad. And it's better to have opposite formal charges right next to each other (so you get a formal "ionic bond"), and like formal charges farther from each other.
Hybrid Orbitals
Skills to Develop
• Describe how to construct hybrid orbitals
Hybrid orbitals are combinations of atomic orbitals that fit well with the 3D structures of molecules (that point in the same directions as the bonds or lone pairs). Because chemical bonding is based on Coulomb forces between electrons and nuclei, we expect that bonding electrons should spend a lot of their time between the nuclei. This is what makes a chemical bond. But what orbitals are they in when they do this? Because of the observed shapes of molecules, it seems like valence electron pairs each occupy their own area, called a domain. But the positions of these areas that we learned in the previous section don't match the positions of the atomic orbitals we studied before. We can fix this by making combinations of the atomic orbitals that have new shapes, called hybrids.
When are Hybrids Useful?
We mostly use hybrid orbitals to describe the bonding in organic chemistry, for compounds composed mainly of C and H. Hybrids aren't so good for molecules like SF6
Building Hybrid Orbitals
We can build hybrid orbitals using wave interference. First, let's think about a molecule like BeF2 (in the gas phase, where it is a molecule and not an ionic solid). This molecule is predicted to be linear, so how do we put the 4 bonding electrons into 2 orbitals pointed the right directions? We can imagine taking the Be 2s orbital and 2pz orbital, and combining them in 2 ways, adding and subtracting. This gives us 2 new orbitals. (Any time you combine orbitals, you get as many out as you put in.) This is first shown using "1-D" waves, where the x-axis is radius and the y-axis is Ψ. It's important to think about the phases of the orbitals during this combination, and think about constructive and destructive interference.
Once you are comfortable with the wave mixing pattern in the diagram above, look at the 2-D version.
You can see that by combining an s orbital and a p orbital on the same atom, we get 2 new orbitals, which point opposite directions. These orbitals are good for making 2 bonds in a linear shape, like in the molecule BeF2. Each of the big black lobes with make a bond with an orbital or hybrid orbital on F. In the same way, we can mix 1s orbital and 2p orbitals to get sp2 hybrids, or 1 s and 3 p orbitals to get sp3 hybrids, shown below. We use sp2 hybrids for trigonal planar molecules like BH3 and sp3 orbitals for tetrahedral shapes like in CH4.
Important Hybrids
Although other hybrids are sometimes mentioned, the most important ones are sp, sp2 and sp3. (Hybrid orbitals using d orbitals are usually not a good or useful description of what's going on.) Just remember the general shape of each of these, and what geometry molecule they match. | textbooks/chem/General_Chemistry/General_Chemistry_Supplement_(Eames)/Valence_Bond_Theory/Formal_Charges_in_Lewis_Structures.txt |
Skills to Develop
• Define multiple bonds (name, characteristics)
• Describe and draw resonance structures
• Predict hybridization and numbers of σ and π bonds
You already know that sometimes we have to use multiple bonds (double or triple bonds) to draw good Lewis structures. Lewis' idea of multiple bonds suggests that they should be shorter than single bonds (because the tetrahedra share an edge or a face, not just a point). This turns out to match the data from crystal structures. In the previous section, we used hybrid orbitals to show how we can make bonds using overlapping orbitals. This actually matches Lewis' image of the tetrahedra pretty well! So we could think about multiple bonds being made with hybrid orbitals just in the way that Lewis pictured, which are now called "bent bonds" because the electron density doesn't seem to be on the line between the nuclei. This isn't a bad way to think for some molecules. However, it becomes very confusing in cases of multiple bond resonance!
Multiple Bond Resonance
Let's think about Lewis structures for the nitrate ion, NO3. We have to draw 1 double bond and 2 single bonds between N and O to give N exactly 8 electrons. This suggests that there should be 1 short bond and 2 long bonds in nitrate ion. However, there is no difference between the O atoms, so how would we choose which one gets the double bond? We could put it in any position, giving 3 different structures with the same energy. If you remember the idea of resonance energy you might expect that the availability of 3 different structures with the same energy allows resonance between the structures, so that each bond is actually 4/3 of a bond, and the molecule is much more stable because of the resonance energy.
When we study the structure of nitrate, we can't find any difference between the oxygens. If we do X-ray crystallography on CsNO3, say, we can check what all the bond lengths and bond angles are, and they appear equal. Other methods give the same result. Molecules or ions in which multiple bond resonance is important (just like bonds in which ionic resonance is important) are more stable than we would otherwise predict, and they appear to have "fractional" bonds. Benzene, an another important example of resonance, is shown in the figure. Other examples include the sulfate and phosphate ions. Try drawing resonance structures for these ions, using the "resonance arrow" ↔.
Orbitals for Multiple Bond Resonance
It's hard to imagine how multiple bond resonance can happen using the "bent bond" idea. However, we can use a different model of multiple bonds that makes it much clearer. Think about a molecule of oxygen. Each atom makes 2 bonds and has 2 lone pairs. We could think about doing this with sp3 hybrids, but there's another way as well if we assume that the second bond is different from the first. Imagine using sp2 hybrids on each atom. One p orbital is leftover, perpendicular to the 3 hybrids. We make a normal straight bond with one hybrid from each atom, which we will call a σ bond. We put the lone pairs in the other hybrids. The last bond pair goes in a bond made of the leftover p orbitals. This is called a π bond. This is illustrated in the figure.
Now imagine orbitals for nitrate ion. N makes 4 bonds and has no lone pairs. We want to make 3 normal bonds, and one more bond shared over all 3 oxygen atoms. Each oxygen atom has a normal bond to N, and 2 or 3 lone pairs. Let's use sp2 hybrids on each atom. N makes 3 σ bonds using the sp2 hybrids and has 1 p orbital leftover. Each O makes a normal σ bond with one hybrid and puts lone pairs in the other 2. All the hybrid orbitals on each atom are in the same plane, and the leftover p orbitals on each atom are sticking up and down from this plane. The 4 leftover p orbitals hold 3 electron pairs, and all these orbitals can mix together. We can't easily tell from this model which electrons are lone pair electrons and which are bonding electrons, but we can see that the bond is shared equally over all 3 O atoms.
Ethylene, C2H4, is a planar molecule (all atoms are in the same plane). This makes sense using either the bent bond description or the σ and π bond description. In the latter case, if we twist the molecule around the double bond, the p orbitals won't line up with each other, so the double bond would break. Because double bonds can't rotate easily, the molecule can be different if there are different groups attached to each end of the double bond. These are called isomers.
Predicting Hybridization and Numbers of σ and π Bonds
Only 1 σ bond can be made between 2 atoms. If more bonds or partial bonds are present, they must be π bonds. So it's easy to count the number of σ and π bonds in a Lewis structure. Once you've done that, you can easily find what type of hybrids each atom makes (only for atoms with 8 total electrons; you shouldn't use hybrids for atoms that have more than 8). Each π bond needs a p orbital. The hybridization will be spx, where
$x=3-(number\; of\; \pi \; bonds)$
For example, an atom that makes 2 double bonds makes 2 π bonds, so it will be sp hybridized, just like an atom that makes 1 triple bond. An atom that makes 1 double bond is sp2 hybridized. | textbooks/chem/General_Chemistry/General_Chemistry_Supplement_(Eames)/Valence_Bond_Theory/Multiple_Bonds.txt |
Skills to Develop
• List some of Pauling's contributions to chemistry
• Describe some of the limitations of valence bond theory
Pauling's father died when he was young and his mother had some medical and psychological problems, so he started working to support the family when he was 13. His mother wanted him to keep working to support her instead of going to college, but he didn't listen. He worked over the summer to earn tuition, but before his third year his mother took the money he had saved for tuition. Luckily, the college gave him a job to cover his tuition. He read Lewis and Langmuir's papers on chemical bonding, and applied to graduate school at Berkeley, where Lewis was. Unfortunately, Berkeley lost his application letter, so he went to Caltech. (After that, Lewis changed the graduate admission process so there would be no more mistakes!)
In graduate school, Pauling studied X-ray crystallography, which gave him a good understanding of the structures of molecules, such as bond lengths and bond angles, that was very useful for his later work. However, he had some trouble because he didn't want to give the usual credit to his teachers when he published papers. Pauling wanted to work with Lewis after he finished his doctorate, but the head of Caltech's chemistry department, Noyes, convinced him to go to Europe instead. The reason was that Noyes wanted Pauling to be a professor at Caltech, and was afraid Lewis would give him a job at Berkeley if they were allowed to meet. In Europe, Pauling studied the new quantum mechanics.
Pauling's big contribution to chemistry was valence bond theory, which combined his knowledge of quantum mechanical theory with his knowledge of basic chemical facts, like bond lengths and and bond strengths and shapes of molecules. Valence bond theory, like Lewis's bonding theory, provides a simple model that is useful for predicting and understanding the structures of molecules, especially for organic chemistry. Later Pauling applied his understanding of molecular shapes and bonding to a pretty good explanation of protein structures. Later still, he became famous for peace activism and promoting ascorbic acid (vitamin C) as a cure for colds and cancer.
Valence bond theory is good for explaining the "ground state" properties of molecules, or the properties of molecules that only involve the lowest energy states. However, just like a hydrogen atom can be "excited" by light, moving the electron from n = 1 to n = 2, 3, 4... states, molecules have excited states that can be very important for reactions, colors, etc. Valence bond theory is not good for anything involving excited states, so we will talk about Molecular Orbital Theory as our next big topic.
Resonance
Skills to Develop
• Describe the 3 types of resonance
We previously discussed a different kind of resonance, which is the tendency of some systems, such as the strings of instruments, to vibrate easily at particular frequencies. Now we will talk about quantum mechanical resonance. First, it may be easier to understand if you think about Fourier analysis. Perhaps you have heard before that any function can be written as a sum of sin and cos waves, like this:
$f(x) = \Sigma_{n} a_{n}sin(nx) + \Sigma_{m} b_{m} cos(mx)$
When we try to find wavefunctions for real chemical systems, it is too complicated to find an exact solution like the solutions for single hydrogen atoms. Instead, we usually describe the real wavefunction ψ(x,y,z,t) using series of functions (call them φ(x,y,z,t)) like a Fourier transform.
$\Psi = \Sigma_{n} a_{n} \varphi_{n} (x, y, z, t)$
Often the functions φn used to build the real wavefunction ψ are the hydrogen wavefunctions we talked about before, including s, p, d, and f orbitals in each shell. There is a principle that says that if you choose the coefficients (an) so that the energy of the total wavefunction is minimized, those are the best coefficients that get closest to the real wavefunction ψ. In other words, real life finds the lowest possible energy (highest stability), so the lowest energy we can find is the closest to the real thing. If we imagine a system that might be described by φ1 or φ2,
$\Psi = a \varphi_{1} + b \varphi_{2}$
we can calculate the ratio a:b that minimizes the energy of ψ. If a:b is very big, φ1 is a good description of the system (at least compared to φ2). If a:b is very small, φ2 is a good description. If a:b is close to 1, then the real system is somewhere in between. The system is described as resonating between the two states. This doesn't mean that it alternates between them. It's like the difference between blue and yellow stripes (alternating between colors), and green (resonance, a mix of blue and yellow). The energy calculated for the combination will be lower than either single energy, and this difference is called the resonance energy.
Example
We can use different Lewis structures to represent the trial wavefunctions φ. For instance, imagine the formation of a bond between a H atom and a H+ ion (H nucleus). We can consider 2 trial wavefunctions corresponding to the following structures:
Structure 1: HA+ • HB
Structure 2: HA • HB+
If we calculate energy as a function of distance between the nuclei, for either structure 1 or structure 2 we don't predict a bond to form (instead, we expect the nucleus and atom to repel each other). If we allow resonance between structure 1 and structure 2, then we find that at a certain distance, 1.06 Å, the energy is a minimum. This means a bond can form. The resonance between the 2 structures means that the electron spends time near both nuclei. Since it has to move back and forth (very quickly, but maybe in a random motion, we don't know exactly), it must spend more of its time between the nuclei. When it is between them, we can expect that both nuclei are attracted to the electron, so it holds them together in a bond.
Ionic Resonance in H2
In the hydrogen molecule, we can use a similar model. We use the 2 structures HA(1) • HB(2) + HA(2) • HB(1) (where HA and HB are the 2 nuclei, and 1 and 2 are the electrons) to represent the normal covalent bond, H—H. We get closer to the experimental data when we include the ionic structures [HA+][:HB] + [:HA][HB+]. At the normal bond length, the Coulomb attraction between the ions makes these structures stable enough to contribute about 2% to the full description of the molecule.
Ionic Resonance in Other Bonds
Ionic resonance structures are much more important in cases where the bond is between different elements. For instance, in HF, we expect the structure [F][H+] to be very important, perhaps as important as H—F, because F pulls much harder on electrons. We will keep discussing this in the next sections. | textbooks/chem/General_Chemistry/General_Chemistry_Supplement_(Eames)/Valence_Bond_Theory/Overview_of_Pauling_and_Valence_Bond_Theory.txt |
Skills to Develop
• Describe the significance of bond dissociation energy
Bond dissociation energy or bond energy is the energy required to break a bond. Usually we mean bond dissociation enthalpy, the enthalpy change for this reaction:
$A—B \rightarrow A(g) + B(g)$
For example, H—Cl → H• (g) + Cl• (g)
The • represents an unpaired electron, called a radical. When measuring the bond enthalpy, we break the bond so that half the bonding electrons go to each atom. Breaking a bond is always an endothermic process, so the bond energy is always positive. We use the symbol D(A—B) to represent the bond enthalpy of an A—B bond.
Bond energy is similar, but is the average energy required to break all the bonds in a molecule. For instance, C-H bond energy is the enthalpy of atomization of methane divided by 4. This is different from the energy required to break the first C-H bond in CH4 because breaking the second bond might be easier or harder than breaking the first.
Some Approximate Average Bond Enthalpies (kcal/mol)
Bond D (kcal/mol) Bond D (kcal/mol)
H-H 103 H-F 135
H-Cl 102 H-Br 87
H-I 70 H-C 98
H-O 110 H-N 92
C-C 83 C-F 116
C-Cl 78 C-Br 68
C-O 86 C-N 73
N-N 40 O-O 34
C=C 144 C≡C 200
C=O 191 C=N 147
O=O 118 N≡N 225
Cl-O 52 Cl-N 72
N-O 48 N=O 145
How do we Measure Bond Dissociation Energy?
As you might guess from the discussion above about bond energy vs. bond dissociation energy, it's actually quite complicated to measure these! There are a lot of complications to consider, and you can't always measure directly the heat of reaction that you want.
Why does Bond Dissociation Energy Matter?
Having a general knowledge of bond strengths lets us understand the structures of molecules and predict reactions. For instance, when we guess the arrangement of atoms to make a Lewis structure, we will probably do better if we avoid making weak bonds. N-N and O-O single bonds are pretty weak, so usually we try not to put them in Lewis structures, unless nothing else makes sense. The structure will be better if we can make a multiple bond or put some other atom in between.
We can also use bond strength knowledge to predict what direction reactions will go and how fast they will go. Recall that we can calculate the enthalpy of a reaction by comparing the atomization energies of the reactants and products. (This is the same idea as when we calculated enthalpies of reaction using standard enthalpies of formation.) If the product has stronger bonds than the reactants, the reaction is likely to be exothermic, and exothermic reactions often go forward. If the reactant has strong bonds, you might need to use a lot of energy to get the reaction started, even if it is exothermic. Combustion reactions are an example: C-C and C-H bonds are pretty strong, so we need a spark or a match to start the burning. Another example is "nitrogen fixation": we need N to make proteins and other important molecules in our bodies, but it is very hard to get N from N2 because of the strong triple bond. This is why we have to use fertilizer in agriculture, and making nitrogen fertilizer uses roughly 1% of the world's energy. Another example is teflon or PTFE, which makes a very slippery and non-reactive coating that is used in non-stick pans, containers for reactive chemicals, etc. PTFE has only C-C and C-F bonds, both of which are very strong, so it doesn't easily react even at high temperature when you cook with it. In contrast, if a reactant has some very weak bonds, it might be very easy to start a reaction. For instance, hydrogen peroxide (HOOH) can be used as a bleach or disinfectant because the weak O-O single bond can easily break, helping it oxidize whatever is nearby. Hydrogen peroxide and hydrazine (N2H4) can both be used as rocket fuel because they react easily and exothermically (they have weak bonds, and they form products with stronger bonds). | textbooks/chem/General_Chemistry/General_Chemistry_Supplement_(Eames)/Valence_Bond_Theory/Strengths_of_Covalent_Bonds.txt |
Day 1: Chemistry, Matter, Energy, Models
Welcome to Chemistry 109!
If you have not yet worked through the Introduction, please do so before beginning this section.
Today’s work begins with a review of important fundamentals of chemistry. We expect that you are familiar with these ideas and are ready to use them to help you learn more chemistry. Then we introduce fundamental ideas about energy and matter. These ideas will recur throughout this course and we expect they will serve you well during your entire career
D1.1 Substances and Chemical Reactions
A fundamental aspect of chemistry is that substances change into other substances. A process in which one or more substances changes into one or more different substances is called a chemical reaction. This course aims to enhance your understanding of substances and chemical reactions. For a specific substance, what properties can you expect? Given one or two substances, is a reaction likely? When a reaction occurs, what new substances are formed? How is energy related to chemical reactions? Can we make new substances that have properties we want (such as alleviating disease)? Ability to answer such questions is valuable in a broad range of fields, from physics to pharmacy.
Activity 1: Chemical Reactions
Here are videos of some chemical reactions. Watch each reaction carefully and write down your observations.
Reaction of copper with air and of copper oxide with hydrogen:
https://mediaspace.wisc.edu/id/1_ly43nvus
Reaction of lead nitrate aqueous solution with potassium iodide solution:
https://mediaspace.wisc.edu/id/1_init9r1n
Reactions of Li, Na, and K with air and of Li, Na, K, Rb, and Cs with water:
https://mediaspace.wisc.edu/id/1_k9nbvl8a
For each chemical reaction, write a few sentences in your class notebook describing the reaction: How do you know that a reaction occurred? What are similarities and differences among the reactions you observed? How do temperature and other variables affect the reactions?
Query $1$
Saying that one substance changes into another is not very precise without a definition of what a substance is. In chemistry, a substance is matter that, when purified, has specific, characteristic properties and composition. For example, all samples of pure copper have an orange, lustrous surface; all conduct electricity equally well; all react with dry air to produce a black substance; and all consist solely of copper. The American Chemical Society currently lists more than 160 million chemical substances in a data base. New substances are being synthesized every day.
Because there are so many chemical substances, it is useful to develop categories and classifications to help guide our thinking. The most important is that chemical substances are composed of only a few chemical elements (118 of which have been discovered so far). A chemical element is a substance that cannot be changed by chemical reaction into two or more different substances. Elements combine to form chemical compounds, substances that can be decomposed by chemical reactions into two or more new substances. If you know something about the chemical elements and how they combine to form chemical compounds, you can predict properties of a wide range of substances and often predict which substances are likely to react to form what products.
Another useful classification is to divide the elements into metals and nonmetals. Metals conduct electricity as both solids and liquids, have lustrous surfaces when pure, can be pounded into different shapes and drawn into wires, and conduct heat well. Nonmetals have very small electrical conductivity and have a broad range of other properties. Six elements, the metalloids, have properties intermediate between metals and nonmetals.
Chemical reactions and the classifications of substances discussed so far are based on what are called macroscopic observations. Macroscopic refers to things large enough to be seen and manipulated in a laboratory (or anywhere else). Enhancing your understanding of the macroscopic world is a goal of this course.
D1.2 Atoms, Molecules, and Ions
A different fundamental aspect of chemistry is that we can better understand elements, compounds, and reactions if we build models based on tiny particles that are constantly in motion: atoms and molecules.
An atom is the tiniest particle of an element that has the chemical properties of that element. Atoms are extremely small, with sizes on the order of 100 pm to 350 pm (1 pm = 10−12 m). The smallest thing discernible with the most powerful optical microscope is more than 1000 times bigger than an atom. A molecule is two or more atoms connected by chemical bonds. Atomic scale or sub-microscopic scale refers to things about the same size as atoms and molecules. Chemists build atomic-scale models to interpret and predict macroscopic phenomena. For example, many substances consist of molecules; the properties of these molecular substances can be predicted if we know which atoms make up their molecules and how those atoms are arranged.
Atoms contain three kinds of particles: protons, neutrons, and electrons. Atoms are identified by the number of protons in the nucleus, the atomic number. Protons are positively charged. Neutrons have no electric charge. Electrons have negative charge with the same magnitude as a proton’s positive charge. Protons and neutrons have much greater mass than electrons and are found in the nucleus, a very small volume in the center of an atom that contains most of the atom’s mass. Electrons, which constitute less than 1/1000 the mass of an atom occupy 99.9999999999999% of the space the atom occupies. When a chemical reaction occurs, atoms remain unchanged, except that a few electrons in the outer part of one atom may transfer to the outer part of another atom. Before and after any chemical reaction the same number of atoms of each type is present; this is known as the law of conservation of matter. When copper metal reacted with oxygen from air to form copper oxide in the first video in Activity 1, the copper oxide included all the copper atoms that originally were in the copper that reacted. When hydrogen reacted with the copper oxide those copper atoms remained on the surface as copper metal.
Ions can form from atoms or molecules. An ion is an atom or molecule that has gained or lost one or more electrons and therefore has a negative or positive electric charge. For example, when copper reacts with oxygen, the product consists of copper(II) ions and oxide ions. In a copper(II) ion a Cu atom has lost two electrons to form an ion with two units of excess positive charge, Cu2+; in an oxide ion an O atom has gained two electrons to form an ion with two units of excess negative charge, O2−. When a molecule gains or loses electrons, polyatomic ions form: the molecule NO2 forms both a positive ion, NO2+ and a negative ion, NO2, by loss and gain of one electron.
D1.3 Chemical Symbols, Formulas, and Equations
It is convenient to deal with macroscopic properties and atomic-scale models by defining symbols to represent elements, compounds, atoms, molecules, and structures. For example, the symbol Li can represent the element lithium or it can represent a lithium atom. When Li is used to represent the element, it should bring to mind various properties: Li is a metal that can be cut with a knife; Li reacts fairly quickly with air; Li reacts vigorously with water. When Li is used to represent a lithium atom, we can use numbers to indicate how many Li atoms are present. For example, 2 Li represents two lithium atoms. In Li2O, the formula for one substance that forms when lithium reacts with oxygen, the subscript “2” indicates that there are two Li atoms for every one O atom. Li2O also represents the macroscopic substance lithium oxide, which has specific properties including high melting point (1438 °C) and high solubility in water.
Symbols can also represent chemical reactions. When lithium reacts with water the chemical equation is
2 Li(s) + 2 H2O(l) → H2(g) + 2 LiOH(aq)
The letters in parentheses (s, l, g, and aq) indicate that lithium, water, hydrogen, and lithium hydroxide are solid, liquid, gas, and an aqueous solution. The chemical symbols and formulas indicate that lithium and water are reactants and hydrogen and lithium hydroxide solution are products. The coefficients indicate how much of each reactant reacts away and how much of each product forms. The quantities can be expressed on the atomic scale as two lithium atoms reacting with two water molecules to give one hydrogen molecule and two lithium ions and two hydroxide ions in solution. (Lithium hydroxide consists of lithium ions, each with one unit of positive charge, and hydroxide ions, each with one unit of negative charge.) The quantities can be scaled up by a factor of 6.02214076 × 1023 (Avogadro’s number) and the equation says that two moles of solid lithium reacts with 2 moles of liquid water to give one mole of gaseous hydrogen and two moles of lithium hydroxide dissolved in water.
For molecules, symbolism can indicate which atoms are bonded to which other atoms, what types of bonds are present, and how the atoms are arranged in three-dimensional space. These structures illustrate some of the possibilities:
The chemical formula tells only the number of each kind of atom in the molecule. The Lewis structure indicates which atoms are bonded to which. The wedge-dash structure indicates that the molecule is three-dimensional, which is shown more clearly in the ball-and-stick model. (To see and manipulate a three-dimensional ball-and-stick model, click here.) Finally, in the space-filling model the sizes of all atoms and the molecule as a whole are shown; it is clear that a chlorine atom is bigger than a carbon atom, which is bigger than a hydrogen atom. The last two models are more pictorial than symbolic, but they are still representations of something we cannot see. None of these representations is the molecule itself; all provide useful information about its properties.
Chemists use representations such as these all the time and move effortlessly from one to the other as they think about molecules. Some imagination and much experience using these symbolic representations will enable you to make predictions about properties of substances and chemical reactions.
D1.4 The Periodic Table
As you saw in the videos in Activity 1, the elements Li, Na, K, Rb, and Cs have similar properties and react with air and water in similar ways. In addition, all these elements react with fluorine, chlorine, bromine, and iodine to form similar compounds: LiF, NaF, KF, RbF, CsF, LiCl, NaCl etc. (These compounds are referred to collectively as “salts” because their properties are similar to those of table salt, NaCl.) The reaction of Na(s) with Cl2(g) to form NaCl(s) is shown in this video.
https://mediaspace.wisc.edu/id/1_zygntgnp
The elements Li, Na, K, Rb, and Cs are called alkali metals. The elements F, Cl, Br, and I are called halogens. (“Halogen” comes from Greek hals, halo—”salt”.) All halogens consist of diatomic molecules, such as Cl2. Another group of elements, Be, Mg, Ca, Sr, and Ba also have similar physical properties, react with air and water (but more slowly than alkali metals), and react with halogens to form compounds with formulas like BeF2, BeCl2, MgCl2, etc. Be, Mg, Ca, Sr, and Ba are known as alkaline earth metals or alkaline earths. Just before the turn of the 20th century, chemists discovered another group of similar elements: He, Ne, Ar, Kr, and Xe. All are gases that undergo almost no chemical reactions, so He, Ne, Ar, Kr, and Xe are called noble gases. (One meaning of “noble” is “unreactive”.)
Here is a list of the first twenty elements in order of increasing atomic number:
H He Li Be B C N O F Ne Na Mg Al Si P S Cl Ar K Ca
The noble gases, alkali metals, alkaline earth metals, and halogens are color coded. Notice that the colors repeat periodically—every eight elements. About 150 years ago, Russian chemist Dmitri Mendeleev recognized this periodicity and created a table with horizontal rows of elements in atomic-weight order and groups of similar elements in vertical columns. Since Mendeleev’s time many more elements have been discovered: there are 118 in the modern periodic table below. But the idea of rows (periods) across which properties of elements vary and columns (groups) of elements with similar properties remains.
Each group (column) is identified by two numbers. The alkali-metal group is group 1 or IA; the alkaline-earth group is group 2 or IIA; the halogen group is group 17 or VIIA; and the noble-gas group is group 18 or VIIIA. The first number is designated by the International Union of Pure and Applied Chemistry; the second is more commonly used in the United States.
Down the left side of the table, the periods (rows) are numbered from 1 to 7. The first period contains only two elements; the second and third periods contain eight elements each; the fourth and fifth periods each contain 18 elements; and the sixth and seventh periods contain 32 elements. (Some elements in these latter two periods have been moved to the bottom of the table so it can fit on a printed page and be big enough to read. The curved arrows show where these elements should fit.)
The table is color coded to indicate whether an element is a metal, a metalloid, or a nonmetal. Metals conduct electricity, have lustrous surfaces when pure, combine with other metals to form alloys, and are malleable and ductile. The electrical conductivity of metals increases as temperature decreases. Non-metals are electrical insulators, usually are brittle as solids but may be liquids or gases at room temperature, and combine with other nonmetals by forming covalent bonds. Metalloids are intermediate between metals and nonmetals, Metalloids have lower electrical conductivity than metals but their conductivity increases as temperature increases.
Most elements are metals (the large blue-gray area); only 20 elements are nonmetals and six are metalloids. Notice that some groups, such as groups 14 (IVA) and 15 (VA) contain nonmetals, metalloids, and metals. This means that properties (electrical conductivity, for example) of some elements in these groups are significantly different from properties of other elements in the group; that is, in some groups elements are not as similar to each other as in the alkali-metal group. In all groups, however, the formulas of compounds are similar. For example, in group 14 (IVA), all elements form oxides with formulas XO2: CO2, SiO2, GeO2, SnO2, and PbO2. (Fl, flerovium, has been formed in quantities of only a few atoms in a particle accelerator, so the formula of its oxide has not been determined experimentally.)
Similarity of chemical formulas for elements in the same periodic group involves valence, the combining power of an atom. For example, sodium has a valence of 1 and so does chlorine; therefore, sodium and chlorine form the compound NaCl in which sodium atoms and chlorine atoms combine in a 1:1 ratio. Oxygen has a valence of 2; therefore the formula for sodium oxide is Na2O—it takes two valence 1 sodium atoms to satisfy oxygen’s valence of 2. From this you should be able to predict that the valence of C (or Si, Ge, Sn, or Pb) is 4 because the formula of carbon dioxide is CO2.
D1.5 Matter, Energy, Models
In this course you will be asked to examine data and draw conclusions, to explain phenomena by applying basic principles, and to build models from which you can predict physical and chemical properties. Two important and interconnected ideas are fundamental:
• The spatial arrangement of atomic-level particles (structure) can predict macroscopic properties and chemical reactivity;
• Energies of atomic-level particles can be used to explain atomic-level structures and macroscopic energy changes.
Atomic-scale particles adopt structures with minimum energy, unless energy is transferred to them from an external source. Therefore it is useful to be able to calculate quantitatively or predict qualitatively whether one situation has higher or lower energy than another. Chemistry often involves electrically-charged atomic-scale particles, such as protons, electrons, or ions. The potential energy of two point electric charges (charges that occupy a single geometric point) can be calculated using an equation derived from Coulomb’s law:
$E_{\text{p}} = k_e\dfrac{Q_1Q_2}{r} \nonumber$
In this equation ke is a proportionality constant equal to 8.99 × 109 J m C−2,Q1 and Q2 are electric charge values, and r is the distance between the charges. Thus, the magnitude of the potential energy of two charged particles is proportional to the size of each charge and is inversely proportional to the distance between the charges. The energy is positive if the charges of the two particles have the same sign (both positive or both negative). The energy is negative if the charges are opposite—opposite charges attract and lower potential energy is the result. The direct proportionality to electric charge and inverse proportionality to distance enable qualitative predictions: Larger opposite charges closer together result in lower energy and hence greater stability.
Exercise 1: Potential Energy of Charged Particles
Activity 2: Potential Energy and Distance between Ions
Query $3$
Activity 3: Evaluating and Modifying a Model
Query $4$
At the atomic level, particles are most stable when energy is minimum. This happens for a sodium ion and a chloride ion when the ions are 276 pm apart (lowest point in the blue curve in Activity 3). Curves like this can be used to describe attractive forces between atoms, molecules, or ions. Particles attract each other so their potential energy decreases as they get closer, but eventually there are repulsive forces that prevent them from being in the same place at the same time. The balance of these forces results in a curve with a minimum at some distance of separation. The depth of the minimum indicates how strongly the particles attract.
D 1.6 Structure, Energy, and States of Matter
The idea that atomic-scale particles attract one another and adopt a low-energy arrangement unless energy is supplied from an external source can be applied to changes from solid to liquid to gas.
The kinetic-molecular theory states that atomic-level particles are in constant, random motion. How fast the particles move depends on temperature. As temperature increases the average speed of the particles increases; hence their kinetic energies also increase. The average of the energies of the particles is proportional to the absolute temperature (in units of kelvins). Near 0 K the particles have very little kinetic energy and adopt a structure with minimum potential energy.
Activity 4: Atomic-level View of Solids, Liquids, and Gases
Temperature and motion of particles affect whether a substance is a solid, liquid, or gas. View this simulation of the behavior of noble gas atoms in solids, liquids, and gases. Choose “States”, then choose “Neon” (upper right corner). Click on each of the boxes: “Solid”, “Liquid”, and “Gas” (or use “Heat” (below the container of molecules) to raise the temperature).
Based on the simulation, write in your notebook a description of the differences in position and movement of the molecules in solids, liquids, and gases.
Query $5$
Let’s apply these ideas to sodium chloride, which is a solid at low temperature. Sodium ions and chloride ions are packed closely together in a regular pattern, as seen in the diagram at the right. This arrangement minimizes the potential energy by bringing oppositely charged particles closer together. A little above 0 K the ions are in motion: each vibrates a bit around its specific location, but no ion has sufficient kinetic energy to overcome the Coulomb’s-law attractions that hold it in place. Thus, no ion exchanges place with any other ion. This atomic-scale structure is consistent with macroscopic properties: a solid is rigid because the regular pattern of its structure does not change.
As temperature increases the average kinetic energy of the ions increases: the ions move more, vibrating farther from their average positions. On the macroscopic scale, the solid expands because each vibrating ion pushes against neighboring ions. Pushing neighboring ions away enlarges the space occupied by each ion and the crystal’s volume increases. Eventually, the vibrations are large enough that ions can move relative to each other and the regular arrangement becomes much more random: the solid melts. Ions are still close together but they can move past one another, so the liquid has no specific shape. The liquid is fluid and can be poured. For sodium chloride, because the potential energy is lowered a lot by Coulomb’s-Law attractions among the ions, the temperature required for melting is high: 1074 K (801 °C).
Further increase in temperature makes the motion of the atoms faster and at 1738 K (1465 °C) the liquid sodium chloride boils, forming a gas. The sodium ions and chloride ions move much further apart, which means they bump into each other much less often. The gas is fluid but expands to fill whatever volume is available and has much lower density than the liquid or solid.
D1.7 What’s Ahead?
Throughout this course you will build on the fundamental ideas of chemistry developed in this section. You will begin by studying the properties of atoms and how those properties vary depending on position in the periodic table. This will lead to how atoms lose or gain electrons to form ions and how atoms share electrons to form molecules, both of which are related to valence. Based on the properties of atomic-scale atoms and ions, you will be able to understand properties of metals and ionic compounds. The number and variety of molecules is very large so a lot of your study will involve how atoms bond to each other to form molecules and how molecular structures affect properties and reactivity of molecular substances. Then we will consider how fast chemical reactions go and to what extent products of those reactions can be formed from a given set of reactants. These ideas will then be applied to two important classes of reactions: acid-base reactions and oxidation-reduction reactions.
Bon voyage!
Activity 5: Wrap-up
In your notebook, write a summary of the important ideas in this day’s work. Write your summary so that you can refer back to it, should you need to refresh your memory, and so that you can use it to review for exams.
Day 1 Pre-Class Podia Problem: Describe Chemistry
In most pre-class assignments we will include a Podia question that may require numerical, text, and/or image-based responses. Each pre-class Podia question is based on the pre-class material and working through the pre-class material will help you formulate your response. Consider the problem and write down/draw out your solution in your class notebook.
Two days before the next whole-class session, the pre-class Podia question will become visible in Podia (within the “Whole Class” section), where you can submit your solution. [For example, the Podia question in a Wednesday pre-class assignment will open by Monday evening.]
When you come to the next whole-class session, the class leader will choose one or more student-submitted solutions and explain why they are (or are not) effective scientific statements. These questions are designed to hone your skills so that you can solve mastery problems you will encounter on exams.
Click on the relevant open Podia question and submit your answer there. Your submission to instructor-opened questions will be anonymous to your classmates, and you can only view your own submission. (If you make a separate post, you will not be anonymous.)
You may find Podia sign-in instructions here.
Your first pre-class Podia question is designed to help you learn how Podia works. Write a short sentence about chemistry. Accompany your sentence with an image (sketched by you) that you think is a good example of chemistry. See whether your class leader selects your submission as a good one to show (anonymously) to the class.
Comments. If you found any inconsistencies, errors, or other things you would like to report about this module, please use this link to report them. A similar link will be included in each day’s material. We appreciate your comments. | textbooks/chem/General_Chemistry/Interactive_Chemistry_(Moore_Zhou_and_Garand)/01%3A_Unit_One/1.01%3A_Day_1-_Chemistry_Matter_Energy_Models.txt |
Day 2: Atomic Spectra and Atomic Orbitals
If you have not yet worked through the Introduction and Day 1, please do so before beginning this section.
D2.1 Electromagnetic Radiation
The periodic table (Section D1.4) summarizes much information about chemical elements. That information can be better understood by assuming that both physical and chemical properties of the different elements depend on differences in the underlying structures of their atoms. An important way to learn about the structures of atoms is to study how energy, in the form of electromagnetic radiation, interacts with matter.
Activity 1: Preparation—Atomic Spectra and Atomic Structure
In your course notebook, make a heading for Atomic Spectra and Atomic Structure. After the heading write down what you remember about atomic spectra from courses you have already taken. Also write what you recall about the relation of spectra to atomic structure—how electrons are arranged in atoms. If there is anything you remember being puzzled about, write that down as well. We will ask you to refer back to what you have written when you complete this section.
Electromagnetic radiation consists of oscillating, perpendicular electric and magnetic fields that travel through space and can transfer energy. The oscillating fields (waves) are characterized by wavelength (λ, measured in meters, m) and frequency (ν, measured in hertz, Hz or s−1). In a vacuum, electromagnetic radiation travels at the speed of light (c):
${\lambda }\nu = c = 2.998\;\times\;10^{8}\;\dfrac{m}{s} \nonumber$
Electromagnetic radiation occurs in small, indivisible quantities of energy called photons. The energy of a photon, Ephoton, can be determined from either its frequency or its wavelength:
$E_{\rm{photon}} = h\nu = \dfrac{hc}{\lambda } \nonumber$
In this equation h represents Planck’s constant; h = 6.626 × 10−34 J s.
Exercise 1: Photon Energy from Wavelength
Your calculation in Exercise 1 showed that the energy of a single photon is quite small. Most interactions of electromagnetic radiation and matter involve lots of photons and lots of atoms. The total energy transferred is proportional to the number of photons, N. If all photons have the same frequency,
$E_{\rm{electromagnetic\ radiation}} = N\times E_{\rm{photon}} = Nh\nu = N\dfrac{hc}{\lambda } \nonumber$
Notice that electromagnetic radiation has been described as involving wave motion and also as a number of particles (photons). Originally, scientists thought that electromagnetic radiation could be described entirely by a wave model, but that model was unable to predict all experimental observations. Consequently, both wave and particle models need to be combined for full understanding of electromagnetic radiation.
Figure 1 shows the enormous range of all types of electromagnetic radiation: Frequencies of 105 Hz to 1020 Hz, that is, wavelengths of 103 m (km) to 10−12 m (pm) have been observed. What we can see, visible light, is only a tiny portion (380-740 nm) of that range.
Different parts of the electromagnetic spectrum typically use different units: Low-energy photons, such as microwaves and radio waves, are specified in frequencies (MHz or GHz); mid-energy photons, such as infrared and visible light, are specified in wavelengths (μm, nm, pm, or Å); high-energy photons, such as x-rays and gamma-rays, are specified in energies (keV or MeV).
Our eyes detect visible-range photons, allowing us to see the world around us. But scientific instruments allow us to “see” lots more by detecting photons over a much wider range of energies. For example, studies of atomic spectra, experiments involving interaction of gaseous matter with visible, ultraviolet, and infrared light, led to better understanding of the structure of atoms.
D2.2 Atomic Spectra
Heating a gaseous element at low pressure or passing an electric current through the gas imparts additional energy into the atoms. These higher energy atoms can then release the additional energy by emitting photons. For instance, the colors of “neon” signs are produced by passing electric current through low-pressure gases. Interestingly, the photons emitted by the higher-energy atoms have only a few specific energies, thereby producing a line spectrum consisting of very sharp peaks (lines) at a few specific frequencies. Line spectra were intriguing because there was no reason to expect that some frequencies would be preferred over others.
Each element displays its own characteristic set of lines. For example, when electricity passes through a tube containing H2 gas at low pressure, the H2 molecules are broken apart into separate H atoms and the H atoms emit a purple color. Passing the purple light through a prism produces the uppermost line spectrum shown in Figure 2: the purple color consists of four discrete visible wavelengths: 656.4 nm, 486.2 nm, 434.1 nm, and 410.2 nm.
The H-atom emission spectrum also contains lines in the ultraviolet and infrared ranges. In 1888, Johannes Rydberg developed an equation that predicts wavelengths for all hydrogen emission lines:
$\dfrac{1}{\lambda } = R_{\infty} \left(\dfrac{1}{n_2^2} - \dfrac{1}{n_1^2}\right) \nonumber$
Here, n1 and n2 are positive integers with n2 < n1, and the Rydberg constant R = 1.09737316 × 107 m−1.
Because the wavelengths of hydrogen emission lines were measured to very high accuracy, the Rydberg constant could be determined very precisely. That a simple formula involving integers could account for such precise measurements seemed astounding at the time.
Exercise 2: Hydrogen-Atom Emission
Use the Rydberg equation to calculate the wavelength of the photon emitted by a hydrogen atom when n2 = 3 and n1 = 7.
D2.3 Atomic Energy Levels
Why should a hydrogen atom emit only four specific colors of visible light? To understand this better, we need to know more about the energy of the atom and how energy depends on atomic structure. Any atom consists of a tiny nucleus surrounded by one or more electrons. The simplest atom, a hydrogen atom, has one proton as the nucleus and one electron outside the nucleus. According to Coulomb’s law, the electron and proton attract.
Exercise 3: Electrostatic Potential Energy
Activity 2: Line Spectra and Energies
Think about the implications of line spectra. If a hydrogen atom emits only four specific wavelengths in the visible region, what does this imply regarding the energies of the emitted photons? Why should only these four wavelengths be emitted, but none of the other possible wavelengths? Write your explanation in your notebook.
Query $5$
The model currently used to describe the distribution of electrons in an atom has these attributes:
• The energies of electrons in an atom are restricted to energy levels, which are specific allowed energies.
• Each line in the spectrum of an element results when an electron’s energy changes from one energy level to another; a change from one electronic energy level to another is called an electronic transition.
• Electrons are distributed in regions centered on the nucleus, called shells; each shell has a different average distance from the nucleus.
• As described by Coulomb’s law, an electron’s energy increases with increasing average distance from the nucleus; that is, with increasing size of an electron shell.
• Both energy levels and shells are described by quantum numbers, numbers restricted to specific allowed values; the electron energies are said to be quantized, restricted to discrete energy levels.
An atom is most stable when it has the lowest possible energy. The lowest energy electronic state of an atom is called its electronic ground state (or simply ground state). Any higher energy state of an atom is called an electronic excited state (or simply an excited state).
Figure 3 shows the first few energy levels of a hydrogen atom. The atom is in its ground state when its electron is in the n = 1 (lowest energy) level. When a photon is absorbed by a ground state hydrogen atom, as shown on the left side of Figure 3, the energy of the photon moves the electron to a higher n (higher energy) level, and the atom is now in an excited state.
An atom in an excited state can release the extra energy as a single photon if the electron returns to its ground state (say, from n = 5 to n = 1), or the energy can be released as two or more lower energy photons if the electron falls to an intermediate state then to the ground state (say, from n = 5 to n = 2, emitting one photon, then from n = 2 to n = 1, emitting a second photon).
D2.4 The Quantum Mechanical Model of the Hydrogen Atom
If we could calculate the energy for each energy level, we could predict the emission spectrum for hydrogen. In 1926, Erwin Schrödinger applied quantum mechanics ,a model that uses both wave and particle analogies to describe atomic-scale matter, to the hydrogen atom. Instead of viewing the electron as a particle, Schrödinger applied mathematics appropriate for three-dimensional stationary waves constrained by electrostatic potential (Coulomb’s-law attraction between electron and nucleus). For each wave he derived a mathematical function to describe the wave, a wave function. The wave function is typically designated by the Greek letter ψ.
Schrödinger showed that these wave functions could be used to calculate allowed energies of a hydrogen atom. The calculated energies are given by this equation:
$E_n = -\dfrac{k}{n^2}, n = 1, 2, 3, . . . \nonumber$
where the proportionality constant k = 2.179 × 10−18 J, and n is a quantum number restricted to positive integer values.
In an electronic transition, an electron moves from one energy level to a different energy level. The energy of the corresponding photon is the energy difference between the two energy levels, an initial level with energy Ei and a final level with energy Ef.
$\Delta E = E_f - E_i = -k \left(\dfrac{1}{n_f^2} - \dfrac{1}{n_i^2}\right) \nonumber$
A positive ΔE means that the atom’s energy increased, corresponding to absorption of a photon: the photon’s energy has been added to the atom’s initial energy. Similarly, a negative ΔE means that the atom has lost energy through emission of a photon.
Conservation of energy requires that the energy of the photon, Ephoton = hc/λ, equals the absolute value of the energy difference, |ΔE|, for emission or absorption. The sign of ΔE indicates whether the photon was absorbed (+) or emitted (−).
Exercise 4: Hydrogen-Atom Electronic Transition Energy
The equation Schrödinger obtained is equivalent to the equation Rydberg used to calculate hydrogen emission lines:
$E_{\rm{photon}} = \dfrac{hc}{\lambda } = |\Delta E| = \left|-k \left(\dfrac{1}{n_f^2} - \dfrac{1}{n_i^2}\right) \right| \nonumber$
$\dfrac{1}{\lambda} = \dfrac{k}{hc} \left(\dfrac{1}{n_f^2} - \dfrac{1}{n_i^2}\right) = R_{\infty} \left(\dfrac{1}{n_2^2} - \dfrac{1}{n_1^2}\right) \nonumber$
and thus, R = k/(hc). Substituting values for k, h, and c into this equation gives R = 1.097× 107 m−1, which is the same as the experimental Rydberg constant to four significant figures. That a wave model could reproduce these highly accurate energy levels was strong evidence that quantum mechanics is an appropriate atomic-level model.
D2.5 Wave-Particle Duality
The combined wave and particle model is referred to as wave-particle duality. Its impact in describing atomic-scale particles (protons and electrons) is dramatic: wave-particle duality implies that we should not just think of an electron as located at a specific position around a nucleus or moving in a specific direction and with a specific speed: the wave-like electron seems to be all around the nucleus at once! In other words, the wave nature of an electron is just as important in describing its properties as its particle nature.
Query $7$
An important consequence of wave-particle duality is the Heisenberg uncertainty principle: it is impossible to determine the exact position and the exact momentum of an atomic-level particle simultaneously. This means that, if we know exactly where an electron is at one instant, we have no idea where it will be an instant later; or, if we know an electron’s exact speed and direction, we have no information about where it is. As a result of the uncertainty principle, the best we can do is determine the probability of finding an electron at a specific position; that probability is proportional to the square of the wave function, ψ2.
Query $8$
The electron-density distribution (or just electron density) is the three-dimensional distribution of electron probability, which can be derived from the square of a wave function. Wave functions and their electron-density distributions are called orbitals. It is these orbitals that help us understand atomic properties, chemical bonds, and forces between molecules.
A graphic way of showing electron-density distribution (depicting an orbital) is by the density of shading or stippling. That is, we draw lots of dots or darker shading where the probability is high and we draw fewer dots where the probability is low. For example, consider the wave shown below in red along with its square shown in blue. Wherever there is a maximum in the square of the wave function (blue curve), there are many dots. Where the wave function (and therefore also its square) approaches zero, there are few dots.
Such a way to visualize electron density is quite useful for a 3-D view of the electron density surrounding a nucleus. For example, you can see the electron-density distribution of a hydrogen atom in its ground state in this video, where the hydrogen atom is rotated around its nucleus.
Activity 3: Electron Density Distribution for H Atom
In your class notebook write a description of (1) the shape of the electron density distribution for a ground-state H atom and (2) how electron density changes with distance from the nucleus. (The nucleus is located at the intersection of the three axes: x, y, and z axes).
Query $9$
In the electron-density diagram below, click on the white cross that corresponds to the lowest electron density.
Query $10$
Based on the density of dots in the diagram, make a graph with electron density on the vertical axis and distance from the nucleus on the horizontal axis. How is a graph of wave function versus distance from the nucleus related to the graph you made?
Query $11$
Similar to dot-density diagrams, but visually simpler while conveying a little less information, are 3-D boundary-surface plots, which show a surface that has the same shape as the electron density distribution and encloses some fraction, such as 90%, of the electron density. That is, if we could repeatedly locate the electron exactly, nine times out of ten the electron would be located inside the boundary surface. The figure below shows how a boundary surface is related to the dot-density plot you have already viewed. Move the slider at the bottom to change from dot-density to boundary-surface diagram.
Query $12$
Visualizing electron-density distributions confirms an idea mentioned earlier: there are shells of electron density, concentric spheres each farther from the nucleus. The orbitals that belong to a given shell have the same quantum number, n and their electron density is approximately the same average distance from the nucleus. The principal quantum number n dictates the overall size of the orbital.
As n gets larger, the radius, r, of the electron shell gets larger, and En becomes less negative—closer to zero. Therefore, the limits n ⟶ ∞ and r ⟶ ∞ imply that En = 0 corresponds to ionization of the atom, complete separation of the electron from the nucleus. Thus, for a hydrogen atom in the ground state, the ionization energy is:
$\Delta E = E_{n \rightarrow \infty} - E_1 = 0 - \dfrac{-k}{1^2} = k = 2.179\;\times\;10^{-18}\;\text{J} \nonumber$
Another interesting aspect is that while we know En, which is the total energy of the electron, we cannot measure the electron’s kinetic energy (or its velocity) and potential energy separately. We know from Coulomb’s law that the potential energy changes as a function of r. Since the electron density is distributed over a range of r, there’s a range of potential energy, and hence a range of kinetic energy—when the electron’s potential energy is higher, its kinetic energy would be lower. This uncertainty is again related to the wave nature of the electron. However, knowing the total energy, En, is quite sufficient for us!
Activity 4: Wrap-up—Atomic Spectra and Atomic Structure
Review what you wrote about atomic spectra and atomic structure at the beginning of this section. Update the information based on what you have learned. Write a summary that will be a good study aid when you review for an exam.
D2.6 Atomic Orbitals and Quantum Numbers
Quantum Numbers: n, ℓ, mℓ
The atomic wave functions can be defined using three quantum numbers: n, ℓ, and m. Each wave function corresponds to an atomic orbital. Each atomic orbital defines a region in the atom within which electron probability density is large.
The energy of a hydrogen atom orbital depends on the principal quantum number, n:
$E_n = - \dfrac{2.179\;\times\;10^{-18}\;\text{J}}{n^2}, n = 1, 2, 3, . . . \nonumber$
As discussed above, the size of the orbital (the average distance of the electron from the nucleus) increases as n increases. The farther the electron is from the nucleus, the higher (less negative) the energy is.
The second quantum number, , is the orbital quantum number (sometimes called azimuth or angular-momentum quantum number). It determines the shape of the electron-density distribution. ℓ is an integer with values ranging from 0 to n – 1; that is, ℓ = 0, 1, 2, …, n – 1. Thus, 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.
Values of ℓ are designated using letters:
ℓ = 0 s orbitals
ℓ = 1 p orbitals
ℓ = 2 d orbitals
ℓ = 3 f orbitals
ℓ = 4 g orbitals
ℓ = 5 h orbitals
Activity 5: Shapes of Hydrogen-Atom Orbitals
View the shapes of s, p, and d orbitals shown in these videos:
1s orbital 2p orbital 3d orbital
In your class notebook write a description of the shape of the electron density distribution for each orbital. Also, make a rough drawing of the corresponding boundary-surface diagram for each dot-density diagram.
Query $13$
The magnetic quantum number, m, specifies the spatial orientation of a particular orbital. Values of m can be any integer from – to . For example, for an s orbital, = 0, and the only value of m is 0; thus there is only a single s orbital. For p orbitals, = 1, and m can be equal to –1, 0, or +1; thus there are three different p orbitals, which are commonly shown as px, py, and pz, oriented along the x, y, and z axes.
The total number of possible orbitals with the same value of is 2 + 1. There are five d orbitals, seven f orbitals, and so forth. The sizes and shapes of s, p, and d orbitals in the first three shells are shown in Figure 6. Each of these orbitals corresponds to a different set of n, , and m values.
Exercise 5: Electron-Density Distance from Nucleus
Query $14$
All orbitals with the same value of n and ℓ form a subshell. For example, the 2px, 2py, and 2pz orbitals constitute the 2p subshell because each of these orbitals has n = 2 and ℓ = 1. The number of orbitals in a subshell equals the number of different values for the m quantum number. Each shell contains n subshells: for example, when n = 3, there is a 3s, a 3p, and a 3d subshell, corresponding to the three possible values of 0, 1 and 2.
Orbital Phases and Nodes
Figure 6 shows some atomic orbitals as either all blue or all red (for example, 1s and 2s), while other orbitals contain both colors (for example, the 2p orbitals). The blue and red colors show the mathematical sign of the wave function, a property called the phase of the wave function. Most wave functions are positive in some regions and negative in others. Phase is important when wave functions on two atoms interact to form a chemical bond.
The relationship between wave function and electron density is depicted more explicitly in Figure 7. In the middle row, boundary-surface diagrams from Figure 6 show the sizes and shapes of electron-density distributions for the 1s, 2s, and one of the 2p orbitals. In the top row are the corresponding wave functions (ψ); the bottom row shows electron density, ψ2.
The wave functions and electron-density functions are color coded to show the regions where ψ > 0 as blue and the areas where ψ < 0 as red. A surface where the wave function changes sign is called a node. At a node, ψ = 0 and therefore the probability of finding the electron in that surface is also 0.
For the 1s orbital, there are no nodes; its wave function only has positive values. For the 2s orbital, there is one node at r = 1.05 Å; the wave function goes from positive values near the nucleus to negative values farther out. (You cannot see this node in the boundary-surface diagram, because the node is a sphere that is within the volume enclosed by the boundary surface.) A node that occurs at a specific value of r is a radial node, and therefore radial nodes have a spherical shape. The 3s orbital, for example, has two radial nodes enclosed within the boundary-surface diagram.
Each 2p orbital has one node. Look at the boundary-surface diagram to verify that this node is planar instead of spherical. Non-spherical nodes are angular nodes. For example, the 2px orbital has one angular node: the yz plane; ψ is equal to 0 at all points in the yz plane, rather than at any specific r value. You can see the presence of angular nodes clearly in boundary-surface diagrams.
An orbital has n – 1 nodes, of these, n – 1 – ℓ are radial nodes, and the number of angular nodes corresponds to the value of ℓ. A generalization that is true for all kinds of waves is that the greater the total number of nodes, the higher the energy. The phase of the wave function differs on either side of a node, because the wave function changes from one mathematical sign to the other as it crosses a node. Visually, we represent these different phases, and the presence of nodes, with different colors in the boundary-surface diagram.
Activity 6: Radial and Angular Nodes
Fourth Quantum Number, ms
The fourth quantum number, called the spin quantum number, ms, differs from the other three quantum numbers in that it describes the electron rather than the orbital. An electron has a very small magnetic field. Moving a macroscopic charge in a circular path produces a magnetic field, so initially the electron’s magnetic field was attributed to its spinning like a top; hence the name “spin quantum number”. The electron’s magnetic field can have two quantized states: either “up” or “down”. This gives two possible ms values, ms = +½ or ms = -½. When an electron occupies an orbital, we typically depict these two possible ms values as an ↑ or ↓ arrow.
Exercise 6: Quantum Numbers
Query $16$
Activity 7: Summary of Quantum Numbers
In your class notebook make a table summarizing the names, symbols, allowed values, and important properties of the four quantum numbers. (You can use this table for studying later.) If you are studying with someone else, make your tables independently and then compare.
When you are satisfied with your table, compare it with ours.
Day 2 Pre-Class Podia Problem: Atomic Orbitals
This Podia problem is based on today’s pre-class material; working through that material will help you solve the problem.
Consider these two sentences:
• The surface shown in this illustration of an atomic orbital is chosen so that the entirety of the electron’s wave function is enclosed.
• The colors in the diagram represent regions of opposite electron spin.
Decide whether each sentence is correct. Rewrite all incorrect text so that it is correct. Be brief but include all important ideas and use scientifically appropriate language.
Query $17$
Two days before the next whole-class session, this Podia question will become live on Podia, where you can submit your answer.
Comments. If you found any inconsistencies, errors, or other things you would like to report about this module, please use this link to report them. A similar link will be included in each day’s material. We appreciate your comments. | textbooks/chem/General_Chemistry/Interactive_Chemistry_(Moore_Zhou_and_Garand)/01%3A_Unit_One/1.02%3A_Day_2-_Atomic_Spectra_and_Atomic_Orbitals.txt |
D3.1 Atoms with More than a Single Electron
Activity 1: Reflection
In your course notebook, make a heading for electron configurations and write down what you remember about electron configurations for atoms from your previous experience. Also note any aspects of electron configurations that puzzle you. We will ask you to refer back to what you have written when you complete this section.
The ideas already developed about quantum numbers, orbitals, and sizes and shapes of electron-density distributions apply to all atoms. However, when there are two or more electrons in an atom there are repulsive forces between the electrons as well as attractive forces between electrons and the nucleus. These repulsions affect electron energies.
For example, the energy levels in a He+ ion (which, like H, has a single electron) are significantly lower than in a H atom because of the stronger Coulomb’s law attraction between the one electron and the 2+ charge of the He nucleus. However, in a He atom, which has two electrons, electron-electron repulsions between the electrons raise energy levels significantly compared to He+, and a He atom is not as stable as we might have expected.
For atoms with many electrons, the effect of electron-electron repulsions differs for different subshells. Therefore orbital energy depends on both n and quantum numbers. For the same value of n (the same shell), as increases the energy also increases. Thus s-subshell electrons have lower energy than p-subshell electrons, which are lower than d-subshell electrons, and so forth. Orbitals within the same subshell (for example 2px, 2py, and 2pz) all have the same energy; orbitals that have the same energy are said to be degenerate.
The Austrian physicist Wolfgang Pauli formulated what is now called the Pauli exclusion principle:
• Each electron in an atom must have a different set of values for the four quantum numbers.
• If two electrons share the same orbital (have the same n, , and m), then their spin quantum numbers ms must have different values; we say the two electrons have opposite spin.
• Because ms can only have two values, +½ or -½, no more than two electrons can occupy the same orbital.
By applying the Pauli exclusion principle, the arrangement of electrons in any multi-electron atom can be determined by recognizing that the ground state of an atom has all its electrons in orbitals with the lowest energies possible.
Activity 2: Arrangement of Electrons in Li
D3.2 Orbital Energy Level Diagrams
An orbital energy level diagram (or just orbital diagram)shows the relative energies of orbitals and how electrons are distributed among orbitals within a subshell. In an orbital energy level diagram, individual orbitals are usually represented by horizontal lines whose vertical position conveys the relative energies of the orbitals. The electrons are represented as arrows with the direction of the arrow communicating the sign of ms (the ↑ arrow represents ms = +½ and the ↓ arrow represents ms = -½).
For example, a boron atom has this orbital energy level diagram:
As illustrated by the boron example, an orbital diagram includes all orbitals in all subshells within a partially occupied shell, even if some orbitals are unoccupied. Note that the three 2p orbitals are degenerate, so the electron can occupy any one of them.
A carbon atom has six electrons, so there are two electrons in the 2p subshell. These two electrons could (1) pair in a single 2p orbital or (2) occupy separate orbitals but with opposite spin or (3) occupy separate orbitals but with parallel spin. All three possibilities are valid based on quantum numbers and the Pauli principle, but only one is lowest in energy. Electrons having parallel spins cannot occupy the same space (the same orbital), so repulsions between them must be smaller than if they had opposite spins. Thus option 3 is lowest in energy and therefore represents the ground state of a carbon atom; options 1 and 2 represent excited states.
According to Hund’s rule, the lowest energy configuration has the maximum number of unpaired electrons with parallel spin within a set of degenerate orbitals. Thus, the orbital diagram for the ground state of carbon is
D3.3 Electron Configurations
The specific arrangement of electrons in atomic orbitals is called the electron configuration of the atom. It determines many physical and chemical properties of that atom. The periodic table, which is arranged in accordance with the properties of the elements, can therefore be used to predict the ground state electron configurations of atoms.
An electron configuration is written symbolically to provide three pieces of information: the principal quantum number (shell number), n; a letter that designates the subshell (s, p, d, etc.); a superscript showing the number of electrons in that particular subshell. For example, the notation 2p4 indicates 4 electrons in a p subshell ( = 1) with a principal quantum number (n) of 2.
For any element, the ground state electron configuration can be built up by starting with hydrogen and following the atomic-number order through the periodic table. To go from one element to the next, add one proton (and one or more neutrons) to the nucleus and one electron to the lowest energy subshell that has an incompletely filled orbital. Repeat until you reach the desired element. This process of filling electrons into orbitals is called the aufbau principle, from the German word Aufbauen (“to build up”). Watch the video in Figure 1 to see how to use the aufbau principle to determine the electron configurations of oxygen and chromium.
https://mediaspace.wisc.edu/id/0_sr9om3nc?playerId=25717641
Figure 1. View this video to see how to use the periodic table to predict electron configurations.
Writing the complete electron configuration all the time can be cumbersome, so chemists often abbreviate by using the noble-gas notation. For example, the ground-state electron configuration of vanadium (V) is 1s22s22p63s23p64s23d3. The noble gas that immediately precedes V is argon (Ar); it has a ground-state electron configuration of 1s22s22p63s23p6, which can be represented as [Ar]. Thus the ground-state electron configuration of V can be shortened to [Ar]4s23d3, and it communicates the same information as the complete electron configuration. A list of ground state electron configurations for all elements in the appendix uses noble-gas notation.
Exercise 2: Electron Configuration
The aufbau principle is based on the concept that for ground-state electron configurations, an electron occupies a lower energy atomic orbital rather than occupying a higher energy orbital. Hence, the fact that we observe the 4s orbital fill before the 3d orbitals indicates that the 4s orbital is lower in energy. Similarly, a 6s orbital is lower in energy compared to a 4f orbital.
The energy difference between s, p, d, and f subshells causes orbitals with different n values to have similar energies. In many cases these energies are so similar that there are exceptions to the periodic-table prediction of electron configuration. These exceptions occur for d-block and f-block elements, but not for s-block and p-block elements.
D3.4 Valence Electrons
Valence, the combining power of an atom, was defined near the end of Section D1.4. Ground state electron configurations of atoms provide insights into valence: for example, why does sodium oxide have formula Na2O but magnesium oxide is MgO?. When two atoms approach and form a chemical bond, the electron density farthest from the nucleus of each atom, in the higher-energy orbitals, interacts with electron density in the other atom. Electrons in lower-energy orbitals, whose electron density is nearer the nucleus, are less important.
Electrons can be separated into two groups: valence electrons occupy the outermost orbitals of an atom; core electrons occupy inner orbitals, with electron density closer to the nucleus. When an electron configuration is written using the noble-gas notation, all electrons represented by the noble-gas symbol in brackets are core electrons. Electrons beyond the noble-gas configuration are valence electrons if they are in the outermost shell of the atom (have the highest n value) or if they are in incompletely filled subshells. For example, consider vanadium, V: [Ar]4s23d3. There are five valence electrons: two 4s electrons and three 3d electrons. There are 18 core electrons in the 1s, 2s, 2p, 3s, and 3p subshells. The fact that V has five valence electrons results in V forming compounds in which the valence of vanadium ranges from 2 to 5. For example, fluorides of vanadium have formulas VF2, VF3, VF4, and VF5.
Exercise 3: Valence Electrons
An American chemist, G. N. Lewis, suggested a simple way to keep track of the number of valence electrons: draw dots around the symbol of an element to represent the valence electrons. The element symbol then represents the nucleus and core electrons of an atom. A diagram in which dots represent valence electrons is called a Lewis diagram. Lewis diagrams are most useful for the main-group (representative) elements. Here are Lewis diagrams for atoms of elements in the third row of the periodic table:
When drawing a Lewis diagram dots are added one at a time to each of the four sides of the element symbol. If there are more than four dots to add, dots are paired. Lewis originated the idea that when an atom bonds to another atom the valence electrons rearrange to form an octet, a stable configuration of valence electrons (s2p6) that corresponds to each noble gas at the right side of a row in the periodic table. Thus electron configurations and Lewis diagrams for atoms can predict how an atom forms chemical bonds, an idea that we will explore later.
Exercise 4: Lewis Diagrams
In your notebook write a Lewis diagram for each element:
B Ge Br K Sr Se Xe Sc
D3.5 Effective Nuclear Charge
Periodic trends in atomic properties can be predicted by applying these ideas about electron-nucleus attraction and electron-electron repulsion:
• Electron-density distributions are in shells that increase in size as the principal quantum number, n, increases. Electrons in larger shells are, on average, farther from the nucleus and less strongly attracted.
• Electrons repel other electrons, raising electrostatic potential energy. This partly counteracts the lowering of energy due to attraction of an electron by the nucleus. Electrons are said screen or shield other electrons from nuclear charge.
Exercise 5: Electrostatic Attraction and Repulsion
The electron density for a core electron (an electron in an inner shell) is, on average, closer to the nucleus than the electron density for a valence electron. Thus, core electrons can significantly counteract the effect of nuclear attraction. Consider a lithium atom (Li, 1s22s1), which has three protons in the nucleus. Because the 2s orbital is larger than the 1s orbital, the 1s electron density is mostly located between the nucleus and the 2s electron density. (Move the slider in the middle of Figure 2 to see how much of the 1s electron density lies between the nucleus and 2s electron density.) Thus, the two 1s electrons repel the 2s electron away from the nucleus, counteracting part of the 3+ charge of the nucleus.
Query \(7\)
Figure 2. Electron-density distribution of Li 1s electrons and a Li 2s electron. Use the slider in the center of the diagram to alternate between the two distributions. (The nucleus is at the intersection of the axes.)
To account for such electron-electron repulsions, we use an effective nuclear charge, Zeff,the positive nuclear charge (given by the atomic number) reduced by the repulsion of a specific electron by all the other electrons. In the case of the Li 2s electron, quantum mechanics calculates that the repulsions from the two 1s electrons reduces the nuclear charge by 1.72; that is, Zeff for the 2s electron is 3 − 1.72 = 1.28. If all the electron density of the 1s electrons were between the nucleus and the 2s electron, Zeff would be reduced to 1.
Activity 3: Effective Nuclear Charge
D3.6 Periodic Variation in Atomic Radius
Given that electron density is distributed throughout space but concentrated near the nucleus, it is hard to define the size of an atom. Typically, chemists think of atoms as spheres with radii on the order of tens to hundreds of picometers. One way to determine atomic radii is to measure the distance between atomic nuclei in homonuclear diatomic molecules. (Homonuclear means two atoms of the same element bonded to each other.) The radius of one atom is half the internuclear distance. A second way is to measure the distance between the nuclei of two atoms in a solid metal, where each atom touches several nearest neighbors. Once a set of atomic radii has been determined, these values can be used to estimate the lengths of bonds that have not yet been measured.
Query \(9\)
Figure 3. Atomic radius as a function of atomic number. Click on each “+” for a description of an important trend.
Activity 4: Periodic Variation of Atomic Radii
Exercise 7: Atomic Radii and Periodic Table
Day 3 Pre-Class Podia Problem: Quantum Numbers and the Periodic Table
This Podia problem is based on today’s pre-class material; working through that material will help you solve the problem.
In a hypothetical parallel universe, the quantum numbers are defined differently: n, ℓ, and ms obey the same rules as in our universe, but m cannot have negative values. Thus, a p subshell contains only two p orbitals, a d subshell contains three d orbitals, and so forth. The energies of the subshells are the same as in our universe. In your notebook, answer each question below and explain each answer clearly, concisely, and with scientifically appropriate language.
1. Draw a periodic table for the hypothetical universe.
2. Determine the atomic number of the fourth noble gas element.
3. Write the electron configuration of a stable ion formed by element number 6.
4. Write the electron configuration of a stable ion formed by element number 18.
Query \(13\)
Two days before the next whole-class session, this Podia question will become live on Podia, where you can submit your answer.
Comments. If you found any inconsistencies, errors, or other things you would like to report about this module, please use this link to report them. A similar link will be included in each day’s material. We appreciate your comments. | textbooks/chem/General_Chemistry/Interactive_Chemistry_(Moore_Zhou_and_Garand)/01%3A_Unit_One/1.03%3A_Day_3-_Orbital_Energy_and_Electron_Configuration.txt |
Day 4: Periodic Trends; Forces between Atoms
In addition to atomic radius, several other properties vary depending on position in the periodic table. Like atomic radius, these properties depend on effective nuclear charge and size of atomic shell.
D4.1 Periodic Variation in Ionization Energies
The quantum mechanics model for the hydrogen atom (Section D2.4) discussed ionization of the hydrogen atom: excitation of the electron from the n = 1 level to the limit of n → ∞:
$\Delta E = E_{n \rightarrow \infty} - E_1= 0 – \dfrac{-k}{1^2} = k = 2.179\;\times\;10^{–18}\;\text{J} \nonumber$
Ionization energy (IE), the minimum energy required to remove an electron from the atomic orbital it occupies (exciting it to n = ∞), also applies to multi-electron atoms. As shown in this table of ionization energies, there are successive ionization energies, one for each electron.
The first ionization energy (IE1) is the minimum energy required to remove the least tightly bound electron, i.e. the electron in the highest energy orbital. It corresponds to the process:
X(g) → X+(g) + e ΔE = IE1
The second ionization energy (IE2) is the minimum energy required for removing an electron from the 1+ cation, corresponding to:
X+(g) → X2+(g) + e ΔE = IE2
And so forth.
Electrons in atoms have lower potential energy than when separated from the atom, so energy is always required to remove the electrons from atoms or ions. Ionization is an endothermic process and IE values are always positive.
Ionization energies can be determined experimentally from atomic spectra or by shining light on a gas-phase sample and successively increasing the photon energy until ejection of an electron is observed. Such experiments also give us direct information about the atomic orbital the electron was occupying.
Activity 1: Periodic Variation of First Ionization Energy
Exercise 1: Periodic Variation of Ionization Energy
In the preceding activity, you developed a general rule that across a period, IE1 increases with increasing atomic number, Z; down a group, IE1 decreases with increasing Z. Look at Figure 1 below and make certain you see how the data in the figure support both trends. There are a few systematic deviations from these trends, which also can be seen in Figure 1.
Activity 2: Deviations from IE Trends
Examine data presented in Figure 1. Identify all deviations from the trend that ionization energy increases across a period and decreases down a group. Click on each deviation involving an element in Period 1, Period 2, or Period 3.
Query $3$
Figure 1. The first ionization energies of elements in the first five periods are plotted against atomic numbers.
Activity 3: Explaining Deviations from IE Trends
Query $4$
Another deviation occurs when a subshell becomes more than half filled. For example, oxygen’s IE1 is slightly lower than that for nitrogen. The orbital diagram of oxygen shows that the last electron added (red) is forced to pair with another electron, because the 2p subshell is more than half full.
Loss of the electron (colored red in the orbital diagram) yields a greater reduction of electron–electron repulsion because there are no longer two electrons in the same orbital. This makes it easier to lose one electron from O and the ionization energy is smaller than expected. The same explanation applies to the dip for sulfur after phosphorus in Figure 1.
Looking at the successive IEs in the appendix, we see that for any element, IE1 < IE2 < IE3, and so forth, and this is due to the greater Zeff in successively more positive ions. However, sometimes the increase in successive IE is larger than expected. For example, consider the data in Table 1:
Element IE1 IE2 IE3 IE4 IE5 IE6 IE7
K 419 3069 4438 5876 7975 9620 11385
Ca 590 1145 4941 6465 8142 10496 12350
Sc 633 1244 2388 7130 8877 10720 13314
Ga 579 1982 2962 6194 8299 10874 13585
Ge 760 1537 3301 4409 9012 11183 13981
As 947 1949 2731 4834 6040 12302 14183
Table 1. Successive Ionization Energies (IEs) for Some Period-4 Elements (kJ/mol)
Activity 4: Successive Ionization Energies
Query $5$
These observed ionization energies are consistent with the idea of valence electrons—electrons in outer shell(s) that are less bound and more energetically accessible, allowing them to participate in chemical transformations.
D4.2 Periodic Variation in Electron Affinities
Electron affinity (EA) is defined as the change in energy when an electron is added to an atom to form an anion. The first EA corresponds to adding one electron to an atom:
X(g) + e → X(g) ΔE = EA1
Exercise 2: Ionization Energy vs Electron Affinity
Many elements have negative EA1, which means that the energy of the 1− anion is lower than the energy of the atom plus the free electron; that is, the anion is more stable.
For other elements, EA1 is positive, meaning that the anion is less stable compared to the parent atom plus a free electron. For these elements, an input of energy is required to form the anion, and, in the gas phase, the anion dissociates to yield the neutral atom and a free electron because the latter are lower in energy.
Activity 5: Electron Affinity Trends
D4.3 Electron Configurations of Monoatomic Ions
When atoms lose electrons or gain electrons, Ions form. It is useful to know what kinds of ions form and what their properties are. A monoatomic ion is a single atom that has gained or lost one or more electrons. A positively charged ion, a cation, forms when an atom loses one or more electrons. A negatively charged ion, an anion, forms when an atom gains one or more electrons.
Nonmetallic elements on the far right side of the periodic table (except the noble gases) have higher ionization energies and more negative electron affinities. It is energetically more favorable for them to gain electrons and form anions and less energetically favorable for them to form cations. For example, elements in groups VIIA, VIA, and some in VA can gain 1, 2, or 3 electrons, respectively, to achieve the electron configuration of a noble gas (a full octet, s2p6 in the outermost shell).
Metallic elements on the left side of the periodic table have lower ionization energies and less negative (or positive) electron affinities. It is energetically more favorable for them to form cations. For examples, elements in groups IA, IIA, and IIIA can lose 1, 2, or 3 electrons, respectively, to achieve a full octet (plus additional filled d and f subshells).
To find the ground state electron configuration of a monoatomic ion, start with the electron configuration of the corresponding atom and remove (or add) an appropriate number of electrons from (or to) the valence orbital(s) of the atom. Here are some examples:
K([Ar]4s1) ⟶ K+([Ar]) + e‾
Ga([Ar]3d104s24p1) ⟶ Ga3+([Ar]3d10) + 3e‾
O(1s22s22p4) + 2e‾ ⟶ O2-(1s22s22p6 or [Ne])
When transition elements and inner transition elements form cations, electron(s) in the outer-most shell (largest n) are removed before any d or f electrons. For example, when Fe loses two electrons to form Fe2+, the two 4s electrons are lost:
Fe([Ar]3d64s2) ⟶ Fe2+([Ar]3d6) + 2e‾
This happens because electrons in the 3d subshell are very effective at screening the 4s electrons from the nucleus but much less effective at screening each other. Quantum mechanics calculates that the effective nuclear charge experienced by a 4s electron in Fe is 2.1, while the effective nuclear charge experienced by a 3d electron is 4.3. After 4s electrons have been removed, some 3d electrons can also ionize. Here are more examples:
V([Ar]3d34s2) ⟶ V5+([Ar]) + 5e‾
Re([Xe]4f145d56s2) ⟶ Re2+([Xe]4f145d5) + 2e‾
Exercise 3: Electron Configurations for Monoatomic Ions
In your notebook write the correct electron configuration for each ion listed here:
Sr2+ Te2 Al3+ Fe3+ Nd4+
Exercise 4: Identify Element from Ion Electron Configuration
Ions and atoms that have the same electron configuration are isoelectronic. For example, the isoelectronic Na+, Ne, and F all have ground state electron configuration of 1s22s22p6 (or [Ne]). For main-group elements, the most commonly formed ions are isoelectronic with a noble gas; that is, these ions have complete octets.
D4.4 Unpaired Electrons and Magnetism
One way chemists have discovered information about electron configurations of ions involves magnetism. You are probably familiar with refrigerator magnets, iron magnets, or neodymium (rare earth) magnets. These exhibit ferromagnetism, a strong attraction to a magnetic field that is easily observable and sometimes can be made permanent. All substances exhibit diamagnetism, a very weak repulsion from a magnetic field that can only be observed in extremely large magnetic fields. Because diamagnetism is so weak, it is usually too small to notice.
Query $11$
In addition to diamagnetism, some substances also exhibit paramagnetism, an attraction to a magnetic field that is not as strong as ferromagnetism but much stronger than diamagnetic repulsion. Paramagnetism arises due to the presence of unpaired electrons. Each electron has a tiny magnetic field; that is, each electron is a tiny magnet. When a macroscopic magnet comes near a paramagnetic substance, most of the unpaired electron magnets align with the magnetic field, causing the substance to be attracted to the magnet (Figure 2).
Figure 3 shows how ferromagnetism differs from paramagnetism.
When electrons are paired in an orbital, their opposite magnetic moments create opposite magnetic fields that cancel each other: substances with all electrons paired are neither paramagnetic nor ferromagnetic. They exhibit only diamagnetism and are said to be diamagnetic. For example, argon has a ground state electron configuration of 1s22s22p63s23p6, where each subshell is full and each orbital therein contains a pair of electrons with opposite spins. Hence, a sample of argon is diamagnetic.
Paramagnetic and diamagnetic materials do not act as permanent magnets. A magnetic field is required to align the magnetic moments and make them magnetic. The magnitude of paramagnetism can be measured by weighing a sample with a highly sensitive balance and then weighing it again with a strong magnet just below the sample. A sample that is attracted to the magnet appears heavier because the magnet attracts it downward. The increase in apparent weight is proportional to the number of unpaired electrons.
Activity 6: Ionization of Transition Metals
D4.5 Ionic Radii
Ionic radius is the radius of a sphere representing a cation or an anion; it can be determined from structures of ionic crystals.
Cations have fewer electrons than the uncharged atoms from which they are derived. Hence, there is less electron-electron repulsion (that is, larger Zeff), which makes a cation’s radius smaller than the corresponding uncharged atom’s radius. For example, the atomic radius of Al ([Ne]3s23p1) is 143 pm, which is more than twice as large as the 68 pm ionic radius of Al3+ ([Ne]). Often, as in the case of Al, formation of a cation involves removal of all electrons from the outermost shell of an atom, which means the remaining electrons are in a smaller shell—another reason why cations are smaller than the atoms from which they form.
For the same element, cations with larger positive charges are smaller than cations with smaller charges. For example, V2+ has an ionic radius of 93 pm, while that of V3+ is 78 pm. (An uncharged V atom has an atomic radius of 135 pm.)
Anions have more electrons and therefore greater electron-electron repulsion (that is, smaller Zeff) than the neutral atoms from which they are derived. Thus, an anion’s radius is larger than the parent atom’s radius. For example, the ionic radius of S2- ([Ne]3s23p6) is 170 pm, larger than the 104 pm atomic radius of S ([Ne]3s23p4).
Periodic trends in radii of a set of anions (or cations) with the same charge are similar to the atomic-radius trends. For instance, proceeding down a group, radii of 1+ cations generally increase as atomic number increases, corresponding to the increase in the principal quantum number, n.
For isoelectronic species (ions or atoms with the same electron configuration, Section D4.3), the greater the nuclear charge (number of protons), the smaller the atomic/ionic radius. This implies that isoelectronic anions are larger than the neutral atom which is larger than the cations.
Exercise 7: Atomic and Ionic Radii
In Days 2 and 3 and above, we have discussed atomic structure and electron density as well as periodic trends in effective nuclear charge, size of atoms, ionization energies, and electron affinities. All of these are central to understanding properties and chemical reactivity of the elements. Next, we apply those ideas to the noble gases and to metals.
Applying Core Ideas: Two Similar Atoms, Two Very Different Substances
Section D1.5 introduced the idea that atoms, molecules, and oppositely charged ions attract and that their potential energy can be described by a curve that starts at zero when the particles are far apart, falls to a minimum, and increases when the particles are very close together. The depth of the minimum in such a curve can be related to physical properties such as boiling points, because atomic-level particles gain energy as temperature increases.
Consider the boiling points of the first five noble gases in the table below.
Noble Gas Atomic Number Atomic Radius (pm) Boiling Point (K)
He 2 31 4.22
Ne 10 68 27.1
Ar 18 91 87.4
Kr 36 108 119.9
Xe 54 126 165
Based on the boiling point data, for which noble gas is the attraction between particles greatest? Which noble gas has the deepest minimum in its curve of potential energy vs distance between atoms? In your notebook, answer these questions. Then use one set of axes and sketch the potential-energy curve for each of the five noble gases in the table, describe the curves in words and explain why you drew the curves as you did.
Query $16$
Based on the experimental data and the curves you drew, correlate the size of the attraction between atoms with the number of electrons and the size of each atom. Write several sentences in your notebook describing the correlation.
Query $17$
Now consider iron, which has atomic number 26 and atomic radius 126 pm (the same radius as Xe, but fewer electrons). Based on this information, sketch the potential energy curve for iron atoms on the graph you made for the noble gases. Describe the curve in words and explain why you drew the curve as you did.
Does what you predicted for iron make sense?
D4.6 Forces Between Atoms
Based only on experimental data for noble gases, one might predict that iron would be a gas at room temperature, but iron is a solid. Attractive forces between iron atoms must be a lot stronger than attractive forces between noble-gas atoms. To make sense of the difference between xenon and iron, we need a better model for forces between atoms.
Let’s begin by thinking about attractions between xenon atoms. On average the electron density distribution of the 54 electrons surrounding a xenon nucleus is spherically symmetric. That is, no matter which direction you go from the nucleus, the electron density is the same at the same distance from the nucleus. However, there can be very brief deviations or fluctuations from this average. In 1928 German-American physicist Fritz London used quantum mechanics to show how such fluctuations could lead to attractive forces between atoms and other atomic-scale particles.
Here is a simplified explanation using a Xe atom as an example. Consider a fluctuation in which there is slightly more electron density on one side of the nucleus than on the other. Such a brief deviation creates a dipole, a distribution of electric charge where one side is more positive and the other side is more negative. The dipole only occurs for an instant and is therefore called an instantaneous dipole. This is shown in Figure 4, parts a and b, where δ+ and δ indicate tiny fractions of the charge of an electron.
If a second Xe atom is close to the first one when the instantaneous dipole forms, for example, on the right of the first one as in Figure 4, part c, the excess negative charge on the right of the first Xe atom repels the electrons on the second Xe atom. This forms a second dipole, again for only an instant. The second dipole is said to be induced by the first one. The positively charged end of the second dipole is attracted to the negatively charged end of the first dipole. For the instant that the dipoles exist, there is a weak attraction between the two atoms.
These weak attractive forces due to instantaneous fluctuations in electron density are called London dispersion forces; we will often refer to them as LDFs. LDFs are present between all atomic-scale particles: atoms, molecules, and ions. The size of the attractive force depends on the number of electrons in a particle and how easily the electron density distribution can be distorted from its average shape.
Query $19$
Activity 7: LDFs and Noble Gas Boiling Points
D4.7 Metals
Based on iron’s boiling point of 3135 K (2862 °C), the forces between iron atoms must be much greater than just LDFs. What you have already learned about atomic structure, effective nuclear charge, and valence electrons can be used to make sense of the much larger forces that attract iron atoms, as well as atoms of other metals.
Metals, which include most of the elements in the periodic table (Figure 5), have characteristic macroscopic properties: they conduct heat and electricity well when they are solids or liquids; they have lustrous (metallic-looking) solid surfaces; and they deform, rather than shatter, when hammered (they are malleable). Many metals are hard, quite strong, and useful as construction materials. What gives rise to these characteristics?
Metallic substances are made of atoms whose valence electron(s) can be removed with relatively low quantities of energy; that is, elements with relatively low ionization energies. Valence electrons of metals experience lower effective nuclear charge than valence electrons of nonmetals; hence, metals tend to form positive ions and nonmetals tend to form negative ions.
In a metallic solid, atoms are packed closely (Figure 6). One model describes solid metals at the atomic scale as atoms surrounded by mobile valence electrons—a so-called “sea” of electrons. Each “atom” consists of the nucleus and core electrons; that is, a positively charged ion that has lost all its valence electrons. The loosely-bound valence electrons are shared (delocalized) among many different atoms. Electrostatic attractions between the sea of shared valence electrons and the metal ions is known as metallic bonding. The strength of metallic bonding increases as the number of valence electrons shared in the electron sea increases.
Because the valence electrons are delocalized over many metal atoms, if one atom moves relative to others, the attraction between valence electrons and atomic cores remains; that is metallic bonding is not specific to a particular direction. This characteristic accounts for many of the observed physical properties of metals.
Activity 8: Metallic Bonding
Query $21$
Day 4 Pre-Class Podia Problem: Chemical Combinations
This Podia problem is based on today’s pre-class material; working through that material will help you solve the problem.
When nonmetals combine with other nonmetals, compounds form with subscripts that are small whole numbers, such as CO and CO2. When metals combine with nonmetals, formulas of compounds are analogous, such as FeCl2 and FeCl3. When metals combine with other metals, however, they form alloys, substances that have metallic properties but can consist of many different compositions. For example, sodium and potassium can form alloys in which the ratio of sodium atoms to potassium atoms varies from all sodium to all potassium. Thus, if you wrote a formula for sodium-potassium alloy as NaxK, x could have any value from zero to infinity.
Think about why metals should be different in this way. Then write an explanation for the different behavior.
Query $22$
Two days before the next whole-class session, this Podia question will become live on Podia, where you can submit your answer.
Comments. If you found any inconsistencies, errors, or other things you would like to report about this module, please use this link to report them. A similar link will be included in each day’s material. We appreciate your comments. | textbooks/chem/General_Chemistry/Interactive_Chemistry_(Moore_Zhou_and_Garand)/01%3A_Unit_One/1.04%3A_Day_4-_Periodic_Trends_Forces_between_Atoms.txt |
Day 5: Ionic Compounds; Covalent Bonding
Applying Core Ideas: Two Substances, Some Similarities, Some Big Differences
Compare iron, which melts at 1811 K (1538 °C), and calcium fluoride, which melts at 1691 K (1418 °C). Are the forces between atomic-level particles similar in iron and calcium fluoride? Write in your notebook an explanation of your answer to this question.
Query $1$
Consider some other properties of iron and calcium fluoride.
Iron has typical metallic properties: pure solid iron has metallic luster, is malleable, and conducts electricity.
Calcium fluoride has significantly different properties: as a solid it is brittle, does not look metallic, and does not conduct electricity; when molten, calcium fluoride does conduct electricity though.
It appears that these two substances have similar attractions between atomic-level particles, but quite different properties. We need a better atomic-scale model to make sense of these differences.
D5.1 Ionic Compounds
In metals, which have low effective nuclear charges (and low ionization energies), it is relatively easy to form positive ions within a sea of electrons so that attractions among the ions and the electrons hold the atoms together. What happens when an atom with a low ionization energy interacts with an atom with a large negative electron affinity? In such a case, transfer of one or more electrons from the atom with the low ionization energy to the one with the high electron affinity can be energetically favorable. For example, consider transfer of an electron from a Li atom to a F atom to form a cation and anion close together: an ion pair, Li+F.
Activity 1: Analyzing Formation of an Ion Pair
Query $2$
From the activity above, you’ve worked out that the formation of the ion pair is associated with a negative ΔE: the joined ion pair is lower in total energy (more stable) than the two atoms from which it formed.
The gain and loss of electron(s) in forming an ion-pair typically results in a full octet for the cation and anion. For example:
Na([Ne]3s1) + Cl([Ne]3s23p5) ⟶ Na+([Ne]) + Cl‾([Ar])
Mg([Ne]3s2) + O(1s22s22p4) ⟶ Mg2+([Ne]) + O2-([Ne])
Ca([Ar]4s2) + 2F(1s22s22p5) ⟶ Ca2+([Ar]) + 2F‾([Ne])
When a large number of ions form, anions and cations form a structure, called an ionic crystal lattice, where there are equal numbers of anions and cations (so there is zero total electric charge); in the lattice each anion has several cations as its nearest neighbors, and each cation has several anions as its nearest neighbors (see example in Figure 1). This arrangement maximizes anion-cation attractions (smaller r results in lower Coulomb’s law energy) and minimizes anion-anion and cation-cation repulsions (greater r between anions and between cations). A compound made up of anions and cations is called an ionic compound and the overall attraction making an ionic crystal lattice stable is called ionic bonding.
Query $3$
Figure 1. Ionic crystal lattice. Cations (smaller spheres) and anions (larger spheres) are arranged so that there is a net attraction that makes the collection of ions stable. Move the slider to change the diameters of the ions and show lines indicating the lattice. (This is a small section of the lattice, which extends millions of times farther in all three dimensions.)
For example, when sodium cations (Na+) and bromide anions (Br) come together, the crystal lattice of the compound sodium bromide, Figure 1, contains one Na+ (green) for each Br (violet). The chemical formula representing the part of the lattice shown in Figure 1 is Na75Br75 because there are 75 Na+ and 75 Br ions. The actual crystal is much larger so the subscripts in the formula of a real crystal would be huge. A formula unit is a group of chemical symbols that indicates the smallest whole number ratio of ions of each kind that make up the substance, and is typically used as the formula for ionic compounds. Thus, the formula unit (and the formula) for sodium bromide is NaBr.
All chemical substances are overall electrically neutral. Thus, in an ionic compound, the positive charge from all the cations must equal to the negative charge from all the anions. In other words, the subscripts in a formula unit of an ionic compound must result in equal quantities of positive and negative charges. If we know the charges of the ions, then we can write the formula.
Activity 2: Predicting Chemical Formulas
Query $4$
Some ions consist of a group of atoms with an overall charge. Examples are sulfate dianion, SO42, dihydrogen phosphate anion, H2PO4, and ammonium cation, NH4+. It is quite useful to be able to recognize such polyatomic ions and their charges. If you need to refresh your knowledge of polyatomic ions, use the reference table as you do the next two exercises.
D5.2 Lattice Energy
Refer back to Activity 1. The third step involved bringing one Li+ ion close to one F ion, which lowered the energy. When a large number of oppositely charged ions are brought close together to form a crystal lattice (Figure 1), the energy is lowered even more, because more ions are close together in a structure where attractive forces predominate over repulsive forces.
The lattice energy of an ionic compound is defined as the decrease in energy when the ionic crystal lattice forms from the separated ions. Lattice energy can be calculated using a modified form of Coulomb’s law; some calculated lattice energies are in Table 1. (Note that because lattice energy is defined as a decrease in energy, a large positive lattice energy corresponds to a large negative ΔE, that is, a very stable (low energy) ionic solid.)
Substance Lattice Energy (kJ/mol)
NaI 682
CaI2 1971
MgI2 2293
NaOH 887
Na2O 2481
NaNO3 755
Ca3(PO4)2 10,602
CaCO3 2804
Table 1. Representative calculated lattice energies. Data from CRC Handbook of Chemistry and Physics (2004).
Lattice energies have a wide range of values depending on which ions make up a compound. Consider, for example, why the lattice energy increases as you go from NaI, to CaI2, to MgI2 (first three rows in Table 1).
According to Coulomb’s law,
$E_{\text{p}} = k_e\frac{Q_1Q_2}{r} \nonumber$
the lattice energy is directly proportional to the sizes of the charges on the ions and inversely proportional to the distance between them (the sum of their ionic radii). Thus, we can compare NaI, CaI2, and MgI2 with respect to magnitudes of charges and sizes of ions.
Magnitudes of Charges: NaI consists of 1+ ions and 1− ions; CaI2 and MgI2 consist of 2+ ions and 1− ions.
Because the magnitude of Q1Q2 is larger, the lattice energies of CaI2 and MgI2 should be larger than for NaI.
Sizes of ions: All three compounds contain I anions, so we only need to compare cation radii.
The radii of Ca2+ and Na+ are similar. This can be estimated from the periodic table: Na+ and Ca2+ are both smaller than K+, Na+ because it is above K+ in the periodic table, and Ca2+ because it is isoelectronic with K+ but has more protons in the nucleus. It is reasonable to assume that the radii are not very different, and the ionic-radius table confirms this: Na+ 116 pm; Ca2+ 114 pm. Thus, r is not a major factor affecting lattice energy for NaI and CaI2.
The radius of Mg2+ is significantly smaller than for Ca2+ (and Na+) because Mg is directly above Ca in the periodic table and the ions have the same charge. Thus, MgI2 should have a larger lattice energy than CaI2, which is consistent with the values in the table.
Activity 3: Comparing Lattice Energies
Query $7$
When looking at Table 1, you may have noticed that lattice energies for compounds containing 2+ and 2− ions are nearly quadruple those for similar compounds containing 1+ and 1− ions of similar size (compare NaNO3 and CaCO3). This observation is consistent with the general rule that lattice energies are highest for substances with small, highly charged ions.
D5.3 Properties of Ionic Compounds
The physical and chemical properties of an ionic compound are determined by the ions that that constitute the compound. The compound’s properties are quite different from the properties of the elements that reacted to form the compound. For example, both sodium and chlorine react with water, but sodium chloride (NaCl) dissolves in water without reacting. In other words, sodium ions and chloride ions do not react with water, but sodium atoms and Cl2molecules do.
https://mediaspace.wisc.edu/id/0_zns...&st=137&ed=154
Figure 2. Sodium reacts vigorously with water, producing a gas and making the water basic (as indicated by the pink color of phenolphthalein in the water). Adding sodium ions to water, in the form of table salt, NaCl, is much less exciting!
Ionic compounds have many physical properties in common:
• they usually have melting points and boiling points well above room temperature.
• they are crystalline solids with distinct crystal shapes
• they are brittle, will shatter if struck by a hammer, and can easily be cleaved (cleave means to break along smooth planes, as shown in this video).
• they are electrical insulators when solid but conduct electricity when molten (liquid). (See this video.)
• when an ionic compound dissolves in water, the solution conducts electricity much more effectively than pure water.
Because all ionic compounds have similar properties, it is useful to be able to identify an ionic compound from its chemical formula. In general, ionic compounds contain cations of metals from the left side of the periodic table and anions of nonmetals from the right side of the periodic table. If a compound’s chemical formula contains a polyatomic ion, then the compound is an ionic compound.
Exercise 3: Identifying Ionic Compounds
The properties of ionic compounds can be interpreted in terms of ions. In the solid phase, ions are essentially fixed in their positions in a crystal lattice. Lattice shapes depend on the number and type of ions in the formula, but each lattice has a distinctive shape that results in the lowest energy (maximum lattice energy). Because distances between ions are small, lattice energies are large. It requires a lot of energy to overcome attractions among the ions, making the crystal hard. The lattice shape is related to the shape of the macroscopic crystal. When a crystal cleaves, the cleavage planes are parallel to planes in the crystal lattice. (To see why, watch this video.)
When an ionic solid melts, the anions and cations are free to move randomly among each other. Although the average minimum distance between ions is still small (just a little larger than in the crystalline solid), some anions may be next to other anions and some cations near other cations. Hence, the electrostatic attractions are not as strong and the electrostatic repulsions are greater than in the solid phase; that is, part of the lattice energy needs to be overcome before an ionic compound melts.
When an ionic compound boils to form a gas, the ions continue to move randomly but also are so far apart that their electrostatic attractions are negligible. All of the lattice energy must be overcome for an ionic compound to boil.
Electric current is the movement of electric charge from one place to another. Electric charge is carried by any moving charged particle. In a liquid ionic compound, where ions can move independently, electric charge is conducted by ions. When a solid ionic compound dissolves in water, the ions separate and can move independently throughout the solution. Movement of ions conducts electricity through the solution, just as it does in the molten liquid.
Activity 4: Explaining Properties of Ionic Compounds
Query $9$
Activity 5: Reflection
In your course notebook, write down as many properties of metals and of ionic compounds as you can remember. Then check to make certain your list is complete. Then write your explanation for each property in terms of ions, crystal lattices, attractive forces, and energy. Make sure your list provides a good summary you can use to review later for an exam.
Applying Core Ideas: Comparing Hydrogen Atoms and Helium Atoms
The boiling point of helium is 4.22 K (−268.93 °C), a consequence of very weak London dispersion forces (LDFs) between He atoms. Even when two He atoms are only a short distance apart, there is very little attraction.
When two H atoms get close there is a much stronger attraction: the two atoms become a H2 molecule by forming a covalent chemical bond. For the H2 molecules to have enough energy to break that bond, separating the two H atoms, the temperature needs to be similar to the temperature of the Sun (5778 K)!
Why do He atoms behave so differently from H atoms? We need to refine our models of structure and energy still further. This is done in the next sections.
D5.4 Covalent Bonding: Molecular Orbitals
Activity 6: Covalent Chemical Bonds
In your notebook, write a few sentences describing your understanding of covalent chemical bonds. What constitutes a covalent bond? How does it form? What holds the atoms together in a molecule? As you work through this section, make additional notes to review later.
If you wrote that a covalent bond involves sharing electrons between atoms, that’s a good start. But how does sharing electrons hold atoms together?
Just as a balance of Coulomb’s-law attractions and repulsions affects the electron(s) and nucleus in an atom, coulombic interactions affect electrons and nuclei in a molecule. The major difference is that in a molecule, each electron interacts with two or more nuclei; this additional interaction provides the stabilization that allows the molecule to exist. If a collection of separate atoms were more stable than a molecule, the molecule would simply fall apart into individual atoms.
Just as electron(s) occupy stable atomic orbitals around a single nucleus in an atom, electrons occupy stable orbitals around multiple nuclei in a molecule. An orbital that extends over two or more atomic nuclei is called a molecular orbital (MO); in the MO model, a molecular orbital forms when atomic orbitals (AOs) overlap (or interpenetrate).
To understand this better, think about the formation of a H2 molecule from two ground-state hydrogen atoms (labeled as HA and HB in Figure 3) . When HA is sufficiently far away from HB—for example, more than 400 pm apart (top of Figure 3)—its electron (eA) occupies the lowest-energy atomic orbital, the 1s orbital. The same is true for HB.
When HA and HB are closer, eA begins to experience significant Coulomb’s-law attraction towards HB‘s nucleus. For example, at 250 pm apart (Figure 3, second from top) the orbital occupied by eA is no longer spherical around HA‘s nucleus but rather is distorted—there is more electron density on the side toward HB. Although HA‘s nucleus attracts eA more strongly than HB‘s nucleus does (because HB‘s nucleus is screened by eB), there is a net increase in the positive charge interacting with eA. This lowers eA‘s potential energy. The same argument applies to eB.
When the two atoms are even closer—for example, 100 pm apart—eA and eB become delocalized over both nuclei: both electrons occupy a molecular orbital, which surrounds both nuclei. Notice that in this H2 molecular orbital, there is significant electron density in between the two nuclei, attracting both nuclei. This lowers the overall energy relative to the energy of the two separate hydrogen atoms.
As the two atoms get closer still, there is a minimum in the overall energy for the molecule. For H2 the minimum is at 74 pm (bottom of Figure 3). At internuclear distances smaller than 0.74 Å, the energy rises again due to increasing nucleus-nucleus repulsion. In a ground-state H2 molecule the distance between nuclei of two bonded atoms, the bond length, is 74 pm.
Figure 4 allows you to explore how the electron density changes as two H atoms interact to form H2 (or how an H2 molecule dissociates into two separate H atoms). The blue dot on the potential energy curve shows the corresponding internuclear distance (bond length) and potential energy of the molecule. The ground-state H2 molecule is more stable than the two separated ground-state H atoms. (The energy of the separated atoms is set to be 0 kJ/mol in Figure 4.) The magnitude of the energy reduction, 436 kJ/mol, is called the bond energy, the energy required to break the bond, completely separating the atoms.
Query $10$
Figure 4. A graph of energy versus distance between two hydrogen atoms shows minimum energy at a separation of 74 pm (the bond length). At the minimum, the energy is 436 kJ/mol less than the energy of the separated atoms; this is the bond energy.
What we have just described is one of two ways that the two H 1s atomic orbitals can overlap/combine to form an H2 molecular orbital. It occurs when the two 1s atomic orbitals are in-phase. In-phase means that wave functions describing electrons A and B both have maxima in the same direction at the same time.
The second possibility is that the two overlapping 1s atomic orbitals are out-of-phase, which means that one atomic wave function is positive and the other atomic wave function is negative. This is shown in Figure 5, where the two phases (positive and negative) are shown in green and yellow.
The result is a molecular orbital with a node halfway between the two H atoms. (The node, where electron density is zero, is a plane perpendicular to a line connecting the two nuclei, the internuclear axis). Note that the bottom depiction in Figure 5 is a single molecular orbital with one node, just as the bottom depiction in Figure 3 is a single molecular orbital but without a node.
Query $11$
As is true of atomic orbitals, a molecular orbital with more nodes has higher energy, which makes sense because the presence of the node reduces electron density between the nuclei (compared to the electron densities of the individual atoms). This reduction in electron density results in less electron-nucleus attraction than if the atomic orbitals had not interacted at all.
D5.5 Molecular Orbital (MO) Diagram
When two H atoms come together to form H2, both in-phase overlap and out-of-phase overlap of 1s atomic orbitals are possible and both occur. Hence, two 1s atomic orbitals (AOs) overlap to form two molecular orbitals (MOs). This is true in general: the number of MOs formed equals the total number of AOs from which they are derived.
A MO diagram shows the relative energies of MOs and AOs and indicates which MOs form from overlap of which AOs. For example, Figure 6 shows the MO diagram of the H2 molecule. A MO lower in energy than the AOs from which it is derived is called a bonding MO. An electron occupying this bonding MO increases the stability of the bond (strengthens the bond). The H2 bonding MO is labeled “σ1s“; the subscript “1s” designates that it is derived from the 1s AOs.
A MO higher in energy than the AOs from which it is derived is called an antibonding MO—an electron occupying this MO decreases the stability of the bond (weakens the bond). The H2 antibonding MO is labeled as “$\sigma^*_{1s}$.”; the “*” denotes that it is an antibonding MO.
The Greek letter in the MO labels denotes the symmetry of the MO with respect to the internuclear axis (the line connecting the two nuclei). Greek letters σ, π, δ are analogous to the Roman letters s, p, d that designate shapes of atomic orbitals. A MO that is cylindrically symmetric is called a sigma MO and designated σ. (Cylindrically symmetric means that if you stood the molecule on end and spun it around the internuclear axis it would look the same all the time—just as a cylinder would look if you stood it on end and spun it).
Chemists often use simple 2D drawings to show both the shape and phase of AOs and MOs. So the sphere of a 1s AO is drawn as a circle and its phase is indicated by two different colors or shading/no shading or plus/minus sign. For example, formation of the σ1s MO from two 1s AOs can be drawn as:
where the black dots represent H nuclei, the circles with hatching (or + sign) represent boundary-surface plots of H 1s AOs, each with positive phase, and the oval with hatching (or + sign) represents the H2 σ1s MO, also with positive phase.
D5.6 Electron Configurations and Bond Order
Just as an atom has an electron configuration, so does a molecule. The molecular ground-state electron configuration can be obtained by filling MOs, facilitated by MO diagrams such as the one shown in Figure 6, following rules similar to those for atomic electron configurations:
• Sum all the electrons from all the atoms that make up the molecule.
• Each MO can hold at most two electrons. (Two electrons occupying the same orbital must have opposite spin.)
• In the MO diagram the MOs are arranged in order of increasing energy.
• Add electrons to the lowest-energy MO first and then to successively higher-energy MOs (Aufbau principle)
• If two or more MOs have the same energy (are degenerate), apply Hund’s rule: assign one electron to each degenerate MO before pairing any electrons
For the H2 molecule: there is one electron from each H atom, so there are two electrons in H2. There is only one lowest-energy MO, σ1s, therefore the two electrons, with opposite spin, will fill it. The ground state electron configuration of H2 is (σ1s)2.
$\sigma_{1s}$
Activity 7: Molecular Orbitals and Electromagnetic Radiation
Query $14$
The two electrons in the σ1s MO, which are shared between the two H atoms, form a covalent bond. The configuration (σ1s)2 indicates a single, sigma covalent bond. In general, a covalent bond involves overlap of two or more AOs that leads to increased electron densities close to atomic nuclei so that the electrons’ energies are lowered. Electrons are shared between pairs of atomic nuclei, or more widely among several nuclei within a molecule.
In some cases more than two electrons can be shared between two nuclei. This gives rise to the idea of bond order, which is the number of shared electron pairs between two atoms. Using a MO diagram, bond order can be defined as:
$\text{Bond Order} = \dfrac{1}{2}\left(n_{\text{bonding e}^-} - n_{\text{antibonding e}^-}\right) = \dfrac{1}{2}(n_\text{b} - n_\text{a}) \nonumber$
where nbonding e‾ is the number of electrons in bonding MOs and nantibonding e‾ is the number of electrons in antibonding MOs. There is a factor 1/2 because two shared electrons make one bond. In general a greater number of shared electrons means a stronger bond, so the larger the bond order is, the stronger the bond is.
Exercise 5: Bond Order
Activity 8: Molecular Orbitals for He
Query $16$
Day 5 Pre-Class Podia Problem: Melting Points of Salts
This Podia problem is based on today’s pre-class material; working through that material will help you solve the problem.
The formulas and melting points of four salts are listed below. All four compounds have the same crystal lattice as NaCl. Write a clear, concise explanation of the differences in melting points using scientifically appropriate language and including any relevant data.
NaF 996 °C NaCl 801 °C KCl 770 °C CaO 2572 °C
Query $7$
Two days before the next whole-class session, this Podia question will become live on Podia, where you can submit your answer.
Comments. If you found any inconsistencies, errors, or other things you would like to report about this module, please use this link to report them. A similar link will be included in each day’s material. We appreciate your comments. | textbooks/chem/General_Chemistry/Interactive_Chemistry_(Moore_Zhou_and_Garand)/01%3A_Unit_One/1.05%3A_Day_5-_Ionic_Compounds_Covalent_Bonding.txt |
D6.1 Second-Row Diatomic Molecules
Let’s consider some slightly more complex examples of molecular orbitals. F2, O2, and N2 are diatomic molecules formed by elements from the second row of the periodic table. These molecules contain many more electrons than H2, and their molecular orbitals are derived from p atomic orbitals as well as s atomic orbitals.
Exercise 1: Shapes and Phases of Atomic Orbitals
Additional Practice
Query \(2\)
Now think about what happens when two atoms containing 2p atomic orbitals approach each other. Assume that the internuclear axis is the z axis. This means that the 2pz atomic orbitals are aligned along the internuclear axis while the 2px and 2py atomic orbitals are oriented perpendicular to the internuclear axis.
When the two atoms approach, the bonding and antibonding overlap of the two 2pz atomic orbitals occurs along the internuclear axis (z axis):
MOs derived from these two combinations are labeled σ₂ₚₓ and σ*₂ₚₓ. If you don’t understand why these MOs are have the label σ, review Section D5.5.
It is also possible to make bonding and antibonding combinations from the two 2px AOs and from the two 2py AOs. Here is a diagram for the two ways the 2px AOs can overlap. Notice that the orbitals overlap side-by-side, not end-on, because the 2px AOs are aligned perpendicular to the internuclear axis (z axis).
Exercise 2: Molecular Orbitals Involving Atomic Orbitals
Additional Practice
Query \(4\)
When two 2p AOs overlap side-by-side, the bonding MO formed is not symmetric with respect to rotation around the internuclear axis. Thus, the bond formed is not a σ bond. If you look down the internuclear (bond) axis, the “side view” of the MO looks similar to a 2p atomic orbital; this MO is called a π orbital.
Think about all atomic orbitals that are occupied in a fluorine atom, F: 1s, 2s, 2pz, 2px, and 2py. For each pair of AOs (such as 2s on atom A with 2s on atom B), overlap produces one bonding and one antibonding MO. There is σ and σ* for 1s – 1s, 2s – 2s, and 2pz – 2pz overlaps. There is π and π* for 2px – 2px, and 2py – 2py overlaps. These ideas result in the MO energy-level diagram shown here:
From the ten AOs (five from each F atom), ten MOs are formed in F2. Note that whether a MO is bonding or antibonding is dependent on whether it is lower or higher in energy than the AOs from which it is derived. Hence, even though σ*₁ₛ is lower in energy than σ₁ₛ, it is still an antibonding MO.
We also see in Figure 1 that the two π bonding MOs, π₂ₚₓ and π₂ₚy, are degenerate (have the same energy). This is because the side-by-side overlap of two 2px AOs is identical to the side-by-side overlap of two 2py AOs. They differ only in that π₂ₚₓ and π₂ₚy are perpendicular to each other, because the 2px AO is perpendicular to the 2py AO. Similar reasoning leads to the conclusion that the π*₂ₚₓ and π*₂ₚy MOs are also degenerate. Recognizing degenerate MOs is important when applying Hund’s rule to determine molecular electron configurations.
Figure 2 shows the formation of the two perpendicular π bonds as two N atoms approach each other. (The two molecular depictions in the figure represent the same N2 molecule: one shows the 2px – 2px orbital overlap and the other shows the 2py – 2py orbital overlap.)
Query \(5\)
Figure 2. A graph of energy versus distance between two nitrogen atoms. Boundary surface plots are shown for the two degenerate π bonding MOs formed from the in-phase overlap of the 2px AOs and the in-phase overlap of the 2py AOs. To watch the bond form, place the slider all the way to the right and then move it step-by-step toward the left.
Exercise 3: Molecular Orbital Electron Configurations
Additional Practice
Query \(7\)
Based on the electron configuration for the F2 molecule, the π and π* MOs are all filled; there is no net π bond in the molecule. This is reflected in the bond order calculation: F2 has a bond order of 1, corresponding to a single σ bond.
D6.2 Bond Length and Bond Enthalpy
Formation of a stable covalent molecule involves sharing electrons between two or more nuclei (the electrons occupy molecular orbitals that increase electron density between nuclei). The bond length between any two adjacent nuclei in such a covalent molecule is the distance between the two nuclei at the minimum energy in a graph of energy versus nuclear separation. For example, the bond length in a H2 molecule is 74 pm, as shown in Figure 3.
Bond enthalpy (also called bond energy) is the enthalpy change when a chemical bond is broken; that is, when two bonded atoms are completely separated. For example, Figure 3 shows that the bonded hydrogen atoms have energy of −436 kJ/mol relative to the separated hydrogen atoms. This means that the energy of the molecule must be increased by 436 kJ/mol to separate the atoms (break the bond). So the bond enthalpy for H2 is 436 kJ/mol.
Exercise 4: Bond Enthalpy
Lengths of single bonds can be roughly estimated by using the covalent radii of the bonded atoms. For example, adding the covalent radius of C (77 pm) to that of O (74 pm), estimates the length of a C—O bond to be 151 pm. This is quite close to the average C—O bond length of 143 pm. (Both of these values are estimates because covalent radii and bond lengths are averaged over many molecules and therefore are not exact for any specific molecule.)
In general, the bigger the atoms are, the longer the bond between them is. For example, consider the following trend in the average C-X bond lengths, where X is a halogen:
C-F 141 pm
C-Cl 176 pm
C-Br 191 pm
C-I 210 pm
Bond lengths are also dependent on bond order. For example, the C—C single bond has an average length of 154 pm, while a C=C double bond is 134 pm long, and a C≡C triple bond has an average length of 121 pm.
Exercise 5: Bond Length and Atomic Radius
Comparisons of the average bond length and bond enthalpy values show a general trend: a covalent bond with a shorter bond length generally has a larger bond enthalpy.
D6.3 Bonding in Molecules with More Than Two Atoms
Molecules with three or more atoms have molecular orbitals that span the entire molecule. The MOs are derived from the overlap of AOs from all the atoms in the molecule. Both the MO wave functions and the structure of the energy-level diagram are much more complicated than for diatomic molecules, but mathematical techniques exist for calculating and displaying the electron densities that form chemical bonds. In this course we will not delve deeply into these more complicated cases except to make several general points:
• The number of MOs for a molecule equals the number of AOs on the atoms that make up the molecule.
• The energies of the MOs increase as the number of nodes in the MO increases.
• MOs can extend over the entire molecular structure; they are not necessarily confined to pairs of atoms.
While MOs provide accurate physical information about the molecule, such as energies of the valence electrons involved in a reaction, their visualization does not always provide easy chemical understanding. It is possible to “re-combine” the MOs in such a way that the electron densities are displayed as being localized between pairs of atoms or on individual atoms; this allows us to correlate MO derived electron densities with the more familiar Lewis structures, which represent electrons in chemical bonds as lines between pairs of atoms and electrons on a single atom as dots. We will discuss Lewis structures in more depth on Day 7.
Exercise 6: MOs for Polyatomic Molecules
Some of the molecular orbitals for a water molecule are shown here. Based only on what you know about the appearance of bonding and antibonding orbitals, rank these MOs from lowest-energy to highest-energy. (Click on each image for a rotatable 3D view of the MO.)
Query \(10\)
Activity 1: Reflection
Stop to think about molecular orbitals and bonding. Make a list of the main things you learned as you studied this topic both in today’s work and the previous day’s work. Your list should provide a summary you can use to review later for an exam.
D6.4 Lewis Structures for Covalent Molecules
Molecular orbitals for molecules with three or more atoms are complicated and hard to draw. Thus, although MOs would convey a more descriptive and accurate picture of electron distribution within a molecule, chemists often rely on simpler diagrams to depict the covalent bonding. It will aid your understanding of chemistry if you can connect these simpler diagrams mentally with the more complete picture given by MOs.
The most commonly used hand-drawn depiction is the Lewis structure:a diagram that represents atomic nuclei and core electrons by chemical symbols and valence electrons as dots or lines. A Lewis structure is built by combining Lewis diagrams (Section D3.4) of the constituent atoms.
Activity 2: Lewis Diagrams
Consider each element listed below. In your course notebook write the electron configuration for an atom of each element, determine the number of valence electrons, and write a Lewis diagram (Section D3.4). How are the Lewis diagrams related to the position of each element in the periodic table?
N C S As O Br F Si H
Query \(11\)
In Lewis structures, a single covalent bond is drawn as a pair of electron dots shared between two adjacent atoms, a bond pair. Valence electrons that are not in a bond are shown as pairs of dots associated with individual atoms, lone pairs. For example:
In the Cl2 molecule, each Cl atom has three lone pairs and the two Cl atoms share one bond pair. Hence, each Cl atom in Cl2 has formed an octet (is surrounded by eight valence electrons).
For simplicity and clarity, a bond pair is typically represented by a line instead of a pair of dots:
Activity 3: Lewis Structure and Electron Sharing
The Octet Rule
The octet rule states that atoms of main-group elements tend to gain, lose, or share enough electrons to form an octet (eight valence electrons). Such noble-gas electron configurations with completely filled valence shells are more stable, and therefore should correspond to how the electrons are arranged in a molecule.
The Lewis diagram for an atom can be used to predict the number of bonds the atom will form. For example, a carbon atom has four valence electrons and therefore requires four more electrons to reach an octet:
Query \(13\)
It is important to keep in mind that it is impossible to exceed an octet for atoms in the second period. This is particularly relevant because you will encounter numerous molecules containing the elements C, N and O.
Finally, because a hydrogen atom needs only two electrons to fill its valence shell, H is an important exception to the octet rule and forms only one bond.
Double and Triple Bonds
Two atoms may need to share more than one pair of electrons to achieve the requisite octet. In other words, the bond order is greater than 1. A double bond consists of two pairs of electrons being shared between two atoms. For example:
A triple bond forms when three pairs of electron are shared between two atoms. For example:
Activity 4: Double and Triple Bonds
Write answers to these questions in your course notebook:
Write a Lewis structure for N2 and a Lewis structure for O2. Describe the type of bond in each case.
Do the N atoms in N2 and the O atoms in O2 follow the rule for number of bonds in Exercise 7?
Use the molecular-orbital energy level diagram in Figure 1 to calculate the bond order for N2 and for O2. How do the bond orders relate to the Lewis structures?
D6.5 General Guidance for Drawing Lewis Structures
Often you can draw a Lewis structure based on the number of bonds formed by each kind of atom. in more complicated cases, here is a step-by-step procedure for drawing Lewis structures of molecules and polyatomic ions:
1. Determine the total number of valence electrons by summing the number of valence electrons on all atoms.
• For a polyatomic cation, subtract one electron for each positive charge.
• For a polyatomic anion, add one electron for each negative charge.
2. Choose one or more central atoms; a central atom bonds to several other atoms and is usually the atom that forms the greatest number of bonds.
• Usually the central atom is written first in a chemical formula, such as P in PCl3.
• If there are two or more central atoms, connect them using single bond lines.
3. Draw a skeleton structure of the molecule by arranging the other atoms (which are called terminal atoms) around the central atom or atoms.
• Connect terminal atoms to the central atom(s) by single bond lines.
4. Distribute the remaining electrons as lone pairs on the terminal atoms (except hydrogen), completing an octet around each atom.
5. If there are still valence electrons available, place them on the central atom(s).
6. If the number of electrons around a central atom is less than an octet, rearrange the electrons to make multiple bonds with the central atom(s) until each atom has an octet.
Let’s apply these rules to a simple molecule, ammonia, NH3.
Query \(16\)
Here is a more complicated case: ethene (ethylene), C2H4.
Query \(17\)
Exercise 8: Lewis Structures and Valence Electrons
If you were to build a Lewis structure for nitrate ion (NO3‾), how many electrons would you need to allocate in your structure? (In other words, how many non-core electrons have to be in your structure?)
Query \(18\)
Exercise 9: Identifying Incorrect Lewis Structures
For each Lewis structure, determine whether the structure is correct. If the structure is incorrect, identify the error made in the representation.
D6.6 Exceptions to the Octet Rule
Some stable covalent molecules contain one or more atoms that do not have an octet. Such molecules fall into three categories.
Odd-electron Molecules
A molecule that contains an odd number of electrons must have at least one electron unpaired and therefore must have an atom with fewer than eight electrons in its valence shell (typically it’s seven electrons). A molecule with at least one unpaired electron is called a free radical. Nitric oxide, NO, which is produced in internal combustion engines when oxygen and nitrogen react at high temperatures, is an example.
To draw the Lewis structure for an odd-electron molecule, follow the same steps as outlined previously, but recognize that an odd-electron molecule must have less than an octet on some atom. For example, the Lewis structure for NO is:
Forming a triple bond would cause either oxygen or nitrogen to exceed an octet, which is a very unlikely electron arrangement.
Molecules with an Incomplete Octet on a Central Atom
Some molecules contain a central atom that does not have a filled valence shell. Usually, these central atoms are from groups 2 (IIA) and 13 (IIIA). For example, the Lewis structure of beryllium chloride, BeCl2, shows beryllium with only four electrons, and that of boron trifluoride, BF3, shows boron with only six electrons.
It is possible to draw a structure for each of these molecules where there are double bonds to the central atom and therefore an octet. For example, BF3 with one B=F double bond would satisfy the octet rule. However, experimental evidence tells us that the bond lengths in BF3 are closer to B–F single bonds. The observed reactivity of BF3 is also consistent with less than an octet on boron: BF3 reacts readily with NH3, with the lone pair on nitrogen (red) forming a bond (red) that completes the octet on the boron atom:
Molecules with More Than an Octet on a Central Atom
Some Lewis structures show more than four pairs of electrons around the central atom. This usually happens when the central atom is in the third or a higher period (n ≥ 3). Molecules with more than an octet around a central atom are called hypervalent molecules. For example, in the Lewis structure for PCl5, the central phosphorus atom is surrounded by five pairs of electrons. In SF6, sulfur has six pairs of electrons.
Exercise 10: Exceptions to the Octet Rule
Day 6 Pre-class Podia Problem: Molecular Orbital Energy Levels
This Podia problem is based on today’s pre-class material; working through that material will help you solve the problem.
The molecule C2 is quite unstable and can only be isolated and studied experimentally in the gas phase. Molecular spectroscopy of C2 reveals that a C2 molecule in the ground state has no unpaired electrons.
1. Use the MO energy-level diagram in Figure 1 to predict the ground state electron configuration of C2. Does your result agree with the experimental observation?
2. Consider the MO diagram below. Is this a better diagram than Figure 1 for C2? Use scientifically appropriate language to explain which diagram is better.
Query \(21\)
Two days before the next whole-class session, this Podia question will become live on Podia, where you can submit your answer.
Comments. If you found any inconsistencies, errors, or other things you would like to report about this module, please use this link to report them. A similar link will be included in each day’s material. We appreciate your comments. | textbooks/chem/General_Chemistry/Interactive_Chemistry_(Moore_Zhou_and_Garand)/01%3A_Unit_One/1.06%3A_Day_6-_Molecular_Orbitals_Lewis_Structures.txt |
Day 7: Covalent Molecular Substances; Hydrocarbons
Applying Core Ideas: Comparing Hydrogen Molecules and Helium Atoms
The boiling point of helium is 4.22 K (−268.93 °C). The boiling point of hydrogen is 20.28 K (−252.87 °C). However, the attractive force between two hydrogen atoms 100 pm apart is almost 5000 times stronger than the attractive force between two helium atoms 100 pm apart.
Think about helium and hydrogen at the atomic scale. Then write in your notebook an explanation for the fact that both helium and hydrogen have very low boiling points but hydrogen’s is higher.
D7.1 Covalent Molecular Substances
A substance made of molecules is called a covalent molecular substance. An important point in the activity you just completed is this: unlike ionic compounds, metals, or noble gases, where boiling involves freeing ions or atoms from each other, boiling a covalent molecular substance involves freeing molecules from each other. No covalent bonds are broken during the boiling process and the same molecules are present in the gas phase as were present in the liquid phase. The same reasoning applies to melting: covalent bonds between atoms within molecules are not broken, but forces between the molecules must be partially overcome.
Because there are many different kinds of nonmetal atoms that can form covalent bonds, and because molecules can consist of anywhere from two to many thousands of atoms, the range of properties of covalent molecular substances is much broader than for ionic compounds or metals. Many covalent molecular substances are liquids or gases: they melt (and some boil) below room temperature or not too far above. Covalent molecular substances do not conduct electricity well as solids or liquids, the solids may be weak and brittle or soft and waxy, and many are insoluble in water. We will begin to explore this broad range of molecules and properties in Unit 2. For now, we consider a single class of covalent molecular substances: hydrocarbons.
D7.2 Hydrocarbons
Many properties of substances that consist of covalently bonded molecules are exemplified by hydrocarbons, compounds that contain only the elements carbon and hydrogen. Carbon is unique among the elements in that carbon atoms can form long chains of carbon-carbon bonds. This happens because carbon atoms form strong single bonds to other carbon atoms and because carbon’ has four valence electrons, resulting in four bonds per carbon atom. When all other bonds in a chain of carbon atoms are to hydrogen atoms, the molecule is a hydrocarbon. In addition to long chains, hydrocarbon molecules can have chains with branches; chains folded back on themselves to form rings; and chains, branched chains, or rings that include double or triple bonds. Many hydrocarbons are found in plants, animals, and their fossils; other hydrocarbons have been prepared in the laboratory.
The most important reaction of hydrocarbons is combustion. The simplest example is combustion of methane:
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) ΔrH = −802.3 kJ/mol
Combustion of hydrocarbons is highly exothermic so hydrocarbons are excellent fuels. For example, methane (CH4) is the principal component of natural gas, LP gas is mainly propane (C3H8) and butane (C4H10) . Other hydrocarbon fuels are acetylene, diesel fuel, jet fuel, and heating oil. The familiar plastics polyethylene, polypropylene, and polystyrene are also hydrocarbons.
D7.3 Alkanes
The simplest hydrocarbons, alkanes, contain only single bonds between carbon atoms. Each of the carbon atoms in an alkane is bonded to four other atoms, each of which is either carbon or hydrogen. The bond enthalpies for C–C and C–H bonds are 346 and 436 kJ/mol, respectively. These strong bonds are difficult to break so alkanes are relatively unreactive.
Alkanes that do not contain rings of carbon atoms are called acyclic alkanes; they have the general molecular formula CnH2n+2. A molecular formula, such C6H14 for the alkane named hexane, specifies how many atoms of each type are in the molecule; in C6H14 there are 6 C atoms and 14 H atoms in each molecule. These alkanes are also called saturated hydrocarbons because each C atom is bonded to the maximum possible number of H atoms. The alkane molecule is said to be “saturated” with hydrogen.
Query \(3\)
Figure 1. Pictured are the Lewis structures, ball-and-stick models, and space-filling models for molecules of methane, ethane, and pentane. Move the slider to see the various representations.
Unbranched acyclic alkanes are often called “straight-chain” alkanes because the carbon atoms can be drawn in a single, straight row in the Lewis structure. However, the Lewis structure represents only some aspects of the molecule, not all. As the ball-and-stick and space-filling models of pentane in Figure 1 show, the C atoms do not lie in a completely straight line. Why there is a zigzag row of atoms is a topic for Unit 2.
The structures of alkanes may also be represented by condensed structural formulas, such as CH3CH3 for ethane and CH3CH2CH2CH2CH3 for pentane. Condensed structural formulas indicate how many H atoms are bonded to each C atom; they are related to Lewis structures, but all the bond symbols have been removed.
Branched alkanes contain more than one carbon chain. For example, 3-methylpentane has a branch at the third carbon atom along the chain. (The C atoms are numbered with subscripts to show that the branch occurs at the third C atom. The number of the C atom with the branch appears in the name. “Methyl” refers to the group H3C—, which is a methane molecule lacking one H nucleus.)
A cycloalkane has at least one ring of carbon atoms. Comparing a linear alkane with a cycloalkane shows that an additional C—C bond must be formed, which removes two hydrogen atoms. Consequently, the general formula for a cycloalkane containing one ring is CnH2n.
We refer to this reduction in number of hydrogen atoms as degree of unsaturation, One degree of unsaturation corresponds to having two hydrogen atoms fewer; a cycloalkane containing one ring has one degree of unsaturation. As in the case of the zigzag “straight-chain” alkane, the planar Lewis structure does not represent all aspects of the cyclohexane structure. The ring in cyclohexane is puckered and does not lie flat in a plane, as shown in this 3D model. The H atoms also are not in the plane of the screen.
D7.4 Alkenes
Unsaturated hydrocarbons that contain one or more double bonds are called alkenes. The general molecular formula for alkenes with one double bond is CnH2n. The formula has two hydrogen atoms fewer than the corresponding alkane with same number of carbon atoms, and hence 1 degree of unsaturation. It is possible to have a ring of carbon atoms that contains a double bond. Cyclic alkenes have one degree of unsaturation from each cyclic structure and one from each C=C double bond.
The carbon atoms involved in the double bond are sp2 hybridized, and therefore the local geometry there is trigonal planar. The presence of these double bond(s) is signified by the suffix -ene in the name. Ethene, C2H4, commonly called ethylene, is the simplest alkene. Some other examples include:
The presence of the π bond makes alkenes much more reactive than alkanes because a carbon-carbon π bond is usually weaker and more easily disrupted than a σ bond. The double bond is therefore a functional group, a specific structure that has similar chemical behavior in every molecule where it occurs. For example, all alkenes can undergo a characteristic reaction, called an addition reaction, in which the π bond is broken and replaced by two additional σ bonds. Hydrogen and halogen molecules can undergo addition reactions with alkenes, for example:
In this reaction, the π bond in the C=C double bond and the Cl-Cl σ bond are broken, and two C-Cl σ bonds are formed. The σ bond in the C=C double bond remains intact throughout the reaction. The reaction can occur relatively easily because the π bond is not as strong as the σ bond—that is, because the C=C double bond (bond enthalpy = 598 kJ/mol) is not twice as strong as the C-C single bond (bond enthalpy = 346 kJ/mol).
Activity 1: Analyzing an Addition Reaction
Think about the reaction of chlorine with ethane, CH3CH3. Can this be an addition reaction? Explain why or why not. What bonds need to be broken and formed if chlorine reacts with ethane and how does the reaction differ from the reaction of chlorine with ethene shown above? Do you expect the reaction of chlorine with ethane to be faster or slower than the reaction with ethene? Why?
D7.5 Alkynes
An alkyne is a hydrocarbon with one or more carbon-carbon triple bonds. Two carbon atoms joined by a triple bond are bound together by one σ bond and two π bonds. The general molecular formula of an alkyne with one triple bond is CnH2n-2. The alkyne has four hydrogen atoms fewer than the corresponding alkane with same number of carbons, and hence 2 degrees of unsaturation.
The suffix -yne is used to indicate the presence of a triple bond. The simplest alkyne is ethyne, C2H2, commonly called acetylene. The Lewis structure for ethyne is:
Chemically, alkynes have reactivity similar to alkenes. Since the C≡C functional group has two π bonds, alkynes can react with twice as much reagent in an addition reaction. For example, acetylene can react with bromine in the following reaction:
D7.6 Petroleum Chemistry
Petroleum (from Latin, petra: “rock”, oleum: “oil”) consists primarily of naturally occurring hydrocarbons, predominantly alkanes and cycloalkanes. The alkane chains can be quite long, and properties such as melting point and boiling point usually vary smoothly and predictably as a function of the number electrons in various alkane molecules (Table 1).
Alkane Molecular Formula Number of Electrons Melting Point (°C) Boiling Point (°C) Phase at Room Temperature
methane CH4 10 –182.5 –161.5 gas
ethane C2H6 18 –183.3 –88.6 gas
propane C3H8 26 –187.7 –42.1 gas
butane C4H10 34 –138.3 –0.5 gas
pentane C5H12 42 –129.7 36.1 liquid
hexane C6H14 50 –95.3 68.7 liquid
heptane C7H16 58 –90.6 98.4 liquid
octane C8H18 66 –56.8 125.7 liquid
nonane C9H20 74 –53.6 150.8 liquid
decane C10H22 82 –29.7 174.0 liquid
tetradecane C14H30 114 5.9 253.5 solid
octadecane C18H38 146 28.2 316.1 solid
Table 1. Melting and boiling points of alkanes.
Petroleum is the main source of hydrocarbon fuels, such as LP gas, gasoline, and fuel oil. These are separated by fractional distillation, a process in which petroleum is boiled and the different hydrocarbons condense to liquids at different temperatures (Figure 2). Fractional distillation takes advantage of the boiling-point differences of the various component substances (Table 1). The different boiling points arise from the differences in the London dispersion forces between molecules.
Gasoline is a liquid mixture of linear and branched alkanes, each containing five to twelve carbon atoms. Gasoline contains various additives to improve its performance as a fuel. Kerosene, diesel fuel, motor oil, and fuel oil are primarily mixtures of alkanes made of larger molecules with more electrons than gasoline molecules. Provided there is plenty of oxygen available, combustion converts almost all the carbon in the alkane fuel to carbon dioxide and water. Thus combustion of alkanes invariably adds water vapor and CO2 to the atmosphere—a human contribution to global warming.
Because there is greater demand for gasoline than for other components of petroleum, catalytic cracking is used in petroleum refining to break up larger molecules into smaller molecules, some of which are within the gasoline range of 5–12 carbon atoms. Catalytic cracking involves temperatures of 480–550 °C and a catalyst—conditions that can break (crack) carbon-carbon bonds and rearrange molecular structures. The hydrocarbon molecules are broken up in a fairly random way to produce mixtures of smaller molecules, some of which have carbon-carbon double bonds. One possible reaction involving C15H32 might be:
C15H32 → 2 C2H4 (ethene) + C3H6 (propene) + C8H18 (octane)
The alkene products, ethene and propene, are important for producing other organic chemicals. The octane product is a component of gasoline. Note that catalytic cracking involves temperatures higher than fractional distillation as well as a catalyst in order to break carbon-carbon bonds (as opposed to overcoming LDFs between hydrocarbon molecules).
D7.7 Attractions Between Atomic-scale Particles
Consider the different ways atomic-scale particles attract other atomic-scale particles and how strengths of those attractions affect macroscopic properties. Noble-gas atoms have only weak London dispersion forces (LDFs) between them, leading to very low boiling points.
Metals consist of positive ions surrounded by valence electrons that are not associated with any specific positive ion (that is, with any specific metal-atom core). Attractions between metal atoms involve positive ions and electrons and therefore are much larger than attractions between noble-gas atoms. Metallic bonding becomes stronger as the number of valence electrons in the metal atom increases. Metal atoms are attracted strongly enough that most metals are solids at room temperature.
When an atom with low attraction for electrons (a metal atom with low ionization energy) approaches an atom with greater attractions for electrons (a nonmetal with large negative electron affinity), electron density can transfer from the metal atom to the nonmetal atom to form ions. Coulomb’s law attractions between ions are large and depend on the charges of the ions and the distance between the ions. This results in formation of ionic crystal lattices that require significant increase in temperature for melting or boiling.
When two nonmetal atoms approach, the typical result is a covalent chemical bond, although there are cases such as He2 where there are enough electrons to fill antibonding as well as bonding molecular orbitals and give a bond order of zero. The characteristics of covalent bonds depend on properties of the bonded atoms, such as size and number of electrons. Covalent bonds connect atoms to form molecules. The strengths of bonds in molecules are typically as large as or larger than the strengths of metallic bonds and ionic bonds. Thus, when a molecular substance melts or boils, atoms remain bonded and the atomic-scale particles in the liquid or gas phase are molecules.
For example, when NaCl melts, Na+ ions and Cl‾ ions break from the crystal lattice and move freely around each other. Breaking the ionic bonds requires significant energy, and hence NaCl has a high melting point of 801 °C. In contrast, when methane (CH4) melts, individual methane (CH4) molecules stay intact, but they can move freely around other methane molecules. The attractions between methane molecules must be partially overcome, but those attractions are not as strong as covalent bonds or ionic bonds. Hence, the melting point of methane is much lower than for NaCl. Methane melts at −182 °C.
Day 7 Pre-Class Podia Problem: Forces between Atomic-Scale Particles
This Podia problem is based on today’s pre-class material; working through that material will help you solve the problem.
Forces between atoms, ions, and molecules typically result in a graph of energy versus distance between particles that looks like this:
For each pair of substances listed below make a graph of energy versus distance between atomic-scale particles. Use a single set of coordinate axes for each pair. Then draw a curve like the one above for each substance in the pair. Your curves should be to scale relative to each other, but there is no need to put values on the tick marks along the axes. The relative depths of the minima and the relative positions of the minima along the horizontal axis are important.
1. NaCl and CaS
2. CH4 and C6H14
3. Cr and Cs
4. Ne and CH4
Query \(8\)
Two days before the next whole-class session, this Podia question will become live on Podia, where you can submit your answer.
Comments. If you found any inconsistencies, errors, or other things you would like to report about this module, please use this link to report them. A similar link will be included in each day’s material. We appreciate your comments. | textbooks/chem/General_Chemistry/Interactive_Chemistry_(Moore_Zhou_and_Garand)/01%3A_Unit_One/1.07%3A_Day_7-_Covalent_Molecular_Substances_Hydrocarbons.txt |
D9.1 Bonds, Molecules, and Structures
In a molecule, atoms are connected by covalent bonds that result from overlap of atomic orbitals. Molecules can consist of a small or large number of atoms and may involve all the same kind or a dozen different kinds of atoms. The atoms can be connected by bonds in a variety of different ways, so there is a broad range of different molecular structures. Thus, there are many things a chemist needs to know about a molecule:
• What kinds of atoms and how many of each are in the molecule?
• Which atoms are bonded to which other atoms?
• What are the bond lengths between atoms?
• How strong are the bonds?
• What are the angles between the bonds?
• How are the atoms arranged in three-dimensional space?
• What kinds of attractions are there between molecules?
• How strong are the attractions between molecules?
• Are there noncovalent attractive forces within a large molecule, holding one part of the molecule to another?
At this point we have introduced the first four of these, using molecular formulas, Lewis structures, atomic radii/bond lengths, and bond enthalpies. For example, in a water molecule, the molecular formula H2O indicates that there are two H atoms and one O atom. A Lewis structure, H–O–H shows that both H atoms are bonded to the O atom. The O–H bond length (94 pm) and bond enthalpy (467 kJ/mol) verify that the bonds are strong—separating the atoms is difficult.
But there is more to a water molecule than that. A ball-and-stick or space-filling model shows that the angle between the two O–H bonds is 104.5°—somewhat more than a right angle. Angles larger or smaller than 104.5° result in higher energy (lower stability). As a result of the shape and type of atoms in the water molecule, there are stronger forces between water molecules than between methane (CH4) molecules, although both contain the same number of electrons.
As the number of atoms in a molecule increases the last four factors in the list above become more and more important. You will learn about them throughout this Unit, beginning with additional properties of chemical bonds in the next sections.
D9.2 Bond Polarity
If the two atoms that form a covalent bond are identical, as in H2 or Cl2, then the electrons in the bond must be shared equally between the two atoms. In a pure covalent bond, shared electrons have an equal probability of being near each nucleus.
On the other hand, if the two atoms are different, they may have different attractions for the shared electrons. When the bonding electrons are attracted by one atom more than the other atom the bond is called a polar covalent bond. For example, in HCl, the Cl atom attracts the bonding pair of electrons more than the H atom does, and electron density of the H–Cl bond is shifted toward the chlorine atom. Quantum mechanics calculations show that the chlorine atom, which has 17 protons, has electron density equivalent to 17.28 electrons and therefore a partial negative charge, δ = −0.28. The hydrogen atom has a partial positive charge, δ+ = +0.28.
This unequal distribution of electron density on two bonded atoms produces a bond dipole moment, the magnitude of which is represented by µ (Greek letter mu). The dipole moment is equal to:
μ = Qr
where Q is the magnitude of the partial charges (for HCl this is 0.28 times the charge of an electron) and r is the distance between the charges (the bond length). Bond dipole moments are measured in units of debyes (D); 1 D = 3.336 × 10-30 coulomb meter.
The bond dipole moment has both direction and magnitude and can be represented as a vector (Figure 3). A dipole vector is drawn as an arrow, with the arrowhead pointing to the partially negative end, and a small + sign on the partially positive end. The length of the arrow is proportional to the magnitude of the dipole moment.
D9.3 Electronegativity
The polarity of a covalent bond is determined by the difference between the electronegativities of the bonded atoms. Electronegativity (EN) is the tendency of an atom in a molecule to attract bonding electron density. Thus, in a bond, the more electronegative atom is the one with the δ charge.
The greater the difference in electronegativity between two bonded atoms is, the larger the shift of electron density in the bond toward the more electronegative atom is. Greater electronegativity difference, greater Δ(EN), gives larger partial charges on the atoms. Electronegativity values for most elements are shown in the periodic table in Activity 1; they also are tabulated in the appendix.
Activity 1: Periodic Trends in Electronegativity
Electronegativity, Electron Affinity, and Ionization Energy
These three properties are all associated with an atom gaining/losing electrons. Electron affinity and ionization energy are experimentally measurable physical quantities.
Electron affinity (EA) is the energy change when an isolated gas-phase atom acquires an electron; it is usually expressed in kJ/mol.
X(g) + e → X(g) ΔE = EA1
Ionization energy (IE) is the energy that must be transferred to an isolated gas-phase atom to remove an electron; it is also typically expressed in kJ/mol.
X(g) → X+(g) + e ΔE = IE1
Electronegativity describes how strongly an atom attracts electron density in a bond. It is calculated, not measured, has an arbitrary relative scale, and has no units.
Activity 2: Electronegativity, Electron Affinity, and Ionization Energy
Electronegativity and Bond Type
The difference in electronegativity, Δ(EN), of two bonded atoms provides a rough estimate of polarity of the bond, and thus of the bond type. When Δ(EN) is very small (or zero), the bond is covalent and nonpolar. When Δ(EN) is large, the bond is polar covalent or ionic. (In a joined pair of ions, such as Na+Cl, there is nearly complete transfer of valence electrons from one atom to another to produce a positive ion and a negative ion. The Na+ and Cl ions form a dipole with δ+ approximately equal to +1 and δ approximately −1.)
Δ(EN) spans a continuous scale and serves as a general guide; there is no definitive cutoff that defines a bond type. For example, HF has Δ(EN) = 1.8 and is considered a polar covalent molecule. On the other hand, NaI has a Δ(EN) of 1.7 but forms an ionic compound. When considering the covalent or ionic character of a bond, you should also take into account the types of atoms involved and their relative positions in the periodic table. Bonds between two nonmetals are usually described as covalent; bonding between a metal and a nonmetal is often ionic.
Some compounds contain both covalent and ionic bonds. For example, potassium nitrate, KNO3, contains the K+ cation and the polyatomic NO3 anion, which has covalent bonds between N and O.
D9.4 Formal Charge
It is useful to consider how valence electrons are distributed in a molecule. Formal charge, the charge an atom would have if the electron density in the bonds were equally shared between the atoms, is one way to do this. For each atom in a Lewis structure, half of the electrons in bonds are assigned to the atom, and all lone-pair electrons (which are not shared with other atoms) are assigned to the atom.
An atom’s formal charge is calculated as the difference between its number of valence electrons (in the unbonded, free atom) and its assigned number of electrons in the molecule:
formal charge = number of valence e‾ – [number of lone-pair e‾ + ½(number of bond-pair e‾)]
• If the assigned number of electrons equals the number of valence electrons, the atom has zero formal charge.
• If the assigned number of electrons exceeds the number of valence electrons, the atom has a negative formal charge.
• If the assigned number of electrons is less than the number of valence electrons, the atom has a positive formal charge.
Because formal charge counts all valence electrons in a molecule, the sum of the formal charges of all the atoms in a molecule or ion must equal the actual charge of the molecule or ion.
The formal charge for any given atom is not the same as its actual partial charge, such as those calculated in Section D9.2 above. This is because formal-charge calculations assume all covalent bonds are nonpolar, which is seldom the case except for homonuclear molecules.
Using Formal Charge to Predict the Most Likely Lewis Structure
While formal charges do not portray the true electron density distribution within a molecule, they nonetheless account for electron arrangement in a Lewis structure in units of a whole electron. Therefore, if following the steps for drawing Lewis structures lead to more than one possible arrangement of electrons and/or atoms for a given molecule, formal charges can help to decide which arrangement is likely to be the most stable, and hence the most likely Lewis structure for the given molecule.
• For an uncharged molecule, a Lewis structure in which all atoms have a formal charge of zero is preferable.
• The fewer atoms with nonzero formal charges, the better.
• The smaller the magnitude of the formal charges, the better.
• A Lewis structure with formal charges of the same sign (both + or both −) on adjacent atoms is less likely.
• Lewis structures with negative formal charges on more electronegative atoms are preferable.
For example, consider these three possible Lewis structures of carbon dioxide, CO2:
All structures have octets on each atom, but the structure on the left is preferable because all atoms have zero formal charge. The structure on the right is least likely because of the larger formal charges.
D9.5 Resonance Structures
In a single Lewis structure, a pair of electrons can only be depicted as shared between two atoms or localized to a single atom. However, as mentioned in Section D6.3, the molecular orbitals of a polyatomic molecule often span the entire molecule. For example, such delocalized electron distributions in π bonds can have a direct effect on molecular properties and chemical reactivity. Therefore, it is important to be able to use Lewis structures to indicate electron delocalization.
For example, two Lewis structures can be drawn for the nitrite anion, NO2, both of which satisfy the guidelines for the best Lewis structure for NO2‾:
Note that in these two Lewis structures, each of the three atoms is in the same position. The difference is in the location of electrons. In other words, these two Lewis structures convey the idea that the π bond may be between left O and central N or between central N and right O.
If the NO2‾ molecule were correctly described by either one of the Lewis structures, we would expect one N-O bond to be longer than the other. However, experiments show that both bonds in NO2 are the same length. Moreover, the bonds are longer than a N=O double bond and shorter than a N-O single bond. Hence, neither Lewis structure is a correct depiction of the actual molecule, and the best representation of NO2 is an average of these two Lewis structures.
When the actual distribution of electrons is a weighted average of a set of Lewis structures, those Lewis structures are called resonance structures. The actual electronic structure of the molecule (the average of the resonance forms) is called a resonance hybrid. A double-headed arrow between Lewis structures indicates that resonance structures are being depicted:
A molecule does not fluctuate between resonance structures; rather, the actual electronic structure is always the weighted average of the resonance structures. In other words, a single Lewis structure is insufficient to correctly represent the molecule (a shortcoming of a simple diagram), and a set of resonance structures (a resonance hybrid) is a better representation of electron density distribution in the molecule.
In the specific case of NO2‾, the two resonance structures above are needed to correctly depict two π-bond electrons that are delocalized over the entire molecule (click on the image below for a rotatable 3D view of the π molecular orbital occupied by these two electrons):
The carbonate anion, CO32, provides another example of insufficiency of a single Lewis structure and the need for a set of resonance structures:
Experiments show that all three C–O bonds are exactly the same. In other words, the two electrons in the π bond are delocalized over the entire molecule, as opposed to being only between one oxygen atom and the carbon atom.
To summarize, in a single Lewis structure, bonding (σ or π) is always between two atoms. Hence, two or more Lewis structures are needed to properly describe a molecule with delocalized electrons (spread over three or more atoms). When drawing a set of resonance structures:
• Each resonance structure should have the same number of electrons.
• Total formal charge is a useful tool for checking the number of electrons.
• Between resonance structures, atom locations are fixed: only the electrons move.
• The skeleton structure of the molecule remains the same in all resonance structures.
• However, you can draw a set of resonance structures in any perspective. For example, you could also draw the CO32 resonance structures as
• Double-headed arrows between Lewis structures communicate that what is drawn is a set of resonance structures.
In NO2‾, the two major resonance structures contribute equally to the resonance hybrid. Similarly, the three major resonance structures of CO32 contribute equally to the resonance hybrid. However, it is possible for some structures in a resonance hybrid to be more important than others. For example, consider these three resonance structures of cyanate ion (OCN):
The atoms in each resonance structure have a full octet, but they differ in their formal charges. This implies that certain electron arrangements may be a bit more stable than others, and hence they do not contribute equally to the resonance hybrid.
From the formal-charge rules, we can estimate that the resonance structure on the right would contribute the least; that arrangement of electrons is the least stable of the three. The resonance structure on the left would contribute more than the middle structure because it has the -1 formal charge on the more electronegative O atom. (For OCN, high level quantum mechanics calculations show that the left structure contributes 61% to the resonance hybrid, the middle structure contributes 30%, and the right resonance structure contributes only 4%.)
D9.6 Aromatic Molecules
Benzene, C6H6, is representative of a large number of aromatic compounds. These compounds contain ring structures and exhibit bonding that must be described using resonance structures. The resonance structures for benzene are:
All six C-C bonds are equivalent and exhibit bond lengths that are intermediate between those of a C–C single bond and a C=C double bond.
The chemical reactivity of aromatic compounds differs from the reactivity of alkenes. For example, aromatic compounds do not undergo addition reactions. Instead, with the aid of a catalyst, they can undergo substitution reactions where one of the hydrogen atoms is replaced by a substituent: another atom or group of atoms. A substitution reaction leaves the delocalized double bonds intact.
D9.7 Valence Bond Theory
Lewis structures are easy-to-draw, planar representations of bonding in molecules. They help us to figure out and think about which atoms are bonded to which and whether bonds are single or multiple. However, by default, they do not represent the 3D geometry of a molecule, nor the molecular orbitals (MOs) that determine electron-density distributions.
You have probably used VSEPR to predict the 3D shapes of molecules. VSEPR involves counting electron regions (pairs) around a central atom, assuming that electron regions repel and stay as far apart as possible, and bonding terminal atoms to electron regions. VSEPR is often good at predicting the arrangement of bonds around an atom, and it is OK to use it to predict idealized linear, trigonal planar, and tetrahedral arrangements of bonds that you will encounter in this course, but VSEPR has significant limitations:
• VSEPR has little or no basis in modern quantum theory; you have just spent significant time studying quantum theory and we want you to be able to use that experience.
• It is often difficult to apply VSEPR to molecules described by two or more resonance structures (that is, molecules with delocalized electrons). Thus VSEPR makes it more difficult to understand many molecular structures—for example, structures of protein molecules.
• VSEPR assumes that lone pairs occupy more space than bond pairs, but there is no evidence, experimental or theoretical, to support that assumption; in fact, there is some evidence to the contrary.
• VSEPR assumes that all lone pairs are equivalent, but there is experimental evidence that they are not. For example, the two lone pairs in a water molecule do not have the same ionization energy and do not have equivalent probability distributions (Journal of Chemical Education 1987, Vol. 64, pp 124-128.).
• VSEPR often cannot explain relative bond angles. For example, why is the H-P-H angle in PH3 93.5° while the H-N-H angle in NH3 is 107.5°? (If the decrease in bond angle from the tetrahedral angle of 109.5° to 107.5° for NH3 is due to a “fatter” lone pair, why does the angle decrease so much more for the larger P atom? A “fatter” lone pair should be less likely to repel the other bonds because they are farther apart.)
For these reasons, VSEPR is a model that has limited applicability. In this course we will use a better model—valence bond theory—which is consistent with modern quantum theory, makes more accurate and more comprehensive predictions than VSEPR, and is a better basis for understanding more advanced bonding topics. If you want to, it is OK to use VSEPR to predict idealized shapes, but applying the ideas presented in this section and sections D10.1 through D10.6 will allow you to describe structures with delocalized electrons better and predict bond angles more accurately.
Valence bond theory is a model that focuses on the formation of individual chemical bonds, such as the formation of a σ bond between two atoms within a polyatomic molecule. Like molecular orbital theory, valence bond theory deals with how atomic orbitals (AOs) change and combine when a molecule forms, but instead of forming MOs that span the whole molecule, valence bond theory combines valence orbitals of each atom individually so that the combination gives stronger bonding in specific directions. Hence, valence bond theory allows us to derive idealized 3D geometries for molecules based only on their Lewis structures, without having to perform any computation.
Valence bond theory uses the extent of orbital overlap to infer the strengths of chemical bonds: greater overlap leads to bonds that are stronger and hence a molecule that is more stable. For a given atom in a molecule, overlap with orbitals on other atoms can be greater when some or all of the atom’s AOs form hybrid orbitals. Hybrid orbitals are combinations of valence atomic orbitals that emphasize concentration of electron density in specific directions. A hybrid orbital’s greater electron density in a specific direction provides greater overlap with an orbital from another atom when forming a σ bond.
For an example of how orbital hybridization works, consider combining a single 2s AO with a single 2p AO, both on the same atom (Figure 4). The 2s AO is spherically symmetric, so it has the same phase (mathematical sign) on either side of the nucleus, but the 2p AO changes sign at the nucleus. Thus, on one side of the nucleus, the 2s and 2p AOs are in phase, while on the other side they are out of phase.
If we add the two AOs, the new hybrid orbital will be larger on the side where the AOs are in phase and smaller on the other side where the AOs are out of phase. If we subtract them, the resultant hybrid orbital will be larger on the side where the AOs are out of phase and smaller where they are in phase. Hence, from one 2s AO and one 2p AO, we can derive two sp hybrid orbitals.
Activity 3: Orbital Hybridization
Query \(9\)
Day 9 Pre-class Podia Problem: Covalent Bonds
1. Consider these chemical bonds:
C–H C=C C–C C–Br C–F
Choose a pair of bonds from the list, predict which is longer, and write an explanation of your prediction.
Choose a pair of bonds from the list, predict which is stronger, and write an explanation of your prediction.
Choose a pair of bonds from the list, predict which is more polar, and write an explanation of your prediction.
2. NO is a molecule with an odd number of electrons. Write a Lewis structure for NO. Are there resonance structures? Is one resonance structure more dominant than another? If so, identify the more dominant structure and explain why it is more dominant.
Query \(10\)
Two days before the next whole-class session, this Podia question will become live on Podia, where you can submit your answer.
Comments. If you found any inconsistencies, errors, or other things you would like to report about this module, please use this link to report them. A similar link will be included in each day’s material. We appreciate your comments. | textbooks/chem/General_Chemistry/Interactive_Chemistry_(Moore_Zhou_and_Garand)/02%3A_Unit_Two/2.01%3A_Day_9-_Bond_Properties_Valence_Bond_Theory.txt |
Day 10: Hybrid Orbitals; Molecular Geometry
Valence bond theory and hybrid orbitals were introduced in Section D9.7. The ideas summarized here will be developed further in today’s work:
• Hybrid orbitals are derived by combining two or more atomic orbitals from the valence shell of a single atom.
• AOs are the most stable arrangement of electrons in isolated atoms.
• Hybrid orbitals are important in molecules because they result in stronger σ bonding.
• Most σ bonds are formed from overlaps of hybrid orbitals. Most π bonds are formed from overlap of unhybridized AOs.
• The number of hybrid orbitals equals the number of valence AOs that were combined to produce the hybrid orbitals.
• The type of hybrid orbitals for each bonded atom in a molecule correlates with the local 3D geometry of that atom.
• The type of hybrid orbitals for each atom can be determined from the Lewis structure (or resonance structures) of a molecule.
• From the local 3D geometry of each atom, we can obtain the overall 3D geometry of the molecule.
D10.1 Types of Hybrid Orbitals
sp Hybrid Orbitals
Combining the valence s AO with one of the valence p AOs yields two degenerate sp hybrid orbitals, as shown in Figure 1 for the case of 2s and 2p AOs. The two sp hybrid orbitals are oriented at 180° to each other—a linear geometry. After the hybridization, there are two unhybridized 2p AOs left on the atom.
Because these hybrid orbitals are formed from one s AO and one p AO, they have a 1:1 ratio of “s” and “p” characteristics, hence the name “sp“. One of the ways in which the hybrid orbitals exhibit their mixed “s” and “p” characteristics is in their energy. Specifically, the sp hybrid orbitals’ relative energies are about half-way between the 2s and 2p AOs, as illustrated in Figure 1.
Query \(1\)
Figure 1. Hybridization of the 2s and one of the 2p AOs forms two sp hybrid orbitals, oriented 180° with respect to each other; the two other 2p AOs remain unhybridized. (Move the slider to see the before/after of hybridization).
The hybridized orbitals are not energetically favorable for an isolated atom. For example, a beryllium atom is lower in energy with its two valence electrons in the 2s AO than if the electrons were in the two sp hybrid orbitals. However, in a covalent molecule, the one large lobe of each sp hybrid orbital gives greater overlap with another orbital from another atom, yielding σ bonds that lower the molecule’s energy.
sp2 Hybridization
Combining one valence s AO and two valence p AOs produces three degenerate sp2 hybrid orbitals, as shown in Figure 2 for the case of 2s and 2p AOs. The three sp2 hybrid orbitals are oriented at 120° with respect to each other and are in the same plane—a trigonal planar (or triangular planar) geometry. Each hybrid orbital is pointed toward a different corner of an equilateral triangle. After hybridization, there is one unhybridized 2p AO left on the atom.
The sp2 hybrid orbitals have twice as much “p” character as “s” character; this is indicated by the superscript “2” in sp2. Energetically, sp2 hybrid orbitals lie closer to the p AO than the s AO, as illustrated in Figure 2 (the sp2 hybrid orbitals are higher in energy than the sp hybrid orbitals).
Query \(2\)
Figure 2. Hybridization of the 2s and two of the 2p AOs forms three sp2 hybrid orbitals, oriented 120° with respect to each other in the same plane; one of the 2p AOs remain unhybridized (move the slider around to see the before/after of hybridization).
sp3 Hybridization
Combining one valence s AO and all three valence p AOs produces four degenerate sp3 hybridized orbitals, as shown in Figure 4 for the case of 2s and 2p AOs. The four sp3 hybridized orbitals are oriented at 109.5° with respect to each other, each pointing toward a different corner of a tetrahedron—a tetrahedral geometry.
A tetrahedron is a three-dimensional object that has four equilateral triangular faces and four apexes (corners). All four corners are equivalent. See Figure 3.
An sp3 hybrid orbital has 75% “p” character and 25% “s” character, a 3:1 ratio, hence the superscript “3” in its name. The sp3 hybrid orbitals are higher in energy than the sp2 hybrid orbitals, as illustrated in Figure 4.
Query \(3\)
Figure 4. Hybridization of the 2s and all three 2p AOs forms four sp3 hybrid orbitals, oriented 109.5° with respect to each other (move the slider around to see the before/after of hybridization).
Formation of a σ bond
When a σ bond forms between two atoms, a hybrid orbital with one unpaired electron from one atom overlaps with a hybrid orbital with one unpaired electron from the other atom. The resulting σ bond is an orbital that contains a pair of electrons (just as a line in a Lewis structure represents two electrons in a σ bond). For example, Figure 5 shows the formation of a C-C σ bond from two sp3 hybridized carbon atoms.
The σ bond thus formed by two hybrid orbitals (valence bond theory) is similar to a σ bond formed in a diatomic molecule as described by MO theory (Section D5.5). In polyatomic molecules with more than three atoms, the MOs are not localized between two atoms like this, but in valence bond theory, the bonds are described individually, between each pair of bonded atoms. Notice that in either MO or valence bond theory, the σ bond has a cylindrical symmetry with respect to the bonding axis.
If a hybridized orbital on an atom in a molecule has two electrons but is not pointing at another atom, the filled hybrid orbital is not involved in bonding. This corresponds to a lone pair on an atom in a Lewis structure.
D10.2 Predicting the Geometry of Bonds Around an Atom
Query \(4\)
Once you have drawn the best Lewis structure (or a set of resonance structures) for a molecule, you can use the structure(s) to assign hybridization to each atom, predict the geometric arrangement of bonds around each atom, and then predict the 3D structure for the molecule. This and the next few sections explain how this works.
For each atom in a molecule, determine the number of AOs that are hybridized, nhyb, and use this value to predict hybridization.
• Count the number of σ bonds (nσ) the atom forms.
• A double (or triple) bond contains 1 σ bond and 1 (or 2) π bond(s).
• Are there any lone pairs on the atom?
• If yes: nhyb = nσ + 1
• If no: nhyb = nσ
• Is an atom’s nhyb different in one resonance structure from another?
• If yes, use the smaller nhyb to determine hybridization.
• Use the value of nhyb to determine the number of AOs combined and hence the type of hybridization:
• For nhyb = 2, the atom is sp hybridized (two AOs are combined);
• for nhyb = 3, the atom is sp2 hybridized (three AOs are combined);
• for nhyb = 4, the atom is sp3 hybridized (four AOs are combined);
• An H atom in a molecule has nhyb = 1. It is not hybridized; its electron is in the 1s AO when forming a σ bond.
These rules derive from the idea that hybridized orbitals form stronger σ bonds. Therefore, the more σ bonds to an atom, the more atomic orbitals are combined to form hybrid orbitals.
Activity 1: Molecular geometry of BeCl2
Query \(5\)
Activity 2: Molecular geometry of BF3
D10.3 Three-dimensional Bond Geometry
The two examples so far were a linear (one-dimensional) molecule, BeCl2, and a planar (two-dimensional) molecule, BF3. What happens when a molecule is three dimensional?
Activity 3: Molecular geometry of CH4
Query \(7\)
Activity 4: Molecular geometry of NH3
Query \(8\)
Wedge-dash Notation
The Lewis structures in the activities above are drawn using wedge and dash notation.Straight lines represent bonds in the plane of the page/screen, solid wedges represent bonds coming toward you out of the plane, and dashed wedges represent bonds going away from you behind the plane.
Examine this 3D model of NH3 and rotate it until it looks like the Lewis structure drawn in the answer in Activity 4. The NH3 molecule has trigonal pyramidal geometry because the lone pair on nitrogen occupies one of the corners of a tetrahedron, leaving the three N-H bonds occupying the other three corners; this gives a three-cornered pyramid.
Redraw the Lewis structure you drew for ammonia in Activity 4 using wedge-dash notation. Then, rotate the 3D model until it matches your drawing. If you can find an orientation that matches, your wedge-dash Lewis structure is probably correct; if you cannot find a match, your Lewis structure is probably incorrect.
For simplicity, a wedge-dash Lewis structure draws as many as possible of a molecule’s bonds in a plane. However, as is the case with CH4 and NH3, most molecules do not have all bonds in the same plane.
D10.4 Molecules with More Than One Central Atom
Larger molecules have more than one “central” atom with several other atoms bonded to it. The arrangement of bonds for each central atom can be predicted as described in the preceding sections. The way these local structures are oriented with respect to each other influences the overall molecular shape.
Activity 5: Molecules with Several Central Atoms
Double and Triple Bonds
The Lewis structure of ethene, C2H4, shows that each carbon atom is surrounded by one other carbon atom and two hydrogen atoms:
Each carbon atom has nhyb = 3 and therefore is sp2 hybridized. Around each C atom there are three bonds in a plane. Two of the sp2 orbitals form two C–H σ bonds and the third sp2 orbital forms a C-C σ bond.
The double bond between the two C atoms contains a π bond as well as a σ bond. The π bond results from overlap of the unhybridized 2p AO on each carbon atom. The unhybridized 2p AO is perpendicular to the plane of the sp2 hybrid orbitals (Figure 6). Thus when the 2p AOs overlap in a side-by-side fashion to form a π bond, the electron densities in the π bond are above and below the plane of the molecule (the plane containing the σ bonds).
If the plane containing the sp2 hybrid orbitals of one carbon atom were rotated 90° relative to the other carbon, the two 2p AOs would also be rotated 90° to each other (Figure 7). The 2p AOs would no longer be able to overlap and the π bond cannot form.
This is a significant difference between σ and π bonds: one atom rotating around the internuclear axis with respect to the other atom does not change the extent to which the σ bonding orbitals overlap because the σ bond is cylindrically symmetric about the bond axis (see Figure 5); in contrast, rotation by 90° about the internuclear axis breaks the π bond entirely because the p orbitals can no longer overlap.
In acetylene, H−C≡C−H, each carbon atom has nhyb = 2 and therefore is sp hybridized with two unhybridized 2p orbitals. One sp hybrid orbital from each C atom overlaps to form a C-C σ bond, the other sp hybrid orbital forms a C-H σ bond with a hydrogen atom. The unhybridized 2p AOs overlap to form two perpendicular C-C π bonds (Figure 8). The two carbon atoms of acetylene are thus bound together by one σ bond and two π bonds, giving a triple bond.
Query \(10\)
Figure 8. In the C2H2 molecule, there are two C–H σ bonds and a C≡C triple bond. the triple bond involves one σ bond and two π bonds. The two π bonds are perpendicular to each other. Slide back and forth to see all the bonds; the molecule is fixed in the same perspective throughout.
Because π bonds are formed from unhybridized p AOs, an atom that is involved in π bonding cannot be sp3 hybridized.
Exercise 1: Identifying Hybridization
Activity 6: Predicting Structure of a Molecule with Several Central Atoms
D10.5 Hybridization and Bond Angles
Think back to the example molecules CH4 and NH3 in Section D9.4. Both involve sp3 hybridized orbitals on the central atom. In the case of CH4, a 1s orbital on each of the four H atoms overlaps with each of the four sp3 hybrid orbitals to form four bonds. All angles between pairs of C–H bonds are 109.5°. In NH3 the situation is different in that there are only three H atoms. Three of the four sp3 hybrid orbitals form three bonds to H atoms, but the fourth sp3 hybrid orbital contains the lone pair. The lone pair is different from the H atoms, and this is important.
In earlier sections we described each of a set of four sp3 hybridized orbitals as having ¼ s character and ¾ p character. This is what happens in CH4. Each of the four C–H bonds involves a hybrid orbital that is ¼ s and ¾ p. Summing over the four bonds gives 4 × ¼ = 1 s orbital and 4 × ¾ = 3 p orbitals—exactly the number and type of AOs from which the hybrid orbitals were formed.
In NH3, however, three of the four sp3 hybrids form bonds to H atoms and the fourth involves a lone pair. In this and similar situations, the partial s and p characters must still sum to 1 and 3 but each hybrid orbital does not have to be the same as all the others. That is, a hybrid orbital forming an N–H bond could have more p character (and less s character) compared to the hybrid orbital involving the lone pair. The only requirement is that the total s character and the total p character, summed over all four hybrid orbitals, must be one s and three p.
A different ratio of s character and p character gives a different bond angle. For example, in sp2 hybridized orbitals (with one-third s character and two-thirds p character) the angle between bonds is 120°, whereas, for sp3 the angle is 109.5°. More p character results in a smaller bond angle. (This makes sense, because for the maximum p character, that is, for two unhybridized p orbitals, the bond angle would be 90° because the p orbitals are at 90°.)
How can you tell how much s character and how much p character is in a specific hybrid orbital? A MO-theory calculation can provide this information, but, for our purposes, a qualitative rule that indicates where there will be more p character is sufficient. Bent’s rule says that a hybrid orbital on a central atom has greater p character the greater the electronegativity of the other atom forming a bond. A lone pair is assigned zero electronegativity because there is no atom attracting electrons in the bond away from the central atom.
Applying Bent’s rule to NH3, the three bonded H atoms have higher electronegativity than the lone pair (no atom) so we expect more p character in the hybrid orbitals that form the bond pairs. Thus, the angle between any two N–H bonds should be less than the tetrahedral angle. The experimentally measured angle is 106.7°, a bit less than the expected 109.5°.
Activity 7: Molecular geometry of H2O
Query \(13\)
In the H2O molecule, two of the O’s sp2 hybrid orbitals are involved in forming the O-H σ bonds. One of O lone pairs is in the other sp2 hybrid orbital; the other O lone pair is in the unhybridized 2p AO. The overall molecular geometry is bent. If O had perfect sp2 hybridization, the H-O-H angle would be 120°, but because the three hybrid orbitals are not equivalent, the angle deviates from ideal. To obtain an accurate bond angle requires an experiment or a high-level MO calculation.
D10.6 Hybridization in Resonance Hybrids
The assignment of hybridization and molecular geometry for molecules that have two or more major resonance structures is similar to the process discussed above, but remember that a set of resonance structures describes a single molecule. Hence the hybridization (and molecular geometry) assigned to one resonance structure must be the same as all other resonance structures in the set. Hence, when assigning hybridization, you should consider all the major resonance structures.
Exercise 2: Molecular Structure of Resonance Hybrid
Consider Figure 9:
The delocalized π MO extends over the oxygen, carbon, and nitrogen atoms. One of the three AOs contributing to this π MO is an unhybridized 2p AO on the N atom. This is also described by the set of resonance structures, where there is double-bond character between O and C and between C and N. Therefore the nitrogen atom must have sp2 hybridization (it forms three σ bonds) and a trigonal planar local geometry. This is more obvious when looking at the right resonance structure. When looking at the left resonance structure, you might be tempted to assign sp3 hybridization to N given its similarity to ammonia (NH3). However, this is a resonance structure; the set of resonance structures describes a molecule that cannot be described correctly by a single Lewis structure. Hence, the lone pair on N in the left resonance structure is in an unhybridized 2p AO.
All atoms must remain in the same positions from one resonance structure to another in a set of resonance structures. There cannot be a N atom that is trigonal pyramidal in one resonance structure and trigonal planar in another resonance structure, because the atoms attached to the N would have to change positions. Experimental evidence and high-level MO calculations show that formamide is a planar molecule.
Exercise 3: Molecular Structure of Resonance Hybrid
Day 10 Pre-class Podia Problem: Three-dimensional Models and Lewis Structures
Here are three links to 3-D models of molecules. For each molecule rotate the model to observe the structure. Then draw three 3-D Lewis structures of each molecule, using wedge and dash notation. Each wedge-dash structure should be viewed from a different perspective. If there are any lone pairs and/or formal charges, be sure to include them.
Glycine is an amino acid, a component of protein molecules.
Acrolein is used to kill algae and weeds in irrigation ditches and other natural waters.
Methyl formate is used mainly in the manufacture of other chemicals.
Query \(16\)
Two days before the next whole-class session, this Podia question will become live on Podia, where you can submit your answer.
Comments. If you found any inconsistencies, errors, or other things you would like to report about this module, please use this link to report them. A similar link will be included in each day’s material. We appreciate your comments. | textbooks/chem/General_Chemistry/Interactive_Chemistry_(Moore_Zhou_and_Garand)/02%3A_Unit_Two/2.02%3A_Day_10-_Hybrid_Orbitals_Molecular_Geometry.txt |
D11.1 Line Structures
Think about the importance of structures of covalent molecular substances. Hydrocarbons (Section D7.2) or closely related molecules are good examples because there are many different hydrocarbon molecules and many hydrocarbons have lots of atoms. Knowing which atoms are bonded to which and the different ways they are bonded enables predictions about the properties of the corresponding compounds.
For example, molecules of hydrocarbons in motor oil for automobiles contain from 16 to 20 carbon atoms and more than twice as many associated hydrogen atoms. Drawing Lewis structures for such large molecules takes time and can be overly complicated, so a simpler kind of drawing, called a line structure, is often used. In a line structure (also called a skeletal structure) for a hydrocarbon, only the C–C bonds are shown; element symbols and C–H bonds are omitted. Carbon atoms are represented by the end of a line or a juncture between two lines. Figure 1 shows an example line structure. Decide where each C atom and each H atom is in the line structure; then count the number of C and H atoms to verify that the line structure represents the same molecule as the Lewis structure.
Query \(1\)
Figure 1. The molecule 3-ethyl-2-methylhexane can be represented as a full Lewis structure or a line structure. Move the slider to see either representation. Note that in the full Lewis structure, the methyl and ethyl groups are stretched out a bit to accommodate drawing all the bonds in those areas.
Line structures are much easier to draw, and the general molecular structure of the carbon chain or chains is clearer. However, it is also much easier to make mistakes when drawing and reading line structures: an extra line is a whole extra CHn group and it is also easier to miss some of the implied C-H bonds.
Exercise 1: Interpreting Line Structures
In line structures, atoms other than carbon and hydrogen are represented by their elemental symbols, and any H atoms bonded to them are shown explicitly (for example, see molecule C in the exercise above). In line structures, lone pairs on atoms are often omitted as well. Because C-H bonds and lone pairs are not shown, it is necessary to include the formal charge on any atom that has a nonzero formal charge when drawing a line structure. Otherwise, there would be a miscount of bonds or electrons.
D11.2 Isomeric Structures
When more than one molecular structure corresponds to the same molecular formula the two or more structures are called isomeric structures, or isomers. For example, there are two structures corresponding to the formula CHN. They are H–C≡N: (hydrogen cyanide) and H–N≡C: (hydrogen isocyanide). At room temperature and especially at lower temperatures, both structures exist as separate compounds. That is, the substance represented by one structure can be purified and separated from the other and each substance has different properties.
When molecular formulas are different, the formulas must describe substances with different properties because the formula tells how many atoms of each kind are in a molecule and chemical bonds would have to be broken to change the number or atoms or type of atoms. Breaking covalent bonds requires energy, and at room temperature very few molecules have enough energy for bond breaking to occur. Isomers occur when two different molecular structures with the same number of atoms of each type do not have sufficient energy to change from one structure to the other. For example, H–N≡C: does not change into H–C≡N: because the change requires breaking a N–H bond. At room temperature the H–N≡C: molecules do not have enough energy to overcome the bond enthalpy of the N–H bond, so they do not change; thus, H–N≡C: can be purified and separated from H–C≡N:.
Typically isomers are defined by whether they can interchange at room temperature. At room temperature molecules do not have sufficient energy to break most covalent bonds. Thus, if interchanging two structures requires breaking a covalent bond, the two structures are isomers. If the temperature is raised, the molecules have greater average energy and it becomes more likely that one structure can change into another. If the temperature is lowered, it becomes harder to interchange structures; that is, the average energy of the molecules is lower and structures that could interchange at room temperature no longer have sufficient energy.
D11.3 Conformations
At room temperature, molecules are in constant motion both with respect to other molecules and as a result of internal motions such as rotation of one part of a molecule relative to another part of the same molecule. Below is an animation of a butane molecule at room temperature. Notice that there is rotation around the central C–C single bond (which is kept stationary in the animation to show the rotation better) and also around the other C–C bonds. Such rotations in a molecule lead to different conformations or conformers, structures that differ only because of rotations around single bonds.
A link to an interactive elements can be found here.
Figure 2. Rotation around the C-C bonds in butane leads to various conformations of the molecule. The C-C bond in the center of the molecule is kept in the same position in this animation so that rotations around it are more obvious, but rotation occurs around all three C-C bonds. To see the rotations more clearly. stop the animation and drag the slider at the bottom. You can change the perspective from which you view the molecule by holding the mouse button down and moving the mouse. (Animation by Michael Aristov.)
Figure 2a. Rotation around the C-C bonds in butane leads to various conformations of the molecule. Here four conformations are projected onto a plane and drawn as wedge-dash structures. You can change the perspective from which you view the molecule by holding the mouse button down and moving the mouse. (Animation by Michael Aristov.)
Activity 1: Depicting Conformers
Query \(4\)
Rotations around C–C bonds require little energy because the molecular orbital of a σ covalent bond has cylindrical symmetry along the internuclear axis (Section D5.5, Section D10.1). Cylindrical symmetry means that regardless of how two σ-bonded atoms are rotated with respect to each other around the internuclear axis, the σ bond remains unbroken between them. For example, consider the rotation around the C-C bond in 1,2-dichloroethane:
Figure 3. Rotation around the C-C bond in 1,2-dichloroethane. Energy (indicated by the readout and the green dot on the graph) varies with angle of rotation. Expanding the animation to full screen makes the graph easier to see. You can pause the video and adjust to various angles of rotation to find the corresponding energy. (Animation by Michael Aristov.)
Notice in Figure 3 that complete rotation around the C–C bond requires an energy change of 40 kJ/mol. The bond enthalpy of a C–C bond is 346 kJ/mol, so the energy required for rotation is a bit more than one-tenth the energy required to break the bond. Because the energy required for rotation around single bonds is small, such rotations occur readily at room temperature for most molecules. Hence, at room temperature, one conformer cannot be isolated from another and chemists consider that conformers represent the same chemical compound, with the same name and the same physical properties. No bond breaking is needed to go from one conformer to another.
If you could see all molecular structures in a room-temperature sample of 1,2-dichloroethane at a specific instant, you would find all three of these structures:
Each structure corresponds to one of the minima in the energy curve in Figure 3. If, instead, you followed one molecule over time, you would see it go from one structure to the next as the C-C bond rotates (as seen in Figure 3). Because experiments are typically done at room temperature, we consider all three structures as conformers of the same molecule. You can draw any of them as a representation of 1,2-dichloroethane.
From this animation of a nonane molecule you can see that different conformers of the same molecule can adopt, at a first glance, dramatically different molecular shapes. If you change your viewpoint by moving the molecule with your mouse, you can see that the molecule also looks quite different from different perspectives.
Given two or more Lewis structures, it is important that you can recognize whether they are conformers or isomers. Lewis structures or line structures that look quite different may be different conformers or the same conformer drawn from different perspectives; in either case, the structures represent the same substance.
A good way to tell whether two structures represent the same substance is to work out the correct name of each structure. If the names are the same, the structures represent conformers, not different substances. You will not be explicitly tested on naming compounds (nomenclature) in this course, but we strongly recommend that you study the section about alkanes in the appendix IUPAC nomenclature. IUPAC names are a systematic way of naming chemical compounds that can also help you when thinking about molecular structure. Going forward, if you encounter a name for a chemical compound that’s unfamiliar to you, the IUPAC nomenclature appendix can help.
Activity 2 Identifying Conformers
Query \(5\)
Chemically, different conformers can affect the reactivity of a molecule. And on a larger scale, huge protein molecules adopt only a few of many possible conformations; the shapes of protein molecules are essential to their biological functions.
D11.4 Constitutional Isomers
Compounds with the same molecular formula but different atomic connectivity are called constitutional isomers (or structural isomers). For example, there are two alkanes with the formula C4H10:
The n– in n-butane stands for normal, meaning an unbranched carbon chain. Typically the “n-” is omitted and the compound is just named “butane”; it is assumed that you know that no prefix refers to the unbranched chain.
The traditional name of 2-methylpropane is isobutane. Many times, the “2-” is omitted and the compound is just called “methylpropane” because that’s the only location for the methyl group to bond to (if the methyl is bonded to either end, the molecule is n-butane); hence there’s no ambiguity by omitting the “2-“.
Activity 3 Analyzing Constitutional Isomers
In your course notebook write an answer to each question and an explanation of your answer:
• Describe the shape of the n-butane molecule and how it differs from the shape of the 2-methylpropane molecule.
• Analyze the bonding in each molecule. Are all C atoms bonded to the same number of H atoms? If not, is the number of similarly bonded C atoms the same in each of the two structures ?
• To convert from n-butane to 2-methylpropane, would one or more chemical bonds need to be broken so that atoms could be re-arranged?
• Would you expect the physical properties (such as melting point and boiling point) to be different for the two substances?
Query \(6\)
To change from one constitutional isomer to another requires breaking and re-forming chemical bonds, which requires significant input of energy. At room temperature, very few molecules have that much energy. Therefore, constitutional isomers can be synthesized and separated from one another: they are different substances.
In addition to examining the connectivity in Lewis structures, another way to tell whether a Lewis structure represents the same molecule as another Lewis structure is to work out the IUPAC names for both structures. IUPAC names are designed so that the same structure always has the same name; if two structures are different, naming them correctly will result in different names.
Exercise 3: Naming Alkanes
Alkenes and alkynes can also exhibit constitutional isomerism. For example, the four carbon atoms in the chain of butene give rise to three different constitutional isomers:
It’s not just the π bond location that differs, be sure that you can deduce all the C-H and C-C bonds that need to be broken and re-formed going from one isomer to the another by looking at just the line structures. See the 3D models of 1-butene, 2-butene, and 2-methylpropene to verify.
And similarly, butyne can have two different constitutional isomers:
Exercise 4: Isomers of Substituted Benzenes
The structure of ortho-xylene is shown here:
Click on each structure below that is a constitutional isomer of ortho-xylene.
D11.5 Stereoisomers: Geometric Isomers
Stereoisomers are molecules that have the same molecular formula and the same atomic connectivity, but differ in the orientation of the atoms in 3D space. Hence, molecules that are stereoisomers of each other stem from the same structural isomer. There are several different types of stereoisomers. First we will discuss the geometric isomers.
The two carbon atoms in a C=C double bond cannot freely rotate with respect to each other because such rotation requires breaking the π bond (see Section D10.4). Once the π bond is broken there would be free rotation around the remaining σ bond, but at room temperature most molecules do not have sufficient energy to break a π bond. Thus, there is no free rotation around a C=C double bond. This gives rise to geometric isomers, stereoisomers that differ in the orientation of the groups connected to a C=C bond.
For example, there are two geometric isomers of 2-butene, cis-2-butene and trans-2-butene:
The isomer with both methyl groups on the same side of the double bond is called a cis isomer (above, the methyl groups are shown as both being above the double bond). The one with the methyl groups on opposite sides is called a trans isomer. Geometric isomers have different physical properties, such as boiling point, that make separating them possible. Hence they are different substances, and in nomenclature, they are distinguished by the “cis” and “trans” prefix. Figure 5 shows an animation of a cis-2-butene molecule showing conformations that are accessible at room temperature.
A link to an interactive elements can be found here.
Figure 5. At room temperature a molecule of cis-2-butene can adopt many conformations, but none of them involve full rotation around the C=C bond to form trans-2-butene. (Animation by Michael Aristov.)
Cis-trans isomerism is only possible when there are two different groups at each end of a double bond. For example, in an alkene molecule like this
there can be geometric isomers if a and b are different groups and c and d are different groups. If both groups on either end of the double bond were the same (a the same as b, or c the same as d), rotating around the double bond would not produce a different molecular structure.
D11.6 Stereoisomers: Enantiomers
Another type of stereoisomer is a pair of molecules that are mirror images that cannot be superimposed on each other. Such stereoisomers are called enantiomers (or optical isomers), and are described as being chiral (from the Greek word cheir, χειρ, meaning “hand”; your left and right hands exhibit chirality—they are non-superimposable mirror images of each other).
Most physical, chemical, and physiological properties of two enantiomeric substances are identical. Differences between enantiomers only become evident when the substances interact with other chiral molecules or environments. (An infamous example of a physiological difference is the drug thalidomide, where one isomer causes birth defects but the other does not.)
Most chiral molecules have at least one atom that is bonded to four different groups, a chiral center. Chiral centers are often marked with an asterisk (*) in molecular structures. A carbon atom that is a chiral center is referred to as an asymmetric carbon atom or a chiral carbon atom. For example, the carbon atom in bromochloroiodomethane is a chiral center and there are two enantiomeric isomers: the two mirror-image molecules of CHBrClI are not superimposable on each other (Figures 6 and 7).
A link to an interactive elements can be found here.
Figure 6. Bromochloroiodomethane is a chiral molecule. Use the mouse to orient the molecules. At the beginning of the animation, the two enantiomers are shown as mirror images. then one molecule is rotated and moved on top of the other to show that these two mirror-image molecules are not superimposable. (Animation by Michael Aristov.)
Query \(10\)
Figure 7. Bromochloroiodomethane is a chiral molecule. The two enantiomers are shown as mirror images in one view. When you move the slider, the right molecule has been rotated so that the C-H and C-I bonds are aligned; rotation shows that these two mirror image molecules are not superimposable.
Activity 4 Analyzing Enantiomers
Study Figure 6 and Figure 7 carefully. Make certain that you understand what superimpose means and why these two mirror-image molecules cannot be superimposed.
• Based on your observations, devise a method for distinguishing one structure from the other; that is, describe how you can tell that one molecule is different from the other. Write your method in your notebook.
• To convert from (R)-bromochloroiodomethane to (S)-bromochloroiodomethane, would one or more chemical bonds need to be broken so that atoms could be re-arranged? If so, describe which bonds could be broken and which could be formed to interchange the structures.
Query \(11\)
The difference between the two enantiomers may not appear to be significant, but you would have to at least partially break and re-form two σ bonds in order to go from one enantiomer to the other. Hence, enantiomeric structures represent different substances that can be separated from one another. Separation is often difficult and requires a chiral environment: a different chiral substance that interacts differently with the left-hand molecule compared to the right-hand molecule. A macroscopic analogy is this: if all the gloves you own were mixed up, you could separate left-hand gloves from right-hand gloves by how they interact with your right hand; all gloves that fit are right-handed.
All molecules have a mirror image, but only chiral molecules have nonsuperimposable mirror images. Contrast CHBrClI with CHCl2I (dichloroiodomethane), which is shown in Figure 8. CHCl2I and its mirror image are superimposable. Thus CHCl2I is not a chiral molecule (it is said to be achiral). Two groups (the Cl atoms) bonded to the carbon atom in CHCl2I are the same, so there is no chiral center.
A link to an interactive elements can be found here.
Figure 8. Dichlorofluoromethane is not chiral. The molecule and its mirror image are superimposable.
Generally, the easiest way to spot a chiral center is to look for four different groups bonded to an atom. The “groups” are considered in their entirety, not just the atom that is directly bonded to the chiral center. For example, a methyl group is a different group from an ethyl group, which is different from a propyl group and so forth. Hence, alkanes can be chiral, for example:
Query \(12\)
Just because you see dashed and solid wedges in a structure, do not automatically assume that you are looking at a chiral center. For example, lactic acid has an asymmetric (chiral) carbon atom (denoted by *) and is a chiral molecule, while isobutyric acid is not a chiral molecule.
Make sure that you can recognize from the wedge-dash Lewis structure that the chiral carbon atom in lactic acid has four different groups around it but the corresponding carbon atom in isobutyric acid does not. Use these three links to access 3D rotatable models of the two enantiomers of lactic acid and the single structure of isobutyric acid. Lactic acid link 1. Lactic acid link 2. Isobutyric acid.
D11.7 Intermolecular Forces
Whether a covalent molecular substance is a solid, liquid, or gas at room temperature (or any other temperature) depends on the strengths of the attractive forces between the molecules that make up the substance. Any attractive forces between molecules are referred to as intermolecular forces (IMFs). Strengths of intermolecular forces vary widely, but IMFs are usually weaker than covalent bonds. One form of IMF, London dispersion forces, has already been discussed in Unit 1, where we indicated that boiling points reflect the energy needed to overcome intermolecular forces as molecules go from being in close proximity to each other in the liquid phase to being far apart in the gas phase. Thus, boiling points are good indicators of the relative strengths of IMFs of different molecular substances.
The boiling points of linear alkanes (CnH2n+2), with n = 1-20 are shown in Figure 9. The boiling points gradually increase with increasing length of the molecule. You should be able to explain this trend in boiling points based on LDFs.
LDFs increase with number of electrons and also with ease of distortion of electron probability distribution in an atom or molecule. Because longer linear alkanes have more electrons, the LDFs increase and boiling points also increase.
Exercise 6: Match Boiling Points
Another factor that affects the strength of LDFs is molecular geometry. For example, the boiling points for n-pentane, isopentane (2-methylbutane), and neopentane (2,2-dimethylpropane) are 36 °C, 27 °C, and 9.5 °C, respectively, indicating that LDFs are greatest for n-pentane and least for neopentane.
Query \(14\)
Figure 10. The strength of the LDFs increases with larger contact surface area between molecules, as demonstrated by the boiling points of these C5H12 isomers. Use the slider to see total electron density views of the molecules and simple line drawings of the molecules. Click on the links to see rotatable models: 2,2-dimethylpropane; 2-methylbutane;n-pentane. Compare all three molecules in separate tabs in your browser to see that they are isomers and become less compact from left to right.
These three molecules are constitutional isomers and therefore have same number of electrons. In n-pentane, the open linear shape provides a greater surface area between molecules when they come in contact, resulting in stronger LDFs between the molecules. Neopentane has the most compact shape of the three, yielding the smallest surface area for intermolecular contact and, hence, the weakest LDFs.
Exercise 7: Predicting Boiling Points
Another physical property that is influenced by IMFs is the viscosity of a liquid, which is a measure of the liquid’s resistance to flow. The stronger the IMFs, the more difficult it is for molecules to move past each other and the greater is the viscosity of the liquid. Alkane molecules can get quite large and Figure 9 shows that the strength of London dispersion forces is quite significant in larger alkanes (the boiling point of pentadecane C15H32 is 270°C). Hence, alkanes consisting of larger molecules also have higher viscosity. Some examples of uses for these long-chain alkanes are lubricating oil and paraffin wax. Alkanes with a chain length of 35 or more carbon atoms are found in asphalt (or bitumen), which is a sticky and highly viscous substance.
Day 11 Pre-class Podia Problem: Conformers and Isomers
Here is a structure of 2,5-dimethylhept-3-ene.
• There is one chiral carbon center in this molecule. Draw the structure and identify the chiral carbon with an asterisk (*).
• Draw a different conformation of the structure above.
• Draw the enantiomer of this molecule.
• Draw the geometric isomer of this molecule.
• Draw a constitutional isomer of this molecule that would be likely to have a higher boiling point.
• Explain why you think this constitutional isomer would have a higher boiling point.
• Draw a different conformation of this constitutional isomer.
Query \(16\)
Two days before the next whole-class session, this Podia question will become live on Podia, where you can submit your answer.
Comments. If you found any inconsistencies, errors, or other things you would like to report about this module, please use this link to report them. A similar link will be included in each day’s material. We appreciate your comments. | textbooks/chem/General_Chemistry/Interactive_Chemistry_(Moore_Zhou_and_Garand)/02%3A_Unit_Two/2.03%3A_Day_11-_Molecular_Structure-_Isomers.txt |
Day 12: Intermolecular Forces; Functional Groups
Applying Core Ideas: Comparing Propane and Dimethyl Ether
D12.1 Intermolecular Forces: Dipole-Dipole Attractions
The additional IMF alluded to in the Applying Core Ideas box is called dipole-dipole attraction, attractive electrostatic forces between polar molecules. The attractive force arises when the positive end of one molecular dipole interacts with the negative end of another molecular dipole (Figure 1).
The more polar a molecule is (that is, the larger its molecular dipole moment is), the stronger the dipole-dipole attractions are between molecules of that substance. Molecular polarity depends both on the sizes of the bond dipoles (that is, on electronegativity differences between pairs of bonded atoms) and the shape of the molecule. Physical properties of a substance are influenced by all IMFs between molecules of the substance, so it is important to consider both LDFs and dipole-dipole attractions when predicting properties such as boiling points.
How do we know whether a molecule has a dipole moment? In Section D9.2 we described polar covalent bonds—bonds in which there is an unequal distribution of electron density on two bonded atoms and hence a bond dipole moment. The sum of all bond dipole moments in a molecule gives a molecular dipole moment. Molecules that have a molecular dipole moment are called polar molecules; molecules that have a zero (or near zero) molecular dipole moment are called nonpolar molecules.
To predict whether a molecule is polar, first determine whether there are polar bonds by comparing electronegativities of each pair of bonded atoms. If electronegativity differences are small or zero, there are no polar bonds and the molecule must be nonpolar. If there are polar bonds, the molecule might be polar, but it is also possible that the bond dipoles might cancel. For example, both carbon dioxide (CO2) and sulfur dioxide (SO2) have polar bonds, but only SO2 is polar. In CO2, the central carbon has two σ bonds, it is sp hybridized, and therefore the molecule is linear. The bond dipoles are large (ΔEN = 3.5 − 2.6 = 0.9) and equal in magnitude , but they are pointing in exact opposite directions. This results in a molecular dipole moment of zero. In SO2, where the central S atom has two σ bonds and one lone pair, the S atom is sp2 hybridized and the molecule is bent. Thus, the bond dipoles are also large (ΔEN = 3.5 − 2.6 = 0.9), but they are at an angle and the resultant molecular dipole is not zero.
Bond dipoles behave as vectors, so if you are familiar with vector addition you can predict when bond dipoles cancel and when they do not. Another way to predict is this: molecules with all terminal atoms the same and no lone pairs on the central atom are nonpolar because of cancellation of bond dipoles. (In the case of a molecule with an odd number of electrons, a single electron on the central atom counts as a lone pair.) For multi-centered molecules, predicting molecular dipoles is trickier. Generally, if atoms have similar electronegativities, then bond dipoles are weak and the molecular dipole moment is small. For example, because C and H have similar electronegativity, C-H bonds have small bond polarity, and hydrocarbon molecules are nonpolar. The dipole moment of propane, for example, is less than 0.1 D—essentially negligible.
D12.2 Functional Groups
There are more than twenty million known organic compounds, so it would be impossible to memorize chemical properties for each one. Fortunately, we can make use of functional groups to deduce the likely chemical and physical properties of a molecule. A functional group is an atom or group of atoms that has similar chemical properties whenever it is present in a molecule. Even if other parts of a molecule are quite different, a specific functional group usually reacts the same way.
Because C–C and C–H bonds are strong, alkanes are unreactive at room temperature; they are used primarily as fuels (Section D7.2). The alkane parts of molecules usually don’t participate in reactions and are not defined as functional groups. An alkyl group is a portion of an alkane molecule bonded to something else. Examples of alkyl groups are -CH3 (methyl), -CH3CH2 (ethyl), and -CH(CH3)2 (2-propyl). (See alkane nomenclature for more examples.) We often use R (for the Rest of the molecule) to designate any alkyl group (or sometimes another type of group) in a molecule. When there are two or more different alkyl groups, we use R, R’, R”, etc. For example, R and R’, are trans to each other in the alkene structure below:
When a molecule is drawn using R or R’ for alkyl groups, greater focus is put on a specific functional group, in this case, the alkene C=C bond.
Sections D7.3 and D7.4 described the functional groups in alkenes and alkynes. The aromatic functional group was discussed in Section D9.6. The next few sections consider functional groups that contain heteroatoms: atoms other than carbon and hydrogen. Each of these functional groups has its own specific reactivity. Each functional group can also affect the types of intermolecular forces, giving rise to differing physical properties.
D12.3 Aldehydes and Ketones
An aldehyde or a ketone contains a carbonyl group, a carbon atom double bonded to an oxygen atom. The carbon atom in a carbonyl group is called the carbonyl carbon. In an aldehyde functional group, the carbonyl carbon is also bonded to a hydrogen atom. Hence, an aldehyde group can only bond to one R group (another carbon atom or a H atom), and the aldehyde group is always at the end of a chain of carbon atoms (click on the image below for a 3D model.)
A ketone functional group consists solely of the carbonyl group. It bonds to two R groups, which may be the same or different, and is found partway along of a chain of carbon atoms. (click on the image below for a 3D model.)
The reactivity of both aldehydes and ketones are directly related to the reactivity of the carbonyl group.
Formaldehyde is a colorless gas with a pungent and irritating odor. It is sold in an aqueous solution called formalin, which contains about 37% formaldehyde by mass.
Formaldehyde causes coagulation of proteins, so it kills bacteria (and any other living organism) and stops many of the biological processes that cause tissue to decay. Thus, formaldehyde is used for preserving tissue specimens and embalming bodies. It is also used to sterilize soil or other materials. Formaldehyde is used in the manufacture of Bakelite, a hard plastic having high chemical and electrical resistance.
Dimethyl ketone, CH3COCH3, commonly called acetone, is the simplest ketone. It is a colorless liquid that can be made commercially by fermenting corn or molasses. Among its many uses are as a solvent for lacquer (including fingernail polish), cellulose acetate, cellulose nitrate, acetylene, plastics, and varnishes; as a paint and varnish remover; and as a solvent in the manufacture of pharmaceuticals and chemicals.
Activity 1: Ketone Hybridization and Local Bond Geometry
D12.4 Ethers
An ether functional group contains the group –O–, which bonds to two different R groups and is found in the middle of a molecule.
Diethyl ether, the most widely used compound of this class, is a colorless, volatile liquid that is highly flammable. It was first used in 1846 as an anesthetic, but better anesthetics have now largely taken its place. Diethyl ether and other ethers are now used primarily as solvents for gums, fats, waxes, and resins. Methyl tert-butyl ether (abbreviated MTBE) is used as an additive for gasoline. MTBE belongs to a group of chemicals known as oxygenates due to their capacity to increase the oxygen content of gasoline.
Activity 2: Ether Hybridization and Local Bond Geometry
D12.5 Esters
An ester functional group contains a carbonyl group with a second oxygen atom single bonded to the carbonyl carbon and also single bonded to another carbon atom. A general ester structure has an R group bonded to the carbonyl carbon atom and another R group bonded to the second oxygen. It is a functional group that is found in the middle of a molecule. (Click on the image below for a 3D model.)
The ester functional group’s carbon atom is sp2 hybridized with a trigonal planar local geometry. Its carbonyl oxygen is sp hybridized, and one of its unhybridized 2p AOs forms the π bond with the carbon’s unhybridized 2p AO. This oxygen also has two lone pairs: one occupies a sp hybrid orbital; the other occupies a 2p AO that is perpendicular to the π bond. The second oxygen (non-carbonyl oxygen) is sp2 hybridized and has a bent local geometry. It also has two lone pairs, one in a sp2 hybrid orbital, the other in the unhybridized 2p AO. See Figure 2 below.
Query \(6\)
Figure 2. The ethyl acetate molecule, showing the orbitals of the carbonyl O lone pairs, carbonyl π bond, and non-carbonyl O lone pairs.
As shown in Figure 2, the 2p lone pair on the non-carbonyl O is aligned parallel to the p orbitals that form the π bond. This leads to some delocalization of the lone pair electron densities, which can be expressed by resonance structures:
While the resonance structure on the right makes only a minor contribution to the description of the ester molecule, that structure is important in understanding the ester’s chemical and physical properties. For example, the -COO- ester group is planar, and the non-carbonyl C-O bond is not as freely rotatable as a typical single bond. Moreover, an ester’s reactivity is quite different from that of a ketone or an ether, and hence an ester is a distinct functional group.
Activity 3: Ester Hybridization and Local Bond Geometry
Query \(7\)
The odors of ripe bananas and many other fruits are due to the presence of esters.
Day 12 Pre-class Podia Problem: Predicting Boiling Points
The observed boiling points for CCl4 and CHCl3 are:
CCl4 : 77 °C CHCl3 : 61 °C
Use your knowledge of intermolecular forces to write an explanation for why CCl4 has a higher boiling point.
Query \(8\)
Two days before the next whole-class session, this Podia question will become live on Podia, where you can submit your answer.
Comments. If you found any inconsistencies, errors, or other things you would like to report about this module, please use this link to report them. A similar link will be included in each day’s material. We appreciate your comments. | textbooks/chem/General_Chemistry/Interactive_Chemistry_(Moore_Zhou_and_Garand)/02%3A_Unit_Two/2.04%3A_Day_12-_Intermolecular_Forces_Functional_Groups.txt |
D13.1 Alcohols
An alcohol functional group contains a –OH (hydroxyl) group covalently bonded to a carbon atom. (Note that the –OH group is covalently bonded—the OH is not a hydroxide ion.) The names of alcohols are derived from the names of the corresponding alkanes by removing the final “e” and adding “ol”. For example, the structure of the simplest alcohol is derived from the structure of methane by replacing one H atom with an OH group (CH3OH). It is named methanol .
Ethanol is produced by some species of yeast that catalyze fermentation of various sugars:
Ethanol can also be synthesized using the addition reaction of ethene and water using an acid as a catalyst:
Exercise 1: Structures of Alcohols
Knowing how to systematically name compounds can be helpful, especially when trying to decipher whether the line structures shown are constitutional isomers or not. See the nomenclature appendix for additional descriptions on how to name alcohol compounds. (You will not be explicitly tested on nomenclature in this course.)
Query $1$
Alcohol molecules can be classified according to the number of alkyl groups attached to the carbon with the -OH group, as shown in Figure 1. If one alkyl group (or only hydrogen) is attached to that carbon, the alcohol is a primary (1º) alcohol. If two alkyl groups are attached, the alcohol is a secondary (2º) alcohol. If three alkyl groups are attached, the alcohol is a tertiary (3º) alcohol.
D13.2 Carboxylic Acids
A carboxylic acid functional group, -COOH, has a carbonyl and a hydroxyl (-OH) group linked to the same carbon atom. It differs from an ester in that the non-carbonyl oxygen is bonded to a hydrogen atom rather than an R group. Hence, carboxylic acid groups are found at one end of a molecule.
The simplest carboxylic acid is formic acid, known since 1670. Its name comes from the Latin word for ant, formicus. It is partially responsible for the pain and irritation of ants’ and wasps’ stings. Acetic acid is a main component (>4% by volume) of vinegar. Cider vinegar is produced by allowing apple juice to ferment without oxygen present; yeast changes sugar to ethanol, which is then converted to acetic acid via biological oxidation. Butyric acid, a component of rancid butter and Limburger cheese, has a vile odor.
Exercise 2: Identifying Alcohols from Structure
Carboxylic acid groups are acidic because their O-H bond can break relatively easily to yield -COO and H+, allowing them to donate a proton, H+, to another molecule or ion. This acidity arises from the relative stability of the carboxylate anion, RCOO.
Activity 1: Resonance and Acidity of a Carboxylic Acid
D13.3 Amines
An amine functional group is a derivative of ammonia that contains one or more carbon-nitrogen bonds.
You can classify amine molecules by the number of C-N bonds. In the above example, methylamine is a primary (1º) amine, dimethylamine is a secondary (2º) amine, and trimethylamine is a tertiary (3º) amine. In this figure methyl groups are shown but any R group, such as ethyl, could replace any of the methyl groups.
Like ammonia, amines are basic due to the lone pair on the nitrogen atom, and can undergo acid-base reactions to form protonated amine cations analogous to the ammonium ion NH$_4^+$:
D13.4 Amides
An amide functional group contains a nitrogen atom connected to the carbon atom of a carbonyl group. Like amines, amides can be classified by the number of carbon atoms bonded to the nitrogen:
Activity 2: Amines, Amides, and Resonance Structures
Query $4$
Although the resonance structure with formal charges on O and N does not contribute as much to the resonance hybrid as the resonance structure without formal charges, it is crucial in understanding the chemical and physical properties of amide molecules. For example, the partial double bond character gives rise to a significant energy barrier for rotation of the C-N bond. And because the lone pair on the N atom is part of the π bonding network, the N atom in an amide is about 1010 times less basic than the N atom in an amine.
D13.5 Reactions of Alcohols, Amines, and Carboxylic Acids
Oxidation Reactions
All substances whose molecules contain hydrocarbon sections are combustible—the hydrocarbon parts can be oxidized completely to carbon dioxide and water. For many compounds controlled oxidation is more important than combustion because it can convert one type of functional group into another functional group, giving us chemical compounds useful in various applications. An example is oxidation of alcohols, which can be converted to aldehydes, ketones, or carboxylic acids.
Oxidation of an organic compound can usually be recognized as either an addition of oxygen atom(s) to or removal of hydrogen atoms from the reactant molecule. The ease with which an alcohol can be oxidized and the extent of the oxidation depends on whether the alcohol is primary, secondary, or tertiary.
For a primary alcohol, controlled oxidation first produces an aldehyde. If there is an excess of oxidant, the aldehyde is further oxidized to a carboxylic acid:
Common oxidizing agents used in the laboratory for controlled oxidation are aqueous solution of potassium permanganate, KMnO4(aq), or aqueous acidic potassium dichromate, K2Cr2O7(aq).
For a secondary alcohol, the oxidation product is a ketone:
Ketones are difficult to oxidize further because, without breaking C–C bonds, there is no obvious way to add one more oxygen atom to the carbonyl carbon and there are no hydrogen atoms to remove from that carbon either.
Tertiary alcohols, with no hydrogen atoms attached to the carbon atom that is bonded to the –OH group, are difficult to oxidize. Tertiary alcohols do undergo combustion (to yield CO2 and H2O), but they usually do not undergo controlled oxidation.
Condensation Reactions
In a condensation reaction, two molecules join to form a larger molecule and a small, stable molecule such as H2O or HCl. For example, ethyl acetate, CH3COOCH2CH3, is formed when acetic acid reacts with ethanol in the presence of an acid catalyst:
In general, condensation of a carboxylic acid and an alcohol produces an ester.
Condensation reactions are reversible: not all reactant molecules are converted to product molecules and, if product molecules are mixed, some reactant molecules form. The reverse reaction of a condensation that produces water is called a hydrolysis reaction. In hydrolysis, water breaks apart into H and OH with H attaching to one part the larger molecule and OH to another: hydrolysis comes from hydro, “water” and lysis, “breaking apart”.
Ethers can be obtained from condensation reactions involving two alcohol molecules. If the two alcohol molecules are the same, a symmetric ether forms. For example, when ethanol is treated with a limited quantity of sulfuric acid and heated to 140 °C, diethyl ether and water are formed:
If the condensation reaction involves two different alcohols, an asymmetric ethers can form. For example:
Alternatively, addition reactions between alkenes and alcohols can form ethers. For example:
Amides can be produced from carboxylic acids and primary or secondary amines (or ammonia) via condensation reactions:
It is through this reaction that amino acids (molecules containing both an amine and a carboxylic acid functional group) link together in a polymer to form peptides and proteins.
Exercise 4: Recognizing Functional Groups
Applying Core Ideas: Comparing Propane, Dimethyl Ether, and Ethanol
When we previously compared the boiling points of propane and dimethyl ether, we saw the effect of dipole-dipole attractions. Let’s add ethanol to that comparison:
D13.6 Intermolecular Forces: Hydrogen Bonding
The additional IMF that exists between ethanol molecules, but does not exist between propane molecules or between dimethyl ether molecules, is called hydrogen bonding. It is the interaction between an X–H covalent bond (X denotes a highly electronegative atom) and the lone pair on an electron rich atom, Z. The hydrogen bonding interaction can be designated as X-H···Z (the “···” is the hydrogen bond itself).
Strong hydrogen bonding occurs between F-H, O-H, or N-H bond and an electron lone pair on another F, O, or N atom. F, O, and N, are among the most electronegative elements in the periodic table, a characteristic necessary for strong hydrogen bonding.
A hydrogen bond has about 5-10% the strength of a typical covalent bond. Part of this strength comes from dipole-dipole interactions, but more important is partial electron sharing that resembles the formation of a covalent bond. This is illustrated in Figure 2, which shows the hydrogen bond interaction between two water molecules. The left water molecule provides the O-H bond (we will denote this as OL-H); the right water molecule provides the O atom with the lone pair (we will denote this as OR).
The empty OL-H antibonding σ* orbital (yellow/green) overlaps with the orbital of the OR atom’s lone pair (blue/red). This overlap allows some electron density to be shared between the two water molecules, thus forming a hydrogen bond. If more electron density were to move from the OR lone pair to the OL-H σ* orbital, the antibonding orbital would be filled, the OL-H bond would break, and a chemical reaction would have occurred—a new OR-H bond would form, resulting in the right water molecule becoming H3O+:
H2O + H2O ⟶ OH + H3O+
Thus, forming a hydrogen bond resembles the formation of a covalent bond.
Activity 3: IMFs between Water Molecules and between Hydrogen Fluoride Molecules
Query $8$
For amines, 1º and 2º amines are capable of strong hydrogen bonding due to the presence of a N-H bond and the lone pair on the N atom, but 3º amines cannot form hydrogen bonds. For example, for the three amine isomers shown below, the boiling point of the 3º amine is significantly lower than the 1º and 2º isomers.
For amides, because the lone pair on N atom is part of the π bonding network, it is not available for hydrogen bonding. However, hydrogen bonds can still form in 1º and 2º amides between the N-H bond(s) and the lone pairs on the O atom.
Molecules containing the carboxylic acid functional group can also form hydrogen bonds. Pure acetic acid is called glacial acetic acid because its melting point of 16.6 °C is high enough that it can freeze in a cold laboratory. It is also quite thick and syrupy because the many hydrogen-bonding attractions between molecules result in high viscosity. The acidity of the carboxylic acid group enhances the O-H···O hydrogen-bond strength, such that hydrogen-bonding between carboxylic acid molecules is usually greater than between alcohol molecules. For example:
D13.7 Intermolecular Forces: Water Solubility
IMFs directly affect the water-solubility of a compound. Let’s consider the alcohol functional group as an example.
The -OH end of an alcohol molecule is hydrophilic (“water-loving”), capable of strong intermolecular interactions with water molecules. The dipole-dipole and hydrogen-bond interactions between alcohol and water molecules are similar in strength to those between alcohol and alcohol molecules or between water and water molecules. The hydrocarbon (alkyl) end of an alcohol molecule is essentially nonpolar and is hydrophobic (“water-fearing”), where the intermolecular interactions between hydrophobic molecules and water molecules are weak compared to IMFs between water-water molecules or hydrocarbon-hydrocarbon molecules.
Hence, although the size of the alkyl group has little influence on the reactivity of an alcohol molecule, it has a significant impact on the solubility of alcohols in water. Alcohols with small alkyl groups, e.g. methanol and ethanol, are completely miscible with water, which means they have infinite mutual solubility in water. Alcohols with larger alkyl groups, such as 1-octanol, are immiscible in water, which means that when 1-octanol is added to water two layers form. One layer is nearly pure water and the other layer nearly pure 1-octanol, because the solubility of each substance in the other is very low.
This is similarly true for the other polar and hydrogen-bonding functional groups. For example, acetic acid is miscible with water, while octanoic acid is immiscible with water.
Day 13 Pre-class Podia Problem: Functional Groups, Reactions, Solubility
Consider the structures shown below. List all substances that fall into each of the five categories below and explain why each substance fits each category you placed it in. Also explain why the other substances do not fit each category.
1. Contains a secondary alcohol functional group.
2. Is less soluble in water than 1-hexanol.
3. Has a higher boiling point than 1-hexanol.
4. Reacts with K2Cr2O7(aq) to form a carboxylic acid.
5. Undergoes addition of water (with an acid catalyst) to form a secondary alcohol.
Query $9$
Two days before the next whole-class session, this Podia question will become live on Podia, where you can submit your answer.
Comments. If you found any inconsistencies, errors, or other things you would like to report about this module, please use this link to report them. A similar link will be included in each day’s material. We appreciate your comments. | textbooks/chem/General_Chemistry/Interactive_Chemistry_(Moore_Zhou_and_Garand)/02%3A_Unit_Two/2.05%3A_Day_13-_Alcohols_Carboxylic_Acids_Amines_Amides_Hydrogen_Bonding.txt |
Day 14: Macromolecules
Applying Core Ideas: Making Really Large Molecules
DNA is an example of a macromolecule—a huge molecule. A typical DNA molecule from the human genome is about 4 cm long and contains about 4 billion atoms. Suppose that you want to make a really, really big molecule like this. How might you go about it? What element or elements would likely be involved? What kind of chemical reaction might be used? Construct a model process, based on molecular structure and types of chemical reactions, that would enable synthesis of very large molecules. Write a description of the process in your notebook.
Query \(1\)
Polymers are large molecules made by covalently linking many, many small molecules. The small molecules that link to form a polymer are called monomers. Polymers can be natural (such as starch and proteins) or synthetic (such as nylon and polypropylene). Because they include a variety of structures and functional groups, polymers have a broad range of properties and uses that make polymer-based plastics integral parts of our everyday lives.
Polymers provide good examples of how we can apply models involving molecular structure, functional groups, and intermolecular attractions to explain and predict chemical and physical properties of materials.
D14.1 Addition Polymers
Addition polymers are made by addition reactions, wherein two molecules combine to form a single product molecule (see examples of addition reactions involving alkenes in Section D7.4). Typical monomers for addition polymerization have at least one double bond.
Figure 1 shows polyethylene, an addition polymer, forming from ethylene (ethene, H2C=CH2) monomers
The reaction can be initiated by a molecule with an unpaired electron: a free radical. An example initiator is an organic peroxide, which can form two free radicals when the relatively weak O-O bond breaks:
Activity 1: Analyzing Peroxide Decomposition
Query \(2\)
When a radical encounters an ethylene monomer, the ethylene π bond breaks. One electron from the π bond goes to pair with the electron from the radical and form a σ bond. The other electron from the π bond remains a radical, and can go on to react with another ethylene monomer.
Query \(3\)
Figure 2. Move the slider at the bottom to show steps in polymerization of ethylene. The curved arrows represent movement of π-bond electrons in the Lewis structure (that is, rearrangement of electron density). One π-bond electron pairs with another electron forming a new σ bond and the other becomes an unpaired electron in a new free radical.
The process in Figure 2 repeats with many, many more monomers, sometimes as many as 100,000 units, yielding a long polymer chain of carbon atoms.
D14.2 Polymer Structure and Properties
In polyethylene, the addition reactions convert all ethylene double bonds to single bonds and join the monomers without losing any atoms. Hence, this polymer at a molecular level consists of a collection of long-chain alkane molecules, most of which contain tens of thousands of carbon atoms. Many of polyethylene’s properties are what we would expect from this molecular composition.
If the intermolecular forces between the chains are smaller, so that it is easier for the molecules to move past one another, the polymer will be softer and easier to scratch. If the intermolecular forces between the chains are sufficiently strong to prevent motion of the molecules past one another, the polymer will be harder or more rigid. Because polyethylene is a mixture of long alkane molecules, each of slightly different chain length, it softens over a range of temperatures rather than having a single melting point. You may have noticed that some plastics, when wrapped around something in a microwave oven, soften and change shape but never become liquid.
This leads to an important idea in the field of materials science: materials can be “tuned” to give exactly the properties desired. By adjusting the strengths of the intermolecular forces one can obtain plastic materials with a range of properties.
For polyethylene, the extent of branching of a polymer strand can be varied. Depending on the production process, a polymer strand can be a very long linear chain or there can be branching in the chain:
Exercise 1: Polymer Structure and Properties
Polymer properties can also be affected by cross linking. A cross-link is a covalent σ bond between two separate polymer chains that is not at the end of either chain.
Cross links increase the molecular weight and limit the motions of the chains with respect to one another. With enough cross links, a sample of a polymer can become a three-dimensional network held by σ bonds—a single gigantic molecule. Another way to think about cross links: because they are covalent bonds, they are stronger than intermolecular forces between polymer chains and therefore amplify the effect of increasing intermolecular forces on polymer properties.
D14.3 Various Addition Polymers
Table 1 below lists monomers for some well-known addition polymers and also some of their uses. You can probably find at least one example of each of them in your home. Each monomer is a variation of the ethylene (ethene) structure in which one or more H atoms has been replaced by another group (highlighted in the table). Such substitution(s) in the monomer allow the physical properties of the polymer, such as density, melting range, strength, and hydrophobicity, to be precisely controlled, and the polymer tailored for specialized uses. Note that in the equation below the polymer structure is specified by enclosing a single repeating unit in brackets and specifying the repeating nature by a subscript n.
Table 1. Monomers for Some Common Addition Polymers
Monomer Common Name Polymer Some Typical Uses
Ethylene Polyethylene Film for packaging and bags, toys, bottles, coatings
Propylene Polypropylene (Herculon) Milk cartons, rope, outdoor carpeting
Styrene Polystyrene (Styrofoam, Styron) Transparent containers, plastic glasses, refrigerators, styrofoam
Vinyl chloride Polyvinyl chloride (PVC) Pipe and tubing, raincoats, curtains, phonograph records, luggage, floor tiles
Vinylidene chloride Polyvinylidene chloride (Saran) Clinging food wrap
Acrylonitrile Polyacrylonitrile (Orlon, Acrilan) Textiles, rugs
Tetrafluoroethylene Polytetrafluoroethylene (Teflon) Nonstick pan coatings, bearings, gaskets
Vinyl acetate Polyvinyl acetate Elmer’s glue, wood glue
Methyl methacrylate Polymethyl methacrylate (Plexiglass, Lucite) Stiff, clear, plastic sheets, blocks, tubing, and other shapes
The notation where a repeating unit in a polymer chain is shown in brackets emphasizes that it is important to be able to recognize a repeating unit within a polymer chain. It is also important to be able to generate the full polymer structure from a repeating unit. Enclosed within the bracket can be one repeating unit (equivalent of a monomer), or two, or three, etc. Let’s use polyvinylidene chloride as an example:
Shown on the left is a polyvinylidene chloride polymer strand. The individual repeating units are color-coded. However, there are two ways to consider repeating units: starting at the CH2 (top) or the CCl2 (bottom). Either way is fine.
You can see that the repeating pattern in this polymer is alternating CH2 and CCl2. When you put a bracket around a repeating unit, you are communicating that what is outside of the bracket exactly repeats what is inside, starting immediately with the bonds that are divided by the bracket. Therefore, the leftmost carbon atom of the next repeating unit is directly bonded to the rightmost carbon of the unit shown.
In other words, something like the following is incorrect for depicting polyvinylidene chloride:
It is depicting a polymer where there are CH2 unit bonded to CH2 unit, which does not occur in polyvinylidene chloride. (this polymer is …-CH2-CH2-CCl2-CH2-CH2-CCl2-CH2-CH2-CCl2-…)
D14.4 Conjugated Diene Polymers
Conjugated dienes (alkenes with two double bonds and a single bond in between) can be polymerized to form important substances, such as rubber. This occurs in nature as well as in the laboratory. The simplest conjugated diene is 1,3-butadiene; Figure 4 shows the 1,4-polymerization of this monomer. In the resulting polymer, a new σ bond (highlighted in red) is formed between carbon 1 of one monomer and carbon 4 of another monomer, and within each monomer, a π bond is moved to between carbon 2 and carbon 3.
Let’s consider the reaction in a bit more detail. During the polymerization reaction: one electron from the C1=C2 π bond in monomer B pairs with an electron from an adjacent monomer A to form a new σ bond involving C1; similarly, one electron from the C3=C4 π bond in monomer B pairs with an electron from another adjacent monomer C to form a new σ bond involving C4; the other electron from each π bond moves to the center of the molecule, and forms a new π bond between C2 and C3.
A link to an interactive elements can be found here.
Query \(6\)
Figure 5. 1,4-polymerization of 1,3-butadiene to form the cis polymer. Upper: Run the animation to see the orbitals involved in the double bonds. The boxes represent a source of many monomer molecules (right) and the growing polymer chain (left). Lower: Move the slider to see the electron movements described in the text above. Each arrow represents movement of one electron and the electron movements are repeated for three monomers. Rotation around the sigma bond between C2 and C3 enables formation of either cis or trans polymer.
Activity 2: 1,4-Addition Polymerization
Query \(7\)
Chemical reactions involving double bonds on adjacent polymer chains can lead to cross-linking, which enhances elasticity of the polymer. In 1839, Charles Goodyear discovered that when natural rubber was heated to 140–160 °C in the presence of sulfur, the rubber became tougher, more resistant to heat and cold, and more elastic. This process was later called vulcanization after Vulcan, the Roman god of fire and volcanos. Above 140 °C, S–S bonds in sulfur molecules, S8, break, and linear chains of sulfur atoms form. These chains then react with some of the remaining double bonds in the polymer, forming cross links. The development of vulcanized rubber for automobile tires greatly aided the automobile industry.
Another important conjugated diene used in synthetic rubber is chloroprene (2-chloro-1,3-butadiene). Polymerized chloroprene was developed by DuPont and given the trade name Neoprene. Cross-linking in polychloroprene involves combination of two chlorine atoms from adjacent chains with a Zn2+ ion to form ZnCl2. The C–Cl bonds in the uncross-linked polymer become C–C bonds—the cross-link. Cross-linking contributes to the overall elasticity of neoprene.
D14.5 Copolymers
Some of the most commercially important addition polymers are copolymers, made by polymerizing a mixture of two or more monomers. For example, styrene-butadiene rubber (SBR), which is a copolymer of 1,3-butadiene and styrene mixed in about a 3:1 ratio.
The properties of a copolymer are distinct from those of a mixture of the single polymers. For instance, the properties of SBR are different from a mixture containing polybutadiene and polystyrene, no matter what the ratio of the mixture is.
SBR was developed in the U.S. during World War II when important supplies of natural rubber were cut off. It is more resistant to abrasion and oxidation than natural rubber and can also be vulcanized. More than 40% of the synthetic rubber production is SBR, which is used in tire production. Several other types of rubber are copolymers, such as butyl rubber, which is copolymerized from 2-methylpropene (H2C=C(CH3)2) and a small percentage of isoprene.
Exercise 3: Monomer(s) from Polymer Structure
Day 14 Pre-class Podia Problem: Polymers and Intermolecular Forces
The strength or toughness of a polymeric material depends on strengths of intermolecular forces between polymer chains and several other factors. Assuming other factors are equal, compare the toughness/strength of polyethylene and polyacrylonitrile. Explain clearly which is stronger and why.
Query \(9\)
Two days before the next whole-class session, this Podia question will become live on Podia, where you can submit your answer.
Comments. If you found any inconsistencies, errors, or other things you would like to report about this module, please use this link to report them. A similar link will be included in each day’s material. We appreciate your comments. | textbooks/chem/General_Chemistry/Interactive_Chemistry_(Moore_Zhou_and_Garand)/02%3A_Unit_Two/2.06%3A_Day_14-_Macromolecules.txt |
D15.1 Condensation Polymers
A condensation polymer is a polymer formed via a condensation reaction (see Section D13.5). The very small molecule produced in each condensation can be H2O, HCl, or some other simple molecule. Condensation polymerizations often, but not always, combines two different monomers in an alternating structure.
Condensation polymers usually grow by forming ester or amide linkages, where new C-O or C-N σ bonds form to link monomers. A typical monomer has a functional group at each end of the molecule. When one end of the monomer reacts and is added onto a polymer chain, the functional group at the other end remains and allows for further reaction to lengthen the polymer chain.
D15.2 Polyesters
A polyester is a polymer where the individual units are held together by ester linkages. For example, the common polyester polyethylene terephthalate (abbreviated as PET or PETE; has brand names such as Dacron or Mylar) consists of polymer chains formed from terephthalic acid (benzene-1,4-dicarboxylic acid) and ethylene glycol (ethane-1,2-diol) monomers.
The condensation reaction that results in the formation of the ester linkage in PET is shown in Figure 1. Repeated condensation reactions result in the step-wise growth of a polymer chain.
Query \(1\)
Figure 1. The condensation reaction between terephthalic acid and ethylene glycol, forming an ester linkage, is shown. Click on the buttons for more information.
The repeating unit of PET is
The presence of ester groups, a polar functional group, on the polymer increases the intermolecular forces between polymer chains, and thereby increases the crystallinity and tensile strength of the polymer. PET makes for excellent fibers and are used in many fabrics. It is also used to make bottles for soda and other beverages. Polyester is biologically inert, so a knitted polyester tube can be used in surgery to repair or replace diseased sections of blood vessels.
D15.3 Polyamides
A polyamide is a polymer in which the individual units are held together by amide linkages. For example, nylon 66 is obtained from the monomers 1,6-hexanediamine and hexanedioic acid. The condensation reaction between these two monomers forms an amide linkage (highlighted):
On one end, the product molecule has a carboxylic acid group, which can undergo a condensation reaction with another 1,6-hexanediamine molecule. On the other end there is an amine group, which can react with another hexanedioic acid molecule. Such continuous condensation reactions lead to the formation of a nylon 66 polymer strand, the repeating unit of which is also shown above.
The “66” in nylon 66 stands for the six carbon atoms in each of the monomer molecules. Other nylons have different numbers of carbon atoms in the monomer molecules, such as nylon 510, which has 5 carbon atoms in the diamine (1,5-pentanediamine) and 10 carbon atoms in the diacid (decanedioic acid).
Nylon makes extremely strong threads and fibers because in addition to London dispersion forces and dipole-dipole attractions, there are hydrogen bonds between the polymer chains. Specifically, a hydrogen bond can form between a N-H in one strand and a carbonyl O lone pair in a neighboring strand.
If you pull on both ends of a nylon thread, after a slight stretch, it will resist breaking because of the strong IMFs that hold neighboring chains together.
Kevlar is a synthetic polymer made from 1,4-phenylenediamine and terephthaloyl chloride monomers. The byproduct of this condensation reaction is HCl.
The material has a high tensile-strength to weight ratio (it is about 5 times stronger than an equal weight of steel), making it useful for many applications from bicycle tires to sails to body armor.
Similar to nylon, part of Kevlar’s strength is due to dipole-dipole interactions and hydrogen bonding (Figure 3) that increase intermolecular forces between polymer strands.
Notice that compared to nylon 66, the density of hydrogen bonds in Kevlar is higher, imparting stronger intermolecular forces.
Furthermore, Kevlar also has stronger London dispersion forces. Consider the repeating unit of Kevlar highlighted in Figure 3. It is a planar structure, with planar benzene rings connected by planar amide groups. Hence a single layer of Kevlar, as shown in Figure 3, is planar. When layers of Kevlar are placed on top of each other, they can be stacked very efficiently, allowing for maximal London dispersion forces.
In addition to its better-known uses, Kevlar is used in cryogenic applications because of its very low thermal conductivity. Kevlar maintains its high strength when cooled to liquid nitrogen temperatures (–196 °C). Many other plastics become brittle and break at that temperature.
D15.4 Proteins
Proteins are condensation polymers made from amino acids. An amino acid consists of a carbon atom (called the α carbon, shown in red below) bonded to a hydrogen atom, an amine group, a carboxylic acid group, and an R group, often called the side chain .
Notice that the α carbon is chiral when the “R” group differs from the other three groups. About 500 naturally occurring amino acids are known, although only twenty are found in protein molecules in humans. Of these twenty amino acids, nineteen have a chiral α carbon.
Exercise 1: Features of Amino-Acid Molecules
Amino acids are linked together in proteins by amide linkages formed via condensation reactions. An amide linkage in a protein is often called a peptide bond.
Note that the reaction does not involve the α carbon atom, and the chirality at the α carbon is the same after the reaction as it was before. (In the equation above, R2 is depicted with a dashed bond because the amino acid shown in blue has been flipped—rotated 180° around a horizontal axis—from the reactant to the product.)
Like all amide functional groups, the amide linkages in a protein strand are planar. The α carbon atoms, however, have tetrahedral local geometry. This molecular chain of α carbons connected by amide groups, is referred to as the protein backbone. It is the main chain of a strand of protein. The amino acid R groups are called side chains because they are attached to, but branch off from, the main chain.
15.5 Amino Acids
The twenty amino acids are often sorted into four groups—hydrophobic, polar but uncharged, basic, and acidic—depending on the nature of their side chains. The nine amino acids that have hydrophobic side chains are shown in Figure 4:
These hydrophobic side chains are composed mostly of carbon and hydrogen, have very small dipole moments, and tend not to mix with water. As we will see, this characteristic has important implications for a protein’s structure.
The six amino acids shown in Figure 5 have side chains that are polar but neither acidic nor basic.
The three amino acids in Figure 6 have side chains containing functional groups that are basic at neutral pH:
Activity 1: What Part of the Histidine Side Chain Is Most Basic?
Query \(3\)
And finally, Figure 7 shows two amino acids that have acidic side chains containing the carboxylic acid functional group:
D15.6 Protein Structure
Protein molecules are complicated: there are 20 different kinds of monomers that can be linked to form a huge number of different polymer chains. Thus protein structures are classified in several ways. Fundamental is the order in which the monomers are linked: the primary structure. Long protein backbones can adopt unique three-dimensional secondary and tertiary structures, both of which are dependent on the primary structure. When two or more protein strands of proteins are held together by intermolecular forces, they can also adopt specific quaternary structures.
Primary Structure
A protein molecule is a polyamide (polypeptide), but because there are 20 different amino-acid monomers, a protein is a copolymer on steroids. Although the repeating unit is the amino-acid structure, there are 20 possibilities for each side chain (highlighted in red below):
Thus a large number of different protein-molecule structures are possible. Each possible copolymer structure has a unique sequence of side chains. This specific sequence of amino acids is called the protein’s primary structure.
By convention, protein sequences are written from the end with the free amine group (the N terminus) to the end with the free carboxyl group (the C terminus). For example, Ala-Ser (alanine-serine) differs from Ser-Ala (serine-alanine).
Exercise 2: Amino Acids in a Tripeptide
A table of amino acid structures can be found at this link.
Secondary Structure
A protein’s secondary structure is the three dimensional shape of small segments of proteins. In secondary structures amino acids that are not next to each other along the backbone are close enough to each other for hydrogen bonding. The type of secondary structure is defined by the pattern of hydrogen bonds between amide groups that are not adjacent in the protein’s backbone. Two common elements of secondary structure, which are present in over 60% of known proteins, are the α-helix and β-sheet.
The α-helix has a helical structure where there are, on average, 3.6 amino acids per turn of the helix. All the side chains project out from the helical axis.
An α-helix has a regular pattern of hydrogen bonding between backbone amide groups, where the nth amino acid is hydrogen-bonded to the (n+4)th amino acid. For example, as shown in Figure 9, the C=O of amino acid 1 forms a hydrogen bond with N-H of amino acid 5; the C=O of amino acid 2 forms a hydrogen bond with N-H of amino acid 6; etc. This regularly repeating hydrogen bonding is a prominent characteristic of an α-helix.
The β-sheet or β-pleated sheet consists of stretches of the protein chain connected side-by-side by hydrogen bonds between backbone amide groups.
Such backbone hydrogen bonding gives rise to a general pleated sheet appearance, hence the name. The side chains of the amino acids alternate above and below the β-sheet (Figure 11).
The numerous backbone hydrogen bonds stabilize both the α helix and β sheet structures. These structures can combine to form various larger structures, called motifs, that are found in more than one protein structure.
Tertiary Structure
The tertiary structure of a protein is the overall three-dimensional shape of the protein. It is how α-helices, β-sheets, and motifs come together to form the overall structure. Tertiary structures are often defined by various intermolecular interactions involving the protein’s side-chains as well as the environment the protein is in.
Two general kinds of proteins are found in cells, water soluble and water insoluble proteins. Water soluble proteins, which include enzymes and transport proteins, are found free in cellular compartments such as the cytoplasm, nucleus, or endoplasmic reticulum.
In the tertiary structure of water-soluble proteins, most hydrophobic side-chain groups are in the interior of the protein and away from water, while most polar side-chain groups are mostly kept on the exterior and therefore exposed to water. (The backbone amide groups, while polar and capable of forming hydrogen bonds, do not interact much with water because they are already engaged in hydrogen bonding with each other in α helices and β sheets.)
The water-insoluble proteins include membrane channels, pumps, and receptors. These proteins are found in lipid bilayers. A description of how and why these proteins are associated with membranes is in Section D16.7.
One important determinant of tertiary structure in some proteins is the disulfide bond. It is similar to the cross-linking we’ve discussed for addition polymers. A protein can form disulfide bonds when cysteines in different parts of its strand are linked by a covalent bond between the sulfur atoms in their side-chains (Figure 13).
Quaternary Structure
Quaternary structure arises when more than one protein chain is present and the chains form a larger structure. For example, the hemoglobin in your blood is composed of four chains (colored magenta, blue, aqua, and tan below) which form two subunits.
When oxygen binds to one subunit of hemoglobin, the interactions between the subunits change. In other words, oxygen binding affects the quaternary structure. Quaternary structure is the usual mechanism for the processes known as allostery (in which binding of a molecule at one site on a protein has effects at other sites far away from that site) and cooperativity (in which binding of a molecule at one site in a multi-subunit protein increases or decreases the likelihood that another molecule will bind at another site of the protein).
D15.7 Protein Folding and Denaturation
How do proteins fold into the complicated three-dimensional structures? This is a very active area of research in chemistry and biochemistry right now. The first hint came from the work of Christian Anfinsen on the protein ribonuclease, which breaks down RNA molecules. Anfinsen discovered that after treating ribonuclease with high concentrations of certain chemicals that cause proteins to unfold and lose their tertiary and secondary structure, the ribonuclease no longer broke down RNA. Moreover, if the chemicals were removed, the ribonuclease would spontaneously recover nearly all its RNA-hydrolyzing activity, without needing any other cellular components. Anfinsen concluded that the primary structure of a protein completely determines its three-dimensional structure at the secondary, tertiary, and quaternary levels.
Scientists are still trying to learn how the primary structure of a protein determines its other levels of structure. They have determined the primary forces that stabilize a protein’s three-dimensional structure are:
• Sequestration of hydrophobic side-chains away from water (for example, in the interior of water-soluble proteins)
• Maximizing London dispersion interactions (minimizing open spaces) in the interior of proteins
• Maximizing hydrogen bonding (for example, in α-helices or β-sheets)
• Attractions between negatively and positively charged sites formed when acidic and basic side-chains lose and gain H+ ions
The process that Anfinsen used is called denaturation, in which proteins lose the quaternary, tertiary, and secondary structures which is present in their native state. Proteins can be denatured by application of some external stress or compound such as a strong acid or base, a concentrated inorganic salt, an organic solvent (e.g., alcohol or chloroform), radiation, or heat. Denatured proteins can exhibit a wide range of characteristics, from conformational change and loss of solubility to aggregation due to the exposure of hydrophobic groups.
Protein folding is key to whether a protein can do its job correctly; it must be folded into the right shape to function. However, hydrogen bonds, which play a big part in folding, are weak compared to covalent bonds and thus can easily be affected by heat, acidity, varying salt concentrations, and other stresses. This is one reason why homeostasis is physiologically necessary in many life forms.
https://mediaspace.wisc.edu/id/0_o4v8ja01?playerId=25717641
Day 15 Pre-class Podia Problem: Nylon
The repeating unit of nylon-6,10 has this structure.
Answer these questions:
1. Write the structures of two monomers that could be combined to form nylon-6,10.
2. List all the intermolecular forces that exist between polymer chains in nylon-6,10.
3. Depending on the orientation of adjacent polymer chains, intermolecular forces between nylon-6,10 chains might be stronger or weaker. Draw structures of two nylon-6,10 repeating units, one above and one below the structure given. Orient the structures you draw to maximize intermolecular forces.
4. Indicate in your drawing which types of intermolecular forces are important between which parts of the molecules. Also, write an explanation in words describing which forces are important where and why.
Query \(5\)
Two days before the next whole-class session, this Podia question will become live on Podia, where you can submit your answer.
Comments. If you found any inconsistencies, errors, or other things you would like to report about this module, please use this link to report them. A similar link will be included in each day’s material. We appreciate your comments. | textbooks/chem/General_Chemistry/Interactive_Chemistry_(Moore_Zhou_and_Garand)/02%3A_Unit_Two/2.07%3A_Day_15-_Condensation_Polymers_Proteins.txt |
D16.1 DNA
Deoxyribonucleic acid (DNA) consists of two polymer chains that coil around each other, forming a double helix. The monomer units of DNA strands are called nucleotides. A nucleotide consists of a phosphate group and a nucleobase attached to a sugar (deoxyribose).
There are four different nucleobases found in DNA. They are called adenine (abbreviated A), thymine (T), guanine (G) and cytosine (C). The nucleotides containing those nucleobases are called deoxyadenosine monophosphate, deoxythymidine monophosphate, deoxyguanosine monophosphate, and deoxycytidine monophosphate (click on each name to see a rotatable molecular model).
The polymerization reaction forming a DNA strand is a condensation reaction between the phosphate group and deoxyribose (specifically the -OH group), which results in the formation of phosphodiester bonds joining the nucleotides. In organisms containing DNA the pH is about 7, and at that pH the phosphate groups are ionized. Thus, DNA is a huge polyatomic anion with one negative charge for each monomer unit along the chain. The negative charge is balanced by positive charges on metal ions, protein molecules, and other cations surrounding the DNA.
Each DNA polymer chain has a backbone of alternating sugar group and phosphodiester bond, with aromatic nitrogen-containing nucleobases forming side-chains bonded to the backbone.
A DNA strand has a free phosphate group at one end (called the 5′ end because the phosphate is attached to the carbon-5 position in the ribose) and a free hydroxyl (-OH) group at the other end (called the 3′ end because the hydroxyl is attached to the carbon-3 position in the ribose). By convention, the sequences are named from 5′ to 3′, and only the nucleobases are included in the name. For example, the molecule shown above is ATGC (not CGTA).
Exercise 1: Forces within DNA
Two single strands of DNA combine to make the double-helix molecular anion (typically called just the DNA molecule). The strands run in opposite directions: at one end of the double helix is the 5′ end of one strand and the 3′ end of the other strand. The two strands wrap around each other to form two intertwined helixes. The rotating figure shows how the two helixes are oriented. Note that one of the gaps between the entwined strands is wider than the other gap. The two separate strands are held in this orientation by hydrogen bonding between nucleobases in the center of the molecule.
D16.2 Base Pairing
One of the most remarkable things about DNA is that the quantity of adenine always equals the quantity of thymine and the quantity of guanine always equals the quantity of cytosine. From one organism to another, the quantity of adenine can vary, but the quantity of thymine always varies the same way. These equalities imply that the nucleobases occur in pairs: adenine paired with thymine and guanine paired with cytosine. These pairs of nucleobases are called complementary base pairs.
Out of six possible combinations, why are these two the complementary pairs? First, the nucleobases have different sizes: T and C have single rings and are smaller; A and G have double rings and are larger. If A and G paired, their larger size would force the double helix to bulge out. If T and C paired, their smaller size would cause pinching of the double helix. Only a pairing between a single-ringed and a double-ringed nucleobase has the proper consistent spacing.
This leaves four possible combinations: A-C, A-T, G-C and G-T. Of these, A-T and G-C pairs maximize the number of hydrogen bonds across the shared helical axis:
The A-C and G-T pairings, on the other hand, would have fewer hydrogen bonding interactions. To achieve any hydrogen-bonding, there would have to be awkward positioning of the nucleobases relative to each other.
Therefore, intermolecular interactions, specifically the efficient hydrogen-bonding, between adenine and thymine, and between guanine and cytosine, result in the specificity observed for the complementary base pairs.
Because of the nature of complementary base pairing, if you know the sequence of one strand of DNA, you can predict the sequence of the strand that will pair with, or “complement” it.
During DNA replication, the strands of the double helix of DNA first separate. Then, enzymes catalyze the synthesis of new DNA in the 5′ to 3′ direction, using the two original strands as models, or templates, for the complementary strands. Two complete DNA molecules, each an exact copy of the original, are the result.
Because of the specificity of base pairing and a proofreading activity of the enzyme that synthesizes new DNA, less than one error in 100 million is incorporated into the new DNA strands. Thus genetic information is passed accurately on to the next generation.
D16.3 Lipids
Fats and oils are part of a class of biomolecules called lipids, which are loosely defined as biomolecules that are insoluble in water but soluble in organic solvents like hexane or chloroform. Here, we discuss two important classes of lipids: glycerolipids and phospholipids. The molecular structures of lipids give rise to different intermolecular interactions that influence their physical properties and biological functions.
D16.4 Glycerolipids
Glycerolipids are composed of glycerol and fatty acids. Glycerol has three carbon atoms, each of which has a hydroxyl (-OH) group bonded to it:
Fatty acids are long, unbranched hydrocarbon chains with a carboxylic acid group at one end.
Fatty acids are systematically named based on the number of carbon atoms and C=C double bonds in the chain. But their common names, such as lauric acid, are often used to identify them.
Note that there are only C-C single bonds in lauric acid; similar to hydrocarbons, such a fatty acid is a saturated fatty acid. In contrast, palmitoleic acid has a C=C double bond, i.e. it has one degree of unsaturation. The π bond can undergo addition of H2 and form a saturated fatty acid. Lipids made from saturated and unsaturated fatty acids have different properties.
Glycerolipids are formed when three fatty acids are joined to glycerol. Catalyzed by enzymes, each hydroxyl group on glycerol can undergo condensation with the carboxylic acid group of a fatty acid, forming an ester linkage.
The lengths of the fatty acid chains (e.g. R1, R2, R3) and the number of C=C double bonds within them determine the melting point of a glycerolipid. Longer chains have stronger London dispersion forces than shorter chains, leading to a higher melting point. Saturated chains can pack against each other regularly, giving rise to stronger London dispersion forces between the chains and higher melting points. Unsaturated chains, especially with cis double bonds, which force the chain to bend, cannot stack together as well, so the London dispersion forces are weaker.
Glycerolipids with a higher melting point that are usually solids at room temperature are called fats or waxes. Glycerolipids with a lower melting point that are usually liquids at room temperature are called oils.
An important role of glycerolipids is in energy storage. The carbon atoms in fatty acids are mostly involved in C-H bonds, and therefore can be oxidized and in the process release energy. In living organisms, this is carried out in a controlled manner via enzymes. But we can also burn fats and waxes (for example, in a candle) to obtain energy in the form of light.
D16.5 Phospholipids
If one of the fatty acids in a glycerolipid is replaced by a hydrophilic phosphate group, the result is a phospholipid molecule.
Below the line structure is a common, simplified way of representing a phospholipid molecule. The phosphate group, together with the polar ester groups, is often called the hydrophilichead of the molecule (red circle). The fatty-acid hydrocarbon chains are often called the hydrophobic tails of the molecule (black squiggly lines). Molecules with both hydrophobic and hydrophilic parts are called amphiphiles or amphiphilic molecules.
In an aqueous environment, such as a living cell, amphiphilic molecules spontaneously arrange into ordered structures. The most important of these is a lipid bilayer, two layers of phospholipid molecules with the hydrophobic tails intermingled and the hydrophilic heads contacting the water. As the diagram below shows, water is excluded from the center of the bilayer where the fatty acid tails are, while the polar head-groups have access to the water. Phospholipid bilayers form the basis for cell membranes and all the other organelle boundaries found in living cells. Membranes found in cells also contain many other components, such as free fatty acids, other lipids, and proteins.
Since the center of the bilayer is nonpolar, only nonpolar molecules, such as O2, CO and CO2, can cross the bilayer. Ions, such as K+, Na+, and Cl, are completely blocked by bilayers. Because of the large number of atoms that would have to be rearranged to allow passage of large molecules, fatty acids, complex carbohydrates, proteins, and nucleic acids are also blocked by bilayers.
16.6 Proteins, Lipids, and Fatty Acids
Passage of large molecules or ions through lipid bilayers, can be controlled by protein molecules that are embedded in a bilayer and extend outside the bilayer on both sides. In these membrane-spanning proteins, side chains facing the lipid membranes are usually hydrophobic; side chains contacting the aqueous space on either side of the membrane can be polar.
Many proteins are anchored to cell membranes by forming a covalent bond to a fatty acid. For example, the amine group (-NH2) on the N-terminus of a protein or in a lysine side chain can undergo condensation with the carboxylic acid group (-COOH) on a fatty acid found in a cell membrane to form an amide linkage. Such a covalent bond keeps the protein associated with the membrane.
Several vitamins necessary for higher animal life are only soluble in lipids. These vitamins are composed almost entirely of carbon and hydrogen, meaning they are hydrophobic like the fatty acid tails of lipids. A diet completely lacking in fat will be deficient in these vitamins unless the vitamins are taken as pills.
For example, vitamin K1 is necessary for blood clotting:
α-Tocopherol (Vitamin E) prevents the inappropriate oxidation of membrane lipids. Deficiency in Vitamin E also causes infertility in rats.
Retinol (Vitamin A) is necessary for normal juvenile growth. Vitamin A deficiency also leads to night blindness.
The four-ringed lipid cholesterol forms the basis for another class of lipids. Cholesterol is a building block for many important molecules used for various purposes in higher organisms. It is partially oxidized in the liver to bile salts, which act like detergents to solubilize fatty acids in foods and allow them to be absorbed by the body. Many hormones, including estrogens and testosterone, are also derived from cholesterol. The diagram below shows the structures of cholesterol, glycocholic acid (the most common bile acid), estradiol and testosterone (two well-known hormones that are derivatives of cholesterol). As you can see, all four are molecules are amphiphiles, since they have hydrophobic and hydrophilic regions.
Lipids also serve a role in intracellular signaling. Some stimuli, such as the smell of bell peppers in olfactory receptor cells, activate an enzyme that breaks up phospholipids into smaller molecules. These molecules, derivatives of glycerol and fatty acids, then act on other enzymes in the cell.
Day 16 Pre-class Podia Problem: Polyfluoroalkyl Substances
The presence of per- and polyfluoroalkyl substances (PFASs) in the environment is a significant public health issue. Some sites in Wisconsin and many in Michigan and other states have been identified as being contaminated by these compounds. They’ve been dubbed “forever chemicals” because they don’t degrade naturally. See the Science article at this link for more information. The Science article says that one of the problems of PFASs is the fact that some of the molecules involved have structures that enable water and airborne droplets to carry them long distances away from their sources.
1. One of the PFASs is perfluorooctanoic acid, which is used in the manufacture of teflon polymers. (Perfluoro means that all H atoms in a hydrocarbon chain have been replaced by F atoms.) Write a Lewis structure for perfluorooctanoic acid.
2. Consider the structure of perfluorooctanoic acid and the statement above that “some of the molecules involved have structures that enable water and airborne droplets to carry them long distances.” Use models of molecular structure and intermolecular forces to construct an explanation of this fact.
3. Other perfluoroalkyl substances have longer chains of carbon atoms than perfluorooctanoic acid. In what type of body tissue would you expect to find these substances? Use models of molecular structure and intermolecular forces to answer the question and construct an explanation.
Query \(3\)
Two days before the next whole-class session, this Podia question will become live on Podia, where you can submit your answer.
Comments. If you found any inconsistencies, errors, or other things you would like to report about this module, please use this link to report them. A similar link will be included in each day’s material. We appreciate your comments. | textbooks/chem/General_Chemistry/Interactive_Chemistry_(Moore_Zhou_and_Garand)/02%3A_Unit_Two/2.08%3A_Day_16-_DNA_and_Lipids.txt |
D18.1 Reaction Rate
The rate of a chemical reaction is usually defined as the change in concentration of a reactant or a product per unit time. Rates can be determined by measuring the concentration of a reactant or product at each of a series of times after the reaction is started.
Often, it is easier to measure some property related to a substance’s concentration instead. For example, for a reaction involving a colored reactant, light absorption can be measured at different times after the start of the reaction. Then the reactant’s concentration at a given time can be calculated from the proportionality between light absorption and concentration.
Consider the decomposition reaction of cyclobutane to ethene in the gas phase:
C4H8(g) ⟶ 2 C2H4(g)
The rate at which cyclobutane decomposes can be expressed in terms of the rate of change of its concentration:
$\begin{array}{rcl} \text{rate of decomposition of C}_4\text{H}_8 &= & - \dfrac{\text{change in concentration of reactant}}{\text{time interval}} \[1em] &= & - \dfrac{[\text{C}_4\text{H}_8]_{t_2}\;-\;[\text{C}_4\text{H}_8]_{t_1}}{t_2\;-\;t_1} \[1em] &= & - \dfrac{{\Delta}[\text{C}_4\text{H}_8]}{{\Delta}t} \end{array} \nonumber$
In this equation, square brackets represent concentration in mol/L (M), so Δ[C4H8] represents the change in concentration of cyclobutane during the time interval Δt (that is, t2t1). The minus sign in front of the fraction is there because reaction rate is defined to be positive. The reactant concentration decreases as the reaction proceeds, making Δ[C4H8] a negative quantity, so a negative sign is needed to make the calculated rate positive.
Table 1 provides an example of data collected during the decomposition of C4H8.
Table 1. concentrations of cyclobutane measured at 40 °C
Time (s) [C4H8] (M) Δ[C4H8] (M) Δt (s) Rate of decomposition (M/s)
0.0 0.240
20.0 0.120 -0.120 20.0 0.00600
40.0 0.060 -0.060 20.0 0.0030
60.0 0.030 -0.030 20.0 0.0015
80.0 0.015 -0.015 20.0 0.00075
Notice that the reaction rate varies with time, decreasing as the reaction proceeds and [C4H8] decreases. An average rate over a given time period can be calculated using the concentrations at the beginning and end of the period. For example, the average rate for the first and last 20-second period are:
$\text{rate =} \dfrac{-\Delta[\text{C}_4\text{H}_8]}{\Delta t} = \dfrac{-(0.120\;\text{M}\;-\;0.240\;\text{M})}{(20.0\;\text{s}\;-\;0.00\;\text{s})} = 0.00600\;\dfrac{\text{M}}{\text{s}} \nonumber$
$\text{rate =} \dfrac{-\Delta[\text{C}_4\text{H}_8]}{\Delta t} = \dfrac{-(0.015\;\text{M}\;-\;0.030\;\text{M})}{(80.0\;\text{s}\;-\;60.0\;\text{s})} = 0.00075\;\dfrac{\text{M}}{\text{s}} \nonumber$
The rate of reaction atany specific time is known as the instantaneous rate. The instantaneous rate when the reaction starts (at t0), is the initial rate. The instantaneous rate of a reaction may be determined one of two ways:
• If concentration changes can be measured over very short time intervals, then average rates over these very short time intervals provide reasonably good approximations of instantaneous rates.
• If we plot concentration vs. time, the instantaneous rate at any time t is given by the negative of the slope of a straight line that is tangent to the curve at that time (Figure 1).
Query $1$
Figure 1. Graph of [C4H8] versus time. The reaction rate at any instant is equal to the negative of the slope of a line tangent to this curve at that time. Tangents are shown at t = 0 s (magenta; initial rate) and at t = 40 s (orange); click on “+” signs for more information.
D18.2 Relative Rates of Reaction
The reaction rate can be expressed in terms of the change in concentration of any reactant or product, and therefore depends on the stoichiometry of the reaction. Let’s use the ammonia decomposition reaction as an example:
2 NH3(g) ⟶ N2(g) + 3 H2(g)
From the balanced reaction, we can see that one N2 molecule is produced for every two NH3 molecules that have reacted. Therefore, the formation of N2 is half as fast as disappearance of ammonia:
$\text{rate =} \dfrac{{\Delta}[\text{N}_2]}{{\Delta}t} = -\dfrac{1}{2}\dfrac{{\Delta}[\text{NH}_3]}{{\Delta}t} \nonumber$
The negative sign accounts for the fact that NH3 (reactant) concentration is decreasing while N2 (product) concentration is increasing. The fraction ½ accounts for the stoichiometry.
Similarly, because 3 mol H2 forms during the time required for formation of 1 mol N2:
$\text{rate =} \dfrac{{\Delta}[\text{N}_2]}{{\Delta}t} = \dfrac{1}{3}\dfrac{{\Delta}[\text{H}_2]}{{\Delta}t} \nonumber$
Figure 2 plots concentrations vs. time for this reaction. At any time, the instantaneous rates for reactants and products are related by the reaction stoichiometry. For example, at 500 s the rate of H2 production is three times greater than that for N2 production.
Query $3$
Figure 2. Concentrations of reactants and products during the reaction 2NH3 → N2 + 3H2 as a function of time at 1100 °C. The rates of change of the three concentrations are related by the reciprocals of their stoichiometric coefficients. An example of this is shown by the different slopes (click on the “+” signs) of the tangents at t = 500 s.
The rate of a reaction is therefore defined by taking the change in concentration per unit time of a reactant or product and multiplying by the reciprocal of the stoichiometric coefficient for that reactant or product. The reaction rate determined this way is the same regardless of which reactant or product is measured during an experiment. For a generic reaction:
a A + b B → c C + d D
where lower-case letters are stoichiometric coefficients and upper-case letters represent chemical formulas, the rate of the reaction is:
$\text{rate =} -\dfrac{1}{a}\;\dfrac{{\Delta}[\text{A}]}{{\Delta}t} = -\dfrac{1}{b}\;\dfrac{{\Delta}[\text{B}]}{{\Delta}t} = \dfrac{1}{c}\;\dfrac{{\Delta}[\text{C}]}{{\Delta}t} = \dfrac{1}{d}\;\dfrac{{\Delta}[\text{D}]}{{\Delta}t} \nonumber$
D18.3 Factors Affecting Reaction Rates
During a chemical reaction, reactant molecules are changed into product molecules. This involves changes in bonding (i.e. bonds broken and bonds formed), and typically this requires molecules to come into close contact; that is, atomic-scale particles must collide to react. Anything that affects the number and/or effectiveness of those collisions will affect the rate of reaction.
Chemical Nature of the Reacting Substances
Some substances react faster than others. For example, potassium and calcium, which are next to each other in the fourth row of the periodic table, both react with water to form H2 gas and a basic solution. Yet calcium reacts at a moderate rate, whereas potassium reacts so rapidly that the reaction is almost explosive. One factor affecting these different rates is that the reactions involve loss of electrons from potassium or calcium atoms, and potassium has a smaller first ionization energy, making loss of an electron easier.
https://mediaspace.wisc.edu/id/1_jkttq85m
Video 1. Different substances react at different rates. Left: reaction of potassium with water. Right: reaction of calcium with water. The reaction of potassium with water is faster. At the end of the video the calcium reaction is enlarged so that bubbles of H2(g) can be seen more easily.
Temperature
Chemical reactions typically occur faster at higher temperatures. At higher temperatures atomic-scale particles move faster, so they collide harder and more often, both of which increase the probability that they will react. For example, methane (CH4) does not react rapidly with air at room temperature, but strike a match and POP!
https://mediaspace.wisc.edu/id/1_c52zzadj
Video 2. Temperature affects the rate of a chemical reaction. Natural gas coming out of a burner does not combust rapidly until its temperature is raised by a burning match.
Concentrations
Reaction rates usually increase when the concentration of one or more of the reactants increases. In some cases, rates depend on the concentrations of products as well. For example, calcium carbonate (CaCO3) deteriorates as a result of its reaction with the pollutant sulfur dioxide (SO2). Specifically, sulfur dioxide reacts with water vapor to produce sulfurous acid:
SO2(g) + H2O(g) → H2SO3(aq)
Sulfurous acid then reacts with calcium carbonate:
CaCO3(s) + H2SO3(aq) → CaSO3(aq) + CO2(g) + H2O(l)
The rate of the overall reaction depends on the concentration of sulfur dioxide in the air. In a more polluted atmosphere where the concentration of sulfur dioxide is higher, calcium carbonate deteriorates more rapidly (Figure 3).
In another example, a cigarette burns slowly in air, which contains about 21% oxygen by volume, but burns much more rapidly in pure oxygen, as shown in the video below.
https://mediaspace.wisc.edu/id/0_obzl1l3a
Video 3. Higher concentration usually increases reaction rate. The flask contains pure oxygen, which reacts with paper and tobacco in the cigarette. The cigarette burns much faster in pure oxygen than in air, which contains about one-fifth the oxygen concentration.
Presence and Concentration of a Catalyst
A catalyst is a substance that increases the rate of a chemical reaction by providing an alternative reaction pathway but is not consumed by the reaction. The greater the concentration of a catalyst the more the catalyst can speed up a reaction. How catalysts work will be discussed in detail later on in this course. Watch the video below to see how a catalyst can speed up the decomposition of hydrogen peroxide to oxygen and water.
https://mediaspace.wisc.edu/id/0_8ps7n4sa
Video 4. Effect of a catalyst. Manganese dioxide catalyzes decomposition of aqueous hydrogen peroxide, forming water and oxygen. Until the MnO2 is added there is no perceptible reaction. A rapid reaction occurs as soon as the MnO2 is added to the aqueous solution of H2O2. The reaction is exothermic so the temperature goes up. The white cloud is water droplets that condense when heated water vapor cools after escaping from the soda bottle.
Surface Area
The factors discussed so far apply to homogeneous reactions, reactions that occur in a single phase (solid, liquid, or gas). If a reaction occurs at a surface, an increase in surface area of the intersection of two phases (such as the surface of a solid in contact with a gas) can increase the rate. Reactions that take place at a surface are called heterogeneous reactions. A finely divided solid (like a powder) has more surface area available for reaction than one large solid piece of the same substance. For example, large pieces of wood smolder, smaller pieces burn rapidly, and sawdust burns explosively. The video below shows how large pieces of iron can be held in a burner flame for a long time and hardly react, whereas iron powder blown into the flame sparkles as the tiny particles burn.
https://mediaspace.wisc.edu/id/0_r05fieh4
Video 5. Surface area affects the rate of a heterogeneous reaction. A bar of iron held in a flame does not oxidize perceptibly, but when the iron is powdered and blown into the flame from a plastic bottle, a rapid reaction with oxygen occurs.
D18.4 Effect of Concentration: Rate Laws
Rate laws or rate equations are mathematical expressions that relate the rate of a chemical reaction to the concentrations of reactants (and sometimes products or catalysts). Often the rate of reaction is proportional to the concentration, or to a power of the concentration, of a substance involved in the reaction:
rate = k[A]m[B]n[C]p
Here, k is the rate constant, a proportionality constant independent of reactant concentrations that is specific for a particular reaction at a particular temperature.
Each exponent, m, n, or p, defines the order of a reaction with respect to each reactant, A, B, or C. It is the power to which a concentration must be raised to correctly calculate the rate. For example, if m = 1, the reaction is first-order with respect to A; if n = 2, the reaction is second-order with respect to B; if p = 0, the reaction is zeroth-order with respect to C, which means that the rate of the reaction is not affected by the concentration of C, because [C]0 = 1. The overall reaction order is the sum of the individual orders, m + n + p + … Reaction orders are usually positive integers, although they can be fractions or negative numbers.
Activity 1: Order of a Reaction and Rate Law
D18.5 Method of Initial Rates
The rate constant and the reaction orders must be determined experimentally by observing how the rate of a reaction changes as the concentrations of the reactants are changed. One way to do this is the method of initial rates. To use this method, select two sets of rate data where all concentrations but one are the same and set up a ratio of the two rates and the two rate laws. This will yield an equation that contains only one unknown: the reaction order of the substance whose concentration varies.
Activity 2: Rate Law from Initial Rates
Query $7$
Activity 3: Rate Law from Initial Rates, Mathematical Approach
D18.6 Reaction Order and Rate Constant Units
It is often true that, as in the last activity, the reaction orders in the rate law are different from the coefficients in the chemical equation for the reaction. It is important to note that rate laws must be determined by experiment and are not reliably predicted by reaction stoichiometry.
Reaction orders play a role in determining the units for the rate constant—the units for k are whatever is needed so that substituting into the rate law expression affords the appropriate units for the rate.
The units for the rate constant for common reaction orders are summarized below.
Overall Reaction Order (m+n+…) Units of k (M1-(m+n+…)s-1)
zeroth M/s or M s-1
first 1/s or s-1
second 1/M s or M-1 s-1
third 1/M2 s or M-2 s-1
Day 18 Pre-class Podia Problem: Determining a Rate Law
This Podia problem is based on today’s pre-class material; working through that material will help you solve the problem.
Consider these data for the hydrolysis of benzene sulfonyl chloride (abbreviated BSC) in aqueous solution containing fluoride ions at 15 °C. The concentration of BSC was 2 × 10−4 M in all trials and from other experiments where no fluoride ions were present the reaction is known to be first order in BSC. Determine the rate law for the overall reaction. Also determine numeric values for all rate constants and express them in appropriate units.
Trial [F] (M) Initial Rate (M/s)
1 0 2.4 × 10−7
2 0.5 × 10−2 5.4 × 10−7
3 2.0 × 10−2 13.9 × 10−7
4 5.0 × 10−2 32.0 × 10−7
Query $9$
Two days before the next whole-class session, this Podia question will become live on Podia, where you can submit your answer.
Comments. If you found any inconsistencies, errors, or other things you would like to report about this module, please use this link to report them. A similar link will be included in each day’s material. We appreciate your comments. | textbooks/chem/General_Chemistry/Interactive_Chemistry_(Moore_Zhou_and_Garand)/03%3A_Unit_Three/3.01%3A_Day_18-_Reaction_Rate.txt |
Instead of using the method of initial rates, you can use the integrated rate law method to determine the rate law and rate constant from experimental data. An integrated rate law relates the concentration of a reactant or product to the elapsed time of the reaction. Thus, it can be used to determine the concentration of a reactant or product present after a certain period of time or to estimate the time required for a reaction to proceed to a certain extent.
For a given rate equation, calculus can be used to derive an appropriate integrated rate law. If you are not familiar with calculus and integration, don’t worry—you can still use the integrated rate law for a reaction without deriving it yourself.
For a generic reaction:
A ⟶ products
The rate of the reaction can be expressed as:
$\text{rate} = - \dfrac{\Delta[\text{A}]}{{\Delta}t} = k[\text{A}]^m \nonumber$
In calculus, the definition of a derivative is:
$\dfrac{d[\text{A}]}{dt} = \lim_{{\Delta}t \rightarrow 0} \dfrac{\Delta [\text{A}]}{{\Delta}t} \nonumber$
So a rate law for the reaction can be written as:
$\text{rate} = -\dfrac{d[\text{A}]}{dt} = k[\text{A}]^m \nonumber$
Rearranging gives:
$-\dfrac{d[\text{A}]}{[\text{A}]^m} = k\;dt \nonumber$
Integrating both sides of this equation gives:
$\displaystyle{\int^{[\text{A}]_t}_{[\text{A}]_0} -\dfrac{d[\text{A}]}{[\text{A}]^m} = \int^t_0 k\;dt} \nonumber$
Assuming that the reaction starts at t = 0, the right side integration simply becomes kt. And then multiplying both sides by -1 gives:
$\displaystyle{\int^{[\text{A}]_t}_{[\text{A}]_0} \dfrac{d[\text{A}]}{[\text{A}]^m} = -kt} \nonumber$
[A]0 is the initial reactant concentration and [A]t is the reactant concentration at time t. The solution to the left side integration has a different mathematical form depending on the order of the reaction with respect to A (value of m). We will explore cases where m = 1, 2, and 0, in the next three sections.
D19.2 First-Order Reaction
When m = 1, the integrated rate law for the reaction “A ⟶ products” becomes:
$\displaystyle{\int^{[\text{A}]_t}_{[\text{A}]_0} \dfrac{d[\text{A}]}{[\text{A}]^m} = \int^{[\text{A}]_t}_{[\text{A}]_0} \dfrac{d[\text{A}]}{[\text{A}]} = \text{ln}[\text{A}]_t - \text{ln}[\text{A}]_0 = -kt} \nonumber$
This integrated rate law for a first order reaction can be alternatively expressed as:
$\text{ln}\left(\dfrac{[\text{A}]_t}{[\text{A}]_0}\right) = -kt \nonumber$
It is easier to use this form of this equation when trying to calculate the time required for a reaction to proceed to a certain extent.
On the other hand, if you raise e (the base of the natural logarithm system) to the power of each side of the equation, it gives:
$\dfrac{[\text{A}]_t}{[\text{A}]_0} = e^{-kt} \;\;\;\;\;\text{or}\;\;\;\;\; [\text{A}]_t = [\text{A}]_0e^{-kt} \nonumber$
It is easier to use this form when trying to determine the concentration of reactant left after a certain period of time.
Exercise 1: Concentration of Reactant after Given Time
The integrated rate law for a first order reaction can be rearranged to have a standard linear equation format:
$\begin{array}{rcl} \text{ln}[\text{A}]_t &=& -kt + \text{ln}[\text{A}]_0 \[0.5em] y &=& mx + b \end{array} \nonumber$
Hence, if a reaction “A ⟶ products” is first order in [A], a plot of “ln[A]t vs. t” must give a straight line. The slope of such a plot would be −k and the y-intercept would correspond to ln[A]0. If the plot is not a straight line, the reaction is not first order in [A].
Activity 1: First-order Rate Constant from Graph
D19.3 Second-Order Reaction
When m = 2, the integrated rate law for the reaction “A ⟶ products” becomes:
$\displaystyle\int^{[\text{A}]_t}_{[\text{A}]_0} \dfrac{d[\text{A}]}{[\text{A}]^m}=\int^{[\text{A}]_t}_{[\text{A}]_0} \dfrac{d[\text{A}]}{[\text{A}]^2}=-\dfrac{1}{[\text{A}]_t}-\left(-\dfrac{1}{[\text{A}]_0}\right)=-kt \nonumber$
This integrated rate law for a second order reaction can be alternatively expressed as:
$\dfrac{1}{[\text{A}]_t} = kt\;+\;\dfrac{1}{[\text{A}]_0} \nonumber$
Exercise 2: Calculate Concentration from Time
The integrated rate law for second-order reaction also has a standard linear equation format:
$\begin{array}{rcl} \dfrac{1}{[\text{A}]_t} &=& kt + \dfrac{1}{[\text{A}]_0} \[0.5em] y &=& mx + b \end{array} \nonumber$
Hence, if a reaction “A ⟶ products” is second order in [A], a plot of
$\frac{1}{[\text{A}]_t} \nonumber$
vs t should be a straight line, where the slope equals k and the y-intercept is
$\frac{1}{[\text{A}]_0} \nonumber$
If the plot is not a straight line, then the reaction is not second order with respect to [A].
Activity 2: Order from Integrated Rate Law
D19.4 Zeroth-Order Reaction
The “A ⟶ products” reaction that is zeroth-order (m = 0) in [A] exhibits a constant reaction rate regardless of the concentration of A:
$\text{Rate} = k[A]^0 = k \nonumber$
The integrated rate law for such a zeroth-order reaction is:
$\displaystyle\int^{[\text{A}]_t}_{[\text{A}]_0} \dfrac{d[\text{A}]}{[\text{A}]^m}=\int^{[\text{A}]_t}_{[\text{A}]_0} d[\text{A}] = [\text{A}]_t - [\text{A}]_0 = -kt \nonumber$
This integrated rate law also has a standard linear equation format:
$\begin{array}{rcl} [\text{A}]_t &=& -kt + [\text{A}]_0 \[0.5em] y &=& mx + b \end{array} \nonumber$
A plot of [A]t vs t for a zeroth-order reaction is a straight line with a slope of −k and a y-intercept of [A]0.
Figure 1 shows two plots, both are for the decomposition reaction of ammonia. One reaction occurred on a hot tungsten (W) surface, while the other reaction occurred on a hot quartz (SiO2) surface.
We can see from this set of data that the reaction on tungsten is zeroth-order; the plot of [NH3] vs t is a straight line. From the slope, we find that the rate constant for this reaction under the experimental conditions is:
$-\text{slope} = k = 1.25 \times 10^{-6}\frac{M}{s} \nonumber$
The decomposition on hot quartz, on the other hand, is not zeroth-order (analysis of the data shows that it is first order).
Equations for zeroth-, first-, and second-order reactions are summarized in Table 1.
Property Zeroth-Order First-Order Second-Order
rate law rate = k rate = k[A] rate = k[A]2
units of rate constant M/s 1/s 1/M s
integrated rate law [A]t = −kt + [A]0 ln[A]t = −kt + ln[A]0 1/[A]t = kt + 1/[A]0
linear plot [A] vs. t ln[A] vs. t 1/[A]t vs. t
relationship between slope of linear plot and rate constant k = −slope k = −slope k = +slope
Table 1. Summary of Rate Laws for Zeroth-, First-, and Second-Order Reactions
D19.5 Flooding Method: Pseudo-Order Reaction
The integrated rate laws are quite useful when determining the reaction order and rate constant using all the data from a single experimental trial. However, thus far we have only considered example reactions involving one reactant, whereas most reactions involve two or more reactants. How can we use integrated rate laws to find the reaction order and rate constants for those reactions?
Let’s consider a generic reaction:
A + B → products
where:
Rate = k[A]m[B]n
Flooding refers to running a reaction that involves two or more reactants with a large excess of all but one reactant. For example, we can run the reaction above with a large excess of B so that A is the limiting reactant by a significant amount. In that case, [B]0 >> [A]0, and the concentration of B would effectively remain constant during the course of the reaction. (If [B]0 = 0.100 M and [A]0 = 0.00100M, then at a time t when all the A has reacted, [B]t = (0.100-0.00100) M = 0.099 M, which is essentially no change in the concentration of B.)
Under this condition of [B]t ≈ [B]0 = constant, the rate law becomes:
Rate = k[A]m[B]0n = kobs[A]m
where kobs, the rate constant we observe during the flooded experiment, is kobs = k[B]0n. This new rate equation allows us to use the integrated rate laws we’ve just discussed to determine the reaction order with respect to [A], and also to determine kobs and k. The order of the reaction, m, is called a pseudo order because it is obtained under flooding conditions and is not necessarily the overall order of the reaction. If m = 1, we say the reaction is pseudo first order and kobs is called a pseudo-first-order rate constant.
Here is an example experiment: For trial 1, flood the reaction mixture with a large excess of reactant B
${[B]_{0}}_{1} \nonumber$
and measure [A]t as the reaction progresses. Then plot ln[A]t vs t. If the graph is a straight line, then the reaction is first-order with respect to [A]. (In this case, the flooded reaction a pseudo-first-order reaction.) The slope of this graph is
$-k_{obs_1} \nonumber$
For trial 2, flood the reaction mixture with a different large excess of B
${[B]_{0}}_{2} \nonumber$
The plot of ln[A]t vs t has a different slope corresponding to
$-k_{obs_2} \nonumber$
The ratio of the two kobs allows us to determine n, the reaction order with respect to [B]:
$\dfrac{k_{obs_2}}{k_{obs_1}} = \dfrac{k([B]_{0_2})^n}{k([B]_{0_1})^n} = \left(\dfrac{[B]_{0_2}}{[B]_{0_1}}\right)^n \nonumber$
In the above equation, all the variables aside from n are known or have been experimentally determined.
Finally, the actual rate constant for the reaction, k, can be determined from the relationship kobs = k[B]0n. The data all trials should be averaged to get the best value of k.
Activity 3: Determining a Rate Constant Using Flooding
Query $5$
Day 19 Pre-class Podia Problem: Integrating a Rate Law
In Activity 1, earlier in today’s work, data are given for the decomposition of hydrogen peroxide. Hydrogen peroxide decomposes to form oxygen and water. Write a balanced chemical equation for decomposition of hydrogen peroxide. How is the equation you wrote for hydrogen peroxide different from the chemical equation A → products that was used to derive the integrated rate law in Section D19.1?
Now define the rate of reaction for decomposition of hydrogen peroxide using the method given in Section D18.2 and derive the integrated first-order rate law, following the derivation in Sections D19.1 and D19.2. Use your mathematical work to show that the rate constant calculated in Activity 1 for the first-order decomposition of hydrogen peroxide is twice as big as it should be.
Query $6$
Two days before the next whole-class session, this Podia question will become live on Podia, where you can submit your answer.
Comments. If you found any inconsistencies, errors, or other things you would like to report about this module, please use this link to report them. A similar link will be included in each day’s material. We appreciate your comments. | textbooks/chem/General_Chemistry/Interactive_Chemistry_(Moore_Zhou_and_Garand)/03%3A_Unit_Three/3.02%3A_Day_19-_Integrated_Rate_Law.txt |
D20.1 Radioactive Decay
Determining the rate law and rate constant of a reaction has numerous real world applications, one of which involves the rate of radioactive decay, the spontaneous change of an unstable nuclide into a different nuclide. (A nuclide is an atomic species of a specific isotope; all examples of a given nuclide have the same number of protons and the same number of neutrons.) The rate of radioactive decay allows scientists to learn about the histories of our world in geological and archaeological studies. Before we get to the kinetics of radioactive decay, let’s first consider what reactions are involved in radioactive decay.
During a radioactive decay process, the unstable nuclide is called the parent nuclide; the nuclide that results from the decay is called the daughter nuclide (Figure 1). The daughter nuclide may be stable, or it may decay itself.
The atomic representations used when discussing radioactive decay reactions have two numbers written to the left of the atomic symbol, for example,
$^{238}_{\;92}\text{U} \nonumber$
and its isotope
$^{235}_{\;92}\text{U} \nonumber$
The subscript denotes the atomic number, Z, of the element (number of protons), and the superscript denotes the mass number, A, of the isotope (number of protons + number of neutrons). Both subscript and superscript are necessary for balancing nuclear equations.
D20.2 Types of Radioactive Decay
Experiments involving the interaction of radiation with a magnetic or electric field (Figure 2) indicated that one type of radioactive decay particle consists of positively charged and relatively massive α particles, which are high-energy helium nuclei
$(_2^4\text{He},\;\text{or}\;_2^4{\alpha}) \nonumber$
A second type consists of negatively charged and much lighter β particles, which are high-energy electrons
$(_{-1}^{\;\;0}{\beta},\;\text{or}\;_{-1}^{\;\;0}\text{e}) \nonumber$
A third consists of uncharged electromagnetic waves, γ rays, which are very high energy photons. (Note that for β particles, which do not contain any protons, the subscript denotes the charge of the particle.) These three types of radioactive decays are the most commonly observed ones.
α decay is the emission of an α particle from the nucleus. For example:
$_{\;84}^{210}\text{Po}\;{\longrightarrow}\;_2^4{\alpha}\;+\;_{\;82}^{206}\text{Pb} \nonumber$
α decay occurs primarily in heavy nuclei (A > 200, Z > 83). The loss of an α particle gives a daughter nuclide with a mass number four units smaller and an atomic number two units smaller than those of the parent nuclide. Note that the sum of protons (subscripts) on the product side is equal to that on the reactant side. The same is true for the mass number (superscripts). This nuclear reaction is balanced.
β decay is the emission of an electron from a nucleus. For example:
$_{\;53}^{131}\text{I}\;{\longrightarrow}\;_{-1}^{\;\;0}{\beta}\;+\;_{\;54}^{131}\text{Xe} \nonumber$
Beta decay involves the conversion of a neutron into a proton and a β particle. The β particle emitted is from the atomic nucleus and is not one of the electrons surrounding the nucleus. Emission of a β particle does not change the mass number of the nuclide but does increase its number of protons and decrease its number of neutrons.
γ decay is the emission of a γ-ray photon from a nucleus. The energies of γ rays are quite high, in the range of 105-107 kJ/mol, so γ rays can easily break chemical bonds if they are absorbed by matter. (In comparison, x-rays have energies roughly 10 times smaller than γ rays.) Cobalt-60 is a nuclide that decays via β emission as well as γ emission:
$_{27}^{60}\text{Co}\;{\longrightarrow}\;_0^0{\gamma}\;+\;_{-1}^{\;\;0}{\beta}\;+\;_{28}^{60}\text{Ni}^* \nonumber$
The asterisk (*) denotes that the nickel-60 produced in the above radioactive decay is in an excited state. It decays to its ground state with the emission of another γ photon:
$_{28}^{60}\text{Ni}^*\;{\longrightarrow}\;_0^0{\gamma}\;+\;_{28}^{60}\text{Ni} \nonumber$
There is no change in mass number or atomic number during γ emission unless it is accompanied by one of the other modes of decay.
Positron (β+) decay is the emission of a positron, or antielectron, from a nucleus. In the process, a proton is converted into a neutron. For example:
$_{\;8}^{15}\text{O}\;{\longrightarrow}\;_{+1}^{\;\;0}{\beta}\;+\;_{\;7}^{15}\text{N} \nonumber$
Electron capture occurs when one of the inner electrons in an atom is captured by the atom’s nucleus and transforms a proton into a neutron. For example:
$\displaystyle_{19}^{40}\text{K}\;+\;_{-1}^{\;\;0}\text{e}\;{\longrightarrow}\;_{18}^{40}\text{Ar} \nonumber$
Electron capture has the same effect on the nucleus as does β+ decay, and both are sometimes considered as a type of β decay.
The table below summarizes the various types of radioactive decay.
D20.3 Half-Life of a Reaction
An important aspect of the kinetics of radioactive decay is the half-life (t½) of the reaction, which is the time required for the concentration of a reactant (e.g. the parent nuclide) to be reduced to half of its initial value. In each succeeding half-life, the remaining concentration of the reactant is again halved.
The half-life of a reaction can be derived from the integrated rate law. Hence, there is a general equation for half-life for zeroth-order, first-order, and second-order reaction.
First-Order Reaction
The integrated rate law gives:
$\begin{array}{rcl} \text{ln}\left(\dfrac{[\text{A}]_0}{[\text{A}]_t}\right) &=& kt \[0.2em] t &=& \text{ln}\left(\dfrac{[\text{A}]_0}{[\text{A}]_t}\right) \times \dfrac{1}{k} \end{array} \nonumber$
When t = t½:
$[\text{A}]_{t_{\frac{1}{2}}} = \dfrac{1}{2}[\text{A}]_0 \nonumber$
Therefore:
$\begin{array}{rcl} t_{\frac{1}{2}} &=& \text{ln}\left(\dfrac{[\text{A}]_0}{\frac{1}{2}[\text{A}]_0}\right) \times \dfrac{1}{k} \[0.2em] &=& \text{ln}(2) \times \dfrac{1}{k} \[0.2em] &=& \dfrac{0.693}{k} \end{array} \nonumber$
The half-life of a first-order reaction is inversely proportional to the rate constant k: A fast reaction (larger k) has a shorter half-life; a slow reaction (smaller k) has a longer half-life. Moreover, the half-life is, conveniently, independent of the concentration of the reactant. Therefore, you do not need to know the initial concentration to calculate the rate constant from the half-life, or vice versa.
Second-Order Reactions
The integrated rate law is:
$\dfrac{1}{[\text{A}]_t} - \dfrac{1}{[\text{A}]_0} = kt \nonumber$
When t = t½,
$[\text{A}]_{t_{\frac{1}{2}}} = \dfrac{1}{2}[\text{A}]_0 \nonumber$
so:
$\begin{array}{rcl} \dfrac{1}{\frac{1}{2}[\text{A}]_0} - \dfrac{1}{[\text{A}]_0} &=& kt_{\frac{1}{2}} \[0.2em] \dfrac{2}{[\text{A}]_0} - \dfrac{1}{[\text{A}]_0} &=& kt_{\frac{1}{2}} \[0.2em] \dfrac{1}{[\text{A}]_0} &=& kt_{\frac{1}{2}} \end{array} \nonumber$
Therefore:
$t_{\frac{1}{2}} = \dfrac{1}{k[\text{A}]_0} \nonumber$
For a second-order reaction, t½ is inversely proportional to the rate constant and the concentration of the reactant, and hence is not constant throughout the reaction. The half-life increases as the reaction proceeds due to decreasing concentration of reactant. Consequently, unlike the situation with first-order reactions, the rate constant of a second-order reaction cannot be calculated directly from the half-life unless the initial concentration is known.
Zeroth-Order Reactions
For a zeroth-order reaction:
$[\text{A}]_t = -kt + [\text{A}]_0 \nonumber$
When t = t½,
$[\text{A}]_{t_{\frac{1}{2}}} = \dfrac{1}{2}[\text{A}]_0 \nonumber$
so:
$\begin{array}{rcl} \dfrac{[\text{A}]_0}{2} & = & -kt_{\frac{1}{2}} + [\text{A}]_0 \[0.2em] kt_{\frac{1}{2}} &=& \dfrac{[\text{A}]_0}{2} \[0.2em] t_{\frac{1}{2}} &=& \dfrac{[\text{A}]_0}{2k} \end{array} \nonumber$
The half-life of a zeroth-order reaction is inversely proportional to the rate constant and directly proportional to the concentration of the reactant. Therefore, t½ decreases as the reaction progresses and the reactant concentration decreases.
Activity 1: Half-life and Order from Concentration-Time Data
D20.4 Radioactive Half-Lives
Radioactive decay occurs in atomic nuclei and therefore does not depend on chemical bonding or collisions of molecules. Therefore, most radioactive decay processes follow first-order kinetics, and have a characteristic, constant half-life. A radioactive isotope’s half-life allows us to determine how long a sample of a useful isotope will be available, or how long a sample of an undesirable or dangerous isotope must be stored before it decays to a sufficiently low radiation level.
For example, coba source, both the amount of
$\text{Co}_{27}^{60} \nonumber$
and the intensity of the radiation emitted is cut in half every 5.27 years.
The number of nuclear transformations per unit time is called the activity, symbol A, of the radioactive sample. The activity of a sample can be measured with an instrument, such as a Geiger counter. The activity is directly proportional to the number of radioactive nuclei present, which is symbolized by N. The rate constant for nuclear decay is called the decay constant, symbol λ. Hence, the rate expression for radioactive decay becomes:
Rate of radioactive decay = A = λN
This is analogous to the first-order rate law for a chemical reaction, X → products,
$\text{Rate} = -\dfrac{\Delta \text{[X]}}{\Delta t} = k[X] \nonumber$
but the rate is called the activity and the concentration of reactant is replaced by the number of radioactive nuclei. Because the radioactivity is defined in terms of the number of radioactive nuclei, the activity increases as the mass of a sample of radioactive material increases.
The other kinetic equations for first-order reactions similarly apply to radioactive decay, but with the radioactive-decay-specific symbols:
$t_{\frac{1}{2}} = \dfrac{\text{ln}(2)}{\lambda} = \dfrac{0.693}{\lambda} \nonumber$
$\text{N}_t = \text{N}_0\left(e^{-{\lambda}t}\right)\;\;\;\;\text{or}\;\;\;\;t = -\dfrac{1}{\lambda}\text{ln}\left(\dfrac{\text{N}_t}{\text{N}_0}\right) \nonumber$
Note that rather than concentration units (M or mol/L), N0 and Nt have the units of moles or number of atoms.
Activity 2: Radioactive Decay and Half-life
D20.5 Radiometric Dating
Several radioisotopes have half-lives and other properties that make them useful for radiometric dating, the process of determining an object’s date of origin. The objects of interest can be geological formations, archeological artifacts, or formerly living organisms. Radiometric dating has been responsible for many breakthrough scientific discoveries about the geological history of the earth, the evolution of life, and the history of human civilizations.
Radiometric Dating Using Carbon-14
Radiocarbon dating or carbon-14 dating can provide reasonably accurate dating of carbon-containing substances up to ~50,000 years old.
Naturally occurring carbon consists of three isotopes: 12C, which constitutes 99% of the carbon on earth; 13C, about 1% of the total; and 1 part per trillion (0.0000000001%) of 14C. 14C is formed in the upper atmosphere by the reaction of nitrogen atoms with neutrons:
$_{\;7}^{14}\text{N} + \;_0^1\text{n} \;\longrightarrow\; _{\;6}^{14}\text{C} + \;_1^1\text{H} \nonumber$
All isotopes of carbon react with oxygen to produce CO2. In the atmosphere, the ratio of 14CO2 to 12CO2 has been nearly constant for a long time, as shown by analysis of carbon-containing gas samples trapped in Greenland ice sheets.
The incorporation of 14CO2 and 12CO2 into plants is a regular part of photosynthesis. Therefore, a living plant is constantly exchanging carbon with its environment, meaning that the 14C/12C ratio found in a living plant is the same as the 14C/12C ratio in the atmosphere. But when a plant dies, the carbon exchange with the atomosphere stops. Because 12C is a stable isotope and does not undergo radioactive decay, the amount of 12C in the dead plant does not change. However, 14C undergoes β decays with a half-life of 5730 years:
$_{\;6}^{14}\text{C} \;\longrightarrow\; _{\;7}^{14}\text{N} +\; _{-1}^{\;0}\beta \nonumber$
Thus, the 14C/12C ratio gradually decreases after the plant dies, and this provides a measure of the time that has elapsed since the death of the plant. For example, if the 14C/12C ratio in a wooden object found in an archaeological site is half what it is in a living tree, this indicates that 5730 years have elapsed since the death of the tree from which the wooden object was made. Highly accurate determinations of 14C/12C ratios can be obtained from very small samples (as little as a milligram) by the use of a mass spectrometer.
This process works for other organisms as well, since they also constantly exchange carbon (by eating plants or eating others that eat plants) until they die.
Activity 3: Radiocarbon Dating
The accuracy of radiocarbon dating assumes that the 14C/12C ratio in a living plant is the same now as it was in an earlier era, but this is not always valid. For example, right now, due to the increasing accumulation of CO2 (largely 12CO2) in the atmosphere, caused by combustion of fossil fuels in which essentially all of the 14C has decayed, the 14C/12C ratio in our atmosphere is decreasing. This in turn affects the 14C/12C ratio in currently living organisms on the earth. Fortunately, we can use other data, such as tree dating via examination of annual growth rings, to calculate correction factors. With these correction factors, more accurate dates can be determined. In general, radioactive dating only works for about 10 half-lives; therefore, the limit for carbon-14 dating is about 57,000 years.
Radiometric Dating Using Nuclides Other than Carbon-14
Radiometric dating using radioactive nuclides with half-lives longer than that of carbon-14 can date events older than 57,000 years ago.
For example, uranium-238 can be used to estimate the ages of some of the oldest rocks on earth. Since 238U has a half-life of 4.5 billion years, it takes that amount of time for half of the original 238U to decay by a series of nuclear reactions into 206Pb. In a rock sample that does not contain appreciable qunatities of 208Pb, the most abundant isotope of lead, we can assume that lead was not present when the rock was formed. Therefore, by measuring and analyzing the ratio of 238U/206Pb, we can determine the age of the rock. This assumes that all of the 206Pb present came from the decay of 238U. If there is 206Pb from sources other than 238U, which is often indicated by the co-presence of other lead isotopes in the sample, it is necessary to make adjustments in the calculations.
Potassium-argon dating uses a similar method. 40K decays to form 40Ar with a half-life of 1.25 billion years. If a rock sample is crushed and the amount of 40Ar gas that escapes is measured, determination of the 40K/40Ar ratio yields the age of the rock. Other methods, such as rubidium-strontium dating (87Rb decays into 87Sr with a half-life of 48.8 billion years), operate on the same principle.
To estimate the lower limit for the earth’s age, scientists determine the age of various rocks and minerals, making the assumption that the earth is older than the oldest rocks and minerals in its crust. As of 2014, the oldest known rocks on earth are the Jack Hills zircons from Australia, found by uranium-lead dating to be almost 4.4 billion years old.
Day 20 Pre-class Podia Problem: Radiocarbon Dating
A sample of an igneous rock contains 9.58 × 10–5 g of 238U, 2.51 × 10–5 g of 206Pb, and a negligible quantity of 208Pb. No other lead isotopes are detected. Determine the approximate time at which the rock was formed. (238U decays into 206Pb with a half-life of 4.5 × 109 y.)
Query $11$
Two days before the next whole-class session, this Podia question will become live on Podia, where you can submit your answer.
Comments. If you found any inconsistencies, errors, or other things you would like to report about this module, please use this link to report them. A similar link will be included in each day’s material. We appreciate your comments. | textbooks/chem/General_Chemistry/Interactive_Chemistry_(Moore_Zhou_and_Garand)/03%3A_Unit_Three/3.03%3A_Day_20-_Rate_of_Radioactive_Decay.txt |
D21.1 Factors that Affect the Rate Constant
Section D18.3 listed several factors that affect reaction rate: the chemical nature of the reactants; temperature; concentrations; catalysts; and, for heterogeneous reactions, surface area. The rate law, and the order of reaction with respect to each concentration in the rate law, indicate how rate depends on concentration. Chemical nature of reactants and temperature both affect the rate constant k. We will discuss catalysts and heterogeneous reactions later.
It appears obvious to say that different substances are likely to react at different rates. Thus a rate constant value applies to specific reactant and product molecules; that is, to a particular balanced chemical equation. Less obvious is that the rate “constant” is only constant at a given temperature; that is, a reaction has many different rate constants, each constant at a specified temperature. The value of the rate constant is determined by three main factors, the first two of which depend on temperature:
1. To react, molecules must come close enough to each other to exchange energy and perhaps to break and form bonds; that is, molecules must collide. The rate constant is proportional to the rate of collisions:
$k\;{\propto}\;\dfrac{\text{number of collisions}}{\text{time}} \nonumber$
2. For a reaction to occur, there must be sufficient energy in the reactant molecule or molecules to allow electrons to rearrange (to break and/or form bonds). The smaller the energy required, the larger the rate constant is.
3. The reacting molecules must collide in an orientation that allows the reaction to proceed; that is, even though the molecules have enough energy, a collision is more likely to result in reaction when the molecules are arranged in certain ways.
Let’s consider these factors in more detail.
D21.2 Reaction Energy Diagrams
The rate of a chemical reaction depends on the energies of the reacting atoms and/or molecules. Bonds often need to be broken along the course of the reaction as new bonds are formed, and energy is required to break those bonds. Because energy is crucial for understanding the kinetics of a reaction, it is useful to diagram the energy changes that occur as a reaction proceeds.
Let’s use a simple reaction as an example. Consider cistrans isomerization—changing a cis isomer into a trans isomer or vice versa. We’ve learned before that these geometric isomers can be separated at room temperature because there is a barrier to rotation around an alkene double bond. This “barrier” is the energy needed to rotate around the double bond until the π bond is completely broken (when the two 2p AOs that form the π bond are oriented at 90° to each other).
Query $1$
Figure 1. The π bonding orbital in ethene is shown. Move the slider to see the two 2p AOs from the two C atoms that overlap side-by-side to form the π bond. As the front C atom is rotated 90° relative to the back C atom, the p-orbital overlap is reduced to zero, breaking the π bond.
For a cistrans isomerization, the angle of rotation around the double bond is a good measure of the progress of the reaction. Figure 2 shows energy versus reaction progress for conversion of cis-1,2-difluoroethene to trans-1,2-difluoroethene.
As one end of the molecule rotates relative to the other, energy increases because the π bond is partially broken. When the angle of rotation reaches 90° the π bond is completely broken and the energy of the molecule has increased by the bond energy of the π bond. Once the angle exceeds 90°, a new π bond begins to form and the molecule’s energy decreases. Finally, when the angle of rotation is 180° the new π bond is completely formed and the molecule’s energy has decreased almost to the same value it was initially. If a molecule does not have enough energy, which in this case is 262 × 10-21 J, the rotation cannot occur. Thus, only molecules whose energy is more than 262 × 10-21 J can react. As we shall see, at room temperature only a tiny fraction of all cis-1,2-difluoroethene molecules has this much energy, so rotation around the double bond is very slow at room temperature. Raising the temperature increases the fraction of molecules with enough energy to react and therefore increases the rate.
A diagram that shows energy as a function of reaction progress (such as Figure 2) is called a reaction energy diagram. In Figure 2, the reactant (cis isomer) and product (trans isomer) are connected by a single transition state, which is a maximum on the reaction energy diagram. The transition-state structure (or activated complex) is the structure that corresponds to the transition state. Because a transition-state structure has greater energy than either reactant molecule(s) or product molecule(s), it is unstable and exists for a very short time (typically < 10-15 s). This makes it very difficult to observe experimentally, unlike the stable reactant(s) and product(s).
The difference in energy between the transition state and the reactant(s) is the activation energy (Ea) of the reaction going forward (from reactant to product). If all other factors affecting the rate are the same, the greater the activation energy is, the slower the reaction is.
The difference in energy between products and reactants is the reaction energy change, ΔrE:
ΔrE = ∑Eproducts – ∑Ereactants
Figure 2 shows that the cistrans isomerization of 1,2-difluoroethene has an activation energy of 262 × 10-21 J/molecule (or 158 kJ/mol) and a reaction energy change of +6 × 10-21 J/molecule (or +4 kJ/mol; the positive sign indicates that the trans isomer is slightly higher in energy than the cis isomer).
D21.3 Temperature and Maxwell-Boltzmann Distribution
How do we think about the energies available to each molecule in a reaction mixture, and how does the temperature of the system play into this?
For a collection of molecules, say a 1-L container of gaseous cis-1,2-difluoroethene at 1 bar, the average kinetic energy of all the molecules, KEavg, is directly proportional to the observed temperature:
$KE_{\text{avg}} = \frac{3}{2}RT \nonumber$
R is the gas constant, 8.314 J/K·mol.
But the individual molecules do not all have the same KE nor are they all traveling at the same velocity. Recall that the kinetic energy of a molecule of mass (m) and velocity (v) is:
KE = ½mv2
At any given temperature (or KEavg), there is a distribution of individual molecular velocities (or kinetic energies). This distribution is known as the Maxwell-Boltzmann distribution. It shows the fraction of molecules that have a particular velocity. For example, Figure 3 shows the Maxwell-Boltzmann distribution for cis-1,2-difluoroethene gas at various temperatures.
The fraction of cis-1,2-difluoroethene molecules moving at 1200 m/s is much larger when the sample is at 1000 K (red curve) than at 100 K (blue curve). In general, if the temperature of a gas sample increases, its KEavg increases, and the Maxwell-Boltzmann distribution shifts toward higher velocity.
Because the molecules are in constant motion, they will occasionally collide against each other. In a collision, energy can be transferred from one molecule to another, so that one molecule speeds up after the collision while the other slows down. Hence, even while the temperature of the sample and KEavg remain constant, the kinetic energy of any given molecule is continually changing over time.
D21.4 Activation Energy and Temperature
A reaction will not proceed unless the kinetic energy available to the reactant(s) is at least as high as the activation energy (Ea). If the activation energy is much larger than the average kinetic energy of the reactants, Ea >> KEavg, the reaction will occur slowly. This is the case when Ea is large (e.g., the transition state structure involves numerous covalent bonds breaking but few bonds forming) and/or when the temperature of the sample is low. On the other hand, if Ea << KEavg, the reaction will likely proceed rapidly.
For example, Figure 2 indicates that the cistrans isomerization reaction of cis-1,2-difluoroethene has Ea = 262 × 10-21 J/molecule, corresponding to the energy needed to break the π bond. At 500 K, the Maxwell-Boltzmann distribution (Figure 3, orange curve) shows that nearly all the molecules have KE < 15 × 10-20 J, that is, 150 × 10-21 J, much smaller than Ea. Very few reactant molecules have enough energy to reach the transition state.
Nevertheless, at 500 K, a very small fraction of the molecules is moving very fast (v > 1200 m/s). If such a rapidly moving molecule hits another molecule in a way that causes one carbon to start rotating around the other carbon, and if the hit is hard enough so that after the collision the second molecule has energy greater than Ea , that is, > 262 × 10-21 J/molecule, then the reaction can occur. But this happens very rarely at 500 K.
If the temperature increases, for example to 1000 K, then there is a larger fraction of molecules moving at sufficiently high speed, and more molecules can overcome the Ea barrier and change from a cis isomer into a trans isomer at any given time. Hence, the reaction is occurring faster (rate is higher). On the other hand, at room temperature (300 K), the fraction of molecules that can have sufficient energy to overcome the barrier is so small that the reaction can essentially be considered as not occurring (rate ≈ 0). Hence, at room temperature we can separate the cis and trans isomers and they are considered as different substances.
The fraction of reactant molecules with sufficient energy to react depends on the activation energy of the reaction and the temperature of the reaction mixture. The rate constant is proportional to the fraction of molecules with sufficient energy and that fraction involves an exponential (a power of e, the base of the natural logarithm system).:
k ∝ e-Ea/RT
Here, R is the ideal gas constant (as 8.314 J mol-1 K-1), T is absolute temperature (in kelvins), and Ea is expressed in J/mol; hence units in the exponent cancel. This proportionality relationship tells us that k is larger when Ea is smaller as well as when T is larger. Hence, increasing the temperature of a reaction has a similar effect on reaction rate as lowering the reaction’s activation energy. This is illustrated in Figure 4.
The number of cis-1,2-difluoroethene molecules exceeding the Ea = 2.62 × 10-19 J barrier at 3000 K (violet curve) is shaded in violet. This is the same fraction of molecules (shaded in cyan) that exceeds a lower Ea = 7 × 10-20 J barrier at room temperature (300K, cyan curve). This lower barrier, 7 × 10-20 J/molecule (or 40 kJ/mol), is typical for rotation around a C-C single bond (conformational isomers). Hence, we cannot separate one conformer from another at room temperature, because the reaction for converting between them has a fast rate at room temperature (conformers are not considered as distinct substances).
D21.5 Steric Factor
The steric factor is the fraction of collisions energetic enough to react that actually results in reaction. To see why some sufficiently energetic collisions do not always result in a reaction proceeding forward, consider this reaction:
CO(g) + O2(g) → CO2(g) + O(g)
The reaction energy diagram is shown in Figure 5. In order to proceed forward, an O2 molecule needs to collide with a CO molecule with sufficient energy to reach the transition state, where a new C=O double bond begins to form while the O=O double bond breaks and one of the C≡O π bonds also breaks.
Activity 1: Steric Factor
D21.6 Arrhenius Equation and Arrhenius Plot
Taking all that we have discussed above, the relationship between the activation energy, temperature, steric factor and the rate constant for a given reaction is summarized by the Arrhenius equation :
k = Ae-Ea/RT
The constant A is called the frequency factor and it depends on the rate at which collisions occur and the fraction of collisions that have the correct orientation (steric factor). The exponential term, e-Ea/RT, is the fraction of the total number of collisions that has sufficient energy to overcome the activation-energy barrier of the reaction.
The Arrhenius equation describes quantitatively much of what we have already discussed about reaction rates. For example, for two similar reactions occurring at the same temperature, the reaction with the higher Ea has the smaller rate constant and hence is slower. Also, a change in conditions that increases the number of collisions with a favorable orientation results in an increase in A and, consequently, an increase in k.
Exercise 1: Calculating Rate Constants
If we take the natural logarithm of both sides of the Arrhenius equation, we have:
$\text{ln}(k)\; =\; \text{ln}(A)\; +\; \left(-\dfrac{E_a}{RT}\right) \nonumber$
which has a standard linear equation format:
$\begin{array}{rcccl} \text{ln}(k) & = & \left(-\dfrac{E_a}{R}\right)\left(\dfrac{1}{T}\right) &+& \text{ln}(A) \[0.5em] y &=& m\ \ \ \ x &+& b \end{array} \nonumber$
This provides a convenient way to experimentally determine Ea and A for a reaction. By measuring k at different temperatures and plotting ln(k) versus 1/T, we can obtain a straight line where slope = –Ea/R and y-intercept = ln(A).
Activity 2: Determining Ea
Query $4$
You can estimate the activation energy without constructing the Arrhenius plot if the rate constant was determined at only two temperatures.
The slope of an Arrhenius plot is:
$\text{Slope} = \dfrac{\Delta(\text{ln}\;k)}{\Delta\left(\dfrac{1}{T}\right)} = -\dfrac{E_a}{R} \nonumber$
Therefore:
$\dfrac{\text{ln}(k_1)-\text{ln}(k_2)}{\dfrac{1}{T_1}-\dfrac{1}{T_2}} = -\dfrac{E_{\text{a}}}{R} \nonumber$
which can be rearranged as:
$\text{ln}\left(\dfrac{k_1}{k_2}\right) = -\dfrac{E_{\text{a}}}{R}\left(\dfrac{1}{T_1}\;-\;\dfrac{1}{T_2}\right) \nonumber$
Query $5$
Day 21 Pre-class Podia Problem
The rate constant for the rate of decomposition of N2O5 to NO and O2 in the gas phase:
2 N2O5(g) → 4 NO(g) + 3 O2(g)
is 1.66 M-1s-1 at 650. K and 7.39 M-1s-1 at 700. K. Assuming this reaction obeys the Arrhenius equation, calculate the activation energy.
Query $6$
Two days before the next whole-class session, this Podia question will become live on Podia, where you can submit your answer.
Comments. If you found any inconsistencies, errors, or other things you would like to report about this module, please use this link to report them. A similar link will be included in each day’s material. We appreciate your comments. | textbooks/chem/General_Chemistry/Interactive_Chemistry_(Moore_Zhou_and_Garand)/03%3A_Unit_Three/3.04%3A_Day_21-_Reaction_Energy_Diagram_and_Arrhenius_Equation.txt |
D22.1 Elementary Reactions
A balanced equation for a chemical reaction indicates which substances react, which substances are produced when the reaction is over, and how the amounts of reactants and products are related. But it does not necessarily show what happens on the atomic scale as the reaction takes place. Although it is not obvious on the laboratory scale, most chemical reactions occur as a series of atomic-scale steps; that is, as a sequence of simpler reactions, each of which often involving collisions between molecules. The step-by-step sequence of simple reactions by which an overall reaction occurs is called a reaction mechanism.
For example, ozone in the stratosphere protects Earth’s surface from harmful ultraviolet radiation. Ultraviolet photons cause ozone molecules to decompose to oxygen molecules. The overall reaction equation is:
$2\;\text{O}_3(g) \xrightarrow{h\nu} 3\;\text{O}_2(g) \nonumber$
However, at the atomic scale this reaction does not involve collision and reaction between two O3 molecules. Rather, there are two steps that occur one after the other:
$\begin{array}{rcl} &\text{step 1}:& \;\;\;\;\;\;\;\;\;\text{O}_3(g) &\xrightarrow{h\nu}& \text{O}_2(g) + \text{O}(g)\[0.5em]&\text{step 2}:& \;\;\;\;\;\;\;\;\;\text{O}(g) + \text{O}_3(g) &\longrightarrow& 2\;\text{O}_2(g)\[0.5em] &\text{overall}:& \;\;\;\;\;\;\;\;\;2\;\text{O}_3(g) &\longrightarrow& 3\;\text{O}_2(g)\end{array} \nonumber$
In step 1, upon absorption of an UV photon, a bond breaks in an O3 molecule producing an O2 molecule and an O atom:
In step 2, the O atom formed in step 1 reacts with a second O3 molecule, producing two O2 molecules:
The overall reaction is the sum of the two steps. Write reactants from all steps to the left of a reaction arrow; write products from all steps to the right of the arrow; then cancel formulas that appear on both sides of the arrow. This gives
O3(g) + O(g) + O3(g) ⟶ O2(g) + O(g) + 2O2(g)
which is the overall reaction equation given above: 2 O3(g) ⟶ 3 O2(g). An atom or molecule that is a product in an earlier step and reacts away in a later step of a reaction mechanism is called a reaction intermediate.
Each step in a reaction mechanism is called an elementary reaction, which is a chemical reaction that has only a single transition state. An elementary reaction shows which atomic-scale particles collide, break apart, or rearrange their structures to form reaction products and/or intermediates.
The equation for an elementary reaction specifies exactly which atoms or molecules are involved in that reaction. For example, step 2 in the ozone decomposition reaction mechanism above states that one O atom reacts with one O3 molecule and two O2 molecules are formed. That is, for this elementary reaction to occur, one O atom must collide with (come very close to) one O3 molecule.
In contrast, the overall reaction equation does not necessarily specify which atoms or molecules collide and react. Even though the overall ozone reaction is 2 O3(g) ⟶ 3 O2(g), two O3 molecules do not have to collide for products to appear.
Because each elementary reaction has a single transition state, in an overall reaction that consists of several sequential elementary reaction steps there is a series of transition states, one for each step in the mechanism. For example, Figure 1 shows the overall reaction energy diagram for the ozone decomposition reaction.
Query $1$
Figure 1. Reaction energy diagram for the overall 2 O3(g) → 3 O2(g) reaction. This reaction proceeds via a mechanism consisting of two elementary reaction steps. Click on each + sign to identify each chemical species in the diagram.
Note that although one of the O3 molecules does not react until step 2, you still need to include it as a reactant in the reaction energy diagram. In other words, in a reaction energy diagram, all the atoms and molecules that are involved in the reaction are accounted for from the very beginning to the end. Hence, step 1 could be written 2 O3(g) ⟶ O2(g) + O(g) + O3(g), to emphasize that one O3 molecule has not yet reacted, and step 2 becomes O2(g) + O(g) + O3(g) ⟶ 3 O2(g). This is because there is a significant quantity of energy associated with each atom or molecule; to omit one, or suddenly add one in the middle of the diagram would change the energy (y-axis) a lot.
D22.2 Unimolecular Elementary Reactions
A unimolecular elementary reaction involves the rearrangement of a single reactant molecule to produce one or more product molecules. The isomerization of cis-1,2-difluoroethene (Section 21.2), step 1 in the mechanism for ozone decomposition given above, and many radioactive decay reactions (Day 20), are all examples of unimolecular elementary reactions.
A general equation for a unimolecular elementary reaction is:
A → products
Suppose that a 1.0-L flask contains 0.0010 mol of reactant A. The concentration of A is 0.0010 mol/1.0 L = 0.0010 M. At any given temperature a tiny fraction of the A molecules has enough energy to overcome the activation-energy barrier and react. If we add another 0.0010 mol A to the flask, the number of A molecules doubles and the concentration of A doubles to 0.0020 M. If the temperature remains the same, the fraction of molecules that has enough energy to react remains the same, but now there are twice as many molecules so the number of molecules reacting (and the rate of reaction) doubles; that is, the rate of reaction is directly proportional to the concentration of A. Thus, the rate equation for a unimolecular reaction is:
rate = k[A]
If a reaction is known to be a unimolecular elementary reaction, its reaction rate is directly proportional to the concentration of the reactant and the reaction is overall first-order. We see here that because an elementary reaction describes exactly the reaction that is occurring, it is possible to determine the order of an elementary reaction solely by looking at the equation. This is true for all elementary reactions.
D22.3 Bimolecular Elementary Reactions
The collision and reaction of two molecules or atoms in an elementary reaction is a bimolecular elementary reaction. There are two general types of bimolecular elementary reactions. In one type, the two reactants are different:
A + B → products
The rate law for such a reaction is first-order in [A] and [B]:
rate = k[A][B]
In the other type of bimolecular elementary reaction, the two reactants are the same:
A + A → products or 2 A → products
The rate law for such a reaction is second-order in [A]:
rate = k[A]2
If a reaction is known to be a bimolecular elementary reaction, then its rate law can be derived by considering how concentration of each reactant affects the number of collisions. This is shown in Figure 3. Assuming that the fraction of collisions that results in reaction is the same in all three cases, the rate of reaction doubles when the number (and concentration) of each type of molecule doubles. The rate law derived from number of collisions agrees with the rate law derived from the reaction equation. A bimolecular reaction is overall second-order.
Some chemical reactions have mechanisms that consist of a single bimolecular elementary reaction. An example is the gas-phase reaction of 1,3-butadiene with ethene to form cyclohexene, shown in Figure 4.
C4H6(g) + C2H4(g) → C6H10(g) rate = k[C4H6][C2H4]
Another example of a bimolecular elementary reaction is step 2 in the ozone decomposition mechanism given earlier:
O(g) + O3(g) → 2O2(g) rate = k[O][O3]
D22.4 Trimolecular Elementary Reactions
An trimolecular (or termolecular) elementary reaction involves the simultaneous collision of three atoms or molecules. Trimolecular elementary reactions are very unlikely, because the probability of three particles colliding simultaneously is less than 0.1% of the probability of two particles colliding. When a reaction involves three reactant molecules, it is much more likely for it to proceed via a multi-step mechanism involving unimolecular and/or bimolecular elementary reaction steps.
While there are rare examples of trimolecular elementary reaction occurring under certain conditions, in general, the individual steps in reaction mechanisms for gas-phase reactions are restricted to unimolecular and bimolecular reactions.
Day 22 Pre-class Podia Problem: Activated Complex (Transition State)
Consider the gas-phase decomposition of cyclobutane, C4H8, to ethylene, C2H4, which occurs via a unimolecular elementary reaction:
1. Write the rate law for this reaction. Express the rate of reaction in terms of concentration. (Don’t just write “rate =”, use Δ[A]/Δt.)
2. Describe in words which bonds must break and which bonds must form when this reaction occurs.
3. Based on your description in part 2, suggest a structure for the activated complex (transition state) for this reaction. Use dotted lines to indicate existing bonds that are partially broken or new bonds that are partially formed.
4. At the temperature of the kinetic study the reaction energy change, ΔrE = – 300 kJ/mol. The activation energy, Ea = 261 kJ/mol. Use this information to draw a reaction energy diagram. Make the vertical axis scale as accurate as possible. Draw the structures of the reactant, the transition state, and the product molecules at appropriate positions on the diagram.
Query $7$
Two days before the next whole-class session, this Podia question will become live on Podia, where you can submit your answer.
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D23.1 Multi-step Reactions and Rate-Determining Step
A valid mechanism for a multi-step reaction has these characteristics:
• The mechanism should consist of a series of unimolecular and/or bimolecular elementary reaction steps.
• The sum of the reaction steps should agree with the overall balanced reaction equation.
• The mechanism must agree with the experimentally observed rate law.
For elementary reactions, rate laws (and reaction order) can be derived directly from the stoichiometry of the chemical equations, but this is not true for a multi-step reaction where the balanced overall equation is not an elementary reaction. For some multi-step mechanisms, it is possible to derive the overall rate law from the known rate laws of the individual elementary steps. The overall rate law can also be determined from experimental data. If the experimental rate law agrees with the theoretical rate law (derived from the mechanism), the mechanism is a plausible theory for how the reaction occurs. Other experimental data can also support the plausibility of a mechanism. For example, if an intermediate proposed in the mechanism can be detected, that would support the mechanism.
Deriving rate law from a reaction mechanism can be a complex task. However, for many multi-step reactions, one elementary reaction step is significantly slower than the other steps, and this step limits the rate at which the overall reaction occurs. This slowest step in a mechanism is called the rate-determining step (or rate-limiting step), and it allows for some simplifying approximations.
As an example of a rate-determining step, consider the oxidation of iodide ions by hydrogen peroxide in aqueous solution:
$2\;\text{I}^{-}(aq) + \text{H}_2\text{O}_2(aq) + 2\;\text{H}^{+}(aq) \longrightarrow \text{I}_2(aq) + \text{H}_2\text{O}(l) \nonumber$
The currently accepted mechanism for this reaction has three steps. The third step occurs twice each time the first and second steps take place, so it is written twice. For simplicity the states of matter have been omitted.
$\begin{array}{rcl} \text{Step 1:}\;\;\;\;\; \text{H}_2\text{O}_2 + \text{I}^{-} &\xrightarrow{k_1}& \text{HOI} + \text{OH}^{-}\;\;\;\;\; \text{Slow}\[0.5em] \text{Step 2:} \;\;\;\;\; \text{HOI} + \text{I}^{-}\;\; &\xrightarrow{k_2}& \text{I}^2 + \text{OH}^{-} \;\;\;\;\;\;\;\;\; \text{Fast}\[0.5em] \text{Step 3:} \;\;\;\;\; \text{OH}^{-} + \text{H}^{+} &\xrightarrow{k_3}& \text{H}_2\text{O}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \text{Fast}\[0.5em] \text{OH}^{-} + \text{H}^{+} &\xrightarrow{k_3}& \text{H}_2\text{O}\end{array} \nonumber$
The first step is labeled slow, which means that the rate constant k1 is smaller than the other two rate constants. Steps 2 and 3 are labeled fast, because rate constants k2 and k3 are larger than k1. Initially steps 2 and 3 cannot occur, because the concentration of one of their reactants is zero or very low. For example, the concentration of HOI is zero before the reaction begins because HOI is not a reactant in the overall reaction. Thus, step 2 cannot occur until there is some HOI available, which means that until step 1 produces some HOI and raises [HOI], step 2 has zero rate. No matter how big the rate constant for step 2 might be, step 2 cannot go any faster than step 1. We say that the rate of step 2 is limited by the rate of step 1. Similarly, step 3 cannot occur until steps 1 and 2 produce some OH, so the rate of step 3 also limited by the rate of step 1. Thus, in this case, step 1 is the rate-limiting step.
Activity 1: Reaction Energy Diagram
D23.2 First Step is Rate-Determining
When the rate-determining step is the first step in a mechanism, the rate law for the overall reaction can be approximated as the rate law for the first step. The reaction of NO2 and CO provides an illustrative example:
NO2(g) + CO(g) ⟶ CO2(g) + NO(g)
At temperatures below 225 °C, the experimentally observed rate law is:
rate = k[NO2]2
This is consistent with a mechanism that involves these two elementary reaction steps:
$\begin{array}{rcl} &\textcolor{teal}{\text{Step 1}}:& \text{NO}_2(g) + \text{NO}_2(g) &\xrightarrow{k_1}& \text{NO}_3(g) + \text{NO}(g)\;\;\;\;\;\textcolor{teal}{\text{slow (larger E}_a)}\[0.5em] &\textcolor{blue}{\text{Step 2}}:& \text{NO}_3(g) + \text{CO}(g) &\xrightarrow{k_2}& \text{CO}_2(g) + \text{NO}_2(g)\;\;\;\;\textcolor{blue}{\text{fast (smaller E}_a)}\[0.5em] &\text{Overall}:& \text{NO}_2(g) + \text{CO}(g) &\longrightarrow& \text{CO}_2(g) + \text{NO}(g)\end{array} \nonumber$
Both steps in this mechanism are bimolecular elementary reactions, and the sum of the two steps agrees with the overall reaction (the NO3 that is formed in step 1 reacts away in step 2, and one of the NO2 molecules that reacts in step 1 is reformed in step 2, so only one NO2 molecule reacts overall).
Figure 1 shows the corresponding reaction energy diagram for this mechanism. Ea,1 is the activation energy for step 1, and it reflects the energy difference between the reactants and the first transition state. Ea,2 is the activation energy for step 2, and it reflects the energy difference between the intermediary minimum and the second transition state.
Query $2$
Figure 1. Reaction energy diagram for the NO2(g) + CO(g) → CO2(g) + NO(g) reaction. Click on each “i” in the figure for additional information.
With Ea,1 being much larger than Ea,2, step 1 has a smaller rate constant than step 2 (k1< k2, assuming that the frequency factor (A) for the two steps are similar). However, step 2 cannot occur until step 1 produces some amount of NO3 (NO3 is a reaction intermediate, and therefore its concentration is zero at the beginning of the reaction). Because k2 is larger, as soon as some NO3 molecules are formed, they readily react away (step 2 reaction). Hence, the concentration of NO3 is always very small, and the rate of step 2 cannot go faster than the rate of step 1.
Step 1 limits the rate of step 2, and is therefore the rate-determining step in this mechanism. Typically, the slowest reaction step is the one with the largest Ea and/or the highest energy transition state, as illustrated in Figure 1.
The stoichiometry of step 1, which is an elementary reaction, gives this rate law:
rate1 = k1[NO2]2
which can be approximated as the rate law for the overall reaction, and it is in agreement with the experimentally observed rate law.
D23.3 Equilibrium Approximation
When the rate-determining step is not the first step, the rate law can be approximated as the rate law for the rate-determining step. However, one or more of the reactants involved in the rate-determining step is a reaction intermediate formed from a previous step. Hence, the rate law for the rate-determining step includes the concentration of one or more reaction intermediates.
Experimentally, the concentration of a reaction intermediate is rarely measurable. (Even if it can be measured, the accuracy is usually low.) Therefore, experimentally determined rate laws are always expressed in terms of concentrations of reactants and/or products, for which accurate measurements are much easier to obtain.
In order to compare the theoretical rate law derived from the mechanism to the experimentally determined rate law, we must express the mechanism rate law only in terms of reactant and product concentrations, that is, we cannot simply use the rate law of the rate-determining step as is. If a preceding step that forms the reaction intermediate is at equilibrium, then we can make use of the equilibrium approximation to express the concentration of the intermediate in terms of concentrations of reactants.
For example, consider the following multi-step reaction:
2 NO(g) + Cl2(g) ⟶ 2 NOCl(g)
The currently accepted mechanism for this reaction is:
$\begin{array}{rcl} &\text{Step 1}:& \text{NO}(g) + \text{Cl}_2(g) &\xrightleftharpoons[k_{-1}]{k_1}& \text{NOCl}_2(g)\;\;\;\;\;\text{fast}\[0.5em]&\text{Step 2}:& \text{NOCl}_2(g) + \text{NO}(g) &\xrightarrow{k_2}& \text{2NOCl}(g)\;\;\;\;\;\text{slow}\[0.5em]&\text{Overall}:& \text{2NO}(g) + \text{Cl}_2(g) &\longrightarrow& \text{2NOCl}(g)\end{array} \nonumber$
Query $3$
Figure 2. Reaction energy diagram for the 2 NO(g) + Cl2(g) → 2 NOCl(g) reaction. Click on each “i” for more info.
Step 2 is the rate-determining step, and its rate law is:
rate2 = k2 [NOCl2] [NO]
However, this rate law involves the concentration of an intermediate, [NOCl2], so it cannot be compared to experimental data. We need to express [NOCl2] in terms of concentrations of reactants.
Step 1 in this mechanism is much faster than step 2 and it is a reversible reaction. Reversible means that not only can NO and Cl2 react to form NOCl2, but NOCl2 rapidly reacts to form NO and Cl2. That is, step 1 can go forward with rate constant k1 and backward with rate constant k-1; this is represented by the two arrows pointing in opposite directions.
Because step 2 is slow, reaction of NOCl2 with NO is much slower than reaction of NOCl2 to form NO and Cl2 (the reverse of step 1). Thus, when NOCl2 molecules are formed, they are more likely to decompose back to NO and Cl2 than to react in step 2. When enough NOCl2 has formed, we can approximate that the rate of its decomposition (rate-1) is equal to the rate at which it forms (rate1) and step 1 is at equilibrium (rate1 = rate-1). This approximation allows us to express the rate laws of step 1 as:
rate1 = rate-1
k1[NO][Cl2] = k-1[NOCl2]
This relationship can be rearranged to solve for the concentration of NOCl2:
$[\text{NOCl}_2] = \left(\dfrac{k_1}{k_{-1}}\right)[\text{NO}][\text{Cl}_2] \nonumber$
And substituting this into the rate law for step 2, we have:
$\text{rate}_2 = k_2[\text{NOCl}_2][\text{NO}] = k_2\left(\dfrac{k_1}{k_{-1}}\right)[\text{NO}][\text{Cl}_2][\text{NO}] = \left(\dfrac{k_1k_2}{k_{-1}}\right)[\text{NO}]^2[\text{Cl}_2] \nonumber$
If we make:
$k' = \dfrac{k_1k_2}{k_{-1}} \nonumber$
then the rate law for the overall reaction becomes:
rate = k’[NO]2[Cl2]
This rate law can be compared with experimental data to determine whether the proposed mechanism is a plausible one.
Activity 2: Rate Law from Mechanism
D23.4 Catalysts and Reaction Mechanisms
A catalyst increases the rate of a reaction by altering the mechanism, allowing the reaction to proceed via a pathway with lower activation energy than for the uncatalyzed reaction. A catalyzed mechanism must involve at least two steps, one where the catalyst interacts with a reactant to form an intermediate substance, and one where the intermediate then decomposes or reacts with another reactant in one or more steps to regenerate the original catalyst and form product. Hence, the catalyst is involved in the reaction mechanism but is not consumed by the reaction.
An extremely important example of catalysis involves the catalytic destruction of ozone in Earth’s stratosphere, 10 to 40 km above the surface (Section D22.1). Stratospheric ozone intercepts ultraviolet radiation from the Sun that otherwise would reach Earth’s surface, damaging many forms of life including humans. The reaction mechanism for this process is
$\begin{array}{rcl} &\text{Step 1: (fast)}& \text{O}_3(g) &\xrightarrow{h\nu}& \text{O}_2(g) + \text{O}(g)\[0.5em] &\text{Step 2: (slow)}& \text{O}_3(g) + \text{O}(g) &\longrightarrow& 2\;\text{O}_2(g)\[0.5em] &\text{Overall:}& 2\;\text{O}_3(g) &\xrightarrow{h\nu}& 3\;\text{O}_2(g)\end{array} \nonumber$
In the first step, a photon of ultraviolet (UV) radiation with wavelength between 200 and 310 nm breaks a bond in the ozone molecule, forming O2 and an O atom; the UV radiation is indicated by above the reaction arrow. In the second step the O atom from step 1 reacts with a second O3 molecule to form two oxygen molecules; this second step has higher activation energy and is the rate-limiting step.
Ozone is formed in the stratosphere by short-wavelength UV photons (wavelength below 240 nm) that break the double bonds in O2 molecules. The O atoms thus formed react with O2 molecules to form O3. The concentration of O3 in the stratosphere is a small, constant value because the rate of formation of O3 equals the rate at which it reacts away according to the mechanism above. Anything that speeds up the 2 O3 → 3 O2 reaction will reduce the concentration of ozone and allow more UV radiation to reach Earth’s surface.
One catalyst for the ozone decomposition reaction is chlorine atoms, which can be generated in the stratosphere from chlorofluorocarbon molecules, which at one time were used in air conditioners and cans of aerosol sprays. An example chlorofluorocarbon is CF2Cl2. An ultraviolet photon can break a C–Cl bond in CF2Cl2, producing Cl atoms, which react with ozone via this simplified mechanism:
Step 1: (slow) Cl(g) + O3(g) O2(g) + ClO(g)
Step 2: (fast) O3(g) + ClO(g) 2 O2(g) + Cl(g)
Overall: 2 O3(g) 3 O2(g)
Notice that Cl is a reactant in the first step and a product in the second step, so Cl participates in the mechanism but is not consumed by the overall reaction; that is, Cl is a catalyst. Because Cl is not reacted away, a single Cl atom can destroy as many as 100,000 O3 molecules before the Cl atom reacts with something else and is removed from the stratosphere. Discovery of the catalytic effect of Cl atoms led to an international agreement, the Montreal Protocol, halting production of chlorofluorocarbons and banning their use. The Montreal Protocol now has 197 signatory countries—an essentially unanimous international agreement that is reversing the trend to lower ozone concentrations in the stratosphere and greater UV radiation at Earth’s surface.
Exercise 1: Catalyst and Reaction Rate
Reactants or products can also be catalysts in a reaction. When a product catalyzes a reaction the reaction is called autocatalytic. An autocatalytic reaction can be dangerous because the reaction can “run away”; that is, it can speed up a lot as product is formed and produce products way too fast. An example of a reaction catalyzed by a reactant was in Day 23:
Step 1: NO2(g) + NO2(g) NO3(g) + NO(g)
Step 2: NO3(g) + CO(g) CO2(g) + NO2(g)
Overall: NO2(g) + CO(g) CO2(g) + NO(g)
The second NO2 molecule in step 1 (highlighted in green) is a catalyst, because it is reformed as a product in step 2. This catalyzed reaction has the first step as the rate-determining step, which yields a reaction rate of:
ratecatalyzed = kcatalyzed[NO2]2
Without the catalytic action, this reaction (NO2(g) + CO(g) ⟶ CO2(g) + NO(g)) would be a bimolecular elementary reaction (with only one transition state) with a rate law of:
rateuncatalyzed = kuncatalyzed[NO2][CO]
Figure 3 shows the reaction energy diagrams of the catalyzed and uncatalyzed reactions. Because the reactants and products involved in both reactions are exactly the same, they are at the same energies (a catalyst has no effect on the relative energies of the reactants and products).
Query $6$
Figure 3. Reaction energy diagrams of catalyzed (green) and uncatalyzed (brown) NO2(g) + CO(g) ⟶ CO2(g) + NO(g) reaction. Click on each “i” for more info.
The transition states, and therefore the activation energies, of the two pathway differ. The lower Ea in the catalyzed pathway results in kcatalyzed > kuncatalyzed, and the reaction proceeds almost entirely via the faster pathway.
In this particular example, the catalyzed pathway involves a two-step mechanism (note the presence of two transition states) and an intermediate species (represented by the valley between the two transitions states). Other catalyzed reactions might have more than two steps. Usually there are more mechanistic steps in a catalyzed reaction than in the uncatalyzed mechanism.
Podia Question
The important industrial chemical hydrazine, N2H4, is produced by the Raschig process. Here is a proposed mechanism for the process:
Step 1 (slow): NH3(aq) + OCl(aq) ⟶ NH2Cl(aq) + OH(aq)
Step 2 (fast): NH2Cl(aq) + NH3(aq) ⟶ N2H5+(aq) + Cl(aq)
Step 3 (fast): N2H5+(aq) + OH(aq) ⟶ N2H4(aq) + H2O(l)
1. Determine which is the rate-limiting step. Describe how this step limits the overall reaction rate.
2. Determine the overall reaction equation for the Raschig process.
3. Identify all reaction intermediates in the proposed mechanism.
4. Another student tells you that the reaction is second-order in ammonia and does not depend on any other concentration. The student says this is consistent with the mechanism written above. Explain why the student is, or is not, correct.
Query $7$
Two days before the next whole-class session, this Podia question will become live on Podia, where you can submit your answer.
Comments. If you found any inconsistencies, errors, or other things you would like to report about this module, please use this link to report them. A similar link will be included in each day’s material. We appreciate your comments. | textbooks/chem/General_Chemistry/Interactive_Chemistry_(Moore_Zhou_and_Garand)/03%3A_Unit_Three/3.06%3A_Day_23-_Reaction_Mechanisms.txt |
D24.1 Enzymes
Millions of chemical reactions occur within each of our cells every minute. Without catalysts, many of these reactions proceed at an extremely slow rate. For example, the dissociation of carbonic acid that takes place in the lungs:
H2CO3(aq) → CO2(g) + H2O(l)
only proceeds at a rate of ~10-7 M/s at room temperature. However, CO2 needs to be produced in our body at a much higher rate than this. While it’s theoretically possible to accelerate the reaction by raising the temperature, unfortunately, most life is only compatible with a limited temperature range.
Biological catalysts, known as enzymes, are crucial for life as they can accelerate reactions by factors of 106-1020. For example, the enzyme carbonic anhydrase accelerates the above reaction at room temperature by more than a million times the rate of the uncatalyzed reaction.
Like all catalysts, enzymes act by combining with reactant(s) and thereby forming lower energy transition states. In an enzyme-catalyzed reaction, the reactant(s) with which the enzyme combines is called the substrate(s). Unlike chemical catalysts, an enzyme’s interactions with substrate molecules are often partly or entirely noncovalent; that is, the interactions involve hydrogen bonding, ionic attractions, or dipole-dipole attractions.
In addition, enzyme-catalyzed reactions are highly specific to a particular substrate or a particular category of substrate molecules. Enzymes are said to “recognize” the substrate, or class of substrates, for which the enzyme serves as a catalyst. Enzyme interactions with substrates are sometimes so specific that an enzyme does not recognize a molecule that differs from its preferred substrate by as little as a single methyl (-CH3) group, and many enzymes will recognize one enantiomer but not its mirror image.
How do enzymes accelerate chemical reactions and achieve their specificity? The answer to both questions lies in how enzymes interact with their substrates. Typically an enzyme is a protein molecule—a macromolecule. Usually only part of the large enzyme molecule interacts with a substrate. The part of an enzyme molecule that interacts with a substrate is called the active site.
In 1890, the chemist Emil Fischer proposed that the substrate fits into the enzyme’s active site as a key fits into a lock. The key (substrate) has a specific molecular shape (arrangement of functional groups and other atoms) that allows it, and no other key, to fit into the lock (the enzyme).
In 1958, Daniel E. Koshland Jr. modified this lock-and-key model by proposing that binding of the substrate to the enzyme alters the configuration of both, providing a better fit. This induced-fit model explains both the large increase in reaction rates and the specificity of enzyme-catalyzed reactions:
• As the substrate interacts with the enzyme, the substrate is distorted (atoms are shifted, bonds are stretched, and reactive groups are brought close together) to a structure closer to the transition state of the reaction. This lowers the energy of the transition state, accelerating the reaction.
• Only molecules with the correct functional groups in the correct configurations are able to be induced to fit the active site of the enzyme.
For many years, it was believed that the only enzymes were proteins. Recently, however, several RNA enzymes have been discovered. These enzymes are called ribozymes and exhibit all the same features as protein enzymes (Figure 2).
D24.2 Enzyme Kinetics: Michaelis-Menten Mechanism
The Michaelis-Menten mechanism is a two-step reaction mechanism that applies to many enzyme-catalyzed reactions. An enzyme (E) binds with a substrate (S) to form an enzyme-substrate complex (ES), which then separates to give the product (P) and regenerates the enzyme. The overall reaction is:
S → P
with the simple two-step mechanism:
$\begin{array}{rcl}&\text{step 1: (fast)}&\;\;\;\;\;\;\;\;\;\;\;E + S &\xrightleftharpoons[k_{-1}]{k_1}& ES\[0.5em] &\text{step 2: (slow)}&\;\;\;\;\;\;\;\;\;\;\;ES &\xrightarrow{k_2}& P + E\end{array} \nonumber$
The second step is rate determining. Enzyme kinetics studies are typically carried out by measuring the reaction rate as a function of the concentration of substrate and the concentration of enzyme. The rate of the catalyzed reaction can be measured by observing the rate of the production of the product:
$rate=\dfrac{\Delta[P]}{\Delta t}=k_2[ES] \nonumber$
but ES is a reactive intermediate whose concentration is not easily determined. We need to express the reaction rate in terms of concentrations that are easily measured. This can be done by assuming that, once the reactive intermediate, ES, forms, its concentration remains approximately constant throughout the rest of the reaction; this is called the steady-state approximation. It differs from the equilibrium approximation we discussed in Section D23.3 in that it is not necessary to assume step 1 has reached equilibrium. The steady-state approximation is reasonable, because once [ES] builds up enough, the reactions that consume ES become faster and prevent the concentration from increasing further.
When the concentration of the enzyme-substrate complex reaches a steady state:
$\dfrac{d[ES]}{dt}=0 \nonumber$
The ES complex is formed by the forward reaction in step 1, and is reacted away in the reverse reaction of step 1 and forward reaction of step 2. Summing the rates of these reactions:
$\dfrac{d[ES]}{dt}=0=k_1[E][S]-k_{-1}[ES]-k_2[ES] \nonumber$
Rearranging the equation to solve for [ES] gives:
$(k_{-1}+k_2)[ES]=k_1[E][S] \nonumber$
$[ES]=\dfrac{k_1[E][S]}{k_{-1}+k_2} \nonumber$
or:
$[ES]=\dfrac{[E][S]}{K_M} \nonumber$
where KM is the Michaelis constant:
$K_M=\dfrac{k_{-1}+k_2}{k_1} \nonumber$
It is difficult to know the exact concentration of the free enzyme ([E]) at any given moment. However, we do know the total concentration of enzyme, [E]total (the enzyme must be either in its unreacted form (E) or combined with substrate (ES)):
[E]total = [E] + [ES]
Substituting into the above equation gives:
$[ES]=\dfrac{([E]_{\text{total}}-[ES])[S]}{K_M} \nonumber$
Rearranging the variables gives:
$[ES]=\dfrac{[E]_{\text{total}}[S]}{K_M}-\dfrac{[ES][S]}{K_M} \nonumber$
$\dfrac{[ES][S]}{K_M}+[ES]=\dfrac{[E]_{\text{total}}[S]}{K_M} \nonumber$
$\left(\dfrac{[S]+K_M}{K_M}\right)[ES]=\dfrac{[E]_{\text{total}}[S]}{K_M} \nonumber$
$[ES]=\left(\dfrac{[E]_{\text{total}}[S]}{K_M}\right)\left(\dfrac{K_M}{[S]+K_M}\right)=\dfrac{[E]_{\text{total}}[S]}{[S]+K_M} \nonumber$
Substituting the expression for [ES] into the expression for the reaction rate in terms of [ES} gives
$rate=\dfrac{\Delta[P]}{\Delta t}=\dfrac{k_2[E]_{\text{total}}[S]}{[S]+K_M} \nonumber$
or
$rate=\dfrac{V_{\text{max}}[S]}{[S]+K_M}\;\;\;\text{Michaelis-Menten equation} \nonumber$
where Vmax = k2[E]total is the maximum rate the reaction can achieve. There are two limiting cases for this equation, depending on the relative sizes of [S] and KM.
If there is a lot of substrate, [S] ≫ KM, and therefore [S] + KM ≈ [S]. In this case the Michaelis-Menten equation becomes
$\text{rate}=\dfrac{V_{\text{max}}[S]}{[S]}=V_{\text{max}}=k_2[E]_{\text{total}} \nonumber$
The reaction reaches its maximum rate, rate = Vmax. The reaction rate depends only on [E]total and is independent of [S]. This happens when almost all of the enzyme molecules have a substrate molecule in the active site, which means that increasing the concentration of substrate can no longer increase the concentration of enzyme-substrate complex (and thus can no longer increase the rate).
If there is a very small concentration of substrate, [S]≪KM, we have [S] + KMKM, and the Michaelis-Menten equation becomes first-order in [S]:
$rate=\left(\dfrac{V_{\text{max}}}{K_M}\right)[S] \nonumber$
The Michaelis constant is an inverse measure of the substrate’s affinity for the enzyme—a small KM indicates high affinity, meaning that the rate approaches Vmax more quickly. The value of KM depends on both the enzyme and the substrate, as well as reaction conditions such as temperature and pH.
D24.3 Enzyme Denaturation and Inhibitors
Denaturation is a process in which proteins lose their quaternary, tertiary and secondary structure (see Section D15.6). Enzymes must be folded into the right 3D shape to function. But hydrogen bonds, which play a big part in protein folding, are rather weak, and it does not take much heat, acidity, or other stress to break some hydrogen bonds and denature the enzyme.
Hence, enzyme-catalyzed reactions exhibit an unusual temperature dependence. At relatively low temperatures, the reaction rate increases with temperature, as expected. However, at higher temperatures, the reaction rate drops dramatically due to denaturation of the enzyme, as shown in Figure 4.
Usually enzymes will not denature at temperatures typically encountered by the living organism in which they are found. As a result, enzymes from bacteria living in high-temperature environments such as hot springs are prized by industrial users for their ability to function at high temperatures.
Exercise 1: Enzyme Characteristics
An inhibitor (I) interacts with an enzyme to decreases the enzyme’s catalytic efficiency. An irreversible inhibitor covalently binds to the enzyme’s active site, producing a permanent loss in catalytic efficiency even if the inhibitor’s concentration is later decreased. A reversible inhibitor forms a noncovalent complex with the enzyme, resulting in a temporary decrease in catalytic efficiency. Reducing the concentration of a reversible inhibitor returns the enzyme’s catalytic efficiency to its normal level.
There are several kinds of reversible inhibition. In competitive inhibition the substrate and the inhibitor compete for the same active site on the enzyme. Because the substrate cannot bind to an enzyme–inhibitor complex (EI), the concentration of enzyme available to form enzyme-substrate complex is lower and the enzyme-catalyzed reaction is slower. With uncompetitive inhibition the inhibitor binds to the enzyme-substrate complex but not to the active site, forming an enzyme–substrate–inhibitor (ESI) complex. The formation of an ESI complex decreases catalytic efficiency because some of the enzyme-substrate complex has reacted with inhibitor. This reduces the concentration of ES, which reduces the rate of the rate-limiting step in the mechanism. Finally, in noncompetitive inhibition the inhibitor binds to both the enzyme itself and the enzyme–substrate complex at a site different from the active site. As in uncompetitive inhibition, this forms an inactive ESI complex and reduces the concentration of ES.
Query $3$
Podia Question
Consider the reaction mechanism for Michaelis-Menten enzyme kinetics in Section D24.2:
1. Draw a reaction energy diagram for this reaction. If you need to make any assumptions to draw the diagram, describe each assumption. Also, describe which parts of the diagram you can draw definitively given the information in Section D24.2.
2. Suppose that a reversible, competitive inhibitor is present during the enzyme-catalyzed reaction. How does the presence of this inhibitor affect the reaction energy diagram?
Query $4$
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D25.1 Homogeneous Catalysis
Reactions that are facilitated by catalysts can be divided into two major classes: homogeneous catalysis and heterogeneous catalysis. A homogeneous catalyst is present in the same phase as the reactants. Gas-phase reactions and reactions in solution are homogeneous reactions and we have already discussed several examples where catalyst and reactants are all in the same phase. Here is one more.
In Section D13.5 we described condensation reactions and hydrolysis reactions that occur in aqueous solutions. (Hydrolysis is the reverse of a condensation that produces water as the small molecule.) We also mentioned that condensation and hydrolysis can be catalyzed by strong acids (that is, by H+ ions). This is a homogeneous catalytic reaction because reactants, products, and catalyst are all in aqueous solution.
An example of ester hydrolysis is the acid-catalyzed decomposition of methyl acetate to form acetic acid and methanol. The reaction mechanism is shown in Figure 1.
Activity 1: Analyzing a Reaction Mechanism
D25.2 Heterogeneous Catalysts
A heterogeneous catalyst is present in a different phase from the reactants. Such catalysts are usually solids, and often function by furnishing an active surface upon which one or more steps in the reaction can occur.
A heterogeneous catalytic reaction has at least four steps in its reaction mechanism:
1. Adsorption of the reactant(s) onto the surface of the catalyst
2. Activation of the adsorbed reactant(s)
3. Reaction of the adsorbed reactant(s)
4. Diffusion of the product(s) from the surface into the gas or liquid phase (desorption)
Any one of these steps may be slow and thus may serve as the rate determining step. But the overall rate of the reaction is still faster than it would be without the catalyst. Figure 1 illustrates the reaction of alkenes with hydrogen on a nickel catalyst.
The uncatalyzed C2H4(g) + H2(g) ⟶ C2H6(g) reaction would necessitate a transition state where the C=C π bond and the H-H σ bond are breaking while the C-H σ bonds form. Such a transition state is so high in energy that without a catalyst, H2 is considered as being unreactive towards alkenes under most conditions.
Nickel is a catalyst often used in the hydrogenation of polyunsaturated fats and oils to produce saturated fats and oils. Other significant industrial processes that involve the use of heterogeneous catalysts include the preparation of sulfuric acid, the preparation of ammonia, the oxidation of ammonia to nitric acid, and the synthesis of methanol. Heterogeneous catalysts are also used in the catalytic converters found on most gasoline-powered automobiles.
Query \(5\)
Podia Question
Platinum metal is a heterogeneous catalyst for this reaction:
2 NO(g) → N2(g) + O2(g)
The reaction rate varies with concentration of NO as shown in the graph.
Explain each of these observations. Include in your explanation an atomic-level description of NO molecules, the platinum surface, and how the two interact.
• The graph is linear with positive slope at low concentrations of NO.
• The graph is horizontal at high concentrations of NO.
Suggest an experiment that could be done to support or contradict your explanation of the horizontal graph. Describe the hypothesis you propose for what will happen in the experiment if your explanation is correct; also describe what experimental results would contradict your hypothesis.
Query \(6\)
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D27.1 Energy, Temperature, and Heat
Thermal energy is kinetic energy associated with the random motion of atoms and molecules. When thermal energy is transferred into an object, its atoms and molecules move faster on average (higher KEaverage), the object’s temperature increases, and we say that the object is “hotter”. When thermal energy is transferred out of an object, its atoms and molecules move more slowly on average (lower KEaverage), the object’s temperature decreases, and we say that the object is “colder”.
Heating (or heat), represented by q, is the transfer of thermal energy between two bodies at different temperatures (Figure 1).
The heat capacity (C) of an object is the heating required when the object’s temperature changes by 1 °C. (Because 1 K is the same size as 1 °C, ΔT has the same numeric value whether expressed in K or °C.) It typically has units of J/°C. Heat capacity depends on both the type and the quantity of substance, and therefore is an extensive property—its value is proportional to the quantity of the substance.
The specific heat capacity (c) of a substance is the heating required to raise the temperature of 1 g of a substance by 1 °C. It typically has units of J/g·°C. Specific heat capacity depends only on the type of substance and therefore is an intensive property. The molar heat capacity, also an intensive property, is the heat capacity per mole of a particular substance and typically has units of J/mol·°C.
Specific heat capacities of some common substances are listed in Table 1.
Substance Symbol (state) Specific Heat Capacity (J/g·°C)
helium He(g) 5.193
water H2O(l) 4.184
ethanol C2H6O(l) 2.376
ice H2O(s) 2.093 (at −10 °C)
water vapor H2O(g) 1.864
nitrogen N2(g) 1.040
oxygen O2(g) 0.918
aluminum Al(s) 0.897
carbon dioxide CO2(g) 0.853
argon Ar(g) 0.522
iron Fe(s) 0.449
copper Cu(s) 0.385
lead Pb(s) 0.130
gold Au(s) 0.129
silicon Si(s) 0.712
Table 1. Specific Heat Capacities of Common Substances at 25 °C and 1 bar
If we know the mass, m, of a sample and its specific heat capacity, c, we can calculate the heat transferred to or from the sample by measuring the temperature change during heating or cooling:
q = m·c·ΔT = m·c·(TfinalTinitial)
The sign of ΔT tells us whether the substance is being heated (positive value for q) or cooled (negative q ).
D27.2 Calorimetry
Calorimetry is an experimental technique used to quantitatively measure heat transfer of energy. Energy is exchanged with a calorimeter, a device with known heat capacity that therefore can relate ΔT to q. A calorimeter must be insulated thermally so that energy does not transfer beyond its physical boundaries. In calorimetry it is useful to define a system, the substance(s) undergoing the chemical or physical change, and the surroundings, everything else that can exchange energy with the system.
For example, if we place a piece of hot metal (M) into cool water (W), heat transfer of energy occurs from metal to water until the two substances reach the same temperature (Figure 2).
If this occurs in a well insulated calorimeter, this heat transfer would ideally occur only between the two substances. Thus, the magnitude of q is the same for both substances.
qM + qW = 0
The arithmetic sign of q is determined by whether the substance loses or gains energy. In our example, energy is transferred from the metal (qM is negative) to the water (qW is positive).
Exercise 4: Energy Transfer in a Calorimeter
The same principles apply when we apply calorimetry to determine the heat transfer of energy involved in chemical reactions:
qreaction + qsurroundings = 0
Here, qreaction is defined as the change in energy of all atoms present in reactants and products.
A reaction in which there is heat transfer from the reacting substances to their surroundings (a reaction that heats the surroundings) is an exothermic reaction. For example, the combustion reaction that occurs in the flame of a lit match is exothermic. A reaction in which there is heat transfer from the surroundings to the system (a reaction that cools the surroundings) is an endothermic reaction. For example, when the substances in a cold pack (water and a salt such as ammonium nitrate) are mixed, the resulting process transfers energy from the surroundings, making the surroundings colder.
Exercise 5: Energy Transfer to a Cold Pack
If the heat capacity of a calorimeter is too large to neglect or if we require more accurate results, then we must take into account energy transferred to or from the calorimeter as well as energy transfers within the the calorimeter. For example, a type of calorimeter that operates at constant volume, colloquially known as a bomb calorimeter because it involves a strong, steel container that will not explode when an exothermic reaction occurs inside it (the “bomb”). Bomb calorimetry is used to measure energy transfers for reactions such as combustion reactions. The reaction occurs inside the bomb (Figure 3), which is immersed in a water bath and the temperature change of the water bath is measured. Energy has to be transferred to the bomb so the bomb can heat the water, so you need to account for the heat capacity of the bomb as well as the water. That is,
$q_{\text{reaction}} = -[q_{\text{water}} + q_{\text{bomb}}] \nonumber$
D27.3 Enthalpy
Chemical thermodynamics deals with the relationships between heat, work, and other means of energy transfer in the context of chemical and physical processes. Work, represented by w, is a process that transfers kinetic energy to or from a macroscopic object. When a golf club strikes a golf ball, for example, the club does work on the ball, accelerating the ball to a high speed.
Substances act as reservoirs of energy. The total of all possible kinds of energy present in a substance is called the internal energy (U). (The symbol U, instead of E, represents a sum over several different kinds of energy within the substance.) Energy is transferred into a system when it is heated (q) by the surroundings or when the surroundings does work (w) on the system. For example, energy can be transferred into room-temperature metal wire if it is immersed in hot water (the water heats the wire), or if you rapidly bend the wire back and forth (the wire becomes warmer because of the work done on it). Both processes increase the internal energy of the wire, which is reflected in an increase in the wire’s temperature.
The relationship between internal energy, heat, and work can be represented by the equation:
ΔU = q + w
assuming there are no energy transfers other than heat and work. This is one version of the first law of thermodynamics, the law of conservation of energy. The equation shows that the internal energy of a system changes through heat transfer into or out of the system (positive q is heat transfer in; negative q is heat transfer out) or work done on or by the system. The work, w, is positive if it is done on the system (increases the system’s internal energy) and negative if it is done by the system. This is summarized in the figure below.
Chemists ordinarily use a property known as enthalpy (H) to describe the thermodynamics of chemical and physical processes. Enthalpy is defined by this equation:
H = U + PV
where P is pressure and V is volume. Enthalpy is closely related to energy, differing by the quantity PV.
Enthalpy values for specific substances cannot be measured directly. Only enthalpy changes for chemical or physical processes can be determined. For processes that take place at constant pressure (a common condition for many chemical and physical changes that take place at atmospheric pressure), the enthalpy change (ΔH) is:
ΔH = ΔU + Δ(PV) = ΔU + PΔV
(where P is factored out of Δ(PV) because P is constant).
The mathematical product PΔV represents pressure-volume work, also called expansion work. For example, consider a gas (the system) confined in a cylinder by a piston, as shown in the figure. The pressure exerted by the gas results from many collisions of gas molecules with the bottom of the piston and the walls of the cylinder. If the pressure of the gas causes the piston to move upward, the gas does work on the piston (part of the surroundings). This reference shows that the work done by the system on the surroundings equals PΔV. Work done by the system is the negative of w, which is defined as work done on the system, so
PΔV = –w
Substituting yields:
ΔH = ΔU + PΔV
= (qP + w) – w
= qP
where qp is the heat of reaction under constant pressure. Therefore, if a chemical or physical process is carried out at constant pressure with the only work done caused by expansion or contraction, then the heat transfer of energy (qp) equals the enthalpy change (ΔH) for the process.
The heat transfer to the surroundings when you operate a Bunsen burner is equal to the enthalpy change of the methane combustion reaction that takes place, because the reaction occurs at the essentially constant pressure of the atmosphere. Chemists usually perform experiments under normal atmospheric conditions, at constant external pressure with qp = ΔH, which makes enthalpy the most convenient choice for determining heat transfer of energy as a result of a chemical reaction. The enthalpy change for a chemical reaction is symbolized ΔrH, where the subscript r indicates that the change is for a reaction.
On the other hand, the heat transfer of energy by a reaction measured in a bomb calorimeter is not equal to ΔH because the closed, constant-volume metal container prevents expansion work from occurring. In a constant-volume system w = 0 and ΔU = qv + w= qv, where the subscript v indicates constant volume.
D27.4 Standard-state Reaction Enthalpy Change, ΔrH°
Heat effects of a chemical reaction are summarized in thermochemical expressions, balanced chemical equations together with values of ΔrH°, the standard-state reaction enthalpy change, and a temperature. The standard-state reaction enthalpy change, ΔrH°, is the standard-state enthalpy of pure, unmixed products minus the standard-state enthalpy of pure, unmixed reactants; that is, the enthalpy change for the reaction under standard-state conditions.
ΔrH° = (products) – (reactants)
A standard state is a commonly accepted set of conditions used as a reference point. For chemists, the standard state refers to substances under a pressure of 1 bar and solutions at a concentration of 1 mol/L (1 M). (Note that some thermochemical tables may list values with a standard state of 1 atm. Because 1 bar = 0.987 atm, thermochemical values are nearly the same under both sets of standard conditions; however, for accurate work the standard state should be checked.)
The standard state does not specify a temperature. Because ΔrH° can vary with temperature, temperature is typically specified in a thermochemical expression.
We will include a superscript “º” to designate standard state. Thus, the symbol ΔrH°₂₉₈ ₖ indicates an enthalpy change for a process occurring under standard-state conditions and at 298 K.
For example, consider this thermochemical expression:
2 H2(g) + O2(g) ⟶ 2 H2O(g) ΔrH° = 8.031 × 10−22 kJ = −483.6 kJ/mol (25 °C)
This refers to reaction of two molecules of hydrogen with 1 molecule of oxygen to form two molecules of water, all in the gas phase at 1 bar pressure. If this reaction equation took place once, the two hydrogen molecules and the one oxygen molecule would react to form two water molecules and 8.031 × 10−22 kJ would be transferred to the surroundings. Because we are interested in laboratory-scale reactions, where moles of reactants are involved, ΔrH° is always reported per mole of reaction rather than for a single reaction event. A mole of reaction involves a chemical reaction equation happening 6.022 × 1023 times; in this case that is 2 mol H2(g) reacting with 1 mol O2(g) to give 2 mol H2O(g). The heat transfer of energy to the surroundings is then 8.031 × 10−22 kJ × 6.022 × 1023 mol−1 = 483.6 kJ/mol. Because the energy transfer is from system to surroundings, the sign is negative and ΔrH° = −483.6 kJ/mol.
These conventions apply to thermochemical expressions:
1. In a thermochemical expression, the listed ΔrH° value indicates the heat transfer of energy for the coefficients in the chemical equation. If the coefficients are multiplied by some factor, ΔrH° must be multiplied by that same factor. In other words, ΔrH° is an extensive property. For example:
2 H2(g) + O2(g) ⟶ 2 H2O(g) ΔrH° = −483.6 kJ/mol
two-fold increase: 4 H2(g) + 2 O2(g) ⟶ 4 H2O(g) ΔrH° = 2(−483.6 kJ/mol) = -967.2 kJ/mol
two-fold decrease: H2(g) + ½ O2(g) ⟶ H2O(g) ΔrH° = ½(−483.6 kJ/mol) = -241.8 kJ/mol
2. ΔrH° of a reaction depends on the physical state of the reactants and products (whether we have gases, liquids, solids, or aqueous solutions), so physical states must be shown.
3. A negative ΔrH° indicates an exothermic reaction; a positive ΔrH° indicates an endothermic reaction. If the direction of a chemical equation is reversed, the arithmetic sign of its ΔrH° is changed. (A process that is endothermic when reactants change into products is exothermic when products change into reactants).
Be sure to take both stoichiometry and limiting reactants into account when determining the ΔrH° for a chemical reaction.
D27.5 Bond Enthalpy and Reaction Enthalpy Change
The enthalpy change for a gas-phase chemical reaction, ΔrH, equals the sum of the enthalpy required to break each of the bonds in the reactant molecules (energy in, positive sign) plus the sum of the enthalpy released when each of the bonds in the product molecules forms (energy out, negative sign). This can be expressed mathematically as:
ΔrH = ∑Ebonds broken – ∑Ebonds formed
Because the bond enthalpy values provided in the Appendix are averaged over many different molecules for each type of bond, this calculation is not exact, but it provides a good estimate for the enthalpy change of any specific reaction. Also, bond enthalpy calculations assume that all molecules are far from each other (which means that reactants and products must be in the gas phase). Additional enthalpy changes occur when a gas condenses to a liquid or solid or dissolves into a solution; these transformations are not accounted for by bond enthalpies.
Consider this balanced reaction:
H2(g) + Cl2(g) → 2 HCl(g)
One H–H bond (436 kJ/mol) and one Cl–Cl bond (242 kJ/mol) are broken; two H-Cl bonds (431 kJ/mol each) are formed. Representing bond enthalpies by Dbond, we have:
ΔrH = ∑Ebonds broken – ∑Ebonds formed = [DH-H + DCl-Cl] – [DH-Cl + DH-Cl]
ΔrH = [436 kJ/mol + 242 kJ/mol] – 2(431 kJ/mol) = -184 kJ/mol
There are the same number of bonds formed as were broken. Because the bonds in the products are stronger than those in the reactants, the reaction has a net release (negative sign) of about 184 kJ for every mole of reaction as written. The energy released increases the temperature of the surroundings (the reaction is exothermic).
Here are two general rules to predict whether a chemical reaction releases energy (is exothermic):
• If there are more bonds in the product molecules than in the reactant molecules and the bonds have about the same strength, the reaction is exothermic.
• If there are stronger bonds in the product molecules than in the reactant molecules and the number of bonds is the same in reactants and products, the reaction is exothermic.
Exercise 8: Enthalpy Change from Bond Enthalpies
Check Your Learning
D27.6 Hess’s Law
The enthalpy change for a particular reaction is the same regardless of which or how many steps the reaction is carried out in. This conveniently allows us to calculate the heat transfer for a chemical change from other experimentally determined enthalpy changes. This type of calculation usually involves Hess’s law: If a process can be written as the sum of several stepwise processes, the enthalpy change of the total process equals the sum of the enthalpy changes of the various steps.
For example, we can think of this reaction occurring in a single step:
C(s) + O2(g) ⟶ CO2(g) ΔrH° = -394 kJ/mol
or in a two-step process:
step 1: C(s) + ½O2(g) CO(g) ΔrH° = -111 kJ/mol
step 2: CO(g) + ½O2(g) CO2(g) ΔrH° = -283 kJ/mol
According to Hess’s law, the ΔrH° of the one-step reaction equals the sum of the ΔrH° of the two steps in the two-step reaction:
step 1: C(s) + ½O2(g) CO(g) ΔrH° = -111 kJ/mol
step 2: CO(g) + ½O2(g) CO2(g) ΔrH° = -283 kJ/mol
Sum: C(s) + O2(g) CO2(g) ΔrH° = -394 kJ/mol
This concept is illustrated in Figure 4.
D27.7 Standard Enthalpy of Formation
Standard enthalpy of formation, ΔfH°, is the enthalpy change for a reaction in which exactly one mole of a pure substance in a specified state (s, l, or g) is formed from free elements in their most stable states under standard-state conditions. ΔfH° is also referred to as the standard heat of formation.
For example, ΔfH° of CO2(g) at 25 °C is −393.5 kJ/mol. This is the enthalpy change for the exothermic reaction:
C(s, graphite) + O2(g) ⟶ CO2(g) ΔfH° = -393.5 kJ/mol (25 °C)
The gaseous reactant and product are at a pressure of 1 bar, the carbon is present as graphite, which is the most stable form of carbon under standard conditions.
For nitrogen dioxide, NO2(g), ΔfH° is 33.2 kJ/mol at 25 °C:
½N2(g) + O2(g) ⟶ NO2(g) ΔfH° = +33.2 kJ/mol (25 °C)
A reaction equation with ½ mole of N2 and 1 mole O2 is appropriate in this case because the standard enthalpy of formation always refers to formation of 1 mol NO2(g).
By definition, the standard enthalpy of formation of an element in its most stable form is 0 kJ/mol under standard conditions. A table of ΔfH° values for many common substances can be found in the Appendix.
Exercise 9: Equations for Standard Formation Enthalpy
Hess’s law can be used to determine the ΔrH° of any reaction if the ΔfH° of the reactants and products are available. In other words, we can think of any reaction as occurring via step-wise decomposition of the reactants into their component elements followed by re-combination of the elements to give the products. (Almost no reaction would actually occur via such a mechanism because some of the reactions would have large activation-energy barriers and consequently very small rates, but we can imagine such a path for the sole purpose of using Hess’s law to calculate ΔrH°.)
The ΔrH° of the overall reaction is therefore equal to:
ΔrH° = ∑ΔfH°(products) – ∑ΔfH°(reactants)
Exercise 10: Using Standard Formation Enthalpies
Activity 1: Applying Hess’s Law to Standard Formation Enthalpies
Query $12$
Podia Question
Chlorine monofluoride can react with fluorine to form chlorine trifluoride:
$(i)\;\text{ClF}(g) + \text{F}_2(g) \longrightarrow \text{ClF}_3(g) \;\;\;\;\; \Delta _\text{r}H\text{°}= \;? \nonumber$
Use the reactions here to determine the ΔrH° for reaction (i):
$(ii)\;2\text{OF}_2(g) \longrightarrow \text{O}_2(g) + 2\text{F}_2(g) \;\;\;\;\; \Delta _\text{r}H\text{°}_{(ii)} = -49.4 \;\text{kJ/mol} \nonumber$
$(iii)\;2\text{ClF}(g) + \text{O}_2(g) \longrightarrow \text{Cl}_2 \text{O}(g) + \text{OF}_2(g) \;\;\;\;\; \Delta _\text{r}H\text{°}_{(iii)} = +205.6 \;\text{kJ/mol} \nonumber$
$(iv)\;\text{ClF}_3(g) + \text{O}_2(g) \longrightarrow \dfrac{1}{2}\text{Cl}_2 \text{O}(g) + \dfrac{3}{2} \text{OF}_2(g) \;\;\;\;\; \Delta_\text{r}H\text{°}_{(iv)} = +266.7 \;\text{kJ/mol} \nonumber$
Query $13$
Two days before the next whole-class session, this Podia question will become live on Podia, where you can submit your answer.
Comments. If you found any inconsistencies, errors, or other things you would like to report about this module, please use this link to report them. A similar link will be included in each day’s material. We appreciate your comments. | textbooks/chem/General_Chemistry/Interactive_Chemistry_(Moore_Zhou_and_Garand)/04%3A_Unit_Four/4.01%3A_Day_27-_Thermochemistry_and_Enthalpy.txt |
D28.1 Entropy and Microstates
An important goal of chemical thermodynamics is to predict whether reactants are changed to products or products are changed to reactants. For a given reaction, such a prediction requires knowledge of specific conditions of temperature and of concentrations or partial pressures of reactants and products. When reactants change to products we say that the reaction is spontaneous, or spontaneous in the forward direction. When products change to reactants, we say the reaction is not spontaneous, or that it is spontaneous in the reverse direction. In this context, the word “spontaneous” does not imply that the reaction is fast or slow, just that reactants change to products. Even if it takes millions of years for a process to occur, if there is an overall change of reactants to products we call the process spontaneous.
It is useful to define parallel terms that relate to a convenient reference point—standard-state conditions. If, when all substances are at the standard-state pressure of 1 bar or at the standard-state concentration of 1 M, reactants change to products, we call a reaction product-favored. In contrast, if products change to reactants under standard-state conditions, a process is reactant-favored. That is, a product-favored process is spontaneous under the specific conditions of standard-state pressures or concentrations and a reactant-favored process is not spontaneous under those specific conditions..
Deciding whether a process is spontaneous requires knowledge of the system’s enthalpy change, but that is not sufficient. It also requires knowledge of change in another property: entropy. (Similarly, deciding whether a process is product-favored requires knowledge of the system’s standard enthalpy change and standard entropy change.) The entropy change, ΔS, at constant temperature is defined as:
${\Delta}S = \dfrac{q_{\text{rev}}}{T} \nonumber$
Here, qrev is the heat transfer of energy for a reversible process, a theoretical process that takes place at such a slow rate that it is always at equilibrium and its direction can be changed (it can be “reversed”) by an infinitesimally small change in some condition. An example of a reversible process is melting water at 0 °C and 1 bar, where liquid water and ice are at equilibrium. Raising the temperature a tiny bit causes the ice to melt; lowering the temperature a tiny bit reverses the process, causing liquid water to freeze.
On a molecular scale, the entropy of a system can be related to the number of possible microstates (W). A microstate is a specific configuration of the locations and energies of the atoms or molecules that comprise a system. The relationship is:
S = kB·ln(W)
where kB is the Boltzmann constant with a value of 1.38×10−23 J/K.
Similar to enthalpy, the change in entropy for a process is the difference between its final (Sf) and initial (Si) values:
${\Delta}S = S_{\text{f}}\;-\;S_{\text{i}} = k_\text{B}\cdot\text{ln}(W_{\text{f}})\;-\;k_\text{B}\cdot\text{ln}(W_{\text{i}}) = k_\text{B}\cdot\text{ln}\left(\dfrac{W_{\text{f}}}{W_{\text{i}}}\right) \nonumber$
For processes involving an increase in the number of microstates, Wf > Wi, the entropy of the system increases, ΔS > 0. Conversely, processes that reduce the number of microstates, Wf < Wi, yield a decrease in system entropy, ΔS < 0.
Consider the general case of a system comprised of N particles distributed among n boxes. The number of microstates possible for such a system is nN. For example, suppose four atoms, one atom each of He, Ne, Ar, and Kr, are distributed among two boxes as illustrated in Figure 1. There are 24 = 16 different microstates. Microstates with equivalent particle arrangements (not considering individual particle identities) are grouped together and are called distributions. The probability that a system exists as a given distribution is proportional to the number of microstates within the distribution. Because entropy increases logarithmically with the number of microstates, the most probable distribution is the one of greatest entropy.
For the system in Figure 1, the most probable distribution is (c), where the particles are evenly distributed between the boxes. The probability of finding the system in this configuration is $\frac{6}{16}$ or 37.5%. The least probable configuration of the system is distributions (a) and (e), where all four particles are in one box, and each distribution has a probability of $\frac{1}{16}$.
As you add more particles to the system, the number of possible microstates increases exponentially (2N). A macroscopic system typically consists of moles of particles (N ≈ 1023), and the corresponding number of microstates is staggeringly huge. Regardless of the number of particles in a system, the distributions with uniform dispersal of particles between the boxes have the greatest number of microstates and are the most probable.
Activity 1: Entropy Change and Microstates
Query $1$
Consider another system as shown in Figure 2. This system consists of two objects, AB and CD, and two units of energy (represented as “*”). Distribution (a) shows the three microstates possible for the initial state of the system, where both units of energy are contained within the hot AB object. If one energy unit is transferred, the result is distribution (b) consisting of four microstates. If both energy units are transferred, the result is distribution (c) consisting of three microstates.
Hence, we may describe this system by a total of ten microstates. The probability that there is no heat transfer of energy when the two objects are brought into contact (the system remains in distribution (a)) is 30%. It is much more likely for heat transfer to occur and yield either distribution (b) or (c), the combined probability being 70%. The most likely result, with 40% probability, is heat transfer to yield the uniform dispersal of energy represented by distribution (b).
This simple example supports the common observation that placing hot and cold objects in contact results in heat transfer that ultimately equalizes the objects’ temperatures. Such a process is characterized by an increase in system entropy.
D28.2 Predicting the Sign of ΔS
The relationships between entropy, microstates, and matter/energy dispersal make it possible for us to assess relative entropies of substances and predict the sign of entropy changes for chemical and physical processes.
Consider the phase changes illustrated in Figure 3. In the solid phase, the molecules are restricted to nearly fixed positions relative to each other and only oscillate a little about these positions; the number of microstates (Wsolid) is relatively small. In the liquid phase, the molecules can move over and around each other, though they remain in relatively close proximity. This increased freedom of motion results in a greater variation in possible particle locations, Wliquid > Wsolid. As a result, Sliquid > Ssolid and the process of melting is characterized by an increase in entropy, ΔS > 0.
Solids and liquids have surfaces that define their volume, but in the gas phase the molecules occupy the entire container; therefore, for the same sample, each molecule in a gas can be found in many more locations (and there are many more microstates) than in the liquid or solid phase. Consequently, Sgas > Sliquid > Ssolid, and the processes of vaporization and sublimation involve increases in entropy, ΔS > 0.
We have already discussed that the temperature of a substance is proportional to the average kinetic energy of its particles. Raising the temperature results in more extensive vibrations of the particles in solids and more rapid movements of the particles in liquids and gases. At higher temperatures, the Maxwell-Boltzmann distribution of molecular kinetic energies is also broader than at lower temperatures; that is, there is a greater range of energies of the particles. Thus, the entropy for any substance increases with temperature (Figure 4).
The entropy of a substance is also influenced by the structure of the particles that comprise the substance. For atomic substances, heavier atoms possess greater entropy at a given temperature than lighter atoms. For heavier atoms the energy levels corresponding to movement from one place to another are closer together, which means that at a given temperature there are more occupied energy levels and more microstates. For molecules, greater number of atoms (regardless of their masses) within a molecule gives rise to more ways in which the molecule can vibrate, which increases the number of possible microstates. Thus, the more atoms there are in a molecule the greater the entropy is.
Finally, variations in the types of particles affects the entropy of a system. Compared to a pure substance, in which all particles are identical, the entropy of a mixture of different particle types is greater due to the additional orientations and interactions that are possible. For example, when a solid dissolves in a liquid, the particles of the solid experience both a greater freedom of motion and additional interactions with the solvent particles. This corresponds to a more uniform dispersal of matter and energy and a greater number of microstates. All other things being equal, the process of dissolution therefore involves an increase in entropy, ΔS > 0.
D28.3 Second Law of Thermodynamics
The second law of thermodynamics enables predictions of whether a process, such as a chemical reaction, is spontaneous and whether the process is product-favored.
In thermodynamic models, the system and surroundings comprise everything (in other words, the universe), and so
ΔSuniv = ΔSsys + ΔSsurr
To illustrate this relation, consider the process of heat transfer between two objects, one identified as the system and the other as the surroundings. There are three possibilities for such a process:
1. The objects are at different temperatures, and energy transfers from the hotter object to the cooler object. This is always observed to occur. Designating the hotter object as the system and invoking the definition of entropy yields:
${\Delta}S_{\text{sys}} = \dfrac{-q_{\text{rev}}}{T_{\text{sys}}}\;\;\;\;\;\;\;\text{and}\;\;\;\;\;\;\;{\Delta}S_{\text{surr}} = \dfrac{q_{\text{rev}}}{T_{\text{surr}}} \nonumber$
The arithmetic signs of qrev denote the loss of energy by the system and the gain of energy by the surroundings. Since Tsys > Tsurr in this scenario, ΔSsurr is positive and its magnitude is greater than the magnitude of ΔSsys. Thus, ΔSsys and ΔSsurr sum to a positive value for ΔSuniv. This process involves an increase in the entropy of the universe.
2. The objects are at different temperatures, and energy transfers from the cooler object to the hotter object. This is never observed to occur. Again designating the hotter object as the system:
${\Delta}S_{\text{sys}} = \dfrac{q_{\text{rev}}}{T_{\text{sys}}}\;\;\;\;\;\;\;\text{and}\;\;\;\;\;\;\;{\Delta}S_{\text{surr}} = \dfrac{-q_{\text{rev}}}{T_{\text{surr}}} \nonumber$
The magnitude of ΔSsurr is again greater than that for ΔSsys, but in this case, the sign of ΔSsurr is negative, yielding a negative value for ΔSuniv. This process involves a decrease in the entropy of the universe. (Note also that possibility 1, which is the reverse of this process, always occurs.)
3. The temperature difference between the objects is infinitesimally small, TsysTsurr, and so the heat transfer is thermodynamically reversible. In this case, the system and surroundings experience entropy changes that are equal in magnitude and therefore sum to a value of zero for ΔSuniv. This process involves no change in the entropy of the universe.
These results lead to the second law of thermodynamics: all changes that take place of their own accord (are spontaneous) involve an increase in the entropy of the universe.
ΔSuniv > 0 spontaneous (takes place of its own accord)
ΔSuniv < 0 not spontaneous (reverse reaction would occur)
ΔSuniv = 0 system is at equilibrium
For many realistic applications, the surroundings are vast in comparison to the system. In such cases, the heat transfer of energy to or from the surroundings as a result of some process is a nearly infinitesimal fraction of its total thermal energy. For example, combustion of a hydrocarbon fuel in air involves heat transfer from a system (the fuel and oxygen molecules reacting to form carbon dioxide and water) to surroundings that are significantly more massive (Earth’s atmosphere). As a result, qsurr is a good approximation of qrev, and the second law may be stated as:
ΔSuniv = ΔSsys + ΔSsurr= ΔSsys + qsurr/T
We can use this equation to predict whether a process is spontaneous.
Activity 2: Enthalpy, Entropy, and Spontaneous Reactions
D28.4 Third Law of Thermodynamics
Consider the entropy of a pure, perfectly crystalline solid possessing no kinetic energy (that is, at a temperature of absolute zero, 0 K). This system may be described by a single microstate, as its purity, perfect crystallinity and complete lack of motion means there is but one possible location for each identical molecule comprising the crystal (W = 1). According to the Boltzmann equation, the entropy of this system is zero:
S = kB·ln(W) = kB·ln(1) = 0
This limiting condition for a system’s entropy represents the third law of thermodynamics: the entropy of a pure, perfect crystalline substance at 0 K is zero.
Starting with zero entropy at absolute zero, it is possible to make careful calorimetric measurements (qrev/T) to determine the temperature dependence of a substance’s entropy and to derive absolute entropy values at higher temperatures. (Note that, unlike enthalpy values, the third law of thermodynamics identifies a zero point for entropy; thus, there is no need for formation enthalpies and every substance, including elements in their most stable states, has an absolute entropy.) Standard entropy (S°) values are the absolute entropies per mole of substance at a pressure of 1 bar or a concentration of 1 M. The standard entropy change (ΔrS°) for any chemical process may be computed from the standard entropy of its reactant and product species:
ΔrS° = ∑S°(products) − ∑S°(reactants)
The thermodynamics table in the appendix lists standard entropies of select compounds at 298.15 K.
Exercise 2: Standard Entropy Change
Suppose an exothermic chemical reaction takes place at constant atmospheric pressure. There is a heat transfer of energy from the reaction system to the surroundings, qsurr = –qsys. The heat transfer for the system is the enthalpy change of the reaction because, at constant pressure, ΔrH° = q (Section D27.3). Because the energy transfer to the surroundings is reversible, the entropy change for the surroundings can also be expressed as:
${\Delta}S_{\text{surr}} = \dfrac{q_{\text{rev}}}{T} = \dfrac{q_{\text{surr}}}{T} = \dfrac{-q_{\text{sys}}}{T} = \dfrac{-\Delta H_{\text{sys}}}{T} = \dfrac{-\Delta _{\text{r}}H^{\circ}}{T} \nonumber$
The same reasoning applies to an endothermic reaction: qsys and qsurr are equal but have opposite sign.
Also, for a chemical reaction system, ΔSsys = ΔrS° (the standard entropy change for the reaction). Hence, ΔSuniv can be expressed as:
ΔSuniv = ΔSsys + ΔSsurr= ΔrS° − ΔrH°/T
The convenience of this equation is that, for a given reaction, ΔSuniv can be calculated from thermodynamics data for the system only. That is, from data found in the Appendix.
D28.5 Gibbs Free Energy
A new thermodynamic property was introduced in the late 19th century by American mathematician Josiah Willard Gibbs. The property is called the Gibbs free energy (G) (or simply the free energy), and is defined in terms of a system’s enthalpy, entropy, and temperature:
Gsys = Hsys − TSsys
The change in Gibbs free energy (ΔG) at constant temperature may be expressed as:
ΔGsys = ΔHsysTΔSsys
ΔG is related to whether a process is spontaneous. This relationship can be seen by comparing to the second law of thermodynamics:
ΔSuniv= ΔSsys + ΔSsurr
ΔSuniv = ΔSsys – ΔHsys/T
Multiplying both sides of this equation by −T, and rearranging, yields:
TΔSuniv = ΔHsysTΔSsys
Which can be compared to the equation:
ΔGsys = ΔHsysTΔSsys
Hence:
ΔGsys= TΔSuniv
For a process that is spontaneous, ΔSuniv must be positive. Because thermodynamic temperature is always positive (values are in kelvins), ΔG must be negative for a process that proceeds forward of its own accord.
ΔSuniv > 0 ΔGsys < 0 spontaneous (takes place of its own accord)
ΔSuniv < 0 ΔGsys > 0 not spontaneous (reverse reaction would occur)
ΔSuniv = 0 ΔGsys = 0 at equilibrium
D28.6 Calculating ΔG°
A convenient and common approach to calculate ΔrG° for reactions is by using standard state thermodynamic data. One method involves determining the ΔrH° and ΔrS° first, then using the equation:
ΔrG° = ΔrH° TΔrS°
Exercise 3: Standard Gibbs Free Energy Change from Enthalpy and Entropy Changes
It is also possible to use the standard Gibbs free energy of formationfG°) of the reactants and products involved in the reaction to calculate ΔrG°. ΔfG° is the Gibbs free energy change that accompanies the formation of one mole of a substance from its elements in their standard states. It is by definition zero for the most stable form of an elemental substance under standard state conditions.
For a generic reaction:
mA + nB ⟶ xC + yD
ΔrG° from ΔfG° would be:
ΔrG° = ∑ΔfG°(products) − ∑ΔfG°(reactants) = [xΔfG°(C) + yΔfG°(D)] – [mΔfG°(A) + nΔfG°(B)]
Exercise 4: Calculating Standard Gibbs Free Energy Change
Podia Question
Use data from the Appendix to calculate ΔrS° for this reaction at 298 K:
2 NO(g) + O2(g) → 2 NO2(g)
colorless gas colorless gas → red-brown gas
Predict whether the reaction is product-favored (spontaneous under standard-state conditions) or reactant-favored (not spontaneous under standard-state conditions).
After making your prediction, watch this video where NO(g) is added to O2(g) in a flask:
https://mediaspace.wisc.edu/id/1_5g4s60ey
Does the video validate your prediction? Explain why or why not.
If the video does not validate your prediction, try to figure out why your prediction did not work and re-do it in a different way.
If the video does validate your prediction, explain what thermodynamic aspect of the reaction accounts for the observations in the video.
Query $8$
Two days before the next whole-class session, this Podia question will become live on Podia, where you can submit your answer.
Comments. If you found any inconsistencies, errors, or other things you would like to report about this module, please use this link to report them. A similar link will be included in each day’s material. We appreciate your comments. | textbooks/chem/General_Chemistry/Interactive_Chemistry_(Moore_Zhou_and_Garand)/04%3A_Unit_Four/4.02%3A_Day_28-_Entropy_Gibbs_Free_Energy.txt |
D29.1 Temperature Dependence of Gibbs Free Energy
Whether a reaction is product-favored, that is, whether the reactants are converted to products under standard-state conditions, is reflected in the arithmetic sign of its ΔrG°. This equation
ΔrG° = ΔrH° − TΔrS°
shows that the sign of ΔrG° depends on the signs of ΔrH° and ΔrS°, and, in some cases, the absolute temperature (which can only have positive values). Four possibilities exist:
1. Both ΔrH° and ΔrS° are positive—an endothermic process with an increase in system entropy. ΔrG° is negative if TΔrS° > ΔrH°, and positive if TΔrS° < ΔrH°. Such a process is product-favored at high temperatures and reactant-favored at low temperatures.
2. Both ΔrH° and ΔrS° are negative—an exothermic process with a decrease in system entropy. Δr is negative if |TΔrS°| < |ΔrH°| and positive if |TΔrS°| > |ΔrH°|. Such a process is product-favored at low temperatures and reactant-favored at high temperatures. (Remember that |TΔrS°| represents the magnitude of TΔrS°, ignoring mathematical sign.)
3. ΔrH° is positive and ΔrS° is negative—an endothermic process that with a decrease in system entropy. ΔrG° is positive regardless of the temperature. Such a process is reactant-favored at all temperatures.
4. ΔrH° is negative and ΔrS° is positive—an exothermic process with an increase in system entropy. ΔrG° is negative regardless of the temperature. Such a process is product-favored at all temperatures.
These four scenarios are summarized in Figure 1.
Activity 1: Temperature and Product-favored or Reactant-favored Reactions
Query $1$
Figure 2 illustrates the four scenarios graphically, where ΔrG° is plotted versus temperature:
ΔrG° = ΔrS°(T) + ΔrH°
y = m(x) + b
Query $2$
Figure 2. These plots show the variation in ΔrGo with temperature for the four possible combinations of arithmetic sign for ΔrGo. Click on each “+” for more information.
For most reactions, neither ΔrH° nor ΔrS° change significantly as temperature changes. Thus, in Figure 2, the lines representing ΔrG° are linear because the slope of each line (−ΔrS°) is the same at all temperatures. The orange and green plots (representing examples of scenario 1 and 2, respectively) cross from product-favored to reactant-favored (as reflected by the sign of ΔrG°) at a temperature that is characteristic of the specific process. This temperature is represented by the x-intercept, the value of T for which ΔrG° is zero:
ΔrG° = 0 = ΔrH°TΔrS°
$T_{\Delta_{r}G^{\circ}=0} = \dfrac{\Delta_{r}H^{\circ}}{\Delta_{r}S^{\circ}} \nonumber$
Hence, saying a process is product-favored at “high” or “low” temperatures is simply indicating whether the temperature is above or below TΔrG°=0. These relative terms are reaction-specific, that is, what is a “high” temperature for one reaction may very well be a “low” temperature for another reaction.
D29.2 Chemical Equilibrium
A chemical reaction is usually written with a single arrow, which suggests it proceeds in one direction, the direction of the arrow. But all chemical reactions are reversible, and both the forward and reverse reaction occur simultaneously. When reactions involve gases or solutions, where concentrations change as the reaction proceeds, the reaction eventually reaches a dynamic chemical equilibrium.
In a chemical equilibrium, the forward and reverse reactions occur at the same rates, and the concentrations of products and reactants remain constant over time. This implies that, if a reaction occurs in a closed system so that the products cannot escape, the reaction often does not yield 100% products. Instead, some reactants remain after the concentrations stop changing. At this point, when there is no further change in concentrations of reactants and products, we say the reaction is at equilibrium.
For example, when we place a sample of dinitrogen tetraoxide (N2O4, a colorless gas) in a glass vessel, the color becomes darker as N2O4 is converted to nitrogen dioxide (NO2, a red-brown gas) by the reaction:
$\text{N}_2\text{O}_4(g)\;\xrightleftharpoons[k_r]{k_f}\;2\text{NO}_2(g) \nonumber$
Activity 2: Reaction Energy Diagram
Query $4$
At the beginning of this reaction, there is pure N2O4. As soon as the forward reaction produces some NO2, at rate = kf[N2O4]t, the reverse reaction begins to occur at rate = kr[NO2]t2, and NO2 starts to react to form N2O4. (The subscripts, t, indicate a time before equilibrium is reached.) As the reaction proceeds, the rate of the forward reaction decreases as [N2O4]t decreases and the rate of the reverse reaction increases as [NO2]t increases. When the system reaches equilibrium, both N2O4 and NO2 are present.
Query $5$
Figure 3. A container of N2O4(g) reacts toward equilibrium at 75°C. Colorless N2O4 converts to brown NO2. As the reaction proceeds toward equilibrium, the color of the mixture darkens due to the increasing concentration of NO2. Once equilibrium has been reached, there is no further darkening of color.
At equilibrium, [N2O4] and [NO2] no longer change over time because the rate of NO2 formation is exactly equal to the rate of NO2 consumption, and the rate of N2O4 formation is exactly equal to the rate of N2O4 consumption. Chemical equilibrium is a dynamic process: the numbers of reactant and product molecules remain constant, but the forward and reverse reactions do not stop.
We use the ⇌ arrow when writing an equation for a reversible reaction. Such a reaction may or may not be at equilibrium. When we wish to speak about one particular aspect of a reversible reaction, we use a single arrow. For example, when the reaction in Figure 3 is at equilibrium, the rate of the forward reaction:
N2O4(g) → 2 NO2(g)
is equal to the rate of the reverse reaction:
2 NO2(g) → N2O4(g)
An equilibrium can be established for a physical change as well as for a chemical reaction. For example:
Br2(l) ⇌ Br2(g)
Figure 4 shows a sample of liquid Br2 at equilibrium with Br2 vapor in a closed container. When we pour liquid Br2 into an empty bottle in which there is no bromine vapor, some liquid evaporates: the amount of liquid decreases and the amount of vapor increases. If we seal the container so no vapor escapes, the amount of liquid and vapor will eventually stop changing; at that point an equilibrium between the liquid and the vapor has been established. If the container were not sealed, the bromine vapor would escape and no equilibrium would be reached.
D29.3 Concentration Equilibrium Constants
A concentration equilibrium constant (Kc ) is a ratio of equilibrium concentrations of products and reactants that is constant for a given reaction at a given temperature. For example, consider this generic reversible reaction:
m A + n B ⇌ x C + y D
For this reaction, the concentration equilibrium constant, Kc, is:
$K_c = \dfrac{[\text{C}]_e^{\;x}\;[\text{D}]_e^{\;y}}{[\text{A}]_e^{\;m}\;[\text{B}]_e^{\;n}} \nonumber$
This mathematical expression is called the equilibrium constant expression. The “[…]e” expression indicates explicitly equilibrium concentration of a reactant or product.
Query $6$
As a reaction approaches equilibrium, concentrations of reactants and products need to change until the rates of forward and reverse reactions are equal. Therefore, only substances whose concentrations can change as a reaction occurs are included in an equilibrium constant expression. For example, consider the following reaction at equilibrium:
2 HgO(s) ⇌ 2 Hg(l) + O2(g) Kc = [O2]e
Because HgO is a pure solid, the number of HgO formula units in a given volume of HgO is the same throughout the reaction; it depends only on the density of HgO at the temperature of the reaction. Similarly, the number of Hg atoms per unit volume of pure Hg(l) is constant throughout the reaction. Thus, these concentrations are not included in the Kc expression. It is necessary for some HgO(s) and some Hg(l) to be present for the equilibrium to be maintained, but the quantity of each does not matter.
In general, Kc expressions do not contain terms for pure solids or pure liquids. In addition, for dilute solutions, the concentration of solvent remains constant throughout an equilibrium reaction and is also not included in the Kc expression, even though the solvent may appear in the reaction equation.
A homogeneous equilibrium is one in which all of the reactants and products are present in a single phase. Examples of homogeneous equilibria are reactions in the gas phase and reactions in liquid solutions. For example:
C2H2(g) + 2 Br2(g) ⇌ C2H2Br4(g)
$K_c = \dfrac{[\text{C}_2\text{H}_2\text{Br}_4]_e}{[\text{C}_2\text{H}_2]_e[\text{Br}_2]_e^{\;2}} \nonumber$
I2(aq) + I‾(aq) ⇌ I3‾(aq)
$K_c = \dfrac{[\text{I}_3^{-}]_e}{[\text{I}_2]_e[\text{I}^{-}]_e} \nonumber$
Hg22+(aq) + NO3‾(aq) + 3 H3O+(aq) ⇌ 2 Hg2+(aq) + HNO2(aq) + 4H2O(l)
$K_c = \dfrac{[\text{Hg}^{2+}]_e^{\;2}[\text{HNO}_2]_e}{[\text{Hg}_2^{2+}]_e[\text{NO}_3^{-}]_e[\text{H}_3\text{O}^{+}]_e^{\;3}} \nonumber$
HF(aq) + H2O(l) ⇌ H3O+(aq) + F‾(aq)
$K_c = \dfrac{[\text{H}_3\text{O}^{+}]_e[\text{F}^{-}]_e}{[\text{HF}]_e} \nonumber$
NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH‾(aq)
$K_c = \dfrac{[\text{NH}_4^{+}]_e[\text{OH}^{-}]_e}{[\text{NH}_3]_e} \nonumber$
In the aqueous equilibrium systems, H2O(l) is the solvent. Its concentration does not appear in the Kc expression.
A heterogeneous equilibrium is a system in which reactants and products are found in two or more phases. Some heterogeneous equilibria involve chemical changes, for example:
PbCl2(s) ⇌ Pb2+(aq) + 2Cl‾(aq) Kc = [Pb2+]e[Cl‾]e2
CaO(s) + CO2(g) ⇌ CaCO3(s)
$K_c = \dfrac{1}{[\text{CO}_2]_e} \nonumber$
C(s) + 2S(g) ⇌ CS2(g)
$K_c = \dfrac{[\text{CS}_2]_e}{[\text{S}]_e^{\;2}} \nonumber$
Other heterogeneous equilibria involve phase changes, for example:
Br2(l) ⇌ Br2(g) Kc = [Br2(g)]e
Working with Kc
When all the coefficients in a balanced chemical equation are multiplied by some factor n, then the new Kc is the original Kc raised to the nth power. For example:
2 NO2(g) ⇌ N2O4(g)
$K_{c,1} = \dfrac{[\text{N}_2\text{O}_4]_e}{[\text{NO}_2]_e^{\;2}} \nonumber$
NO2(g) ⇌ ½ N2O4(g)
$K_{c,2} = \dfrac{[\text{N}_2\text{O}_4]_e^{\;1/2}}{[\text{NO}_2]_e} = (K_{c,1})^{1/2} \nonumber$
When a reaction’s direction is reversed, the Kc for the new reaction is the reciprocal (inverse) of the original reaction Kc. For example:
A + 2B ⇌ AB2
$K_{c,1} = \dfrac{[\text{AB}_2]_e}{[\text{A}]_e[\text{B}]_e^{\;2}} \nonumber$
AB2 ⇌ A + 2B
$K_{c,2} = \dfrac{[\text{A}]_e[\text{B}]_e^{\;2}}{[\text{AB}_2]_e} = \dfrac{1}{K_{c,1}} \nonumber$
When two reactions occur sequentially to yield a new overall reaction, the Kc for the overall reaction is the product of the Kc values for the individual steps. For example:
A + C ⇌ AC
$K_{c,1} = \dfrac{[\text{AC}]_e}{[\text{A}]_e[\text{C}]_e} \nonumber$
AC + C ⇌ AC2
$K_{c,2} = \dfrac{[\text{AC}_2]_e}{[\text{AC}]_e[\text{C}]_e} \nonumber$
A + 2C ⇌ AC2
$K_{c,3} = \dfrac{[\text{AC}_2]_e}{[\text{A}]_e[\text{C}]_e^{\;2}} = K_{c,1}\times K_{c,2} \nonumber$
D29.4 Equilibrium Constant and Partial Pressure
Reactions in which all reactants and products are in the gas phase are another class of homogeneous equilibria. In these cases, the partial pressure of each gas can be used instead of its concentration in the equilibrium constant equation. At constant temperature, the partial pressure of a gas is directly proportional to its molar concentration (c), which is the amount of a substance (moles) per unit volume (liters): c = n/V. This proportionality of pressure and concentration can be derived from the ideal gas equation:
$\begin{array}{rcl} PV & = & nRT \[0.5em] P & = & (\dfrac{n}{V})\;RT \[1em] P & = & cRT \end{array} \nonumber$
For an example, consider the following reaction:
C2H6(g) ⇌ C2H4(g) + H2(g)
We can write the equilibrium constant, Kp, using the equilibrium partial pressures of the gases, by following the same guidelines as for Kc expressions:
$K_{\text{p}} = \dfrac{P_{\text{C}_2\text{H}_4}P_{\text{H}_2}}{P_{\text{C}_2\text{H}_6}} \nonumber$
The two equilibrium constants, Kc and Kp, are directly related to each other. For the generic gas-phase reaction:
m A + n B ⇌ x C + y D
$\begin{array}{rcl} K_{\text{p}} & = & \dfrac{(P_C)^x(P_D)^y}{(P_A)^m(P_B)^n} \[1em] & = & \dfrac{([\text{C}]\;\times\;RT)^x([\text{D}]\;\times\;RT)^y}{([\text{A}]\;\times\;RT)^m([\text{B}]\;\times\;RT)^n} \[1em] & = & \dfrac{[\text{C}]^x[\text{D}]^y}{[\text{A}]^m[\text{B}]^n}\;\times\;\dfrac{(RT)^{x+y}}{(RT)^{m+n}} \[1em] & = & K_{\text{c}}(RT)^{(x+y)\;-\;(m+n)} \[0.5em] K_{\text{p}} & = & K_{\text{c}}(RT)^{{\Delta}n} \end{array} \nonumber$
where Δn is the difference between the sum of the coefficients of the gaseous products and the sum of the coefficients of the gaseous reactants in the reaction (that is, the change in amount of gas between the reactants and the products).
Note that the gas constant, R, can be expressed in different units. Use the R value and associated units that match the partial pressure units used in the Kp expression.
For heterogeneous equilibria that involve gases, equilibrium constants can also be expressed using partial pressures instead of concentrations. Two examples are:
CaO(s) + CO2(g) ⇌ CaCO3(s)
$K_p = \dfrac{1}{P_{CO_2}} \nonumber$
C(s) + 2 S(g) ⇌ CS2(g)
$K_p = \dfrac{P_{CS_2}}{(P_S)^2} \nonumber$
D29.5 Calculations Involving Equilibrium Constants
One way to determine the value for an equilibrium constant is to measure the concentrations (or partial pressures) of all reactants and all products at equilibrium.
Exercise 5: Calculating an Equilibrium Constant
Calculate the equilibrium constant Kc for the decomposition of PCl5 at 250 °C.
PCl5(g) ⇌ PCl3(g) + Cl2(g)
At equilibrium, [PCl5]e = 4.2 × 10-5 M, [PCl3]e = 1.3 × 10-2 M, [Cl2]e = 3.9 × 10-3 M
Query $10$
A known equilibrium constant can be used to calculate an unknown equilibrium concentration, provided the equilibrium concentrations of all other reactants and products are known.
D29.6 Equilibrium Constants and Product-favored Reactions
The magnitude of an equilibrium constant is a measure of the yield of a reaction when it reaches equilibrium. A very large value for Kc (Kc >> 1) indicates that product concentrations are much larger than reactant concentrations when equilibrium has been achieved: nearly all reactants have been converted into products. If Kc is large enough, the reaction has gone essentially to completion when it reaches equilibrium.
Earlier we defined a product-favored reaction as one that proceeds spontaneously in the forward direction when all concentrations (or partial pressures) have the standard-state value of 1 M. If Kc > 1,so that the concentrations of products are greater than the concentrations of reactants, then when all concentrations are 1 M the reaction needs to produce greater concentrations of products to reach equilibrium. That is, the reaction needs to proceed in the forward direction. Thus, Kc > 1 (or Kp > 1) means that a reaction is product-favored.
A very small value of Kc, (Kc << 1) indicates that equilibrium is achieved when only a small fraction of the reactants has been converted into products. Such a reaction is reactant-favored. If Kc is small enough, essentially no reaction has occurred when equilibrium is reached. When Kc ≈ 1, both reactant and product concentrations are significant and it is necessary to use the equilibrium constant to calculate equilibrium concentrations.
By an argument similar to the one in the next to last paragraph above, Kc < 1 (or Kp < 1) means that a reaction is reactant-favored.
Exercise 7: Identifying Reactant-Favored and Product-Favored Processes
Podia Question
One way to remove silver oxide tarnish from silver is to heat the silver to a high temperature.
https://mediaspace.wisc.edu/id/0_p9myzb11
The reaction is:
2 Ag2O → 4 Ag(s) + O2(g)
At 298 K, the reaction is reactant-favored. Obtain data from the Appendix and determine at what temperature the reaction becomes product-favored. What assumptions need to be made to solve this problem?
Query $13$
Two days before the next whole-class session, this Podia question will become live on Podia, where you can submit your answer.
Comments. If you found any inconsistencies, errors, or other things you would like to report about this module, please use this link to report them. A similar link will be included in each day’s material. We appreciate your comments. | textbooks/chem/General_Chemistry/Interactive_Chemistry_(Moore_Zhou_and_Garand)/04%3A_Unit_Four/4.03%3A_Day_29-_Gibbs_Free_Energy_Chemical_Equilibrium.txt |
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