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Water, even pure water, has an amphiprotic nature. This means that a small amount of ions will form in pure water. Some molecules of H2O will act as acids, each donating a proton to a corresponding H2O molecule that acts as a base. Thus, the proton-donating molecule becomes a hydroxide ion, OH-, while the proton-accepting molecule becomes a hydronium ion, H3O+.
Introduction
Water molecules are amphiprotic and can function as both acids and bases. One water molecule (acting as a base) can accept a hydrogen ion from a second water molecule (acting as an acid). This will be happening anywhere there is even a trace of water - it does not have to be pure. A hydronium ion and a hydroxide ion are formed.
However, the hydroxonium (used unambiguously with hydronium) ion is a very strong acid, and the hydroxide ion is a very strong base; in fact, they are the strongest in water. As fast as they are formed, they react to produce water again. The net effect is that an equilibrium is set up:
$2H_2O\,(l) \rightleftharpoons H_3O^+ \,(aq) + OH^- \, (aq) \tag{1}$
At any one time, there are incredibly small numbers of hydroxonium ions and hydroxide ions present. Further down this page, we shall calculate the concentration of hydroxonium ions present in pure water. It turns out to be 1.00 x 10-7 mol dm-3 at room temperature. This equilibrium written in a simplified form:
$H_2O\,(l) \rightleftharpoons H^+ \,(aq) + OH^- \, (aq) \tag{2}$
with H+(aq) actually referring to a hydronium ion.
It is important to remember that water contains VERY low concentration of these ions. In the reversible reaction:
$\underset{\text{base 1}}{H_2O} + \underset{\text{acid 2}}{H_2O} \rightleftharpoons \underset{\text{acid 1}}{H_3O^+} + \underset{\text{base 2}}{OH^-} \tag{4}$
the reaction proceeds by far to the left. Pure water will dissociate to form equal concentrations (here, we are using molarities) of hydronium and hydroxide ions, thus:
$[H_3O^+] = [OH^-] \tag{5}$
For this equation, we can find K, the equilibrium constant:
$K= [H_3O^+][OH^-] \tag{6}$
At standard temperature and pressure (STP), the equilibrium constant of water, $K_w$, is equal to:
$K_w= [H_3O^+][OH^-] \tag{7}$
$K_w=[1.0 \times 10^{-7}][1.0 \times 10^{-7}]\tag{8}$
$K_w=1.0 \times 10^{-14} \tag{9}$
In this equation, [H3O+] is the concentration of hydronium ions which, in a chemical equation, is the acid dissociation constant, $K_a$. The [OH-] is the concentration of hydroxide ions which, in a chemical equation, is the base constant, $K_b$. If given a pH, then you can easily calculate the [H3O+] by:
$[H_3O^+] = 10^{-pH}. \tag{10}$
The same formula applies to obtaining [OH-] from the pOH:
$[OH-]=10^{-pOH} \tag{11}$
Adding the pH's gives you the $pK_w$
$pK_w= pH + pOH =14.00 \tag{11}$
Since the reaction proceeds so heavily to the left, the concentration of these hydroxide and hydronium ions in pure water is extremely small. When making calculations determining involving acids and bases in solution, you do not need to take into account the effects of water's autoionization unless the acid or base of interest is incredibly dilute. However, it is interesting to note that water's self-ionization is significant in that it makes the substance electrically conductive.
Example 1
In the equation depicting the autoionization of water,
$H_2O + H_2O \rightleftharpoons H_3O^+ + OH^-$
The reaction proceeds far to the __________.
Solution
left. The concentration of hydroxide and hydronium ions in pure water is very, very small. Although it is rarely something you need to worry about when looking at acids and bases in solution, it does help account for certain properties of water, such as electrical conductivity.
Example 2
If a solution has a pH of 2.1, determine the concentration of hydroxide ions, [OH-].
Solution
To solve for this, you must first determine the concentration of the hydronium ion, [H3O+]:
[H3O+] = 10-pH
= 10-2.1
= 7.94 x 10-3
Then, you solve for [OH-] using the Kw constant:
Kw = [H3O+] [OH-]
1.0 x 10-14 = [OH-][7.94 x 10-3]
[OH-] = (1 x 10-14) / (7.94 x 10-3)
= 1.26 x 10-12
Example 3
If an aqueous solution has a pOH of 11.2, determine the concentration of hydronium ions.
Solution
To solve for this, you must first determine the concentration of the hydroxide ion, [OH-]:
[OH-] = 10-pOH
= 10-11.2
= 6.31 x 10-12
Then, you solve for [H3O+] using the Kw constant:
Kw = [H3O+] [OH-]
1.0 x 10-14 = [H3O+][6.31 x 10-12]
[H3O+]= (1 x 10-14)/ (6.31 x 10-12) = 0.00158
References
1. Petrucci, et al. General Chemistry: Principles & Modern Applications. 9th ed. Upper Saddle River, New Jersey: Pearson/Prentice Hall, 2007.
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A buffer is a solution that can resist pH change upon the addition of an acidic or basic components. It is able to neutralize small amounts of added acid or base, thus maintaining the pH of the solution relatively stable. This is important for processes and/or reactions which require specific and stable pH ranges. Buffer solutions have a working pH range and capacity which dictate how much acid/base can be neutralized before pH changes, and the amount by which it will change.
• Blood as a Buffer
Buffer solutions are extremely important in biology and medicine because most biological reactions and enzymes need very specific pH ranges in order to work properly.
• Henderson-Hasselbalch Approximation
The Henderson-Hasselbalch approximation allows us one method to approximate the pH of a buffer solution.
• How Does A Buffer Maintain pH?
A buffer is a special solution that stops massive changes in pH levels. Every buffer that is made has a certain buffer capacity, and buffer range. The buffer capacity is the amount of acid or base that can be added before the pH begins to change significantly. It can be also defined as the quantity of strong acid or base that must be added to change the pH of one liter of solution by one pH unit.
• Introduction to Buffers
A buffer is a solution that can resist pH change upon the addition of an acidic or basic components. It is able to neutralize small amounts of added acid or base, thus maintaining the pH of the solution relatively stable. This is important for processes and/or reactions which require specific and stable pH ranges. Buffer solutions have a working pH range and capacity which dictate how much acid/base can be neutralized before pH changes, and the amount by which it will change.
• Preparing Buffer Solutions
When it comes to buffer solution one of the most common equation is the Henderson-Hasselbalch approximation. An important point that must be made about this equation is it's useful only if stoichiometric or initial concentration can be substituted into the equation for equilibrium concentrations.
Thumbnail: Simulated titration of an acidified solution of a weak acid (pKa = 4.7) with alkali. (Public Domain; Lasse Havelund).
Buffers
Buffer solutions are extremely important in biology and medicine because most biological reactions and enzymes need very specific pH ranges in order to work properly.
Blood
Human blood contains a buffer of carbonic acid ($\ce{H2CO3}$) and bicarbonate anion ($\ce{HCO3^{-}}$) in order to maintain blood pH between 7.35 and 7.45, as a value higher than 7.8 or lower than 6.8 can lead to death. In this buffer, hydronium and bicarbonate anion are in equilibrium with carbonic acid. Furthermore, the carbonic acid in the first equilibrium can decompose into $\ce{CO2}$ gas and water, resulting in a second equilibrium system between carbonic acid and water. Because $\ce{O2}$ is an important component of the blood buffer, its regulation in the body, as well as that of $\ce{O2}$, is extremely important. The effect of this can be important when the human body is subjected to strenuous conditions.
In the body, there exists another equilibrium between hydronium and oxygen which involves the binding ability of hemoglobin. An increase in hydronium causes this equilibrium to shift towards the oxygen side, thus releasing oxygen from hemoglobin molecules into the surrounding tissues/cells. This system continues during exercise, providing continuous oxygen to working tissues.
In summation, the blood buffer is:
$\ce{H_3O^+ + HCO_3^- <=> H_2CO_3 + H_2O} \nonumber$
With the following simultaneous equilibrium:
$\ce{H_2CO_3 <=> H_2O + CO_2} \nonumber$
Buffers are used often in biological research to maintain pH of specific processes. This can be especially useful when culturing bacteria, as their metabolic waste can affect the pH of their medium, consequently killing the sample. For example, a buffer of cacodylic acid ($\ce{C2H7AsO2}$) and its conjugate base is used to make samples which will undergo electron microscopy. Another buffer, tricine ($\ce{C6H13NO5}$), is used to buffer chloroplast reactions.
Outside Links
• Urbansky, Edward T.; Schock, Michael R. "Understanding, Deriving, and Computing Buffer Capacity." J. Chem. Educ. 2000 771640.
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The Henderson-Hasselbalch approximation allows us one method to approximate the pH of a buffer solution. The basic equation is as follows:
$pH \approx pK_a + \log_{10} \dfrac{[A^-]}{[HA]} \nonumber$
We have straightforward calculations for strong acids and bases, but the computations behind buffers are rather complex and time consuming. By using the fact that weak acids and bases barely ionize, allowing us to approximate the pH of buffer solutions using initial concentrations. Though the approximation has a few restrictions, it simplifies a lengthy calculation into a simple equation derived from $K$.
Lawrence Joseph Henderson and Karl Albert Hasselbalch
Lawrence Joseph Henderson (1878-1942) was a talented biochemist, among many other titles, who spent most of his career at Harvard. He was responsible for developing the components of the equation after studying equilibrium reactions that took place within blood as a result of respiration (specializing in "fatigue"). His equation was incomplete without the solid calculations into it provided by Hasselbalch.
Karl Albert Hasselbalch (1874-1962) was a chemist who studied pH closely. He also studied blood and reactions that took place with oxygen, to put in the simplest of terms. He eventually modified Henderson's equation by putting mathematical logs into it creating a solid relationship.
Derivation for Buffers from Weak Acids
For a weak acid $HA$ and its conjugate base $A^-$:
$HA + H_2O\leftrightharpoons H^+ + A^-$
which has an acid ionization constant $K_a$. The Henderson-Hasselbalch approximation is derived from this acid ionization constant.
\begin{align} K_a &= \dfrac{[H^+][A^-]}{[HA]} \[4pt] -\log_{10} K_a &= -\log_{10} \dfrac{[H^+][A^-]}{[HA]} \[4pt] - \log_{10} K_a &= - \log_{10} [H^+] - \log_{10} \dfrac{[A^-]}{[HA]} \[4pt] pK_a &= pH - \log_{10} \dfrac{[A^-]}{[HA]} \[4pt] pH &= pK_a + \log_{10} \dfrac{[A^-]}{[HA]} \label{HH1} \end{align}
Equation \ref{HH1} is formulated in terms of equilibrium concentrations in solution. Since $HA$ is a weak acid and weakly dissociates and we can introduce two approximations
$[HA]\approx[HA]_i \nonumber$
and
$[A^-]\approx[A^-]_i \nonumber$
Hence, we can use the initial concentrations rather than equilibrium concentrations because
$pK_a + \log_{10} \dfrac{[A^-]_i}{[HA]_i} \approx pK_a + \log_{10} \dfrac{[A^-]}{[HA]} = pH \nonumber$
Example $1$: Nitrous acid buffer
Find $[H^+]$ in a solution of 1.0 M $HNO_2$ and 0.225 M $NaNO_2$. The $K_a$ for $HNO_2$ is $5.6 \times 10^{−4}$ (Table E1).
Solution
$pK_a = -\log_{10} K_a = -\log_{10} (7.4 \times 10^{-4}) = 3.14 \nonumber$
$pH = pK_a + \log_{10} \left( \dfrac{[NO_2^-]}{[HNO_2]}\right)\nonumber$
$pH = 3.14 + \log_{10} \left( \dfrac{1}{0.225} \right)\nonumber$
$pH = 3.14 + 0.648 = 3.788\nonumber$
$[H^+] = 10^{-pH} = 10^{-3.788} = 1.6 \times 10^{-4} \nonumber$
Example $2$: Acetic Acid buffer
What ratio $\frac{[A^-]}{[HA]}$ will create an acetic acid buffer of pH 5.0? ($K_a$ acetic acid is $1.75 \times 10^{−5}$ (Table E1).
Solution
$pKa = -log Ka = -log(1.8 \times10^{-5}) = 4.74\nonumber$
$pH = pK_a + \log_{10} \left( \dfrac{[A^-]}{[HA]}\right) \nonumber$
$5.0 = 4.74 + \log_{10} \left( \dfrac{[A^-]}{[HA]}\right)\nonumber$
$0.26 = \log_{10} \left( \dfrac{[A^-]}{[HA]}\right)\nonumber$
$10^{0.26} = \dfrac{[A^-]}{[HA]}\nonumber$
$1.8 = \dfrac{[A^-]}{[HA]}\nonumber$
Derivation for Buffers from Weak Bases
Similarly, for a weak base $B$ and its conjugate acid $HB^+$:
$B + H_2O \leftrightharpoons OH^- + HB^+ \nonumber$
which has an base ionization constant $K_b$. The Henderson-Hasselbalch approximation for basic buffers is derived from this base ionization constant.
\begin{align} K_b &= \dfrac{[OH^-][HB^+]}{[B]} \[4pt] -\log_{10} K_b &= -\log_{10} \dfrac{[OH^-][HB^+]}{[B]} \[4pt] -\log_{10} K_b &= - \log_{10} [OH^-] - \log_{10} \dfrac{[HB^+]}{[B]} \[4pt] pK_b &= pOH - \log_{10} \dfrac{[HB^+]}{[B]} \[4pt] pOH &= pK_b + \log_{10} \dfrac{[HB^+]}{[B]} \end{align} \nonumber
Note that $B$ is a weak base and does not dissociate completely and we can say $[B] \approx [B]_i$ and $[HB^+] \approx [HB^+]_i$. Hence, we can use the initial concentrations because
$pK_b + \log_{10} \dfrac{[HB^+]_i}{[B]_i} \approx pK_b + \log_{10} \dfrac{[HB^+]}{[B]} = pOH \nonumber$
Example $3$: Ammonium Buffer
You prepare a buffer solution of 0.323 M $NH_3$ and $(NH_4)_2SO_4$. What molarity of $(NH_4)_2SO_4$ is necessary to have a pH of 8.6? The $pK_b$ for $NH_3$ is 4.74 (Table E2).
Solution
$pK_a+ pK_b = 14\nonumber$
$pK_a= 14 - 4.74 = 9.26\nonumber$
$pH = pK_a + \log_{10} \left( \frac{[A^-]}{[HA]}\right)\nonumber$
$8.6 = 9.26 + \log_{10} \left( \dfrac{0.323}{[NH_4^+]}\right)\nonumber$
$-0.66 = \log_{10} \left( \dfrac{0.323}{[NH_4^+]}\right)\nonumber$
$[NH_4^+] = 1.48 M\nonumber$
Example $4$: Buffering Capability
What is the pH of a buffer 0.500 moles acetic acid and 0.500 moles acetate ion and the total volume is 5 L when you add 0.350 moles HCl? The $K_a$ for acetic acid is $1.75 \times 10^{−5}$ (Table E1).
Solution
$pK_a = -\log_{10} K_a = -\log_{10} (1.75 \times 10^{−5}) = 4.756\nonumber$
$pH = pK_a + \log_{10} \left( \dfrac{[acetate]}{[\text{acetic acid}] + [HCl]} \right)\nonumber$
note that mole ratio also works in place of concentrations as both the acid and base are in the same solution
$pH = 4.756 + \log_{10} (0.588)\nonumber$
$pH = 4.756 - 0.230 = 4.52\nonumber$
Example $5$: effective buffer range
What is the range of an acetic acid buffer described in problem $4$ without the added $\ce{HCl}$?
Solution
The effective buffer range is of magnitude 2 pH units with the $pK_a$ as the midpoint.
$pK_a = -\log_{10} K_a = -\log_{10} (1.75 \times 10^{−5}) = 4.756\nonumber$
Hence, the buffer range would be 3.75 to 5.75.
Assumptions in the Henderson-Hasselbach Approximation
• $-1 < \log_{10} \dfrac{[A^-]}{[HA]} < 1 \nonumber$
• The molarity of the buffer(s) should be 100x that of the acid ionization constant, or Ka.
• This equation will give poor or inaccurate results if there are strong acids or bases. pKa values between 5 and 9 will give good approximations, but when we are out of this range there is a strong chance that the pH value will be incorrect.J In other words near the end or the beginning of a titration where the "relative concentrations of the acid or base differ substantially[,]... approximate calculations break down."J
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A is a special solution that stops massive changes in pH levels. Every buffer that is made has a certain buffer capacity, and buffer range. The buffer capacity is the amount of acid or base that can be added before the pH begins to change significantly. It can be also defined as the quantity of strong acid or base that must be added to change the pH of one liter of solution by one pH unit. The buffer range is the pH range where a buffer effectively neutralizes added acids and bases, while maintaining a relatively constant pH.
Introduction
The equation for pH also shows why pH does not change by much in buffers.
$K_a = \dfrac{[H^+][A^-]}{[HA]} \tag{1}$
$pH = pK_a + \log\dfrac{[A^-]}{[HA]} \tag{2}$
Where,
• $A^-$ is the concentration of the conjugate base
• $HA$ is the concentration of the acid
When the ratio between the conjugate base/ acid is equal to 1, the pH = pKa. If the ratio between the two is 0.10, the pH drops by 1 unit from pKa since log (0.10) = -1. If a ratio increases to a value of 10, then the pH increases by 1 unit since \log (10) = 1. The buffer capacity has a range of about 2. This means when a buffer is created, the pH can be changed by -1 by acid or +1 by base before the pH begins to change substantially. After the addition of base to raise the pH by 1 or more, most of the conjugate acid will have been depleted to try to maintain a certain pH, so the pH will be free to increase faster without the restraint of the conjugate acid. The same goes for the addition of acid, once the conjugate base has been used up, the pH will drop faster since most of the conjugate base has been used up.
Example 1
What is the effect on the pH of adding 0.006 mol HCL to 0.3L of a buffer solution that is 0.250M HC2H3O2 and 0.560 M NaC2H3O2? pKa= 4.74
Solution
$pH = 4.74 + \log\dfrac{0.560}{0.250} = 4.74 + 0.35 = 5.09 \nonumber$
$C_2H_3O_2^- + H_3O^+ \rightleftharpoons HC_2H_3O_2 + H_2O \nonumber$
Calculate the starting amount of C2H3O2-
$0.300 \; L \times 0.560 \; M = 0.168 \; mol \; C_2H_3O_2^- \nonumber$
Calculate the starting amount of HC2H3O2
$0.300 \; L \times 0.250 \; M = 0.075 \; mol \; HC_2H_3O_2 \nonumber$
C2H3O2- H3O- HC2H3O2
Original Buffer 0.168 mol 0.075 mol
Add 0.006 mol
Change -0.006 mol -0.006 mol +0.006 mol
Final Amount 0.162 mol 0.081 mol
Now calculate the new concentrations of C2H3O2- and HC2H3O2:
$\dfrac{0.162 \; mol}{0.300 \; L} = 0.540 \; M \; C_2H_3O_2^- \nonumber$
$\dfrac{0.081 \; mol}{0.300 \; L} = 0.540 \; M \; HC_2H_3O_2 \nonumber$
Using the new concentrations, we can calculate the new pH:
$pH = 4.74 + \log\dfrac{0.540}{0.270} = 4.74 + 0.30 = 5.04 \nonumber$
Calculate the pH change:
$pH_{final} - pH_{initial} = 5.04 - 5.09 = -0.05 \nonumber$
Therefore, the pH dropped by 0.05 pH units.
Example 2
What is the effect on the pH of adding 0.006 mol NaOH to 0.3L of a buffer solution that is 0.250M HC2H3O2 and 0.560 M NaC2H3O2? pKa= 4.74
Solution
$pH = 4.74 + \log\dfrac{0.560}{0.250} = 4.74 + 0.35 = 5.09 \nonumber$
$HC_2H_3O_2 + OH^- \rightleftharpoons C_2H_3O_2^- + H_2O \nonumber$
Calculate the starting amount of HC2H3O2
$0.300 \; L \times 0.250 \; M = 0.075 \; mol \; HC_2H_3O_2 \nonumber$
Calculate the starting amount of C2H3O2-
$0.300 \; L \times 0.560 \; M = 0.168 \; mol \; C_2H_3O_2^- \nonumber$
HC2H3O2 OH- C2H3O2-
Original Buffer 0.075 mol 0.168 mol
Add 0.006 mol
Change -0.006 mol -0.006 mol +0.006 mol
Final Amount 0.069 mol 0.174 mol
Now calculate the new concentrations of HC2H3O2 and C2H3O2-:
$\dfrac{0.069 \; mol}{0.300 \; L} = 0.230 \; M \; HC_2H_3O_2 \nonumber$
$\dfrac{0.174 \; mol}{0.300 \; L} = 0.580 \; M \; C_2H_3O_2^- \nonumber$
Using the new concentrations, we can calculate the new pH:
$pH = 4.74 + \log\dfrac{0.580}{0.230} = 4.74 + 0.40 = 5.14 \nonumber$
Calculate the pH change:
$pH_{final} - pH_{initial} = 5.14 - 5.09 = +0.05 \nonumber$
Therefore, the pH increased by 0.05 pH units.
Buffers in the Human Body
Blood contains large amounts of carbonic acid, a weak acid, and bicarbonate, a base. Together they help maintain the bloods pH at 7.4. If blood pH falls below 6.8 or rises above 7.8, one can become sick or die. The bicarbonate neutralizes excess acids in the blood while the carbonic acid neutralizes excess bases.
Another example is when we consume antacids or milk of magnesia. After eating a meal with rich foods such as sea food, the stomach has to produce gastric acid to digest the food. Some of the acid can splash up the lower end of the esophagus causing a burning sensation. To relieve this burning, one would take an antacid, which when dissolved the bases buffer the excess acid by binding to them.
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A buffer is a solution that can resist pH change upon the addition of an acidic or basic components. It is able to neutralize small amounts of added acid or base, thus maintaining the pH of the solution relatively stable. This is important for processes and/or reactions which require specific and stable pH ranges. Buffer solutions have a working pH range and capacity which dictate how much acid/base can be neutralized before pH changes, and the amount by which it will change.
What is a buffer composed of?
To effectively maintain a pH range, a buffer must consist of a weak conjugate acid-base pair, meaning either a. a weak acid and its conjugate base, or b. a weak base and its conjugate acid. The use of one or the other will simply depend upon the desired pH when preparing the buffer. For example, the following could function as buffers when together in solution:
• Acetic acid (weak organic acid w/ formula CH3COOH) and a salt containing its conjugate base, the acetate anion (CH3COO-), such as sodium acetate (CH3COONa)
• Pyridine (weak base w/ formula C5H5N) and a salt containing its conjugate acid, the pyridinium cation (C5H5NH+), such as Pyridinium Chloride.
• Ammonia (weak base w/ formula NH3) and a salt containing its conjugate acid, the ammonium cation, such as Ammonium Hydroxide (NH4OH)
How does a buffer work?
A buffer is able to resist pH change because the two components (conjugate acid and conjugate base) are both present in appreciable amounts at equilibrium and are able to neutralize small amounts of other acids and bases (in the form of H3O+ and OH-) when the are added to the solution. To clarify this effect, we can consider the simple example of a Hydrofluoric Acid (HF) and Sodium Fluoride (NaF) buffer. Hydrofluoric acid is a weak acid due to the strong attraction between the relatively small F- ion and solvated protons (H3O+), which does not allow it to dissociate completely in water. Therefore, if we obtain HF in an aqueous solution, we establish the following equilibrium with only slight dissociation (Ka(HF) = 6.6x10-4, strongly favors reactants):
$HF_{(aq)} + H_2O_{(l)} \rightleftharpoons F^-_{(aq)} + H_3O^+_{(aq)} \nonumber$
We can then add and dissolve sodium fluoride into the solution and mix the two until we reach the desired volume and pH at which we want to buffer. When Sodium Fluoride dissolves in water, the reaction goes to completion, thus we obtain:
$NaF_{(aq)} + H_2O_{(l)} \rightarrow Na^+_{(aq)} + F^-_{(aq)} \nonumber$
Since Na+ is the conjugate of a strong base, it will have no effect on the pH or reactivity of the buffer. The addition of $NaF$ to the solution will, however, increase the concentration of F- in the buffer solution, and, consequently, by Le Chatelier’s Principle, lead to slightly less dissociation of the HF in the previous equilibrium, as well. The presence of significant amounts of both the conjugate acid, $HF$, and the conjugate base, F-, allows the solution to function as a buffer. This buffering action can be seen in the titration curve of a buffer solution.
As we can see, over the working range of the buffer. pH changes very little with the addition of acid or base. Once the buffering capacity is exceeded the rate of pH change quickly jumps. This occurs because the conjugate acid or base has been depleted through neutralization. This principle implies that a larger amount of conjugate acid or base will have a greater buffering capacity.
If acid were added:
$F^-_{(aq)} + H_3O^+_{(aq)} \rightleftharpoons HF_{(aq)} + H_2O_{(l)} \nonumber$
In this reaction, the conjugate base, F-, will neutralize the added acid, H3O+, and this reaction goes to completion, because the reaction of F- with H3O+ has an equilibrium constant much greater than one. (In fact, the equilibrium constant the reaction as written is just the inverse of the Ka for HF: 1/Ka(HF) = 1/(6.6x10-4) = 1.5x10+3.) So long as there is more F- than H3O+, almost all of the H3O+ will be consumed and the equilibrium will shift to the right, slightly increasing the concentration of HF and slightly decreasing the concentration of F-, but resulting in hardly any change in the amount of H3O+ present once equilibrium is re-established.
If base were added:
$HF_{(aq)} + OH^-_{(aq)} \rightleftharpoons F^-_{(aq)} + H_2O_{(l)} \nonumber$
In this reaction, the conjugate acid, HF, will neutralize added amounts of base, OH-, and the equilibrium will again shift to the right, slightly increasing the concentration of F- in the solution and decreasing the amount of HF slightly. Again, since most of the OH- is neutralized, little pH change will occur.
These two reactions can continue to alternate back and forth with little pH change.
Selecting proper components for desired pH
Buffers function best when the pKa of the conjugate weak acid used is close to the desired working range of the buffer. This turns out to be the case when the concentrations of the conjugate acid and conjugate base are approximately equal (within about a factor of 10). For example, we know the Ka for hydroflouric acid is 6.6 x 10-4 so its pKa= -log(6.6 x 10-4) = 3.18. So, a hydrofluoric acid buffer would work best in a buffer range of around pH = 3.18.
For the weak base ammonia (NH3), the value of Kb is 1.8x10-5, implying that the Ka for the dissociation of its conjugate acid, NH4+, is Kw/Kb=10-14/1.8x10-5 = 5.6x10-10. Thus, the pKa for NH4+ = 9.25, so buffers using NH4+/NH3 will work best around a pH of 9.25. (It's always the pKa of the conjugate acid that determines the approximate pH for a buffer system, though this is dependent on the pKb of the conjugate base, obviously.)
When the desired pH of a buffer solution is near the pKa of the conjugate acid being used (i.e., when the amounts of conjugate acid and conjugate base in solution are within about a factor of 10 of each other), the Henderson-Hasselbalch equation can be applied as a simple approximation of the solution pH, as we will see in the next section.
Example 1: HF Buffer
In this example we will continue to use the hydrofluoric acid buffer. We will discuss the process for preparing a buffer of HF at a pH of 3.0. We can use the Henderson-Hasselbalch approximation to calculate the necessary ratio of F- and HF.
$pH = pKa + \log\dfrac{[Base]}{[Acid]} \nonumber$
$3.0 = 3.18 + \log\dfrac{[Base]}{[Acid]} \nonumber$
$\log\dfrac{[Base]}{[Acid]} = -0.18 \nonumber$
$\dfrac{[Base]}{[Acid]} = 10^{-0.18} \nonumber$
$\dfrac{[Base]}{[Acid]} = 0.66 \nonumber$
This is simply the ratio of the concentrations of conjugate base and conjugate acid we will need in our solution. However, what if we have 100 ml of 1 M HF and we want to prepare a buffer using NaF? How much Sodium Fluoride would we need to add in order to create a buffer at said pH (3.0)?
We know from our Henderson-Hasselbalch calculation that the ratio of our base/acid should be equal to 0.66. From a table of molar masses, such as a periodic table, we can calculate the molar mass of NaF to be equal to 41.99 g/mol. HF is a weak acid with a Ka = 6.6 x 10-4 and the concentration of HF is given above as 1 M. Using this information, we can calculate the amount of F- we need to add.
The dissociation reaction is:
$HF_{(aq)} + H_2O_{(l)} \rightleftharpoons F^-_{(aq)} + H_3O^+_{(aq)} \nonumber$
We could use ICE tables to calculate the concentration of F- from HF dissociation, but, since Ka is so small, we can approximate that virtually all of the HF will remain undissociated, so the amount of F- in the solution from HF dissociation will be negligible. Thus, the [HF] is about 1 M and the [F-] is close to 0. This will be especially true once we have added more F-, the addition of which will even further suppress the dissociation of HF.
We want the ratio of Base/Acid to be 0.66, so we will need [Base]/1M = 0.66. Thus, [F-] should be about 0.66 M. For 100 mL of solution, then, we will want to add 0.066 moles (0.1 L x 0.66 M) of F-. Since we are adding NaF as our source of F-, and since NaF completely dissociates in water, we need 0.066 moles of NaF. Thus, 0.066 moles x 41.99 g/mol = 2.767 g.
Note that, since the conjugate acid and the conjugate base are both mixed into the same volume of solution in the buffer, the ratio of "Base/Acid" is the same whether we use a ratio of the "concentration of base over concentration of acid," OR a ratio of "moles of base over moles of acid." The pH of the solution does not, it turns out, depend on the volume! (This is only true so long as the solution does not get so dilute that the autoionization of water becomes an important source of H+ or OH-. Such dilute solutions are rarely used as buffers, however.)
Adding Strong Acids or Bases to Buffer Solutions
Now that we have this nice F-/HF buffer, let's see what happens when we add strong acid or base to it. Recall that the amount of F- in the solution is 0.66M x 0.1 L = 0.066 moles and the amount of HF is 1.0 M x 0.1L = 0.10 moles. Let's double check the pH using the Henderson-Hasselbalch Approximation, but using moles instead of concentrations:
pH = pKa + log(Base/Acid) = 3.18 + log(0.066 moles F-/0.10 moles HF) = 3.00
Good. Now let's see what happens when we add a small amount of strong acid, such as HCl. When we put HCl into water, it completely dissociates into H3O+ and Cl-. The Cl- is the conjugate base of a strong acid so is inert and doesn't affect pH, and we can just ignore it. However, the H3O+ can affect pH and it can also react with our buffer components. In fact, we already discussed what happens. The equation is:
$F^-_{(aq)} + H_3O^+_{(aq)} \rightleftharpoons HF_{(aq)} + H_2O_{(l)} \nonumber$
For every mole of H3O+ added, an equivalent amount of the conjugate base (in this case, F-) will also react, and the equilibrium constant for the reaction is large, so the reaction will continue until one or the other is essentially used up. If the F- is used up before reacting away all of the H3O+, then the remaining H3O+ will affect the pH directly. In this case, the capacity of the buffer will have been exceeded - a situation one tries to avoid. However, for our example, let's say that the amount of added H3O+ is smaller than the amount of F- present, so our buffer capacity is NOT exceeded. For the purposes of this example, we'll let the added H3O+ be equal to 0.01 moles (from 0.01 moles of HCl). Now, if we add 0.01 moles of HCl to 100 mL of pure water, we would expect the pH of the resulting solution to be 1.00 (0.01 moles/0.10 L = 0.1 M; pH = -log(0.1) = 1.0).
However, we are adding the H3O+ to a solution that has F- in it, so the H3O+ will all be consumed by reaction with F-. In the process, the 0.066 moles of F- is reduced:
0.066 initial moles F- - 0.010 moles reacted with H3O+ = 0.056 moles F- remaining
Also during this process, more HF is formed by the reaction:
0.10 initial moles HF + 0.010 moles from reaction of F- with H3O+ = 0.11 moles HF after reaction
Plugging these new values into Henderson-Hasselbalch gives:
pH = pKa + log (base/acid) = 3.18 + log (0.056 moles F-/0.11 moles HF) = 2.89
Thus, our buffer did what it should - it resisted the change in pH, dropping only from 3.00 to 2.89 with the addition of 0.01 moles of strong acid.
Outside Links
• Urbansky, Edward T.; Schock, Michael R. "Understanding, Deriving, and Computing Buffer Capacity." J. Chem. Educ. 2000 1640..
Preparing Buffer Solutions
When it comes to buffer solution one of the most common equation is the Henderson-Hasselbalch approximation. An important point that must be made about this equation is it's useful only if stoichiometric or initial concentration can be substituted into the equation for equilibrium concentrations.
Origin of the Henderson-Hasselbalch Equation
Where the Henderson-Hasselbalch approximation comes from
$HA + H_2O \rightleftharpoons H_3O^+ + A^- \label{1}$
where,
• $A^-$ is the conjugate base
• $HA$ is the weak acid
We know that $K_a$ is equal to the products over the reactants and, by definition, H2O is essentially a pure liquid that we consider to be equal to one.
$K_a = [H_3O^+][A^-] \label{2}$
Take the $-\log$ of both sides:
$-\log \; K_a = -\log([H_3O^+][A^-]) \label{3}$
$-\log \; K_a = -\log[H_3O^+] \; -\log[A^-] \label{4}$
Using the following two relationships:
$-\log[K_a] = pK_a \label{5}$
$-\log[H_3O^+] = pH \label{6}$
We can simplify the above equation:
$pK_a = pH - \log[A^-] \label{7}$
If we add $\log[A^-]$ to both sides, we get the Henderson-Hasselbalch approximation:
$pH = pK_a + \log[A^-] \label{8}$
This approximation is only valid when:
1. The conjugate base / acid falls between the values of 0.1 and 10
2. The molarity of the buffers exceeds the value of the Ka by a factor of at least 100
Example $1$
Suppose we needed to make a buffer solution with a pH of 2.11. In the first case, we would try and find a weak acid with a pKa value of 2.11. However, at the same time the molarities of the acid and the its salt must be equal to one another. This will cause the two molarities to cancel; leaving the $\log [A^-]$ equal to $\log(1)$ which is zero.
$pH = pK_a + \log[A^-] = 2.11 + \log(1) = 2.11 \nonumber$
This is a very unlikely scenario, however, and you won't often find yourself with Case #1
Example $2$
What mass of $NaC_7H_5O_2$ must be dissolved in 0.200 L of 0.30 M HC7H5O2 to produce a solution with pH = 4.78? (Assume solution volume is constant at 0.200L)
Solution
$HC_7H_5O_2 + H_20 \rightleftharpoons H_3O^+ + C_7H_5O_2$
$K_a =6.3 \times 10^{-5}$
$K_a = \dfrac{[H_3O^+][C_7H_5O_2]}{[HC_7H_5O_2]} = 6.3 \times 10^{-5}$
$[H_3O^+] = 10^{-pH} = 10^{-4.78} = 16.6 \times 10^{-6}\;M\;[HC_7H_5O_2] = 0.30\;M\;[C_7H_5O_2] =$
$[C_7H_5O_2^-] = K_a \times \dfrac{[HC_7H_5O_2]}{[H_3O^+]} \nonumber$
$1.14 \; M = 6.3 \times 10^{-5} \times \dfrac{0.30}{16.6 \times 10^{-6}} \nonumber$
Mass = 0.200 L x 1.14 mol C7H5O2- / 1L x 1mol NaC7H5O2 / 1 mol C7H5O2- x 144 g NaC7H5O2 / 1 mol NaC7H5O2 = 32.832 g NaC7H5O2
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Acids_and_Bases/Buffers/Introduction_to_Buffers.txt
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• Acid and Base Strength
All acids and bases do not ionize or dissociate to the same extent. This leads to the statement that acids and bases are not all of equal strength in producing H+ and OH- ions in solution. The terms "strong" and "weak" give an indication of the strength of an acid or base. The terms strong and weak describe the ability of acid and base solutions to conduct electricity. If the acid or base conducts electricity strongly, it is a strong acid or base.
• Calculating a Ka Value from a Known pH
The quantity pH, or "power of hydrogen," is a numerical representation of the acidity or basicity of a solution. It can be used to calculate the concentration of hydrogen ions [H+] or hydronium ions [H3O+] in an aqueous solution. Solutions with low pH are the most acidic, and solutions with high pH are most basic.
• Calculating Equilibrium Concentrations
\(K_a\) is an acid dissociation constant, also known as the acid ionization constant. It describes the likelihood of the compounds and the ions to break apart from each other. As we already know, strong acids completely dissociate, whereas weak acids only partially dissociate. A big \(K_a\) value will indicate that you are dealing with a very strong acid and that it will completely dissociate into ions.
• Fundamentals of Ionization Constants
A simple way to understand an ionization constant is to think of it in a clear-cut way: To what degree will a substance produce ions in water? In other words, to what extent will ions be formed?
• Weak Acids and Bases
Unlike strong acids/bases, weak acids and weak bases do not completely dissociate (separate into ions) at equilibrium in water, so calculating the pH of these solutions requires consideration of a unique ionization constant and equilibrium concentrations. Although this is more difficult than calculating the pH of a strong acid or base solution, most biochemically important acids and bases are considered weak, and so it is very useful to understand how to calculate the pH of these substances.
• Weak Acids and Bases
Weak acids and bases are only partially ionized in their solutions, whereas strong acids and bases are completely ionized when dissolved in water. The ionization of weak acids and bases is a chemical equilibrium phenomenon. The equilibrium principles are essential for the understanding of equilibria of weak acids and weak bases. In this connection, you probably realize that conjugate acids of weak bases are weak acids and conjugate bases of weak acids are weak bases.
Thumbnail: Variation of the % formation of a monoprotic acid, AH, and its conjugate base, A−, with the difference between the pH and the pKa of the acid. (Public Domain; Krishnavedala).
Ionization Constants
All acids and bases do not ionize or dissociate to the same extent. This leads to the statement that acids and bases are not all of equal strength in producing H+ and OH- ions in solution. The terms "strong" and "weak" give an indication of the strength of an acid or base. The terms strong and weak describe the ability of acid and base solutions to conduct electricity. If the acid or base conducts electricity strongly, it is a strong acid or base. If the acid or base conducts electricity weakly, it is a weak acid or base.
Demonstration of Acid and Base Conductivity
The instructor will test the conductivity of various solutions with a light bulb apparatus. The light bulb circuit is incomplete. If the circuit is completed by a solution containing a large number of ions, the light bulb will glow brightly indicating a strong ability to conduct electricity as shown for HCl. If the circuit is completed by a solution containing large numbers of molecules and either no ions or few ions, the solution does not conduct or conducts very weakly as shown for acetic acid.
An acid or base which strongly conducts electricity contains a large number of ions and is called a strong acid or base and an acid or base which conducts electricity only weakly contains only a few ions and is called a weak acid or base.
Conductivity Behavior of Acids and Bases
Compounds Appearance of light bulb Classification
Weak or Strong
Inference of Ions or Molecules
H2O no light weak molecules
HCl bright strong ions
HC2H3O2 dim weak molecules
H2SO4 bright
H2CO3 dim
NaOH bright
KOH bright
NH4OH dim
Bond Strength
The bond strengths of acids and bases are implied by the relative amounts of molecules and ions present in solution. The bonds are represented as:
acid base
H-A M-OH
where A is a negative ion, and M is a positive ion
• Strong acids have mostly ions in solution, therefore the bonds holding H and A together must be weak. Strong acids easily break apart into ions.
• Weak acids exist mostly as molecules with only a few ions in solution, therefore the bonds holding H and A together must be strong. Weak acids do not readily break apart as ions but remain bonded together as molecules.
Bond Strength Principle
Acids or bases with strong bonds exist predominately as molecules in solutions and are called "weak" acids or bases. Acids or bases with weak bonds easily dissociate into ions and are called "strong" acids or bases.
Table 1: Summary List of Characteristics for Strong and Weak Acids and Bases. All characteristics of acids and bases are related to whether the predominate forms are molecules and ions.
Characteristic Strong Acid or Base Weak Acid or Base
Molecules few large number
Ions large number small number
Conductivity strong weak
Bond Strength weak strong
Acids and bases behave differently in solution based on their strength. Acid or base "strength" is a measure of how readily the molecule ionizes in water.
Introduction Again
Some acids and bases ionize rapidly and almost completely in solution; these are called strong acids and strong bases. For example, hydrochloric acid (HCl) is a strong acid. When placed in water, virtually every HCl molecule splits into a H+ ion and a Cl- ion in the reaction.1
$\ce{HCl(aq) + H2O(l) <=> H3O^{+}(aq) + Cl^{-}(aq)} \nonumber$
For a strong acid like HCl, if you place 1 mole of HCl in a liter of water, you will get roughly 1 mole of H30+ ions and 1 mole of Cl- ions. In a weak acid like hydrofluoric acid (HF), not all of the HF molecules split up, and although there will be some H+ and F- ions released, there will still be HF molecules in solution1. A similar concept applies to bases, except the reaction is different. A strong base like sodium hydroxide (NaOH) will also dissociate completely into water; if you put in 1 mole of NaOH into water, you will get 1 mole of hydroxide ions.1
$\ce{NaOH(aq) + H2O(l) <=> Na^{+}(aq) + OH^{-}(aq) + H2O(l)} \nonumber$
The terms "strong" and "weak" in this context do not relate to how corrosive or caustic the substance is, but only its capability to ionize in water. The ability of a substance to eat through other materials or damage skin is more of a function of the properties of that acid, as well as its concentration. Although, strong acids are more directly dangerous at lower concentrations a strong acid is not necessarily more dangerous than a weak one. For example, hydrofluoric acid is a weak acid1, but it is extremely dangerous and should be handled with great care. Hydrofluoric acid is particularly dangerous because it is capable of eating through glass, as seen in the video in the links sectionV1. The percent dissociation of an acid or base is mathematically indicated by the acid ionization constant (Ka) or the base ionization constant (Kb)1. These terms refer to the ratio of reactants to products in equilibrium when the acid or base reacts with water. For acids the expression will be
Ka = [H3O+][A-]/[HA]
where HA is the concentration of the acid at equilibrium, and A- is the concentration of its conjugate base at equilibrium and for bases the expression will be
$K_b = \dfrac{[\ce{OH^{-}}][\ce{HB^{+}}]}{\ce{B}}$
where B is the concentration of the base at equilibrium and HB+ is the concentration of its conjugate acid at equilibrium
The stronger an acid is, the lower the pH it will produce in solution. pH is calculated by taking the negative logarithm of the concentration of hydronium ions. For strong acids, you can calculate the pH by simply taking the negative logarithm of its molarity as it completely dissociates into its conjugate base and hydronium. The same goes for strong bases, except the negative logarithm gives you the pOH as opposed to the pH. For weak acids and bases, the higher the Ka or Kb, the more acidic or basic the solution. To find the pH for a weak acid or base, you must use the K equation and a RICE table to determine the pH.
All acids have a conjugate base that forms when they react with water, and similarly, all bases have a conjugate acid that reacts when they form with water.1 You can judge the relative strength of a conjugate by the $K_a$ or $K_b$ value of the substance because $K_a \times K_b$ is equal to the ionization constant of water, Kw which is equal to $1 \times 10^{-14}$ at room temperature. The higher the Ka, the stronger the acid is, and the weaker its conjugate base is. Similarly, the higher the Kb, the stronger the substance is as a base, and the more weakly acidic its conjugate acid is.1
Calculation of Ka
For an acid that reacts with water in the reaction
$HA_{(aq)} + H_2O_{(l)} \rightleftharpoons H_3O^+_{(aq)} + A^-_{(aq)}$
$K_a = \dfrac{[H_3O^+][A^-]}{[HA]}$
where each bracketed term represents the concentration of that substance in solution.
Relation of Kw, Kb, Ka
$K_w = K_a \times K_b \nonumber$
Partial List of Strong Acids: Hydrochlroic acid (HCl), Nitric Acid (HNO3), Perchloric Acid (HClO4), Sulfuric Acid (H2SO4)
Partial List of Strong Bases: Sodium Hydroxide (NaOH), Barium Hydroxide (Ba(OH)2), Calcium Hydroxide (Ca(OH)2), Lithium Hydroxide (LiOH) (Hydroxides of Group I and II elements are generally strong bases)
Partial List of Weak Acids: Acetic Acid (CH3COOH), Carbonic Acid (H2CO3), Phosphoric Acid (H3PO4)
Partial List of Weak Bases: Ammonia (NH3), Calcium Carbonate (CaCO3), Sodium Acetate (NaCH3COO)
Example $1$
Find the pH of 0.5 grams of HCl disolved into 100 ml of water:
Solution
First find moles of acid:
grams / molar mass = moles
0.5 grams / (36.5 g/mole) = 0.014 moles HCl
Then find molarity:
moles / volume = molarity
0.014 moles / 0.100 L = 0.14 M
HCl is a strong acid and completely dissociates in water, therefore the pH will be equal to the negative logarithm of the concentration of HCl
pH = -log(H3O+)
pH = -log(0.14) = 0.85
Example $2$
The Ka value for acetic acid is 1.76*10-5, and the Ka value for benzoic acid is 6.46*10-5, if two solutions are made, one from each acid, with equal concentrations, which one will have the lower pH?
Solution
The Ka value is a measure of the ratio between reactants and products at equilibrium. For an acid, the reaction will be HA + H2O --> A- + H3O+ . PH is based on the concentration of the hydronium ion (H3O+) which is a product of the reaction of acid and water. A higher Ka value means a higher ratio of reactants to products, and so the acid with the higher Ka value will be producing more hydronium, and therefore have a lower pH. Therefore the solution of benzoic acid will have a lower pH.
Example $3$
The Ka value of ammonium (NH4+) is 5.6*10-10, the Kb value of ammonia (NH3) 1.8*10-5, is ammonium more strongly acidic than ammonia is basic?
Solution
The relative strength of an acid or base depends on how high its Ka or Kb value is, in this case, the Ka value is far lower than the Kb value so the ammonia is more strongly basic than ammonium is acidic.
Contributors and Attributions
• Charles Ophardt, Professor Emeritus, Elmhurst College; Virtual Chembook,. Lloyd McCarthy (UCD)
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Acids_and_Bases/Ionization_Constants/Acid_and_Base_Strength.txt
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The quantity pH, or "power of hydrogen," is a numerical representation of the acidity or basicity of a solution. It can be used to calculate the concentration of hydrogen ions [H+] or hydronium ions [H3O+] in an aqueous solution. Solutions with low pH are the most acidic, and solutions with high pH are most basic.
Definitions
Although pH is formally defined in terms of activities, it is often estimated using free proton or hydronium concentration:
$pH \approx -\log[H_3O^+] \label{eq1}$
or
$pH \approx -\log[H^+] \label{eq2}$
$K_a$, the acid ionization constant, is the equilibrium constant for chemical reactions involving weak acids in aqueous solution. The numerical value of $K_a$ is used to predict the extent of acid dissociation. A large $K_a$ value indicates a stronger acid (more of the acid dissociates) and small $K_a$ value indicates a weaker acid (less of the acid dissociates).
For a chemical equation of the form
$HA + H_2O \leftrightharpoons H_3O^+ + A^-$
$K_a$ is express as
$K_a = \dfrac{[H_3O^+][A^-]}{[HA]} \label{eq3}$
where
• $HA$ is the undissociated acid and
• $A^-$ is the conjugate base of the acid.
Since $H_2O$ is a pure liquid, it has an activity equal to one and is ignored in the equilibrium constant expression in (Equation \ref{eq3}) like in other equilibrium constants.
Howto: Solving for $K_a$
When given the pH value of a solution, solving for $K_a$ requires the following steps:
1. Set up an ICE table for the chemical reaction.
2. Solve for the concentration of $\ce{H3O^{+}}$ using the equation for pH: $[H_3O^+] = 10^{-pH}$
3. Use the concentration of $\ce{H3O^{+}}$ to solve for the concentrations of the other products and reactants.
4. Plug all concentrations into the equation for $K_a$ and solve.
Example $1$
Calculate the $K_a$ value of a 0.2 M aqueous solution of propionic acid ($\ce{CH3CH2CO2H}$) with a pH of 4.88.
$\ce{CH_3CH_2CO_2H + H_2O \leftrightharpoons H_3O^+ + CH_3CH_2CO_2^- } \nonumber$
Solution
ICE TABLE
ICE $\ce{ CH_3CH_2CO_2H }$ $\ce{ H_3O^+ }$ $\ce{ CH_3CH_2CO_2^- }$
Initial Concentration (M) 0.2 0 0
Change in Concentration (M) -x +x +x
Equilibrium Concentration (M) 0.2 - x x x
According to the definition of pH (Equation \ref{eq1})
\begin{align*} -pH = \log[H_3O^+] &= -4.88 \[4pt] [H_3O^+] &= 10^{-4.88} \[4pt] &= 1.32 \times 10^{-5} \[4pt] &= x \end{align*}
According to the definition of $K_a$ (Equation \ref{eq3}
\begin{align*} K_a &= \dfrac{[H_3O^+][CH_3CH_2CO_2^-]}{[CH_3CH_2CO_2H]} \[4pt] &= \dfrac{x^2}{0.2 - x} \[4pt] &= \dfrac{(1.32 \times 10^{-5})^2}{0.2 - 1.32 \times 10^{-5}} \[4pt] &= 8.69 \times 10^{-10} \end{align*}
Contributors and Attributions
• Paige Norberg (UCD) and Gabriela Mastro (UCD)
Calculating Equilibrium Concentrations
$K_a$ is an acid dissociation constant, also known as the acid ionization constant. It describes the likelihood of the compounds and the ions to break apart from each other. As we already know, strong acids completely dissociate, whereas weak acids only partially dissociate. A big $K_a$ value will indicate that you are dealing with a very strong acid and that it will completely dissociate into ions. A small $K_a$ will indicate that you are working with a weak acid and that it will only partially dissociate into ions.
General Guide to Solving Problems involving $K_a$
Generally, the problem usually gives an initial acid concentration and a $K_a$ value. From there you are expected to know:
1. How to write the $K_a$ formula
2. Set up in an ICE table based on the given information
3. Solve for the concentration value, x.
4. Use x to find the equilibrium concentration.
5. Use the concentration to find pH.
How to write the $K_a$ formula
The general formula of an acid dissociating into ions is
$HA_{(aq)} + H_2O_{(l)} \rightleftharpoons H_3O^+_{(aq)} + A^-_{(aq)} \label{1}$
with
• $HA$ is the acid,
• $A^-$ is the conjugate base and
• $H_3O^+$ is the the hydronium ion
By definition, the $K_a$ formula is written as the products of the reaction divided by the reactants of the reaction
$K_a = \dfrac{[Products]}{[Reactants]} \label{2}$
Based off of this general template, we plug in our concentrations from the chemical equation. The concentrations on the right side of the arrow are the products and the concentrations on the left side are the reactants. Using this information, we now can plug the concentrations in to form the $K_a$ equation. We then write:
$K_a = \dfrac{[H_3O^+][A^-]}{[HA]} \label{3}$
The concentration of the hydrogen ion ($[H^+]$) is often used synonymously with the hydrated hydronium ion ($[H_3O^+]$).
To find a concentration of hydronium ions in solution from a pH, we use the formula:
$[H_3O^+]= 10^{-pH}$
This can be flipped to calculate pH from hydronium concentration:
$pH = -\log[H_3O^+]$
• An acidic solution is one that has an excess of $H_3O^+$ ions compared to $OH^-$ ions.
• An basic (or alkaline) solution is one that has an excess of $OH^-$ ions compared to $H_3O^+$ ions.
• A neutral solution is one that has equal concentrations of $OH^-$ ions and $H_3O^+$ ions.
At 25 °C, we can correlate whether a solution is acidic, basic, or neutral based off of the measured pH of the solutions:
• pH = 7 is neutral
• pH > 7 is basic
• pH < 7 is acidic
However, these relationships are not valid at temperatures outside 25 °C.
cid
Calculate the pH of a weak acid solution of 0.2 M HOBr, given:
$HOBr + H_2O \rightleftharpoons H_3O^+ + OBr^-$
$K_a = 2 \times 10^{-9}$
Solution
Step 1: The ICE Table
Since we were given the initial concentration of HOBr in the equation, we can plug in that value into the Initial Concentration box of the ICE chart. Considering that no initial concentration values were given for H3O+ and OBr-, we can assume that none was present initially, and we indicate this by placing a zero in the corresponding boxes. M stands for molarity.
HOBr H3O+ OBr-
Initial Concentration 0.2 M 0 0
Change in Concentration
Equilibrium Concentration
Because we started off without an initial concentration of H3O+ and OBr-, it has to come from somewhere. In the Change in Concentration box, we add a +x because while we do not know what the numerical value of the concentration is at the moment, we do know that it has to be added and not taken away. In contrast, since we did start off with a numerical value of the initial concentration, we know that it has to be taken away to reach equilibrium. Because of this, we add a -x in the HOBr box.
HOBr H3O+ OBr-
Initial Concentration 0.2 M 0 0
Change in Concentration -x M +x M +x M
Equilibrium Concentration
Now its time to add it all together! Go from top to bottom and add the Initial concentration boxes to the Change in concentration boxes to get the Equilibrium concentration.
HOBr H3O+ OBr-
Initial Concentration 0.2 M 0 0
Change in Concentration -x M +x M +x M
Equilibrium Concentration (0.2 - x) M x M x M
Step 2: Create the $K_a$ equation using this equation: $K_a = \dfrac{[Products]}{[Reactants]}$
$K_a = \dfrac{[H_3O^+][OBr-]}{[HOBr-]}$
Step 3: Plug in the information we found in the ICE table
$K_a = \dfrac{(x)(x)}{(0.2 - x)}$
Step 4: Set the new equation equal to the given Ka
$2 \times 10^{-9} = \dfrac{(x)(x)}{(0.2 - x)}$
Step 5: Solve for x
$x^2 + (2 \times 10^{-9})x - (4 \times 10^{-10}) = 0$
To solve for x, we use the quadratic formula
• $a=1$
• $b=2 \times 10^{-9}$
• $c=-4 \times 10^{-10}$
$x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}= \dfrac{-2 \times 10^{-9} \pm \sqrt{(2 \times10^{-9})^2 - 4(1)(-4 \times 10^{-10})}}{2(1)}$
$x=2.0 \times 10^{-5}$
Step 6: Plug x back into the ICE table to find the concentration
$x= [H_3O^+] = 2 \times 10^{-5} \; M$
Step 7: Use the formula using the concentration to find pH
$pH = -\log[H_3O^+] = -\log(2 \times 10^{-5}) = -(-4.69) = 4.69$
$pH= 4.69$
Example $2$: Concentrated Solution of Acetic Acid (Vineger)
For acetic acid, HC2H3O2, the $K_a$ value is $1.8 \times 10^{-5}$. Calculate the concentration of H3O+ in a 0.3 M solution of HC2H3O2.
Solution
Step 1: The ICE Table
Since we were given the initial concentration of HC2H3O2 in the original equation, we can plug in that value into the Initial Concentration box of the ICE chart. Considering that no initial concentration values were given for $H_3O^+$ and $C_2H_3O_2^-$, we assume that none was present initially, and we indicate this by placing a zero in the corresponding boxes.
HC2H3O2 H3O+ C2H3O2
Initial Concentration 0.3 M 0 0
Change in Concentration
Equilibrium Concentration
Because we started off without any initial concentration of H3O+ and C2H3O2-, is has to come from somewhere. For the Change in Concentration box, we add a +x because while we do not know what the numerical value of the concentration is at the moment, we do know that it has to be added and not taken away. In contrast, since we did start off with a numerical value of the initial concentration, we know that it has to be taken away to reach equilibrium. Because of this, we add a -x in the $HC_2H_3O_2$ box.
HC2H3O2 H3O+ C2H3O2
Initial Concentration 0.3 M 0 0
Change in Concentration -x M +x M +x M
Equilibrium Concentration
Now its time to add it all together! Go from top to bottom and add the Initial concentration boxes to the Change in concentration boxes to get the Equilibrium concentration.
HC2H3O2 H3O+ C2H3O2
Initial Concentration 0.3 M 0 0
Change in Concentration -x M +x M +x M
Equilibrium Concentration (0.3 - x) M x M x M
Step 2: Create the $K_a$ equation using this equation: $K_a = \dfrac{[Products]}{[Reactants]}$
$K_a = \dfrac{[H_3O^+][C_2H_3O_2]}{[HC_2H_3O_2]}$
Step 3: Plug in the information we found in the ICE table
$K_a = \dfrac{(x)(x)}{(0.3 - x)}$
Step 4: Set the new equation equal to the given Ka
$1.8 x 10^{-5} = \dfrac{(x)(x)}{(0.3 - x)}$
Step 5: Solve for x
$(x^2)+ (1.8 \times 10^{-5}x)-(5.4 \times 10^{-6})$
To solve for x, we use the quadratic formula
• $a=1$
• $b=1.8 \times 10^{-5}$
• $c=-5.4 \times 10^{-6}$
$x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}= \dfrac{-1.8 \times 10^{-5} \pm \sqrt{(1.8 \times10^{-5})^2 - 4(1)(-5.4 \times 10^{-6})}}{2(1)}$
$x=0.0023$
Step 6: Plug x back into the ICE table to find the concentration
$x= [H_3O^+] = 0.0023\; M$
Example $3$: Concentrated Solution of Benzoic Acid
Find the equilibrium concentration of HC7H5O2from a 0.43 M solution of Benzoic Acid, HC7H5O2.
Solution
Given: $K_a$ for HC7H5O2= 6.4 x 10-5
Step 1: The ICE Table
HC7H5O2 H3O+ C7H5O2-
Initial Concentration 0.43 M 0 0
Change in Concentration -x +x +x
Equilibrium Concentration (0.43-x)M x M x M
Step 2: Create the $K_a$ equation using this equation :$K_a = \dfrac{[Products]}{[Reactants]}$
$K_a = \dfrac{[H_3O^+][C_7H_5O_2-]}{[HC_7H_5O_2]}$
Step 3: Plug in the information we found in the ICE table
$K_a = \dfrac{(x)(x)}{(0.43 - x)}$
Step 4: Set the new equation equal to the given Ka
$6.4 x 10^{-5} = \dfrac{(x)(x)}{(0.43 - x)}$
Step 5: Solve for x.
x=0.0052
Step 6: Plug x back into the ICE table to find the concentration
[HC7H5O2]= (0.43-x)M
[HC7H5O2]= (0.43-0.0052)M
Answer: [HC7H5O2]= 0.425 M
Example $4$: Concentrated Solution of Hypochlorous acid
For a 0.2 M solution of Hypochlorous acid, calculate all equilibrium concentrations.
Solution
Given: $K_a = 3.5 \times 10^{-8}$
Step 1: The ICE Table
HOCl H3O+ OCl-
Initial Concentration 0.2 M 0 0
Change in Concentration -x M +x M +x M
Equilibrium Concentration (0.2 - x) M x M x M
Step 2: Create the $K_a$ equation using this equation: $K_a = \dfrac{[Products]}{[Reactants]}$
$K_a = \dfrac{[H_3O^+][OCl-]}{[HOCl-]}$
Step 3: Plug in the information we found in the ICE table
$K_a = \dfrac{(x)(x)}{(0.2 - x)}$
Step 4: Set the new equation equal to the given Ka
$3.5 x 10^{-8} = \dfrac{(x)(x)}{(0.2 - x)}$
Step 5: Solve for x
x=8.4 x 10-5
Step 6: Plug x back into the ICE table to find the concentration
[HOCl]= [(.2)-(8.4 x 10-5)]=.199
[H3O+]=8.4 x 10-5
[OCl-]=8.4 x 10-5
Example $5$: pH
Calculate the pH from the equilibrium concentrations of [H3O+] in Example $4$.
Solution
Given:
[HOCl]=0.199
[H3O+]=8.4 x 10-5
[OCl-]=8.4 x 10-5
Step 1: Use the formula using the concentration of [H3O+] to find pH
$pH = -\log[H3O+] = -\log(8.4 x 10^{-5}) = 4.08$
Contributors and Attributions
• Kellie Berman (UCD), Alysia Kreitem (UCD)
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Acids_and_Bases/Ionization_Constants/Calculating_A_Ka_Value_From_A_Measured_Ph.txt
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A simple way to understand an ionization constant is to think of it in a clear-cut way: To what degree will a substance produce ions in water? In other words, to what extent will ions be formed?
Introduction
Water has a very low concentration of ions that are detectable. Water undergoes self-ionization, where two water molecules interact to from a hydronium ion and a hydroxide ion.
$H_2O + H_2O \rightleftharpoons H_3O^+ + OH^- \label{1}$
Even though water does not form a lot of ions, the existence of them is evident in pure water by the electrical conductivity measurements. Water undergoes ionization because the powerfully electronegative oxygen atom takes the electron from a hydrogen atom, making the proton dissociate.
$\ce{H-O-H} \rightleftharpoons H^+ + OH^- \label{2}$
with two ions are formed:
• hydrogen ions $H^+$
• hydroxyl (hydroxide) ions $OH^-$
The hydrogen ions then react with water to form hydronium ions:
$H^+ + H_2O \rightarrow H_3O^+ \label{3}$
Typically, hydrogen atoms are bonded with another water molecule resulting in a hydration that forms a hydronium and hydroxide ion.
$H_2O + H_2O \rightarrow H_3O^+ + OH^- \label{4}$
In 1 L of pure water at 25 degrees Celsius, there is 10-7 moles of hyrodronium ions or hydroxide ions at equilibrium. Let’s come back to the question of to what degree will a substance form ions in water? For water, we have learned that it will occur until 10-7 moles of either ion will ionize at the previous given conditions. Since this is during equilibrium, a constant can be formed.
$H_2O \rightleftharpoons H^+ + OH^- \label{5}$
$K_{eq}= \dfrac{[H^+] [OH^-]}{[H_2O]} \label{6}$
This equilibrium constant is commonly referred to as $K_w$.
What about acids and bases?
Now that we have an idea of what an ionization constant is, let’s take a look at how acids and bases play in this scenario. Strong acids and bases are those that completely dissociate into ions once placed in solution. For example:
$KOH + H_2O → K^+ + OH^-$
So, a 2 M solution of KOH would have 2 M concentration of of OH- ions (also 2 M concentration of $K$).
Weak acids and bases, however do not behave the same way. Their amounts cannot be calculated as easily. This is because the ions do not fully dissociate in the solution. Weak acids have a higher pH than strong acids, because weak acids do not release all of its hydrogens. The acid dissociation constant tell us the extent to which an acid dissociates
$HCN_{(aq)} + H_2O \rightleftharpoons H_3O^+ + CN^- \label{8}$
$K_a= \dfrac{[H_3O^+] [CN^-]}{[HCN]} \label{9}$
This equation is used fairly often when looking at equilibrium reactions. During equilibrium, the rates of the forward and backward reaction are the same. However, the concentrations tend to be varied. Since concentration is what gives us an idea of how much substance has dissociated, we can relate concentration ratios to give us a constant. K is found by first finding out the molarity of each substance. Then, just as shown in the equation, we divide the products by the reactants, excluding solids and liquids. Also when there is more than one product or reactant, their concentrations must be multiplied together. Even though you will not see a multiplication sign, if there are two molecules associated, remember to multiply them. If there is a coefficient in front of a molecule, the concentration must be raised to that power in the calculations.
A weaker acid tends to have a smaller $K_a$ because the concentration on the bottom of the equation is larger. There is more of the acid, and less of the ions. You can think of Ka as a way of relating concentration in order to find out other calculations, typically the pH of a substance. A pH tells you how basic or acidic something is, and as we have learned that depends on how much ions become dissociated.
Example $1$: Calculate Ionization Constant
Calculate the ionization constant of a weak acid. Solve for Ka given 0.8 M of hydrogen cyanide and 0.0039 M for hyrodronium and cyanide ions:
$HCN_{(aq)} + H_2O → H_3O^+ + CN^-$
Solution
• [HCN] is 0.8 M
• [H_3O^+] is 0.0039 M
• [CN^-] is 0.0039 M.
We can assume here that [H3O+] = [CN-]
The equilibrium constant would be
$K_a= \dfrac{0.0039^2}{0.8} = 1.9 \times -^{5}$
We can use the similar method to find the Kb constant for weak bases. Again, an ionization constant is a measure of the extent a base will dissociate. Kb relates these molarity quantities. A smaller Kb corresponds to a weaker base, as a higher one a stronger base. Some common weak bases: NH3, NH4OH, HS-, C5H5N
$C_5H_5N_{(aq)} + H_2O_{(l)} → C_5H_5N^+_{(aq)} + OH^-_{ (aq)} \label{10}$
We would find Kb the same way we did Ka. However, most problems are not as simple and obvious. Let’s do an example that is a little more challenging to help you understand this better.
Example $2$: Calculate Concentration
Calculate the concentration of [OH-] in a 3M Pyridine solution with Kb= 1.5 * 10-9.
Solution
Since we already have our equation, let’s write the expression for the constant as discussed earlier (concentration of products divided by concentration of reactants)
$K_b= \dfrac{[OH^-] [C_5H_5NH^+]}{[C_5H_5N]} = 1.5 \times 10^{-9}$
Because pyridine is a weak base, we can assume that not much of it will disassociate. And since we have 3M, let’s make 3M-X our calculation. Subtracting X is going to be the change in molarity due to dissociation. Since our ions are unknown and they are both one mole, we can solve for them as being X. THUS: (X2/3-X) = 1.5 * 10-9
To make the calculation simple, we can estimate 3-X to be approximately 3 because of the weak base. Now our equation becomes: (X2/3) = 1.5 * 10-9. When we solve for X our answer is $4.7 \times 10^{-5}\; M$
This approximation was effective because x is small, which must be less than 5% of the initial concentration for that estimation to be justified.
Summary
We have learned that an ionization constant quantifies the degree a substance will ionize in a solution (typically water). Ka, Kb, and Kw are constants for acids, bases, and finally water, respectively and are related by
$K_a \times K_b = K_w$
However, these constants are also used to find concentrations as well as pH.
Contributors and Attributions
• Tatiana Litvin (UCD)
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Acids_and_Bases/Ionization_Constants/Fundamentals_of_Ionization_Constants.txt
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Unlike strong acids/bases, weak acids and weak bases do not completely dissociate (separate into ions) at equilibrium in water, so calculating the pH of these solutions requires consideration of a unique ionization constant and equilibrium concentrations. Although this is more difficult than calculating the pH of a strong acid or base solution, most biochemically important acids and bases are considered weak, and so it is very useful to understand how to calculate the pH of these substances. The same basic method can be used to determine the pH of aqueous solutions of many different weak acids and bases.
Introduction
An aqueous solution of a weak acid or base contains both the protonated and unprotonated forms of the compound, so an ICE table can be made and used to plug in concentrations into an equilibrium constant expression. The ionization constant for the acid (Ka) or base (Kb) is a measure of how readily the acid donates protons or how readily a base accepts protons. Because you are calculating pH, you must solve for the unknown concentration of hydronium ions in solution at equilibrium.
The first step in calculating the pH of an aqueous solution of any weak acid or base is to notice whether the initial concentration is high or low relative to 10-7 M (the concentration of hydronium and hydroxide ions in water due to the autoionization of water). If the concentration of the acid or base is very close to or less than 10-7 M, then the solution is considered dilute and additional steps must be taken to calculate pH.
Weak Acids and Bases
You must first be familiar with equilibrium constant expressions and how to write them for a chemical reaction. Then, by making an ICE table, you can find unknown concentration values that can be plugged into this equilibrium expression.
Example $1$: Vinegar as a Weak Acid
What is the pH of 1.5 L of a vinegar that is 3% acetic acid by mass? (Ka = 1.8 x 10-5)
Solution
To start, you must find the initial concentration of acetic acid in the vinegar. Assume that the vinegar is really just a solution of acetic acid in water, and that density = 1 g/mL.
So if the vinegar is 3% acetic acid by mass and the molar mass of HC2H3O2 = 60.05 g/mol, then
$\dfrac{1.5L,vinegar}{} \times \dfrac{1000mL}{1L} \times \dfrac{1g}{1mL} \times \dfrac{3g,acetic acid}{100g,vinegar} \times \dfrac{1mol,acetic acid}{60.05g,acetic acid} = 0.75 \;mol\; HC_2H_3O_2 \nonumber$
Divide 0.75 mol by 1.5 L to get an initial concentration of 0.50 M.
Now make an ICE table, considering the ionization of acetic acid in water into acetate ion and hydronium ion. Because only solutes and gases are incorporated into the equilibrium expression, you can ignore the concentration of water (a pure liquid) in our calculations.
$HC_2H_3O_{2(aq)} + H_2O_{(l)} \rightleftharpoons C_2H_3O^-_{2(aq)} + H_3O^+_{(aq)} \nonumber$
HC2H3O2 H2O C2H3O2- H3O+
Initial 0.5 --- 0 0
Change -x --- +x +x
Equilibrium 0.5 - x --- x x
For every acetic acid molecule that dissociates, one acetate ion and one hydronium ion is produced. This can be represented by subtracting "x" from the original acetic acid concentration, and adding "x" to the original concentrations of the dissociated ions.
You can create a modified equilibrium constant expression
$K_a = \dfrac{[C_2H_3O_2^-][H_3O^+]}{[HC_2H_3O_2]} \nonumber$
and then plug in the concentration values you found in the ICE table
$1.8 \times 10^{-5} = \dfrac{x^2}{0.5 - x} \nonumber$
so
$x^2 + (1.8 \times 10^{-5})x - (9 \times 10^{-6}) = 0 \nonumber$
then use the quadratic formula to calculate
$x = 0.0030\; M = [H_3O^+] \nonumber$
which can be plugged into the formula
$pH = -\log[H_3O^+] \nonumber$
$-\log(0.0030) = pH = 2.5 \nonumber$
The same thing can be done for calculating the pH of a weak base.
Example $2$: Ammonia as a Weak Base
What is the pH of a $7.0 \times 10^{-3}$ M NH3 solution? (pKb = 4.74)
Solution
$NH_{3(aq)} + H_2O_{(l)} \rightleftharpoons NH^+_{4(aq)} + OH^-_{(aq)} \nonumber$
NH3 H2O NH4+ OH-
I 7.0 x 10-3 --- 0 0
C -x --- +x +x
E 7.0x10-3 - x --- x x
Instead of Kb, you were given pKb. So to get Kb
pKb = -log(Kb) = 4.74
Kb = 10-4.74 = 1.8 x 10-5
Plug these values into the equilibrium expression to get
$K_b = \dfrac{[NH_4^+][OH^-]}{[NH_3]} = 1.8 \times 10^{-5} = \dfrac{x^2}{7 \times 10^{-3} - x} \nonumber$
and use the quadratic formula to find that
$x = 3.46 \times 10^{-4}\; M = [OH^-] \nonumber$
so
$pOH = -\log(3.46 \times 10^{-4}) = 3.46 \nonumber$
and in water at 25 degrees
$pH + pOH = 14 \nonumber$
Hence:
$14 - 3.46 = pH = 10.54 \nonumber$
Weak Polyprotic Acids and Bases
Polyprotic acids have more than one proton to donate to water, and so they have more than one ionization constant (Ka1, Ka2, etc) that can be considered. Polyprotic bases take more than one proton from water, and also have more than one ionization constant (Kb1, Kb2, etc).
Most often the first proton exchange is the only one that considerably affects pH. This is discussed more at the end of the first example.
Example $3$: Citric Acid as a Polyprotic Acid
What is the pH of a grapefruit that contains 0.007 M citric acid solution (C6H8O7)?
(Ka1 = 7.5 x 10-4, Ka2 = 1.7 x 10-5, Ka3 = 4.0 x 10-7)
Solution
Make an ICE table for the first dissociation
$C_6H_8O_{7(aq)} + H_2O_{(l)} \rightleftharpoons C_6H_7O_7^- + H_3O^+_{(aq)} \nonumber$
C6H8O7 H2O C6H7O7- H3O+
I 0.007 --- 0 0
C -x --- +x +x
E 0.007 - x --- x x
$K_{a1} = \dfrac{[C_6H_7O_7^-][H_3O^+]}{[C_6H_8O_7]} = 7.5 \times 10^{-4} = \dfrac{x^2}{0.007 - x} \nonumber$
and use the quadratic formula to find that
$x = 0.00195 \;M = [H_3O^+] \nonumber$
Then a second ICE table can be made for the second dissociation
$C_6H_7O^-_{7(aq)} + H_2O_{(l)} \rightleftharpoons C_6H_6O^{2-}_7 + H_3O^+_{(aq)} \nonumber$
C6H7O7- H2O C6H6O72- H3O+
I 0.00195 --- 0 0.00195
C -x --- +x +x
E 0.00195 - x --- x 0.00195 + x
Remember that, for the first dissociation, x = [H3O+] = [C6H7O7-], so you can plug in the first value of x in for the initial concentrations of C6H7O7- and H3O+.
$K_{a2} = \dfrac{[C_6H_6O_7^{2-}][H_3O^+]}{[C_6H_7O_7^-]} = 1.7 \times 10^{-5} = \dfrac{(x)(0.00195 + x)}{0.00195- x} \nonumber$
and use the quadratic formula to find that
$x = 1.67 \times 10^{-5} \nonumber$
$[H_3O^+] = 0.00195 + 1.67 \times 10^{-5} = 0.00197 \;M \nonumber$
$-\log(0.00197) = pH = 2.71 \nonumber$
Note that if you ignored the addition of hydronium from the second dissociation, then [H3O+] = 0.00195 M, and using this value to calculate pH still gives you the answer of 2.71.
So even though you made two ICE tables (you could even make a third table for Ka3), the protons donated in the second dissociation were negligible compared to the first dissociation. So you can see that it is really only the first dissociation that affects pH. Most often this is the case, and only one ICE table is necessary. It is up to you how certain you want to be and how many ICE tables you want to make when you calculate these problems.
Example $4$: Soda Ash as a Polyprotic Base
What is the pH of a saturated solution of sodium carbonate (Na2CO3)?
(solubility in water is 21.6 g/100mL at room temperature and for carbonic acid, H2CO3, Ka1 = 4.5 x 10-7, Ka2 = 4.7 x 10-11
Solution
First, you have to find the find the initial concentration of CO32- which can be found from
$\dfrac{21.6g,Na_2CO_3}{} \times \dfrac{1\;mol}{105.99\;g} = 0.204\; mol\; Na_2CO_3 = 0.204\; mol\; CO_3^{2-} \nonumber$
then divide 0.204 mol by 0.100 L to get 2.04 M CO32-
Plug into an ICE table
$CO^{2-}_{3(aq)} + H_2O_{(l)} \rightleftharpoons HCO^-_{3(aq)} + OH^-_{(aq)} \nonumber$
CO32- H2O HCO3- OH-
I 2.04 --- 0 0
C -x --- +x +x
E 2.04 - x --- x x
But notice that the equilibrium constants are for carbonic acid. If you were considering the dissociation of carbonic acid, you would write the following expressions
$K_{a1} = \dfrac{[HCO_3^-][H_3O^+]}{[H_2CO_3]} = 4.5 \times 10^{-7} \nonumber$
for
$H_2CO_3 + H_2O \rightleftharpoons HCO_3^- + H_3O^+ \nonumber$
$K_{a2} = \dfrac{[CO_3^{2-}][H_3O^+]}{[HCO_3^-]} = 4.7 \times 10^{-11} \nonumber$
for
$HCO_3^- + H_2O \rightleftharpoons CO_3^{2-} + H_3O^+ \nonumber$
The second acid ionization constant corresponds to the first base ionization constant (because the base reactions go backwards). To convert the second acid ionization constant to the first base ionization constant, you use the equation
$K_a \times K_b = K_w = 10^{-14} \nonumber$
so that
$K_{a2} \times K_{b1} = 10^{-14} \nonumber$
$K_{b1} = \dfrac{10^{-14}}{4.7 \times 10^{-11}} = 2.13 \times 10^{-4} \nonumber$
Use the same equation to convert the first acid ionization constant to the second base ionization constant
$K_{a1} \times K_{b2} = 10^{-14} \nonumber$
$K_{b2} = \dfrac{10^{-14}}{4.5 \times 10^{-7}} = 2.22 \times 10^{-8} \nonumber$
The expressions for the protonation of carbonate are now known to be
$K_{b1} = \dfrac{[HCO_3^-][OH^-]}{[CO_3^{2-}]} = 2.13 \times 10^{-4} \nonumber$
for
$CO_3^{2-} + H_2O \rightleftharpoons HCO_3^- + OH^- \nonumber$
$K_{b2} = \dfrac{[H_2CO_3][OH^-]}{[HCO_3^-]} = 2.22 \times 10^{-8} \nonumber$
for
$HCO_3^- + H_2O \rightleftharpoons H_2CO_3 + OH^- \nonumber$
Plug the ICE tables values into the first equilibrium expression
$K_{b1} = \dfrac{x^2}{2.04 - x} = 2.13 \times 10^{-4} \nonumber$
and use the quadratic formula to solve
$x = 0.0207\; M = [OH^-] \nonumber$
You can ignore the second base ionization constant because it removes a negligible amount of protons from water. If you want to test this by making an ICE table, you should get that the total hydroxide concentration is 0.0207000222 M \approx 0.0207 M
so pOH = -log(0.0207) = 1.68
pH + pOH = 14
14 - 1.68 = pH = 12.32
Dilute Weak Acids and Bases
'Dilute' refers to the concentration of the acid or base in water. If the concentration is close to or below 10-7 M, then you must consider the donation of hydronium ions from water as well as from your acid or base. This is done by making an ICE table to find the protonation of the acid or base, while also incorporating the ion product of water.
Example $5$: Dilute Weak Acid
An average bee sting contains 5 micrograms of formic acid ($HCO_2H$). What is the pH of a 500 mL solution of formic acid? (pKa = 3.75)
Solution
First calculate the number of moles of formic acid was excreted.
$\dfrac{5.00 \mu g}{} \times \dfrac{1 g}{10^6 \mu g} \times \dfrac{1\;mol,HCO_2H}{46.03g,HCO_2H} = 1.09 \times 10^{-7} mol,HCO_2H \nonumber$
$\dfrac{1.09 \times 10^{-7} mol}{0.500 L} = 2.17 \times 10^{-7}\; M \nonumber$
$HCO_2H_{(aq)} + H_2O_{(l)} \rightleftharpoons CO_2H^-_{(aq)} + H_3O^+_{(aq)} \nonumber$
HCO2H H2O CO2H- H3O+
I $2.17 \times 10^{-7}$ --- 0 0
C $-x$ --- +x +x
E $2.17 \times 10^{-7} - x$ --- x x
A second ICE table can be made for the autoionization of water
$2H_2O_{(l)} \rightleftharpoons H_3O^+_{(aq)} + OH^-_{(aq)} \nonumber$
2H2O H3O+ OH-
I --- 0 0
C --- +y +y
E --- y y
Notice that total [H3O+] = x + y
$pK_a = 3.75 \nonumber$
$K_a = 10^{-3.75} = 1.78 \times 10^{-4} \nonumber$
So you can simultaneously solve both equations
$K_a = \dfrac{[CO_2H^-][H_3O^+]}{[HCO_2H]} = \dfrac{x(x + y)}{2.17 \times 10^{-7} - x} = 1.78 \times 10^{-4} \nonumber$
and
$K_w = [H_3O^+][OH^-] = (x + y)(y) = 10^{-14} \nonumber$
These calculations can be tricky, and it is very easy to make mistakes. It is usually easier to use variables to solve these problems instead of handling awkward numbers.
For this problem use
• a = Ka = 1.78 x 10-4
• c = initial [HCO2H] = 2.17 x 10-7
• w = Kw = 10-14
$a = \dfrac{x(x+y)}{c-x} \nonumber$
and
$(x+y)(y) = w \nonumber$
$ac-ax = x^2 +xy \nonumber$
and
$x=\dfrac{w}{y}-y \nonumber$
$ac-a\left(\dfrac{w}{y}-y\right)=\left(\dfrac{w}{y}-y\right)^2+\left(\dfrac{w}{y}-y\right)y \nonumber$
$ac-\dfrac{aw}{y}+ay=\dfrac{w^2}{y^2}-\dfrac{2wy}{y}+y^2+\dfrac{wy}{y}-y^2 \nonumber$
$ac-\dfrac{aw}{y}+ay=\dfrac{w^2}{y^2}-2w+y^2+w-y^2 \nonumber$
$ac-\dfrac{aw}{y}+ay=\dfrac{w^2}{y^2}-w \nonumber$
$ac-\dfrac{aw}{y}+ay+w=\dfrac{w^2}{y^2} \nonumber$
$acy^2-awy+ay^3+wy^2=w^2 \nonumber$
$ay^3+(ac+w)y^2-awy-w^2=0 \nonumber$
so when we plug back in the values
$(1.78 \times 10^{-4})y^3+((1.78 \times 10^{-4})(2.17 \times 10^{-7})+(10^{-14})) y^2-(1.78 \times 10^{-4})(10^{-14})y-(10^{-14})^2=0 \nonumber$
$(1.78 \times 10^{-4})y^3+(3.86 \times 10^{-11})y^2-(1.78 \times 10^{-18})y-10^{-28}=0 \nonumber$
and use a graphing calculator to find that
$y = 3.91 \times 10^{-8} \nonumber$
and
$x=\dfrac{w}{y}-y = \dfrac{10^{-14}}{3.91 \times 10^{-8}}-3.91 \times 10^{-8} = 2.17 \times 10^{-7} \nonumber$
and total
$[H_3O^+] = (x + y) = (2.17 \times 10^{-7} + 3.91 \times 10^{-8}) = 2.56 \times 10^{-7} \;M \nonumber$
so
$-\log[H_3O^+] = -\log(2.56 \times 10^{-7}) = pH = 6.59 \nonumber$
Compare this value to pH = 6.66, which is what would have been calculated if the autoprotonization of water was not considered.
Buffer Solutions
Buffer solutions resist pH change when more acid or base is added. They are made from a weak acid and its conjugate base or a weak base and its conjugate acid. The Henderson-Hasselbalch approximation can be used to find the pH of a buffer solution, and is derived from the acid equilibrium expression.
$HA_{(aq)} + H_2O_{(l)} \rightleftharpoons A^-_{(aq)} + H_3O^+_{(aq)} \nonumber$
$K_a = \dfrac{[A^-][H_3O^+]}{[HA]} \nonumber$
$\log(K_a) = log\left( \dfrac{[A^-][H_3O^+]}{[HA]} \right) \nonumber$
$\log(K_a) = \log[H_3O^+] + \log \left( \dfrac{[A^-]}{[HA]} \right) \nonumber$
$-pK_a = -pH + \log \left( \dfrac{[A^-]}{[HA]} \right) \nonumber$
$pH = pK_a + \log \left( \dfrac{[A^-]}{[HA]} \right) \nonumber$
A similar equation can be used for bases
$pOH = pK_b + \log \left( \dfrac{[HB^+]}{[B]} \right) \nonumber$
Example $6$: Blood
The pH of blood plasma is 7.40, and is maintained by a carbonic acid/hydrogen carbonate buffer system. What mass of sodium bicarbonate (NaHCO3) should be added to a one liter solution of 0.250 M H2CO3 to maintain the solution at pH of 7.40? (pKa = 6.35)
Solution
$pH = pK_a + \log \left( \dfrac{[HCO_3^-]}{[H_2CO_3]} \right) \nonumber$
$7.40 = 6.35 + \log \left( \dfrac{[HCO_3^-]}{[0.250]} \right) \nonumber$
$1.05 = \log[HCO_3^-] -\log(0.250) \nonumber$
$0.448 = \log[HCO_3^-] \nonumber$
$[HCO_3^-] = 0.356\; M \nonumber$
$(0.356\; M) \times (1\; L\; solution) = 0.356\; mol\; HCO_3^- \nonumber$
$[0.356 mol\,HCO_3^{-} \times \dfrac{1 mol\,NaHCO_3}{1 mol\,HCO_3^-} \times \dfrac{84.01g}{1 mol,NaHCO_3} = 29.9 g \text{sodium bicarbonate} \nonumber$
It is also possible to use the Henderson-Hasselbalch equation to find pKa, pH, or [HA] if the other variables are given or calculated. Also notice that because $\dfrac{[A^-]}{[HA]}$ will cancel out the unit of volume, moles of $HA$ and $A^-$ can be used instead of molarity.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Acids_and_Bases/Ionization_Constants/Weak_Acids_and_Bases.txt
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Learning Objectives
• Define a weak acid or base.
• Calculate pH and pOH of a weak acid or base solution using simple formula, quadratic equation, and including autoionization of water.
• Calculate the pH or pOH quickly.
Weak acids and bases are only partially ionized in their solutions, whereas strong acids and bases are completely ionized when dissolved in water. Some common weak acids and bases are given here. Furthermore, weak acids and bases are very common, and we encounter them often both in the academic problems and in everyday life. The ionization of weak acids and bases is a chemical equilibrium phenomenon. The equilibrium principles are essential for the understanding of equilibria of weak acids and weak bases. In this connection, you probably realize that conjugate acids of weak bases are weak acids and conjugate bases of weak acids are weak bases.
Common Weak Acids Common Weak Bases
Acid Formula Base Formula
Formic $\ce{HCOOH}$ ammonia $\ce{NH3 }$
Acetic $\ce{CH3COOH}$ trimethyl ammonia $\ce{N(CH3)3 }$
Trichloroacetic $\ce{CCl3COOH }$ pyridine $\ce{C5H5N }$
Hydrofluoric $\ce{HF }$ ammonium hydroxide $\ce{NH4OH }$
Hydrocyanic $\ce{HCN }$ water $\ce{H2O }$
Hydrogen sulfide $\ce{H2S }$ $\ce{HS-}$ ion $\ce{HS-}$
Water $\ce{H2O }$ conjugate bases of weak acids e.g.: $\ce{HCOO-}$
Conjugate acids of weak bases $\ce{NH4+ }$
Ionization of Weak Acids
Acetic acid, $\ce{CH3COOH}$, is a typical weak acid, and it is the ingredient of vinegar. It is partially ionized in its solution.
$\ce{CH3COOH \rightleftharpoons CH3COO- + H+} \nonumber$
The structure of the acetate ion, $\ce{CH3COO-}$, is shown below.
Example 1
In a solution of acetic acid, the equilibrium concentrations are found to be $\mathrm{[CH_3COOH] = 1.000}$; $\mathrm{[CH_3COO^-] = 0.0042}$. Evaluate the pH of this solution and the equilibrium constant of ionization of acetic acid.
Solution
From the ionization of acetic acid,
\begin{alignat}{3} \ce{&CH3COOH \:\rightleftharpoons\: &&CH3COO- + &&\:\:\:\:H+}\ &\:\:\:\:\:\:\:0.100 &&\:\:\:\:0.0042 &&0.0042 \end{alignat}
we conclude that
\begin{align} \ce{[H+]} &= \ce{[CH3COO- ]}\ &= 0.0042 \end{align}
Thus, $\mathrm{pH = -\log0.0042 = 2.376}$.
The equilibrium constant of ionzation,
$K = \dfrac{(0.0042)^2}{1.000} = 1.78\times10^{-5} \nonumber$
DISCUSSION
The equilibrium constant of an acid is represented by Ka; and similar to the pH scale, a pKa scale is defined by
$\mathrm{p\mathit K_a = - \log \mathit K_a} \nonumber$
and for acetic acid, pKa = 4.75. Note that Ka = 10-pKa
The pH and pKa of Weak Acid
There are many weak acids, which do not completely dissociate in aqueous solution. As a general discussion of weak acids, let $\ce{HA}$ represent a typical weak acid. Then its ionization can be written as:
$\ce{HA \rightleftharpoons H+ + A-} \nonumber$
In a solution whose label concentration is $C = \ce{[HA]} + \ce{[A- ]}$, l et us assume that x is the concentration that has undergone ionization. Thus, at equilibrium, the concentrations are
$\ce{[HA]} = C - x \nonumber$
$\ce{[H+]} = \ce{[A- ]} = x \nonumber$
Make sure you understand why they are so, because you will have to set up these relationships in your problem solving. In summary, we formulate them as
$\begin{array}{cccccl} \ce{HA &\rightleftharpoons &H+ &+ &A- &}\ C & & & & &\leftarrow \textrm{ initial concentration, assume }\textit{x}\textrm{ M ionized}\ C-x & &x\:\:\: & &x &\leftarrow \textrm{ equilibrium concentration} \end{array}$
$K_{\ce a} = \dfrac{x^2}{C - x}$
$\mathrm{p\mathit K_a = - \log \mathit K_a} \nonumber$
$\mathrm{\mathit K_a = 10^{-\large{p\mathit K}_{\Large a}}} \nonumber$
The pKa values of many weak acids are listed in table form in handbooks, and some of these values are given in Table E1.
Example 2
The pKa of acetic acid is 4.75. Find the pH of acetic acid solutions of labeled concentrations of 1.0 M, 0.010 M, and 0.00010 M.
Solution
Assume the label concentration as C and x mole ionized, then the ionization and the equilibrium concentrations can be represented by the ICE table below.
ICE $\ce{CH3COOH}$ $\rightleftharpoons$ $\ce{CH_3COO^-}$ $\ce{H+}$
Initial C 0 0
Change -x +x +x
Equilibrium C - x x x
with
$K_{\ce a} = \dfrac{x^2}{C - x} \nonumber$
The equation is then
$x^2 + K_{\ce a} x = C K_{\ce a} = 0 \nonumber$
The solution of x is then
$x = \dfrac{- K_{\ce a} + \sqrt{K_{\ce a}^2 + 4 C K_{\ce a}}}{2} \nonumber$
Recall that Ka - 1.78e-5, the values of x for various C are given below:
C = 1.0 0.010 0.00010 M
x = 0.0042 0.00041 0.0000342 M
pH = 2.38 3.39 4.47
DISCUSSION
In the above calculations, the following cases may be considered:
1. $x = \sqrt{K_{\ce a} \times C} \nonumber$ Note that you are comparing x with C here. If C > 100*Ka, the above method gives satisfactory results.
2. $x^2 + K_{\ce a} x - C K_{\ce a} = 0 \nonumber$ and the solution for x, which must not be negative, has been given above.
3. Both cases 1 and 2 neglect the contribution of $\ce{[H+]}$ from the ionization of water. However, if the pH calculated from cases 1 and 2 falls in the range between 6 and 7, the concentration from self-ionization of water cannot be neglected. When the contribution of pH due to self-ionization of water cannot be neglected, there are two equilibria to be considered.
$\begin{array}{cccccl} \ce{HA &\rightleftharpoons &H+ &+ &A- &}\ C-x & &x & &x &\ \ \ce{H2O &\rightleftharpoons &H+ &+ &OH- &}\ 55.6 &&y &&y &\leftarrow (\ce{[H2O]} = 55.6) \end{array}$
Thus,
\begin{align} \ce{[H+]} &= (x+y), \ \ce{[A- ]} &= x, \ \ce{[OH- ]} &= y, \end{align}
and the two equilibria are
$K_{\ce a} = \dfrac{(x+y) x}{C - x}\label{1}$
and
\begin{align} K_{\ce w} &= (x+y)\, y, \label{2} \ (K_{\ce w} &= \textrm{1E-14}) \end{align} \nonumber
There are two unknown quantities, x and y in two equations, and (1) may be rearranged to give
$x^2 + (y + K_{\ce a}) x - C K_{\ce a} = 0 \nonumber$
$x = \dfrac{ -(y+K_{\ce a}) + ((y+K_{\ce a})^2 + 4 C K_{\ce a})^{1/2}}{2} \nonumber$
One of the many methods to find a suitable solution for this problem is to use iterations, or successive approximations.
1. Assume that $y = 1 10^{-7}$
2. $x = \dfrac{ -(y+K_{\ce a}) + ((y+K_{\ce a})^2 + 4 CK_{\ce a})^{1/2}}{2} \nonumber$
3. $yn = \dfrac{1 \times 10^{-14}}{x+y} \nonumber$
4. Replace y in step (2) by yn, and recalculate x.
5. Repeat steps (2) and (3) until the new values and the old values differ insignificantly.
The above procedure is actually a general method that always gives a satisfactory solution. This technique has to be used to calculate the pH of dilute weak acid solutions. Further discussion is given in the Exact Calculation of pH.
Calculate pOH of Basic Solutions
The discussion on weak acids provides a paradigm for the discussion of weak bases. For weak base $\ce{B}$, the ionization is
$\ce{B- + H2O \rightleftharpoons HB + OH-} \nonumber$
and
$K_{\ce b} = \ce{\dfrac{[HB] [OH- ]}{[B- ]}} \nonumber$
The pOH can be calculated for a basic solution if Kb is given. In this case, the discussion is similar and parallel to that given above for the calculation of pH of weak acids when Ka is known.
Exercise $1$
A weak acid is a compound that
1. is completely ionized in solution,
2. is not completely ionized in solution,
3. gives a high pH in a solution,
4. gives a low pH in its solution.
Answer
b
Consider... The pH of a solution depends on both the concentration and the degree of ionization, (or using Ka as an indicator). In contrast, a strong acid is completely ionized in solution.
Exercise $2$
The acidity constant, Ka, for a strong acid is
1. infinity,
2. very large,
3. very small,
4. zero.
Answer
b
Consider... Infinity is a concept, it does not represent a definite value. Derive your answer from the definition of equilibrium constant. A strong acid is "completely" ionized in its solution, but the concentration of the conjugate acid is not zero. Thus, a very large Ka is more realistic than infinity.
Exercise $3$
Household vinegar is usually 5% acetic acid by volume. Calculate the molarity of this solution. Assume density of solution to be 1 g/mL. The formula weight of $\ce{CH3COOH}$ is 60.
Answer
0.8 M
Consider...
Assume 1 L solution. You have 50 mL acetic acid in 1 L vinegar. The density is 1 g/mL, thus, you have 50 g acetic acid. There is 50 mL vinegar in 1.0 L of vinegar, $\mathrm{\dfrac{50\: g}{60\: g\: per\: mol} = 0.8\: mol/L}$
Exercise $4$
Acetic acid is a typical and familiar compound that provides a good example for numerical problems. Its acidity constant Ka is $1.85 \times 10^{-5}$. What is the pH of a concentrated vinegar, which is a 1.0 M acetic acid solution?
Answer
4.3e-3
Consider...
Use the approximation of $\ce{[H+]} = \sqrt{K_{\ce a} \times C}$. $\ce{[H+]} = (1.85\textrm{e-}5)^{1/2} = 4.3\textrm{e-}3$. The approximation is justified because 0.0043/1.0 = 0.4%. Note that most textbooks give Ka = 1.75e-5, but we assume a slightly different value in this and the following problems.
Exercise $5$
If you dilute the vinegar 100 times in a soup that you are cooking, the concentration of your soup is 0.010 M in acetic acid. Other ingredients are ignored. What is the pH of this solution?
Answer
3.4
Consider... Use the approximation of $\ce{[H+]} = \sqrt{K_{\ce a} \times C}$. The concentration of $\ce{H+}$ goes from 3.3E-3 in a 1 M solution down to 4.3E-4 M in a 0.01 M solution. The concentration of $\ce{H+}$ decreases 10 times when the concentration of the acid decreases 100 times.
Exercise $6$
What is the pH of a 0.010 M $\ce{HCl}$ solution?
Answer
2
Consider... $\ce{[H+]} = \mathrm{0.010\: M}$. The concentration of $\ce{H+}$ is from $\ce{HCl}$, which is a strong acid. The pH of a 0.01 M $\ce{HCl}$ solution is lower than that of a 1 M acetic acid solution; compare with the previous problem.
Exercise $7$
What is the pH of a $1.0 \times 10^{-4}$ acetic acid solution ($K_a = 1.85\times 10^{-5}$)?
Answer
4.4
Hint...
Note that $\ce{[H+]} = \textrm{4.3E-5}$ is 4% of $\ce{[HAc]}$ (= 1.0E-4). Use the formula $x = \dfrac{ -K_{\ce a} + (K_{\ce a}^2 + 4 C K_{\ce a})^{1/2}}{2}$
and see what you get. You should use the quadratic formula to calculate $\ce{[H+]}$. The value using the quadratic formula is 4.5 rather than 4.4 from $\sqrt{C\times K_{\ce a}}$.
Example $8$
What is the pH of a $1.0\times 10^{-3}$ M chloroacetic acid solution ($K_a = 1.4\times 10^{-3}$)? This is an interesting numerical problem. Make a good effort to solve it.
Solution
3.2
Hint... Even a strong acid with concentration of 1.0E-3 M gives a pH of 3. When C and Ka are comparable, you have to use the quadratic formula.
Exercise $9$
What is the pH of a $1.0\times 10^{-6}$ M chloroacetic acid solution ($K_a = 1.4\times 10^{-3}$)?
Answer
6.0
Consider... At this concentration, the acid is almost completely ionized.
Exercise $10$
What is the pH of a $1.0\times 10^{-7}$ M chloroacetic acid solution ($K_a = 1.4\times 10^{-3}$)?
Answer
6.7
Consider... From the previous question, you know that the chloroacetic acid should have been completely ionized. Thus, $\ce{[H+]}$ is about 2e-7; half of that is contributed by the self-ionization of water. This corresponds to a pH of 6.7.
Exercise $11$
You have done a number of numerical problems involving various concentrations of some weak acids. These problems are inter- related. If you do not yet have the complete picture, you should review all these questions. Better yet, review the module. What is the pH of a $1.0 \times 10^{-9}$ M solution of chloroacetic acid, $K_a = 1.4 \times 10^{-3}$?
Answer
7
Consider... Critical judgment is required. The pH is entirely due to the self-ionization of water at this concentration.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Acids_and_Bases/Ionization_Constants/Weak_Acids_and_Bases_1.txt
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The name "polyprotic" literally means many protons.
• Calculating the pH of the Solution of a Polyprotic Base/Acid
Thus far, we have been discussing problems and answers in equilibria--perhaps the most popular type of problem being how to find the pH of a weak acid solution given a certain concentration of a molecule. However, those problems in particular usually only involve what is called a monoprotic acid. with only one ionizeable hydrogen atom. Polyprotic Acids and Bases allows for two or more ionizeable hydrogen atoms.
• Polyprotic Acids
Polyprotic acids contain more than one mole ionizable hydronium ions per mole of acids.
• Polyprotic Acids And Bases
The name "polyprotic" literally means many protons. Therefore, in this section we will be observing some specific acids and bases which either lose or accept more than one proton. Then, we will be talking about the equations used in finding the degree of dissociation. Finally, with given examples, we will be able to approach problems dealing with polyprotic acids and bases.
• Polyprotic Acids & Bases
According to Brønsted and Lowry an acid is a proton donor and a base is a proton acceptor. A monoprotic acid is an acid that can donate only one proton, while polyprotic acid can donate more than one proton. Similarly, a monoprotic base can only accept one proton, while a polyprotic base can accept more than one proton.
Thumbnail: % species formation calculated with the program HySS for a 10 millimolar solution of citric acid. pKa1 = 3.13, pKa2 = 4.76, pKa3 = 6.40. (Public Domain; Petergans).
Monoprotic Versus Polyprotic Acids And Bases
Thus far, we have been discussing problems and answers in equilibria--perhaps the most popular type of problem being how to find the pH of a weak acid solution given a certain concentration of a molecule. However, those problems in particular usually only involve what is called a monoprotic acid. “Mono” in the word “monoprotic” indicates that there is only one ionizeable hydrogen atom in an acid when immersed in water, whereas the concept of allows for two or more ionizeable hydrogen atoms.
Introduction
Consider the following chemical equation as the molecule acetic acid equilibrates in the solution:
$\ce{ CH_3COOH + H_2O \rightleftharpoons H_3O^{+} + CH_3COO^{-}} \nonumber$
Although acetic acid carries a four hydrogen atoms, only a single becomes ionized. Not to get into too much detail between monoprotic and polyprotic acids, but if you desire to find the pH given a concentration of a weak acid (in this case, acetic acid), you would create and complete anadjusting for how much acetic acid disassociates. However, if you wish to find the pH of a solution after a polyprotic acid disassociates, there are extra steps that would need to be done. Let's first take a look at a unique example:
Example $1$: Finding the pH of sulfuric Acid
What is the pH of 0.75 M sulfuric acid?
Solution
In sulfuric acid (H2SO4), there are two ionizable hydrogen atoms. What makes this molecule interesting is that its ionization constant for the first hydrogen ($K_{a1}$) ionized is significantly larger than is the second ionization constant ($K_{a2}$). The $K_{a1}$ constant for sulfuric acid is conveniently dubbed “very large” while the $K_{a2}$ constant is 1.1 x 10-2. As such, the sulfuric acid will completely disassociate into HSO4- and H3O+ ions (as a strong acid).
$\ce{H2SO4 (aq) + H2O -> HSO4^{-} (aq) + H_3O^{+}} \nonumber$
Since the sulfuric acid completely disassociates in the solution, we can skip the ICE table process for sulfuric acid, and assert that the concentration $\ce{HSO4^{-}}$ and $\ce{H3O^{+}}$ are the same as that of H2SO4, that is 0.75 M. (This neglects the background concentration of $\ce{H_3O^{+}}$ in water of $1 \times 10^{-7}M$).
Equation:
$\ce{HSO4^{-} (aq) + H2O <=> SO4^{2-} (aq) + H_3O^{+}} \nonumber$
ICE Table:
$HSO_4^-$ $SO_4^{2-}$ $H_3O^+$
Initial 0.75 M 0 M 0.75 M
Change -x M +x M +x M
Equilibrium (0.75 - x) M +x M (0.75 + x) M
$K_{a2} = \dfrac{[SO_4^{2-}][H_3O^+]}{[HSO_4^-]} = 1.1 \times 10^{-2} = 0.011 = \dfrac{x(0.75+x)}{0.75 - x}$
Assume $x$ in the denominator is negligible. Therefore,
$x = 0.011 M = [SO_4^{2-}] \nonumber$
Since we know the value of x, we can use the equation from the ICE table to find the value of [HSO4-].
$[HSO_4^-] = 0.75 \; M - x = 0.75 - 0.011 = 0.74 \; M \nonumber$
We can also find [H3O+] using the equation from the ICE table.
$[H_3O+] = 0.75 \; M + x = 0.75 + 0.011 = 0.76 \; M \nonumber$
We can then find the pH from the calculated [H3O+] value.
$pH = -log[H_3O^+] = -log0.76 = 0.119 \nonumber$
Let's say our task is to find the pH given a polyprotic base which gains protons in water. Thankfully, the process is essentially the same as finding the pH of a polyprotic acid except in this case we deal with the concentration of OH- instead of H3O+.
Example $2$: Finding the pH of a polyprotic base
Let's take a look at how to find the pH of C20H24O2N2, a diprotic base with a concentration of 0.00162 M, and a $K_{b1}$ of 10-6 and a $K_{b2}$ of 10-9.8.
Equation:
$C_{20}H_{24}O_2N \; (aq) + H_2O \rightleftharpoons C_{20}H_{24}O_2N_2H^+ + OH^-$
$C_{20}H_{24}O_2N_2$ $C_{20}H_{24}O_2N_2H^+$ $OH^-$
Initial 0.00162 M 0 M 0 M
Change -x M +x M +x M
Equilibrium 0.00162 M x M x M
$K_{b1} = \dfrac{[C_{20}H_{24}O_2N_2H^+][OH^-]}{[C_{20}H_{24}O_2N_2]} = 10^{-6} = 0.011 = \dfrac{(x)(x)}{0.00162 - x} \nonumber$
Again, assume x in the denominator is negligible. Therefore,
$0.011 \approx \dfrac{x^2}{0.00162}$
Then,
$x \approx 4 \times 10^{-5}$
We can then find the pH.
$pOH = -log(4 \times 10^{-5}) = 4.4$
$pH = 14 - 4.4 = 9.6$
As we determine the pH of the solution, we realize that the OH-gained using the second ionization constant is so insignificant that it does not impact the final pH value. For good measure, the following is the process to determine the pH in case the second use of the ICE table would indeed make a difference.
Equation:
$C_{20}H_{24}O_2N_2H^+ + H_2O \rightleftharpoons C_{20}H_{24}O_2N_2H_2^{2+} + OH^-$
ICE Table:
$C_{20}H_{24}O_2N_2H^+$ $C_{20}H_{24}O_2N_2H_2^{2+}$ $OH^-$
Initial 4 x 10-5 M 0 M 4 x 10-5 M
Change -x M +x M +x M
Equilibrium (4 x 10-5 - x) M x M (4 x 10-5 + x) M
$K_{b2} = \dfrac{[C_{20}H_{24}O_2N_2H_2^{2^+}][OH^-]}{[C_{20}H_{24}O_2N_2H^+]} = 10^{-9.8}$
$10^{-9.8} = \dfrac{(4 \times 10^{-5} + x)(x)}{(4 \times 10^{-5} - x)} \nonumber$
$10^{-9.8} = \dfrac{0.00004 \; x + x^2}{0.00004 - x} \nonumber$
$10^{-9.8}(0.00004 - x) = 0.00004 x + x^2 \nonumber$
$x^2 + (4 \times 10^{-5})x - 6.3 \times 10^{-15} = 0 \nonumber$
\begin{align*} x &= \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \[4pt] &= \dfrac{-4 \times 10^{-5} \pm \sqrt{(4 \times 10^{-5})^2 - 4(1)(6.3 \times 10^{-15})}}{2(1)} \[4pt] & = 0 \end{align*}
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Acids_and_Bases/Monoprotic_Versus_Polyprotic_Acids_And_Bases/Calculating_the_pH_of_the_Solution_of_a_Polyprotic_Bas.txt
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Learning Objectives
• Define a polyprotic acid.
• Explain how a polyprotic behaves in its solution.
• Write the equilibrium equations of ionization of polyprotic acids.
• Calculate acidity constants, K1, K2, K3, and the overall K.
• Calculate the concentrations of various species for a given set of data
Polyprotic acids contain more than one mole ionizable hydronium ions per mole of acids. They ionize to give more than one $\ce{H+}$ ions per molecule. Possible forms of three polyprotic acids are given below after their dissociation into $\ce{H+}$ ions.
\begin{align} \ce{&H2S, &&HS- , &&S^2- &&}\ \ce{&H2SO4, &&HSO4- , &&SO4^2- &&}\ \ce{&H3PO4, &&H2PO4- , &&HPO4^2- , &&PO4^3-} \end{align} \nonumber
These acids ionize in several stages, giving out one proton at each stage. The acidity constants for these acids may be written as K1, K2, K3, ... Some polyprotic acids are given in Table $1$ on the right here. Knowing their names and being familiar with their properties (ionization for example) is an asset for you.
Figure $1$: Some Polyprotic Acids
Formula Name
$\ce{H2S}$ Hydrogen sulfide
$\ce{H2SO4}$ Sulfuric acid
$\ce{H2SO3}$ Sulfurous acid
$\ce{H3PO4}$ Phosphoric acid
$\ce{H2C2O4}$ Oxalic acid
$\ce{H2CO3}$ Carbonic acid
$\ce{H2C3H2O4}$ Malonic acid
Consider $\ce{H2S}$,
$\ce{H2S \rightleftharpoons H+ + HS-} \nonumber$
$K_1 = \ce{\dfrac{[H+] [HS- ]}{[H2S]}} \nonumber$
and
$\ce{HS- \rightleftharpoons H+ + S^2-} \nonumber$
$K_2 = \ce{\dfrac{[H+] [S^2- ]}{[HS- ]}} \nonumber$
Obviously, for the overall ionization reaction,
$\ce{H2S \rightleftharpoons 2 H+ + S^2-} \nonumber$
\begin{align} K_{\ce{overall}} &= \ce{\dfrac{[H+]^2 [S^2- ]}{[H2S]}}\ &= K_1 K_2 \end{align} \nonumber
Confirm the above obvious result on a sheet of paper to satisfy yourself.
For polyprotic acids, the following is always true:
$K_1 > K_2 > K_3 > \: ... \nonumber$
For most acids, K1/K2 = 1E5 or 100000, and K2/K3 = 1E5, but oxalic acid is different. For oxalic acid, K1 = 5.6E-2, and K2 = 5.4E-4. The two acidic groups are separated by a $\ce{C-C}$ bond in oxalic acid.
Example $1$: Oxalic Acid
Oxalic acid is an organic compound with the formula $C_2H_2O_4$ and has to ionizable protons (white atoms on structure below).
Calculate the overall equilibrium constant for oxalic acid.
$\ce{H2C2O4 \rightleftharpoons 2 H+ + C2O4^2-}$
$K_1 = \textrm{5.6E-2}$
$K_2 = \textrm{5.4E-5}$
Solution
The calculation is straightforward:
\begin{align} K_{\ce{overall}} &= K_1 K_2\ &= 3.0\textrm{E-}6 \end{align}
Example $2$: Hydrogen Sulfide
A solution is acidified with $\ce{HCl}$ so that its pH is 1.0, and is saturated with $\ce{H2S}$ at 298 K. What is the sulfide $\ce{S^2-}$ ion concentration in this solution? At 298 K, a saturated $\ce{H2S}$ solution has $\mathrm{[H_2S] = 0.10\: M}$ and
$K_{\ce{overall}} = \textrm{1E-20}$
for $\ce{H2S}$.
Solution
$K_{\ce{overall}} = \ce{\dfrac{[H+]^2 [S^2- ]}{[H2S]}}$
Thus,
\begin{align} \ce{[S^2- ]} &= 1\textrm{E-}20 \, \dfrac{0.1}{(0.1)^2}\ &= \textrm{1E-19 F} \end{align}
Example $3$: Sulfuric Acid
Sulfuric acid is a strong acid, and the $\ce pK_{\large\textrm a_{\Large 2}}$ of $\ce{HSO4-}$ is 1.92. What is the pH of a 0.100 M $\ce{NaHSO4}$ solution?
Solution
The salt is completely ionized in its solution.
$\ce{NaHSO4 \rightarrow Na+ + HSO4-}$
The anion further ionizes. If $\ce{[H+]} = x$, then the equilibrium concentrations of various species are:
$\begin{array}{cccccl} \ce{HSO4- &\rightleftharpoons &H+ &+ &SO4^2-} &\hspace{20px} K_{\large\textrm a_{\Large 2}} = 10^{-1.92} = 0.0120\ 0.100 - x &&x, &&x & \end{array}$
$K_{\large\textrm a_{\Large 2}} = \dfrac{x^2}{0.100-x} = 0.0120$
\begin{align} \ce{[H+]}&= x\ &= \dfrac{-0.120 + (0.012^2 + 4\times0.00120)^{1/2}}{2}\ &= \textrm{0.0292 M} \end{align}
Thus,
$\mathrm{pH = - \log 0.0292 = 1.54}$
Exercise $\PageIndex{3A}$
Is the $\ce{NaHSO4}$ salt solution acidic?
Although no concentration is stated, such a solution is acidic because of the acidity of $\ce{HSO4-}$.
Exercise $\PageIndex{3B}$
Given that $\ce{H2SO4}$, $\ce p K_{\large\textrm a_{\Large 2}} = 1.92$
For $\ce{H3SO4}$, $\ce p K_{\large\textrm a_{\Large 1}} = 2.12$; $\ce p K_{\large\textrm a_{\Large 2}} = 7.21$; $\ce p K_{\large\textrm a_{\Large 3}} = 12.67$
Which of the following solutions are acidic, basic, or neutral?
• $\ce{Na2SO4}$
• $\ce{NaH2PO4}$
• $\ce{Na2HPO4}$
• $\ce{Na3PO4}$
• $\ce{NaNO3}$
Work out the answer please; some of these will appear on the examinations. If the concentration of a salt solution is given, you may be required to evaluate the pH or pOH of the solution.
Example $1$: $\ce{NaHSO4}$
What is the pH of a solution containing 0.500 M $\ce{NaHSO4}$ and 0.300 M $\ce{Na2SO4}$?
Solution
The 0.500 M solution of $\ce{NaHSO4}$ supplies 0.500 M $\ce{HSO4-}$ as an acid, and similarly, the solution also contains 0.300 M $\ce{SO4^2-}$.
Using the equation:
\begin{align} \ce{pH} &= \mathrm{p\mathit K_{\large a} - \log \dfrac{[salt]}{[acid]}}\ &= 1.92 + \log \left(\dfrac{0.300}{0.500}\right)\ &= 1.70 \end{align}
Exercise $1$
The two acid ionization constants for sulfurous acid are 1.2E-2 and 6.6E-8 respectively. Calculate the overall equilibrium constant for
$\ce{H2SO3 \rightleftharpoons 2 H+ + SO3^2-}$
Answer
7.9E-10
Consider...
$K_{\ce{overall}} = K_1 \times K_2$
Exercise $2$
What is the pH of a 1.0 M $\ce{H2SO3}$ solution?
Answer
0.98
Consider... Only K1 matters in this calculation. Using the quadratic formula yields a pH of 0.98. When approximation is used, you'll get a pH of 0.96.
Exercise $3$
If the pH of a 1.0 M $\ce{H2SO3}$ solution is 1.0, what is the sulfite ion concentration? (K1 = 1.2E-2, and K2 = 6.6E-8)
Answer
7.1E-8
Consider...
\begin{align} K_{\ce{overall}} = 7.9\textrm{E-}10 &= \ce{\dfrac{[H+]^2 [SO3^2- ]}{[H2SO3]}}\ &= \dfrac{0.12 \ce{[SO3^2- ]}}{0.9} \end{align}
$\ce{[SO3^2- ]} =\: ??$
This is not an easy question to answer!
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Acids_and_Bases/Monoprotic_Versus_Polyprotic_Acids_And_Bases/Polyprotic_Acids.txt
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The name "polyprotic" literally means many protons. Therefore, in this section we will be observing some specific acids and bases which either lose or accept more than one proton. Then, we will be talking about the equations used in finding the degree of dissociation. Finally, with given examples, we will be able to approach problems dealing with polyprotic acids and bases.
Introduction
Polyprotic acids are specific acids that are capable of losing more than a single proton per molecule in acid-base reactions. (In other words, acids that have more than one ionizable H+ atom per molecule). Protons are lost through several stages (one at each stage), with the first proton being the fastest and most easily lost. Contrast with monoprotic acids in section Monoprotic Versus Polyprotic Acids And Bases.
Common Polyprotic Acids Formula Strong/Weak Acid Number of Ionizable Hydrogens Ka1 Ka2 Ka3
Sulfuric acid H2SO4 Strong 2 (diprotic) Very Large 1.1E-2
Sulfurous acid H2SO3 Weak 2 (diprotic) 1.3E-2 6.2E-8
Phosphoric acid H3PO4 Weak 3 (triprotic) 7.1E-3 6.3E-8 4.2E-13
Carbonic acid H2CO3 Weak 2 (diprotic) 4.4E-7 4.7E-11
Hydrosulfuric acid or Hydrogen sulfide H2S Weak 2 (diprotic) 1.0E-7 1E-19
Oxalic acid H2C2O4 Weak 2 (diprotic) 5.4E-2 5.3E-5
Malonic acid H2C3H2O4 Medium Strong 2 (diprotic) 1.5E-3 2.0E-6
From the table above, we see that sulfuric acid is the strongest.
Ionization Constant
It is important to know that K1>K2>K3, where K stands for the acidity constant or acid ionization constant (first, second, and third, respectively). These constants are used to measure the degree of dissociation of hydrogens in the acid. For a more in depth discussion on this, go to Ionization Constants.
Example 1: Hydrosulfuric acid
To find Ka1 of Hydrosulfuric acid (H2S), you must first write the reaction:
$H_2S \rightleftharpoons H^+ + HS^- \nonumber$
Dividing the products by the reactants, we then have:
$K_{a1} = \dfrac{[H^+] [HS^-]}{ [HS-]} \nonumber$
To find Ka2, we start with the reaction:
$HS^- \rightleftharpoons H^+ + S_2^- \nonumber$
Then, like when finding $K_{a1}$, write the products over the reactants:
$K_{a2} = \dfrac{[H^+] [S_2^-]}{[HS^-]} \nonumber$
From these reactions we can observe that it takes two steps to fully remove the H+ ion. This also means that this reaction will produce two equivalence points or stoichiometric points. The equivalence point, by definition, is the point during an acid-base titration in which there has been equal amounts of acid and base reacted. If we were to graph this, we would be able to see exactly just what two equivalence points looks like. Let's check it out:
Note the multiple equivalence points and notice that they are almost straight lines at that point, indicating equal added quantities of acid and base.
Titrations
In strong acid + strong base titrations, the pH changes slowly at first, rapidly through the equivalence point of pH=7, and then slows down again. If it is being titrated in a strong acid, the pH will go up as the base is added to it. Conversely, if it is in a strong base, the pH will fall down as acid is added.
• In strong acid + weak base titrations, the pH changes slowly at the equivalence point and the pH equals the pKa of the acid. The pH is below 7.
• For the weak acid + strong base, the pH is above 7 at the equivalence point.
• If there is strong acid or strong base left over after the equivalence point, this can be used to find the pH of the solution.
Next, let's take a look at sulfuric acid. This unique polyprotic acid is the only one to be completely deprotonated after the first step:
$H_2SO_{4(aq)} + H_2O_{(l)} \rightleftharpoons H_3O^+_{(aq)} + HSO^-_{4(aq)} \nonumber$
Now let's try something a little harder. The ionization of phosphoric acid (three dissociation reactions this time) can be written like this:
Start with H3PO4:
$K_{a1}: H_3PO_{4(aq)} \rightleftharpoons H^+_{(aq)} + H_2PO^-_{4(aq)} \nonumber$
$K_{a2} : H_2PO^-_{4(aq)} \rightleftharpoons HPO_{4(aq)} + H^+_{(aq)} \nonumber$
$K_{a3} : HPO^-_{4(aq)} \rightleftharpoons H^+_{(aq)} + PO^{3-}_{4(aq)} \nonumber$
So from these above reactions we can see that it takes three steps to fully remove the H+ ion. This also means that this reaction will produce three equivalence points. Polyprotic Bases are bases that can accept at least one H+ ion, or proton, in acid-base reactions.
Common Polyprotic Bases Formula Strong/Weak Base Diprotic/Triprotic Base
Phosphate ion PO43- Weak Triprotic
Sulfate ion SO42- Very Weak Diprotic
Carbonate ion CO32- Strong Diprotic
Example 2: Some examples for calculating the constant, Kb
First, start with the reaction A3- + H2O ? HA2- + OH-
Kb1= [OH-][HA2-]/[A3-]=KW/Ka3
Then, we plug in the products over the reactants:
HA2- + H2O ? H2A- + OH-
Kb2 = [OH-][H2A2-]/[HA2-]=KW/Ka2
Finally, we are left with the third dissociation, or Kb3:
H2A- + H2O ? H3A + OH-
Kb3 = [OH-][H3A]/[H2A-]=KW/Ka1
Contributors
• Natalie Kania
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Acids_and_Bases/Monoprotic_Versus_Polyprotic_Acids_And_Bases/Polyprotic_Acids_And_Bases.txt
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According to Brønsted and Lowry an acid is a proton donor and a base is a proton acceptor. This idea of proton donor and proton acceptor is important in understanding monoprotic and polyprotic acids and bases because monoprotic corresponds to the transfer of one proton and polyprotic refers to the transfer of more than one proton. Therefore, a monoprotic acid is an acid that can donate only one proton, while polyprotic acid can donate more than one proton. Similarly, a monoprotic base can only accept one proton, while a polyprotic base can accept more than one proton.
Introduction
One way to display the differences between monoprotic and polyprotic acids and bases is through titration, which clearly depicts the equivalence points and acid or base dissociation constants. The acid dissociation constant, signified by $K_a$, and the base dissociation constant, $K_b$, are equilibrium constants for the dissociation of weak acids and weak bases. The larger the value of either $K_a$ or $K_b$ signifies a stronger acid or base, respectively.
Here is a list of important equations and constants when dealing with $K_a$ and $K_b$:
For the general equation of a weak acid,
$HA_{(aq)} + H_2O_{(l)} \rightleftharpoons H_3O^+_{(aq)} + A^-_{(aq)} \label{1}$
you need to solve for the $K_a$ value. To do that you use
$K_a = \dfrac{[H_3O^+][A^-]}{[HA]} \label{2}$
Another necessary value is the $pK_a$ value, and that is obtained through $pK_a = {-logK_a}$
The procedure is very similar for weak bases. The general equation of a weak base is
$BOH \rightleftharpoons B^+ + OH^- \label{3}$
Solving for the $K_b$value is the same as the $K_a$ value. You use the formula
$K_b = \dfrac{[B^+][OH^-]}{[BOH]} \label{4}$
The $pK_b$ value is found through $pK_b = {-logK_b}$
The $K_w$ value is found with$K_w = {[H3O^+]}{[OH^-]}$.
$K_w = 1.0 \times 10^{-14} \label{5}$
Monoprotic Acids
Monoprotic acids are acids that can release only one proton per molecule and have one equivalence point.
Here is a table of some common monoprotic acids:
Table 1: Common Monoprotic Acids
Name Formula $\pmb{K_a}$
Hydrochloric acid (strong) HCl 1.3 x 106
Nitric acid (strong) HNO3 2.4 x 101
Acetic acid (weak) CH3COOH 1.74 x 10-5
Monoprotic Bases
Monoprotic Bases are bases that can only react with one proton per molecule and similar to monoprotic acids, only have one equivalence point. Here is a list of some common monoprotic bases:
Table 2: Common Monoprotic Bases
Name Formula $\pmb{K_b}$
Sodium hydroxide (strong) NaOH 6.3 X 10-1
Potassium hydroxide (strong) KOH 3.16 X 10-1
Ammonia (weak) NH3 1.80 x 10-5
Example $1$
What is the pH of the solution that results from the addition of 200 mL of 0.1 M CsOH(aq) to 50 mL of 0.2M HNO2(aq)? (pKa= 3.14 for HNO2)
Solution
$\dfrac{0.1 mol}{L}*200 mL* \dfrac{1 L}{1000 mL} = {0.02 mol CsOH} \nonumber$
$\dfrac{0.2 mol}{L}*50 mL* \dfrac{1 L}{1000 mL} = {0.01 mol HNO_2} \nonumber$
Then do an ICE Table for
$CsOH + HNO_2 \rightleftharpoons H_2O + CsNO_2 \nonumber$
yielding $[CsOH]= [OH^-]= 0.01M$
Then to find pH first we find pOH $pOH = {-log[OH^-] = -log[\dfrac{0.01}{0.25}] = 1.4}$
Then $pH = {14 - pOH}$, plugging in
pH = 14 - 1.4 = 12.6
Polyprotic Acids and Bases
So far, we have only considered monoprotic acids and bases, however there are various other substances that can donate or accept more than proton per molecule and these are known as polyprotic acids and bases. Polyprotic acids and bases have multiple dissociation constants, such as $K_{a1}$, $K_{a2}$, $K_{a3}$ or $K_{b1}$, $K_{b2}$, and $K_{b3}$, and equivalence points depending on the number of times dissociation occurs.
Polyprotic Acids
Polyprotic acids are acids that can lose several protons per molecule. They can be further categorized into diprotic acids and triprotic acids, those which can donate two and three protons, respectively. The best way to demonstrate polyprotic acids and bases is with a titration curve. A titration curve displays the multiple acid dissociation constants ($K_a$) as portrayed below.
Here is a list of some common polyprotic acids:
Table 3: Common Polyprotic Acids
Name Formula $\pmb{K_{a1}}$ $\pmb{K_{a2}}$ $\pmb{K_{a3}}$
Sulfuric acid (strong, diprotic) H2SO4 1.0 x 103 1.2 x 10-2 -
Carbonic acid (weak, diprotic) H2CO3 4.2 x 10-7 4.8 x 10-11 -
Phosphoric acid (weak, triprotic) H3PO4 7.1 x 10-3 6.3 x 10-8 4.2 x 10-13
Polyprotic Bases
Polyprotic bases are bases that can attach several protons per molecule. Similar to polyprotic acids, polyprotic bases can be categorized into diprotic bases and triprotic bases. Here is a list of some common polyprotic bases:
Table 4: Common Polyprotic Bases
Name Formula $\pmb{K_b}$
Barium hydroxide (strong, diprotic) Ba(OH)2
Phosphate ion (triprotic) PO43-
Sulfate ion (diprotic) SO42-
Example $2$
For a 4.0 M H3PO4 solution, calculate (a) [H3O+] (b) [HPO42--] and (c) [PO43-].
$H_3PO_4 + H_2O \rightleftharpoons H_3O^+ + H_2PO_4^- \nonumber$
Solution
(a) Using ICE Tables you get:
$K_{a1} = \dfrac{[H_3O^+][H_2PO_4^-]}{[H_3PO_4]} \nonumber$
So,
$x^2$ = .0284
$x$ = 0.17 M
(b) From part (a), $x$ = [H2PO4-] = [H3O+] = 0.17 M
(c) To determine [H3O+] and [H2PO4-], it was assumed that the second ionization constant was insignificant.
The new equation is as follows:
$H_2PO_4^- + H_2O \rightleftharpoons H_3O^+ + HPO_4^{2-}$
Using ICE Tables again:
$K_{a2} = [HPO_4^{2-}] = 6.3 \times 10^{-8}$
Example $3$
The polyprotic acid H2SO4 can ionize two times ( $K_{a1}>>1$, $K_{a2} = 1.1 * 10^-2$). If we start with 9.50*10-3 M solution of H2SO4, what are the final concentrations of H2SO4, HSO4-, SO42-, and H3O+.
Solution
The equation for the first ionization is $H_2SO_4 + H_2O \rightleftharpoons H_3O^+ + HSO_4^-$. This equation goes to completion because H2SO4 is a strong acid and $K_{a1}>>1$.
So since the reaction goes to completion, doing an ICE Table you get [H30+] = 9.50*10-3 M and [HSO4-] = 9.50*10-3 M (after the first ionization).
The equation of the second ionization is $HSO_4- + H_2O \rightleftharpoons H_3O^+ + SO_4^2-$. Using the equation $K_{a2} = \dfrac{[H_3O^+][SO_4^2-^-]}{[HSO_4^-]}$, $K_{a2} = 1.1 * 10^-2$, and an ICE Table to get $x^2 + .0.0205x - 0.0001045 = 0$.
Then you use the quadratic equation to solve for X, to get $x$ = 0.004226.
Now we need to solve for the necessary concentrations
$[H_2S0_4]$ = 0 (because the first ionization reaction went to completion)
$[HS0_4^-]$ = $k_{a1}$ - $k_{a2}$ = 9.50*10-3 M - 0.004226 M = 5.27*10-3 M
$[SO_4^2-]$ = $k_{a2}$ = .004226 M
$[H_3O^+]$ = $k_{a1}$ + $k_{a2}$ = 9.50*10-3 M + 0.004226 M = 1.37*10-2 M
Summary
• Ka and Kb are equilibrium constants and a high value signifies a stronger acid or base.
• Acid are proton donors and bases are proton acceptors.
• Monoprotic acid/base corresponds to the donation/acceptance of only one proton.
• Polyprotic acid/base corresponds to the donation/acceptance of more than one proton.
• Remember diprotic and triprotic.
• $K_{a1}$>$K_{a2}$>$K_{a3}$
Common Errors
Assuming that the [H30+] is the same for all the ionizations. In fact, the pH is dominated by only the first ionization, but the later ionizations do contribute very slightly.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Acids_and_Bases/Monoprotic_Versus_Polyprotic_Acids_And_Bases/Polyprotic_Acids_and_Bases_1.txt
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Atomic theory states that matter is composed of discrete units called atoms, as opposed to the obsolete notion that matter could be divided into any arbitrarily small quantity. It began as a philosophical concept in ancient Greece (Democritus) and India and entered the scientific mainstream in the early 19th century. It was during this time when discoveries in the field of chemistry showed that matter did indeed behave as if it were made up of particles.
• Atomic Mass
Mass is a basic physical property of matter. The mass of an atom or a molecule is referred to as the atomic mass. The atomic mass is used to find the average mass of elements and molecules and to solve stoichiometry problems.
• Atomic Structure
An atom consists of a positively charged nucleus, surrounded by one or more negatively charged particles called electrons. The positive charges equal the negative charges, so the atom has no overall charge; it is electrically neutral. Most of an atom’s mass is in its nucleus; the mass of an electron is only 1/1836 the mass of the lightest nucleus, that of hydrogen. Although the nucleus is heavy, it is quite small compared with the overall size of an atom.
• Atomic Theory
• Dalton's Atomic Theory
John Dalton, a British school teacher, published his theory about atoms in 1808. His findings were based on experiments and the laws of chemical combination.
• Ionization Energies of Diatomic Molecule
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.
• Isotopes
Atoms that have the same atomic number (number of protons), but different mass numbers (number of protons and neutrons) are called isotopes. There are naturally occurring isotopes and isotopes that are artificially produced. Isotopes are separated through mass spectrometry; MS traces show the relative abundance of isotopes vs. mass number (mass : charge ratio).
• Nuclide, Atomic Number, mass number
An atom is the smallest unit of an element that can exist. Every atom is made up of protons, neutrons, and electrons. These particles define a nuclide and its chemical properties and were discovered in the early 20th century and are described by modern atomic theory.
• Simple View of Atomic Structure
• Sizes of Ions
• The Atom
The atom is the smallest unit of matter that is composed of three sub-atomic particles: the proton, the neutron, and the electron. Protons and neutrons make up the nucleus of the atom, a dense and positively charged core, whereas the negatively charged electrons can be found around the nucleus in an electron cloud.
• The Mole and Avogadro's Constant
The mole, abbreviated mol, is an SI unit which measures the number of particles in a specific substance. One mole is equal to $6.02214179 \times 10^{23}$ atoms, or other elementary units such as molecules.
Atomic Theory
Mass is a basic physical property of matter. The mass of an atom or a molecule is referred to as the atomic mass. The atomic mass is used to find the average mass of elements and molecules and to solve stoichiometry problems.
Introduction
In chemistry, there are many different concepts of mass. It is often assumed that atomic mass is the mass of an atom indicated in unified atomic mass units (u). However, the book Quantities, Units and Symbols in Physical Chemistry published by the IUPAC clearly states:
"Neither the name of the physical quantity, nor the symbol used to denote it, should imply a particular choice of unit."
The name "atomic mass" is used for historical reasons, and originates from the fact that chemistry was the first science to investigate the same physical objects on macroscopic and microscopic levels. In addition, the situation is rendered more complicated by the isotopic distribution. On the macroscopic level, most mass measurements of pure substances refer to a mixture of isotopes. This means that from a physical stand point, these mixtures are not pure. For example, the macroscopic mass of oxygen (O2) does not correspond to the microscopic mass of O2. The former usually implies a certain isotopic distribution, whereas the latter usually refers to the most common isotope (16O2). Note that the former is now often referred to as the "molecular weight" or "atomic weight".
Mass Concepts in Chemistry
name in chemistry physical meaning symbol units
atomic mass mass on microscopic scale m, ma Da, u, kg, g
molecular mass mass of a molecule m Da, u, kg, g
isotopic mass mass of a specific isotope Da, u, kg, g
mass of entity mass of a chemical formula m, mf Da, u, kg, g
average mass average mass of a isotopic distribution m Da, u, kg, g
molar mass average mass per mol M = m/n kg/mol or g/mol
atomic weight average mass of an element Ar = m / mu unitless
molecular weight average mass of a molecule Mr = m / mu unitless
relative atomic mass ratio of mass m and and the atomic mass constant mu Ar = m / mu unitless
atomic mass constant mu = m(12C)/12 mu = 1 Da = 1 u Da, u, kg, g
relative molecular mass ratio of mass m of a molecule and and the atomic mass constant mu Mr = m / mu unitless
relative molar mass ? ? ?
mass number nucleon number A nucleons, or unitless
integer mass nucleon number * Da m Da, u
nominal mass integer mass of molecule consisting of most abundant isotopes m Da, u
exact mass mass of molecule calculated from the mass of its isotopes (in contrast of measured ba a mass spectrometer) Da, u, kg, g
accurate mass mass (not normal mass) Da, u, kg, g
These concepts are further explained below.
Average Mass
Isotopes are atoms with the same atomic number, but different mass numbers. A different mass size is due to the difference in the number of neutrons that an atom contains. Although mass numbers are whole numbers, the actual masses of individual atoms are never whole numbers (except for carbon-12, by definition). This explains how lithium can have an atomic mass of 6.941 Da. The atomic masses on the periodic table take these isotopes into account, weighing them based on their abundance in nature; more weight is given to the isotopes that occur most frequently in nature. Average mass of the element E is defined as:
$m(E) = \sum_{n=1} m(I_n) \times p(I_n)$
where ∑ represents a n-times summation over all isotopes $I_n$ of element E, and p(I) represents the relative abundance of the isotope I.
Example 1
Find the average atomic mass of boron using the Table 1 below:
Mass and abundance of Boron isotopes
n isotope In mass m (Da) isotopic abundance p
1 10B 10.013 0.199
2 11B 11.009 0.801
Solution
The average mass of Boron is:
$m(B) = (10.013\ Da)(0.199) + (11.009\ Da)(0.801) = 1.99\ Da + 8.82\ Da = 10.81\ Da$
Relative Mass
Traditionally it was common practice in chemistry to avoid using any units when indicating atomic masses (e.g. masses on microscopic scale). Even today, it is common to hear a chemist say, "12C has exactly mass 12". However, because mass is not a dimensionless quantity, it is clear that a mass indication needs a unit. Chemists have tried to rationalize the omission of a unit; the result is the concept of relative mass, which strictly speaking is not even a mass but a ratio of two masses. Rather than using a unit, these chemists claim to indicate the ratio of the mass they want to indicate and the atomic mass constant mu which is defined analogous to the unit they want to avoid. Hence the relative atomic mass of the mass m is defined as:
$A_r = \dfrac{m}{m_u}$
The quantity is now dimensionless. As this unit is confusing and against the standards of modern metrology, the use of relative mass is discouraged.
Molecular Weight, Atomic Weight, Weight vs. Mass
Until recently, the concept of mass was not clearly distinguished from the concept of weight. In colloquial language this is still the case. Many people indicate their "weight" when they actually mean their mass. Mass is a fundamental property of objects, whereas weight is a force. Weight is the force F exerted on a mass m by a gravitational field. The exact definition of the weight is controversial. The weight of a person is different on ground than on a plane. Strictly speaking, weight even changes with location on earth.
When discussing atoms and molecules, the mass of a molecule is often referred to as the "molecular weight". There is no univerally-accepted definition of this term; however, mosts chemists agree that it means an average mass, and many consider it dimensionless. This would make "molecular weight" a synonym to "average relative mass".
Integer Mass
Because the proton and the neutron have similar mass, and the electron has a very small mass compared to the former, most molecules have a mass that is close to an integer value when measured in daltons. Therefore it is quite common to only indicate the integer mass of molecules. Integer mass is only meaningful when using dalton (or u) units.
Accurate Mass
Many mass spectrometers can determine the mass of a molecule with accuracy exceeding that of the integer mass. This measurement is therefore called the accurate mass of the molecule. Isotopes (and hence molecules) have atomic masses that are not integer masses due to a mass defect caused by binding energy in the nucleus.
Units
The atomic mass is usually measured in the units unified atomic mass unit (u), or dalton (Da). Both units are derived from the carbon-12 isotope, as 12 u is the exact atomic mass of that isotope. So 1 u is 1/12 of the mass of a carbon-12 isotope:
1 u = 1 Da = m(12C)/12
The first scientists to measure atomic mass were John Dalton (between 1803 and 1805) and Jons Jacoband Berzelius (between 1808 and 1826). Early atomic mass theory was proposed by the English chemist William Prout in a series of published papers in 1815 and 1816. Known was Prout's Law, Prout suggested that the known elements had atomic weights that were whole number multiples of the atomic mass of hydrogen. Berzelius demonstrated that this is not always the case by showing that chlorine (Cl) has a mass of 35.45, which is not a whole number multiple of hydrogen's mass.
Some chemists use the atomic mass unit (amu). The amu was defined differently by physicists and by chemists:
• Physics: 1 amu = m(16O)/16
• Chemistry: 1 amu = m(O)/16
Chemists used oxygen in the naturally occurring isotopic distribution as the reference. Because the isotopic distribution in nature can change, this definition is a moving target. Therefore, both communities agreed to the compromise of using m(12C)/12 as the new unit, naming it the "unified atomic mass unit" (u). Hence, the amu is no longer in use; those who still use it do so with the definition of the u in mind. For this reason, the dalton (Da) is increasingly recommended as the accurate mass unit.
Neither u nor Da are SI units, but both are recognized by the SI.
Molar Mass
The molar mass is the mass of one mole of substance, whether the substance is an element or a compound. A mole of substance is equal to Avogadro's number (6.023×1023) of that substance. The molar mass has units of g/mol or kg/mol. When using the unit g/mol, the numerical value of the molar mass of a molecule is the same as its average mass in daltons:
• Average mass of C: 12.011 Da
• Molar mass of C: 12.011 g/mol
This allows for a smooth transition from the microscopic world, where mass is measured in daltons, to the macroscopic world, where mass is measured in kilograms.
Example 2
What is the molar mass of phenol, C6H5OH?
Average mass m = 6 × 12.011 Da + 6 × 1.008 Da + 1 × 15.999 Da = 94.113 Da
Molar mass = 94.113 g/mol = 0.094113 kg/mo
Measuring Masses in the Atomic Scale
Masses of atoms and molecules are measured by mass spectrometry. Mass spectrometry is a technique that measures the mass-to-charge ratio (m/q) of ions. It requires that all molecules and atoms to be measured be ionized. The ions are then separated in a mass analyzer according to their mass-to-charge ratio. The charge of the measured ion can then be determined, because it is a multiple of the elementary charge. The the ion's mass can be deduced. The average masses indicated in the periodic table are then calculated using the isotopic abundances, as explained above.
The masses of all isotopes have been measured with very high accuracy. Therefore, it is much simpler and more accurate to calculate the mass of a molecule of interest as a sum of its isotopes than measuring it with a commercial mass spectrometer.
Note that the same is not true on the nucleon scale. The mass of an isotope cannot be calculated accurately as the sum of its particles (given in the table below); this would ignore the mass defect caused by the binding energy of the nucleons, which is significant.
Table 2: Mass of three sub-atomic particles
Particle SI (kg) Atomic (Da) Mass Number A
Proton 1.6726×10-27 1.0073 1
Neutron 1.6749×10-27 1.0087 1
Electron 9.1094×10-31 0.00054858 0
As shown in Table 2, the mass of an electron is relatively small; it contributes less than 1/1000 to the overall mass of the atom.
Where to Find Atomic Mass
The atomic mass found on the Periodic Table (below the element's name) is the average atomic mass. For example, for Lithium:
The red arrow indicates the atomic mass of lithium. As shown in Table 2 above and mathematically explained below, the masses of a protons and neutrons are about 1u. This, however, does not explain why lithium has an atomic mass of 6.941 Da where 6 Da is expected. This is true for all elements on the periodic table. The atomic mass for lithium is actually the average atomic mass of its isotopes. This is discussed further in the next section.
One particularly useful way of writing an isotope is as follows:
Applications
Applications Include:
1. Average Molecular Mass
2. Stoichiometry
Note: One particularly important relationship is illustrated by the fact that an atomic mass unit is equal to 1.66 × 10-24 g. This is the reciprocal of Avogadro's constant, and it is no coincidence:
$\dfrac{\rm Atomic~Mass~(g)}{1 {\rm g}} \times \dfrac{1 {\rm mol}}{6.022 \times 10^{23}} = \dfrac{\rm Mass~(g)}{1 {\rm atom}}$
Because a mol can also be expressed as gram × atoms,
$1\ u = \frac{M_u\ (molar\ mass\ unit)}{N_A\ (Avogadro's\ Number)} = 1\ \frac{g}{mol\ N_A}$
1u = Mu(molar mass unit)/NA(Avogadro's Number)=1g/mol/NA
NA known as Avogadro's number (Avogadro's constant) is equal to 6.023×1023 atoms.
Atomic mass is particularly important when dealing with stoichiometry.
Practice Problems
1. What is the molecular mass of radium bicarbonate, Ra(HCO3)2?
2. List the following, from least to greatest, in terms of their number of neutrons, and then atomic mass: 14N, 42Cl, 25Na, 10Be
3. A new element, Zenium, has 3 isotopes, 59Ze, 61Ze, and 67Ze, with abundances of 62%, 27%, and 11% respectively. What is the atomic mass of Zenium?
4. An isotope with a mass number of 55 has 5 more neutrons than protons. What element is it?
5. How much mass does 3.71 moles of Fluorine have?
6. How many grams are there in 4.3 × 1022 molecules of POCl3?
7. How many moles are there in 23 grams of sodium carbonate?
Solutions
a) Molecular mass of Ra(HCO3)2
= 226 + 2(1.01 u + 12.01 u + (16.00 u)(3)) = 348 u or g/mol
b) Number of neutrons: 10Be, 14N, 25Na, 42Cl
Atomic Mass: 10Be, 14N, 25Na, 42Cl
Note: It is the same increasing order for both number of neutrons and atomic mass because more neutrons means more mass.
c) Atomic mass of Zenium:
(59 u)(0.62) + (61 u)(0.27) + (67 u)(0.11)
= 37 u + 16 u + 7.4 u
= 60.4 u or g/mol
d) Mn
e) (3.71 moles F2)(19 × 2 g/mol F2)
= (3.71 mol F2)(38 g/mol F2)
= 140 g F2
f) (4.3 × 1022 molecules POCI3)(1 mol/6.022 × 1023 molecules POCI3)(30.97 + 16.00 + 35.45 x 3 g/mol POCI3)
= (4.3 × 1022 molecules POCI3)(mol/6.022 × 1023 molecules POCI3)(153.32 g/mol POCI3)
= 11 g POCI3
g) (23 g Na2CO3)(1 mol/22.99 × 2 + 12.01 + 16.00 × 3 g Na2CO3)
= (23 g Na2CO3)(1 mol/105.99 g Na2CO3)
= (0.22 mol Na2CO3)
Contributors and Attributions
• Gunitika Dandona (UCD)
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Atomic_Theory/Atomic_Mass.txt
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An atom consists of a positively charged nucleus, surrounded by one or more negatively charged particles called electrons. The positive charges equal the negative charges, so the atom has no overall charge; it is electrically neutral. Most of an atom’s mass is in its nucleus; the mass of an electron is only 1/1836 the mass of the lightest nucleus, that of hydrogen. Although the nucleus is heavy, it is quite small compared with the overall size of an atom.
The radius of a typical atom is around 1 to 2.5 angstroms (Å), whereas the radius of a nucleus is about 10-5 Å. If an atom were enlarged to the size of the earth, its nucleus would be only 200 feet in diameter and could easily rest inside a small football stadium. The nucleus of an atom contains protons and neutrons. Protons and neutrons have nearly equal masses, but they differ in charge. A neutron has no charge, whereas a proton has a positive charge that exactly balances the negative charge on an electron. Table \(1\) lists the charges of these three fundamental particles, and gives their masses expressed in atomic mass units.
Table \(1\): Charge and mass of three sub atomic particles
Particle Charge Mass (amu)
Electrons -1 0.000549
Protons +1 1.00782
Neutrons 0 1.00867
The atomic mass unit (amu) is defined as exactly one-twelfth the mass of a carbon atom that has six protons and six neutrons in its nucleus. With this scale, protons and neutrons have masses that are close to, but not precisely, 1 u each (there are 6.022 x 1023 u in 1 gram This number is known as Avogadro’s number, N, and one of the ways this number can be calculated is discussed below). The number of protons in the nucleus of an atom is known as the atomic number, Z. It is equal to the number of electrons around the nucleus, because an atom is electrically neutral. The mass number of an atom is equal to the total number of heavy particles: protons and neutrons.
When two atoms are close enough to combine chemicallyto form chemical bonds with one another—each atom primarily “sees” the outermost electrons of the other atom. These outer electrons are therefore the most important factors in the chemical behavior of atoms. Neutrons in the nucleus have little effect on chemical behavior, and the protons are significant only because they determine how many electrons surround the nucleus in a neutral atom.
All atoms with the same atomic number behave in much the same way chemically, and are classified as the same chemical element. Each element has its own name and a one- or two-letter symbol (usually derived from the element’s English or Latin name). For example, the symbol for carbon is C, and the symbol for calcium is Ca. The symbol for sodium is Na-the first two letters of its Latin (and German) name, natrium, to distinguish it from nitrogen, N, and sulfur, S.
Example \(1\): Bromine
What is the atomic symbol for bromine, and what is its atomic number? Why isn’t the symbol for bromine just the first letter of its name? What other element preempts the symbol B? (Refer to the periodic table)
Solution
Bromine’s atomic number is 35, and its symbol is Br; B is the symbol for boron
Contributors and Attributions
• Dickerson, Richard E. and Gray, Harry B. and Haight, Gilbert P (1979) Chemical principles.
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John Dalton (1766-1844) is the scientist credited for proposing the atomic theory. This theory explains several concepts that are relevant in the observable world: the composition of a pure gold necklace, what makes the pure gold necklace different than a pure silver necklace, and what occurs when pure gold is mixed with pure copper. Before discussing the atomic theory, this article explains the theories that Dalton used as a basis for his theory: the law of conservation of mass and the law of constant composition.
Law of Conservation of Mass: (1766-1844)
The law of conservation of mass states that the total mass present before a chemical reaction is the same as the total mass present after the chemical reaction; in other words, mass is conserved. The law of conservation of mass was formulated by Antoine Lavoisier (1743-1794) as a result of his combustion experiment, in which he observed that the mass of his original substance—a glass vessel, tin, and air—was equal to the mass of the produced substance—the glass vessel, “tin calx”, and the remaining air.
Historically, this was a difficult concept for scientists to grasp. If this law was true, then how could a large piece of wood be reduced to a small pile of ashes? The wood clearly has a greater mass than the ashes. From this observation scientists concluded that mass had been lost. However, the illustration below shows that the burning of word does follow the law of conservation of mass. Scientists did not take into account the gases that play a critical role in this reaction.
Law of Constant Composition
Joseph Proust (1754-1826) formulated the law of constant composition (also called the law of definite proportions). This law states that if a compound is broken down into its constituent elements, the masses of the constituents will always have the same proportions, regardless of the quantity or source of the original substance. Joseph Proust based this law primarily on his experiments with basic copper carbonate. The illustration below depicts this law; 31 grams of H2O and 8 grams of H2O are made up of the same percent of hydrogen and oxygen.
Dalton's Atomic Theory
1. Each chemical element is composed of extremely small particles that are indivisible and cannot be seen by the naked eye, called atoms. Atoms can neither be created nor destroyed. Pictured below is a helium atom. The purple and red dots represent the neutrons and protons in the nucleus. The black area around the nucleus represent the electron cloud. The following sections discuss this further.
2. All atoms of an element are alike in mass and other properties, but the atoms of one element differ from all other elements. For example, gold and silver have different atomic masses and different properties.
Gold Silver
Atomic Mass: 196.97 Atomic Mass: 107.87
Figure 4 (Gold): Courtesy of Chris Ralph that released this image into the public domain. Figure 5 (silver): Courtesy of resourcescommittee.house.gov/.../photogallery/
3. For each compound, different elements combine in a simple numerical ratio. The illustration below describes this rule. The second equation for the reaction is incorrect because half of an atom does not exist.
Atomic theory can be used to answers the questions presented above. A pure gold necklace is made up of atoms. A pure gold necklace and a pure silver necklace are different because they have different atoms. Pure gold mixed with pure copper forms rose gold. The gold and copper atoms combine in a simple numerical ratio.
Dalton's theory has not proven to be correct under all circumstances. The first rule was proven incorrect when scientists divided atoms in a process called nuclear fission. The second rule was proven incorrect by the discovery that not all atoms of the same element have the same mass; there are different isotopes. However, these failures do not justify discarding the atomic theory. It correctly explains the law of conservation of mass: if atoms of an element are indestructible, then the same atom must be present after a chemical reaction as before and, and the mass must constant. Dalton’s atomic theory also explains the law of constant composition: if all the atoms of an element are alike in mass and if atoms unite in fixed numerical ratios, the percent composition of a compound must have a unique value without regards to the sample analyzed. The atomic theory led to the creation of the law of multiple proportions.
Law of Multiple Proportions
The law of multiple proportions states that if two elements form more than one compound between them, the masses of one element combined with a fixed mass of the second element form in ratios of small integers. The illustration of the third rule of the atomic theory correctly depicts this law.
Discovering Electrons
The first cathode-ray tube (CRT) was invented by Michael Faraday (1791-1867). Cathode rays are a type of radiation emitted by the negative terminal, the cathode, and were discovered by passing electricity through nearly-evacuated glass tubes. The radiation crosses the evacuated tube to the positive terminal, the anode. Cathode rays produced by the CRT are invisible and can only be detected by light emitted by the materials that they strike, called phosphors, painted at the end of the CRT to reveal the path of the cathode rays. These phosphors showed that cathode rays travel in straight lines and have properties independent of the cathode material (whether it is gold, silver, etc.). Another significant property of cathode rays is that they are deflected by magnetic and electric fields in a manner that is identical to negatively charged material. Due to these observations, J.J. Thompson (1856-1940) concluded that cathode rays are negatively charged particles that are located in all atoms. It was George Stoney who first gave the term electrons to the cathode rays. The below figures depict the way that the cathode ray is effected by magnetics. The cathode ray is always attracted by the positive magnet and deflected by the negative magnets.
The Plum Pudding Model
After Thompson discovered the electron, he proposed the plum pudding model of an atom, which states that the electrons float in positively-charged material. This model was named after the plum-pudding dessert.
Discovery of the Proton
In 1909, Ernest Rutherford (1871-1937) performed a series of experiments studying the inner structure of atoms using alpha particles. Rutherford knew that alpha particles are significantly more massive than electrons and positively charged. Using the plum-pudding model for reference, Rutherford predicted that particles in an alpha beam would largely pass through matter unaffected, with a small number of particles slightly deflected. The particles would only be deflected if they happened to come into contact with electrons. According to the plum pudding model, this occurrence would be very unlikely. In order to test his hypothesis, Rutherford shot a beam of alpha particles at a thin piece of gold foil. Around the gold foil Rutherford placed sheets of zinc sulfide. These sheets produced a flash of light when struck by an alpha particle. However, this experiment produced results that contradicted Rutherford's hypothesis. Rutherford observed that the majority of the alpha particles went through the foil; however, some particles were slightly deflected, a small number were greatly deflected, and another small number were thrown back in nearly the direction from which they had come. Figure 10 shows Rutherford's prediction based off of the plum-pudding model (pink) and the observed large deflections of the alpha particles (gold).
To account for these observations, Rutherford devised a model called the nuclear atom. In this model, the positive charge is held in an extremely small area called the nucleus, located in the middle of the atom. Outside of the nucleus the atom is largely composed of empty space. This model states that there were positive particles within the nucleus, but failed to define what these particles are. Rutherford discovered these particles in 1919, when he conducted an experiment that scattered alpha particles against nitrogen atoms. When the alpha particles and nitrogen atoms collided protons were released.
The Discovery of the Neutron
In 1933, James Chadwick (1891-1974) discovered a new type of radiation that consisted of neutral particles. It was discovered that these neutral atoms come from the nucleus of the atom. This last discovery completed the atomic model.
Example Problems
1. Basic concept check: When 32.0 grams (g) of methane are burned in 128.0 g of oxygen, 88.0 g of carbon dioxide and 72.0 g of water are produced. Which law is this an example of? (a) Law of Definite Proportions (b) Law of Conservation of Mass or (c) Law of Multiple Proportions.
The answer is (b) Law of Conservation of Mass. The number of grams of reactants (32.0 g of methane and 128.0 g of oxygen = 160.0 g total) is equal to the number of grams of product (88.0 g of carbon dioxide and 72.0 g of water = 160.0 g total).
2. Law of Conservation of Mass: 8.00 grams (g) of methane are burned in 32.00 g of oxygen. The reaction produces 22.00 g of carbon dioxide and an unmeasured mass of water. What mass of water is produced?
The answer is 18.00 g of water. Because the only products are water and carbon dioxide, their total mass must equal the total mass of the reactants, methane and oxygen. 8.00 g of methane + 32.00 g of oxygen = 40.00 total g of reactants. Because the total mass of the reactants equals the total mass of the products, the total mass of the products is also 40.00 g. Thus, 40.00 total g of products = 22.00 g carbon dioxide + unknown mass water. 40.00 total g of products - 22.00 g carbon dioxide = 18.00 g water.
3. Law of Definite Proportions: Two experiments using sodium and chlorine are performed. In the first experiment, 4.36 grams (g) sodium are reacted with 32.24 g of chlorine, using up all the sodium. 11.08 g of sodium chloride was produced in the first experiment. In the second experiment, 4.20 g of chlorine reacted with 20.00 g of sodium, using up all the chlorine. 6.92 g of of sodium chloride was produced in the second experiment. Show that these results are consistent with the law of constant composition.
To solve, determine the percent of sodium in each sample of sodium chloride. There is 4.36 g sodium for every 11.08 g of sodium chloride in the first experiment. The amount of sodium in the sodium chloride for the second experiment must be found. This is found by subtracted the known amount of reacted chlorine (4.20 g) from the amount of sodium chloride (6.92 g). 6.92 g sodium chloride - 4.20 g chlorine = 2.72 g sodium.
Thus, the percent of sodium in each sample is represented below:
% Na = (4.36 g Na)/(11.08 g NaCl) x 100% = 39.4% Na % Na = (2.72 g Na)/(6.92 g NaCl) x 100% = 39.3%
The slight difference in compositions is due to significant figures: each percent has an uncertainty of .01% in either direction. The two samples of sodium chloride have the same composition.
4. Law of Conservation of Mass: 36.0 grams (g) of wood are burned in oxygen. The products of this reaction weigh 74.4 g. (a) What mass of oxygen is needed in this reaction? (b) What mass of oxygen is needed to burn 8.00 lb of wood? 1 lb = 453.59237 g.
1. The answer is 38.4 g of oxygen. The total mass of the products is 74.4 g. Thus, the total mass of the reactants must equal 74.4 g as well. Thus, 74.4 g products - 36.0 g wood reactant = 38.4 g oxygen reactant.
2. The answer is 8.53 lb of oxygen. From, (a) that it takes 38.4 g of oxygen to burn 18.0 g of wood. First, convert both of these values to pounds (alternatively, the 8.00 lb can be converted to grams).
36.0 g wood x (1lb)/(453.59237 g) = .0793664144 lb wood
38.4 g oxygen x (1 lb)/(453.59237 g) = .0846575087 lb oxygen
Now two ratios equal to each other can be set up to determine the unknown mass of oxygen.
(0.0793664144 lb wood)/(.0846575087 lb oxygen) = (8.00 lb wood)/(unknown mass oxygen)
Solving reveals that it requires 8.53 lb of oxygen to burn 8.00 lb of wood.
5. Law of Definite Proportions: A sample of methane contains only carbon and hydrogen, with 3.00 grams (g) of carbon for every 1.00 g of hydrogen. How much hydrogen should be present in a different, 50.0 g same of methane?
The answer is 12.5 g of hydrogen. If there are 3.00 g of carbon present for every 1.00 g of hydrogen, we can assume the smallest whole number combination of these elements in that ratio to be 4.00 g of methane:
50.0 g methane x (1.00 g hydrogen)/(4.00 g methane) = 12.5 g of hydrogen.
Outside links
• Description of Atomic Theory - http://en.Wikipedia.org/wiki/Atomic_theory
• Description of Atomic Theory - http://www.visionlearning.com/librar...wer.php?mid=50
• Cathode Ray Demonstration - www.youtube.com/watch?v=XU8nMKkzbT8
• Theoretical Demonstration of Conservation of Mass - www.youtube.com/watch?v=dExpJAECSL8
• Actual Demonstration of Conservation of Mass - www.youtube.com/watch?v=J5hM1...eature=related
• The alpha scattering experiment-http://www.youtube.com/watch?v=5pZj0u_XMbc
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John Dalton, a British school teacher, published his theory about atoms in 1808. His findings were based on experiments and the laws of chemical combination.
Dalton's Atomic Theory
The concept of the atom (Greek: atomos, "indivisible"), an indivisible particle of matter, goes back to ancient Greece and a man named Democritus, a rival of Aristotle. Democritus held that all matter could be subdivided only until some finite particle was reached. Aristotle had a quite different idea, that matter was a continuous substance, not composed of any fundamental units. And Aristotle was much more famous and admired in his time and so his error persisted into the late 1700's. But eventually work by men such as Lavoisier began to suggest that Aristotle had been seriously wrong.
Dalton's Contributions
In 1808 the first statement of a modern atomic theory was published by John Dalton, a Quaker schoolmaster from Manchester. It may not seem like much, but such a theory was used to explain two of the major laws in chemistry: the Law of Conservation of Mass and the Law of Constant (definite) Composition. And the third postulate led Dalton to formulate the Law of Multiple Proportions.
John Dalton apparently thought that atoms were pretty much little solid spheres without internal structure. His experiments with gases eventually led him to the idea that matter came in small indivisible units but his use of the word "atom" was variable. He described what we now call molecules as "atoms" and was convinced that the simplest binary compound of any two elements would have those atoms in a 1:1 ratio. According to Dalton, water would be HO. This led to a relative mass scale for atoms that was mostly in error.
Dalton also had his own notation for some of the elements he recognized.
Dalton's "modern" atomic theory, in todays terms, would be something like this:
1. all matter consists of indivisible atoms which cannot be created or destroyed
2. atoms of one element cannot be converted into atoms of another element
3. atoms of an element are identical to one another but different from atoms of other elements
4. compounds result from combinations of atoms in specific ratios
Some contemporaries of Dalton remained unimpressed. An eminent organic chemist of the time, Adolf Kolbe, said in 1877, "Dalton's atoms are no more than stupid hallucinations...mere table-tapping and supernatural explanations."
Experimental Evidence
The first evidence for sub-atomic particles came from experiments with the conduction of electricity through gases in sealed glass tubes at low pressures.
Figure : J.J. Thomson (seen here in his lab with one of his many hand-made cathode ray tubes)
Associated with the flow of electricity in such a tube are rays which originate from the negative electrode--the cathode. Thus these tubes have been called cathode ray tubes. From careful experiments with cathode ray tubes J.J. Thomson demonstrated in 1897 that the rays consist of a stream of negatively charged particles which he called electrons. He was able to measure the charge/mass ratio of these particles and found this to be the same regardless of what gas was in the tube or what metal the electrodes were made from.
In 1909 Robert Millikan used the classic oil drop experiment to determine the charge on these particles. Using the smallest charge obtained and Thomson's charge/mass ratio the electron mass is roughly 1/2000 the mass of the lightest atom. Thus there are obviously particles smaller than atoms.
Millikan used his apparatus to make many measurements of the effect of the electric field vs. gravity on the charged oil droplets. His eventual conclusion was that the charges on the drops were either equal to or multiples of one number which he decided was the charge of a single electron. With this value, and the charge/mass ratio that Thomson had measured earlier it was possible to calculate the mass of the electron.
Internal Structure
Thomson's dilemma: how could matter containing electrons be neutral and where was all the mass? Other experiments with discharge tubes suggested the existence of a positive particle with much greater mass (the proton). Based on this evidence Thomson proposed the first atomic model with sub-atomic particles.
Thomson's work (and the work of others) with cathode ray tubes eventually identified a much more massive positive particle apparently common to all matter. These were called protons. With the ingredients needed to keep matter neutral, Thomson speculated and calculated just how it might all hold together. His conclusion has become known as the "plum pudding" model of the atom and was the first to include sub-atomic particles with charges.
Thomson's calculations indicated that to remain stable atoms might need to have hundreds, if not thousands, of electrons. However, the developing techniques for measuring relative atomic masses seemed to contradict this, suggesting instead that less than 100 electrons would be more likely.
Finally in the years between 1908 and 1911 Ernest Rutherford and his students Hans Geiger and Ernest Marsden performed the famous gold-foil experiment in which the nuclear arrangement of the atom was discovered. Rutherford's vision of the atom as a kind of "solar system" in miniature is what most people probably carry around in their heads. As appealingly simple as it is, Rutherford's model contained a serious flaw.
Rutherford's model of the atom was no doubt appealing in part because it seemed to mimic the larger world people could see. But just as the planets slowly lose orbital energy as they circle the sun, the energy supply of the tiny electron should be degraded too.
Physicists had already established in Rutherford's era that a charged particle moving in a circle, but keeping a constant orbital radius must radiate energy. Hence the nuclear model of the atom should render all matter unstable as electrons gradually spiral into nuclei when their orbits "decay" due to loss of energy. Obviously, this does not happen.
The next step in the development of the atomic model changed the way in which scientists looked at and thought about matter, especially very small pieces like electrons and protons.
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This law, formulated by Gay Lussac, states that, "the ratio between the volumes of gaseous reactants and products can be expressed in simple whole numbers."
For example, in the following reaction, the ratio of volumes of hydrogen, chlorine, and hydrogen chloride is 1:1:2 (a simple ratio):
Lavoisier's Law of Conservation of Mass
With the development of more precise ideas on elements, compounds and mixtures, scientists began to investigate how and why substances react. French chemist A. Lavoisier laid the foundation to the scientific investigation of matter by describing that substances react by following certain laws. These laws are called the laws of chemical combination. These eventually formed the basis of Dalton's Atomic Theory of Matter.
Law of Conservation of Mass
According to this law, during any physical or chemical change, the total mass of the products remains equal to the total mass of the reactants.
The law of conservation of mass is also known as the "law of indestructibility of matter."
Example \(1\)
If heating 10 grams of \(\ce{CaCO3}\) produces 4.4 g of \(\ce{CO2}\) and 5.6 g of \(\ce{CaO}\), show that these observations are in agreement with the law of conservation of mass.
Solution
• Mass of the reactants: \(10 \,g\)
• Mass of the products: \(4.4 \,g+ 5.6\, g = 10\, g\).
Because the mass of the reactants is equal to the mass of the products, the observations are in agreement with the law of conservation of mass.
Law of Multiple Proportions
John Dalton (1803) stated, "'When two elements combine with each other to form two or more compounds, the ratios of the masses of one element that combines with the fixed mass of the other are simple whole numbers'.
Example
• Carbon monoxide (\(CO\)): 12 parts by mass of carbon combines with 16 parts by mass of oxygen.
• Carbon dioxide (\(CO_2\)): 12 parts by mass of carbon combines with 32 parts by mass of oxygen.
Ratio of the masses of oxygen that combines with a fixed mass of carbon (12 parts): 16:32 or 1:2
Hydrogen and oxygen are known to form 2 compounds. The hydrogen content in one is 5.93%, and that of the other is 11.2%. Show that this data illustrates the law of multiple proportions.
Solution
In the first compound: hydrogen = 5.93%
Oxygen = (100 -5.93) = 94.07%
In the second compound: hydrogen = 11.2%
Oxygen = (100 -11.2) = 88.88%
Ratio of the masses of oxygen that combine with fixed mass of hydrogen: 15.86:7.9 or 2:1. This is consistent with the law of multiple proportions.
Law of Reciprocal Proportions
The law of reciprocal proportions was proposed by Jeremias Ritcher in 1792. It states that, "If two different elements combine separately with the same weight of a third element, the ratio of the masses in which they do so are either the same or a simple multiple of the mass ratio in which they combine."
Example
Oxygen and sulfur react with copper to give copper oxide and copper sulfide, respectively. Sulfur and oxygen also react with each other to form SO2. Therefore,
in CuS:
Cu:S = 63.5:32
in CuO:
Cu:O = 63.5:16
S:O = 32:16
S:O = 2:1
Now in SO2:
S:O = 32:32
S:O = 1:1
Thus the ratio between the two ratios is the following:
\[\dfrac{2}{1} : \dfrac{1}{1} - 2:1\]
This is a simple multiple ratio.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Atomic_Theory/Dalton%27s_Atomic_Theory/Gay-Lussac%27s_Law_of_Gaseous_Volumes.txt
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John Dalton, a British school teacher, published his theory about atoms in 1808. His findings were based on experiments and the laws of chemical combination.
Postulates
1. All matter consists of indivisible particles called atoms.
2. Atoms of the same element are similar in shape and mass, but differ from the atoms of other elements.
3. Atoms cannot be created or destroyed.
4. Atoms of different elements may combine with each other in a fixed, simple, whole number ratios to form compound atoms.
5. Atoms of same element can combine in more than one ratio to form two or more compounds.
6. The atom is the smallest unit of matter that can take part in a chemical reaction.
Drawbacks of Dalton's Atomic Theory
• The indivisibility of an atom was proved wrong: an atom can be further subdivided into protons, neutrons and electrons. However an atom is the smallest particle that takes part in chemical reactions.
• According to Dalton, the atoms of same element are similar in all respects. However, atoms of some elements vary in their masses and densities. These atoms of different masses are called isotopes. For example, chlorine has two isotopes with mass numbers 35 and 37.
• Dalton also claimed that atoms of different elements are different in all respects. This has been proven wrong in certain cases: argon and calcium atoms each have an atomic mass of 40 amu. These atoms are known as isobars.
• According to Dalton, atoms of different elements combine in simple whole number ratios to form compounds. This is not observed in complex organic compounds like sugar (C12H22O11).
• The theory fails to explain the existence of allotropes; it does not account for differences in properties of charcoal, graphite, diamond.
Merits of Dalton's Atomic Theory
• The atomic theory explains the laws of chemical combination (the Law of Constant Composition and the Law of Multiple Proportions).
• Dalton was the first person to recognize a workable distinction between the fundamental particle of an element (atom) and that of a compound (molecule).
Proust's Law of Constant Proportion
The Law of Constant Composition, discovered by Joseph Proust, is also known as the Law of Definite Proportions. It is different from the Law of Multiple Proportions although both stem from Lavoisier's Law of Conservation of Mass. The French chemist Joseph Proust stated this law the following way: "A chemical compound always contains the same elements combined together in the same proportion by mass."
Joseph Proust
Joseph Proust was a French Chemist best known for his analytical abilities. He was once recommended for a job as a chemistry professor at Segovia's Royal Artillery School by none other than Antoine Lavoisier! His experiments with inorganic binary compounds - mostly sulfates, sulfides, and metallic oxides - led him to formulate the Law of Constant Composition. The law was first published in a paper on iron oxides in 1794.
Proust's law was attacked by the respected French chemist Claude-Louis Berthollet who disagreed that chemical combination was restricted to definite saturation proportions. The confusion was caused by the definition of chemical combination; Berthollet classified solutions as chemical combinations while Proust was careful to distinguish between these and true binary compounds. The conflict lasted until John Dalton, an English chemist, came out with an Atomic Theory that favored Proust's law. Swedish chemist Jons Jacob Berzelius established the relationship between Proust's law and Dalton's theory in 1811.
For example, pure water obtained from different sources such as a river, a well, a spring, the sea, etc., always contains hydrogen and oxygen together in the ratio of 1:8 by mass. Similarly, carbon dioxide (CO2) can be obtained by different methods such as,
• Burning of carbon
• Heating of lime stone
• Applying dilute HCl to marble pieces
Each sample of CO2 contains carbon and oxygen in a 3:8 ratio.
Example $1$: Reduction of Cupric Oxide
When 1.375 g of cupric oxide is reduced on heating in a current of hydrogen, the weight of copper remaining 1.098 g. In another experiment, 1.179 g of copper is dissolved in nitric acid and resulting copper nitrate converted into cupric oxide by ignition. The weight of cupric oxide formed is 1.476 g. Show that these results illustrate the law of constant proportion.
Solution
First experiment
• Copper oxide = 1.375 g
• Copper left = 1.098 g
• Oxygen present = 1.375 - 1.098 = 0.277 g
$\text{Percentage of oxygen in CuO} = \dfrac{(0.277)(100\%)}{1.375} = 20.15\% \nonumber$
Second Experiment
• Copper taken = 1.179 g
• Copper oxide formed = 1.476 g
• Oxygen present = 1.476 - 1.179 = 0.297 g
$\text{Percentage of oxygen in CuO} = \dfrac{(0.297)(100\%)}{1.476} = 20.12\% \nonumber$
Percentage of oxygen is approximately (within significant figures) the same in both the above cases. So the law of constant composition is illustrated.
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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:
$H_2 (g) \rightarrow H_2^+(g) + e^-(g)$
The measured ionization energy of H2 is 1488 kJ mol-1. This number is primarily important in comparison to the ionization energy of a hydrogen atom, which is 1312 kJ mol-1. Therefore, it requires more energy to remove an electron from the hydrogen molecule than from the hydrogen atom; the electron therefore has a lower energy in the molecule. To pull the atoms apart, the energy of the electron must be increased. Hence, energy is required to break the bond, and the molecule is bound.
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 is consistent with the view of shared electrons in bonding molecular orbitals.
As a second example, consider the nitrogen molecule, $N_2$. The ionization energy of molecular nitrogen is 1503 kJ mol-1, and that of atomic nitrogen is 1402 kJ mol-1. Once again, 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, consider fluorine, $F_2$. In this case, find that the ionization energy of molecular fluorine is 1515 kJ mol-1, which is smaller than the ionization energy of a fluorine atom, 1681 kJ mol-1. This seems inconsistent with the bonding orbital concept 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 $F_2$ than in F, why is $F_2$ a stable molecule? A more complete description of the molecular orbital concept of chemical bonding is required.
Bond energies are next considered. Recall that the bond energy (or bond strength) is the energy required to separate the bonded atoms. The bond energy of $N_2$ is 956 kJ mol-1; this is much larger than the bond energy of H2, 458 kJ mol-1, and of $F_2$, which is 160 kJ mol-1. The unusually strong bond in nitrogen can be explained using both the valence shell electron pair sharing model and electron orbital descriptions. A nitrogen atom has three unpaired electrons in its valence shell; 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 in a “triple bond.” Intuition suggests that the triple bond in $N_2$ should be much stronger than the single bond in H2 or in $F_2$.
Now consider the molecular orbital description of bonding in $N_2$. Each of the three 2p atomic orbitals in each nitrogen atom must overlap to form a bonding molecular orbital, 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 with cylindrical symmetry and is therefore a σ orbital (a σ* orbital is also formed). The two electrons are paired in the bonding orbital.
This effect is depicted in Fig. 5 in a molecular orbital energy diagram. Each pair of atomic orbitals, one from each atom, overlaps to form a bonding and an anti-bonding orbital. The three 2p orbitals from each atom form one σ and σ* pair and two π and π* pairs. The lowering of the energies of the electrons in the σand π 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 the discussion of bonding was begun with $N_2$ because of the curious result that the ionization energy of an electron in $F_2$ is less than that of an electron in an F atom. Comparing the molecular orbital energy level diagrams for $N_2$ and $F_2$ allows an explanation for 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 $N_2$ and $F_2$.) Two electrons are placed in the σ orbital, four more in the two π orbitals, and four more in the two π* orbitals. Overall, there are six electrons in bonding orbitals and four in anti-bonding orbitals. Because $F_2$ 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 $F_2$ 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 $F_2$ 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 antibonding orbital is higher than that of the atomic orbitals, and the system is stabilized by the electron's removal. Therefore, the ionization energy of molecular fluorine is less than that of atomic fluorine. This clearly demonstrates the physical reality and importance of the antibonding orbitals.
A particularly interesting case is the oxygen molecule, $O_2$. In completing the molecular orbital energy level diagram for oxygen, the last two electrons must either be paired in the same 2p π* orbital or separated into different 2p π* orbitals. To determine which, it is important to note that oxygen molecules are paramagnetic—they are strongly attracted to a magnetic field. To account for this paramagnetism, 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. If 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 $O_2$ molecule is shown in Fig. 7.
In comparing these three diatomic molecules, recall that $N_2$ has the strongest bond, followed by $O_2$ and $F_2$. The comparative bond strengths were previously accounted for with Lewis structures, showing that $N_2$ is a triple bond, $O_2$ is a double bond, and $F_2$ is a single bond. The molecular orbital energy level diagrams in Figs. 5 to 7 add more perspective to 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 $N_2$, 4 for $O_2$, and six for $F_2$. Thus, the strength of a bond must be related to the relative numbers of bonding and antibonding electrons in the molecule. Therefore, the bond order is defined as the following:
$\text{Bond order} = \frac{1}{2} (\# \ of\ bonding\ electrons - \# \ of\ antibonding\ electrons)$
Note that, defined this way, the bond order for $N_2$ is 3, for $O_2$ is 2, and for $F_2$ is 1, which agrees with the conclusions made from Lewis structures. In conclusion, the relative strengths of bond can be predicted by comparing bond orders.
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Atoms that have the same atomic number (number of protons), but different mass numbers (number of protons and neutrons) are called isotopes. There are naturally occurring isotopes and isotopes that are artificially produced. Isotopes are separated through mass spectrometry; MS traces show the relative abundance of isotopes vs. mass number (mass : charge ratio).
Introduction
As mentioned before, isotopes are atoms that have the same atomic number, but different mass numbers. Isotopes are denoted the same way as nuclides, but they are often symbolized only with the mass numbers because isotopes of the same element have the the same atomic number. Carbon, for example, has two naturally occurring isotopes, \(^{12}_6C\) and \(^{13}_6C\). Because both of these isotopes have 6 protons, they are often written as \(^{12}C\) and \(^{13}C\). \(^{12}C\) has 6 neutrons, and \(^{13}C\) has 7 neutrons.
Of all the elements on the periodic table, only 21 are pure elements. Pure, or monotopic, elements are those elements with only one naturally occurring nuclide. The following lists the 21 pure elements:
1. \(^{27}_{13}Al\)
2. \(^{75}_{33}As\)
3. \(^{9}_4Be\)
4. \(^{209}_{83}Bi\)
5. \(^{133}_{55}Cs\)
6. \(^{59}_{27}Co\)
7. \(^{19}_9F\)
8. \(^{197}_{79}Au\)
9. \(^{165}_{67}Ho\)
10. \(^{127}_{53}I\)
11. \(^{55}_{25}Mn\)
12. \(^{93}_{41}Nb\)
13. \(^{31}_{15}P\)
14. \(^{141}_{59}Pr\)
15. \(^{103}_{45}Rh\)
16. \(^{45}_{21}Sc\)
17. \(^{23}_{11}Na\)
18. \(^{159}_{65}Tb\)
19. \(^{232}_{90}Th\)
20. \(^{169}_{69}Tm\)
21. \(^{89}_{39}Y\)
Isotopes of the other elements either occur naturally or are artificially produced.
Natural and Artificial Isotopes
Most elements have naturally occurring isotopes. Percent natural abundances indicate which isotopes of any given element are predominant (occur in greater abundance) and which only occur in trace amounts. Mercury, for example, has seven naturally occurring isotopes: \(^{196}Hg\), \(^{198}Hg\), \(^{199}Hg\), \(^{200}Hg\), \(^{201}Hg\), \(^{202}Hg\), \(^{204}Hg\); these have the percent natural abundances of 0.146%, 10.02%, 16.84%, 23.13%, 13.22%, 29.80%, and 6.85%, respectively. It is clear that \(^{202}Hg\) occurs with greatest abundance, and \(^{200}Hg\) is the next most abundant, but the other isotopes only occur in small traces.
Note: The sum of the percent natural abundances of all the isotopes of any given element must total 100%.
There are 20 elements with only artificially produced isotopes. The majority of these are heavier elements; the lightest elements with artificial isotopes are \(^{43}Tc\) and \(^{61}Pm\). The other elements that only have artificial isotopes are those with atomic numbers of 84-88 and 89-103, otherwise known as the actinoids, but excluding \(^{90}Th\) and \(^{92}U\).
Some naturally occurring and artificially produced isotopes are radioactive. The nucleus of a radioactive isotope is unstable; radioactive isotopes spontaneously decay, emitting alpha, beta, and gamma rays until they reach a stability, usually in the state of a different element. Bismuth (\(^{209}_{83}Bi\)) has the highest atomic and mass number of all the stable nuclides. All nuclides with atomic number and mass number greater than 83 and 209, respectively, are radioactive. However, there are some lighter nuclides that are radioactive. For example, hydrogen has two naturally occurring stable isotopes, \(^{1}H\) and \(^{2}H\) (deuterium), and a third naturally occurring radioactive isotope, \(^{3}H\) (tritium).
Radioisotope Dating
The presence of certain radioisotopes in an object can be used to determine its age. Carbon dating is based on the fact that living plants absorb stable \(^{12}C\), \(^{13}C\) and radioactive \(^{14}C\) from the atmosphere, and animals absorb them from the plants. An organism no longer absorbs carbon after it dies, its age can be determined by measuring the ratio of \(^{13}C\) to \(^{14}C\) in the sample and extrapolating based on its decay rate.
Art forgeries are often detected by similar means. \(^{137}Cs\) and \(^{90}Sr\) do not occur naturally and are only present in the atmosphere today because of nuclear weapons. Any object created before July 1945, then, would have neither of these elements, so finding them through mass spectrometry or other means would indicate that it was created later.
Isotopic Masses, Percent Natural Abundance, and Weighted-Average Atomic Mass
Because most elements occur as isotopes and different isotopes have different masses, the atomic mass of an element is the average of the isotopic masses, weighted according to their naturally occurring abundances; this is the mass of each element recorded on the periodic table, also known as the relative atomic mass (Ar). Treating isotopic masses in weighted averages gives greater importance to the isotope with greatest percent natural abundance. Below is a general equation to calculate the atomic mass of an element based on percent natural abundance and isotopic masses:
* fractional abundance is the percent abundance divided by 100%
Bromine has two naturally occurring isotopes: bromine-79 has a mass of 78.9183 u and an abundance of 50.69%, and bromine-81 has a mass of 80.92 u and an abundance of 49.31%. The equation above can be used to solve for the relative atomic mass of bromine:
atomic mass of Br = (0.5069 x 78.9183 u) + (0.4931 x 80.92 u) = 79.91 u
This is the relative atomic number of bromine that is listed on the periodic table.
Comparing their isotopic masses of any given element to the relative atomic mass of the element reveals that the Ar is very close to the isotope that occurs most frequently. Thus, the isotope whose isotopic mass is closest to the atomic mass of the element is the isotope that occurs in the greatest abundance.
Mass Spectrometry
Mass spectrometry is a technique that can be used to distinguish between isotopes of a given element. A mass spectrometer separates each isotope by mass number. Each isotope is characterized by a peak (of given intensity) according to its relative abundance. The most intense peak corresponds to the isotope that occurs in the largest relative natural abundance, and vice versa. Refer to Mass Spectrometry: Isotope Effects.
Example:
The mass spectrum of strontium has four different peaks, varying in intensity. The four peaks indicate that there are four isotopes of strontium. The four isotopes of strontium have isotopic mass numbers of 84, 86, 87, and 88, and relative abundances of 0.56%, 9.86%, 7.00%, and 82.58%, respectively. The intensity of the peak corresponds to the abundance. \(^{84}Sr\) has the smallest peak, which corresponds to its relative abundance of 0.56%, whereas \(^{88}Sr\) has the largest peak, which corresponds to its relative abundance of 82.58%. This indicates that \(^{88}Sr\) is the isotope that occurs in highest amounts.
Problems
1) Find the number of protons and neutrons in the following isotopes: a) \(^{20}Ne\), \(^{21}Ne\), \(^{22}Ne\); b) \(^{84}Sr\), \(^{86}Sr\), \(^{87}Sr\), \(^{88}Sr\); c) \(^{102}Pd\), \(^{104}Pd\), \(^{105}Pd\), \(^{106}Pd\), \(^{108}Pd\), \(^{110}Pd\)
(Answer: a) \(^{20}Ne\) has 10 p and 10 n; \(^{21}Ne\) has 10 p and 11 n; \(^{22}Ne\) has 10 p and 12 n; b) \(^{84}Sr\) has 38 p and 46 n, \(^{86}Sr\) has 38 p and 48 n, \(^{87}Sr\) has 38 p and 49 n, \(^{88}Sr\) has 38 p and 50 n; c) \(^{102}Pd\) has 46 p and 56 n, \(^{104}Pd\) has 46 p and 58 n, \(^{105}Pd\) has 46 p and 59 n, \(^{106}Pd\) has 46 p and 60 n, \(^{108}Pd\) has 46 p and 62 n, \(^{110}Pd\) has 46 p and 64 n)
2) An isotope of lead has an isotopic mass of 208. Give the appropriate notation for the isotope.
(Answer: \(^{208}Pb\))
3) Thallium has two naturally occurring istotopes, \(^{203}Tl\) and \(^{205}Tl\). Tl-205 has an abundance of 70.48%. What is the percent natural abundance of Tl-203?
(Answer: 29.52%)
4) Chlorine has two naturally occuring isotopes, \(^{35}Cl\) and \(^{37}Cl\), with masses of 34.97 u and 36.97 u, respectively. Which of these two isotopes occur in greater abundance?
(Answer: \(^{35}Cl\))
5. Potassium has three naturally occurring isotopes: \(^{39}K\), \(^{40}K\), and \(^{41}K\), with masses (and abundances) of 38.96 u (93.3%), 39.96 u (0.012%), and 40.96 u (6.7%), respectively. Calculate the relative atomic mass of K.
(Answer: 39.10 u)
Review Questions
1. What do the terms 'monotopic' and 'radioactive' mean?
2. On the mass spectrum of any given atom, what does the number and intensity of the peak(s) tell you?
3. What is the relationship between isotopic masses, percent natural abundance, and weighted-average atomic mass?
Isotopes
Although all atoms of an element have the same number of protons, individual atoms may have different numbers of neutrons. These differing atoms are called isotopes. All atoms of chlorine (Cl) have 17 protons, but there are chlorine isotopes with 15 to 23 neutrons. Only two chlorine isotopes exist in significant amounts in nature: those with 18 neutrons (75.53% of all chlorine atoms found in nature), and those with 20 neutrons (24.47%). To write the symbol for an isotope, place the atomic number as a subscript and the mass number (protons plus neutrons) as a superscript to the left of the atomic symbol. The symbols for the two naturally occurring isotopes of chlorine are written as follows: $^{35}_{17}{\rm Cl}$ and $^{37}_{17}{\rm Cl}$. The subscript is somewhat unnecessary, because all atoms of chlorine have 17 protons; isotope symbols are usually written without the subscript, as in 35Cl and 37Cl. In discussing these isotopes, the terms chlorine-35 and chlorine-37 are used to differentiate between them. In general, for an atom to be stable, it must have more neutrons than protons. Nuclei with too many of either kind of fundamental particle are unstable, and break down radioactively.
Example $1$:
How many protons, neutrons, and electrons are there in an atom of uranium-238? Write the symbol for this isotope.
Solution
The atomic number of uranium (see periodic table) is 92, and the mass number of the isotope is given as 238. Therefore, it has 92 protons, 92 electrons, and 238 — 92 : 146 neutrons. Its symbol is $\ce{^{238}_{92}U}$ (or 238U).
The total mass of an atom is called its atomic weight, and is the approximate sum of the masses of its constituent protons, neutrons, and electrons. When protons, neutrons, and electrons combine to form an atom, some of their mass is converted to energy and is given off. (This is the source of energy in nuclear fusion reactions. Because the atom cannot be broken down into its fundamental particles unless the energy for the missing mass is supplied from outside it, this energy is called the binding energy of the nucleus.
Example $2$:
Calculate the mass that is lost when an atom of carbon-12 is formed from protons, electrons, and neutrons.
Solution
Because the atomic number of every carbon atom is 6, carbon-12 has 6 protons and therefore 6 electrons. To find the number of neutrons, we subtract the number of protons from the mass number: 12 – 6 = 6 neutrons. The data in Table 1-1 can be used to calculate the total mass of these particles:
Protons: 6 x 1.00728 amu = 6.04368 u
Neutrons: 6 x 1.00867 amu = 6.05202 u
Electrons: 6 x 0.00055 amu = 0.00330 u
Total particle mass = 12.09900 u
However, by the definition of the scale of atomic mass units, the mass of one carbon-12 atom is exactly 12 amu. Therefore, 0.0990 u of mass has disappeared in the process of building the atom from its particles.
Each isotope of an element is characterized by an atomic number (the number of protons), a mass number (the total number of protons and neutrons), and an atomic weight (mass of atom in atomic mass units). Because mass losses upon formation of an atom are small, the mass number is usually the same as the atomic weight rounded to the nearest integer (for example, the atomic weight of chlorine-37 is 36.966, which is rounded to 37). If there are several isotopes of an element in nature, then the experimentally observed atomic weight (the natural atomic weight) is the weighted average of the isotope weights. The average is weighted according to the percent abundance of the isotopes. Chlorine occurs in nature as 75.53% chlorine-35 (34.97 u) and 24.47% chlorine-37 (36.97 u), so the weighted average of the isotope weights is (07553 x 34.97 u) + (0.2447 x 36.97 u) = 35.46 u. The atomic weights found in periodic tables are all weighted averages of the isotopes occurring in nature, and these are the figures used for the remainder of this article, except when discussing one isotope specifically. In general, all isotopes of an element behave the same way chemically. Their behaviors differ with regard to mass-sensitive properties such as diffusion rates.
Example $3$:
Magnesium (Mg) has three significant natural isotopes: 78.70% of all magnesium atoms have an atomic weight of 23.985 u, 10.13% have an atomic weight of 24.986 u, and 11.17% have an atomic weight of 25.983 u. How many protons and neutrons are present in each of these three isotopes? How are the symbols for each isotope written? Finally, what is the weighted average of the atomic weights?
Solution
There are 12 protons in all magnesium isotopes. The isotope whose atomic weight is 23.985 u has a mass number of 24 (protons and neutrons), so 24 - 12 protons gives 12 neutrons. The symbol for this isotope is 24Mg. Similarly, the isotope whose atomic weight is 24.986 amu has a mass number of 25, 13 neutrons, and 25Mg as a symbol. The third isotope (25.983 amu) has a mass number of 26, 14 neutrons, and 26Mg as a symbol. The average atomic weight is calculated as follows:
(0.7870 x 23.985) + (0.1013 x 24.986) + (0.1117 x 25.983) = 24.31 u
Example $4$: Boron
Boron has two naturally occurring isotopes, 10B and 11B. In nature, 80.22% of its atoms are 11B, with atomic weight 11.009 u. From the natural atomic weight, calculate the atomic weight of the 10B isotope.
Solution
If 80.22% of all boron atoms are 11B, then 100.00 — 80.22, or 19.78%, are the unknown isotope. In the periodic table the atomic weight of boron is found to be 10.81 u. We can use W to represent the unknown atomic weight in our calculation:
$(0.8022 \times 11.009) + (0.1978 \times W) = 10.81 {\rm u} \quad {\rm (natural~atomic~weight)}$
$W=\dfrac{10.81-8.831}{0.1978}=10.01 {\rm u}$
Contributors and Attributions
• Dickerson, Richard E. and Gray, Harry B. and Haight, Gilbert P (1979) Chemical principles.
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An atom is the smallest unit of an element that can exist. Every atom is made up of protons, neutrons, and electrons. These particles define a nuclide and its chemical properties and were discovered in the early 20th century and are described by modern atomic theory.
Nuclide
Nuclides are specific types of atoms or nuclei. Every nuclide has a chemical element symbol (E) as well as an atomic number (Z) , the number of protons in the nucleus, and a mass number (A), the total number of protons and neutrons in the nucleus. The symbol for the element is as shown below:
$^A_{Z}E$
An example is neon, which has the element symbol Ne, atomic number 10 and mass number 20.
$^{20}_{10}Ne$
A nuclide has a measurable amount of energy and lasts for a measurable amount of time. Stable nuclides can exist in the same state indefinitely, but unstable nuclides are radioactive and decay over time. Some unstable nuclides occur in nature, but others are synthesized artificially through nuclear reactions.They emit energy ($\alpha$, $\beta$, or $\gamma$ emissions) until they reach stability.
Atomic Number
Every element has a defining atomic number, with the symbol "Z". If an atom is neutrally charged, it has the same number of protons and electrons. If it is charged, there may be more protons than electrons or vice versa, but the atomic number remains the same. In the element symbol, the charge goes on the right side of the element. For instance, O2- is an oxygen anion. O2- still has an atomic number of 8, corresponding to the 8 protons, but it has 10 electrons. Every element has a different atomic number, ranging from 1 to over 100. On the periodic table, the elements are arranged in the order of atomic number across a period. The atomic number is usually located above the element symbol. For example, hydrogen has one proton and one electron, so it has an atomic number of 1. Copper has the atomic number of 29 for its 29 protons.
Examples as seen on Periodic Table
Atomic Number 3 4 5
Element's Symbol B C N
Average Atomic Mass of all Element's Isotopes 10.811 12.011 14.007
Atomic Number and Chemical Properties
The atomic number defines an element's chemical properties. The number of electrons in an atom determines bonding and other chemical properties. In a neutral atom, the atomic number, Z, is also the number of electrons. These electrons are found in a cloud surrounding the nucleus, located by probability in electron shells or orbitals. The shell farthest from the nucleus is the valence shell. The electrons in this valence shell are involved in chemical bonding and show the behavior of the atom. The bonding electrons influence the molecular geometry and structure of the atom. They interact with each other and with other atoms in chemical reactions. The atomic number is unique to each atom and defines its characteristics of bonding or behavior or reactivity. Therefore, every atom, with a different atomic number, acts in a different manner.
Mass Number
The mass of an atom is mostly localized to the nucleus. Because an electron has negligible mass relative to that of a proton or a neutron, the mass number is calculated by the sum of the number of protons and neutrons. Each proton and neutron's mass is approximately one atomic mass unit (AMU). The two added together results in the mass number:
$A=p^+ + n$
Elements can also have isotopes with the same atomic number, but different numbers of neutrons. There may be a few more or a few less neutrons, and so the mass is increased or decreased. On the periodic table, the mass number is usually located below the element symbol. The mass number listed is the average mass of all of the element's isotopes. Each isotope has a certain percentage abundance found in nature, and these are added and averaged to obtain the average mass number.
For example, 4He has a mass number of 4. Its atomic number is 2, which is not always included in the notation because He is defined by the atomic number 2.
Problems
1. How many protons, neutrons, and electrons do chlorine atoms have?
2. The mass of gold (Au) is 197, how many neutrons does it have?
3. Carbon has several isotopes. 14C has how many protons, electrons, and neutrons?
4. What is the atomic number of Li+? How many protons and electrons does Li+ have?
5. What does the mass number on the periodic table represent?
Answers
1. Because chlorine has an atomic number of 17, chlorine has 17 protons, 18, neutrons, and 17 electrons
2. 118 neutrons
3. 6 protons, 8 neutrons, and 6 electrons
4. Z=3, 3 protons, 2 electrons
5. The mass number represents the average mass of all of the isotopes of that particular element.
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The sub-atomic particles
The table below gives relative masses and relative charges for protons, neutrons, and electrons:
Relative Mass Relative Charge
Proton 1 +1
Neutron 1 0
Electron 1/1836 -1
In reality, protons and neutrons do not have exactly the same mass; neither of them has a mass of exactly 1 on the carbon-12 scale (the scale on which the relative masses of atoms are measured). On this scale, a proton has a mass of 1.0073, and a neutron a mass of 1.0087. The masses are given as 1 for simplicity and convenience.
Behavior of protons, neutrons and electrons in electric fields
If a beam containing each of these particles is passed between two electrically charged plates—one positive and one negative—the following are observed:
• Protons are positively charged and are thus deflected on a curving path towards the negative plate.
• Electrons are negatively charged and are deflected on a curving path towards the positive plate.
• Neutrons have no charge, and continue on in a straight line.
The exact manner in which these events occur depends on whether the particles have the same energy.
Particles with the same energy
If beams of the three types of particles, all with the same energy, are passed between two electrically charged plates, the following is observed:
• Protons are deflected on a curved path toward the negative plate.
• Electrons are deflected on a curved path toward the positive plate.
• Neutrons continue in a straight line.
If the particles have the same energy, the magnitude of the deflection is exactly the same in the electron beam as in the proton beam, but the deflections occur in opposite directions
If the electric field is strong enough, then the electron and proton beams could curve enough to hit their respective plates.
If the particles have the same speed
If beams of the three particles, all with the same speed, are passed between two electrically charged plates:
• Protons are deflected on a curved path toward the negative plate.
• Electrons are deflected on a curved path toward the positive plate.
• Neutrons continue in a straight line.
If the electrons and protons travel with the same speed, then the lighter electrons are deflected far more strongly than the heavier protons.
The nucleus
The nucleus, located at the center of the atom, contains the protons and neutrons. Protons and neutrons are collectively known as nucleons. Virtually all the mass of the atom is concentrated in the nucleus because electrons weigh so little in comparison to the nucleons.
Determining numbers of protons and neutrons
Number of protons = ATOMIC NUMBER of the atom
The atomic number is also given the more descriptive name of proton number.
Number of protons + number of neutrons = MASS NUMBER of the atom
The mass number is also called the nucleon number.
This information can be expressed in the following form:
The atomic number is the number of protons (9); the mass number counts protons + neutrons (19). If there are 9 protons, there must be 10 neutrons adding up to a total of 19 nucleons in the atom.
The atomic number is tied to the position of the element in the periodic table; the number of protons therefore defines the element of interest. If an atom has 8 protons (atomic number = 8), it must be oxygen. If an atom has 12 protons (atomic number = 12), it must be magnesium.
Similarly, every chlorine atom (atomic number = 17) has 17 protons; every uranium atom (atomic number = 92) has 92 protons.
Isotopes
The number of neutrons in an atom can vary within small limits. For example, there are three kinds of carbon atom: 12C, 13C and 14C. They all have the same number of protons, but the number of neutrons varies.
Protons Neutrons Mass Number
Carbon-12 6 6 12
Carbon-13 6 7 13
Carbon-14 6 8 14
Atoms with the same atomic number but different mass numbers are called isotopes. Varying numbers of neutrons nave no effect on the chemical properties of the atom.
The electrons
Determining the number of electrons
Atoms are electrically neutral, and the positive charge from the protons is balanced by negative charge from the electrons. It follows that in a neutral atom:
number of electrons = number of protons
Therefore, if an oxygen atom (atomic number = 8) has 8 protons, it must also have 8 electrons; if a chlorine atom (atomic number = 17) has 17 protons, it must also have 17 electrons.
The arrangement of the electrons
Electrons are found at considerable distances from the nucleus, arranged in successive energy levels. Each energy level can only hold a certain number of electrons. The first level (nearest the nucleus) holds two electrons, and the second and third levels each hold eight. These levels can be visualized as getting successively further from the nucleus. Electrons always occupy the lowest possible energy level (nearest the nucleus), provided there is space.
Determining the electronic arrangement of an atom
• Look up the atomic number in the periodic table. This is the number of protons, and hence the number of electrons.
• Arrange the electrons in levels, always filling up an inner level before filling an outer one.
This process is demonstrated using chlorine:
• The periodic table gives an atomic number of 17. Therefore, there are 17 protons and 17 electrons.
• The arrangement of the electrons is 2, 8, 7 (i.e. 2 in the first level, 8 in the second, and 7 in the third).
The electronic arrangements of the first 20 elements
After the third level, the pattern is altered due to the transition series.
Two Important Generalizations
Examining the patterns in the table above reveals the following:
• The number of electrons in the outermost level is the same as the group number (helium is the exception because it has only 2 electrons). This pattern extends throughout the periodic table for the main groups (excluding the transition metals and rare earth metals). Barium is in group 2, and it has 2 electrons in its outer level; iodine (group 7) has 7 electrons in its outer level; lead (group 4) has 4 electrons in its outer level.
• Noble gases have full outer levels.
Dots-and-crosses diagrams
Introductory chemistry courses generally depict the electronic structures of hydrogen and carbon in the following way:
This is a simplification and can be misleading. It gives the impression that the electrons are circling the nucleus in orbits like planets around the sun. It is impossible to know exactly how they are actually moving.
The circles show energy levels - representing increasing distances from the nucleus. If the circles are straightened, the electronic structure is shown in a simple energy diagram. The energy diagram for carbon is shown below:
This visualization of the arrangement of the electrons is useful in understanding electronic structure.
Sizes of Ions
Sizes of ions influence:
• packing of ions in ionic lattices, and therefore, the lattice energy
• biological recognition - some ions can pass through certain membrane channels, others may be too large
The size of an ion is influenced by:
• nuclear charge
• number of electrons
• valence orbitals
Cations
• formed by removing one or more valence electrons
• vacates the most spatially extended orbitals
• decreases the total electron-electron repulsion in the outer orbital
Cations are therefore smaller than the parent atom
Anions
• formed by addition of one or more valence electrons
• fills in the most spacially extended orbitals
• increases electron-electron repulsion in outer orbital
Anions are therefore larger than the parent atom
For ions of the same charge (e.g. in the same group) the size increases as we go down a group in the periodic table
As the principle quantum increases the size of both the parent atom and the ion will increase
How does the nuclear charge affect ion size?
Consider the following collection of ions:
ion electrons protons
O2- 10 8
F- 10 9
Na+ 10 11
Mg2+ 10 12
Al3+ 10 13
Each ion:
• contains the same number of electrons (10; with configuration 1s22s22p6) and are thus termed a collection of isoelectronic ions
• varies in the nuclear charge
The radius of each ion decreases with an increase in nuclear charge:
Oxygen and fluorine precede neon and are nonmetals, sodium, magnesium and aluminum come after neon and are metals.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Atomic_Theory/Simple_View_of_Atomic_Structure.txt
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The atom is the smallest unit of matter that is composed of three sub-atomic particles: the proton, the neutron, and the electron. Protons and neutrons make up the nucleus of the atom, a dense and positively charged core, whereas the negatively charged electrons can be found around the nucleus in an electron cloud.
The Structure of Atoms
An atom consists of a positively charged nucleus surrounded by one or more negatively charged particles called electrons. The number of protons found in the nucleus equals the number of electrons that surround it, giving the atom a neutral charge (neutrons have zero charge). Most of an atom’s mass is in its nucleus; the mass of an electron is only 1/1836 the mass of the lightest nucleus, that of hydrogen. Although the nucleus is heavy, it is small compared with the overall size of an atom.
The radius of a typical atom is around 1 to 2.5 angstroms (Å), whereas the radius of a nucleus is about 10-5 Å. If an atom were enlarged to the size of the earth, its nucleus would be only 200 feet in diameter and could easily fit inside a small football stadium.The nucleus of an atom contains protons and neutrons. Protons and neutrons have nearly equal masses, but they differ in charge. A neutron has no charge, whereas a proton has a positive charge that exactly balances the negative charge on an electron. Table 1 lists the charges of these three sub atomic particles, and gives their masses expressed in atomic mass units. The atomic mass unit (amu) is defined as exactly one-twelfth the mass of a carbon atom that has six protons and six neutrons in its nucleus. On this scale, protons and neutrons have masses that are close to, but not precisely, 1 u each. In fact, there are 6.022 x 1023 u in 1 gram. This number is known as Avogadro’s number, N. The number of protons in the nucleus of an atom is known as the atomic number, Z. It is the same as the number of electrons around the nucleus of the electrically-neutral atom. The mass number of an atom is equal to the total number of protons and neutrons.
Table 1: Charge and mass of three sub atomic particles
Particle Charge Mass (grams)
Electrons -1 9.1094x10-28
Protons +1 1.6726x10-24
Neutrons 0 1.6749x10-24
When two atoms are close enough to combine chemically and form chemical bonds with one another, each atom “sees” the outermost electrons of the other atom. These outer electrons, called valence electrons, are the most important factors in the chemical behavior of atoms. Both the neutron and the proton have key roles in the characteristics of an atom. The number of neutrons determine the type of isotope the element could be and the proton determines the number of electrons that surround its nucleus.
All atoms with the same atomic number exhibit the same chemical behaviors and are classified as the same chemical element. Each element has its own name and a one- or two-letter symbol (usually derived from the element’s English or Latin name). For example, the symbol for carbon is C, and the symbol for calcium is Ca. It is important to note that for elements with two letters, the first letter is capital and the second letter is lowercase. The symbol for sodium is Na—the first two letters of its Latin (and German) name, natriumto distinguish it from nitrogen, N, and sulfur, S.
Example
What is the atomic symbol for bromine, and what is its atomic number? Why is it incorrect to use the letter B? What other element preempts the symbol B? (Refer to the periodic table)
Solution
Bromine’s atomic number is 35, and its symbol is Br; B is the symbol for boron
Nucleus
The nucleus of an atom is comprised of protons and neutrons; it is therefore positively charged. The number of protons within the nucleus of a given atom is equal to the atomic number of the corresponding element, which can be found on the periodic table. For example, the atomic number of helium is two. Therefore, the number of protons is also two. The number of neutrons within the nucleus of a given atom can be found by subtracting the atomic number from the atomic mass. The mass number is the sum of protons and neutrons.
Atomic Mass Number = Number of Protons + Number of Neutrons
To find the number of neutrons, subtract the atomic number, the number of protons, from the mass number.
Notation of a specific element follows this format:
$^A_Z {\rm E} ^c$
where E is a specific element, A is mass number, Z is the atomic number, and C is the charge. For helium, the notation is as follows:
$^4_2 {\rm He}$
Helium has 2 protons, 2 neutrons and a charge of zero.
Isotopes
Atoms of the same element that have a different number of neutrons are known as isotopes. Most elements have several naturally occurring isotopes. The atomic mass of a particular element is equal to the average of the relative abundance of all its isotopes found in nature. For example, there are three naturally occurring isotopes of carbon: carbon-12, carbon-13, and carbon-14. Carbon-12 is the most common of these three, making up about 98.89% of all carbon, whereas carbon-13 has 1.11% natural abundance. Carbon-14 occurs rarely in nature. Atomic masses for other elements uses the carbon-12 scale as a reference. Early physicists assigned the atomic mass of 12 to the carbon-12 isotope (which is the most common carbon isotope) so that it would be easier to determine the atomic masses of other atoms. Using this information, we can determine the average atomic mass of carbon. (Use 13 for the approximate mass of carbon-13.)
average atomic mass = mass of carbon-12 x (% natural abundance/100) + mass of carbon-13 x (% natural abundance/100)
= 12 x .9889 + 13 x .0111 = 12.0111
Table 2 gives examples of common isotopes and their percent abundances in nature:
Table 2: Percent natural abundances of common isotopes
Element # of Neutrons Natural Abundance
Hydrogen-1 0 99.985%
Hydrogen-2 1 .015%
Boron-10 5 19.900%
Boron-11 6 80.100%
Carbon-12 6 98.890%
Carbon-13 7 1.110%
Oxygen-16 8 99.762%
Oxygen-17 9 .038%
Oxygen-18 10 .200%
Chlorine-35 18 75.770%
Chlorine-37 20 24.230%
Example
How many protons, neutrons, and electrons are there in an atom of uranium-238? Write the symbol for this isotope.
Solution
The atomic number of uranium (see periodic table) is 92, and the mass number of the isotope is 238. Therefore, it has 92 protons, 92 electrons, 146 neutrons (238 amu - 92 protons). Its symbol is $^{238}_{92}{\rm U}$ (or 238U).
The total mass of an atom is called its atomic weight, and this is approximately the sum of the masses of its constituent protons, neutrons, and electrons. When protons, neutrons, and electrons combine to form an atom, some of their mass is converted to energy and is given off (this is the source of energy in nuclear fusion reactions; because the atom cannot be broken down into its fundamental particles unless the energy for the missing mass is supplied from outside it, this energy is called the binding energy of the nucleus).
Each isotope of an element is characterized by an atomic number (total number of protons), a mass number (total number of protons and neutrons), and an atomic weight (mass of atom in atomic mass units). Because the reduction in mass upon the formation of an atom is small, the mass number is usually the same as the atomic weight rounded to the nearest integer (for example, the atomic weight of chlorine-37 is 36.966, which is rounded to 37). If there are several isotopes of an element in nature, then the experimentally observed atomic weight (the natural atomic weight) is the weighted average of the isotope weights. The average is weighted according to the percent abundance of the isotopes. In general, all isotopes of an element have similar chemical properties. Their behaviors differ with regard to mass-sensitive properties such as diffusion rates.
Example: Magnesium
Magnesium (Mg) has three significant natural isotopes: 78.70% of all magnesium atoms have an atomic weight of 23.985 u, 10.13% have an atomic weight of 24.986 u, and 11.17% have an atomic weight of 25.983 u. How many protons and neutrons are present in each of these three isotopes? What are the symbols for each isotope? What is the weighted average of the atomic weights?
Solution
There are 12 protons in each magnesium isotope. The isotope whose atomic weight is 23.985 u has a mass number of 24 (protons and neutrons), so 24 - 12 protons gives 12 neutrons. The symbol for this isotope is 24Mg. Similarly, the isotope whose atomic weight is 24.986 amu has a mass number of 25, 13 neutrons, and 25Mg as a symbol. The third isotope (25.983 amu) has a mass number of 26, 14 neutrons, and 26Mg as a symbol. We calculate the average atomic weight as follows:
(0.7870 x 23.985) + (0.1013 x 24.986) + (0.1117 x 25.983) = 24.31 u
Example: Boron
Boron has two naturally occurring isotopes, 10B and 11B. We know that 80.22% of its atoms are 11B, atomic weight 11.009 u. From the natural atomic weight given on the inside front cover, calculate the atomic weight of the 10B isotope.
Solution
If 80.22% of all boron atoms are 11B, then 100.00 - 80.22 or 19.78% are the unknown isotope, because the percentage abundance of isotopes must add up to 100%. In the periodic table the atomic weight of boron is found to be 10.81 u. W represents the unknown atomic weight:
$\begin{eqnarray} (0.8022 \times 11.009) + (0.1978 \times W) &=& 10.81 {\rm u} \quad {\rm (natural~atomic~weight)} \ W &=& \frac{10.81-8.831}{0.1978} \ &=& 10.01 {\rm u} \end{eqnarray}$
Electron Cloud
What appears to be a cloud of electrons surrounds the nucleus of an atom. The electron cloud serves as a model to help visualize the location of electrons in an atom. These electrons are held in place by their attraction to protons in the nucleus. This attraction is known as an electromagnetic force. Electrons move around the nucleus very quickly, creating the image of a cloud; however, they actually form electron shells. Each electron shell can hold a certain amount of electrons. The first shell holds two electrons, and subsequent shells hold eight electrons (the movement of electrons is discussed in Unit IV). The number of electrons in a given atom is equal to the number of protons in that atom. However, units of the same element can have different numbers of electrons. Such units are known as ions. Atoms that lose one or more electrons become positively charged and are called cations. Cations are smaller than the original atom due to the loss of an electron. Examples of cations include Ca2+ and Al3+ (note: An H+ cation is simply a proton). Atoms can also gain electrons, forming a negative ion known as an anion. Anions are bigger that the original atoms due to electron gain. Examples of anions include Cl- and O2-. Although some elements may only gain one electron, some atoms can gain up to four electrons.
Problems
1. Complete the chart below.
Element Protons Neutrons Electrons Charge
$^3_2 {\rm He}$ 1 2 0
$^{129}_{53} {\rm I} ^-$ 53 54 -1
$^{30}_{14} {\rm Si}$ 14
18 18 0
$^9_4 {\rm Be} ^{2+}$ 2
3 3 0
$^{80}_{35} {\rm Br} ^+$ 35 45 -1
$^{27}_{13} {\rm Al} ^{3+}$ 13 10
$^{35}_{17} {\rm Cl} ^-$ 18 18 -1
$^{15}_7 {\rm N}$ 7 7
82 126 +2
$^{10}_5 {\rm B}$
2. Boron has two naturally occurring isotopes, boron-10 and boron-11, with the masses of 10.013 g and 11.009 g, respectively. What is the average atomic mass of boron?
3. What is the mass of chlorine-37, if the mass of chlorine-35 is 34.969 g and the atomic mass of chlorine is 35.453 g? (Hint: Refer to table.1.)
4. For the atom 196Pt4+ , indicate the number of protons, neutrons and electrons.
5. Write the notation for an atom containing 24 protons, 28 neutrons and 21 electrons.
Answers:
1.
Element Protons Neutrons Electrons Charge
$^3_2 {\rm He}$ 2 1 2 0
$^{129}_{53} {\rm I} ^-$ 53 76 54 -1
$^{30}_{14} {\rm Si}$ 14 16 14 0
$^{36}_{18} {\rm Ar}$ 18 18 18 0
$^9_4 {\rm Be} ^{2+}$ 4 5 2 2+
$^6_3 {\rm Li}$ 3 3 3 0
$^{80}_{35} {\rm Br} ^+$ 35 45 36 -1
$^{27}_{13} {\rm Al} ^{3+}$ 13 14 10 3+
$^{35}_{17} {\rm Cl} ^-$ 17 18 18 -1
$^{15}_7 {\rm N}$ 7 8 7 0
$^{208}_{82} {\rm Pb} ^{2+}$ 82 126 80 2+
$^{10}_5 {\rm B}$ 5 5 5 0
2. Referring to table.1, the percent natural abundance of boron-10 is 19.9% and the percent natural abundance of boron-11 is 80.1%. Multiply the given masses by their respective percents (in decimal form) to find the average.
(10.0129 x .199) + (11.00931 x .801) = 10.811
The average atomic mass of boron is 10.811 g.
3. If we refer to table.1, we find the percent natural abundance of chlorine-35 is 75.77% and the percent natural abundance for chlorine-37 is 24.23. Using the same general formula as in the previous problem, solve for the mass of chlorine-37 (represented by n)/;
(34.969 x .7577) + (n x .2423) = 35.453
26.496 + .2423n = 35.453
.2423n= 8.957
n = 36.967
The mass of chlorine-37 is 36.967 g.
4. Number of protons = 78.
Number of neutrons = 118.
Number of electrons = 74.
5. 52Cr3+
The Atom
A typical atom consists of three subatomic particles: protons, neutrons, and electrons (as seen in the helium atom below). Other particles exist as well, such as alpha and beta particles (which are discussed below). The Bohr model shows the three basic subatomic particles in a simple manner. Most of an atom's mass is in the nucleus—a small, dense area at the center of every atom, composed of nucleons. Nucleons include protons and neutrons. All the positive charge of an atom is contained in the nucleus, and originates from the protons. Neutrons are neutrally-charged. Electrons, which are negatively-charged, are located outside of the nucleus.
Introduction
The Bohr model is outdated, but it depicts the three basic subatomic particles in a comprehensible way. Electron clouds are more accurate representations of where electrons are found. Darker areas represent where the electrons are more likely to be found, and lighter areas represent where they are less likely to be found.
Particle Electric Charge (C) Atomic Charge Mass (g) Atomic Mass (Au) Spin
Protons +1.6022 x 10-19 +1 1.6726 x 10-24 1.0073 1/2
Neutrons 0 0 1.6740 x 10-24 1.0078 1/2
Electrons -1.6022 x 10-19 -1 9.1094 x 10-28 0.00054858 1/2
• Au is the SI symbol for atomic mass unit.
• The positive charge of protons cancels the negative charge of the electrons. Neutrons have no charge.
• With regard to mass, protons and neutrons are very similar, and have a much greater mass than electrons. Compared with neutrons and protons, the mass of an electron is usually negligible.
• Spin is associated with the rotation of a particle. Protons, neutrons, and electrons each have a total spin of 1/2.
Protons
Protons were discovered by Ernest Rutherford in the year 1919, when he performed his gold foil experiment. He projected alpha particles (helium nuclei) at gold foil, and the positive alpha particles were deflected. He concluded that protons exist in a nucleus and have a positive nuclear charge. The atomic number or proton number is the number of protons present in an atom. The atomic number determines an element (e.g., the element of atomic number 6 is carbon).
Electrons
Electrons were discovered by Sir John Joseph Thomson in 1897. After many experiments involving cathode rays, J.J. Thomson demonstrated the ratio of mass to electric charge of cathode rays. He confirmed that cathode rays are fundamental particles that are negatively-charged; these cathode rays became known as electrons. Robert Millikan, through oil drop experiments, found the value of the electronic charge.
Electrons are located in an electron cloud, which is the area surrounding the nucleus of the atom. There is usually a higher probability of finding an electron closer to to the nucleus of an atom. Electrons can abbreviated as e-. Electrons have a negative charge that is equal in magnitude to the positive charge of the protons. However, their mass is considerably less than that of a proton or neutron (and as such is usually considered insignificant). Unequal amounts of protons and electrons create ions: positive cations or negative anions.
Neutrons
Neutrons were discovered by James Chadwick in 1932, when he demonstrated that penetrating radiation incorporated beams of neutral particles. Neutrons are located in the nucleus with the protons. Along with protons, they make up almost all of the mass of the atom. The number of neutrons is called the neutron number and can be found by subtracting the proton number from the atomic mass number. The neutrons in an element determine the isotope of an atom, and often its stability. The number of neutrons is not necessarily equal to the number of protons.
Identification
Both of the following are appropriate ways of representing the composition of a particular atom:
Often the proton number is not indicated because the elemental symbol conveys the same information.
Consider a neutral atom of carbon: $\ce{^{12}_{6}C}$. The atomic mass number of Carbon is 12 amu, the proton number is 6, and it has no charge. In neutral atoms, the charge is omitted.
Above is the atomic symbol for helium from the periodic table, with the atomic number, elemental symbol, and mass indicated.
Every element has a specific number of protons, so the proton number is not always written (as in the second method above).
• # Neutrons = Atomic Mass Number - Proton Number
• Atomic mass number is abbreviated as A.
• Proton number(or atomic number) is abbreviated Z.
• # Protons = Proton Number or Atomic Number
• In neutral atoms, # Electrons = # Protons
• In ions, # Electrons = # Protons - (Charge)
• Charge is written with the number before the positive or negative sign
• Example, 1+
Note: The atomic mass number is not the same as the atomic mass seen on the periodic table. Click here for more information.
Other Basic Atomic Particles
Many of these particles (explained in detail below) are emitted through radioactive decay. Click here for more information. Also note that many forms of radioactive decay emit gamma rays, which are not particles.
Alpha Particles
Alpha particles can be denoted by He2+,α2+, or just α. They are helium nuclei, which consist of two protons and two neutrons. The net spin on an alpha particle is zero. They result from large, unstable atoms through a process called alpha decay. Alpha decay is the process by which an atom emits an alpha particle, thereby becoming a new element. This only occurs in elements with large, radioactive nuclei. The smallest noted element that emits alpha particles is element 52, tellurium. Alpha particles are generally not harmful. They can be easily stopped by a single sheet of paper or by one's skin. However, they can cause considerable damage to the insides of one's body. Alpha decay is used as a safe power source for radioisotope generators used in artificial heart pacemakers and space probes.
Beta Particles
Beta particles (β) are either free electrons or positrons with high energy and high speed; they are emitted in a process called beta decay. Positrons have the exact same mass as an electron, but are positively-charged. There are two forms of beta decay: the emission of electrons, and the emission of positrons. Beta particles, which are 100 times more penetrating than alpha particles, can be stopped by household items like wood or an aluminum plate or sheet. Beta particles have the ability to penetrate living matter and can sometimes alter the structure of molecules they strike. The alteration usually is considered damage, and can cause cancer and death. In contrast to beta particle's harmful effects, they can also be used in radiation to treat cancer.
Beta- (β-) or Electron Emission
Electron emission may result when excess neutrons make the nucleus of an atom unstable. As a result, one of the neutrons decays into a proton, an electron, and an anti-neutrino. The proton remains in the nucleus, and the electron and anti-neutrino are emitted. The electron is called a beta particle. The equation for this process is given below:
$_{1}^{0}\textrm{n}\rightarrow {_{1}^{1}\textrm{p}}^+ + \textrm{e}^- + \bar{\nu_{e}}$
• n = Neutron
• p+ = Proton
• e- = Electron (beta particle)
• νe = Anti-neutrino
β- Decay
Beta+(β+) or Positron Emission
Position emission occurs when an excess of protons makes the atom unstable. In this process, a proton is converted into a neutron, a positron, and a neutrino. While the neutron remains in the nucleus, the positron and the neutrino are emitted. The positron can be called a beta particle in this instance. The equation for this process is given below:
${ _{1}^{1}\textrm{p}}^+ \rightarrow _{1}^{0}\textrm{n} + \textrm{e}^+ + \nu_{e}$
• n = Neutron
• p+ = Proton
• e+ = Positron (beta particle)
• νe = Neutrino
β+ Decay
Outside Links
• Basic Sub-Atomic Particles: www.youtube.com/watch?v=lP57g...eature=related
• Alpha Particles: en.Wikipedia.org/wiki/Alpha_decay
• Beta Particles: en.Wikipedia.org/wiki/Beta_particle
• What are Sub-Atomic Particles?: www.youtube.com/watch?v=uXcOqjCQzh8
• Atomic Number and Mass Number: www.youtube.com/watch?v=lDo78hPTlgk
Problems
1. Identify the number of protons, electrons, and neutrons in the following atom.
2. Identify the subatomic particles (protons, electrons, neutrons, and positrons) present in the following:
• $\ce{^{14}_6C}$
• $\alpha$
• $\ce{^{35}Cl^-}$
• $\beta^+$
• $\beta^-$
• $\ce{^{24}Mg^{2+}}$
• $\ce{^{60}Co}$
• $\ce{^3H}$
• $\ce{^{40}Ar}$
• $^1_0n$
3. Given the following, identify the subatomic particles present. (The periodic table is required to solve these problems)
• Charge +1, 3 protons, mass number 6.
• Charge -2, 7 neutrons, mass number 17.
• 26 protons, 20 neutrons.
• 28 protons, mass number 62.
• 5 electrons, mass number 10.
• Charge -1, 18 electrons, mass number 36.
4. Arrange the following elements in order of increasing (a) number of protons; (b) number of neutrons; (c) mass.
27Co, when A=59; 56Fe, when Z=26; 11Na, when A=23; 80Br, when Z=35; 29Cu, when A=30; 55Mn, when Z=25
5. Fill in the rest of the table:
Atomic Number Mass Number Number of Protons Number of Neutrons Number of Electrons
2 2
23 11
15 16
85 37
53 74
Solutions and Explanations
1. There are 4 protons, 5 neutrons, and 4 electrons. This is a neutral beryllium atom.
2. Identify the subatomic particles present in the following:
• 146C
• 6 protons, 8 neutrons, 6 electrons
• There are 6 protons in accordance with the proton number in the subscript. There are 6 electrons because the atom is neutral. There are 8 neutrons because 14-6=8. 14 is the atomic mass number in the superscript.
• α
• 2 protons, 2 neutrons, 0 electrons
• This is an alpha particle which can also be written as 4He2+. There are two protons because the element is helium. There are no electrons because 2-2 = 0. There are 2 neutrons because 4-2=2.
• 35Cl-
• 17 protons, 18 neutrons, 18 electrons
• This is a chloride ion. According to the periodic table, there are 17 protons because the element is chlorine. There are 18 electrons due to the negative charge: 17-(-1) = 18. There are 18 neutrons because 35-17=18.
• β+
• 0 protons, 0 neutrons, 0 electrons, 1 positron
• This is a beta+ particle. It can also be written as e+. "e" represents an electron, but when it has as positive charge it is a positron.
• β-
• 0 protons, 0 neutrons, 1 electron
• This is a beta- particle, and can also be written as e-. This is a standard electron.
• 24Mg2+
• 12 protons, 12 neutrons, 10 electrons
• This is a magnesium Ion. There are 12 protons from the magnesium atom. There are 10 electrons because 12-2 = 10. There are 12 neutrons because 24-12 = 12.
• >60Co
• 27 protons, 33 neutrons, 27 electrons
• The cobalt atom has 27 protons as seen in the periodic table. There are also 27 electrons because the charge is 0. There are 33 neutrons because 60-27 = 33.
• 3H
• 1 protons, 2 neutrons, 1 electrons
• There is 1 proton because the element is hydrogen. There is 1 electron because the atom is neutral. There are 2 neutrons because 3-1 = 2.
• 40Ar
• 18 protons, 22 neutrons, 18 electrons
• There are 18 protons from the argon element. There 18 electrons because it is neutral, and 22 neutrons because 40 - 18 = 22.
• n
• 0 protons, 1 neutrons, 0 electrons
• This is a free neutron denoted by the lower case n.
3. Given the following, identify the subatomic particles present. (The periodic table is required to solve these problems)
• Charge +1, 3 protons, mass number 6.
• 3 protons, 3 neutrons, 2 electrons
• Charge -2, 8 neutrons, mass number 17.
• 9 protons, 8 neutrons, 7 electrons
• 26 protons, 20 neutrons.
• 26 protons, 20 neutrons, 26 electrons
• 28 protons, mass number 62.
• 28 protons, 34 neutrons, 28 electrons
• 5 electrons, mass number 10.
• 5 protons, 5 neutrons, 5 electrons
• Charge -1, 18 electrons, mass number 36.
• 17 protons, 19 neutrons, 18 electrons
4. Arrange the following lements in order of increasing (a) number of protons; (b) number of neutrons; (c) atomic mass.
a) Na, Mn, Fe, Co, Cu, Br
• Z=#protons;
• Na: z=11; Mn: Z=25, given; Fe: Z=26, given; Co: Z=27; Cu: Z=29; Br: Z=35, given
b) Na, Cu, Fe, Mn, Co, Br
• A=#protons+#neutrons, so #n=A-#protons(Z);
• Na: #n=23-11=12; Cu: #n=59-29=30; Fe: #n=56-26=30; Mn: #n=55-25=30; Co: #n=59-27=32; Br: #n=80-35=45
Note: Cu, Fe, Mn are all equal in their number of neutrons, which is 30.
c) Na, Mn, Fe, Co, Cu, Br
• Na: 22.9898 amu; Mn: 54.9380 amu; Fe: 55.845 amu; Co: 58.9332 amu; Cu: 63.546 amu; Br: 79.904
Note: This is the same order as the number of protons, because as Atomic Number(Z) increases so does Atomic Mass.
5. Fill in the rest of the table:
Atomic Number Mass Number Number of Protons Number of Neutrons Number of Electrons
2 4 2 2 2
11 23 11 12 11
15 31 15 16 15
37 85 37 48 37
53 127 53 74 53
Note: Atomic Number=Number of Protons=Number of Electrons and Mass Number=Number of Protons+Number of Neutrons
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The number of moles in a system can be determined using the atomic mass of an element, which can be found on the periodic table. This mass is usually an average of the abundant forms of that element found on earth. An element's mass is listed as the average of all its isotopes on earth.
Avogadro's Constant
One mole of oxygen atoms contains $6.02214179 \times 10^{23}$ oxygen atoms. Also, one mole of nitrogen atoms contains $6.02214179 \times 10^{23}$ nitrogen atoms. The number $6.02214179 \times 10^{23}$ is called Avogadro's number ($N_A$) or Avogadro's constant, after the 19th century scientist Amedeo Avogadro.
Each carbon-12 atom weighs about $1.99265 \times 10^{-23}\; g$; therefore,
$(1.99265 \times 10^{-23}\; g) \times (6.02214179 \times 10^{23}\; atoms) = 12\; g\; \text{ of carbon-12} \nonumber$
Applications of the Mole
The mass of a mole of substance is called the molar mass of that substance. The molar mass is used to convert grams of a substance to moles and is used often in chemistry. The molar mass of an element is found on the periodic table, and it is the element's atomic weight in grams/mole (g/mol). If the mass of a substance is known, the number of moles in the substance can be calculated. Converting the mass, in grams, of a substance to moles requires a conversion factor of (one mole of substance/molar mass of substance).
The mole concept is also applicable to the composition of chemical compounds. For instance, consider methane, CH4. This molecule and its molecular formula indicate that per mole of methane there is 1 mole of carbon and 4 moles of hydrogen. In this case, the mole is used as a common unit that can be applied to a ratio as shown below:
$2 \text{ mol H } + 1 \text{ mol O }= 1 \text{ mol } \ce{H2O} \nonumber$
In this this chemical reactions, the moles of H and O describe the number of atoms of each element that react to form 1 mol of $\ce{H_2O}$.
To think about what a mole means, one should relate it to quantities such as dozen or pair. Just as a pair can mean two shoes, two books, two pencils, two people, or two of anything else, a mole means 6.02214179×1023 of anything. Using the following relation:
$\text{1 mole} = 6.02214179 \times 10^{23}$
is analogous to saying:
$\text{1 Dozen} = \text{12 eggs}$
It is quite difficult to visualize a mole of something because Avogadro's constant is extremely large. For instance, consider the size of one single grain of wheat. If all the people who have existed in Earth's history did nothing but count individual wheat grains for their entire lives, the total number of wheat grains counted would still be much less than Avogadro's constant; the number of wheat grains produced throughout history does not even approach Avogadro's Number.
Example $1$: Converting Mass to Moles
How many moles of potassium ($\ce{K}$) atoms are in 3.04 grams of pure potassium metal?
Solution
In this example, multiply the mass of $\ce{K}$ by the conversion factor (inverse molar mass of potassium):
$\dfrac{1\; mol\; K}{39.10\; grams \;K} \nonumber$
39.10 grams is the molar mass of one mole of $\ce{K}$; cancel out grams, leaving the moles of $\ce{K}$:
$3.04\; \cancel{g\; K} \left(\dfrac{1\; mol\; K}{39.10\; \cancel{g\; K}}\right) = 0.0778\; mol\; K \nonumber$
Similarly, if the moles of a substance are known, the number grams in the substance can be determined. Converting moles of a substance to grams requires a conversion factor of molar mass of substance/one mole of substance. One simply needs to follow the same method but in the opposite direction.
Example $2$: Converting Moles to mass
How many grams are 10.78 moles of Calcium ($\ce{Ca}$)?
Solution
Multiply moles of Ca by the conversion factor (molar mass of calcium) 40.08 g Ca/ 1 mol Ca, which then allows the cancelation of moles, leaving grams of Ca.
$10.78 \cancel{\;mol\; Ca} \left(\dfrac{40.08\; g\; Ca}{1\; \cancel{mol\; Ca}}\right) = 432.1\; g\; Ca \nonumber$
The total number of atoms in a substance can also be determined by using the relationship between grams, moles, and atoms. If given the mass of a substance and asked to find the number of atoms in the substance, one must first convert the mass of the substance, in grams, to moles, as in Example $1$. Then the number of moles of the substance must be converted to atoms. Converting moles of a substance to atoms requires a conversion factor of Avogadro's constant (6.02214179×1023) / one mole of substance. Verifying that the units cancel properly is a good way to make sure the correct method is used.
Example $3$: Atoms to Mass
How many atoms are in a 3.5 g sample of sodium (Na)?
Solution
$3.5\; \cancel{g\; Na} \left(\dfrac{1\; mol\; Na}{22.98\; \cancel{g\; Na}}\right) = 0.152\; mol\; Na \nonumber$
$0.152\; \cancel{mol\; Na} \left(\dfrac{6.02214179\times 10^{23}\; atoms\; Na}{1\;\cancel{ mol\; Na}}\right) = 9.15 \times 10^{22}\; atoms\; of\; Na \nonumber$
In this example, multiply the grams of Na by the conversion factor 1 mol Na/ 22.98 g Na, with 22.98g being the molar mass of one mole of Na, which then allows cancelation of grams, leaving moles of Na. Then, multiply the number of moles of Na by the conversion factor 6.02214179×1023 atoms Na/ 1 mol Na, with 6.02214179×1023 atoms being the number of atoms in one mole of Na (Avogadro's constant), which then allows the cancelation of moles, leaving the number of atoms of Na.
Using Avogadro's constant, it is also easy to calculate the number of atoms or molecules present in a substance (Table $1$). By multiplying the number of moles by Avogadro's constant, the mol units cancel out, leaving the number of atoms. The following table provides a reference for the ways in which these various quantities can be manipulated:
Table $1$: Conversion Factors
Known Information Multiply By Result
Mass of substance (g) 1/ Molar mass (mol/g) Moles of substance
Moles of substance (mol) Avogadro's constant (atoms/mol) Atoms (or molecules)
Mass of substance (g) 1/Molar mass (mol/g) × Avogadro's constant (atoms/mol)) Atoms (or molecules)
Example $4$: Mass to Moles
How many moles are in 3.00 grams of potassium (K)?
Solution
$3.00 \; \cancel{g\; K} \left(\dfrac{1\; mol\; K}{39.10\; \cancel{g\; K}}\right) = 0.0767\; mol\; K \nonumber$
In this example, multiply the mass of K by the conversion factor:
$\dfrac{1\; mol\; K}{39.10\; grams\; K} \nonumber$
39.10 grams is the molar mass of one mole of K. Grams can be canceled, leaving the moles of K.
Example $5$: Moles to Mass
How many grams is in 10.00 moles of calcium (Ca)?
Solution
This is the calculation in Example $2$ performed in reverse. Multiply moles of Ca by the conversion factor 40.08 g Ca/ 1 mol Ca, with 40.08 g being the molar mass of one mole of Ca. The moles cancel, leaving grams of Ca:
$10.00\; \cancel{mol\; Ca} \left(\dfrac{40.08\; g\; Ca}{1\;\cancel{ mol\; Ca}}\right) = 400.8\; grams \;of \;Ca \nonumber$
The number of atoms can also be calculated using Avogadro's Constant (6.02214179×1023) / one mole of substance.
Example $6$: Mass to Atoms
How many atoms are in a 3.0 g sample of sodium (Na)?
Solution
Convert grams to moles
$3.0\; \cancel{g\; Na} \left(\dfrac{1\; mol\; Na}{22.98\; \cancel{g\; Na}}\right) = 0.130\; mol\; Na \nonumber$
Convert moles to atoms
$0.130548\; \cancel{ mol\; Na} \left(\dfrac{6.02214179 \times 10^{23}\; atoms \;Na}{1\; \cancel{ mol\; Na}}\right) = 7.8 \times 10^{22} \; atoms\; of\; \; Na \nonumber$
Summary
The mole, abbreviated mol, is an SI unit which measures the number of particles in a specific substance. One mole is equal to $6.02214179 \times 10^{23}$ atoms, or other elementary units such as molecules.
Problems
1. Using a periodic table, give the molar mass of the following:
1. H
2. Se
3. Ne
4. Cs
5. Fe
2. Convert to moles and find the total number of atoms.
1. 5.06 grams of oxygen
2. 2.14 grams of K
3. 0.134 kg of Li
3. Convert the following to grams
1. 4.5 mols of C
2. 7.1 mols of Al
3. 2.2 mols of Mg
4. How many moles are in the product of the reaction
1. 6 mol H + 3 mol O → ? mol H2O
2. 1 mol Cl + 1 mol Cl → ? mol Cl2
3. 5 mol Na + 4 mol Cl → ? mol NaCl
Answers
1. Question 2
1. 1.008 g/mol
2. 78.96 g/mol
3. 20.18 g/mol
4. 132.91g/mol
5. 55.85 g/mol
2. Question 2
2. 5.06g O (1mol/16.00g)= 0.316 mol of O
0.316 mols (6.022x1023 atoms/ 1mol) = 1.904x1023 atoms of O
3. 2.14g K (1mol/39.10g)= 0.055 mol of K
0.055 mols (6.022x1023 atoms/ 1mol) = 3.312x1022 atoms of K
4. 0.134kg Li (1000g/1kg)= 134g Li (1mol/6.941g)= 19.3 mols Li
19.3 (6.022x1023 atoms/ 1mol) = 1.16x1025 atoms of Li
1. Question 3
1. 4.5 mols of C (12.011g/1mol) = 54.05 g of C
2. 7.1 mols of Al (26.98g/1mol) = 191.56 g of Al
3. 2.2 mols of Mg (24.31g/1mol) = 53.48 g of MG
2. Question 4
1. 8. 6 mol H + 3 mol O → 3 mol H2O
2. 9. 1 mol Cl + 1 mol Cl → 1 mol Cl2
3. 10. 5 mol Na + 4 mol Cl → 4 mol NaCl + 1 mol Na (excess)
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A chemical bond is the force that holds atoms together in chemical compounds. There are two idealized types of bonding: (1) covalent bonding, in which electrons are shared between atoms in a molecule or polyatomic ion, and (2) ionic bonding, in which positively and negatively charged ions are held together by electrostatic forces.
• Band Structure
Band Theory was developed with some help from the knowledge gained during the quantum revolution in science. In 1928, Felix Bloch had the idea to take the quantum theory and apply it to solids. In 1927, Walter Heitler and Fritz London discovered bands- very closely spaced orbitals with not much difference in energy.
• Bond Energies
The bond energy is a measure of the amount of energy needed to break apart one mole of covalently bonded gases. Energy is released to generate bonds, which is why the enthalpy change for breaking bonds is positive. Energy is required to break bonds. Atoms are much happier when they are "married" and release energy because it is easier and more stable to be in a relationship (e.g., to generate octet electronic configurations).
• Bond Order and Lengths
Bond order is the number of chemical bonds between a pair of atoms and indicates the stability of a bond. For example, in diatomic nitrogen, N≡N, the bond order is 3; in acetylene, H−C≡C−H, the carbon-carbon bond order is also 3, and the C−H bond order is 1. Bond order and bond length indicate the type and strength of covalent bonds between atoms. Bond order and length are inversely proportional to each other: when bond order is increased, bond length is decreased.
• Chemical Bonds
• Contrasting MO and VB theory
Both the MO and VB theories are used to help determine the structure of a molecule. Unlike the VB theory, which is largely based off of valence electrons, the MO theory describes structure more in depth by taking into consideration, for example, the overlap and energies of the bonding and antibonding electrons residing in a particular molecular orbital.
• Coordinate (Dative Covalent) Bonding
A coordinate bond (also called a dative covalent bond) is a covalent bond (a shared pair of electrons) in which both electrons come from the same atom.
• Covalent Bonding
This page explains what covalent bonding is. It starts with a simple picture of the single covalent bond.
• Covalent Bonds
Covalent bonding occurs when pairs of electrons are shared by atoms. Atoms will covalently bond with other atoms in order to gain more stability, which is gained by forming a full electron shell. By sharing their outer most (valence) electrons, atoms can fill up their outer electron shell and gain stability. Nonmetals will readily form covalent bonds with other nonmetals in order to obtain stability.
• Covalent Bonds vs Ionic Bonds
• Covalent Bond Distance, Radius and van der Waals Radius
• Electrostatic Potential maps
Electrostatic potential maps, also known as electrostatic potential energy maps, or molecular electrical potential surfaces, illustrate the charge distributions of molecules three dimensionally. These maps allow us to visualize variably charged regions of a molecule. Knowledge of the charge distributions can be used to determine how molecules interact with one another.
• Ionic Bonds
Ionic bonding is the complete transfer of valence electron(s) between atoms and is a type of chemical bond that generates two oppositely charged ions. It is observed because metals with few electrons in its outer-most orbital. By losing those electrons, these metals can achieve noble-gas configuration and satisfy the octet rule. Similarly, nonmetals that have close to 8 electrons in its valence shell tend to readily accept electrons to achieve its noble gas configuration.
• Metallic Bonding
A strong metallic bond will be the result of more delocalized electrons, which causes the effective nuclear charge on electrons on the cation to increase, in effect making the size of the cation smaller. Metallic bonds are strong and require a great deal of energy to break, and therefore metals have high melting and boiling points. A metallic bonding theory must explain how so much bonding can occur with such few electrons (since metals are located on the left side of the periodic table and do
• Non-Singular Covalent Bonds
There are three types of covalent bonds: single, double, and triple. The name "Non-singular covalent bonds" speaks for itself. Non-singular covalent bonds are covalent bonds that need to share more then one electron pair, so they create double and triple bonds.
• Valence-Shell Electron-Pair Repulsion Models
The valence-shell electron-pair repulsion (VSEPR) model is used to predict three-dimensional arrangements of atoms or bonds in molecules including bond lengths, bond angles and qualitative bond energies.
Fundamentals of Chemical Bonding
Band Theory was developed with some help from the knowledge gained during the quantum revolution in science. In 1928, Felix Bloch had the idea to take the quantum theory and apply it to solids. In 1927, Walter Heitler and Fritz London discovered bands- very closely spaced orbitals with not much difference in energy.
In this image, orbitals are represented by the black horizontal lines, and they are being filled with an increasing number of electrons as their amount increases. Eventually, as more orbitals are added, the space in between them decreases to hardly anything, and as a result, a band is formed where the orbitals have been filled.
Different metals will produce different combinations of filled and half filled bands.
Sodium's bands are shown with the rectangles. Filled bands are colored in blue. As you can see, bands may overlap each other (the bands are shown askew to be able to tell the difference between different bands). The lowest unoccupied band is called the conduction band, and the highest occupied band is called the valence band.
Bands will follow a trend as you go across a period:
• In Na, the 3s band is 1/2 full.
• In Mg, the 3s band is full.
• In Al, the 3s band is full and the 3p ban is 1/2 full... and so on.
The probability of finding an electron in the conduction band is shown by the equation:
\[ P= \dfrac{1}{e^{ \Delta E/RT}+1} \]
The ∆E in the equation stands for the change in energy or energy gap. t stands for the temperature, and R is a bonding constant. That equation and this table below show how the bigger difference in energy is, or gap, between the valence band and the conduction band, the less likely electrons are to be found in the conduction band. This is because they cannot be excited enough to make the jump up to the conduction band.
ELEMENT ∆E(kJ/mol) of energy gap # of electrons/cm^3 in conduction band insulator, or conductor?
C (diamond) 524 (big band gap) 10-27 insulator
Si 117 (smaller band gap, but not a full conductor) 109 semiconductor
Ge 66 (smaller band gap, but still not a full conductor) 1013 semiconductor
Conductors, Insulators and Semiconductors
A. Conductors
Metals are conductors. There is no band gap between their valence and conduction bands, since they overlap. There is a continuous availability of electrons in these closely spaced orbitals.
B. Insulators
In insulators, the band gap between the valence band the the conduction band is so large that electrons cannot make the energy jump from the valence band to the conduction band.
C. Semiconductors
Semiconductors have a small energy gap between the valence band and the conduction band. Electrons can make the jump up to the conduction band, but not with the same ease as they do in conductors.
There are two different kinds of semiconductors: intrinsic and extrinsic.
i. Intrinsic Semiconductors
An intrinsic semiconductor is a semiconductor in its pure state. For every electron that jumps into the conduction band, the missing electron will generate a hole that can move freely in the valence band. The number of holes will equal the number of electrons that have jumped.
ii. Extrinsic Semiconductors
In extrinsic semiconductors, the band gap is controlled by purposefully adding small impurities to the material. This process is called doping. Doping, or adding impurities to the lattice can change the electrical conductivity of the lattice and therefore vary the efficiency of the semiconductor. In extrinsic semiconductors, the number of holes will not equal the number of electrons jumped. There are two different kinds of extrinsic semiconductors, p-type (positive charge doped) and n-type (negative charge doped).
Problems
1. How do you distinguish between a valence band and a conduction band?
2. Is the energy gap between an insulator smaller or larger than the energy gap between a semiconductor?
3. What two methods bring conductivity to semiconductors?
4. You are more likely to find electrons in a conduction band if the energy gap is smaller/larger? 5. The property of being able to be drawn into a wire is called...
Answers
1. The valence band is the highest band with electrons in it, and the conduction band is the highest band with no electrons in it.
2. Larger
3. Electron transport and hole transport
4. Smaller
5. Ductility
Contributors and Attributions
• Sierra Blair (UCD)
Jim Clark (Chemguide.co.uk)
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Atoms bond together to form compounds because in doing so they attain lower energies than they possess as individual atoms. A quantity of energy, equal to the difference between the energies of the bonded atoms and the energies of the separated atoms, is released, usually as heat. That is, the bonded atoms have a lower energy than the individual atoms do. When atoms combine to make a compound, energy is always given off, and the compound has a lower overall energy.
When a chemical reaction occurs, molecular bonds are broken and other bonds are formed to make different molecules. For example, the bonds of two water molecules are broken to form hydrogen and oxygen.
$2H_2O \rightarrow 2H_2 + O_2$
Energy is always required to break a bond, which is known as bond energy. While the concept may seem simple, bond energy serves a very important purpose in describing the structure and characteristics of a molecule. It can be used to determine which Lewis Dot Structure is most suitable when there are multiple Lewis Dot Structures.
Energy is always required to break a bond. Energy is released when a bond is made.
Although each molecule has its own characteristic bond energy, some generalizations are possible. For example, although the exact value of a C–H bond energy depends on the particular molecule, all C–H bonds have a bond energy of roughly the same value because they are all C–H bonds. It takes roughly 100 kcal of energy to break 1 mol of C–H bonds, so we speak of the bond energy of a C–H bond as being about 100 kcal/mol. A C–C bond has an approximate bond energy of 80 kcal/mol, while a C=C has a bond energy of about 145 kcal/mol. We can calculate a more general bond energy by finding the average of the bond energies of a specific bond in different molecules to get the average bond energy.
Table 1: Average Bond Energies (kJ/mol)
Single Bonds Multiple Bonds
H—H
432
N—H
391
I—I
149
C = C
614
H—F
565
N—N
160
I—Cl
208
C ≡ C
839
H—Cl
427
N—F
272
I—Br
175
O = O
495
H—Br
363
N—Cl
200
C = O*
745
H—I
295
N—Br
243
S—H
347
C ≡ O
1072
N—O
201
S—F
327
N = O
607
C—H
413
O—H
467
S—Cl
253
N = N
418
C—C
347
O—O
146
S—Br
218
N ≡ N
941
C—N
305
O—F
190
S—S
266
C ≡ N
891
C—O
358
O—Cl
203
C = N
615
C—F
485
O—I
234
Si—Si
340
C—Cl
339
Si—H
393
C—Br
276
F—F
154
Si—C
360
C—I
240
F—Cl
253
Si—O
452
C—S
259
F—Br
237
Cl—Cl
239
Cl—Br
218
Br—Br
193
*C == O(CO2) = 799
When a bond is strong, there is a higher bond energy because it takes more energy to break a strong bond. This correlates with bond order and bond length. When the Bond order is higher, bond length is shorter, and the shorter the bond length means a greater the Bond Energy because of increased electric attraction. In general, the shorter the bond length, the greater the bond energy.
The average bond energies in Table T3 are the averages of bond dissociation energies. For example the average bond energy of O-H in H2O is 464 kJ/mol. This is due to the fact that the H-OH bond requires 498.7 kJ/mol to dissociate, while the O-H bond needs 428 kJ/mol.
$\dfrac{498.7\; kJ/mol + 428\; kJ/mol}{2}=464\; kJ/mol$
When more bond energies of the bond in different molecules that are taken into consideration, the average will be more accurate. However,
• Average bonds values are not as accurate as a molecule specific bond-dissociation energies.
• Double bonds are higher energy bonds in comparison to a single bond (but not necessarily 2-fold higher).
• Triple bonds are even higher energy bonds than double and single bonds (but not necessarily 3-fold higher).
Bond Breakage and Formation
When a chemical reaction occurs, the atoms in the reactants rearrange their chemical bonds to make products. The new arrangement of bonds does not have the same total energy as the bonds in the reactants. Therefore, when chemical reactions occur, there will always be an accompanying energy change.
In some reactions, the energy of the products is lower than the energy of the reactants. Thus, in the course of the reaction, the substances lose energy to the surrounding environment. Such reactions are exothermic and can be represented by an energy-level diagram in Figure 1 (left). In most cases, the energy is given off as heat (although a few reactions give off energy as light). In chemical reactions where the products have a higher energy than the reactants, the reactants must absorb energy from their environment to react. These reactions are endothermic and can be represented by an energy-level diagrams like Figure 1 (right).
Technically Temperature is Neither a Reactant nor Product
It is not uncommon that textbooks and instructors to consider heat as a independent "species" in a reaction. While this is rigorously incorrect because one cannot "add or remove heat" to a reaction as with species, it serves as a convenient mechanism to predict the shift of reactions with changing temperature. For example, if heat is a "reactant" ($\Delta{H} > 0$), then the reaction favors the formation of products at elevated temperature. Similarly, if heat is a "product" ($\Delta{H} < 0$), then the reaction favors the formation of reactants. A more accurate, and hence preferred, description is discussed below.
Exothermic and endothermic reactions can be thought of as having energy as either a "product" of the reaction or a "reactant." Exothermic reactions releases energy, so energy is a product. Endothermic reactions require energy, so energy is a reactant.
Example $1$: Exothermic vs. Endothermic
Is each chemical reaction exothermic or endothermic?
1. $2H_{2(g)} + O_{2(g)} \rightarrow 2H_2O_{(ℓ)} + \text{135 kcal}$
2. $N_{2(g)} + O_{2(g)} + \text{45 kcal} \rightarrow 2NO_{(g)}$
Solution
No calculates are required to address this question. Just look at where the "heat" is in the chemical reaction.
1. Because energy is released; this reaction is exothermic.
2. Because energy is absorbed; this reaction is endothermic.
Exercise $1$
If the bond energy for H-Cl is 431 kJ/mol. What is the overall bond energy of 2 moles of HCl?
Answer
Simply multiply the average bond energy of H-Cl by 2. This leaves you with 862 kJ (Table T3).
Example $2$: Generation of Hydrogen Iodide
What is the enthalpy change for this reaction and is it endothermic or exothermic?
$H_2(g)+I_2(g) \rightarrow 2HI(g)$
Solution
First look at the equation and identify which bonds exist on in the reactants.
• one H-H bond and
• one I-I bond
Now do the same for the products
• two H-I bonds
Then identify the bond energies of these bonds from the table above:
• H-H bonds: 436 kJ/mol
• I-I bonds: 151 kJ/mol
The sum of enthalpies on the reaction side is:
436 kJ/mole + 151 kJ/mole = 587 kJ/mol.
This is how much energy is needed to break the bonds on the reactant side. Then we look at the bond formation which is on the product side:
• 2 mol H-I bonds: 297 kJ/mol
The sum of enthalpies on the product side is:
2 x 297 kJ/mol= 594 kJ/mol
This is how much energy is released when the bonds on the product side are formed. The net change of the reaction is therefore
587-594= -7 kJ/mol.
Since this is negative, the reaction is exothermic.
Example $2$: Decomposition of Water
Using the bond energies given in the chart above, find the enthalpy change for the thermal decomposition of water:
$2H_2O (g) \rightarrow 2H_2 + O_2 (g)$
Is the reaction written above exothermic or endothermic? Explain.
Solution
The enthalpy change deals with breaking two mole of O-H bonds and the formation of 1 mole of O-O bonds and two moles of H-H bonds (Table T3).
• The sum of the energies required to break the bonds on the reactants side is 4 x 460 kJ/mol = 1840 kJ/mol.
• The sum of the energies released to form the bonds on the products side is
• 2 moles of H-H bonds = 2 x 436.4 kJ/mol = 872.8 kJ/mol
• 1 moles of O=O bond = 1 x 498.7 kJ/mil = 498.7 kJ/mol
which is an output (released) energy = 872.8 kJ/mol + 498.7 kJ/mol = 1371.5 kJ/mol.
Total energy difference is 1840 kJ/mol – 1371.5 kJ/mol = 469 kJ/mol, which indicates that the reaction is endothermic and that 469 kJ of heat is needed to be supplied to carry out this reaction.
This reaction is endothermic since it requires energy in order to create bonds.
Summary
Energy is released to generate bonds, which is why the enthalpy change for breaking bonds is positive. Energy is required to break bonds. Atoms are much happier when they are "married" and release energy because it is easier and more stable to be in a relationship (e.g., to generate octet electronic configurations). The enthalpy change is negative because the system is releasing energy when forming bond.
Contributors and Attributions
• Kim Song (UCD), Donald Le (UCD)
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Chemical_Bonding/Fundamentals_of_Chemical_Bonding/Bond_Energies.txt
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Bond order is the number of chemical bonds between a pair of atoms and indicates the stability of a bond. For example, in diatomic nitrogen, N≡N, the bond order is 3; in acetylene, H−C≡C−H, the carbon-carbon bond order is also 3, and the C−H bond order is 1. Bond order and bond length indicate the type and strength of covalent bonds between atoms. Bond order and length are inversely proportional to each other: when bond order is increased, bond length is decreased.
Introduction
Chemistry deals with the way in which subatomic particles bond together to form atoms. Chemistry also focuses on the way in which atoms bond together to form molecules. In the atomic structure, electrons surround the atomic nucleus in regions called orbitals. Each orbital shell can hold a certain number of electrons. When the nearest orbital shell is full, new electrons start to gather in the next orbital shell out from the nucleus, and continue until that shell is also full. The collection of electrons continues in ever widening orbital shells as larger atoms have more electrons than smaller atoms. When two atoms bond to form a molecule, their electrons bond them together by mixing into openings in each others' orbital shells. As with the collection of electrons by the atom, the formation of bonds by the molecule starts at the nearest available orbital shell opening and expand outward.
Bond Order
Bond order is the number of bonding pairs of electrons between two atoms. In a covalent bond between two atoms, a single bond has a bond order of one, a double bond has a bond order of two, a triple bond has a bond order of three, and so on. To determine the bond order between two covalently bonded atoms, follow these steps:
1. Draw the Lewis structure.
2. Determine the type of bonds between the two atoms.
• 0: No bond
• 1: Single bond
• 2: double bond
• 3: triple bond
If the bond order is zero, the molecule cannot form. The higher bond orders indicate greater stability for the new molecule. In molecules that have resonance bonding, the bond order does not need to be an integer.
Example $1$: $CN^-$
Determine the bond order for cyanide, CN-.
Solution
1) Draw the Lewis structure.
2) Determine the type of bond between the two atoms.
Because there are 3 dashes, the bond is a triple bond. A triple bond corresponds to a bond order of 3.
Example $2$: $H_2$
Determine the bond order for hydrogen gas, H2.
Solution
1) Draw the Lewis structure.
2) Determine the type of bond between the two atoms.
There is only one pair of shared electrons (or dash), indicating is a single bond, with a bond order of 1.
Polyatomic molecules
If there are more than two atoms in the molecule, follow these steps to determine the bond order:
1. Draw the Lewis structure.
2. Count the total number of bonds.
3. Count the number of bond groups between individual atoms.
4. Divide the number of bonds between atoms by the total number of bond groups in the molecule.
Example $3$: $NO_3^-$
Determine the bond order for nitrate, $NO_3^-$.
Solution
1) Draw the Lewis structure.
2) Count the total number of bonds.
4
The total number of bonds is 4.
3) Count the number of bond groups between individual atoms.
3
The number of bond groups between individual atoms is 3.
4) Divide the number of bonds between individual atoms by the total number of bonds.
$\dfrac{4}{3}= 1.33$
The bond order is 1.33
Example $4$: $NO^+_2$
Determine the bond order for nitronium ion: $NO_2^+$.
Solution
1) Draw the Lewis Structure.
2) Count the total number of bonds.
4
The total number of bonds is 4.
3) Count the number of bond groups between individual atoms.
2
The number of bond groups between atoms is 2.
4) Divide the bond groups between individual atoms by the total number of bonds.
$\frac{4}{2} = 2$
The bond order is 2.
A high bond order indicates more attraction between electrons. A higher bond order also means that the atoms are held together more tightly. With a lower bond order, there is less attraction between electrons and this causes the atoms to be held together more loosely. Bond order also indicates the stability of the bond. The higher the bond order, the more electrons holding the atoms together, and therefore the greater the stability.
Trends in the Periodic Table
Bond order increases across a period and decreases down a group.
Bond Length
Bond length is defined as the distance between the centers of two covalently bonded atoms. The length of the bond is determined by the number of bonded electrons (the bond order). The higher the bond order, the stronger the pull between the two atoms and the shorter the bond length. Generally, the length of the bond between two atoms is approximately the sum of the covalent radii of the two atoms. Bond length is reported in picometers. Therefore, bond length increases in the following order: triple bond < double bond < single bond.
To find the bond length, follow these steps:
1. Draw the Lewis structure.
2. Look up the chart below for the radii for the corresponding bond.
3. Find the sum of the two radii.
4
Determine the carbon-to-chlorine bond length in CCl4.
Solution
Using Table A3, a C single bond has a length of 75 picometers and that a Cl single bond has a length of 99 picometers. When added together, the bond length of a C-Cl bond is approximately 174 picometers.
2
Determine the carbon-oxygen bond length in CO2.
Solution
Using Table A3, we see that a C double bond has a length of 67 picometers and that an O double bond has a length of 57 picometers. When added together, the bond length of a C=O bond is approximately 124 picometers.
Trends in the Periodic Table
Because the bond length is proportional to the atomic radius, the bond length trends in the periodic table follow the same trends as atomic radii: bond length decreases across a period and increases down a group.
Problems
1. What is the bond order of $O_2$?
2. What is the bond order of $NO_3^-$?
3. What is the carbon-nitrogen bond length in $HCN$?
4. Is the carbon-to-oxygen bond length greater in $CO_2$ or $CO$?
5. What is the nitrogen-fluoride bond length in $NF_3$?
Solutions
1. First, write the Lewis structure for $O_2$.
There is a double bond between the two oxygen atoms; therefore, the bond order of the molecule is 2.
2. The Lewis structure for NO3- is given below:
To find the bond order of this molecule, take the average of the bond orders. N=O has a bond order of two, and both N-O bonds have a bond order of one. Adding these together and dividing by the number of bonds (3) reveals that the bond order of nitrate is 1.33.
3. To find the carbon-nitrogen bond length in HCN, draw the Lewis structure of HCN.
The bond between carbon and nitrogen is a triple bond, and a triple bond between carbon and nitrogen has a bond length of approximately 60 + 54 =114 pm.
4. From the Lewis structures for CO2 and CO, there is a double bond between the carbon and oxygen in CO2 and a triple bond between the carbon and oxygen in CO.
Referring to the table above, a double bond between carbon and oxygen has a bond length of approximately 67 + 57 = 124 pm and a triple bond between carbon and oxygen has a bond length of approximately 60 + 53 =113 pm. Therefore, the bond length is greater in CO2.
Another method makes use of the fact that the more electron bonds between the atoms, the tighter the electrons are pulling the atoms together. Therefore, the bond length is greater in CO2.
5. To find the nitrogen-to-fluorine bond length in NF3, draw the Lewis structure.
The bond between fluorine and nitrogen is a single bond. From the table above, a single bond between fluorine and nitrogen has a bond length of approximately 64 + 71 =135 pm.
• Wikihow.com
• Anonymous
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Chemical_Bonding/Fundamentals_of_Chemical_Bonding/Bond_Order_and_Lengths.txt
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Learning Objectives
• Define bond length and bond energy and note the relationship between the two.
• Define bond order; explain its relationship to bond length or bond energy.
• Evaluate enthalpies of reactions using bond energies.
• Recognize covalent substances and characterize ionic character as difference in electronegativity.
• Describe trends in bond lengths of a series of related compounds.
Chemical Bond
Learning Objectives
• Explain material structures in terms of chemical bonds.
• Describe various concepts developed for chemical bonds.
• Expand the concept of chemical bond (to be innovative, and creative).
• Describe the octet rule and apply it to write Lewis dot structures for ions and molecules.
Chemical Bond
Chemical bond refers to the forces holding atoms together to form molecules and solids. This force is of an electric nature, and the attraction between electrons of one atom to the nucleus of another atom contributes to what is known as chemical bonds. Although electrons of one atom repel electrons of another, the repulsion is relatively small. So is the repulsion between atomic nuclei.
Various theories regarding chemical bonds have been proposed over the past 300 years, during which our interpretation of the world has also changed. Some old concepts such as Lewis dot structure and valency are still rather useful in our understanding of the chemical properties of atoms and molecules, and new concepts involving quantum mechanics of chemical bonding interpret modern observations very well.
While reading this page, you learn new concepts such as bond length, bond energy, bond order, covalent bond, ionic bond, polar and non-polar bond etc. These concepts help you understand the material world at the molecular level.
Chemical bonds between identical atoms such as those in $\ce{H2}$, $\ce{N2}$, and $\ce{O2}$ are called covalent bonds, in which the bonding electrons are shared. In ionic compounds, such as $\ce{NaCl}$, the ions gather and arrange in a systematic fashion to form a solid. The arrangement of (blue) $\ce{Na+}$ and (green) $\ce{Cl-}$ ions in a solid is shown in on the right here. The attraction force between ions is called ionic bonding. Metals such as sodium, copper, gold, iron etc. have special properties such as being good electric conductors. Electrons in these solids move freely throughout the entire solid, and the forces holding atoms together are called metallic bonds. To some extent, metals are ions submerged in electrons.
A Brief Past on Chemical Bond Concepts
Various concepts or theories have been proposed to explain the formation of compounds. In particular, chemical bonds were proposed to explain why and how one element reacted with another element. In 1852, E. Frankland proposed the concept of valence. He suggested that each element formed compounds with definite amounts of other elements due to a valence connection. Each element has a definite number of valence.
Five years later, F.A. Kekule and others proposed a valence of 4 for carbon. Lines were used to represent valence, and this helped the development of organic chemistry. The structure of benzene was often quoted as an achievement in this development. More than 10 years later, J.H. van 't Hoff and le Bel proposed the tetrahedral arrangement for the four valences around the carbon. These theory helped chemists to describe many organic compounds. In the meantime, chemical bonds were thought to be electric in nature. Since electrons have not been discovered as the negative charge carriers, they were thought to be involved in chemical bonds.
Following the discoveries of electrons by J.J. Thomson and R. A. Millikan, G.N. Lewis proposed to use dots to represent valence electrons. His dots made the valence electrons visible to chemists, and he has been credited as the originator of modern bonding theory. He published a book, in 1923, called Valence and the Structure of Atoms and Molecules.
X-ray diffractions by crystal allow us to calculate details of bond length and bond angles. Using computers, we are able to generate images of molecules from the data provided by X-ray diffraction studies. These data prompted Linus Pauling to look at The Nature of the Chemical Bond, a book that introduced many new concepts such as the resonance, electronegativity, ionic bond, and covalent bond. In England, N.V. Sidgwick and H.E. Powell paid their attention to the lone pairs in a molecule. They developed the valence bond theory, the VSEPR (Valence Shell Electron Pair Repulsion) theory. The application of quantum theory to chemical bonding gave birth to a molecular orbital theory. In this and the few following modules, we will look at some of these concepts in detail.
Lewis Dot Structures
For the elements in the 2nd and 3rd periods, the number of valence electrons range from 1 to 8. Lewis dot structure for them are as indicated:
$\mathrm{ \overset{\Large{.}}Li \hspace{20px} \underset{\Large{.}}{\overset{\Large{.}}Be}\hspace{20px} \cdot \overset{\Large{.}}{B} \cdot\hspace{20px} \cdot \underset{\Large{.}}{\overset{\Large{.}}{C}} \cdot\hspace{20px} \cdot \underset{\Large{.}}{\overset{\Large{.}}{N}} :\hspace{20px} : \underset{\Large{.}}{\overset{\Large{.}}{O}} :\hspace{20px} : \underset{\Large{.}}{\overset{\Large{..}}{F}} :\hspace{20px} : \underset{\Large{..}}{\overset{\Large{..}}{Ne}} :}$
Using dots, Lewis made the valence electron visible. The stability of noble gases is now associated with the 8 valence electrons around them. The stability of 8 valence electrons led him to conclude that all elements strive to acquire 8 electrons in the valence shell, and the chemical reaction takes place due to elements trying to get 8 electrons. This is the octet rule. For the hydrogen and helium atoms, 2 electrons instead of 8 are required.
For example, the octet rule applies to the following molecules:
$\mathrm{H : H}$
(2 electrons)
$\mathrm{H : \underset{\Large{..}}{\overset{\Large{..}}{O}} : H}$
$\mathrm{H : \underset{\Large{..}}{\overset{\Large{..}}{F}} :}$
$\mathrm{H : \underset{\Large{..}}{\overset{\Large{\underset{\huge{..}}H}}{N}} : H}$ $\mathrm{H : \underset{\Large{\overset{\huge{..}}H}}{\overset{\Large{\underset{\huge{..}}H}}{C}} : H}$ $\mathrm{: N ::: N :}$ $\mathrm{: \overset{\Large{..}}O :: \overset{\Large{..}}O :}$ $\mathrm{: \overset{\Large{..}}F : \overset{\Large{..}}F :}$ $\mathrm{: \overset{\Large{..}}O :: C :: \overset{\Large{..}}O : }$
To draw a Lewis dot structure, all the valence electrons are represented. A good way is to draw a type of dot for the valence electrons of one atom different from types in another. To do this on the computer screen using only type fonts is difficult, but you should draw a few by hand on paper.
When a dash is used to represent a bond, it represents a pair of electrons. Thus, in the following representations, a dash represents two electrons, bonding or lone pairs.
_ _ _
:S=O: :O:H :O:H
| _ | | -
O :O:S:O: :O:S:O:
" | " " | "
:O:H :O:H
" "
These structures satisfy
the octet rule. Note
the two ways of drawing
the structures of
$\ce{H2SO4}$.
Exceptions to the Octet Rule
Elements in the 3rd and higher periods may have more than 8 valence electrons. A possible explanation for this is to say that these atoms have d-type atomic orbitals to accommodate more than 8 electrons. In the following molecules, the number of valence electrons in the central atoms are as indicated:
Molecule $\ce{SF6}$ $\ce{PCl5}$ $\ce{ICl3}$ $\ce{XeF4}$
No. of valence
electrons for
central atom
12 10 10 12
OH
|
O=S=O
|
OH
Draw the Lewis dot structures for the above molecules, and count the number of valence electrons for the central atoms. For $\ce{H2SO4}$, the $\ce{S}$ atom has 12 electrons in the structure shown on your right. Each dash represents a chemical bond, which has two electrons. There is a total of 6 bonds around the $\ce{S}$ atom, and therefore 12 electrons.
When $\ce{B}$, $\ce{Be}$, and some metals are the central atoms, they have less than 8 valence electrons. The following compounds do form, but the octet rule is not satisfied. These are electron deficient molecules.
Molecule $\ce{BeCl2}$ $\ce{BF3}$ $\ce{BCl3}$ $\ce{SnCl2}$
No. of valence
electrons for
central atom
4 6 6 6
Another case of exception to the octet rule is molecules with an odd number of valence electrons. For example:
$\ce{NO}$ $\ce{NO2}$ $\ce{ClO2}$
No. of valence
electrons for
central atom
11 17 19
Isoelectronic Molecules and Ions
Counting the number of valence electrons often helps us understand the formation of many molecules and ions. For example, all the following molecules have 11 valence electrons:
$\ce{NO}$ $\ce{CO-}$ $\ce{O2+}$ $\ce{N2-}$
The charged molecules do exist under special circumstances.
The molecules of $\ce{O2}$ are paramagnetic, and thus, they have unpaired electrons. The first dot structure does not agree with this observed fact, but the second one does. However, the second one does not obey the octet rule.
$\mathrm{\overset{\Large{.\,.}}{: O :}\overset{\Large{.\,.}}{: O :}}$ $\mathrm{:\overset{\Large .}{O}:\,:\,:\overset{\Large .}{O}:}$
No unpaired
electron
Violates
octet rule
Later, you will learn that the molecular orbital (MO) theory provides a good explanation for the electronic configuration for $\ce{O2}$.
Confidence Building Questions
1. In 1852, E. Frankland proposed the concept of valence. What is the meaning of valence in this early stage?
1. Number of valence electrons.
2. Number of atoms.
3. Number of chemical bonds an element can form.
4. Number of sticks an atom have.
Hint: c. Number of chemical bonds an element can form.
Skill: Describe a concept. Kekule applied this concept later and suggested four valence for carbon.
2. What do the dots in Lewis dot structures represent?
Hint: valence electrons.
Skill: Describe the Lewis dot structure
The dots represent valence electrons, including those not involved in bonding. Lewis dots made the unshared electrons visible.
3. How many valence electrons do the atoms of oxygen have?
Hint: six or 6.
Skill: Describe the properties of an element based on its group.
Oxygen is the first element of group 6 in the periodic table. How about sulfur, $\ce{S}$?
4. Which one of the following molecules does not satisfy the octet rule?
$\ce{CO2}$, $\ce{CO3^2-}$, $\ce{N2O}$, $\ce{NO2}$, $\ce{NO}$
Hint: $\ce{NO}$
Discussion: Which one of these has an odd number of electrons? The dot structure for $\ce{N2O}$ is $\mathrm{:: N : N : \overset{\Large{.\,.}}{O} :::}$
5. Which of the following molecules satisfy the octet rule?
$\ce{BF3}$, $\ce{BeCl2}$, $\ce{SnCl2}$, $\ce{CO}$, $\ce{SF6}$, $\ce{XeF4}$, $\ce{NO}$
Hint: $\ce{CO}$
Skill: For $\ce{CO}$, you have $\mathrm{:C:\,:\,:O:}$ as the dot structure. The molecule $\ce{NO}$ has one more electron than that of $\ce{CO}$.
6. Which of the following molecules has an odd number of total valence electrons?
$\ce{CO}$, $\ce{CO2}$, $\ce{N2O}$, $\ce{NH3}$, $\ce{H2O}$, $\ce{NO2}$
Hint: Nitrogen dioxide, $\ce{NO2}$
Skill: Elements in the second period are: $\ce{Li}$, $\ce{Be}$, $\ce{B}$, $\ce{C}$, $\ce{N}$, $\ce{O}$, $\ce{F}$, $\ce{Ne}$.
Compare the number of total valence electrons between $\ce{N2O}$ and $\ce{NO2}$.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Chemical_Bonding/Fundamentals_of_Chemical_Bonding/Chemical_Bonds/Bond_Lengths_and_Energies.txt
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Here are some suggested examples.
``` . . .. ..
N N N N
/ \ // \ / \ // \
O O O O O O O O
. .
```
The paired electrons are not shown here.
• What is the formal charge of \(\ce{N}\) in the following dot structure?
``` .
N
/ \
O O
```
Hint: Number of electrons around N = 1 + (6/2)
Skill: Evaluate the formal charge of any atom of any dot structure.
What is the formal charge for the resonance structure having the odd electron on the oxygen atom?
• Hint: Bond order = 1.5
Discussion: Four resonance structures have been given, and all indicate a bond order of 1.5 for \(\ce{[O-N=O]}\) <=> \(\ce{[O=N-O]}\).
• Hint: Huum! This sounds like a real-world problem.
Skill: Solving real-world problems requires you to know what information to look for before the problem is solvable. This is one of those problems.
In an exercise or an open-book test, we can ask you to evaluate the lattice energy of a salt, and you need to find the various data. Since the CAcT quiz is an open-book test, this type of problem should have been given, but you will not have this type of problem in this quiz.
To Quizzes
Electron Density of Sigma and Pi Bonds
hybrid orbitals for the sigma bonds.
```H
\
C = O; but a p orbital is used for the pi bond between C and O.
/
H
```
• What hybrid atomic orbitals are used for the \(\ce{C}\) in \(\ce{CCl4}\)?
Hint: The sp4 hybrid orbitals of \(\ce{C}\) are involved.
Discussion -
The sp3 hybrid AO's have points towards the vertices of a tetrahedron.
• What is the value of the bond angles around the \(\ce{C}\) atom in \(\ce{CH3OH}\)?
Hint: The bond angle is 109.5 degrees.
Discussion -
The tetrahedral arrangement of the bonds around \(\ce{C}\) in methanol gives an average bond angle of 109.5°.
• Hint: The bond order is 1.5 in benzene.
Discussion -
Three double bonds spread over 6 \(\ce{C-C}\) bonds. Thus, the bond order is 1.5.
Enthalpies of Reactions
Since bond energies are given, we use the monoatomic gases as the reference level in this calculation. The energy level diagram shown below illustrates the principle of conservation of energy, and you are expected to have the skill to draw such a diagram.
```------2 H(g) + 2 F(g)-------
| ||436+158 kJ
| ||
|+2*568 kJ ---H2 + F2---
| |
| | \(\Delta H\) = -542 kJ/equation
| ¯
--------2 HF(g)------------
```
This diagram is very similar to those of the Born-Haber cycle used to evaluate lattice energy.
Calculate Enthalpy of Reaction from Enthalpy of Formation
A similar cycle can be devised to calculate energy of a reaction when the standard enthalpies of formation are given. We illustrate this cycle by an an example.
.
```---2 C(graphite) + 3 H2 + 3.5 O2(g)---
| |
| | -228 kJ
| ¯
| ---C2H5OH + 3 O2---
|-394*2 -286*3 |
| |
| | \(\Delta H\) = -394*2 - 286*3 - (-228)
| | = -1418 kJ
¯ ¯
-------------2 CO2 + 3 H2O------------
```
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Chemical_Bonding/Fundamentals_of_Chemical_Bonding/Chemical_Bonds/Chemical_Bonding_-_A_Review.txt
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To put all the information together, we may now draw the cycle:
THE BORN-HABER CYCLE
-----------2Na+ + 2 Br(g)--------
|
|2IP + D |2EA
| ¯
2Na(g) + Br2(g) 2Na+(g) + 2Br-(g)
|
|2Hsub+Hvap |
| |
2Na(s) + Br2(l) |2Ecryst
| |
|2Hf |
¯ ¯
----------2 NaBr(s)---------------
Thus,
$2H_{\large\textrm f} - (2H_{\large\textrm{sub}} + H_{\large\textrm{vap}} + 2IP + D) = 2EA + 2E_{\large\textrm{cryst}}$
Note that we have used two moles of $\ce{NaBr}$ in the above diagram.
This scheme shows that we can calculate the lattice energy of $\ce{NaBr}$ from some known thermodynamic data. The same can be calculated from reaction equations and their associated energies. This is illustrated below
$\ce{2 Na_{\large{(s)}} + Br_{2\large{(l)}} \rightarrow 2 NaBr_{\large{(s)}}}$ $2 H_{\textrm f}$
$\ce{2 Na_{\large{(g)}} \rightarrow 2 Na_{\large{(s)}}}$
or $\ce{2 Na_{\large{(s)}} \rightarrow 2 Na_{\large{(g)}}}$
$- 2 H_{\large\textrm{sub}}$
$\ce{2 Na+_{\large{(g)}} + 2e- \rightarrow 2 Na_{\large{(g)}}}$
or $\ce{2 Na_{\large{(g)}} \rightarrow 2 Na+ + 2e-}$
$- 2\, IP$
$\ce{Br_{2\large{(g)}}\rightarrow Br_{2\large{(l)}}}$
or $\ce{Br_{2\large{(l)}} \rightarrow Br_{2\large{(g)}}}$
$- H_{\large\textrm{vap}}$
$\ce{2 Br_{\large{(g)}} \rightarrow Br_{2\large{(g)}}}$
or $\ce{Br_{2\large{(g)}} \rightarrow 2 Br_{\large{(g)}}}$
$- D$
$\mathrm{2 Br^-_{\large{(g)}} \rightarrow 2 Br_{\large{(g)}} + 2 e^-}$
or $\ce{2 Br_{\large{(g)}} + 2 e- \rightarrow} \mathrm{2 Br^-_{\large{(g)}}}$
$- 2\, EA$
Add all the above equations leading to
$\mathrm{2 Na^+_{\large{(g)}} + 2 Br^-_{\large{(g)}} \rightarrow 2NaBr_{\large{(s)}}}$ $2 E_{\large\textrm{cryst}}$
Thus,
$2 H_{\textrm f} - 2H_{\textrm{sub}} - 2 IP - H_{\textrm{vap}} - D - 2 EA = 2 E_{\textrm{cryst}}$
$E_{\textrm{cryst}} = H_{\textrm f} - H_{\textrm{sub}} - IP - \dfrac{(H_{\textrm{vap}} + D)}{2} - EA$
This is the same result as shown in the diagram.
is shown below:
-----------Na+ + Cl(g)--------
|
| |-349
|496+244/2 ¯
| Na+(g) + Cl-(g)
| |
Na(g) + 0.5Cl2(g) |
|
|108 |
| |Ecryst= -788
Na(s) + 0.5Cl2(g) |
| |
|-411 |
¯ ¯
-------------- NaCl(s) --------------
\begin{align} &E_{\large\textrm{cryst}} = \mathrm{-411-\left(108+496+\dfrac{244}{2}\right)-(-349)\: kJ/mol}\ &\hspace{44px} = \mathrm{-788\: kJ/mol}\ &U = - E_{\large\textrm{cryst}}\ &\hspace{12px} = \textrm{788 kJ/mol (lattice energy)} \end{align}
Discussion
The value calculated for U depends on the data used. Data from various sources differ slightly, and so does the result. The lattice energies for $\ce{NaCl}$ most often quoted in other texts is about 765 kJ/mol.
Compare with the method shown below
$\mathrm{Na_{\large{(s)}} + 0.5 Cl_{2\large{(l)}} \rightarrow NaCl_{\large{(s)}}}$ - 411 Hf
$\mathrm{Na_{\large{(g)}} \rightarrow Na_{\large{(s)}}}$ - 108 -Hsub
$\mathrm{Na^+_{\large{(g)}} + e^- \rightarrow Na_{\large{(g)}}}$ - 496 -IP
$\mathrm{Cl_{\large{(g)}} \rightarrow 0.5\, Cl_{2\large{(g)}}}$ - 0.5 * 244 -0.5*D
$\mathrm{Cl^-_{\large{(g)}} \rightarrow Cl_{\large{(g)}} + 2 e^-}$ 349 -EA
Add all the above equations leading to
$\mathrm{Na^+_{\large{(g)}} + Cl^-_{\large{(g)}} \rightarrow NaCl_{\large{(s)}}}$ -788 kJ/mol = Ecryst
There is a another method based on principles of physics to evaluate the lattice energy, and some examples are given in the discussion enthalpy of hydration and lattice energy.
The Born-Haber cycle enables us to calculate lattice energies of various compounds. For salts containing polyatomic ions, the Born-Haber cycle is not as useful. Some other means have to be used to evaluate the lattice energy or energy of crystallization.
Comparison of Lattice Energies (U in kJ/mol) of Some Salts
Solid U Solid U Solid U Solid U
$\ce{LiF}$ 1036 $\ce{LiCl}$ 853 $\ce{LiBr}$ 807 $\ce{LiI}$ 757
$\ce{NaF}$ 923 $\ce{NaCl}$ 786 $\ce{NaBr}$ 747 $\ce{NaI}$ 704
$\ce{KF}$ 821 $\ce{KCl}$ 715 $\ce{KBr}$ 682 $\ce{KI}$ 649
$\ce{MgF2}$ 2957 $\ce{MgCl2}$ 2526 $\ce{MgBr2}$ 2440 $\ce{MgI2}$ 2327
The lattice energies of some salts are given in the table on the right here.
Among the mono-valent salts, the lattice energies decrease when the sizes of the ions increase.
Comparing the lattice energy for salts with one divalent ion leads to the same conclusion. The lattice energies decrease when the sizes of the ions increase.
The lattice energy of salts involving a divalent ion are much higher than those of monovalent salts, because much more energy is required to separate these ions.
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Learning Objectives
• Draw the Lewis dot structure of a given molecule or ion.
• Draw resonance structures of some molecules.
• Assign formal charge to an atom in a dot structure.
• Assess the stability of a structure by considering formal charges of atoms.
• Give examples for molecules and ions that do not follow the octet rule.
Lewis Dot Structures
Lewis symbols of the main group elements
$\ce{H\cdot}$ $\textrm{He:}$
$\underset{\:}{\ce{Li\cdot}}$
$\underset{\:}{\ce{\cdot Be \cdot}}$
$\ce{ \cdot \underset{\:}{\overset{\Large{\cdot}}{B}} \cdot}$
$\ce{ \cdot \underset{\Large{\cdot}}{\overset{\Large{\cdot}}{C}} \cdot}$
$\underset{\Large{\cdot\,}} {\overset{\Large{\cdot}} {\textrm{:N}\cdot}}$
$\underset{\Large{\cdot\cdot\,}} {\overset{\Large{\cdot}} {\textrm{:O}\cdot}}$
$\underset{\Large{\cdot\cdot}} {\overset{\Large{\cdot\cdot}} {\textrm{:F}\cdot}}$
$\underset{\Large{\cdot\cdot}} {\overset{\Large{\cdot\cdot}}{\textrm{:Ne:}}}$
$\ce{Na}$
$\ce{K}$
$\ce{Rb}$
$\ce{Cs}$
$\ce{Mg}$
$\ce{Ca}$
$\ce{Sr}$
$\ce{Ba}$
$\ce{Al}$
$\ce{Ga}$
$\ce{In}$
$\ce{Tl}$
$\ce{Si}$
$\ce{Ge}$
$\ce{Sn}$
$\ce{Pb}$
$\ce{P}$
$\ce{As}$
$\ce{Sb}$
$\ce{Bi}$
$\ce{S}$
$\ce{Se}$
$\ce{Te}$
$\ce{Po}$
$\ce{Cl}$
$\ce{Br}$
$\ce{I}$
$\ce{At}$
$\ce{Ar}$
$\ce{Kr}$
$\ce{At}$
$\ce{Rn}$
G.N. Lewis used dots to represent the valence electrons in his teaching of chemical bonding. He eventually published his theory of chemical bonding in 1916. He put dots around the symbols so that we see the valence electrons for the main group elements. Formation of chemical bonds to complete the requirement of eight electrons for the atom becomes a natural tendency. Lewis dot symbols of the first two periods are given here to illustrate this point. In fact, the entire group (column) of elements have the same Lewis dot symbols, because they have the same number of valence electrons.
$\ce{CF4}$, $\ce{H2O}$ and $\ce{CO2}$ dot structures
..
: F :
. . . . . .
: F : C : F :
. . . . . .
: F :
' '
.. ..
O
/ \
H H
.. ..
O::C::O
.. ..
or
. . . .
: O=C=O :
Lewis dot structures are useful in explaining the chemical bonding in molecules or ions. When several dot structures are reasonable for a molecule or ion, they all contribute to the molecular or ionic structure making it more stable.
The representation of a molecular or ionic structure by several structures is called resonance. The more stable the dot structure is, the more it contributes to the electronic structure of the molecule or ion.
You need to know what dot structures represent, how to draw them, and what the formal charges for the atoms in the structure are. When several dot structures are possible, consider the resonance structures to interpret the real structure. Apply some simple rules to explain which of the resonance structures are major contributors to the electronic structure.
Drawing Lewis Dot Structures and Resonance Structures
Follow these simple steps to draw Lewis dot structures:
• Draw the atoms on paper and put dots around them to represent valence electrons of the atom. Be sure to have the correct number of electrons.
• If the species is an ion, add or subtract electrons corresponding to the charge of the ion. Add an electron for every negative (-) charge, and subtract an electrons for every positive (+) charge.
• Consider bonding between atoms by sharing electrons; some may come from one atom.
• If possible, apply the octet rule to your structure. Some structures don't obey the octet rule, but explain why.
• Assign formal charges to atoms in the structure.
Exercise
Draw Lewis dot structures for $\ce{CH4}$, $\ce{NH3}$, $\ce{HF}$, $\ce{OF2}$, $\ce{F2}$, $\ce{O2}$, $\ce{N2}$, $\ce{Cl-}$ and some compounds you know.
Formal Charge
The formal charge on any atom in a Lewis structure is a number assigned to it according to the number of valence electrons of the atom and the number of electrons around it. The formal charge of an atom is equal to the number of valence electrons, Nv.e. minus the number of unshared electrons, Nus.e. and half of the bonding electrons, ½ Nb.e..
$Formal\: charge = N_{\large{v.e.}} - N_{\large{us.e.}} - \dfrac{1}{2} N_{\large{b.e.}}$
Some practice of assigning formal charge is necessary before you master this technique. Some examples of drawing Lewis structure and assigning formal charge are given below.
The formal charge is a hypothetical charge from the dot structure. The formal charges in a structure tell us the quality of the dot structure.
Formal charge rules
Often, many Lewis dot structures are possible. These are possible resonance structures, but often we should write a reasonable one, which is stable. The formal charge guides us about the stability of the dot structure. The guidance is called formal charge rules:
• Formulas with the lowest magnitude of formal charges are more stable.
• More electonegative atoms should have negative formal charges.
• Adjacent atoms should have opposite formal charges.
Example 1
Draw Lewis dot structure for $\ce{SO2}$.
Solution
Put down number of valence electrons:
$\mathrm{ :\overset{\Large{..}}O : :\overset{\Large{..}}S : :\overset{\Large{..}}O :}$
Put all atoms together to make a molecule and check to see if it satisfies the octet rule.
\begin{alignat}{1} :&\overset{\Large{..}}{\ce O} : :&&\overset{\Large{..}}{\ce S} : :&&\overset{\Large{..}}{\ce O} : &&\textrm{ <= octet rule not satisfied}\ &\,0 &&\,0 &&\,0 &&\textrm{ formal charge} \end{alignat}
Adjust bonding electrons so that octet rules apply to all the atoms.
\begin{alignat}{1} &:\underset{\Large{..}}{\overset{\Large{..}}{\ce O}} &&:\overset{\Large{..}}{\ce S} : :&&\overset{\Large{..}}{\ce O} : &&\textrm{ <- octet rule satisfied}\ &\,{-1} &&\,{+1} &&\,0 &&\textrm{ formal charge} \end{alignat}
Since the left $\ce{O}$ has 6 unshared plus 2 shared electrons, it effectively has 7 electrons for a 6-valence-electron $\ce{O}$, and thus its formal charge is -1.
Formal charge for $\ce{O}$ = 6 - 6 - (2/2) = -1.
Formal charge for $\ce{S}$ = 6 - 2 - (6/2) = +1.
There is yet another structure that does not satisfy the octet rule, but it's a reasonable structure:
\begin{alignat}{1} &:\underset{\Large{..}}{\overset{\Large{..}}{\ce O}} &&:\overset{\Large{..}}{\ce S} : &&\underset{\Large{..}}{\overset{\Large{..}}{\ce O}} : &&\textrm{ <- octet rule not satisfied}\ &\,{-1} &&\,{+2} &&{-1} &&\textrm{ formal charge} \end{alignat}
Resonance Structures
When several structures with different electron distributions among the bonds are possible, all structures contribute to the electronic structure of the molecule. These structures are called resonance structures. A combination of all these resonance structures represents the real or observed structure. The Lewis structures of some molecules do not agree with the observed structures. For such a molecule, several dot structures may be drawn. All the dot structures contribute to the real structure. The more stable structures contribute more than less stable ones.
For resonance structures, the skeleton of the molecule (or ion) stays in the same relative position, and only distributions of electrons in the resonance structures are different.
Let us return to the $\ce{SO2}$ molecule. The molecule has a bent structure due to the lone pair of electrons on $\ce{S}$. In the last structure that has a formal charge, there is a single $\ce{S-O}$ bond and a double $\ce{S=O}$ bond. These two bonds can switch over giving two resonance structures as shown below.
1 2 3 4
..
S
/ \
:O: :O:
' ' ' '
«
..
S
// \
:O: :O:
' '
«
..
S
/ \
:O: :O:
' '
«
..
S
// \
:O: :O:
In structure 1, the formal charges are +2 for $\ce{S}$, and -1 for both $\ce{O}$ atoms. In structures 2 and 3, the formal charges are +1 for $\ce{S}$, and -1 for the oxygen atom with a single bond to $\ce{S}$. The low formal charges of $\ce{S}$ make structures 2 and 3 more stable or more important contributors. The formal charges for all atoms are zero for structure 4, given earlier. This is also a possible resonance structure, although the octet rule is not satisfied. Combining resonance structures 2 and 3 results in the following structure:
..
S
/.' '' '.\
:O: :O:
Exercise
Draw the Lewis dot structures and resonance structures for the following. Some hints are given.
$\ce{CO2}$ - $\textrm{:O::C::O:}$ (plus two more dots for each of $\ce{O}$)
$\ce{NO2}$ - $\ce{.NO2}$ (bent molecule due to the odd electron)
$\ce{NO2-}$ - $\ce{:NO2-}$ (same number of electron as $\ce{SO2}$)
$\ce{HCO2-}$ - $\ce{H-CO2}$
$\ce{O3}$ - (ozone, $\ce{OO2}$; same number of electrons as $\ce{SO2}$)
$\ce{SO3}$ - (consider $\ce{O-SO2}$, and the resonance structures)
$\ce{NO3-}$ (see Example 2 below)
$\ce{CO3^2-}$ (ditto)
Notice that some of the resonance structures may not satisfy the octet rule. The $\ce{NO2}$ molecule has an odd number of electrons, and the octet rule cannot be satisfied for the nitrogen atom.
Example 2
Draw the resonance structures of $\ce{NO3-}$
Solution
-1
:O:
||
N
/ \
:O: :O:
' ' ' '
«
. . -1
:O:
|
N
// \
:O: :O:
' '
«
. . -1
:O:
|
N
/ \
:O: :O:
' '
The resonance structure is shown on the right here. Note that only the locations of double and single bonds change here. What are the formal charges for the $\ce{N}$ atoms? What are the formal charges for the oxygen atoms that are single bonded and double bonded to $\ce{N}$ respectively? Please work these numbers out.
• Formal charges: $\ce{N}$, +1; $\ce{=O}$, 0; $\ce{-O}$, -1
• The most stable structure has the least formal charge.
• In a stable structure, adjacent atoms should have formal charges of opposite signs.
The more stable the structure, the more it contributes to the resonance structure of the molecule or ion. All three structures above are the same, only the double bond rotates.
Exercise
Draw the Lewis dot structures and resonance structures for
$\ce{HNO3}$
$\ce{H2SO4}$
$\ce{H2CO3}$
$\ce{HClO4}$
$\ce{C5H5N}$
$\ce{NO3-}$
$\ce{SO4^2-}$
$\ce{CO3^2-}$
$\ce{ClO4-}$
$\ce{C6H6}$ (Benzene)
$\ce{Cl2CO}$
You have to do these on paper, because putting dots around the symbols is very difficult using a word processor. The octet rule should be applied to $\ce{HNO3}$, $\ce{NO3-}$, $\ce{H2CO3}$, $\ce{CO3^2-}$, $\ce{C5H5N}$, $\ce{C6H6}$, and $\ce{Cl2CO}$.
Exceptions to the octet rule
We can write Lewis dot structures that satisfy the octet rule for many molecules consisting of main-group elements, but the octet rule may not be satisfied for a number of compounds. For example, the dot structures for $\ce{NO}$, $\ce{NO2}$, $\ce{BF3}$ ($\ce{AlCl3}$), and $\ce{BeCl2}$ do not satisfy the octet rule.
.N:::O:
compare
with
:C:::O:
.
N
// \
:O: :O:
' '
..
:F:
|
B
/ \
:F: :F:
' ' ' '
. . . .
:Cl : Be : Cl:
' ' ' '
The above are structures for the gas molecules. The solids of $\ce{AlCl3}$ and $\ce{BeCl2}$ are polymeric with bridged chlorides.
. . . .
:Cl: :Cl: :Cl:
\ / \ /
Al Al
/ \ / \
:Cl: :Cl: :Cl:
. . . .
Polymeric
solid
structures
:Cl: :Cl: :Cl:
/ \ / \ / \
Be Be
\ / \ / \ /
:Cl: :Cl: :Cl:
Alumunum chloride, $\ce{AlCl3}$, is a white, crystalline solid, and an ionic compound. However, it has a low melting point of 465 K (192°C), and the liquid consists of dimers, $\ce{Al2Cl6}$, whose structure is shown above. It vaporizes as dimers, but further heating gives a monomer that has the same structure as the $\ce{BF3}$.
Arrange dots this way
\begin{alignat}{5} \textrm{::Ex} &= \mathrm{\overset{\:\Large{..}}{:Ex}}\ \textrm{:::Ex} &= \mathrm{\overset{\Large{..}}{:Ex:}} \end{alignat}
In compounds $\ce{PF5}$, $\ce{PCl5}$, $\textrm{:SF}_4$, $\textrm{::ClF}_3$, $\textrm{:::XeF}_2$ and $\textrm{:::I}_3^-$, the center atoms have more than 10 electrons instead of 8. In compounds $\ce{SF6}$, $\ce{IOF5}$, $\textrm{:IF}_5$, $\ce{BrF5}$, $\textrm{::XeF}_4$, $\ce{PF6-}$ etc, the center atoms have 12 electrons.
The formulas given above follow a systematic pattern according to the positions of the elements on the periodic table. As the number of atoms bonded to it decreases, the number of unshared electrons increase.
Confidence Building Questions
1. What is the total number of valence electrons in $\ce{CO2}$?
Hint: Number of valence electrons = 4 + 6 + 6
What about $\ce{NO2}$?
2. What is the formal charge for $\ce{S}$ in $\ce{H2SO4}$ in the following structure?
O-H
| (provide all dots yourself)
O=S=O
|
O-H
Hint: The formal charge is 0 with this structure.
The oxygen double bonded to $\ce{S}$ has a formal charge of 0. What is the formal charge of $\ce{C}$ in $\ce{O-C}\textrm{:::O}$?
3. What is the formal charge for $\ce{S}$ in $\ce{H2SO4}$ in the following structure?
O-H
| (provide all dots yourself)
O-S-O
|
O-H
Hint: The formal charge is +2.
The oxygen double bonded to $\ce{S}$ has formal charge of -1. What is the formal charge of $\ce{C}$ in $\ce{O-C}\textrm{:::O}$?
4. According to the formal charge rule, which structure in the two problems you have just worked on is the best for $\ce{H2SO4}$?
Hint: The one with formal charge = 0 for all atoms.
5. What is the formal charge for $\ce{N}$ in the following structure?
.
N = O
|
:O:
..
Hint: Formal charge = 5 - 1 - (6/2)
What is the formal charge if this is the structure:
.
:O:
| ..
N = O :
..
and which one you think is the best? You can write two resonance structures for each of the two to give 4 resonance structures for $\ce{NO2}$.
6. What is the formal charge for $\ce{B}$ in the following structure?
F
\
B = F (Supply your dots for unshared electrons)
/
F
Hint: Formal charge = 3 - (8/2)
What about $\ce{B(-F)3}$, all single bonds?
7. What is the formal charge of $\ce{I}$ in $\ce{I(-Cl)3}$? (Supply your dots)
Hint: Formal charge = 7 - 4 - (6/2)
What about $\ce{Cl}$ in the same structure?
8. Which one of the following compounds has the same number of valence electrons as $\ce{NO2-}$:
$\ce{CO2}$, $\ce{NO2}$, $\ce{O3}$, $\ce{CO3-}$, or $\ce{CO2-}$?
Hint: ozone
Elements of the 2nd period are: $\ce{Li}$ $\ce{Be}$ $\ce{B}$ $\ce{C}$ $\ce{N}$ $\ce{O}$ $\ce{F}$ $\ce{Ne}$.
Draw Lewis dot structure for $\ce{O3}$ and $\ce{NO2-}$.
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Describe the hydrogen molecule in light of the following:
• \(\ce{H-H}\)
• \(\ce{H:H}\)
• Valence bond theory of \(\ce{H2}\)
• Molecular orbital theory of \(\ce{H2}\)
• Electron configuration of molecules
Valence Bond Theory and Hybrid Atomic Orbitals
However, the application of VSEPR theory can be expanded to complicated molecules such as
H H H O
| | | //
H-C-C=C=C-C=C-C-C
| | \
H N O-H
/ \
H H
By applying the VSEPR theory, one deduces the following results:
• $\ce{H-C-C}$ bond angle = 109o
• $\ce{H-C=C}$ bond angle = 120o, geometry around $\ce{C}$ trigonal planar
• $\ce{C=C=C}$ bond angle = 180o, in other words linear
• $\ce{H-N-C}$ bond angle = 109o, tetrahedral around $\ce{N}$
• $\ce{C-O-H}$ bond angle = 105 or 109o, 2 lone electron pairs around $\ce{O}$
Confidence Building Questions
1. In terms of valence bond theory, how is a chemical bond formed?
Hint: A chemical bond is due to the overlap of atomic orbitals.
Discussion -
Molecular orbital theory considers the energy states of the molecule.
2. When one s and two p atomic orbitals are used to generate hybrid orbitals, how many hybrid orbitals will be generated?
Hint: Using three atomic orbitals generates three hybrid orbitals.
Discussion -
Number of orbitals does not change in hybridization of atomic orbitals.
3. In the structures of $\ce{SO2}$ and $\ce{NO2}$, what are the values of the bond angles?
Hint: The bond angles are expected to be less than 120 degrees.
Discussion -
Since the lone electron pair in $\ce{:SO2}$ and lone electron in $\ce{.NO2}$ take up more space, we expect the structure to distort leaving a smaller angle than 120 between the bonds.
4. What is the geometrical shape of the molecule $\ce{CH4}$, methane?
Hint: Methane molecules are tetrahedral.
Discussion -
The 4 $\ce{H}$ atoms form a tetrahedron, and methane has a tetrahedral shape.
5. What do you expect the bond angles to be in the $\ce{NH4+}$ ion?
Hint: All bond angles are 109.5 degrees, the ideal value for a symmetric tetrahedral structure.
Discussion -
The structure of this ion is very similar to that of $\ce{CH4}$.
6. What hybrid orbitals does the $\ce{C}$ atom use in the compound $\ce{H-C\equiv C-H}$, in which the molecule is linear?
Hint: The sp hybrid orbitals are used by the $\ce{C}$ atom.
Discussion -
Sigma (s) bonds are due to sp hybrid orbitals, and 2 p orbitals are used for pi (p) bonds. The two sigma bonds for each $\ce{C}$ are due to overlap of sp hybrid orbitals of each $\ce{C}$ atom.
7. What hybrid orbitals does $\ce{C}$ use in the molecule:
H
O=C<
H
This is a trigonal planar molecule. It is called formaldehyde, a solvent for preserving biological samples. The compound has an unpleasant smell.
Hint: The $\ce{C}$ atom uses sp2 hybrid orbitals.
Skill -
The $\ce{C}$ atom has 3 sigma (s) bonds by using three sp2 hybrid orbitals and a pi (p) bond, due to one 2p orbital.
8. What is the shape of the molecule $\ce{SF6}$?
Hint: Its shape is octahedral.
Discussion -
Since the $\ce{S}$ atom uses d2sp3 hybrid orbitals, you expect the shape to be octahedral. The $\ce{F}$ atoms form an octahedron around the sulfur.
9. Phosphorus often forms a five coordinated compound $\ce{PX5}$. What hybrid orbitals does $\ce{P}$ use in these compounds?
Hint: The $\ce{P}$ atom uses dsp3 hybrid orbitals.
Discussion -
A total of 5 atomic orbitals are used in the hybridization: one 3d, one 3s and three 3p orbitals. The dsp3 hybrid orbitals of $\ce{P}$ give rise to a trigonal bipyramidal coordination around the $\ce{P}$ atom.
The energy of d orbitals in $\ce{N}$ is not compatible with 2s and 2p orbitals for hybridization. Thus, you seldom encounter a compound with formula $\ce{NX5}$ with $\ce{N}$ as the central atom.
Contrasting MO and VB theory
Both the MO and VB theories are used to help determine the structure of a molecule. Unlike the VB theory, which is largely based off of valence electrons, the MO theory describes structure more in depth by taking into consideration, for example, the overlap and energies of the bonding and antibonding electrons residing in a particular molecular orbital. While MO theory is more involved and difficult, it results in a more complete picture of the structure of a chosen molecule. Despite various shortcomings, complete disregard of one theory and not the other would hinder our ability to describe the bonding in molecules.
• Tony Chhom
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A coordinate bond (also called a dative covalent bond) is a covalent bond (a shared pair of electrons) in which both electrons come from the same atom. A covalent bond is formed by two atoms sharing a pair of electrons. The atoms are held together because the electron pair is attracted by both of the nuclei. In the formation of a simple covalent bond, each atom supplies one electron to the bond - but that does not have to be the case.
The Reaction Between Ammonia and Hydrogen Chloride
If these colorless gases are allowed to mix, a thick white smoke of solid ammonium chloride is formed.
The reaction is
$\ce{NH3 (g) + HCl (g) \rightarrow NH4Cl (s)} \nonumber$
Ammonium ions, NH4+, are formed by the transfer of a hydrogen ion (a proton) from the hydrogen chloride molecule to the lone pair of electrons on the ammonia molecule.
When the ammonium ion, NH4+, is formed, the fourth hydrogen is attached by a dative covalent bond, because only the hydrogen's nucleus is transferred from the chlorine to the nitrogen. The hydrogen's electron is left behind on the chlorine to form a negative chloride ion. Once the ammonium ion has been formed it is impossible to tell any difference between the dative covalent and the ordinary covalent bonds. Although the electrons are shown differently in the diagram, there is no difference between them in reality.
In simple diagrams, a coordinate bond is shown by an arrow. The arrow points from the atom donating the lone pair to the atom accepting it.
Example $1$: Dissolving $HCl_{(g)}$ in Water to make Hydrochloric Acid
Something similar happens. A hydrogen ion (H+) is transferred from the chlorine to one of the lone pairs on the oxygen atom.
\[ \ce{H_2O + HCl \rightarrow H_3O^{+} + Cl^{-}} \nonumber]
The H3O+ ion is variously called the hydroxonium ion, the hydronium ion or the oxonium ion. In an introductory chemistry course, whenever you have talked about hydrogen ions (for example in acids), you have actually been talking about the hydroxonium ion. A raw hydrogen ion is simply a proton, and is far too reactive to exist on its own in a test tube.
If you write the hydrogen ion as $\ce{H^{+}(aq)}$, the "$\ce{(aq)}$" represents the water molecule that the hydrogen ion is attached to. When it reacts with something (an alkali, for example), the hydrogen ion simply becomes detached from the water molecule again. Note that once the coordinate bond has been set up, all the hydrogens attached to the oxygen are exactly equivalent. When a hydrogen ion breaks away again, it could be any of the three.
Reaction between ammonia and boron trifluoride
Boron trifluoride is a compound that does not have a noble gas structure around the boron atom (a notorious "octet violator"). The boron only has three pairs of electrons in its bonding level, whereas there would be room for four pairs. $BF_3$ is described as being electron deficient. The lone pair on the nitrogen of an ammonia molecule can be used to overcome that deficiency, and a compound is formed involving a coordinate bond.
Using lines to represent the bonds, this could be drawn more simply as:
The second diagram shows another way that you might find coordinate bonds drawn. The nitrogen end of the bond has become positive because the electron pair has moved away from the nitrogen towards the boron - which has therefore become negative. We shall not use this method again - it's more confusing than just using an arrow.
The structure of Aluminum Chloride
Aluminum chloride sublimes (phase transition from solid to gas) at about 180°C. If it simply contained ions it would have a very high melting and boiling point because of the strong attractions between the positive and negative ions. The implication is that it when it sublimes at this relatively low temperature, it must be covalent. The dots-and-crosses diagram shows only the outer electrons.
Lewis dot diagram for $AlCl_3$
AlCl3, like BF3, is electron deficient. There is likely to be a similarity, because aluminum and boron are in the same group of the Periodic Table, as are fluorine and chlorine.
Measurements of the relative formula mass of aluminum chloride show that its formula in the vapor at the sublimation temperature is not AlCl3, but Al2Cl6. It exists as a dimer (two molecules joined together). The bonding between the two molecules is coordinate, using lone pairs on the chlorine atoms. Each chlorine atom has 3 lone pairs, but only the two important ones are shown in the line diagram.
The uninteresting electrons on the chlorines have been faded in color to make the coordinate bonds show up better. There's nothing special about those two particular lone pairs - they just happen to be the ones pointing in the right direction.
Energy is released when the two coordinate bonds are formed, and so the dimer is more energetically stable than two separate AlCl3 molecules.
The bonding in hydrated metal ions
Water molecules are strongly attracted to ions in solution - the water molecules clustering around the positive or negative ions. In many cases, the attractions are so great that formal bonds are made, and this is true of almost all positive metal ions. Ions with water molecules attached are described as hydrated ions.
Although aluminum chloride is a covalent compound, when it dissolves in water, ions are produced. Six water molecules bond to the aluminum to give an ion with the formula Al(H2O)63+. It's called the hexaaquaaluminum complex ion with as six ("hexa") water molecules ("aqua") wrapped around an aluminum ion. The bonding in this (and the similar ions formed by the great majority of other metals) is coordinate (dative covalent) using lone pairs on the water molecules.
The electron configuration of aluminum is 1s22s22p63s23px1. When it forms an Al3+ ion it loses the 3-level electrons to leave 1s22s22p6. That means that all the 3-level orbitals are now empty. The aluminum reorganizes (hybridizes) six of these (the 3s, three 3p, and two 3d) to produce six new orbitals all with the same energy. These six hybrid orbitals accept lone pairs from six water molecules.
You might wonder why it chooses to use six orbitals rather than four or eight or whatever. Six is the maximum number of water molecules it is possible to fit around an aluminum ion (and most other metal ions). By making the maximum number of bonds, it releases most energy and is the most energetically stable.
Only one lone pair is shown on each water molecule. The other lone pair is pointing away from the aluminum and so is not involved in the bonding. The resulting ion looks like this:
Because of the movement of electrons towards the center of the ion, the 3+ charge is no longer located entirely on the aluminum, but is now spread over the whole of the ion.
Dotted arrows represent lone pairs coming from water molecules behind the plane of the screen or paper. Wedge shaped arrows represent bonds from water molecules in front of the plane of the screen or paper.
Example $2$: Carbon Monoxide
Carbon monoxide, CO, can be thought of as having two ordinary covalent bonds between the carbon and the oxygen plus a coordinate bond using a lone pair on the oxygen atom.
Example $3$: Nitric Acid
In nitric acid, HNO3, one of the oxygen atoms can be thought of as attaching to the nitrogen via a coordinate bond using the lone pair on the nitrogen atom.
In fact this structure is misleading because it suggests that the two oxygen atoms on the right-hand side of the diagram are joined to the nitrogen in different ways. Both bonds are actually identical in length and strength, and so the arrangement of the electrons must be identical. There is no way of showing this using a dots-and-crosses picture. The bonding involves delocalization.
Covalent Bond Distance Radius and van der Waals Radius
Covalent bond distance, covalent radius, and van der Waals radius are used to describe the size and distance between atoms. Covalent bond distance refers to the distance between the nuclei of two bonded atoms. Covalent radius is half of the internuclear separation between the nuclei of two single-bonded atoms of the same species (homonuclear). While van der Waals radius is used to define half of the distance between the closest approach of two non-bonded atoms of a given element.
Contributors and Attributions
• Andrew Cooley, University of California, Davis
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This page explains what covalent bonding is. It starts with a simple picture of the single covalent bond.
The importance of noble gas structures
At a simple level a lot of importance is attached to the electronic structures of noble gases like neon or argon which have eight electrons in their outer energy levels (or two in the case of helium). These noble gas structures are thought of as being in some way a "desirable" thing for an atom to have. You may well have been left with the strong impression that when other atoms react, they try to achieve noble gas structures. As well as achieving noble gas structures by transferring electrons from one atom to another as in ionic bonding, it is also possible for atoms to reach these stable structures by sharing electrons to give covalent bonds.
Some very simple covalent molecules
Chlorine
For example, two chlorine atoms could both achieve stable structures by sharing their single unpaired electron as in the diagram.
The fact that one chlorine has been drawn with electrons marked as crosses and the other as dots is simply to show where all the electrons come from. In reality there is no difference between them. The two chlorine atoms are said to be joined by a covalent bond. The reason that the two chlorine atoms stick together is that the shared pair of electrons is attracted to the nucleus of both chlorine atoms.
Hydrogen
Hydrogen atoms only need two electrons in their outer level to reach the noble gas structure of helium. Once again, the covalent bond holds the two atoms together because the pair of electrons is attracted to both nuclei.
Hydrogen chloride
The hydrogen has a helium structure, and the chlorine an argon structure. Most of the simple molecules you draw do in fact have all their atoms with noble gas structures. For example:
Even with a more complicated molecule like \(PCl_3\), there's no problem. In this case, only the outer electrons are shown for simplicity. Each atom in this structure has inner layers of electrons of 2, 8. Again, everything present has a noble gas structure.
Cases where the simple view throws up problems
Boron trifluoride, BF3
A boron atom only has 3 electrons in its outer level, and there is no possibility of it reaching a noble gas structure by simple sharing of electrons. Is this a problem? No. The boron has formed the maximum number of bonds that it can in the circumstances, and this is a perfectly valid structure.
Energy is released whenever a covalent bond is formed. Because energy is being lost from the system, it becomes more stable after every covalent bond is made. It follows, therefore, that an atom will tend to make as many covalent bonds as possible. In the case of boron in BF3, three bonds is the maximum possible because boron only has 3 electrons to share.
You might perhaps wonder why boron doesn't form ionic bonds with fluorine instead. Boron doesn't form ions because the total energy needed to remove three electrons to form a B3+ ion is simply too great to be recoverable when attractions are set up between the boron and fluoride ions.
Phosphorus(V) chloride, PCl5
In the case of phosphorus, 5 covalent bonds are possible - as in PCl5. Phosphorus forms two chlorides - PCl3 and PCl5. When phosphorus burns in chlorine both are formed - the majority product depending on how much chlorine is available. We've already looked at the structure of PCl3. The diagram of PCl5 (like the previous diagram of PCl3) shows only the outer electrons.
Notice that the phosphorus now has 5 pairs of electrons in the outer level - certainly not a noble gas structure. You would have been content to draw PCl3 at GCSE, but PCl5 would have looked very worrying.
Why does phosphorus sometimes break away from a noble gas structure and form five bonds? In order to answer that question, we need to explore territory beyond the limits of most current A'level syllabuses. Don't be put off by this! It isn't particularly difficult, and is extremely useful if you are going to understand the bonding in some important organic compounds.
A more sophisticated view of covalent bonding
The bonding in methane, CH4
What is wrong with the dots-and-crosses picture of bonding in methane?
We are starting with methane because it is the simplest case which illustrates the sort of processes involved. You will remember that the dots-and-crossed picture of methane looks like this.
There is a serious mis-match between this structure and the modern electronic structure of carbon, 1s22s22px12py1. The modern structure shows that there are only 2 unpaired electrons to share with hydrogens, instead of the 4 which the simple view requires.
You can see this more readily using the electrons-in-boxes notation. Only the 2-level electrons are shown. The 1s2 electrons are too deep inside the atom to be involved in bonding. The only electrons directly available for sharing are the 2p electrons. Why then isn't methane CH2?
Promotion of an electron
When bonds are formed, energy is released and the system becomes more stable. If carbon forms 4 bonds rather than 2, twice as much energy is released and so the resulting molecule becomes even more stable. There is only a small energy gap between the 2s and 2p orbitals, and so it pays the carbon to provide a small amount of energy to promote an electron from the 2s to the empty 2p to give 4 unpaired electrons. The extra energy released when the bonds form more than compensates for the initial input.
The carbon atom is now said to be in an excited state. Now that we've got 4 unpaired electrons ready for bonding, another problem arises. In methane all the carbon-hydrogen bonds are identical, but our electrons are in two different kinds of orbitals. You aren't going to get four identical bonds unless you start from four identical orbitals.
Hybridization
The electrons rearrange themselves again in a process called hybridization. This reorganizes the electrons into four identical hybrid orbitals called sp3 hybrids (because they are made from one s orbital and three p orbitals). You should read "sp3" as "s p three" - not as "s p cubed".
sp3 hybrid orbitals look a bit like half a p orbital, and they arrange themselves in space so that they are as far apart as possible. You can picture the nucleus as being at the center of a tetrahedron (a triangularly based pyramid) with the orbitals pointing to the corners. For clarity, the nucleus is drawn far larger than it really is.
What happens when the bonds are formed?
Remember that hydrogen's electron is in a 1s orbital - a spherically symmetric region of space surrounding the nucleus where there is some fixed chance (say 95%) of finding the electron. When a covalent bond is formed, the atomic orbitals (the orbitals in the individual atoms) merge to produce a new molecular orbital which contains the electron pair which creates the bond.
Four molecular orbitals are formed, looking rather like the original sp3 hybrids, but with a hydrogen nucleus embedded in each lobe. Each orbital holds the 2 electrons that we've previously drawn as a dot and a cross. The principles involved - promotion of electrons if necessary, then hybridization, followed by the formation of molecular orbitals - can be applied to any covalently-bound molecule.
The bonding in the phosphorus chlorides, PCl3 and PCl5
What's wrong with the simple view of PCl3? This diagram only shows the outer (bonding) electrons.
Nothing is wrong with this! (Although it doesn't account for the shape of the molecule properly.) If you were going to take a more modern look at it, the argument would go like this:
Phosphorus has the electronic structure 1s22s22p63s23px13py13pz1. If we look only at the outer electrons as "electrons-in-boxes":
There are 3 unpaired electrons that can be used to form bonds with 3 chlorine atoms. The four 3-level orbitals hybridise to produce 4 equivalent sp3 hybrids just like in carbon - except that one of these hybrid orbitals contains a lone pair of electrons.
Each of the 3 chlorines then forms a covalent bond by merging the atomic orbital containing its unpaired electron with one of the phosphorus's unpaired electrons to make 3 molecular orbitals. You might wonder whether all this is worth the bother! Probably not! It is worth it with PCl5, though.
What's wrong with the simple view of PCl5?
You will remember that the dots-and-crosses picture of PCl5 looks awkward because the phosphorus doesn't end up with a noble gas structure. This diagram also shows only the outer electrons.
In this case, a more modern view makes things look better by abandoning any pretense of worrying about noble gas structures. If the phosphorus is going to form PCl5 it has first to generate 5 unpaired electrons. It does this by promoting one of the electrons in the 3s orbital to the next available higher energy orbital. Which higher energy orbital? It uses one of the 3d orbitals. You might have expected it to use the 4s orbital because this is the orbital that fills before the 3d when atoms are being built from scratch. Not so! Apart from when you are building the atoms in the first place, the 3d always counts as the lower energy orbital.
This leaves the phosphorus with this arrangement of its electrons:
The 3-level electrons now rearrange (hybridise) themselves to give 5 hybrid orbitals, all of equal energy. They would be called sp3d hybrids because that's what they are made from.
The electrons in each of these orbitals would then share space with electrons from five chlorines to make five new molecular orbitals - and hence five covalent bonds. Why does phosphorus form these extra two bonds? It puts in an amount of energy to promote an electron, which is more than paid back when the new bonds form. Put simply, it is energetically profitable for the phosphorus to form the extra bonds. The advantage of thinking of it in this way is that it completely ignores the question of whether you've got a noble gas structure, and so you don't worry about it.
A non-existent compound - NCl5
Nitrogen is in the same Group of the Periodic Table as phosphorus, and you might expect it to form a similar range of compounds. In fact, it doesn't. For example, the compound NCl3 exists, but there is no such thing as NCl5. Nitrogen is 1s22s22px12py12pz1. The reason that NCl5 doesn't exist is that in order to form five bonds, the nitrogen would have to promote one of its 2s electrons. The problem is that there aren't any 2d orbitals to promote an electron into - and the energy gap to the next level (the 3s) is far too great.
In this case, then, the energy released when the extra bonds are made isn't enough to compensate for the energy needed to promote an electron - and so that promotion doesn't happen. Atoms will form as many bonds as possible provided it is energetically profitable.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Chemical_Bonding/Fundamentals_of_Chemical_Bonding/Covalent_Bonding.txt
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Covalent bonding occurs when pairs of electrons are shared by atoms. Atoms will covalently bond with other atoms in order to gain more stability, which is gained by forming a full electron shell. By sharing their outer most (valence) electrons, atoms can fill up their outer electron shell and gain stability. Nonmetals will readily form covalent bonds with other nonmetals in order to obtain stability, and can form anywhere between one to three covalent bonds with other nonmetals depending on how many valence electrons they posses. Although it is said that atoms share electrons when they form covalent bonds, they do not usually share the electrons equally.
Introduction
Only when two atoms of the same element form a covalent bond are the shared electrons actually shared equally between the atoms. When atoms of different elements share electrons through covalent bonding, the electron will be drawn more toward the atom with the higher electronegativity resulting in a polar covalent bond. When compared to ionic compounds, covalent compounds usually have a lower melting and boiling point, and have less of a tendency to dissolve in water. Covalent compounds can be in a gas, liquid, or solid state and do not conduct electricity or heat well. The types of covalent bonds can be distinguished by looking at the Lewis dot structure of the molecule. For each molecule, there are different names for pairs of electrons, depending if it is shared or not. A pair of electrons that is shared between two atoms is called a bond pair. A pair of electrons that is not shared between two atoms is called a lone pair.
Octet Rule
The Octet Rule requires all atoms in a molecule to have 8 valence electrons--either by sharing, losing or gaining electrons--to become stable. For Covalent bonds, atoms tend to share their electrons with each other to satisfy the Octet Rule. It requires 8 electrons because that is the amount of electrons needed to fill a s- and p- orbital (electron configuration); also known as a noble gas configuration. Each atom wants to become as stable as the noble gases that have their outer valence shell filled because noble gases have a charge of 0. Although it is important to remember the "magic number", 8, note that there are many Octet rule exceptions.
Example: As you can see from the picture below, Phosphorus has only 5 electrons in its outer shell (bolded in red). Argon has a total of 8 electrons (bolded in red), which satisfies the Octet Rule. Phosphorus needs to gain 3 electrons to fulfill the Octet Rule. It wants to be like Argon who has a full outer valence shell.
More examples can be found here.
Single Bonds
A single bond is when two electrons--one pair of electrons--are shared between two atoms. It is depicted by a single line between the two atoms. Although this form of bond is weaker and has a smaller density than a double bond and a triple bond, it is the most stable because it has a lower level of reactivity meaning less vulnerability in losing electrons to atoms that want to steal electrons.
Example 1: HCl
Below is a Lewis dot structure of Hydrogen Chloride demonstrating a single bond. As we can see from the picture below, Hydrogen Chloride has 1 Hydrogen atom and 1 Chlorine atom. Hydrogen has only 1 valence electron whereas Chlorine has 7 valence electrons. To satisfy the Octet Rule, each atom gives out 1 electron to share with each other; thus making a single bond.
Double Bonds
A Double bond is when two atoms share two pairs of electrons with each other. It is depicted by two horizontal lines between two atoms in a molecule. This type of bond is much stronger than a single bond, but less stable; this is due to its greater amount of reactivity compared to a single bond.
2
Below is a Lewis dot structure of Carbon dioxide demonstrating a double bond. As you can see from the picture below, Carbon dioxide has a total of 1 Carbon atom and 2 Oxygen atoms. Each Oxygen atom has 6 valence electrons whereas the Carbon atom only has 4 valence electrons. To satisfy the Octet Rule, Carbon needs 4 more valence electrons. Since each Oxygen atom has 3 lone pairs of electrons, they can each share 1 pair of electrons with Carbon; as a result, filling Carbon's outer valence shell (Satisfying the Octet Rule).
Triple Bond
A Triple bond is when three pairs of electrons are shared between two atoms in a molecule. It is the least stable out of the three general types of covalent bonds. It is very vulnerable to electron thieves!
Example 3: Acetylene
Below is a Lewis dot structure of Acetylene demonstrating a triple bond. As you can see from the picture below, Acetylene has a total of 2 Carbon atoms and 2 Hydrogen atoms. Each Hydrogen atom has 1 valence electron whereas each Carbon atom has 4 valence electrons. Each Carbon needs 4 more electrons and each Hydrogen needs 1 more electron. Hydrogen shares its only electron with Carbon to get a full valence shell. Now Carbon has 5 electrons. Because each Carbon atom has 5 electrons--1 single bond and 3 unpaired electrons--the two Carbons can share their unpaired electrons, forming a triple bond. Now all the atoms are happy with their full outer valence shell.
Polar Covalent Bond
A Polar Covalent Bond is created when the shared electrons between atoms are not equally shared. This occurs when one atom has a higher electronegativity than the atom it is sharing with. The atom with the higher electronegativity will have a stronger pull for electrons (Similiar to a Tug-O-War game, whoever is stronger usually wins). As a result, the shared electrons will be closer to the atom with the higher electronegativity, making it unequally shared. A polar covalent bond will result in the molecule having a slightly positive side (the side containing the atom with a lower electronegativity) and a slightly negative side (containing the atom with the higher electronegativity) because the shared electrons will be displaced toward the atom with the higher electronegativity. As a result of polar covalent bonds, the covalent compound that forms will have an electrostatic potential. This potential will make the resulting molecule slightly polar, allowing it to form weak bonds with other polar molecules. One example of molecules forming weak bonds with each other as a result of an unbalanced electrostatic potential is hydrogen bonding, where a hydrogen atom will interact with an electronegative hydrogen, fluorine, or oxygen atom from another molecule or chemical group.
Example: Water, Sulfide, Ozone, etc.
As you can see from the picture above, Oxygen is the big buff creature with the tattoo of "O" on its arm. The little bunny represents a Hydrogen atom. The blue and red bow tied in the middle of the rope, pulled by the two creatures represents--the shared pair of electrons--a single bond. Because the Hydrogen atom is weaker, the shared pair of electrons will be pulled closer to the Oxygen atom.
Nonpolar Covalent Bond
A Nonpolar Covalent Bond is created when atoms share their electrons equally. This usually occurs when two atoms have similar or the same electron affinity. The closer the values of their electron affinity, the stronger the attraction. This occurs in gas molecules; also known as diatomic elements. Nonpolar covalent bonds have a similar concept as polar covalent bonds; the atom with the higher electronegativity will draw away the electron from the weaker one. Since this statement is true--if we apply this to our diatomic molecules--all the atoms will have the same electronegativity since they are the same kind of element; thus, the electronegativities will cancel each other out and will have a charge of 0 (i.e., a nonpolar covalent bond).
Examples of gas molecules that have a nonpolar covalent bond: Hydrogen gas atom, Nitrogen gas atoms, etc.
As you can see from the picture above, Hydrogen gas has a total of 2 Hydrogen atoms. Each Hydrogen atom has 1 valence electron. Since Hydrogen can only fit a max of 2 valence electrons in its orbital, each Hydrogen atom only needs 1 electron. Each atom has 1 valence electron, so they can just share, giving each atom two electrons each.
Problems
1. Determine the type(s) of bond(s) in
• Benzene (C6H6)
• NO3 (Nitrate)
• F2(Fluorine gas)
Solution:
2. Write the electron configuration and determine how many electrons are needed to achieve the nearest noble-gas configuration for the following:
• Arsenic (As)
• Silicon (Si)
• Tellurium (Te)
Solution:
3. Determine which molecules are polar and which molecules are nonpolar for the following:
• Oxygen gas (O2)
• Hydrochloric acid (HCl)
• Carbon dioxide (CO2)
Solution:
4. Which of the following statements are true? (There can be more than one true statement.)
1. A covalent bond is the same as a ionic bond.
2. The Octet rule only applys to molecules with covalent bonds.
3. A molecule is polar if the shared electrons are equally shared.
4. A molecule is nonpolar if the shared electrons are are equally shared.
5. Methane gas (CH4) has a nonpolar covalent bond because it is a gas.
Solution: Only d) is true.
5. Match each atom or molecule with its corresponding letter(s):
• Nitrogen gas
• Argon
• Carbon monoxide
• Hydrogen gas
a) Nonpolar covalent bond
b) Polar covalent bond
c) Follows the Octet Rule
d) Noble gas
e) Two lone pairs
f) Single bond
Solution:
• Nitrogen gas: a), c), e)
• Argon: c), d)
• Carbon monoxide: b), c), e)
• Hydrogen gas: c), f)
Contributors and Attributions
• Camy Fung (UCD), Nima Mirzaee (UCD)
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Chemical_Bonding/Fundamentals_of_Chemical_Bonding/Covalent_Bonds.txt
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There are two types of atomic bonds - ionic bonds and covalent bonds. They differ in their structure and properties. Covalent bonds consist of pairs of electrons shared by two atoms, and bind the atoms in a fixed orientation. Relatively high energies are required to break them (50 - 200 kcal/mol). Whether two atoms can form a covalent bond depends upon their electronegativity i.e. the power of an atom in a molecule to attract electrons to itself. If two atoms differ considerably in their electronegativity - as sodium and chloride do - then one of the atoms will lose its electron to the other atom. This results in a positively charged ion (cation) and negatively charged ion (anion). The bond between these two ions is called an ionic bond.
Covalent Bonds Ionic Bonds
State at room temperature: Liquid or gaseous Solid
Polarity: Low High
Formation: A covalent bond is formed between two non-metals that have similar electronegativities. Neither atom is "strong" enough to attract electrons from the other. For stabilization, they share their electrons from outer molecular orbit with others An ionic bond is formed between a metal and a non-metal. Non-metals(-ve ion) are "stronger" than the metal(+ve ion) and can get electrons very easily from the metal. These two opposite ions attract each other and form the ionic bond.
Shape: Definite shape No definite shape
Melting point: low High
What is it?: Covalent bonding is a form of chemical bonding between two non metallic atoms which is characterized by the sharing of pairs of electrons between atoms and other covalent bonds. Ionic bond, also known as electrovalent bond, is a type of bond formed from the electrostatic attraction between oppositely charged ions in a chemical compound. These kinds of bonds occur mainly between a metallic and a non metallic atom.
Boiling point: Low High
Examples: Methane (CH4), Hydrochloric acid (HCl) Sodium chloride (NaCl), Sulfuric Acid (H2SO4 )
Occurs between: Two non-metals One metal and one non-metal
Electrostatic Potential maps
Electrostatic potential maps, also known as electrostatic potential energy maps, or molecular electrical potential surfaces, illustrate the charge distributions of molecules three dimensionally. These maps allow us to visualize variably charged regions of a molecule. Knowledge of the charge distributions can be used to determine how molecules interact with one another.
Introduction
Electrostatic potential maps are very useful three dimensional diagrams of molecules. They enable us to visualize the charge distributions of molecules and charge related properties of molecules. They also allow us to visualize the size and shape of molecules. In organic chemistry, electrostatic potential maps are invaluable in predicting the behavior of complex molecules.
The first step involved in creating an electrostatic potential map is collecting a very specific type of data: electrostatic potential energy. An advanced computer program calculates the electrostatic potential energy at a set distance from the nuclei of the molecule. Electrostatic potential energy is fundamentally a measure of the strength of the nearby charges, nuclei and electrons, at a particular position.
To accurately analyze the charge distribution of a molecule, a very large quantity of electrostatic potential energy values must be calculated. The best way to convey this data is to visually represent it, as in an electrostatic potential map. A computer program then imposes the calculated data onto an electron density model of the molecule derived from the Schrödinger equation. To make the electrostatic potential energy data easy to interpret, a color spectrum, with red as the lowest electrostatic potential energy value and blue as the highest, is employed to convey the varying intensities of the electrostatic potential energy values.
Analogous System
Electrostatic potential maps involve a number of basic concepts. The actual process of mapping the electrostatic potentials of a molecule, however, involves factors that complicate these fundamental concepts. An analogous system will be employed to introduce these basic concepts.
Imagine that there is a special type of mine. This mine is simply an explosive with some charged components on top of it. The circles with positive and negative charges in them are the charged components. If the electric field of the electric components are significantly disturbed, the mine triggers and explodes. The disarming device is positively charged. To disarm the mine, the disarming device must take the path of least electric resistance and touch the first charged mine component on this path. Deviating from this minimal energy path will cause a significant disturbance and the mine will explode. The specific charged components within the mine are known.
Q. How do you disarm the following mine?
The mine with positive charge and negative charge.
A. Touch the bottom most portion of negatively charged component, red, with the disarming device.
Introduction to Coulomb's Law and Electrostatic Energy
Coulomb's Law Formula
$F=k \dfrac{q_aq_b}{r^2}$
with
• Molecular electrostatic potential maps also illustrate information about the charge distribution of a molecule. Electrostatic potential maps convey information about the charge distribution of a molecule because of the properties of the nucleus and nature of electrostatic potential energy. For simplicity, consider moving a positively charged test charge along the spherical isosurface of an atom. The positively charged nucleus emits a radially constant electric field. A region of higher than average electrostatic potential energy indicates the presence of a stronger positive charge or a weaker negative charger. Given the consistency of the nucleuses positive charge, the higher potential energy value indicates the absence of negative charges, which would mean that there are fewer electrons in this region. The converse is also true. Thus a high electrostatic potential indicates the relative absence of electrons and a low electrostatic potential indicates an abundance of electrons. This property of electrostatic potentials can be extrapolated to molecules as well.
Here is a simplified visual representation of the relationship between charge distribution and electrostatic potential. Keep in mind the equation used to find the electrostatic potential.
$\text{Total Electrostatic Potential Energy= \sum_i \text{Electrostatic Potential Energy}$
where
$\text{Potential Energy}=K \dfrac{q_1q_2}{r}$
and $K= \text{Coulomb's Constant}$.
Contributors and Attributions
• Thomas Bottyan
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Chemical_Bonding/Fundamentals_of_Chemical_Bonding/Covalent_Bonds_vs_Ionic_Bonds.txt
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Ionic bonding is the complete transfer of valence electron(s) between atoms and is a type of chemical bond that generates two oppositely charged ions. It is observed because metals with few electrons in its outer-most orbital. By losing those electrons, these metals can achieve noble-gas configuration and satisfy the octet rule. Similarly, nonmetals that have close to 8 electrons in its valence shell tend to readily accept electrons to achieve its noble gas configuration.
Introduction
In ionic bonding, electrons are transferred from one atom to another resulting in the formation of positive and negative ions. The electrostatic attractions between the positive and negative ions hold the compound together. The predicted overall energy of the ionic bonding process, which includes the ionization energy of the metal and electron affinity of the nonmetal, is usually positive, indicating that the reaction is endothermic and unfavorable. However, this reaction is highly favorable because of their electrostatic attraction. At the most ideal inter-atomic distance, attraction between these particles releases enough energy to facilitate the reaction. Most ionic compounds tend to dissociate in polar solvents because they are often polar. This phenomenon is due to the opposite charges on each ions.
At a simple level, a lot of importance is attached to the electronic structures of noble gases like neon or argon which have eight electrons in their outer energy levels (or two in the case of helium). These noble gas structures are thought of as being in some way a "desirable" thing for an atom to have. One may well have been left with the strong impression that when other atoms react, they try to organize things such that their outer levels are either completely full or completely empty.
Example: Bonding in NaCl
Sodium Chloride:
• Sodium (2,8,1) has 1 electron more than a stable noble gas structure (2,8). If it gave away that electron it would become more stable.
• Chlorine (2,8,7) has 1 electron short of a stable noble gas structure (2,8,8). If it could gain an electron from somewhere it too would become more stable.
The answer is obvious. If a sodium atom gives an electron to a chlorine atom, both become more stable.
The sodium has lost an electron, so it no longer has equal numbers of electrons and protons. Because it has one more proton than electron, it has a charge of 1+. If electrons are lost from an atom, positive ions are formed. Positive ions are sometimes called cations.
The chlorine has gained an electron, so it now has one more electron than proton. It therefore has a charge of 1-. If electrons are gained by an atom, negative ions are formed. A negative ion is sometimes called an anion.
The nature of the bond
The sodium ions and chloride ions are held together by the strong electrostatic attractions between the positive and negative charges. You need one sodium atom to provide the extra electron for one chlorine atom, so they combine together 1:1. The formula is therefore NaCl.
Example 1: Bonding in MgO
Magnesium Oxide:
Again, noble gas structures are formed, and the magnesium oxide is held together by very strong attractions between the ions. The ionic bonding is stronger than in sodium chloride because this time you have 2+ ions attracting 2- ions. The greater the charge, the greater the attraction. The formula of magnesium oxide is MgO.
2
Calcium Chloride:
This time you need two chlorines to use up the two outer electrons in the calcium. The formula of calcium chloride is therefore CaCl2.
O
Potassium Oxide:
Again, noble gas structures are formed. It takes two potassiums to supply the electrons the oxygen needs. The formula of potassium oxide is K2O.
Some Stable Ions do not have Noble Gas Configurations
You may have come across some of the following ions, which are all perfectly stable, but not one of them has a noble gas structure.
Fe3+ [Ar]3d5
Cu2+ [Ar]3d9
Zn2+ [Ar]3d10
Ag+ [Kr]4d10
Pb2+ [Xe]4f145d106s2
What needs modifying is the view that there is something magic about noble gas structures. There are far more ions which do not have noble gas structures than there are which do.
• Noble gases (apart from helium) have an outer electronic structure ns2np6. Apart from some elements at the beginning of a transition series (scandium forming Sc3+ with an argon structure, for example), all transition metal elements and any metals following a transition series (like tin and lead in Group 4, for example) will have structures like those above.
• That means that the only elements to form positive ions with noble gas structures (apart from odd ones like scandium) are those in groups 1 and 2 of the Periodic Table and aluminum in group 3 (boron in group 3 does not form ions).
• Negative ions are tidier! Those elements in Groups 5, 6 and 7 which form simple negative ions all have noble gas structures.
If elements are not aiming for noble gas structures when they form ions, what decides how many electrons are transferred? The answer lies in the energetics of the process by which the compound is made.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Chemical_Bonding/Fundamentals_of_Chemical_Bonding/Ionic_Bonds.txt
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In the early 1900's, Paul Drüde came up with the "sea of electrons" metallic bonding theory by modeling metals as a mixture of atomic cores (atomic cores = positive nuclei + inner shell of electrons) and valence electrons. Metallic bonds occur among metal atoms. Whereas ionic bonds join metals to non-metals, metallic bonding joins a bulk of metal atoms. A sheet of aluminum foil and a copper wire are both places where you can see metallic bonding in action.
Metals tend to have high melting points and boiling points suggesting strong bonds between the atoms. Even a soft metal like sodium (melting point 97.8°C) melts at a considerably higher temperature than the element (neon) which precedes it in the Periodic Table. Sodium has the electronic structure 1s22s22p63s1. When sodium atoms come together, the electron in the 3s atomic orbital of one sodium atom shares space with the corresponding electron on a neighboring atom to form a molecular orbital - in much the same sort of way that a covalent bond is formed.
The difference, however, is that each sodium atom is being touched by eight other sodium atoms - and the sharing occurs between the central atom and the 3s orbitals on all of the eight other atoms. Each of these eight is in turn being touched by eight sodium atoms, which in turn are touched by eight atoms - and so on and so on, until you have taken in all the atoms in that lump of sodium. All of the 3s orbitals on all of the atoms overlap to give a vast number of molecular orbitals that extend over the whole piece of metal. There have to be huge numbers of molecular orbitals, of course, because any orbital can only hold two electrons.
The electrons can move freely within these molecular orbitals, and so each electron becomes detached from its parent atom. The electrons are said to be delocalized. The metal is held together by the strong forces of attraction between the positive nuclei and the delocalized electrons (Figure $1$).
This is sometimes described as "an array of positive ions in a sea of electrons". If you are going to use this view, beware! Is a metal made up of atoms or ions? It is made of atoms. Each positive center in the diagram represents all the rest of the atom apart from the outer electron, but that electron has not been lost - it may no longer have an attachment to a particular atom, but it's still there in the structure. Sodium metal is therefore written as $\ce{Na}$, not $\ce{Na^+}$.
Example $1$: Metallic bonding in magnesium
Use the sea of electrons model to explain why Magnesium has a higher melting point (650 °C) than sodium (97.79 °C).
Solution
If you work through the same argument above for sodium with magnesium, you end up with stronger bonds and hence a higher melting point.
Magnesium has the outer electronic structure 3s2. Both of these electrons become delocalized, so the "sea" has twice the electron density as it does in sodium. The remaining "ions" also have twice the charge (if you are going to use this particular view of the metal bond) and so there will be more attraction between "ions" and "sea".
More realistically, each magnesium atom has 12 protons in the nucleus compared with sodium's 11. In both cases, the nucleus is screened from the delocalized electrons by the same number of inner electrons - the 10 electrons in the 1s2 2s2 2p6 orbitals. That means that there will be a net pull from the magnesium nucleus of 2+, but only 1+ from the sodium nucleus.
So not only will there be a greater number of delocalized electrons in magnesium, but there will also be a greater attraction for them from the magnesium nuclei. Magnesium atoms also have a slightly smaller radius than sodium atoms, and so the delocalized electrons are closer to the nuclei. Each magnesium atom also has twelve near neighbors rather than sodium's eight. Both of these factors increase the strength of the bond still further.
Note: Transition metals tend to have particularly high melting points and boiling points. The reason is that they can involve the 3d electrons in the delocalization as well as the 4s. The more electrons you can involve, the stronger the attractions tend to be.
Bulk properties of metals
Metals have several qualities that are unique, such as the ability to conduct electricity and heat, a low ionization energy, and a low electronegativity (so they will give up electrons easily to form cations). Their physical properties include a lustrous (shiny) appearance, and they are malleable and ductile. Metals have a crystal structure but can be easily deformed. In this model, the valence electrons are free, delocalized, mobile, and not associated with any particular atom. This model may account for:
• Conductivity: Since the electrons are free, if electrons from an outside source were pushed into a metal wire at one end (Figure $2$), the electrons would move through the wire and come out at the other end at the same rate (conductivity is the movement of charge).
• Malleability and Ductility: The electron-sea model of metals not only explains their electrical properties but their malleability and ductility as well. The sea of electrons surrounding the protons acts like a cushion, and so when the metal is hammered on, for instance, the overall composition of the structure of the metal is not harmed or changed. The protons may be rearranged but the sea of electrons with adjust to the new formation of protons and keep the metal intact. When one layer of ions in an electron sea moves along one space with respect to the layer below it, the crystal structure does not fracture but is only deformed (Figure $3$).
• Heat capacity: This is explained by the ability of free electrons to move about the solid.
• Luster: The free electrons can absorb photons in the "sea," so metals are opaque-looking. Electrons on the surface can bounce back light at the same frequency that the light hits the surface, therefore the metal appears to be shiny.
However, these observations are only qualitative, and not quantitative, so they cannot be tested. The "Sea of Electrons" theory stands today only as an oversimplified model of how metallic bonding works.
In a molten metal, the metallic bond is still present, although the ordered structure has been broken down. The metallic bond is not fully broken until the metal boils. That means that boiling point is actually a better guide to the strength of the metallic bond than melting point is. On melting, the bond is loosened, not broken. The strength of a metallic bond depends on three things:
1. The number of electrons that become delocalized from the metal
2. The charge of the cation (metal).
3. The size of the cation.
A strong metallic bond will be the result of more delocalized electrons, which causes the effective nuclear charge on electrons on the cation to increase, in effect making the size of the cation smaller. Metallic bonds are strong and require a great deal of energy to break, and therefore metals have high melting and boiling points. A metallic bonding theory must explain how so much bonding can occur with such few electrons (since metals are located on the left side of the periodic table and do not have many electrons in their valence shells). The theory must also account for all of a metal's unique chemical and physical properties.
Expanding the Range of Bonding Possible
Previously, we argued that bonding between atoms can classified as range of possible bonding between ionic bonds (fully charge transfer) and covalent bonds (fully shared electrons). When two atoms of slightly differing electronegativities come together to form a covalent bond, one atom attracts the electrons more than the other; this is called a polar covalent bond. However, simple “ionic” and “covalent” bonding are idealized concepts and most bonds exist on a two-dimensional continuum described by the van Arkel-Ketelaar Triangle (Figure $4$).
Bond triangles or van Arkel–Ketelaar triangles (named after Anton Eduard van Arkel and J. A. A. Ketelaar) are triangles used for showing different compounds in varying degrees of ionic, metallic and covalent bonding. In 1941 van Arkel recognized three extreme materials and associated bonding types. Using 36 main group elements, such as metals, metalloids and non-metals, he placed ionic, metallic and covalent bonds on the corners of an equilateral triangle, as well as suggested intermediate species. The bond triangle shows that chemical bonds are not just particular bonds of a specific type. Rather, bond types are interconnected and different compounds have varying degrees of different bonding character (for example, polar covalent bonds).
Using electronegativity - two compound average electronegativity on x-axis of Figure $4$.
$\sum \chi = \dfrac{\chi_A + \chi_B}{2} \label{sum}$
and electronegativity difference on y-axis,
$\Delta \chi = | \chi_A - \chi_B | \label{diff}$
we can rate the dominant bond between the compounds. On the right side of Figure $4$ (from ionic to covalent) should be compounds with varying difference in electronegativity. The compounds with equal electronegativity, such as $\ce{Cl2}$ (chlorine) are placed in the covalent corner, while the ionic corner has compounds with large electronegativity difference, such as $\ce{NaCl}$ (table salt). The bottom side (from metallic to covalent) contains compounds with varying degree of directionality in the bond. At one extreme is metallic bonds with delocalized bonding and at the other are covalent bonds in which the orbitals overlap in a particular direction. The left side (from ionic to metallic) is meant for delocalized bonds with varying electronegativity difference.
The Three Extremes in bonding
In general:
• Metallic bonds have low $\Delta \chi$ and low average $\sum\chi$.
• Ionic bonds have moderate-to-high $\Delta \chi$ and moderate values of average $\sum \chi$.
• Covalent bonds have moderate to high average $\sum \chi$ and can exist with moderately low $\Delta \chi$.
Example $2$
Use the tables of electronegativities (Table A2) and Figure $4$ to estimate the following values
• difference in electronegativity ($\Delta \chi$)
• average electronegativity in a bond ($\sum \chi$)
• percent ionic character
• likely bond type
for the selected compounds:
1. $\ce{AsH}$ (e.g., in arsine $AsH$)
2. $\ce{SrLi}$
3. $\ce{KF}$.
Solution
a: $\ce{AsH}$
• The electronegativity of $\ce{As}$ is 2.18
• The electronegativity of $\ce{H}$ is 2.22
Using Equations \ref{sum} and \ref{diff}:
\begin{align*} \sum \chi &= \dfrac{\chi_A + \chi_B}{2} \[4pt] &=\dfrac{2.18 + 2.22}{2} \[4pt] &= 2.2 \end{align*}
\begin{align*} \Delta \chi &= \chi_A - \chi_B \[4pt] &= 2.18 - 2.22 \[4pt] &= 0.04 \end{align*}
• From Figure $4$, the bond is fairly nonpolar and has a low ionic character (10% or less)
• The bonding is in the middle of a covalent bond and a metallic bond
b: $\ce{SrLi}$
• The electronegativity of $\ce{Sr}$ is 0.95
• The electronegativity of $\ce{Li}$ is 0.98
Using Equations \ref{sum} and \ref{diff}:
\begin{align*} \sum \chi &= \dfrac{\chi_A + \chi_B}{2} \[4pt] &=\dfrac{0.95 + 0.98}{2} \[4pt] &= 0.965 \end{align*}
\begin{align*} \Delta \chi &= \chi_A - \chi_B \[4pt] &= 0.98 - 0.95 \[4pt] &= 0.025 \end{align*}
• From Figure $4$, the bond is fairly nonpolar and has a low ionic character (~3% or less)
• The bonding is likely metallic.
c: $\ce{KF}$
• The electronegativity of $\ce{K}$ is 0.82
• The electronegativity of $\ce{F}$ is 3.98
Using Equations \ref{sum} and \ref{diff}:
\begin{align*} \sum \chi &= \dfrac{\chi_A + \chi_B}{2} \[4pt] &=\dfrac{0.82 + 3.98}{2} \[4pt] &= 2.4 \end{align*}
\begin{align*} \Delta \chi &= \chi_A - \chi_B \[4pt] &= | 0.82 - 3.98 | \[4pt] &= 3.16 \end{align*}
• From Figure $4$, the bond is fairly polar and has a high ionic character (~75%)
• The bonding is likely ionic.
Exercise $2$
Contrast the bonding of $\ce{NaCl}$ and silicon tetrafluoride.
Answer
$\ce{NaCl}$ is an ionic crystal structure, and an electrolyte when dissolved in water; $\Delta \chi =1.58$, average $\sum \chi =1.79$, while silicon tetrafluoride is covalent (molecular, non-polar gas; $\Delta \chi =2.08$, average $\sum \chi =2.94$.
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Non-singular covalent bonds are also known as "multiple covalent bonds." There are three types of covalent bonds: single, double, and triple. The name "Non-singular covalent bonds" speaks for itself. Non-singular covalent bonds are covalent bonds that need to share more then one electron pair, so they create double and triple bonds.
Introduction
The main motive an atom has to bond with other atoms is to fulfill it's need to have eight valence shell electrons (with some exceptions, i.e. Hydrogen). This defines the octet rule. Two different orbital overlaps occur with multiple bonds. The difference between single bonds and multiple bonds is that multiple bonds have one or two pi bonds (one for a double and two for a triple) in addition to the sigma bonds that a single bond creates. The sigma bond with the pi bond is what makes double and triple bonds so strong compared to single bonds. The more bonds there are means there is more overlap between the orbitals. Bond length is also effected by the overlap of the two orbitals, the more overlap the shorter the bond length.
Single Bond= One Sigma bond
Double Bond = One Sigma + One Pi bond
Triple Bond = One Sigma + Two Pi bonds
Sigma "σ"
A Sigma bond "σ" is the strongest chemical covalent bond. It is created by the "end-to-end" overlap of atomic orbitals. Going more in depth, it is in which the region of electron sharing is along the imaginary line which connects the bonded atoms. They can be formed from two s-orbitals, two p-orbitals, one s- and p- orbital, or with sp hybrid orbitals. The sigma bond is like a cylinder pipe connecting the two orbitals. The two electrons can be found somewhere in the region of space within the sigma bond. The sigma bond is symmetric and can freely rotate around the bond axis.
Pi Bonds "π"
Pi Bonds "π" are created by the "side-to-side" overlapping of two parallel p-orbitals (pictured below). A pi bond is a weaker chemical covalent bond than a sigma bond (since π bonds have a smaller overlap between the orbitals), but when it is put with a sigma bond it creates a much stronger hold between the atoms, thus double and triple bonds are stronger then single bonds.The pi bond looks like two macaroni's sandwiching the sigma bond. Above and below the Pi bonds molecular plane has high electron charge densities. Electrons in a pi bond are sometimes referred to as "Pi electrons." The pi bond is a region of space where you can find the two pi electrons that create the bond. Pi bonds create cis-trans isomers since they prevent free rotation around the bond.
Example of Pi bond formation with ethylene, C2H4 :
Delta bonds "δ"
These are where four lobes of one involved electron orbital overlap four lobes of the other involved electron orbital. Of the orbital's node planes, two (and no more), go through both atoms. The Greek letter δ in their name refers to d orbitals, since the orbital symmetry of the delta bond is the same as that of the usual (4-lobed) type of d orbital when seen down the bond axis.
In sufficiently-large atoms, occupied d-orbitals are low enough in energy to participate in bonding. Delta bonds are usually observed in organometallic species. Some ruthenium and molybdenum compounds contain a quadruple bond, which can only be explained by invoking the delta bond. It is possible to excite electrons in acetylene from lower-energy nonbonding orbitals to form a delta bond between the two carbon triple bonds. This is because the orbital symmetry of the pi antibonding orbital is the same as that of the delta bond.
Higher order bonds
Theoretical chemists have conjectured that higher-order bonds (phi bonds and gamma bonds, corresponding to overlap of f and g orbitals) are possible, with even more overlapping lobes of their component atomic orbitals, but no experimental evidence for these has yet been observed.
Outside links
• en.Wikipedia.org/wiki/Pi_bonds
• chemed.chem.wisc.edu/chempath...Bonds-576.html
Problems
1. How many electrons can be found in a pi bond ("pi electrons")?
2. How many sigma bonds and how many pi bonds does carbonic acid, H2CO3, need?
3. What has a longer bondlenght: a compound with one or two pi bonds?
Solutions
1) 2 electrons
2) 5 sigma bonds and 1 pi bond
(-) sigma bonds (=) one sigma and one pi bond
3) A compound with one pi bond would have a longer bond length since it is a double bond.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Chemical_Bonding/Fundamentals_of_Chemical_Bonding/Non-Singular_Covalent_Bonds.txt
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Skills to Be Tested
1. Identify the central atom in a molecule containing more than two atoms as a start.
2. Identify the number of valence electrons of any element. This concept is important, because you need to know the number of valence electrons in order to write a Lewis dot structure for the molecule in question.
3. Count the number of VSEPR pairs or steric number (SN) for the central atom in a molecule. You need this number in order to describe or predict the shape of the molecule in question.
4. Determine the number of lone electron pairs that are not shared with other atoms. Often, a Lewis dot structure is useful to help you count this number.
5. Predict the shape of molecules or ions as the key concept of VSEPR theory. From the shape and by applying the idea that lone electron pairs take up more space, you can predict the bond angles within 5% of the observed values.
6. Predict the values of bond angles and describe the hybrid orbitals used by the central atoms in the molecules or ions.
Valence-Shell Electron-Pair Repulsion (VSEPR) Models
The 3-dimensional structure of $\ce{BF3}$ is different from $\ce{PF3}$, and this is difficult to comphrend by considering their formulas alone. However, the Lewis dot structures for them are different, and the electron pair in $\textrm{:PF}_3$ is the reason for its structure being different from $\ce{BF3}$ (no lone pair). Three-dimensional arrangements of atoms or bonds in molecules are important properties as are bond lengths, bond angles and bond energies. The Lewis dot symbols led us to see the non-bonding electron pairs, whose role in determining the shape of a molecule was examined by N.V. Sidgwick and H.E. Powell in 1940, and later by R.S. Nyholm and R.J. Gillespie. They have developed an extensive rationale called valence-shell electron-pair repulsion (VSEPR) model of molecular geometry.
Molecular shapes and steric numbers (SN)
Example SN Descriptor
$\ce{BeCl2}$, $\ce{CO2}$ 2 Linear
$\ce{BF3}$, $\ce{SO3}$
$\ce{SO2E}$, $\ce{OO2E}$
3 Trigonal planar
bent
$\ce{CH4}$
$\ce{NH3E}$
$\ce{H2OE2}$
4 Tetrahedral
pyramidal
bent
$\ce{PF5}$
$\ce{SF4E}$
$\ce{ClF3E2}$
5 Trigonal bypyramidal
butterfly
T-shape
$\ce{SF6}$, $\ce{OIF5}$
$\ce{BrF5E}$
$\ce{XeF4E2}$
6 octahedral
pyramidal
square planar
$\ce{E}$ represents a lone electron pair. SN is also called the number of VSEPR pairs or number of electron pairs.
The Valence-Shell Electron-Pair Repulsion (VSEPR) models consider the unshared pairs (or lone electron pairs) and the bonding electrons. These considerations of lone and bonding electron pairs give an excellent explanation about the molecular shapes. The VSEPR model counts both bonding and nonbonding (lone) electron pairs, and calls the total number of pairs the steric number (SN). If the element A has m atoms bonded to it and n nonbonding pairs, then
$SN = m + n$
SN is useful for predicting shapes of molecules. If $\ce{X}$ is any atom bonded to $\ce{A}$ (in single, double, or triple bond), a molecule may be represented by $\mathrm{AX_mE_n}$ where $\ce{E}$ denotes a lone electron pair. This formula enables us to predict its geometry. The common SN, descriptor, and examples are given in the table on the right. Note that the SN is also called the number of VSEPR pairs or number of electron pairs. The VSEPR model has another general rule:
Lone pairs of electrons take up more space than bonded pairs making the bond angle, say $\ce{\mathit{H-O-H}}$ for water less than the tetrahedral angle of 109.5 °. Actually, the $\ce{\mathit{H-O-H}}$ angle in water is 105 °.
The geometry of the molecules with their SNs equal to 2 to 6 are given in the Table 1. The first line for each is the shape including the lone electron pair(s). If the lone electron pairs are ignored, the geometry of the molecule is given by another descriptor. To get an idea about the shapes of molecules and ions, three dimensional models are the best to use. However, good computer graphics sometimes also illustrate very well. The link VSEPR Illustration: View and manipulate molecular models gives excellent graphics, and you may enjoy seeing some of the graphics of the molecules.
Confidence Building Questions
1. What is the central atom in $\ce{CCl2O}$?
Hint: Carbon is the central atom.
Skill:
Identifying the central atom is a good start to drawing a structure:
Cl
>C=O
Cl
2. What is the number of valence electrons in carbon?
Hint: Sulfur has 4 valence electrons.
Skill:
Give the number of valence electrons of any main-group element.
3. What is the number of lone pair electrons around the central atom $\ce{S}$ in $\ce{SCl2O}$?
Hint: There is no lone electron pair.
Skill:
Draw a Lewis dot structure, and express it in the form $\ce{CCl2=O}$.
..
:Cl:
. /
:O=C the structure of phosgene.
' \..
:Cl:
This carbonyl chloride is also called phosgene, a colourless, chemically reactive, highly toxic gas having an odour like that of musty hay. It was used during World War I against troops.
4. What is the central atom in $\ce{SCl2O}$?
Hint: Sulfur is the central atom.
Skill:
Identifying the central atom is a good start to drawing a structure:
Cl
>S=O
Cl
5. What is the number of valence electrons in sulfur?
Hint: Sulfur has 6 valence electrons.
Skill:
Give the number of valence electrons of any main-group element.
6. What is the number of lone pair electrons around the central atom $\ce{S}$ in $\ce{SCl2O}$?
Hint: There is one lone electron pair.
Skill:
Draw a Lewis dot structure, and express in the form $\textrm{:SCl}_2\ce{=O}$ or $\ce{SCl2=OE}$.
Cl
/
O=S: the structure.
\
Cl
7. What is the number of lone pairs of electrons around $\ce{O}$ in $\ce{H2O}$?
Hint: There are two lone electron pairs.
Skill:
The structure is
H ..
>O
H ''
8. How many lone electron pairs are there around $\ce{C}$ in $\ce{CO2}$?
Hint: There is no lone electron pair for carbon dioxide.
Skill:
From the structure of $\ce{O=C=O}$, figure out the number of lone electron pairs. $\ce{CO2}$ is linear; so is $\ce{H-CN}$.
9. What is the molecular shape of $\ce{ClF3}$ (chlorine-trifluoride)?
Hint: The molecule has a T shape.
Skill:
Predict the shape of a molecule $\textrm{::ClF}_3$ or $\ce{ClF3E2}$.
F
|
F---Cl---F
.. ..
10. What is the bond angle (in degrees) in a $\ce{CH4}$ molecule?
Hint: The angle for ideal tetrahedral molecule is 109.5 degrees.
Skill:
Calculate bond angles from the geometry.
11. What is the shape of $\ce{SF6}$? Give a geometric term for the configuration of the 6 $\ce{F}$'s around $\ce{S}$.
Hint: The geometrical shape is octahedral.
Skill:
Apply terms linear, T shape, seesaw, trigonal planar, trigonal bipyramidal, tetrahedral, octahedral etc. to describe the geometric shapes of molecules.
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Molecular geometry, also known as the molecular structure, is the three-dimensional structure or arrangement of atoms in a molecule. Understanding the molecular structure of a compound can help determine the polarity, reactivity, phase of matter, color, magnetism, as well as the biological activity.
Introduction
To determine the shapes of molecules, we must become acquainted with the Lewis electron dot structure. Although the Lewis theory does not determine the shapes of molecules, it is the first step in predicting shapes of molecules. The Lewis structure helps us identify the bond pairs and the lone pairs. Then, with the Lewis structure, we apply the valence-shell electron-pair repulsion (VSPER) theory to determine the molecular geometry and the electron-group geometry.
To identify and have a complete description of the three-dimensional shape of a molecule, we need to know also learn about state the bond angle as well. Lewis Electron Dot Structures play crucial role in determining the geometry of molecules because it helps us identify the valence electrons. To learn how to draw a Lewis electron dot structure click the link above.
Valence-Shell Electron-Pair Repulsion Theory
Now that we have a background in the Lewis electron dot structure we can use it to locate the the valence electrons of the center atom. The valence-shell electron-pair repulsion (VSEPR) theory states that electron pairs repel each other whether or not they are in bond pairs or in lone pairs. Thus, electron pairs will spread themselves as far from each other as possible to minimize repulsion. VSEPR focuses not only on electron pairs, but it also focus on electron groups as a whole. An electron group can be an electron pair, a lone pair, a single unpaired electron, a double bond or a triple bond on the center atom. Using the VSEPR theory, the electron bond pairs and lone pairs on the center atom will help us predict the shape of a molecule.
The shape of a molecule is determined by the location of the nuclei and its electrons. The electrons and the nuclei settle into positions that minimize repulsion and maximize attraction. Thus, the molecule's shape reflects its equilibrium state in which it has the lowest possible energy in the system. Although VSEPR theory predicts the distribution of the electrons, we have to take in consideration of the actual determinant of the molecular shape. We separate this into two categories, the electron-group geometry and the molecular geometry.
Electron-group geometry is determined by the number of electron groups.
Number of electron groups Name of electron group geometry
2 linear
3 trigonal-planar
4 tetrahedral
5 trigonal-bipyramidal
6 octahedral
Molecular geometry, on the other hand, depends on not only on the number of electron groups, but also on the number of lone pairs. When the electron groups are all bond pairs, they are named exactly like the electron-group geometry. See the chart below for more information on how they are named depending on the number of lone pairs the molecule has.
VSEPR Notation
As stated above, molecular geometry and electron-group geometry are the same when there are no lone pairs. The VSEPR notation for these molecules are AXn. "A" represents the central atom and n represents the number of bonds with the central atom. When lone pairs are present, the letter Ex is added. The x represents the number of lone pairs present in the molecule. For example, a molecule with two bond pairs and two lone pairs would have this notation: AX2E2.
Geometry of Molecules Chart
Number of Electron Groups Electron-Group Geometry Number of Lone Pairs VSEPR Notation Molecular Geometry Ideal Bond Angles Examples
2 linear 1 AX2 180° BeH2
3 trigonal-planar 0 AX3 120° CO32-
1 AX2E 120° O3
4 tetrahedral 0 AX4
Tetrahedral
109.5° S042-
1 AX3E 109.5° H3O+
2 AX2E2 109.5° H2O
5 trigonal-bipyramidal 0 AX5 90°, 120° PF5
1 AX4Eb 90°, 120° TeCl4
2 AX3E2 90° ClF3
3 AX2E3 180° I3-
6 octahedral 0 AX6
octahedral
90° PF6-
1 AX5E 90° SbCl52-
2 AX4E2 90°
ICl4-
Example $1$:
Lets try determining the geometric structures of H2O and CO2. So starting off by drawing the Lewis structure:
H2O:
Water has four electron groups so it falls under tetrahedral for the electron-group geometry. The four electron groups are the 2 single bonds to Hydrogen and the 2 lone pairs of Oxygen. Since water has two lone pairs it's molecular shape is bent. According to the VSEPR theory, the electrons want to minimize repulsion, so as a result, the lone pairs are adjacent from each other.
CO2:
Carbon dioxide has two electron groups and no lone pairs. Carbon dioxide is therefore linear in electron-group geometry and in molecular geometry. The shape of CO2 is linear because there are no lone pairs affecting the orientation of the molecule. Therefore, the linear orientation minimizes the repulsion forces.
Molecules with More than One Central Atom
The VSEPR theory not only applies to one central atom, but it applies to molecules with more than one central atom. We take in account the geometric distribution of the terminal atoms around each central atom. For the final description, we combine the separate description of each atom. In other words, we take long chain molecules and break it down into pieces. Each piece will form a particular shape. Follow the example provided below:
Butane is C4H10. C-C-C-C is the simplified structural formula where the Hydrogens (not shown) are implied to have single bonds to Carbon. You can view a better structural formula of butane at en.Wikipedia.org/wiki/File:Butane-2D-flat.png
If we break down each Carbon, the central atoms, into pieces, we can determine the relative shape of each section. Let's start with the leftmost side. We see that C has three single bonds to 2 Hydrogens and one single bond to Carbon. That means that we have 4 electron groups. By checking the geometry of molecules chart above, we have a tetrahedral shape. Now, we move on to the next Carbon. This Carbon has 2 single bonds to 2 Carbons and 2 single bonds to 2 Hydrogens. Again, we have 4 electron groups which result in a tetrahedral. Continuing this trend, we have another tetrahedral with single bonds attached to Hydrogen and Carbon atoms. As for the rightmost Carbon, we also have a tetrahedral where Carbon binds with one Carbon and 3 Hydrogens.
Let me recap. We took a look at butane provided by the wonderful Wikipedia link. We, then, broke the molecule into parts. We did this by looking at a particular central atom. In this case, we have 4 central atoms, all Carbon. By breaking the molecule into 4 parts (each part looks at 1 of the 4 Carbons), we determine how many electron groups there are and find out the shapes.
We aren't done, yet! We need to determine if there are any lone pairs because we only looked at bonds. Remember that electron groups include lone pairs! Butane doesn't have any lone pairs. Hence, we have 4 tetrahedrals. Now, what are we going to do with 4 tetrahedrals? Well, we want to optimize the bond angle of each central atom attached to each other. This is due to the electrons that are shared are more likely to repel each other. With 4 tetrahedrals, the shape of the molecule looks like this: en.Wikipedia.org/wiki/File:Butane-3D-balls.png. That means that if we look back at every individual tetrahedral, we match the central Carbon with the Carbon it's bonded to.
Bond Angles
Bond angles also contribute to the shape of a molecule. Bond angles are the angles between adjacent lines representing bonds. The bond angle can help differentiate between linear, trigonal planar, tetraheral, trigonal-bipyramidal, and octahedral. The ideal bond angles are the angles that demonstrate the maximum angle where it would minimize repulsion, thus verifying the VSEPR theory.
Essentially, bond angles is telling us that electrons don't like to be near each other. Electrons are negative. Two negatives don't attract. Let's create an analogy. Generally, a negative person is seen as bad or mean and you don't want to talk to a negative person. One negative person is bad enough, but if you have two put together...that's just horrible. The two negative people will be mean towards each other and they won't like each other. So, they will be far away from each other. We can apply this idea to electrons. Electrons are alike in charge and will repel each other. The farthest way they can get away from each other is through angles. Now, let's refer back to tetrahedrals. Why is it that 90 degrees does not work? Well, if we draw out a tetrahedral on a 2-D plane, then we get 90 degrees. However, we live in a 3-D world. To visualize this, think about movies. Movies in 3D pop out at us. Before, we see movies that are just on the screen and that's good. What's better? 3D or 2D? For bond angles, 3D is better. Therefore, tetrahedrals have a bond angle of 109.5 degrees. How scientists got that number was through experiments, but we don't need to know too much detail because that is not described in the textbook or lecture.
Using the example above, we would add that H2O has a bond angle of 109.5° and CO2 would have a bond angle of 180°.
Steps Used to Find the Shape of the Molecule
To sum up there are four simple steps to apply the VSEPR theory.
1. Draw the Lewis Structure.
2. Count the number of electron groups and identify them as bond pairs of electron groups or lone pairs of electrons. Remember electron groups include not only bonds, but also lone pairs!
3. Name the electron-group geometry. (State whether it is linear, trigonal-planar, tetrahedral, trigonal-bipyramidal, or octahedral.)
4. Looking at the positions of other atomic nuclei around the central determine the molecular geometry. (See how many lone pairs there are.)
Dipole Moments
A molecule is polar when the electrons are not distributed equally and the molecule has two poles. The more electronegative end of the molecule is the negative end and the less electronegative end is the positive end. A common example is HCl. Using the capital sigma + or - as a symbol to show the the positive end and the negative end we can draw the net dipole. So sigma + would be on the hydrogen atom and sigma - would be on the Chlorine atom. Using the cross bow arrow shown below we can show that it has a net dipole. The net dipole is the measurable, which is called the dipole moment. Dipole moment is equal to the product of the partial charge and the distance. The equation for dipole moment is as follows.
$\mu = \delta \times d$
with
• µ = dipole moment (debye)
• δ = partial charge (C)
• d = distance (m)
The units for dipole is expressed in debye which is also known as Coulombs x meter (C x m)
Example of a Dipole
The cross base arrow demonstrates the net dipole.
On the cross-base arrow, the cross represents the positive charge and the arrow represents the negative charge.
Here's another way to determine dipole moments. We need to comprehend electronegativity which is abbreviated EN. What is EN? Well, EN is how much an element really wants an electron. Think about basketball and how two players pass the ball to each other. Each player represent an element and the ball represents the electron. Let's say one player is a ball hog. The player that is the ball hog is more electronegative because he or she wants the ball more.
Here is a link that has all the EN listed: www.green-planet-solar-energy...electroneg.gif
What if we are not given EN? Luckily, there is a trend in the periodic table for EN. From bottom to the top, EN will increase. From left to right, EN will increase. The most electronegative element is Flourine with 4.0.
Now, we are ready to apply EN to determine whether or not molecules are polar. We look back at the picture of H2O above. The EN is given. What do we do with all the EN? We compare the EN between each bond. Oxygen has a greater EN than Hydrogen. Therefore, we can draw a cross bow arrow towards Oxygen. We have two arrows because Oxygen is bonded to two Hydrogens. Since both arrows point toward Oxygen, we can say that there is a net EN. We added the arrows that point to Oxygen and we end up with a new, bigger arrow. This is examplified in the picture above. If arrows are drawn away from each other like <--- and --->, then we are more likely to have no net EN because the molecule is symmetrical. Refer back to the Lewis dot diagram of CO2. The shape is linear and the EN arrows point towards Oxygen. The arrows are opposite of each other and have the same EN difference. Therefore, we have no net charge and the molecule is non-polar.
Summary of Dipole Moments
To recap, when a molecule is polar it means that the electron is not distributed evenly and there is a difference in the electronegativity of the atoms. If a molecule is polar, it means that it had a net dipole which results in having a dipole moment.
Determining Polarity
Is it polar? There are three ways to go about determining whether a molecule is polar or not.
A. If the molecule has a net dipole, then it is polar.
B. If the structure is symmetric, then it is non-polar
C. There are three rules to this part:
1. When there are no lone pairs on the center atom, then the molecule is non-polar
2. If it is linear or square planar, then it is non-polar. (This rule is more important than rule 1, so it overrules it because it has lone pairs.)
3. If it has different terminal atoms, then it is polar. (This rule overrules rule 1 and 2 because it is more important.)
References
1. Petrucci, Ralph H., William S. Harwood, F. Geoffrey Herring, & Jeffry D. Madura, General Chemistry, Principles and Modern Appplications Ninth Edition, Upper Saddle River, New Jersey
2. Tetrahedrality” and the Relationship between Collective Structure and Radial Distribution Functions in Liquid Water P. E. Mason and J. W. Brady J. Phys. Chem. B;2007
3. Inverted geometries at carbon Kenneth B. Wiberg Acc. Chem. Res.; 1984
4. "Molecular Geometries." Chemistry Foundations and Applications. Volume 3. Farmington, MI:Lagowski, J.J., 2004.
Problems
Part I
Draw the Lewis Structure and name the shape of each compound. Also determine the polarity and whether or not it has a dipole moment.
1. HClO3
2. SO3
3. PCl4
4. C2H4
5. SnCl3-
Part II
Name the shape and determine whether they are polar or non-polar.
Solutions
Part I
1.
• Total # of electrons: 1+(3x6)+7=26
• electron group geometry: tetrahedral
• molecular: trigonal pyramidal
• ideal angle: 109.5°
• polar, has a dipole moment
2.
• Total # of electrons: (3x6)+6=24
• electronic group geometry: trigonal planar
• molecular geometry: trigonal planar
• ideal angle: 120°
• polar, has a dipole moment
3.
• Total # of electrons: (4x4)+5=19
• electronic group geometry: trigonal-bi-pyramidal
• molecular geometry: seesaw
• ideal angle: 90°, 120°
• polar, has a dipole moment
4.
• Total # of electrons: (1x4)+(4x2)=12
• electronic group geometry: trigonal planar
• molecular geometry: trigonal planar
• ideal angle: 120°
• non-polar, does not have a dipole moment
5.
• Total # of electrons: (7x3)+4=26
• electronic group geometry: tetrahedral
• molecular geometry: trigonal pyramidal
• ideal angle: 109.5°
• polar, has a dipole moment.
Part II
1. electron group geometry: octahedral
molecular geometry: square planar
non polar because it is symmetrical
2. electron group geometry: octahedral
molecular geometry: square planar
polar because it is not symmetrical
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Lewis Symbols are simplified Bohr diagrams which only display electrons in the outermost energy level. The omitted electrons are those in filled energy levels, which do not contribute to the chemical properties of the species in question. Therefore Lewis Symbols are useful for studying elemental properties and reactions.
Introduction
A Lewis Symbol is constructed by placing dots representing electrons in the outer energy around the symbol for the element. For many common elements, the number of dots corresponds to the element's group number. Below are Lewis Symbols for various elements. Notice the correspondence to each element's group number.
Lewis Diagrams for Molecules
Molecules can be depicted by Lewis Diagrams by placing dots or lines around the constituent elemental symbols. Once again only valence electrons are shown. Lines denote bonded electron pairs, whereas dots are reserved for unbounded electrons. The following algorithm can be used to construct Lewis diagrams of most molecules.
1. Find the total number of electrons: Determine the total number of valence electrons by reading the group number for each element.
2. Draw a first tentative structure: Frequently the central element is the one with least atoms in the molecule. In your first draft, attach the atoms with single bonds.
3. Add electrons as dots to get octets around atoms: Each atom must have its valence shell completely filled. Remember to count bonds and lone electrons.
4. Count the total number of electrons: Ensure the number of electrons displayed in the symbol agree with the calculation from step 1.
5. Cycle through steps 3 and 4: By way of trial and error, repeat the steps above until a conclusion is reached.
Lewis Theory of Bonding
Chemical bonding is at the heart of understanding chemistry. The Lewis Theory of Bonding essentially combined observations at the time about chemical bonding together. Not only was it essential in understanding how elements bonded, it provided a visual representation for them. These Lewis dot structures are a simplistic way of representing the electrons in molecules. This theory also helped to define formal charge and resonance. Understanding of electronegativity and electron affinity will be useful before understanding Lewis Theory of Bonding.
Contributors and Attributions
• Nilpa Shah (UCD)
Violations of the Octet Rule
Three cases can be constructed that do not follow the Octet Rule, and as such, they are known as the exceptions to the Octet Rule. Following the Octet Rule for Lewis Dot Structures leads to the most accurate depictions of stable molecular and atomic structures and because of this we always want to use the octet rule when drawing Lewis Dot Structures. However, it is hard to imagine that one rule could be followed by all molecules. There is always an exception, and in this case, three exceptions. The Octet Rule is violated in these three scenarios:
1. When there are an odd number of valence electrons
2. When there are too few valence electrons
3. When there are too many valence electrons
Reminder: Always use the Octet Rule when drawing Lewis Dot Structures, these exceptions will only occur when necessary.
Exception 1: Species with Odd Numbers of Electrons
The first exception to the Octet Rule is when there are an odd number of valence electrons. An example of this would be the nitrogen (II) oxide molecule ($NO$). Nitrogen atom has 5 valence electrons while the oxygen atom has 6 electrons. The total would be 11 valence electrons to be used. The Octet Rule for this molecule is fulfilled in the above example, however that is with 10 valence electrons. The last one does not know where to go. The lone electron is called an unpaired electron. But where should the unpaired electron go? The unpaired electron is usually placed in the Lewis Dot Structure so that each element in the structure will have the lowest formal charge possible. The formal charge is the perceived charge on an individual atom in a molecule when atoms do not contribute equal numbers of electrons to the bonds they participate in. The formula to find a formal charge is:
Formal Charge= [# of valence e- the atom would have on its own] - [# of lone pair electrons on that atom] - [# of bonds that atom participates in]
No formal charge at all is the most ideal situation. An example of a stable molecule with an odd number of valence electrons would be nitrogen monoxide. Nitrogen monoxide has 11 valence electrons (Figure 1). If you need more information about formal charges, see Lewis Structures. If we were to consider the nitrogen monoxide cation ($NO^+$ with ten valence electrons, then the following Lewis structure would be constructed:
Nitrogen normally has five valence electrons. In Figure 1, it has two lone pair electrons and it participates in two bonds (a double bond) with oxygen. This results in nitrogen having a formal charge of +1. Oxygen normally has six valence electrons. In Figure 1, oxygen has four lone pair electrons and it participates in two bonds with nitrogen. Oxygen therefore has a formal charge of 0. The overall molecule here has a formal charge of +1 (+1 for nitrogen, 0 for oxygen. +1 + 0 = +1). However, if we add the eleventh electron to nitrogen (because we want the molecule to have the lowest total formal charge), it will bring both the nitrogen and the molecule's overall charges to zero, the most ideal formal charge situation. That is exactly what is done to get the correct Lewis structure for nitrogen monoxide:
Free Radicals
There are actually very few stable molecules with odd numbers of electrons that exist, since that unpaired electron is willing to react with other unpaired electrons. Most odd electron species are highly reactive, which we call Free Radicals. Because of their instability, free radicals bond to atoms in which they can take an electron from in order to become stable, making them very chemically reactive. Radicals are found as both reactants and products, but generally react to form more stable molecules as soon as they can. To emphasize the existence of the unpaired electron, radicals are denoted with a dot in front of their chemical symbol as with ${\cdot}OH$, the hydroxyl radical. An example of a radical you may by familiar with already is the gaseous chlorine atom, denoted ${\cdot}Cl$. Interestingly, molecules with an odd number of Valence electrons will always be paramagnetic.
Exception 2: Incomplete Octets
The second exception to the Octet Rule is when there are too few valence electrons that results in an incomplete Octet. There are even more occasions where the octet rule does not give the most correct depiction of a molecule or ion. This is also the case with incomplete octets. Species with incomplete octets are pretty rare and generally are only found in some beryllium, aluminum, and boron compounds including the boron hydrides. Let's take a look at one such hydride, $BH_3$ (Borane).
If one was to make a Lewis structure for $BH_3$ following the basic strategies for drawing Lewis structures, one would probably come up with this structure (Figure 3):
The problem with this structure is that boron has an incomplete octet; it only has six electrons around it. Hydrogen atoms can naturally only have only 2 electrons in their outermost shell (their version of an octet), and as such there are no spare electrons to form a double bond with boron. One might surmise that the failure of this structure to form complete octets must mean that this bond should be ionic instead of covalent. However, boron has an electronegativity that is very similar to hydrogen, meaning there is likely very little ionic character in the hydrogen to boron bonds, and as such this Lewis structure, though it does not fulfill the octet rule, is likely the best structure possible for depicting BH3 with Lewis theory. One of the things that may account for BH3's incomplete octet is that it is commonly a transitory species, formed temporarily in reactions that involve multiple steps.
Let's take a look at another incomplete octet situation dealing with boron, BF3 (Boron trifluorine). Like with BH3, the initial drawing of a Lewis structure of BF3 will form a structure where boron has only six electrons around it (Figure 4).
If you look Figure 4, you can see that the fluorine atoms possess extra lone pairs that they can use to make additional bonds with boron, and you might think that all you have to do is make one lone pair into a bond and the structure will be correct. If we add one double bond between boron and one of the fluorines we get the following Lewis Structure (Figure 5):
Each fluorine has eight electrons, and the boron atom has eight as well! Each atom has a perfect octet, right? Not so fast. We must examine the formal charges of this structure. The fluorine that shares a double bond with boron has six electrons around it (four from its two lone pairs of electrons and one each from its two bonds with boron). This is one less electron than the number of valence electrons it would have naturally (Group seven elements have seven valence electrons), so it has a formal charge of +1. The two flourines that share single bonds with boron have seven electrons around them (six from their three lone pairs and one from their single bonds with boron). This is the same amount as the number of valence electrons they would have on their own, so they both have a formal charge of zero. Finally, boron has four electrons around it (one from each of its four bonds shared with fluorine). This is one more electron than the number of valence electrons that boron would have on its own, and as such boron has a formal charge of -1.
This structure is supported by the fact that the experimentally determined bond length of the boron to fluorine bonds in BF3 is less than what would be typical for a single bond (see Bond Order and Lengths). However, this structure contradicts one of the major rules of formal charges: Negative formal charges are supposed to be found on the more electronegative atom(s) in a bond, but in the structure depicted in Figure 5, a positive formal charge is found on fluorine, which not only is the most electronegative element in the structure, but the most electronegative element in the entire periodic table ($\chi=4.0$). Boron on the other hand, with the much lower electronegativity of 2.0, has the negative formal charge in this structure. This formal charge-electronegativity disagreement makes this double-bonded structure impossible.
However the large electronegativity difference here, as opposed to in BH3, signifies significant polar bonds between boron and fluorine, which means there is a high ionic character to this molecule. This suggests the possibility of a semi-ionic structure such as seen in Figure 6:
None of these three structures is the "correct" structure in this instance. The most "correct" structure is most likely a resonance of all three structures: the one with the incomplete octet (Figure 4), the one with the double bond (Figure 5), and the one with the ionic bond (Figure 6). The most contributing structure is probably the incomplete octet structure (due to Figure 5 being basically impossible and Figure 6 not matching up with the behavior and properties of BF3). As you can see even when other possibilities exist, incomplete octets may best portray a molecular structure.
As a side note, it is important to note that BF3 frequently bonds with a F- ion in order to form BF4- rather than staying as BF3. This structure completes boron's octet and it is more common in nature. This exemplifies the fact that incomplete octets are rare, and other configurations are typically more favorable, including bonding with additional ions as in the case of BF3 .
Example: $BF_3$
Draw the Lewis structure for boron trifluoride (BF3).
Solution
1. Add electrons (3*7) + 3 = 24
2. Draw connectivities:
3. Add octets to outer atoms:
4. Add extra electrons (24-24=0) to central atom:
5. Does central electron have octet?
• NO. It has 6 electrons
• Add a multiple bond (double bond) to see if central atom can achieve an octet:
6. The central Boron now has an octet (there would be three resonance Lewis structures)
However...
• In this structure with a double bond the fluorine atom is sharing extra electrons with the boron.
• The fluorine would have a '+' partial charge, and the boron a '-' partial charge, this is inconsistent with the electronegativities of fluorine and boron.
• Thus, the structure of BF3, with single bonds, and 6 valence electrons around the central boron is the most likely structure
• BF3 reacts strongly with compounds which have an unshared pair of electrons which can be used to form a bond with the boron:
Exception 3: Expanded Valence Shells
More common than incomplete octets are expanded octets where the central atom in a Lewis structure has more than eight electrons in its valence shell. In expanded octets, the central atom can have ten electrons, or even twelve. Molecules with expanded octets involve highly electronegative terminal atoms, and a nonmetal central atom found in the third period or below, which those terminal atoms bond to. For example, $PCl_5$ is a legitimate compound (whereas $NCl_5$) is not:
Expanded valence shells are observed only for elements in period 3 (i.e. n=3) and beyond
The 'octet' rule is based upon available ns and np orbitals for valence electrons (2 electrons in the s orbitals, and 6 in the p orbitals). Beginning with the n=3 principle quantum number, the d orbitals become available (l=2). The orbital diagram for the valence shell of phosphorous is:
Hence, the third period elements occasionally exceed the octet rule by using their empty d orbitals to accommodate additional electrons. Size is also an important consideration:
• The larger the central atom, the larger the number of electrons which can surround it
• Expanded valence shells occur most often when the central atom is bonded to small electronegative atoms, such as F, Cl and O.
There is currently much scientific exploration and inquiry into the reason why expanded valence shells are found. The top area of interest is figuring out where the extra pair(s) of electrons are found. Many chemists think that there is not a very large energy difference between the 3p and 3d orbitals, and as such it is plausible for extra electrons to easily fill the 3d orbital when an expanded octet is more favorable than having a complete octet. This matter is still under hot debate, however and there is even debate as to what makes an expanded octet more favorable than a configuration that follows the octet rule.
One of the situations where expanded octet structures are treated as more favorable than Lewis structures that follow the octet rule is when the formal charges in the expanded octet structure are smaller than in a structure that adheres to the octet rule, or when there are less formal charges in the expanded octet than in the structure a structure that adheres to the octet rule.
The sulfate ion, SO4-2. is an ion that prefers an expanded octet structure. A strict adherence to the octet rule forms the following Lewis structure:
Example 3: The $ICl_4^-$ Ion
Draw the Lewis structure for $ICl_4^-$ ion.
Solution
1. Count up the valence electrons: 7+(4*7)+1 = 36 electrons
2. Draw the connectivities:
3. Add octet of electrons to outer atoms:
4. Add extra electrons (36-32=4) to central atom:
5. The ICl4- ion thus has 12 valence electrons around the central Iodine (in the 5d orbitals)
Expanded Lewis structures are also plausible depictions of molecules when experimentally determined bond lengths suggest partial double bond characters even when single bonds would already fully fill the octet of the central atom. Despite the cases for expanded octets, as mentioned for incomplete octets, it is important to keep in mind that, in general, the octet rule applies.
Practice Problems
1. Draw the Lewis structure for the molecule I3-.
2. Draw the molecule ClF3.
3. The central atom for an expanded octet must have an atomic number larger than what?
4. Draw the Lewis structure for the molecule NO2.
5. Which Lewis structure is more likely?
or
Answers
1.
2.
3. 10 (Sodium and higher)
4.
5.
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Molecular orbital theory is concerned with the combination of atomic orbitals to form new molecular orbitals. These new orbitals arise from the linear combination of atomic orbitals to form bonding and antibonding orbitals. The bonding orbitals are at a lower energy than the antibonding orbitals, so they are the first to fill up. By figuring out the molecular orbitals, it is easy to calculate bond order.
Introduction
The valence bond theory is an extension of the Lewis Structures that considers the overlapping of orbitals to create bonds. The valence bond theory is only limited in its use because it does not explain the molecular geometry of molecules very well. This is where hybridization and the molecular orbital theory comes into place.
Hybridization
Hybridization is a simple model that deals with mixing orbitals to from new, hybridized, orbitals. This is part of the valence bond theory and helps explain bonds formed, the length of bonds, and bond energies; however, this does not explain molecular geometry very well.
• sp An example of this is acetylene (C2H2). This combines one s orbital with one p orbital. This means that the s and p characteristics are equal.
• sp2 An example of this is ethylene (C2H4). This is the combination of one s orbital and two p orbitals.
• sp3 An example of this is methane (CH4). This is the combination of one s orbital and three p orbitals.
If you add the exponents of the hybridized orbitals, you get the amount of sigma bonds associated with that bond. The sp2 hybridized orbital has one p orbitals that is not hybridized and so it can form a pi bond. This means that sp2 orbitals allow for the formation of a double bond. Also, sp hybridized orbitals form a triple bond.
Antibonding vs. Bonding Orbitals
Electrons that spend most of their time between the nuclei of two atoms are placed into the bonding orbitals, and electrons that spend most of their time outside the nuclei of two atoms are placed into antibonding orbitals. This is because there is an increasing in electron density between the nuclei in bonding orbitals, and a decreasing in electron density in antibonding orbitals (Chang 459). Placing an electron in the bonding orbital stabilizes the molecule because it is in between the two nuclei. Conversely, placing electrons into the antibonding orbitals will decrease the stability of the molecule. Electrons will fill according to the energy levels of the orbitals. They will first fill the lower energy orbitals, and then they will fill the higher energy orbitals. If a bond order of zero is obtained, that means that the molecule is too unstable and so it will not exist.
Below are a few examples of bonding and antibonding orbitals drawn out:
• Hydrogen Example:
• Oxygen Example (homonuclear):
• See oxygen file below
• Hydrogen Flouride Example (heteronuclear):
• See hydrogen fluoride file below
Bond Order
Bond order is the amount of bonds formed between two atoms. For example, two bonds are formed between oxygen atoms, so the bond order is 2. The following is the equation to find bond order.
1/2(electrons in bonding molecular orbitals - electrons in antibonding molecular orbitals)
Bond order gives information about bond length and strength. Generally, higher bond order correlates to a shorter bond length. This is due to the greater number of bonds between the atoms. In addition, because of the greater number of bonds between the atoms, the strength should also be greater as bond order increases.
Outside Links
• en.Wikipedia.org/wiki/Antibonding
• <--- This youtube video is a lecture on Molecular Orbital Theory from the open course website at MIT. It is has a good review of antibonding and bonding orbitals.
• Smith, Derek W. "The Antibonding Effect." J. Chem. Educ. 2000 77 780.
Problems
1. Determine if each of the following are paramagnetic or diamagnetic: F2, 02, N2, C2, Ag2.
2. Calculate the bond order for each of the following: H2, H2+, H2-. (remember bond order is bonding minus antibonding electrons times one half)
3. Draw the molecular orbitals for He2and He2+. Are both of these stable molecules?
4. Write out the electron configuration diagram using molecular orbitals and calculate the bond order of HCl, NO-1, and CO. Rank them in order of increasing stability.
Contributors and Attributions
• Mustafa Rupawalla (UCD)
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At an elementary level, we can think about atoms being held together by simple electrostatic attraction. It is a fundamental principal that opposite charges attract. A positively charged ion and a negatively charged ion are held together by this force of attraction.
• This idea works well for ionic compounds, such as sodium chloride.
• It does not work well for compounds in which similar atoms are connected, such as ethane, \(C_2H_6\).
Maybe the carbons are negatively charged and the hydrogens are positively charged, but what is holding the two carbons together if they have like charges? A similar problem is encountered in diatomic molecules such as \(H_2\). Is one of these hydrogen atoms negative while the other hydrogen atom is positive?
In the early twentieth century, G. N. Lewis noticed a trend in the characteristics of compounds that he called "the rule of two." If you were to count up the total number of electrons in any stable compound, you would always come up with an even number -- that is, some number that is divisible by two. Perhaps, Lewis reasoned, this predominance of even numbers arises because electrons need to be in pairs.
What does an element do if it has an odd number of electrons? One solution is to steal an electron from another element, or to allow one to be stolen away; these arrangements lead to ionic bonds. However, those elements not adept at stealing electrons may have a problem; they may need to share them instead. In order to share electrons, elements will have to form close associations with each other. They will become bonded together.
• An "ionic bond" is an electrostatic interaction between an anion and a cation.
• A "covalent bond" is one pair of electrons shared between two atoms.
Lewis took this idea further. If you count up the valence shell electrons around each of the atoms in stable compounds, not only are there even numbers, but there are almost always the same number of electrons as there are in one of the noble gases: He, Ne, Ar, Kr (2, 8, 8, or 18). This observation is sometimes called the Lewis octet rule because so many common atoms that form stable compounds obtain 8 electrons in their outermost shell as a result. Neon is the nearest noble gas to carbon, oxygen and nitrogen, and all of these atoms adopt 8-electron configurations in stable compounds.
Lewis structures illustrate how atoms can maintain these numbers of electrons by sharing with other atoms. These simple structural drawings are used to convey most of our ideas about molecular chemistry. However, additional information can often be found through quantum mechanics and a molecular orbital approach to bonding.
MO7. Experimental Evidence
Experimental Evidence for Molecular Orbital Results
The molecular orbital picture of dioxygen differs from the Lewis picture. Both models predict an oxygen-oxygen double bond, but one model suggests unpaired electrons whereas the other indicates an electron-paired system. Often, there is experimental evidence available to check the reliability of predictions about structure. These data include measurements of bond lengths and bond strengths as well as magnetic properties.
Bond dissociation energy data tell us how difficult it is to separate one atom from another in a molecule. Bond order is one of the factors that influences bond strength. Thus, measuring a bond dissociation energy is one way to confirm that dioxygen really does contain an oxygen-oxygen double bond.
First, we need something to compare it to. Peroxides (such as hydrogen peroxide, H2O2, or sodium peroxide, Na2O2) probably contain oxygen-oxygen single bonds, according to their Lewis structures. These bonds are relatively weak, costing about 35 kcal/mol to break. In contrast, the bond in dioxygen costs about 70 kcal/mol to break. Its bond is about twice as strong; it is a double bond.
• Bond dissociation energies can be used to determine how many bonds there are between two atoms.
Bond dissociation energies can be complicated to measure. They require a comparison of energy changes in numerous chemical reactions so that the energy change resulting from cleavage of a specific bond can be inferred. In contrast, infrared absorption frequencies are easy to measure. They simply require shining infrared light through a sample and measuring what frequencies of the light are absorbed by the material. (A related technique, Raman spectroscopy, gives similar information by measuring subtle changes in the frequency of laser light that is scattered off a sample). The frequencies absorbed depend on what bonds are present in the material. These frequencies vary according to two basic factors: the weights of the atoms at the ends of the bond, and the strength of the bond between them. The stronger the bond, the higher the absorption frequency.
Peroxides absorb infrared light at around 800 cm-1 (this unusual frequency unit is usually pronounced "wavenumbers"). Dioxygen absorbs infrared light around 1300 cm-1. Since the atoms at the ends of the bond in both peroxide and dioxygen are oxygens, we can be sure that this difference in frequency is not due to a difference in mass. It is due to a difference in bond strength. The bond in dioxygen is much stronger than the O-O bond in peroxide, because the former is a double bond and the latter is a single bond.
• Vibrational spectroscopies (IR and Raman spectroscopy) can give information about the bond order between two atoms.
A third measure of bond order is found in bond length measurements. The more strongly bound two atoms are, the closer they are together. An O=O bond should be shorter than an O-O bond. Bond lengths can be measured by microwave spectroscopy (usually for gas-phase molecules), in which frequencies absorbed depend on the distance between the molecules. Alternatively, bond lengths can be measured by x-ray crystallography. X-rays can be diffracted through crystals of solid materials. The interference pattern that is produced can be mathematically decoded to produce a three-dimensional map of where all the atoms are in the material. The distances between these atoms can be measured very accurately.
The O-O bond in peroxides are about 1.49 Angstroms long (an Angstrom is 10-10 m; this unit is often used for bond lengths because it is a convenient size for this task. Covalent bonds are generally one to three Angstroms long). The O-O bond in dioxygen is about 1.21 A long. The O-O bond in dioxygen is shorter and stronger than in a peroxide.
• Bond length data provides insight into the bond order.
In addition to bond order, there is the question of electron pairing in dioxygen. The Lewis structure suggests electrons are paired in dioxygen. The molecular orbital picture suggests two unpaired electrons.
Compounds with paired electrons are referred to as diamagnetic. Those with unpaired electrons are called paramagnetic. Paramagnetic substances interact strongly with magnetic fields.
It turns out that oxygen does interact with a magnetic fields. A sample of liquid-phase oxygen can be held between the poles of a magnet. Oxygen has unpaired electrons. This finding is consistent with molecular orbital theory, but not with simple Lewis structures. Thus, MO theory tells us something that the Lewis picture cannot.
• Magnetic information, and measurements of magnetism, give us experimental evidence of spin states.
• We can tell if electrons are paired, unpaired, and how many unpaired spins there are.
A final important source of experimental data is photoelectron spectroscopy. Photoelectron spectroscopy gives information about the electron energy levels in an atom or compound. In this technique, gas-phase molecules are subjected to high-energy electromagnetic radiation, such as ultraviolet light or X-rays. Electrons are ejected from various energy levels in the molecule, and the binding energies of electrons in those levels is determined. Thus, photoelectron spectroscopy provides verification for exactly the sort of information that quantitative molecular orbital calculations are designed to deliver.
• Photoelectron spectroscopy tells how much energy is needed to remove electrons from various energy levels in a molecule.
• This technique gives us an accurate experimental picture of the energy levels that we predict with molecular orbital calculations.
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Molecular orbitals (MO) are constructed from atomic orbitals. In O2 and F2, there is a crossover of the sigma and the pi ortbials: the relative energies of the sigma orbitals drop below that of the pi orbitals'. Information from the MO diagram justify O2's stability and show that it's bonding order is 2. The LUMO (lowest unoccupied molecular orbital) and HOMO (highest occupied molecular orbital) of difluoride's MO diagram help explain why the molecule is very stable - the diagram also tells us that the bond order is 1.
• Dieu Huynh
MO for HF
Molecular Orbitals for Heterogeneous Diatomic Molecules
A simple approach to molecular orbital (MO) theory for heterogeneous diatomic molecules is to show the energy level diagram. The MO energy levels can be worked out following these steps:
Recall that the energy \(E_n\) for the quantum number n is for an element with atomic Z is approximately
\[E_n = 13.6 \dfrac{Z_{Eff}^2}{n^2} eV\]
We use \(Z_{eff}\) instead of Z to mean that we have to modify the atomic number to get an effective atomic charge for the nucleus. Since we are dealing with approximate values, one may use Z directly. The 1s orbital energy level is -13.6 eV for hydrogen atoms, measured as the ionization energy of H.
Thus, for the quantum number n = 1, the energy level for 1s of He is approximately - 54 eV. Similarly, the 1s energy level for F is - 1101 eV. The 2s and 2p energy levels for He is approximately - 13.6 eV, which is simlar to that of 1s orbital of H.
Thus, the 2s energy level for Li is approximately -6 eV. However, for multi-electron atoms, the p-subshell and s-subshell have different energies due to penetration. At this level, we cannot be precise about it, but simply think that the 2p orbitals are at higher energy than the 2s orbital. Usually, atomic orbitals with energy levels similar to each other will overlap to form molecular orbitals. Thus, we match the energy levels of atomic orbitals, and then make bonding and anti-bonding MOs of them.
However, in case the atomic orbital energy level is very different, we use atomic orbitals of the incomplete subshell to form MOs.
Molecular Orbital Diagram for the HF Molecule
Interaction occurs between the 1s orbital on hydrogen and the 2p orbital in fluorine causing the formation of a sigma-bonding and a sigma-antibonding molecular orbital, as shown below.
Molecular Orbitals of Li to F
Learning Objectives
• Explain how the energy levels of atomic orbitals vary for $\ce{H}$, $\ce{Li}$, $\ce{Be}$, $\ce{B}$, $\ce{C}$, $\ce{N}$, and $\ce{O}$.
• Draw relative energy levels diagrams for homonuclear diatomic molecules of period 2 elements.
• Explain why the relative energy level diagrams for $\ce{Li2}$, $\ce{Be2}$, $\ce{B2}$, $\ce{C2}$, and $\ce{N2}$ are different from those of $\ce{O2}$ and $\ce{F2}$.
• Point out relevant data to support the energy level diagrams of diatomic molecules of period 2 elements.
The molecular orbital theory (MO) has been introduced for the diatomic hydrogen molecules. The same method can be applied to other diatomic molecules, but involving more than the 1s atomic orbitals. For the second period elements, the 2s and 2p orbitals are important for MO considerations. A linear combination of properly oriented atomic orbitals for the formation of sigma s and pi p bonds. The formation of bonds from the linear combination of atomic orbitals is the same as that of the valence bond theory. For simplicity, we are not going into the details of the theory, but simply show you how to construct the MO energy level diagram.
Relative Energy Levels of Atomic Orbitals from Hydrogen to Fluorine
Atomic energy levels E in kJ mol-1
of second group elements
Element E2s E2p E2p-E2s
$\ce{Li}$ -521
$\ce{Be}$ -897
$\ce{B}$ -1350 -801 549
$\ce{C}$ -1871 -1022 849
$\ce{N}$ -2470 -1274 1196
$\ce{O}$ -3116 -1524 1592
$\ce{F}$ -3879 -1795 2084
$\ce{Ne}$ -4680 -2084 2596
In the discussion of electronic configurations of many-electron atoms, the variation of energy levels of the atomic orbitals was given. All the corresponding levels become more negative as the atomic number increases. The energy levels E2s and E2p of the second period are given in the table on the right.
The energy level E2s ranges from -521 to -4680 kJ mol-1 for these elements. The E2p energy levels also become more negative, but the decrease (because they are negative) is not as rapid as that of the E2s levels. Thus, the differences E2p - E2s increase as the atomic numbers increase.
A qualitative diagram showing the changes of energy levels of atomic orbitals is given below:
Variation of energy levels for atomic orbitals of some elements
$\ce{H}$
_2s_ _ _2p
_ 1s
$\ce{Li}$
_ _ _ 2p
_ 2s
_ 1s
$\ce{Be}$
_ _ _ 2p
_ 2s
_ 1s
$\ce{B}$
_ _ _ 2p
_ 2s
_ 1s
$\ce{C}$
_ _ _ 2p
_ 2s
_ 1s
$\ce{N}$
_ _ _ 2p
_ 2s
_ 1s
$\ce{O}$
_ _ _ 2p
_ 2s
_ 1s
$\ce{F}$
_ _ _ 2p
_ 2s
_ 1s
Relative Energy Levels of Molecular Orbitals of O2 and F2
The 2s and 2p energy levels of $\ce{O}$ and $\ce{F}$ are very far apart. The combination of the 2s orbitals from the two atoms form a sigma bonding and sigma antibonding orbitals in a way very similar to the case of the hydrogen molecules, because the 2p orbitals have little to do with the 2s orbitals.
On the other hand, the three 2p orbitals of each $\ce{O}$ (or $\ce{F}$) atom can form one sigma and two pi bonds and their corresponding antibonding molecular orbitals. The interaction of the 2p orbitals for the sigma bond is stronger, and the levels of sigma and anti sigma bonds are farther apart than those of pi and anti pi bonds. Thus, the relative energy level diagram of $\ce{O2}$ and $\ce{F2}$ has the following arrangement:
Relative energy levels of $\ce{O2}$ and $\ce{F2}$ molecules
_ _ _
2p
_
2s
Atomic
orbital
__ s*2p
__ __ p*2p
__ __p2p
__ s2p
__ s*2s
__ s2s
Molecular orbitals
_ _ _
2p
_
2s
Atomic
orbital
The electronic configuration for $\ce{O2}$ is:
s2s2 s*2s2 s2p2 p2p4 p*2p2
This electronic configuration indicates a bond order of 2, and the bond can be represented by $\ce{O=O}$. There is no net bonding from the s2s orbitals, because the number of bonding electrons equals the number of antibonding electrons. The two electons in p*2p2 cancel two of the 6 bonding electrons (s2p2 p2p4). Therefore, there are 4 total bonding electrons. The two electrons in the p*2p2 orbitals have the same spin, and they are responsible for the paramagnetism of oxygen.
As an exercise, please fill electrons in the molecular orbitals of a relative energy level diagram to derive and confirm the above conclusion as well as the conclusion regarding the $\ce{F2}$ molecule.
The electronic configuration for $\ce{F2}$ is:
s2s2 s*2s2 s2p2 p2p4 p*2p4
Bond length (pm) and bond
energy (kJ mol-1) of $\ce{O2}$ and $\ce{F2}$
Bond length Bond energy
$\ce{O=O}$ 121 494
$\ce{F-F}$ 142 155
This electronic configuration shows a single $\ce{F-F}$ bond in the molecule for the reasons given for the $\ce{O2}$ molecule. The bond lengths and bond energies of $\ce{O2}$ and $\ce{F2}$ (shown on the right) correspond to $\ce{O=O}$ and $\ce{F-F}$ respectively. The bond energy is higher for $\ce{O=O}$ than for $\ce{F-F}$ due to the double $\ce{O=O}$ bond, and its $\ce{O=O}$ bond length is shorter than that of $\ce{F-F}$.
Relative Energy Levels of Molecular Orbitals for Li2 to N2
Recently, the study of the energies of electrons in molecules revealed that the relative energy levels of molecular orbitals of $\ce{Li2}$ to $\ce{N2}$ are different from those of $\ce{O2}$ and $\ce{F2}$. The explanation for the difference comes from the consideration of hybrid atomic orbitals. Because the 2s energy levels and 2p energy levels for $\ce{Li}$ to $\ce{N}$ are relatively close, the 2s orbitals are influenced by the 2p orbitals. This influence makes the bonding orbitals stronger than, and the antibonding orbitals weaker than, those formed by pure 2s orbitals. This process is called s p mixing.
Due to s p mixing, the s2p orbital is weakened, and the s*2p2 is also affected. These effects cause the relative order to change, and the typical relative energy levels for $\ce{Li2}$, $\ce{Be2}$, $\ce{B2}$, $\ce{C2}$ and $\ce{N2}$ to have the following diagram:
Relative energy levels of $\ce{Li2}$ to $\ce{N2}$ molecules
_ _ _
2p
_
2s
Atomic
orbital
__ s*2p
__ __ p*2p
__ s2p
__ __p2p
__ s*2s
__ s2s
Molecular orbitals
_ _ _
2p
_
2s
Atomic
orbital
The electronic configurations agree with the experimental bond lengths and bond energies of homonuclear diatomic molecules of second-period elements. They are given in a table below. The argument regarding bond lengths, bond orders, and bond energies given for $\ce{O2}$ and $\ce{F2}$ above applies to all these molecules. Note also that $\ce{B2}$ and $\ce{O2}$ are paramagnetic due to the unpaired electrons in the molecular orbitals. Other molecules in this group are diamagnetic.
Electronic configuration, bond length (pm) and bond energy (kJ mol-1) of $\ce{Li2}$ to $\ce{F2}$
Electronic configuration Bond length Bond energy
$\ce{Li-Li}$ s2s2 267 110
$\textrm{Be..Be}$ s2s2 s*2s2 exist? exist?
$\ce{B-B}$ s2s2 s*2s2 p2p2 159 290
$\ce{C=C}$ s2s2 s*2s2 p2p4 124 602
$\ce{N\equiv N}$ s2s2 s*2s2 p2p4 s2p2 110 942
$\ce{O=O}$ s2s2 s*2s2 s2p2 p2p4 p*2p2 121 494
$\ce{F-F}$ s2s2 s*2s2 s2p2 p2p4 p*2p4 142 155
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Chemical_Bonding/Molecular_Orbital_Theory/MO_bonding_in_F2_and_O2.txt
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The Molecular Orbital Theory, initially developed by Robert S. Mullikan, incorporates the wave like characteristics of electrons in describing bonding behavior. In Molecular Orbital Theory, the bonding between atoms is described as a combination of their atomic orbitals. While the Valence Bond Theory and Lewis Structures sufficiently explain simple models, the Molecular Orbital Theory provides answers to more complex questions. In the Molecular Orbital Theory, the electrons are delocalized. Electrons are considered delocalized when they are not assigned to a particular atom or bond (as in the case with Lewis Structures). Instead, the electrons are “smeared out” across the molecule. The Molecular Orbital Theory allows one to predict the distribution of electrons in a molecule which in turn can help predict molecular properties such as shape, magnetism, and Bond Order.
Introduction
Atoms form bonds by sharing electrons. Atoms can share two, four, or six electrons, forming single, double, and triple bonds respectively. Although it is impossible to determine the exact position of an electron, it is possible to calculate the probability that one will find the electron at any point around the nucleus using the Schrödinger Equation. This equation can help predict and determine the energy and spatial distribution of the electron, as well as the shape of each orbital. The figure below shows the first five solutions to the equation in a three dimensional space for a one electron atom. The colors show the phase of the function. In this diagram, blue stands for negative and red stands for positive. Note, however, that the 2s orbital has 2 phases, one of which is not visible because it is inside the other.
Figure 1: Cartoons of the volume occupied by electrons in the 1s, 2s, and 2p hydrogen-like orbitals.
Principles of Molecular Orbital Theory
In molecules, atomic orbitals combine to form molecular orbitals which surround the molecule. Similar to atomic orbitals, molecular orbitals are wave functions giving the probability of finding an electron in certain regions of a molecule. Each molecular orbital can only have 2 electrons, each with an opposite spin. Each molecular orbital can only have 2 electrons, each with an opposite spin. Once you have the molecular orbitals and their energy ordering the ground state configuration is found by applying the Pauli principle, the aufbau principle and Hund's rule just as with atoms.
The principles to apply when forming pictorial molecular orbitals from atomic orbitals are summarized in the table below:
Principle Details/Examples
Total number of molecular orbitals is equal to the total number of atomic orbitals used to make them.
The molecule H2 is composed of two H atoms. Both H atoms have a 1s orbital, so when bonded together, there are therefore two molecular orbitals.
Bonding molecular orbitals are lower energy than the atomic orbitals from which they were formed.
Antibonding molecular orbitals are higher energy than the atomic orbitals from which they were formed.
Electrons in bonding molecular orbitals help stabilize a system of atoms since less energy is associated with bonded atoms as opposed to a system of unbound atoms. Bonding orbitals are formed by in-phase combinations of atomic orbitals and increase the electron density between the atoms (see figure 2 below)
Electrons in antibonding molecular orbitals cause a system to be destabilized since more energy is associated with bonded atoms than that of a system of unbound atoms. Antibonding orbitals are formed by out-of-phase combinations of atomic orbitals and decrease the electron density between atoms (see figure 2 below).
Following both the Pauli exclusion principle and Hund's rule, electrons fill in orbitals of increasing energy.
Electrons fill orbitals with the lowest energy first. No more than 2 electrons can occupy 1 molecular orbital at a time. Furthermore, all orbitals at an energy level must be filled with one electron before they can be paired. (see figure 3 below)
Molecular orbitals are best formed when composed of Atomic orbitals of like energies.
When Li2 forms the two lowest energy orbitals are the pair of bonding and antibonding orbitals formed from the two possible combinations of the 1s on each atom. The 2s orbitals combine primarily with each other to form another pair of bonding and antibonding orbitals at a higher energy.
Figure 2 (below) shows how bonding and antibonding σ orbitals can be formed by combining s orbitals in-phase (bonding, bottom) and out-of-phase (antibonding, top). If the atomic orbitals are combined with the same phase they interfere constructively and a bonding orbital is formed. Bonding molecular orbitals have lower energy than the atomic orbitals from which they were formed. The lowering of the energy is attributed to the increase in shielding of the nuclear repulsion because of the increase in electron density between the nuclei. If the atomic orbitals are combined with different phases, they interfere destructively and an antibonding molecular orbital is formed. Antibonding molecular orbitals have a higher energy than the atomic orbitals from which they were formed. The higher energy is attributed to the reduced shielding of the nuclear repulsion because of the lower electron probability density between the nuclei.
Figure 2: Combining hydrogen-like s orbitals to generate bonding (bottom) and antibonding (top) orbitals. The dark dot represents the location of the nucleus. Note the decrease in electron density between the nuclei in the antibonding orbital.
$\sigma$ Bonds
Molecular orbitals that are symmetrical about the axis of the bond are called sigma molecular orbitals, often abbreviated by the Greek letter $\sigma$. Figure 2 shows the 1s orbitals of 2 Hydrogen atoms forming sigma orbitals. There are two types of sigma orbitals formed, antibonding sigma orbitals (abbreviated $\sigma^*$), and bonding sigma orbitals (abbreviated $\sigma$). In sigma bonding orbitals, the in phase atomic orbitals overlap causing an increase in electron density along the bond axis. Where the atomic orbitals overlap, there is an increase in electron density and therefore an increase in the intensity of the negative charge. This increase in negative charge causes the nuclei to be drawn closer together. In sigma antibonding orbitals ($\sigma^*$), the out of phase 1s orbitals interfere destructively which results in a low electron density between the nuclei as seen on the top of the diagram.
The diagram below (figure 3) is a representation of the energy levels of the bonding and antibonding orbitals formed in the hydrogen molecule. Two molecular orbitals were formed: one antibonding ($\sigma^*$) and one bonding ($\sigma$).The two electrons in the hydrogen molecule have antiparallel spins. Notice that the $\sigma^*$ orbital is empty and has a higher energy than the $\sigma$ orbital.
Figure 3: An MO energy level diagram for H2. The up and down arrows represent electrons that are spin up or spin down.
Sigma bonding orbitals and antibonding orbitals can also be formed between p orbitals (figure 4). Notice that the orbitals have to be in phase in order to form bonding orbitals. Sigma molecular orbitals formed by p orbitals are often differentiated from other types of sigma orbitals by adding the subscript p below it. So the antibonding orbital shown in the diagram below would be σ*p.
Figure 4: The formation of a σ bonding and antibonding orbital using p-orbitals.
$\pi$ Bonds
The $\pi$ bonding is a side to side overlap of orbitals, which then causes there to be no electron density along the axis, but there is density above and below the axis. The diagram below (figure 5) shows a $\pi$ antibonding molecular orbital and a $\pi$ bonding molecular orbital.
Figure 5: The side on overlap of p orbitals to form pi bonding and antibonding orbitals. Note that there is a second set of p orbitals sticking in and out of the image that can combine in the same way. (see cartoons immediately below)
2py Orbitals
The two 2py atomic orbitals overlap in parallel to form two $\pi$ molecular orbitals which are asymmetrical about the axis of the bond.
2pz orbitals
The two 2pz orbitals overlap to create another pair of pi 2p and pi *2p molecular orbitals. The 2pz-2pz overlap is similar to the 2py-2py overlap because it is just the orbitals of the 2pz rotated 90 degrees about the axis. The new molecular orbitals have the same potential energies as those from the 2py-2py overlap.
In summary the three pairs of p orbitals can combine to form one set of \sigma\ orbitals and two sets of \pi\ orbitals.
Drawing Molecular Orbital Diagrams
1. Determine how many valence electrons you have on each atom (you can ignore the core electrons as core orbitals contribute little to molecular orbitals). This gives you the total number of electrons you will have to distribute among the molecular orbitals you form. For example consider B2 (each atom has an electron configuration of [He]2s22p), which has a total of 6 valence electrons.
2. Draw a cartoon energy level diagram with lines for the valence atomic energy levels (orbitals) of each atom. Put one atom's levels on the left and one on the right. Include the electrons. Leave space in the middle for your molecular energy levels (orbitals). It can help to include cartoons of the atomic orbitals as well. Shown for diboron immediately below.
3. Combine each pair of orbitals of similar energy in-phase and out-of-phase to create molecular orbitals as shown for diboron in the figure following step 4.
4. Move the electrons from the atoms into the molecule to determine the molecular electron configuration as shown immediately below
Note that the σ and π orbitals formed from the p's are not always in the order π energy less than σ energy. For first row diatomics the ordering shown above is valid for Z ≤ 7. Thus for oxygen and fluorine the σ is below the π orbitals.
Bond Orders and Stability of Molecules
Bond Order indicates the strength of the bond with the greater the bond order, the stronger the bond.
$\text{Bond Order}= \dfrac{1}{2} \left(a-b\right)$
where
• $a$ is the number of electrons in bonding molecular orbitals and
• $b$ is the number of electrons in antibondng molecular orbitals.
If the bond order is zero, then no bonds are produced and the molecule is not stable (for example $He_2$). If the Bond Order is 1, then it is a single covalent bond. The higher the Bond Order, the more stable the molecule is. An advantage of Molecular Orbital Theory when it comes to Bond Order is that it can more accurately describe partial bonds (for example in H2+, where the Bond Order=1/2), than Lewis Structures.
References
1. Petrucci, RH et al. (2007). General Chemistry: Principles and Modern Applications. New Jersey: Pearson Prentice Hall.
2. Dingrando, Laurel, Kathleen Tallman, Nicholas Hainen, and Cheryl Wistrom. Chemistry. Glencoe/McGraw-Hill School Pub Co, 2004.
3. Kotz, John C., Paul M. Treichel, and Gabriela C. Weaver. "Bonding and Molecular Structure:Orbital Hybridization and Molecular Orbitals." Chemistry & Chemical Reactivity. Belmont, CA: Thomson Brooks/Cole, 2006. 457-66. Print.
Problems
1. What is the molecular orbital diagram for for the diatomic hydrogen molecule, H2? How stable is the molecule? Is it diamagnetic or paramagnetic?
2. What is the molecular orbital diagram for the diatomic helium molecule, He2? How stable is the molecule? Diamagnetic or paramagnetic?
3. What is the molecular orbital diagram for the diatomic oxygen molecule, O2? How stable is the molecule? Diamagnetic or paramagnetic?
4. What is the molecular orbital diagram for the diatomic neon molecule, Ne2? How stable is the molecule? Diamagnetic or paramagnetic?
5. What is the molecular orbital diagram for the diatomic fluorine molecule, F2? How stable is the molecule? Diamagnetic or paramagnetic?
Solutions
• Bond Order = 1/2(2 - 0) = 1
• The bond order above zero, so H2 is stable.
• Because there are no unpaired electrons, H2 is diamagnetic.
2. The molecular orbital diagram for a diatomic helium molecule, He2, shows the following.
• Bond Order = 1/2(2 - 2) = 0
• bond order is zero so molecule is unstable.
• would be diamagnetic.
3. The molecular orbital diagram for a diatomic oxygen molecule, O2, is
• Bond Order = 1/2(10 - 6) = 2
• The bond order is two so the molecule is stable.
• There are two unpaired electrons, so molecule is paramagnetic.
4.The molecular orbital diagram for a diatomic Neon molecule, Ne2, is
• Bond Order = 1/2(10 - 10) = 0
• bond order is zero, so Ne2 is unstable.
• diamagnetic
5. The molecular orbital diagram for the diatomic fluorine molecule, F2 is
• B.O. = 1/2(10 - 8) = 1
• B.O is one so the fluorine molecule is stable.
• Because all of the electrons are paired, F2 is diamagnetic.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Chemical_Bonding/Molecular_Orbital_Theory/Pictorial_Molecular_Orbital_Theory.txt
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Valence bond (VB) theory assumes that all bonds are localized bonds formed between two atoms by the donation of an electron from each atom. This is actually an invalid assumption because many atoms bond using delocalized electrons. In molecular oxygen VB theory predict that there are no unpaired electrons. VB theory does a good job of qualitatively describing the shapes of covalent compounds. While Molecular Orbital (MO) theory is good for understanding bonding in general. It is more difficult to learn, but predicts the actual properties of molecules better than VB theory. MO theory actually predicts electron transitions because of the differences in the energy levels of orbitals in the molecule. MO theory has been more correct in numerous instances and for this reason it is preferred.
Valence Bond theory describes covalent bond formation as well as the electronic structure of molecules. The theory assumes that electrons occupy atomic orbitals of individual atoms within a molecule, and that the electrons of one atom are attracted to the nucleus of another atom. This attraction increases as the atoms approach one another until the atoms reach a minimum distance where the electron density begins to cause repulsion between the two atoms. This electron density at the minimum distance between the two atoms is where the lowest potential energy is acquired, and it can be considered to be what holds the two atoms together in a chemical bond.
Valence Bond Theory
Learning Objectives
• To introduce the concept of electron delocalization from the perspective of molecular orbitals, to understand the relationship between electron delocalization and resonance, and to learn the principles of electron movement used in writing resonance structures in Lewis notation, known as the curved arrow formalism.
Mobility Of $\pi$ Electrons and Unshared Electron Pairs
Now that we understand the difference between sigma and $\pi$ electrons, we remember that the $\pi$ bond is made up of loosely held electrons that form a diffuse cloud which can be easily distorted. This can be illustrated by comparing two types of double bonds, one polar and one nonpolar. The C=C double bond on the left below is nonpolar. Therefore the $\pi$ electrons occupy a relatively symmetric molecular orbital that’s evenly distributed (shared) over the two carbon atoms. The C=O double bond, on the other hand, is polar due to the higher electronegativity of oxygen. The $\pi$ cloud is distorted in a way that results in higher electron density around oxygen compared to carbon. Both atoms still share electrons, but the electrons spend more time around oxygen. The drawing on the right tries to illustrate that concept.
Using simple Lewis formulas, or even line-angle formulas, we can also draw some representations of the two cases above, as follows.
The dynamic nature of $\pi$ electrons can be further illustrated with the use of arrows, as indicated below for the polar C=O bond:
CURVED ARROW FORMALISM
The CURVED ARROW FORMALISM is a convention used to represent the movement of electrons in molecules and reactions according to certain rules. We’ll study those rules in some detail. For now, we keep a few things in mind:
1. Curved arrows always represent the movement of electrons, not atoms.
2. Electrons always move towards more electronegative atoms or towards positive charges.
We notice that the two structures shown above as a result of “pushing electrons” towards the oxygen are RESONANCE STRUCTURES. That is to say, they are both valid Lewis representations of the same species. The actual species is therefore a hybrid of the two structures. We conclude that:
Curved arrows can be used to arrive from one resonance structure to another by following certain rules.
Just like $\pi$ electrons have a certain degree of mobility due to the diffuse nature of $\pi$ molecular orbitals, unshared electron pairs can also be moved with relative ease because they are not engaged in bonding. No bonds have to be broken to move those electrons. As a result, we keep in mind the following principle:
Curved arrows usually originate with $\pi$ electrons or unshared electron pairs, and point towards more electronegative atoms, or towards partial or full positive charges.
Going back to the two resonance structures shown before, we can use the curved arrow formalism either to arrive from structure I to structure II, or vice versa.
In case A, the arrow originates with $\pi$ electrons, which move towards the more electronegative oxygen. In case B, the arrow originates with one of the unshared electron pairs, which moves towards the positive charge on carbon. We further notice that $\pi$ electrons from one structure can become unshared electrons in another, and vice versa. We’ll look at additional guidelines for how to use mobile electrons later.
Finally, in addition to the above, we notice that the oxygen atom, for example, is $sp^2$ hybridized (trigonal planar) in structure I, but $sp^3$ hybridized (tetrahedral) in structure II. So, which one is it? Again, what we are talking about is the real species. The real species is a hybrid that contains contributions from both resonance structures. In this particular case, the best we can do for now is issue a qualitative statement: since structure I is the major contributor to the hybrid, we can say that the oxygen atom in the actual species is mostly trigonal planar because it has greater $sp^2$ character, but it still has some tetrahedral character due to the minor contribution from structure II. We’ll explore and expand on this concept in a variety of contexts throughout the course.
What about sigma electrons, that is to say those forming part of single bonds? These bonds represent the “glue” that holds the atoms together and are a lot more difficult to disrupt. As a result, they are not as mobile as $\pi$ electrons or unshared electrons, and are therefore rarely moved. There are however some exceptions, notably with highly polar bonds, such as in the case of HCl illustrated below. We will not encounter such situations very frequently.
This representation better conveys the idea that the H–Cl bond is highly polar.
Using Curved Arrows
We now go back to an old friend of ours, $CH_3CNO$, which we introduced when we first talked about resonance structures. We use this compound to further illustrate how mobile electrons are “pushed” to arrive from one resonance structure to another.
The movement of electrons that takes place to arrive at structure II from structure I starts with the triple bond between carbon and nitrogen. We’ll move one of the two $\pi$ bonds that form part of the triple bond towards the positive charge on nitrogen, as shown:
When we do this, we pay close attention to the new status of the affected atoms and make any necessary adjustments to the charges, bonds, and unshared electrons to preserve the validity of the resulting formulas. In this case, for example, the carbon that forms part of the triple bond in structure I has to acquire a positive charge in structure II because it’s lost one electron. The nitrogen, on the other hand, is now neutral because it gained one electron and it’s forming three bonds instead of four.
We can also arrive from structure I to structure III by pushing electrons in the following manner. The arrows have been numbered in this example to indicate which movement starts first, but that’s not part of the conventions used in the curved arrow formalism.
As we move a pair of unshared electrons from oxygen towards the nitrogen atom as shown in step 1, we are forced to displace electrons from nitrogen towards carbon as shown in step 2. Otherwise we would end up with a nitrogen with 5 bonds, which is impossible, even if only momentarily. Again, notice that in step 1 the arrow originates with an unshared electron pair from oxygen and moves towards the positive charge on nitrogen. A new $\pi$ bond forms between nitrogen and oxygen. At the same time, the $\pi$ electrons being displaced towards carbon in step 2 become a pair of unshared electrons in structure III. Finally, the hybridization state of some atoms also changes. For example the carbon atom in structure I is sp hybridized, but in structure III it is $sp^3$ hybridized.
You may want to play around some more and see if you can arrive from structure II to structure III, etc. However, be warned that sometimes it is trickier than it may seem at first sight.
Additional rules for moving electrons to write Resonance Structures:
1. Electron pairs can only move to adjacent positions. Adjacent positions means neighboring atoms and/or bonds.
2. The Lewis structures that result from moving electrons must be valid and must contain the same net charge as all the other resonance structures.
The following example illustrates how a lone pair of electrons from carbon can be moved to make a new $\pi$ bond to an adjacent carbon, and how the $\pi$ electrons between carbon and oxygen can be moved to become a pair of unshared electrons on oxygen. None of the previous rules has been violated in any of these examples.
Now let’s look at some examples of HOW NOT TO MOVE ELECTRONS. Using the same example, but moving electrons in a different way, illustrates how such movement would result in invalid Lewis formulas, and therefore is unacceptable. Not only are we moving electrons in the wrong direction (away from a more electronegative atom), but the resulting structure violates several conventions. First, the central carbon has five bonds and therefore violates the octet rule. Second, the overall charge of the second structure is different from the first. To avoid having a carbon with five bonds we would have to destroy one of the C–C single bonds, destroying the molecular skeleton in the process.
In the example below electrons are being moved towards an area of high electron density (a negative charge), rather than towards a positive charge. In addition, the octet rule is violated for carbon in the resulting structure, where it shares more than eight electrons.
Additional examples further illustrate the rules we’ve been talking about.
(a) Unshared electron pairs (lone pairs) located on a given atom can only move to an adjacent position to make a new $\pi$ bond to the next atom.
(b) Unless there is a positive charge on the next atom (carbon above), other electrons will have to be displaced to preserve the octet rule. In resonance structures these are almost always $\pi$ electrons, and almost never sigma electrons.
As the electrons from the nitrogen lone pair move towards the neighboring carbon to make a new $\pi$ bond, the $\pi$ electrons making up the C=O bond must be displaced towards the oxygen to avoid ending up with five bonds to the central carbon.
c) As can be seen above, $\pi$ electrons can move towards one of the two atoms they share to form a new lone pair. In the example above, the $\pi$ electrons from the C=O bond moved towards the oxygen to form a new lone pair. Another example is:
(d) $\pi$ electrons can also move to an adjacent position to make new $\pi$ bond. Once again, the octet rule must be observed:
One of the most common examples of this feature is observed when writing resonance forms for benzene and similar rings.
Delocalization, Conjugated Systems, and Resonance Energy
The presence of alternating $\pi$ and $\sigma$ bonds in a molecule such as benzene is known as a conjugated system, or conjugated $\pi$ bonds. Conjugated systems can extend across the entire molecule, as in benzene, or they can comprise only part of a molecule. A conjugated system always starts and ends with a $\pi$ bond (i.e. an $sp^2$ or an $sp$-hybridized atom), or sometimes with a charge. The atoms that form part of a conjugated system in the examples below are shown in blue, and the ones that do not are shown in red. Most of the times it is $sp^3$ hybridized atoms that break a conjugated system.
Practically every time there are $\pi$ bonds in a molecule, especially if they form part of a conjugated system, there is a possibility for having resonance structures, that is, several valid Lewis formulas for the same compound. What resonance forms show is that there is electron delocalization, and sometimes charge delocalization. All the examples we have seen so far show that electrons move around and are not static, that is, they are delocalized. Charge delocalization is a stabilizing force because it spreads energy over a larger area rather than keeping it confined to a small area. Since electrons are charges, the presence of delocalized electrons brings extra stability to a system compared to a similar system where electrons are localized. The stabilizing effect of charge and electron delocalization is known as resonance energy.
Since conjugation brings up electron delocalization, it follows that the more extensive the conjugated system, the more stable the molecule (i.e. the lower its potential energy). If there are positive or negative charges, they also spread out as a result of resonance.
The more resonance forms one can write for a given system, the more stable it is. That is, the greater its resonance energy.
Examine the following examples and write as many resonance structures as you can for each to further explore these points:
Let’s look for a moment at the three structures in the last row above. In the first structure, delocalization of the positive charge and the $\pi$ bonds occurs over the entire ring. This becomes apparent when we look at all the possible resonance structures as shown below.
In the second structure, delocalization is only possible over three carbon atoms. This is demonstrated by writing all the possible resonance forms below, which now number only two.
Finally, the third structure has no delocalization of charge or electrons because no resonance forms are possible. Therefore, it is the least stable of the three. This brings us to the last topic. How do we recognize when delocalization is possible? Let’s look at some delocalization setups, that is to say, structural features that result in delocalization of electrons.
Delocalization Setups
There are specific structural features that bring up electron or charge delocalization. The presence of a conjugated system is one of them. Other common arrangements are:
(a) The presence of a positive charge next to a $\pi$ bond. The positive charge can be on one of the atoms that make up the $\pi$ bond, or on an adjacent atom.
(b) The presence of a positive charge next to an atom bearing lone pairs of electrons.
(c) The presence of a $\pi$ bond next to an atom bearing lone pairs of electrons.
Oribital Pictures of Delocalization
The orbital view of delocalization can get somewhat complicated. For now we’re going to keep it at a basic level. We start by noting that $sp^2$ carbons actually come in several varieties. Two of the most important and common are neutral $sp^2$ carbons and positively charged $sp^2$ carbons. Substances containing neutral $sp^2$ carbons are regular alkenes. Species containing positively charged $sp^2$ carbons are called carbocations. The central carbon in a carbocation has trigonal planar geometry, and the unhybridized p orbital is empty. The following representations convey these concepts.
A combination of orbital and Lewis or 3-D formulas is a popular means of representing certain features that we may want to highlight. For example, if we’re not interested in the sp2 orbitals and we just want to focus on what the p orbitals are doing we can use the following notation.
Let’s now focus on two simple systems where we know delocalization of $\pi$ electrons exists. One is a system containing two pi bonds in conjugation, and the other has a pi bond next to a positively charged carbon. We can represent these systems as follows.
If we focus on the orbital pictures, we can immediately see the potential for electron delocalization. The two $\pi$ molecular orbitals shown in red on the left below are close enough to overlap. Overlapping is a good thing because it delocalizes the electrons and spreads them over a larger area, bringing added stability to the system.
It is however time-consuming to draw orbitals all the time. The following representations are used to represent the delocalized system.
A similar process applied to the carbocation leads to a similar picture.
The resonance representation conveys the idea of delocalization of charge and electrons rather well.
Finally, the following representations are sometimes used, but again, the simpler they are, the less accurately they represent the delocalization picture.
There will be plenty of opportunity to observe more complex situations as the course progresses.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Chemical_Bonding/Valence_Bond_Theory/Delocalization_of_Electrons.txt
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Learning Objectives
• Describe the hybrid orbitals used in the formation of bonding for each atom in some carbon containing compounds.
• Calculate formal charge for each atom in some carbon containing compounds.
• Draw resonance structures for some organic compounds.
Diamond crystals such as the one shown here are appreciated by almost everyone, because of their hardness, sparkle, and high value. They are also important in many technical applications. However, in terms of chemistry, diamonds consist of only carbon atoms, except for impurities. Like diamond, the chemistry of carbon is indeed very interesting and valuable.
Carbon atoms have the ability to bond to themselves and to other atoms with sp, sp2, and sp3 hybrid orbitals. This link gives you the basics about the hybrid orbitals, and you are introduced to the various bonding of carbon in this document. Compounds containing carbon-hydrogen bonds are called organic compounds. They may also contain $\ce{C-C}$, $\ce{C=C}$, $\ce{C\equiv C}$, $\ce{C-N}$, $\ce{C=N}$, $\ce{C\equiv N}$, $\ce{C-O}$, and $\ce{C=O}$ bonds. Such a variety is due to the ability of carbon to make use of sp, sp2, and sp3 hybrid orbitals for the bonding. There are also various inorganic compounds such as carbon monoxide, carbon dioxide, calcium carbonate, sodium bicarbonate, etc. involving carbon.
Carbon Atoms Using sp Hybrid Orbitals
When sp hybrid orbitals are used for the sigma bond, the two sigma bonds around the carbon are linear. Two other p orbitals are available for pi bonding, and a typical compound is the acetylene or ethyne $\ce{HC\equiv CH}$. The three sigma and two pi bonds of this molecule from University of Florida: General chemistry are shown below.
Note that molecules $\ce{H-C\equiv C-H}$, $\ce{H-C\equiv N}$, and $\ce{C\equiv O}$ have the same number of electrons. Bonding in these molecules can be explained by the same theory, and thus their formation is no surprise. The $\ce{O=C=O}$ molecule is linear, and the carbon atom in this molecule also involves the sp hybrid orbitals. Two pi bonds are also present in this simple molecule. As an exercise, draw a picture to show the two sigma and two pi bonds for this molecule.
Carbon Atoms Using sp2 Hybrid Orbitals
When carbon atoms make use of sp2 hybrid orbitals for sigma bonding, the three bonds lie on the same plane. One such compound is ethene, in which both carbon atoms make use of sp2 hybrid orbitals. One of the remaining p orbitals for each carbon overlap to form a pi bond. A pi bond consists of two parts where bonding electrons are supposed to be located. A picture depicting the sigma and pi bonds in ethene from the same source as the previous picture is shown on the right.
Carbon atoms make use of sp2 hybrid orbitals not only in ethene, but also in many other types of compounds. The following are some of these compounds:
Formaldhyde Ketene Ethylaldehyde Acetic acid Benzene Others
H
\
C=O
/
H
H
\
C=C=O
/
H
CH3
\
C=O
/
H
CH3
\
C=O
/
O-H
CH
/ \
HC CH
|| |
HC CH
\ //
CH
Arromatic
compounds
Graphite
Fullerenes
During the lecture on covalent bonding, we can illustrate how atomic orbitals overlap in the formation of bonds. Here, we can only show you the nice picture as a result.
Carbon Atoms Using sp3 Hybrid Orbitals
In ethane, the carbon atoms use sp3 hybrid orbitals for the formation of sigma bonds. The four bonds around each $\ce{C}$ atom point toward the vertices of a regular tetrahedron, and the ideal bond angles are 109.5°. The simplest compound is methane, $\ce{CH4}$, which is the first member of the alkane family. The next few members are ethane, $\ce{CH3CH3}$, propane, $\ce{CH3CH2CH3}$, butane, $\ce{CH3CH2CH2CH3}$, etc..
H H
\ /
H--C---C--H
/ \
H H
Diamond is a crystal form of elemental carbon, and the structure is particularly interesting. In the crystal, every carbon atom is bonded to four other carbon atoms, and the bonds are arranged in a tetrahedral fashion. The bonding, no doubt, is due to the sp3 hybrid orbitals. The bond length of 154 pm is the same as the $\ce{C-C}$ bond length in ethane, propane and other alkanes.
An idealized single crystal of diamond is a gigantic molecule, because all the atoms are inter-bonded. The bonding has given diamond some very unusual properties. It is the hardest stone, much harder than anything else in the material world. It is a poor conductor, because all electrons are localized in the chemical bonds. However, diamond is an excellent heat conductor. A stone made of pure carbon is colorless, but the presence of impurities gives it various colors. The index of refraction is very high, and their glitter (sparkle or splendor) has made them the most precious stones.
Comparison of Structural Features
Some typical bonding features of ethane, ethene, and ethyne are summarized in the table below:
Systematic
name
Ethane Ethene Ethyne
Hybrid orbitals
of $\ce{C}$
sp3 sp2 sp
Structural
formula
H H
\ /
H--C---C--H
/ \
H H
H H
\ /
C=C
/ \
H H
H-CC-H
$\ce{C-C}$
Bondlength pm
154 134 120
$\ce{C-H}$
Bondlength pm
112 110 106
$\ce{H-C-C}$
bond angle °
111 121 180
$\ce{C-C}$ bond
energy kJ/mol
368 611 820
$\ce{C-H}$ bond
energy kJ/mol
410 451 536
As the bond order between carbon atoms increases from 1 to 3 for ethane, ethene, and ethyne, the bond lengths decrease, and the bond energy increases. Note that the bond energies given here are specific for these compounds, and the values may be different from the average values for this type of bonds.
Confidence Building Questions
1. What hyprid orbitals are used by the carbon atoms in $\ce{HC\equiv CH}$?
Hint: sp hybrid orbitals
Discussion -
Linear $\ce{-C -}$ bonds due to sp hybridized orbitals. This molecule is linear, and it consists of 3 sigma, s, bonds, and two pi, p, bonds. Compare the bonding of this with $\ce{C\equiv O}$, $\ce{H-C\equiv N}$, and $\ce{CH3-C\equiv N}$.
2. What hyprid orbitals are used by the carbon atoms in $\ce{H2C=CH2}$?
Hint: sp2 hybrid orbitals
Discussion -
Planar $\ce{-C}\textrm{<}$ bonds due to sp2 hybridized orbitals. Another p orbital is used for the pi, p. How many sigma and pi bonds does this molecule have? Do all atoms in this molecule lie on the same plane?
3. What hyprid orbitals are used by the carbon in $\ce{CH3CH3}$?
Hint: sp3 hybrid orbitals
Discussion -
Tetrahedral arrangement around $\ce{C}$ is due to sp3 hybridized orbitals.
4. What hyprid orbitals are used by the oxygen in $\ce{CH3CH2OH}$?
Hint: sp3 hybrid orbitals
Discussion -
Tetrahedral arrangement around $\ce{C}$ is due to sp3 hybridized orbitals.
5. Which carbon to carbon bond length is the shortest in the following molecule: $\ce{CH3CCCH2CHCHCH2OH}$?
Hint: shortest between triple bonded carbon
Discussion -
The bond length decreases as the bond order increases.
6. How many carbon atoms make use of sp2 hybrid orbitals in this molecule: $\ce{CH3CCCH2CHCHCH2COOH}$?
Hint: three
Discussion -
Recognize the type of bonding is important.
7. The molecular formula of caffeine is $\ce{C8H10N4O2}$; draw a reasonable structure for it.
Hint: This is one of the problems for chemists.
Discussion -
The structure is shown below. Can you sketch a bonding structure for caffeine? How many carbon atoms makes use of sp2 hybrid orbitals?
Hybridization
Hybridization is the idea that atomic orbitals fuse to form newly hybridized orbitals, which in turn, influences molecular geometry and bonding properties. Hybridization is also an expansion of the valence bond theory. In order to explore this idea further, we will utilize three types of hydrocarbon compounds to illustrate sp3, sp2, and sp hybridization.
Contributors and Attributions
• Jennifer Lau (all images are made by me)
Hybridization II
For polyatomic molecules we would like to be able to explain:
• The number of bonds formed
• Their geometries
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Chemical_Bonding/Valence_Bond_Theory/Hybrid_Orbitals_in_Carbon_Compounds.txt
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Valence Bond (VB) Theory looks at the interaction between atoms to explain chemical bonds. It is one of the two common theories that helps describe the bonding between atoms. The other theory is Molecular Orbital Theory. Take note that these are theories and should be treated as such; they are not always perfect.
Introduction
Valence Bond Theory has its roots in Gilbert Newton Lewis’s paper The Atom and The Molecule. Possibly unaware that Lewis’s model existed, Walter Heitler and Fritz London came up with the idea that resonance and wavefunctions contributed to chemical bonds, in which they used dihydrogen as an example. Their theory was equivalent to Lewis’s theory, with the difference of quantum mechanics being developed. Nonetheless, Heitler and London's theory proved to be successful, providing Linus Pauling and John C. Slater with an opportunity to assemble a general chemical theory containing all of these ideas. Valence Bond Theory was the result, which included the ideas of resonance, covalent-ionic superposition, atomic orbital overlap, and hybridization to describe chemical bonds.
Atomic Orbital Overlap
Bonds are formed between atoms because the atomic orbitals overlap and that the electrons in those orbitals are localized within that overlap. They have a higher probability of being found within that bond; we will return to this statement when we talk about wavefunctions. Dihydrogen (H2) is a simple diatomic gas that has been used to illustrate this idea; however, let's look at Cl2 as a simple example. Cl has seven valence electrons. From its Lewis structure, one can see that Cl has a radical. That sole radical indicates that Cl can bond once. As a general rule, the number of unpaired electrons denotes how many bonds that atom can make. Since there is only one unpaired electron in each Cl, those electrons interact to bond. In this case, the 3p orbitals overlap. Lone pairs can be seen as the orbitals not interacting with each other. The idea of atomic orbitals overlapping works well for simple molecules, such as diatomic gases, but more complex molecules cannot be explained simply by the overlap of atomic orbitals, especially if they defy the octet rule when drawing out Lewis structures and if they bond beyond their predicted amount.
One small concept to add in here is that an orbital from one atoms can be overlap with the other orbital of the second atom. This will eventually give one of the two results. First, the two orbitals have correct symmetry to interact or "mix" (4). Second, the two orbitals do not have correct symmetry to interact or "mix" (4). In the case of two orbitals do not interact, there will be no bonding interaction, which means one of the atomic orbitals will not contribute into the bonding. Another word, it will not affected because of the presence of other atomic orbitals. For further explaination, when the two wavefunctions are knowns as constructive interaction, which means the orbitals do not interact with one another. And if the two orbitals are interacting, they are knowns as destructive interaction where their wavefunctions have opposite site (4).
In the case of two orbitals do interact with each others, it will become molecular orbitals, and it is knowns as constructive interaction, resulting bonding orbitals. On the other hand, if destructive interaction, it will create an antibonding interaction. The two molecular orbitals are the result from adding and subtracting the two wavefunctions (4).
Atomic Orbital Hybridization
Consider phosphorus pentaflouride (PF5). P, the central atom, has five valence electrons, with three lone electrons. Thus, P should only be able to form three bonds. However, in order for PF5 to exist with P as the central atoms, P must be able to bond five times. This event is described by orbital hybridization. In order to create degenerate hybrid orbitals that allow atoms to bond well beyond their normal amount. Orbitals involved in hybridization are the s, p, and d atomic orbitals. There are certain principles that must be followed:
1. Orbitals are not magically lost or gained. The number of orbitals mixed must match the number of hybrid orbitals acquired.
2. Always start from an s orbital. Build your way up to p orbitals, and then d orbitals as necessary. These are orbitals from the atom in question. In this case, it is the P atom.
Hybrid orbitals consist of sp, sp2, sp3, sp2d, sp3d, and sp3d2 (3). Returning back to PF5, let's see how hybrid orbitals can describe its bonds. Remember that P can only bond three times, but we need it to bond five times. In order to bond five times, five orbitals must be singularly filled in. Bonds occur when orbitals with only one electron are spin paired with the electron from another atom. The five orbitals can be acquired by sp3d orbital hybridization. Starting off with a total of five orbitals will result in five hybrid orbitals. Note that since P is in period 3, it has d orbitals in the same energy level. Thus, 3d orbitals can be used to hybridize, even though electrons do not occupy it. The process is outlined in the original figure below. Remember to follow the aufbau principle, Hund's rule, and the Pauli exclusion principle when assigning electrons to their orbitals. Note that valence electrons are only shown. Other orbitals and their corresponding electrons are not hybridized and are not involved in bonding. Although they are not shown, they still exist.
Now, P has five hybrid orbitals. Remember that P has five valence electrons. Electrons too are not lost or gained, so those electrons transfer over to the hybrid orbitals. Fill them in accordingly, following Hund's rule and the Pauli exclusion principle. Note that the hybrid orbitals are now degenerate, so the aufbau principle doesn't apply for this case. From sp3d hybridization, P has five unpaired electrons and can now bond with five F.
Generally, if two bonds are needed, then use an sp; for three, use sp2; the number of bonds needed equals the number of orbitals that need to be hybridized. Please take into account that lone pairs apply as well. If there is a lone pair when you complete the Lewis structure, then the number of required hybrid orbitals equals the number of lone pairs plus the number of bonds; this is also called the coordinate number. PF5, although not shown, does not have lone pairs. The type of hybrid orbitals also corresponds to the molecular shape. In the case of PF5, its shape as defined by VSEPR is trigonal bypyramidal. If hybrid orbitals are involved, the shape can also be determined by the type of hybridization. As another example, sp is linear, and sp2 is trigonal planar. sp2d is square planar. Think over carefully why the type of hybridization also determines the shape. An example problem will aid you in your thinking in the problems section.
How the wavefunction Applies to VB Theory
The wavefunction describes the state of an electron. From the name of the function, one can derive that the electron can behave like a wave. This wave-like behavior of the electrons defines the shapes of the orbitals. Thus, it makes sense that wavefunctions are related to the Valence Bond Theory. If orbitals overlap to create bonds, and orbital shapes and the state of an electron is described by the wavefunction, then it makes sense that the overlap of orbitals (the bonds) can be described by wavefunctions as well. Thus, covalent and ionic bonds can be described by wavefunctions. Covalent and ionic representations of a bond represent the same bond, but they differ in how the electrons are placed. This is called resonance. Different intermolecular interactions give rise to different wavefunctions. Recall from Hund's Rule and the Pauli exclusion principle that electrons must be spin paired when the right conditions are met. Due to this, there are two separate ways to represent a covalent bond in terms of electron spin, which is related to the wavefunction. The figure shows the two different cases for two electrons in a bond. Thus, it is possible to write two different wavefunctions that describe each case, which can be superimposed to describe the overall covalent bond (2). The superposition of the covalent bond and ionic bond wavefunctions will result in an overall wavefunction that describes the state of the molecule. Due to this module being an overview of Valence Bond Theory, the full details will not be covered.
The wavefunction squared will result in a probability density. The wavefunction alone has no physical significance; however, when the wavefunction is squared, the square wavefunction can determine where the electron is most likely located. Recall that when atomic orbitals overlap, the electrons are localized and more likely to be found within that overlap. In terms of electrostatic interactions, this bond will result in some form of equilibrium between all of the electrostatic forces (between the electrons and the electrons with each nucleus) (1). If the overlap is too far in, there is a net repulsion force; the electrons will also be forced apart. If the atomic orbital overlap is too small, the net attraction force is very small; the electron will have a smaller chance of remaining in the overlap since they will be more attracted to their own nucleus. If the overlap is just right, then the electrons are attracted to both nuclei and more likely to stay in the overlapped area. Thus, wavefunctions and electrostatic forces can both explain why electrons are localized within their bonds.
Reliability of Valence Bond Theory and Uses
As one can see, Valence Bond Theory can help describe how bonds are formed. However, there are some notable failures when it comes to Valence Bond Theory. One such failure is dioxygen. Valence Bond Theory fails to predict dioxygen's paramagnitism; it predicts that oxygen is diamagnetic. A species is paramagnetic if electrons are not spin paired and diamagnetic if the electrons are spin paired. Since Valence Bond theory begins with the basis that atomic orbitals overlap to create bonds and through that reasoning, one can see that electrons are spin paired when bonds overlap, dioxygen is indeed predicted to be diamagnetic if Valence Bond Theory is used. In reality, that is not the case. Also, sp2d and sp3 both have a coordinate number of four. Thus, Valence Bond Theory cannot predict whether the molecule is a square planar or the other shape (3). One must correctly draw the Lewis structure and use VSEPR to determine the shape.
Problems
1. Draw the Lewis structures HF, CO2, and C2H2. Which orbitals are involved with bonding for HF? Now, use hybridization to explain the bonds of CO2 and C2H2. Can you account for the single, double, and triple bonds via orbitals overlapping and hybridization? It is not required to draw the hybridization process, but it might help you in your thinking.
2. Using VSEPR, one can derive that IBr5 has an octahedral parent shape. Now, use Valence Bond Theory to account for all of the bonds and verify that the parent shape is indeed an octahedral. Why can you determine the parent shape of a molecule based on Valence Bond Theory? Show your work.
3. Determine the family shapes correlated with the sp3, sp3d, and sp3d2 hybridization. Does this mean that all molecules with the respectable hybridization will have the same shape? Explain.
4. Predict which of these atoms/molecules is diamagnetic using Valence Bond Theory: H, H2, and NO.
5. Briefly summarize Valence Bond Theory. Is it always reliable?
Solutions
HF: Hydrogen has one electron in the 1s orbital. Flourine has two electrons in the 2s orbital, and five in the 2p orbitals. These can be derived from the electron configurations. We are only interested in the valence electrons, so the lower shell can be ignored. Remember to fill in the orbitals accordingly. The orbitals availiable to bond are Hydrogen's 1s orbital and one of Flourine's 2p orbitals. Thus, the bond must take place between these two orbitals. This is a sigma bond. A bond of this type, or one of similar orientation, such as two s orbitals, are the bonds associated with hybrid orbitals when hybridization takes place.
CO2: C has four valence electrons; two are in 2s and two are in 2p. O has six valence electrons; two in 2s and four in 2p. Notice that there is a double bond. Now, hybridize to explain the bonds. When determining the coordinate number, double bonds and triple bonds count as one bond. Thus, the coordinate number of C is 2 and the corresponding hybridization is sp. The diagram below shows the sp hybridization process for C. Notice that two 2p orbital remains unhybridized. We fill in the hybrid orbitals with two of C's electrons so that C has space available to bond with the two O's. The hybrid orbitals describes two sigma bonds between C and the O's; one 2p orbital from each O is used in this process. Two electrons remain for C, which are allocated to the unhybridized orbitals. Now, C can form the other two bonds for a total of four which is what is expected due to the presence of two double bonds. The unhybridized orbitals overlap O's 2p orbitals and forms what is called a pi bond. There is one pi bond and one sigma bond between C and each O. The lone pairs of the O's are the filled 2s orbital and the final and filled 2p orbital. Thus, a double bond is explained by one sigma bond associated with the hybrid orbitals and a pi bond associated with unhybridized p orbitals.
C2H2: C has four valence electrons (two in 2s and two in 2p) and H has one (one in 1s). There is a total of ten valence electrons. From the Lewis structure, one can see that there is a triple bond between the two C's. For each C, one can explain the bonds through sp hybridization (a triple bond and one single bond). This process is similar to CO2. However, in this case, C's available unhybridized 2p orbitals bond together with the unhybridized p orbitals of the other C. Now, there are two pi bonds and one sigma bond between the C's; there is one sigma bond between each C and H bond. Thus, a triple bond is explained by one sigma bond associated with the hybrid orbitals and two pi bonds associated with unhybridized p orbitals.
2. I is the central atom. Usually, the least electronegative atom is the central atom. Drawing the Lewis structure reveals that I can only form one bond. However, with the use of hybridization, I can form more bonds. Showing that is the goal. First, find out how many valence electrons IBr5 has. This is simply derived by adding all of the valence electrons from each atom; all have 7 valence electrons, so (6)(7)=42. Then, use Lewis structure and allocate all of the electrons into the correct places. You will find that the Lewis structure looks like the picture below.
Notice that there is a lone pair. Remember that the number of hybrid orbitals needed is the summation of lone pairs and bonds. This is a total of six; thus a sp3d2 hybridization is needed. This process is shown below. Take note that the lone pair is the only spin paired electrons in the hybrid orbitals. Since the coordinate number is six, then the family shape of this molecule is an octahedral shape. Even the lone pairs are accounted for when hybridization takes place. It is because we used the coordinate number (determined by using Lewis dot structure first) to determiine the type of hybridization that hybridization does indeed give the correct family shape. There are of course exceptions, such as sp2d and sp3. We cannot predict the shape based on hybridization alone for these.
3. The shapes are tetrahedral, trigonal bypyramidal, and octahedral respectively. Not all molecules will have the shape associated with the hybridization type because lone pairs must be taken into account as well. For example, look at problem number two. The coordinate number is six. That tells you that the parent shape is an octahedral. However, the shape of IBr5 will only be that shape if the lone pair was another atom. The parent shape will not always match the actual shape of the molecule. It gives you a place to start and is only true if lone pairs do not exist. "Removing" one bond away from the octahedral shape will give the result of the square-base pyramidal shape.
4. One can solve this problem simply by counting the valence electrons. H has only one electron; it is not paired with any other electron and thus must be paramagnetic. H2 has two valence electrons; thus, those electrons are spin paired and H2 is diamagnetic. NO has eleven valence electrons; it is paramagnetic. In general, if an atom/molecule has an odd number of electrons, then that atom/molecule is paramagnetic. It is diamagnetic if it has an even number of electrons; however, like the case of O2, this does not always work.
5. Valence Bond Theory looks at the interaction between orbitals to describe bonds. It can also be used to derive the shape of the molecule in question, as well as determining whether or not an atom/molecule is diamagnetic or paramagnetic; however Valence Bond Theory is not always reliable. It fails in some cases. One must always remember that this is a theory.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Chemical_Bonding/Valence_Bond_Theory/Overview_of_Valence_Bond_Theory.txt
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Resonance is a mental exercise within the Valence Bond Theory of bonding that describes the delocalization of electrons within molecules. It involves constructing multiple Lewis structures that, when combined, represent the full electronic structure of the molecule. Resonance structures are used when a single Lewis structure cannot fully describe the bonding; the combination of possible resonance structures is defined as a resonance hybrid, which represents the overall delocalization of electrons within the molecule. In general, molecules with multiple resonance structures will be more stable than one with fewer and some resonance structures contribute more to the stability of the molecule than others - formal charges aid in determining this.
Introduction
Resonance is a way of describing delocalized electrons within certain molecules or polyatomic ions where the bonding cannot be expressed by a single Lewis formula. A molecule or ion with such delocalized electrons is represented by several resonance structures. The nuclear skeleton of the Lewis Structure of these resonance structures remains the same, only the electron locations differ. Such is the case for ozone ($\ce{O3}$), an allotrope of oxygen with a V-shaped structure and an O–O–O angle of 117.5°. Let's motivate the discussion by building the Lewis structure for ozone.
1. We know that ozone has a V-shaped structure, so one O atom is central:
2. Each O atom has 6 valence electrons, for a total of 18 valence electrons.
3. Assigning one bonding pair of electrons to each oxygen–oxygen bond gives
with 14 electrons left over.
4. If we place three lone pairs of electrons on each terminal oxygen, we obtain
and have 2 electrons left over.
5. At this point, both terminal oxygen atoms have octets of electrons. We therefore place the last 2 electrons on the central atom:
6. The central oxygen has only 6 electrons. We must convert one lone pair on a terminal oxygen atom to a bonding pair of electrons—but which one? Depending on which one we choose, we obtain either
Which is correct? In fact, neither is correct. Both predict one O–O single bond and one O=O double bond. As you will learn, if the bonds were of different types (one single and one double, for example), they would have different lengths. It turns out, however, that both O–O bond distances are identical, 127.2 pm, which is shorter than a typical O–O single bond (148 pm) and longer than the O=O double bond in O2 (120.7 pm).
Equivalent Lewis dot structures, such as those of ozone, are called resonance structures. The position of the atoms is the same in the various resonance structures of a compound, but the position of the electrons is different. Double-headed arrows link the different resonance structures of a compound:
The double-headed arrow indicates that the actual electronic structure is an average of those shown, not that the molecule oscillates between the two structures.
When it is possible to write more than one equivalent resonance structure for a molecule or ion, the actual structure is the average of the resonance structures.
The electrons appear to "shift" between different resonance structures and while not strictly correct as each resonance structure is just a limitation of using the Lewis structure perspective to describe these molecules. A more accurate description of the electron structure of the molecule requires considering multiple resonance structures simultaneously.
Delocalization and Resonance Structures Rules
1. Resonance structures should have the same number of electrons, do not add or subtract any electrons. (check the number of electrons by simply counting them).
2. Each resonance structures follows the rules of writing Lewis Structures.
3. The hybridization of the structure must stay the same.
4. The skeleton of the structure can not be changed (only the electrons move).
5. Resonance structures must also have the same number of lone pairs.
"Pick the Correct Arrow for the Job"
Most arrows in chemistry cannot be used interchangeably and care must be given to selecting the correct arrow for the job.
• $\leftrightarrow$: A double headed arrow on both ends of the arrow between Lewis structures is used to show resonance
• $\rightleftharpoons$: Double harpoons are used to designate equilibria
• $\rightharpoonup$: A single harpoon on one end indicates the movement of one electron
• $\rightarrow$: A double headed arrow on one end is used to indicate the movement of two electrons
Example $2$: Carbonate Ion
Identify the resonance structures for the carbonate ion: $\ce{CO3^{2-}}$.
Solution
1. Because carbon is the least electronegative element, we place it in the central position:
2. Carbon has 4 valence electrons, each oxygen has 6 valence electrons, and there are 2 more for the −2 charge. This gives 4 + (3 × 6) + 2 = 24 valence electrons.
3. Six electrons are used to form three bonding pairs between the oxygen atoms and the carbon:
4. We divide the remaining 18 electrons equally among the three oxygen atoms by placing three lone pairs on each and indicating the −2 charge:
5. No electrons are left for the central atom.
6. At this point, the carbon atom has only 6 valence electrons, so we must take one lone pair from an oxygen and use it to form a carbon–oxygen double bond. In this case, however, there are three possible choices:
As with ozone, none of these structures describes the bonding exactly. Each predicts one carbon–oxygen double bond and two carbon–oxygen single bonds, but experimentally all C–O bond lengths are identical. We can write resonance structures (in this case, three of them) for the carbonate ion:
The actual structure is an average of these three resonance structures.
Like ozone, the electronic structure of the carbonate ion cannot be described by a single Lewis electron structure. Unlike O3, though, the actual structure of CO32− is an average of three resonance structures.
Using Formal Charges to Identify viable Resonance Structures
While each resonance structure contributes to the total electronic structure of the molecule, they may not contribute equally. Assigning Formal charges to atoms in the molecules is one mechanism to identify the viability of a resonance structure and determine its relative magnitude among other structures. The formal charge on an atom in a covalent species is the net charge the atom would bear if the electrons in all the bonds to the atom were equally shared. Alternatively the formal charge on an atom in a covalent species is the net charge the atom would bear if all bonds to the atom were nonpolar covalent bonds. To determine the formal charge on a given atom in a covalent species, use the following formula:
$\text{Formal Charge} = (\text{number of valence electrons in free orbital}) - (\text{number of lone-pair electrons}) - \frac{1}{2} (\text{ number bond pair electrons}) \label{FC}$
Rules for estimating stability of resonance structures
1. The greater the number of covalent bonds, the greater the stability since more atoms will have complete octets
2. The structure with the least number of formal charges is more stable
3. The structure with the least separation of formal charge is more stable
4. A structure with a negative charge on the more electronegative atom will be more stable
5. Positive charges on the least electronegative atom (most electropositive) is more stable
6. Resonance forms that are equivalent have no difference in stability and contribute equally (eg. benzene)
Example $3$: Thiocyanate Ion
Consider the thiocyanate ($CNS^-$) ion.
Solution
1. Find the Lewis Structure of the molecule. (Remember the Lewis Structure rules.)
2. Resonance: All elements want an octet, and we can do that in multiple ways by moving the terminal atom's electrons around (bonds too).
3. Assign Formal Charges via Equation \ref{FC}.
Formal Charge = (number of valence electrons in free orbital) - (number of lone-pair electrons) - ( $\frac{1}{2}$ number bond pair electrons)
Remember to determine the number of valence electron each atom has before assigning Formal Charges
C = 4 valence e-, N = 5 valence e-, S = 6 valence e-, also add an extra electron for the (-1) charge. The total of valence electrons is 16.
4. Find the most ideal resonance structure. (Note: It is the one with the least formal charges that adds up to zero or to the molecule's overall charge.)
5. Now we have to look at electronegativity for the "Correct" Lewis structure.
The most electronegative atom usually has the negative formal charge, while the least electronegative atom usually has the positive formal charges.
It is useful to combine the resonance structures into a single structure called the Resonance Hybrid that describes the bonding of the molecule. The general approach is described below:
1. Draw the Lewis Structure & Resonance for the molecule (using solid lines for bonds).
2. Where there can be a double or triple bond, draw a dotted line (-----) for the bond.
3. Draw only the lone pairs found in all resonance structures, do not include the lone pairs that are not on all of the resonance structures.
Example $4$: Benzene
Benzene is a common organic solvent that was previously used in gasoline; it is no longer used for this purpose, however, because it is now known to be a carcinogen. The benzene molecule ($\ce{C6H6}$) consists of a regular hexagon of carbon atoms, each of which is also bonded to a hydrogen atom. Use resonance structures to describe the bonding in benzene.
Given: molecular formula and molecular geometry
Asked for: resonance structures
Strategy:
1. Draw a structure for benzene illustrating the bonded atoms. Then calculate the number of valence electrons used in this drawing.
2. Subtract this number from the total number of valence electrons in benzene and then locate the remaining electrons such that each atom in the structure reaches an octet.
3. Draw the resonance structures for benzene.
Solution:
A Each hydrogen atom contributes 1 valence electron, and each carbon atom contributes 4 valence electrons, for a total of (6 × 1) + (6 × 4) = 30 valence electrons. If we place a single bonding electron pair between each pair of carbon atoms and between each carbon and a hydrogen atom, we obtain the following:
Each carbon atom in this structure has only 6 electrons and has a formal charge of +1, but we have used only 24 of the 30 valence electrons.
B If the 6 remaining electrons are uniformly distributed pairwise on alternate carbon atoms, we obtain the following:
Three carbon atoms now have an octet configuration and a formal charge of −1, while three carbon atoms have only 6 electrons and a formal charge of +1. We can convert each lone pair to a bonding electron pair, which gives each atom an octet of electrons and a formal charge of 0, by making three C=C double bonds.
C There are, however, two ways to do this:
Each structure has alternating double and single bonds, but experimentation shows that each carbon–carbon bond in benzene is identical, with bond lengths (139.9 pm) intermediate between those typically found for a C–C single bond (154 pm) and a C=C double bond (134 pm). We can describe the bonding in benzene using the two resonance structures, but the actual electronic structure is an average of the two. The existence of multiple resonance structures for aromatic hydrocarbons like benzene is often indicated by drawing either a circle or dashed lines inside the hexagon:
Example $5$: Nitrate Ion
Draw the possible resonance structures for the Nitrate ion $\ce{NO_3^{-}}$.
Solution
1. Count up the valence electrons: (1*5) + (3*6) + 1(ion) = 24 electrons
2. Draw the bond connectivities:
3. Add octet electrons to the atoms bonded to the center atom:
4. Place any leftover electrons (24-24 = 0) on the center atom:
5. Does the central atom have an octet?
• NO, it has 6 electrons
• Add a multiple bond (first try a double bond) to see if the central atom can achieve an octet:
6. Does the central atom have an octet?
• YES
• Are there possible resonance structures? YES
Note: We would expect that the bond lengths in the $\ce{NO_3^{-}}$ ion to be somewhat shorter than a single bond.
Problems
1. True or False, The picture below is a resonance structure?
1. Draw the Lewis Dot Structure for SO42- and all possible resonance structures. Which of the following resonance structure is not favored among the Lewis Structures? Explain why. Assign Formal Charges.
2. Draw the Lewis Dot Structure for CH3COO- and all possible resonance structures. Assign Formal Charges. Choose the most favorable Lewis Structure.
3. Draw the Lewis Dot Structure for HPO32- and all possible resonance structures. Assign Formal Charges.
4. Draw the Lewis Dot Structure for CHO21- and all possible resonance structures. Assign Formal Charges.
5. Draw the Resonance Hybrid Structure for PO43-.
6. Draw the Resonance Hybrid Structure for NO3-.
Answers
1. False, because the electrons were not moved around, only the atoms (this violates the Resonance Structure Rules).
2. Below are the all Lewis dot structure with formal charges (in red) for Sulfate (SO42-). There isn't a most favorable resonance of the Sulfate ion because they are all identical in charge and there is no change in Electronegativity between the Oxygen atoms.
3. Below is the resonance for CH3COO-, formal charges are displayed in red. The Lewis Structure with the most formal charges is not desirable, because we want the Lewis Structure with the least formal charge.
4. The resonance for HPO32-, and the formal charges (in red).
5. The resonance for CHO21-, and the formal charges (in red).
6. The resonance hybrid for PO43-, hybrid bonds are in red.
7. The resonance hybrid for NO3-, hybrid bonds are in red.
Contributors and Attributions
• Sharon Wei (UCD), Liza Chu (UCD)
d-orbital Hybridization is a Useful Falsehood
For main group molecules, chemists (like Pauling) thought a long time ago that hypervalence is due to expanded s2p6 octets. The consensus is now clear that d orbitals are NOT involved in bonding in molecules like SF6 any more than they are in SF4 and SF2. In all three cases, there is a small and roughly identical participation of d-orbitals in the wavefunctions. This has been established in both MO and VB theory. However using Hybrid orbitals with d-orbital contributions equips us with a language which can pragmatically describe the geometries of highly coordinated substances.
While hybrid orbitals are a powerful tool to describe the geometries and shape of molecules and metal complexes. However, in "real" molecules, their significance may be debated. Often with a more realistically molecular orbitals approach is needed. However, from an epistemologically simple point of view, bonding theories can only be judged by their predictions. To the extent that hybridization can explain the shapes of PF5 and SF6, valence bond theory is a perfectly good theory. To the extent that if you write out the valence bond wavefunction using hybridized orbitals and calculate energies and other properties à la Pauling (i.e., ionization energy and electron affinities) and find them to be off from experimental results (by tens of kcals/mol), then valence bond theory is not accurate.
Bonding theories can only be judged by their predictions.
A simple explanation that can be given is that molecular wavefunctions constructed out of hybridized atomic orbitals is accurate enough to predict some things, but not others. Predictions of any theory must be compared with empirical evidence to assess when they work and when they fail. When a theory gives the wrong answer, at least one assumption must not hold. In this case, the valence bond wavefunction is not accurate enough to capture some important features of a system's electronic structure. It may not be the most intellectually satisfying answer, but to say more would result in a much more complicated answer and certainly far beyond the level reasonably expected from general chemistry discussions.
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Electronic structure is the state of motion of electrons in an electrostatic field created by stationary nuclei. The term encompass both the wavefunctions of the electrons and the energies associated with them.
• Atomic Orbitals
This page discusses atomic orbitals at an introductory level. It explores s and p orbitals in some detail, including their shapes and energies. d orbitals are described only in terms of their energy, and f orbitals are only mentioned in passing.
• Bohr Diagrams of Atoms and Ions
Bohr diagrams show electrons orbiting the nucleus of an atom somewhat like planets orbit around the sun. In the Bohr model, electrons are pictured as traveling in circles at different shells, depending on which element you have.
• Electronic Configurations
The electron configuration of an atom is the representation of the arrangement of electrons distributed among the orbital shells and subshells. Commonly, the electron configuration is used to describe the orbitals of an atom in its ground state, but it can also be used to represent an atom that has ionized into a cation or anion by compensating with the loss of or gain of electrons in their subsequent orbitals. Many of the physical and chemical properties of elements can be correlated to electro
• Evaluating Spin Multiplicity
Spin-multiplicity value and its corresponding spin state was first discovered by Friedrich Hund in 1925. The formula which is generally used for the prediction of spin multiplicity value is (2S+1).
• Hydrogen's Atomic Emission Spectrum
This page introduces the atomic hydrogen emission spectrum, showing how it arises from electron movements between energy levels within the atom. It also explains how the spectrum can be used to find the ionization energy of hydrogen.
• Magnetic Behavior of Diatomic Species
• Predicting the Bond-Order of Diatomic Species
In this article text based learning approaches have been highlighted by innovative and time economic way to enhance interest of students’ who belong to paranoia zone in chemical bonding. In this pedagogical survey, I have tried to hub one time economic pedagogy by including four (04) new formulae in the field of chemical education. This article explores the results and gives implications for context based teaching, learning and assessment in a time economic way.
• Predicting the Bond-Order of Oxides based Acid Radicals
• Predicting the Hybridization of Simple Molecules
• Prediction of Aromatic, Anti Aromatic and Non Aromatic Character of Heterocyclic Compounds along with their Omission Behavior- Innovative Mnemonics
Electronic Structure of Atoms and Molecules
This page discusses atomic orbitals at an introductory level. It explores s and p orbitals in some detail, including their shapes and energies. d orbitals are described only in terms of their energy, and f orbitals are only mentioned in passing.
Orbitals and orbits
When a planet moves around the sun, its definite path, called an orbit, can be plotted. A drastically simplified view of the atom looks similar, in which the electrons orbit around the nucleus. The truth is different; electrons, in fact, inhabit regions of space known as orbitals. Orbits and orbitals sound similar, but they have quite different meanings. It is essential to understand the difference between them.
The impossibility of drawing orbits for electrons
To plot a path for something, the exact location and trajectory of the object must be known. This is imossible for electrons. The Heisenberg Uncertainty Principle states that it is impossible to define with absolute precision, at the same time, both the position and the momentum of an electron. That makes it impossible to plot an orbit for an electron around a nucleus.
Consider a single hydrogen atom: at a particular instant, the position of the electron is plotted. The position is plotted again soon afterward, and it is in a different position. There is no way to tell how it moved from the first place to the second. This process is repeated many times, eventually creating a 3D map of the places that the electron is likely to be found.
In the hydrogen case, the electron can be found anywhere within a spherical space surrounding the nucleus. The figure above shows a cross-section of this spherical space. 95% of the time (or any arbitrary, high percentage), the electron is found within a fairly easily defined region of space quite close to the nucleus. Such a region of space is called an orbital, and it can be thought of as the region of space the electron inhabits. It is impossible to know what the electron is doing inside the orbital, so the electron's actions are ignored completely. All that can be said is that if an electron is in a particular orbital, it has a particular, definable energy.
Each orbital has a name.
The orbital occupied by the hydrogen electron is called a 1s orbital. The number "1" represents the fact that the orbital is in the energy level closest to the nucleus. The letter "s" indicates the shape of the orbital: s orbitals are spherically symmetric around the nucleus—they look like hollow balls made of chunky material with the nucleus at the center.
The orbital shown above is a 2s orbital. This is similar to a 1s orbital, except that the region where there is the greatest chance of finding the electron is further from the nucleus. This is an orbital at the second energy level. There is another region of slightly higher electron density (where the dots are thicker) nearer the nucleus ("electron density" is another way of describing the likelihood of an electron at a particular place).
2s (and 3s, 4s, etc.) electrons spend some of their time closer to the nucleus than might be expected. The effect of this is to slightly reduce the energy of electrons in s orbitals. The nearer the nucleus the electrons get, the lower their energy. 3s, 4s (etc.) orbitals are progressively further from the nucleus.
p orbitals
Not all electrons inhabit s orbitals (in fact, very few electrons live occupy s orbitals). At the first energy level, the only orbital available to electrons is the 1s orbital, but at the second level, as well as a 2s orbital, there are 2p orbitals. A p orbital is shaped like 2 identical balloons tied together at the nucleus. The orbital shows where there is a 95% chance of finding a particular electron.
Imagine a horizontal plane through the nucleus, with one lobe of the orbital above the plane and the other beneath it; there is a zero probability of finding the electron on that plane. How does the electron get from one lobe to the other if it can never pass through the plane of the nucleus? At an introductory level, it must simply be accepted. To find out more, read about the wave nature of electrons.
At any one energy level it is possible to have three absolutely equivalent p orbitals pointing mutually at right angles to each other. These are arbitrarily given the symbols px, py and pz. This is simply for convenience; the x, y, and z directions change constantly as the atom tumbles in space.
The p orbitals at the second energy level are called 2px, 2py and 2pz. There are similar orbitals at subsequent levels: 3px, 3py, 3pz, 4px, 4py, 4pz and so on. All levels except for the first level have p orbitals. At the higher levels the lobes are more elongated, with the most likely place to find the electron more distant from the nucleus.
d and f orbitals
In addition to s and p orbitals, there are two other sets of orbitals that become available for electrons to inhabit at higher energy levels. At the third level, there is a set of five d orbitals (with complicated shapes and names) as well as the 3s and 3p orbitals (3px, 3py, 3pz). At the third level there are nine total orbitals. At the fourth level, as well the 4s and 4p and 4d orbitals there are an additional seven f orbitals, adding up to 16 orbitals in all. s, p, d and f orbitals are then available at all higher energy levels as well.
Fitting electrons into orbitals
An atom can be pictured as a very strange house (somewhat an inverted pyramid), with the nucleus living on the ground floor, and then various rooms (orbitals) on the higher floors occupied by the electrons. On the first floor there is only 1 room (the 1s orbital); on the second floor there are 4 rooms (the 2s, 2px, 2py and 2pz orbitals); on the third floor there are 9 rooms (one 3s orbital, three 3p orbitals and five 3d orbitals); and so on. However, the rooms are not large: each orbital can only hold 2 electrons.
"Electrons-in-boxes"
A convenient way of showing the orbitals that the electrons live in is to draw "electrons-in-boxes". Orbitals can be represented as boxes with the electrons depicted with arrows. Often an up-arrow and a down-arrow are used to show that the electrons are different. The need for all electrons in an atom to be different originates from quantum theory. If the electrons inhabit different orbitals, they can have identical properties, but if they are both in the same orbital there must be a distinction between them. Quantum theory allocates them a property known as "spin," represented by the direction the arrow is pointing.
A 1s orbital holding 2 electrons is drawn as shown on the right, but it can be written even more quickly as 1s2. This is read as "one s two," not as "one s squared." Do not confuse the energy level with the number of electrons in this notation.
The Aufbau Principle: the order of filling orbitals
Aufbau is a German word meaning building up or construction. Moving from one atom to the next in the periodic table, the electronic structure of the next atom can be determined by fitting an extra electron into the next available orbital. Electrons fill low energy orbitals (closer to the nucleus) before they fill higher energy ones. If there is a choice between orbitals of equal energy, they fill the orbitals singly as far as possible before pairing up.
This filling of orbitals singly where possible is known as Hund's rule. It only applies to orbitals with exactly the same energies (as with p orbitals, for example), and helps to minimize the repulsions between electrons and so makes the atom more stable. The diagram (not to scale) summarizes the energies of the orbitals up to the 4p level that you will need to know when you are using the Aufbau Principle.
Notice that the s orbital always has a slightly lower energy than the p orbitals at the same energy level, so the s orbital always fills with electrons before the corresponding p orbitals do. The oddity is the position of the 3d orbitals. They are at a slightly higher level than the 4s, so the 4s orbital fills first, followed by all the 3d orbitals and then the 4p orbitals.
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Objectives
• Recall the stability associated with an atom that has a completely-filled valence shell
• Construct an atom according to the Bohr model
Key Terms
• Octet rule: A rule stating that atoms lose, gain, or share electrons in order to have a full valence shell of 8 electrons. (Hydrogen is excluded because it can hold a maximum of 2 electrons in its valence shell. )
• Electron shell: The collective states of all electrons in an atom having the same principal quantum number (visualized as an orbit in which the electrons move).
Electron Shells
Niels Bohr proposed an early model of the atom as a central nucleus containing protons and neutrons being orbited by electrons in shells. As previously discussed, there is a connection between the number of protons in an element, the atomic number that distinguishes one element from another, and the number of electrons it has. In all electrically-neutral atoms, the number of electrons is the same as the number of protons. Each element, when electrically neutral, has a number of electrons equal to its atomic number.
An early model of the atom was developed in 1913 by Danish scientist Niels Bohr (1885–1962). The Bohr model shows the atom as a central nucleus containing protons and neutrons with the electrons in circular orbitals at specific distances from the nucleus (Figure \(1\)). These orbits form electron shells or energy levels, which are a way of visualizing the number of electrons in the various shells. These energy levels are designated by a number and the symbol "n." For example, the 1n shell represents the first energy level located closest to the nucleus.
An electron normally exists in the lowest energy shell available, which is the one closest to the nucleus. Energy from a photon of light can bump it up to a higher energy shell, but this situation is unstable and the electron quickly decays back to the ground state.
Bohr Diagrams
Bohr diagrams show electrons orbiting the nucleus of an atom somewhat like planets orbit around the sun. In the Bohr model, electrons are pictured as traveling in circles at different shells, depending on which element you have. Figure \(2\) contrast the Bohr diagrams for lithium, fluorine and aluminum atoms. The shell closest to the nucleus is called the K shell, next is the L shell, next is the M shell.
Each shell can only hold certain number of electrons. K shell can have 2, L can have 8 , M can have 18 electrons and so on.
• Lithium has three electrons:
• two go to K shell and
• the remaining one goes to the L shell.
• Its electronic configuration is K(2), L(1)
• Fluorine has nine electrons:
• two go to K shell and
• the remaining seven go to the L shell.
• Its electronic configuration is K(2), L(7). Note that L can have 8 electrons.
• Aluminum has thirteen electrons:
• two go to the K shell,
• eight go to the L shell, and
• remaining three go to the M shell.
• Its electronic configuration is K(2), L(8), M(3). Note that the M shell can have 18 electrons.
Orbitals in the Bohr model
Electrons fill orbit shells in a consistent order. Under standard conditions, atoms fill the inner shells (closer to the nucleus) first, often resulting in a variable number of electrons in the outermost shell. The innermost shell has a maximum of two electrons, but the next two electron shells can each have a maximum of eight electrons. This is known as the octet rule which states that, with the exception of the innermost shell, atoms are more stable energetically when they have eight electrons in their valence shell, the outermost electron shell. Examples of some neutral atoms and their electron configurations are shown in Figure \(3\). As shown, helium has a complete outer electron shell, with two electrons filling its first and only shell. Similarly, neon has a complete outer 2n shell containing eight electrons. In contrast, chlorine and sodium have seven and one electrons in their outer shells, respectively. Theoretically, they would be more energetically stable if they followed the octet rule and had eight.
Bohr diagrams
Bohr diagrams indicate how many electrons fill each principal shell. Group 18 elements (helium, neon, and argon are shown) have a full outer, or valence, shell. A full valence shell is the most stable electron configuration. Elements in other groups have partially-filled valence shells and gain or lose electrons to achieve a stable electron configuration.
An atom may gain or lose electrons to achieve a full valence shell, the most stable electron configuration. The periodic table is arranged in columns and rows based on the number of electrons and where these electrons are located, providing a tool to understand how electrons are distributed in the outer shell of an atom. As shown in , the group 18 atoms helium (He), neon (Ne), and argon (Ar) all have filled outer electron shells, making it unnecessary for them to gain or lose electrons to attain stability; they are highly stable as single atoms. Their non-reactivity has resulted in their being named the inert gases (or noble gases). In comparison, the group 1 elements, including hydrogen (H), lithium (Li), and sodium (Na), all have one electron in their outermost shells. This means that they can achieve a stable configuration and a filled outer shell by donating or losing an electron. As a result of losing a negatively-charged electron, they become positively-charged ions. When an atom loses an electron to become a positively-charged ion, this is indicated by a plus sign after the element symbol; for example, Na+. Group 17 elements, including fluorine and chlorine, have seven electrons in their outermost shells; they tend to fill this shell by gaining an electron from other atoms, making them negatively-charged ions. When an atom gains an electron to become a negatively-charged ion this is indicated by a minus sign after the element symbol; for example, \(F^-\). Thus, the columns of the periodic table represent the potential shared state of these elements' outer electron shells that is responsible for their similar chemical characteristics.
Lewis Symbols
Lewis Symbols are simplified Bohr diagrams which only display electrons in the outermost energy level.
Summary
• In the Bohr model of the atom, the nucleus contains the majority of the mass of the atom in its protons and neutrons.
• Orbiting the positively-charged core are the negatively charged electrons, which contribute little in terms of mass, but are electrically equivalent to the protons in the nucleus.
• In most cases, electrons fill the lower-energy orbitals first, followed by the next higher energy orbital until it is full, and so on until all electrons have been placed.
• Atoms tend to be most stable with a full outer shell (one which, after the first, contains 8 electrons), leading to what is commonly called the "octet rule".
• The properties of an element are determined by its outermost electrons, or those in the highest energy orbital.
• Atoms that do not have full outer shells will tend to gain or lose electrons, resulting in a full outer shell and, therefore, stability.
Contributors and Attributions
Boundless (www.boundless.com)
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• Basic Electronic Structure of Atoms
This page explores how you write electronic structures for atoms using s, p, and d notation. It assumes that you know about simple atomic orbitals - at least as far as the way they are named, and their relative energies.
• Case Study: Electron Configuration of Mn vs. Cu
• d-orbital Occupation and Electronic Configurations
• Electronic Configurations Intro
The electron configuration of an atom is the representation of the arrangement of electrons distributed among the orbital shells and subshells. Commonly, the electron configuration is used to describe the orbitals of an atom in its ground state, but it can also be used to represent an atom that has ionized into a cation or anion by compensating with the loss of or gain of electrons in their subsequent orbitals. Many of the physical and chemical properties of elements can be correlated to electro
• Electronic Structure and Orbitals
• Hund's Rules
Hund's rule states that: Every orbital in a sublevel is singly occupied before any orbital is doubly occupied. All of the electrons in singly occupied orbitals have the same spin (to maximize total spin). When assigning electrons to orbitals, an electron first seeks to fill all the orbitals with similar energy (also referred to as degenerate orbitals) before pairing with another electron in a half-filled orbital. Atoms at ground states tend to have as many unpaired electrons as possible.
• Pauli Exclusion Principle
The Pauli Exclusion Principle states that, in an atom or molecule, no two electrons can have the same four electronic quantum numbers. As an orbital can contain a maximum of only two electrons, the two electrons must have opposing spins. This means if one is assigned an up-spin ( +1/2), the other must be down-spin (-1/2).
• Spin Pairing Energy
Spin pairing energy refers to the energy associated with paired electrons sharing one orbital and its effect on the molecules surrounding it. Electron pairing determining the direction of spin depends on several laws founded by chemists over the years such as Hund's law, the Aufbau principle, and Pauli's exclusion principle. An overview of the different types laws associated with the electron pairing rules.
• Spin Quantum Number
The Spin Quantum Number (s) is a value (of 1/2) that describes the angular momentum of an electron. An electron spins around an axis and has both angular momentum and orbital angular momentum. Because angular momentum is a vector, the Spin Quantum Number (s) has both a magnitude (1/2) and direction (+ or -). This vector is called the magnetic spin quantum number (ms).
• The Aufbau Process
The Aufbau model lets us take an atom and make predictions about its properties. All we need to know is how many protons it has (and how many electrons, which is the same as the number of protons for a neutral atom). We can predict the properties of the atom based on our vague idea of where its electrons are and, more importantly, the energy of those electrons.
• The Octet Rule
The octet rule refers to the tendency of atoms to prefer to have eight electrons in the valence shell. When atoms have fewer than eight electrons, they tend to react and form more stable compounds.
• The Order of Filling 3d and 4s Orbitals
This page looks at some of the problems with the usual way of explaining the electronic structures of the d-block elements based on the order of filling of the d and s orbitals. The way that the order of filling of orbitals is normally taught gives an easy way of working out the electronic structures of elements. However, it does throw up problems when you come to explain various properties of the transition elements.
Electronic Configurations
Aufbau comes from the German word "Aufbauen" which means "to build". In essence when writing electron configurations we are building up electron orbitals as we proceed from atom to atom. As we write the electron configuration for an atom, we will fill the orbitals in order of increasing atomic number. The Aufbau principle originates from the Pauli’s exclusion principle which says that no two fermions (e.g., electrons) in an atom can have the same set of quantum numbers, hence they have to "pile up" or "build up" into higher energy levels. How the electrons build up is a topic of electron configurations.
Example \(1\)
If we follow the pattern across a period from B (Z=5) to Ne (Z=10) the number of electrons increase and the subshells are filled. Here we are focusing on the p subshell in which as we move towards Ne, the p subshell becomes filled.
• B (Z=5) configuration: 1s2 2s2 2p1
• C (Z=6) configuration: 1s2 2s2 2p2
• N (Z=7) configuration: 1s2 2s2 2p3
• O (Z=8) configuration: 1s2 2s2 2p4
• F (Z=9) configuration: 1s2 2s2 2p5
• Ne (Z=10) configuration: 1s2 2s2 2p6
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This page explores how you write electronic structures for atoms using s, p, and d notation. It assumes that you know about simple atomic orbitals - at least as far as the way they are named, and their relative energies.
The first period
Hydrogen has its only electron in the 1s orbital - 1s1, and at helium the first level is completely full - 1s2.
The second period
Now we need to start filling the second level, and hence start the second period. Lithium's electron goes into the 2s orbital because that has a lower energy than the 2p orbitals. Lithium has an electronic structure of 1s22s1. Beryllium adds a second electron to this same level - 1s22s2. Now the 2p levels start to fill. These levels all have the same energy, and so the electrons go in singly at first.
Example
B 1s22s22px1
C 1s22s22px12py1
N 1s22s22px12py12pz1
The next electrons to go in will have to pair up with those already there.
Example
O 1s22s22px22py12pz1
F 1s22s22px22py22pz1
Ne 1s22s22px22py22pz2
You can see that it is going to get progressively tedious to write the full electronic structures of atoms as the number of electrons increases. There are two ways around this, and you must be familiar with both.vShortcut 1: All the various p electrons can be lumped together. For example, fluorine could be written as 1s22s22p5, and neon as 1s22s22p6.
This is what is normally done if the electrons are in an inner layer. If the electrons are in the bonding level (those on the outside of the atom), they are sometimes written in shorthand, sometimes in full. Don't worry about this. Be prepared to meet either version, but if you are asked for the electronic structure of something in an exam, write it out in full showing all the px, py and pz orbitals in the outer level separately.
For example, although we have not yet met the electronic structure of chlorine, you could write it as 1s22s22p63s23px23py23pz1. Notice that the 2p electrons are all lumped together whereas the 3p ones are shown in full. The logic is that the 3p electrons will be involved in bonding because they are on the outside of the atom, whereas the 2p electrons are buried deep in the atom and are not really of any interest.
Shortcut 2: You can lump all the inner electrons together using, for example, the symbol [Ne]. In this context, [Ne] means the electronic structure of neon - in other words: 1s22s22px22py22pz2 You wouldn't do this with helium because it takes longer to write [He] than it does 1s2. On this basis the structure of chlorine would be written [Ne]3s23px23py23pz1.
The third period
At neon, all the second level orbitals are full, and so after this we have to start the third period with sodium. The pattern of filling is now exactly the same as in the previous period, except that everything is now happening at the 3-level.
Example
short version
Mg 1s22s22p63s2 [Ne]3s2
S 1s22s22p63s23px23py13pz1 [Ne]3s23px23py13pz1
Ar 1s22s22p63s23px23py23pz2 [Ne]3s23px23py23pz2
The beginning of the fourth period
At this point the 3-level orbitals are not all full - the 3d levels have not been used yet. But if you refer back to the energies of the orbitals, you will see that the next lowest energy orbital is the 4s - so that fills next.
Example
K 1s22s22p63s23p64s1
Ca 1s22s22p63s23p64s2
There is strong evidence for this in the similarities in the chemistry of elements like sodium (1s22s22p63s1) and potassium (1s22s22p63s23p64s1). The outer electron governs their properties and that electron is in the same sort of orbital in both of the elements. That wouldn't be true if the outer electron in potassium was 3d1.
s- and p-block elements
The elements in Group 1 of the Periodic Table all have an outer electronic structure of ns1 (where n is a number between 2 and 7). All Group 2 elements have an outer electronic structure of ns2. Elements in Groups 1 and 2 are described as s-block elements. Elements from Group 3 (the boron group) across to the noble gases all have their outer electrons in p orbitals. These are then described as p-block elements.
d-block elements
We are working out the electronic structures of the atoms using the Aufbau ("building up") Principle. So far we have got to calcium with a structure of 1s22s22p63s23p64s2. The 4s level is now full, and the structures of the next atoms show electrons gradually filling up the 3d level. These are known as d-block elements.
Once the 3d orbitals have filled up, the next electrons go into the 4p orbitals as you would expect. d-block elements are elements in which the last electron to be added to the atom using the Aufbau Principle is in a d orbital. The first series of these contains the elements from scandium to zinc, which are called transition elements or transition metals. Technically, the terms "transition element" and "d-block element" do not quite have the same meaning, but it doesn't matter in the present context.
d electrons are almost always described as, for example, d5 or d8 - and not written as separate orbitals. Remember that there are five d orbitals, and that the electrons will inhabit them singly as far as possible. Up to 5 electrons will occupy orbitals on their own. After that they will have to pair up.
d5 means
d8 means
Notice in what follows that all the 3-level orbitals are written together - with the 4s electrons written at the end of the electronic structure.
Example
Sc 1s22s22p63s23p63d14s2
Ti 1s22s22p63s23p63d24s2
V 1s22s22p63s23p63d34s2
Cr 1s22s22p63s23p63d54s1
However, chromium breaks the sequence. In chromium, the electrons in the 3d and 4s orbitals rearrange so that there is one electron in each orbital. It would be convenient if the sequence was tidy - but it is not!
Example
Mn 1s22s22p63s23p63d54s2 (back to being tidy again)
Fe 1s22s22p63s23p63d64s2
Co 1s22s22p63s23p63d74s2
Ni 1s22s22p63s23p63d84s2
Cu 1s22s22p63s23p63d104s1 (another awkward one!)
Zn 1s22s22p63s23p63d104s2
And at zinc the process of filling the d orbitals is complete.
Filling the rest of period 4
The next orbitals to be used are the 4p, and these fill in exactly the same way as the 2p or 3p. We are back now with the p-block elements from gallium to krypton. Bromine, for example, is
1s22s22p63s23p63d104s24px24py24pz1.
Writing the electronic structure of big s- or p-block elements
First work out the number of outer electrons. This is quite likely all you will be asked to do anyway. The number of outer electrons is the same as the group number. (The noble gases are a bit of a problem here, because they are normally called group 0 rather then group 8. Helium has 2 outer electrons; the rest have 8.) All elements in group 3, for example, have 3 electrons in their outer level. Fit these electrons into s and p orbitals as necessary. Which level orbitals? Count the periods in the Periodic Table (not forgetting the one with H and He in it).
Iodine is in group 7 and so has 7 outer electrons. It is in the fifth period and so its electrons will be in 5s and 5p orbitals. Iodine has the outer structure
5s25px25py25pz1.
What about the inner electrons if you need to work them out as well? The 1, 2 and 3 levels will all be full, and so will the 4s, 4p and 4d. That gives the full structure:
1s22s22p63s23p63d104s24p64d105s25px25py25pz1.
Once finished, count all the electrons to make sure that they come to the same as the atomic number. Don't forget to make this check - it is easy to miss an orbital out when it gets this complicated.
Barium is in group 2 and so has 2 outer electrons. It is in the sixth period. Barium has the outer structure 6s2. Including all the inner levels:
1s22s22p63s23p63d104s24p64d105s25p66s2.
It would be easy to include 5d10 as well by mistake, but the d level always fills after the next s level - so 5d fills after 6s just as 3d fills after 4s. As long as you counted the number of electrons you could easily spot this mistake because you would have 10 too many.
Summary
Steps in writing the electronic structure of an element from hydrogen to krypton:
1. Use the Periodic Table to find the atomic number, and hence number of electrons.
2. Fill up orbitals in the order 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p - until you run out of electrons. The 3d is the awkward one - remember that specially. Fill p and d orbitals singly as far as possible before pairing electrons up.
3. Remember that chromium and copper have electronic structures which break the pattern in the first row of the d-block.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Electronic_Structure_of_Atoms_and_Molecules/Electronic_Configurations/Basic_Electronic_Structure_of_Atoms.txt
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Transition metals consist of elements from the d-block found between the group IIa and the group IIb elements of the periodic table. A transition metal is an element that forms one or more stable ions which have incompletely filled d sub-orbitals. While these transition metals contain at most two electrons in their outer shell, they are called d-block elements because their next highest energy shell has incompletely filled d sub-orbitals.
Introduction
The d sub-orbitals in transition metals are filled up, progressively going across the periodic table from left to right. These d sub-orbitals of transition metals are readily available for metallic bonding. They give way to general properties of transition metals such as good electrical and heat conductivity and high melting points. Since both the outer shell and the d sub-orbital shells can be used for bonding, it allows transition metals to bond with a variety of elements in many shapes. It should be noted, however, that not all d-block elements are transition metals. Some of the elements, such as Scandium and Zinc, have either empty or completely filled d sub-orbitals and do not meet the criteria to be considered transition metals.
Electronic Structure
The electronic configuration of transition metal elements are characterized as having full outer sub-orbitals and the second outermost d sub-orbitals incompletely filled, with the exception of Copper which loses one 4s orbital electron to the 3d sub-orbital for increased stability. The electron configuration for copper is 1s22s22p63s23p64s13d10 or given in terms of a noble gas configuration which is [Ar]4s13d10and not [Ar]4s23d9, as would be expected from the Aufbau principle. The loss of the electron from the 4s orbital completes the 3d sub-orbital and leaves it in a more stable, lower energy state. This lowest energy electron configuration is determined by the electron-pairing energy and the exchange energy. The two electrons in the 4s orbital will repel each other slightly, and this will cause a minor increase in the energy of the atom. The removal of one of the electrons from the 4s shell will remove this energy increase which is caused by the pairing energy. When electrons can exchange between orbitals, the energy of the atom is lowered. This electron exchange can occur when there are degenerate electron configurations, and the electrons of the same spin can exchange between orbitals. This exchange energy is defined by the spin-Hamiltonian:
$H_o = -2S_1S_2$
where J is the set of many electron exchange parameters, and S1 and S2 are the spin operators of the two electrons.
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When J is positive, the exchange energy favors electrons with parallel spins. When J is negative, it favors antiparallel spins. This leads to the primary cause of ferro and anti-ferro magnetism. The spin-Hamiltonian is summed over all pair interactions, and it is maximized when the electronic shells are either completely filled or half-filled. Such is the case with copper and its loss of one 4s electron to the 3d sub-orbital to form the noble gas configuration of [Ar]4s13d10.
Manganese, on the other hand, has an electron configuration of 1s22s22p63s23p64s23d5 and a noble gas configuration of [Ar]4s23d5, resulting in one unpaired electron in each 3d sub-orbital. Similar to copper, the exchange energy is maximized since the 3d sub-orbital is exactly half-filled with 5 electrons without the need to steal electrons from the 4s orbital. The electrons in the d sub-orbitals of both copper and manganese are subject to Pauli’s exclusion principle which says that no two fermions in an atom can have the same quantum number.
2
Since electrons are fermions with a spin of 1/2, the wavefunction of the system must be antisymmetric when the electrons are exchanged. The spin-Hamiltonian, that determines the exchange energy interaction, defines the important properties of both manganese and copper.
Copper is found in two common ionic forms, Cuprous Cu(I) with a noble gas electron configuration of [Ar]4s03d10 and Cupric Cu(II) with [Ar]4s03d9. Cu(III) and even Cu(IV) ionic forms of copper have been formed.
4
The unpaired electron in the 4s orbital is lost in Cu(I) which leaves a complete d sub-orbital. However, Cu(II) is actually more stable, due to a higher ionization energy. The relatively unstable Cu(I) ion disproportionates as:
$2Cu(I) \rightarrow Cu(0) + Cu(II)$
Manganese has seven ionic forms from Mn(I) to Mn(VII). The two most common forms are Mn(II), with a noble gas electronic configuration of [Ar]4s03d5 and Mn(VII), with a configuration of [Ar]4s03d0 and a formal loss of all seven electrons from the 3d and 4s orbitals.
Crystal Field Splitting
Crystal field splitting diagrams can be used to determine the way strong and weak ligands will affect the magnitude of ligand field, splitting parameters and determining the number of unpaired electrons are in each transition metal. Using H2O as a weak field ligand, complexed with Mn(II) in the form of [Mn(H2O)6]2+, the low oxidation state of manganese will cause a high spin with five unpaired electrons.
FIGURE HERE
The crystal field splitting of a ligand in the case with water, has an important effect on the d-orbital energy levels. Since the ligand electrons have repulsions with the d sub-orbital electrons, the d-orbital energy levels are raised. However, they are not raised equally. In the case of this octahedral complex ion, ligands approach along the x-, y- and z-axis. When the six ?-donor water molecules are coordinated with the manganese ion, the orbitals ($d_{z^2}$ and $d_{x^2-y^2}$) are head-to-head with the ligand and have a greater repulsion and thus, a higher energy. The energy levels of the ($d_{xy}$, $d_{xz}$, $d_{yz}$) orbitals are also increased, due to this repulsion but to a lesser extent.
Using NH3 as a strong field splitting ligand coordinated to copper, such as [Cu(NH3)4]2+, which has an ionic form of copper Cu(II). The Cu(II) has nine d sub-orbital electrons, and since there are only five orbitals, there can only be one configuration of the electrons which results in one unpaired electron. The complex is paramagnetic, but the structure cannot be determined from magnetic properties.
Spin of Mn
With a Mn(II) ion with a weak field ligand, the energy splitting is low and the cost of placing an electron into a higher energy d orbital is lower than the repulsion energy of pairing two electrons together. Thus, the Mn(II) ion in a weak field ligand setting, produces a high spin as all five electrons are unpaired and the ion is paramagnetic. One electron is placed into each d orbital according to Hund’s rule, and the “high spin” octahedral complex has all five orbitals singly occupied. The magnetic susceptibility is determined to be 5.92, using the formula:
$\chi= [n(n+2)]^{1/2}$
where n is the number of unpaired electrons.
In the case of Cu(II) in a strong ligand field such as NH3, the energy required to pair the electrons is not relevant since there are nine d sub-orbital electrons that need to be placed in five d orbital shells. This results in a single unpaired electron which means that the complex is paramagnetic, but the high spin vs low spin associated energy splittings are not relevant.
5
If the Cu(II) ion were in a weak ligand field setting, the high spin vs low spin energy levels would not be relevant again, due to the number of electrons. This would result in a tetrahedral stereochemistry which typically displays a high spin arrangement.
The ground state electronic term symbols represent a shorthand notation which is predicted, using Hund’s rule for the angular momentum quantum numbers of an atom with multiple electrons. This is related to the energy level of the electronic configurations of the atoms.
The term symbol has the form:
$\large ^{2S+1}L_J$
where S is the spin quantum number; L is the orbital momentum quantum number in spectroscopic notation given in L= 0,1,2,3,4,5 are S,P,D,F,G,H, respectively; and finally, J is the angular momentum quantum number. This symbol assumes that all spins are combined to produce S; all orbital angular momenta are combined to produce L; and all the spin and orbital angular momenta are combined to produce J. The multiplicity of a term is the value of 2S+1, provided that L ? S, and is the number of levels of the term. The values for S, L, and J are constructed by the application of Clebsch-Gordan series. To construct a term symbol arising from an electronic configuration, first determine the possible values of L, and identify the spectroscopic letters for the terms. The possible values of S are determined similarly, using a Clebsch-Gordan series, and the multiplicities are calculated. Finally, the values for J are constructed again, using a Clebsch-Gordan series by combining S and L.
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Clebsch-Gordan Series
$j = j_1+j_2, j_1+j_{2-1},…,|j_1-j_2|$
Using the ground state electronic configuration, the electrons are distributed according to Pauli’s exclusion principle to the available orbitals. The first orbitals filled are those with the highest ml values; S is calculated by added all the ms values; L is calculated by added together the ml values for each electron; and J is then determined, using a set of rules. If the subshell is half-filled, then the L=0 and J will be equal to S. If the subshell is more than half-filled, then J=L+S; if the subshell is less than half-filled, then J=|L-S|. The ground electronic term symbol for copper is thus determined to be 2S1/2. The term symbol for Cu(II), with an electronic configuration [Ar]4s03d9 is calculated to be 2D5/2. The ground electronic term symbol for manganese is thus determined to be 6S5/2. The term symbol for Mn(II) is also 6S5/2, since the orbital with the highest ml value still has 5 electrons in it and thus, L=0.
Outside Links
• Vanquickenborne, L. G.; Pierloot, K.; Devoghel, D. "Transition Metals and the Aufbau Principle." J. Chem. Educ. 1994, 71, 469.
Contributors
• Zane Sterkewolf (UCD)
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Electronic_Structure_of_Atoms_and_Molecules/Electronic_Configurations/Case_Study%3A_Electron_Configuration_of_Mn_vs.txt
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The electron configuration of an atom is the representation of the arrangement of electrons distributed among the orbital shells and subshells. Commonly, the electron configuration is used to describe the orbitals of an atom in its ground state, but it can also be used to represent an atom that has ionized into a cation or anion by compensating with the loss of or gain of electrons in their subsequent orbitals. Many of the physical and chemical properties of elements can be correlated to their unique electron configurations. The valence electrons, electrons in the outermost shell, are the determining factor for the unique chemistry of the element.
Contributors and Attributions
• Sarah Faizi (University of California Davis)
Electronic Structure and Orbitals
In any introductory chemistry course you will have come across the electronic structures of hydrogen and carbon drawn as:
The circles show energy levels - representing increasing distances from the nucleus. You could straighten the circles out and draw the electronic structure as a simple energy diagram.
Atomic Orbitals
Orbits and orbitals sound similar, but they have quite different meanings. It is essential that you understand the difference between them. To plot a path for something you need to know exactly where the object is and be able to work out exactly where it's going to be an instant later. You can't do this for electrons.
The Heisenberg Uncertainty Principle says that you cannot know with certainty both where an electron is and where it's going next. That makes it impossible to plot an orbit for an electron around a nucleus. Is this a big problem? No. If something is impossible, you have to accept it and find a way around it.
Hydrogen's electron - the 1s orbital
Suppose you had a single hydrogen atom and at a particular instant plotted the position of the one electron. Soon afterwards, you do the same thing, and find that it is in a new position. You have no idea how it got from the first place to the second. You keep on doing this over and over again, and gradually build up a sort of 3D map of the places that the electron is likely to be found.
In the hydrogen case, the electron can be found anywhere within a spherical space surrounding the nucleus. The diagram shows a cross-section through this spherical space. 95% of the time (or any other percentage you choose), the electron will be found within a fairly easily defined region of space quite close to the nucleus. Such a region of space is called an orbital. You can think of an orbital as being the region of space in which the electron lives.
What is the electron doing in the orbital? We don't know, we can't know, and so we just ignore the problem! All you can say is that if an electron is in a particular orbital it will have a particular definable energy.
Each orbital has a name. The orbital occupied by the hydrogen electron is called a 1s orbital. The "1" represents the fact that the orbital is in the energy level closest to the nucleus. The "s" tells you about the shape of the orbital. s orbitals are spherically symmetric around the nucleus - in each case, like a hollow ball made of rather chunky material with the nucleus at its center.
The orbital on the left is a 2s orbital. This is similar to a 1s orbital except that the region where there is the greatest chance of finding the electron is further from the nucleus - this is an orbital at the second energy level. If you look carefully, you will notice that there is another region of slightly higher electron density (where the dots are thicker) nearer the nucleus. ("Electron density" is another way of talking about how likely you are to find an electron at a particular place.)
2s (and 3s, 4s, etc) electrons spend some of their time closer to the nucleus than you might expect. The effect of this is to slightly reduce the energy of electrons in s orbitals. The nearer the nucleus the electrons get, the lower their energy. 3s, 4s (etc) orbitals get progressively further from the nucleus.
p orbitals
Not all electrons inhabit s orbitals (in fact, very few electrons live in s orbitals). At the first energy level, the only orbital available to electrons is the 1s orbital, but at the second level, as well as a 2s orbital, there are also orbitals called 2p orbitals. A p orbital is rather like 2 identical balloons tied together at the nucleus. The diagram on the right is a cross-section through that 3-dimensional region of space. Once again, the orbital shows where there is a 95% chance of finding a particular electron.
Unlike an s orbital, a p orbital points in a particular direction - the one drawn points up and down the page. At any one energy level it is possible to have three absolutely equivalent p orbitals pointing mutually at right angles to each other. These are arbitrarily given the symbols px, py and pz. This is simply for convenience - what you might think of as the x, y or z direction changes constantly as the atom tumbles in space.
The p orbitals at the second energy level are called 2px, 2py and 2pz. There are similar orbitals at subsequent levels - 3px, 3py, 3pz, 4px, 4py, 4pz and so on. All levels except for the first level have p orbitals. At the higher levels the lobes get more elongated, with the most likely place to find the electron more distant from the nucleus.
Fitting electrons into orbitals
Because for the moment we are only interested in the electronic structures of hydrogen and carbon, we do not need to concern ourselves with what happens beyond the second energy level. Remember:
• At the first level there is only one orbital - the \(1s\) orbital.
• At the second level there are four orbitals - the \(2s\), \(2p_x\), \(2p_y\) and \(2p_z\) or bitals.
• Each orbital can hold either 1 or 2 electrons, but no more.
"Electrons-in-boxes"
Orbitals can be represented as boxes with the electrons in them shown as arrows. Often an up-arrow and a down-arrow are used to show that the electrons are in some way different. A 1s orbital holding 2 electrons would be drawn as shown on the right, but it can be written even more quickly as 1s2. This is read as "one s two" - not as "one s squared". You mustn't confuse the two numbers in this notation:
The order of filling orbitals
Electrons fill low energy orbitals (closer to the nucleus) before they fill higher energy ones. Where there is a choice between orbitals of equal energy, they fill the orbitals singly as far as possible. The diagram (not to scale) summarizes the energies of the various orbitals in the first and second levels.
Notice that the 2s orbital has a slightly lower energy than the 2p orbitals. That means that the 2s orbital will fill with electrons before the 2p orbitals. All the 2p orbitals have exactly the same energy.
Example 1: The electronic structure of hydrogen
Hydrogen only has one electron and that will go into the orbital with the lowest energy - the 1s orbital. Hydrogen has an electronic structure of 1s1. We have already described this orbital earlier.
Example 2: The electronic structure of carbon
Carbon has six electrons. Two of them will be found in the 1s orbital close to the nucleus. The next two will go into the 2s orbital. The remaining ones will be in two separate 2p orbitals. This is because the p orbitals all have the same energy and the electrons prefer to be on their own if that's the case.
The electronic structure of carbon is normally written 1s22s22px12py10
Contributor
Jim Clark (Chemguide.co.uk)
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The Aufbau section discussed how electrons fill the lowest energy orbitals first, and then move up to higher energy orbitals only after the lower energy orbitals are full. However, there is a problem with this rule. Certainly, 1s orbitals should be filled before 2s orbitals, because the 1s orbitals have a lower value of \(n\), and thus a lower energy. What about filling the three different 2p orbitals? In what order should they be filled? The answer to this question involves Hund's rule.
Hund's rule states that:
1. Every orbital in a sublevel is singly occupied before any orbital is doubly occupied.
2. All of the electrons in singly occupied orbitals have the same spin (to maximize total spin).
When assigning electrons to orbitals, an electron first seeks to fill all the orbitals with similar energy (also referred to as degenerate orbitals) before pairing with another electron in a half-filled orbital. Atoms at ground states tend to have as many unpaired electrons as possible. In visualizing this process, consider how electrons exhibit the same behavior as the same poles on a magnet would if they came into contact; as the negatively charged electrons fill orbitals, they first try to get as far as possible from each other before having to pair up.
Example \(1\): Nitrogen Atoms
Consider the correct electron configuration of the nitrogen (Z = 7) atom: 1s2 2s2 2p3
The p orbitals are half-filled; there are three electrons and three p orbitals. This is because the three electrons in the 2p subshell will fill all the empty orbitals first before pairing with electrons in them.
Keep in mind that elemental nitrogen is found in nature typically as molecular nitrogen, \(\ce{N2}\), which requires molecular orbitals instead of atomic orbitals as demonstrated above.
Example \(2\): Oxygen Atoms
Next, consider oxygen (Z = 8) atom, the element after nitrogen in the same period; its electron configuration is: 1s2 2s2 2p4
Oxygen has one more electron than nitrogen; as the orbitals are all half-filled, the new electron must pair up. Keep in mind that elemental oxygen is found in nature typically as molecular oxygen, \(\ce{O_2}\), which has molecular orbitals instead of atomic orbitals as demonstrated above.
Hund's Rule Explained
According to the first rule, electrons always enter an empty orbital before they pair up. Electrons are negatively charged and, as a result, they repel each other. Electrons tend to minimize repulsion by occupying their own orbitals, rather than sharing an orbital with another electron. Furthermore, quantum-mechanical calculations have shown that the electrons in singly occupied orbitals are less effectively screened or shielded from the nucleus. Electron shielding is further discussed in the next section.
For the second rule, unpaired electrons in singly occupied orbitals have the same spins. Technically speaking, the first electron in a sublevel could be either "spin-up" or "spin-down." Once the spin of the first electron in a sublevel is chosen, however, the spins of all of the other electrons in that sublevel depend on that first spin. To avoid confusion, scientists typically draw the first electron, and any other unpaired electron, in an orbital as "spin-up."
Example \(3\): Carbon and Oxygen
Consider the electron configuration for carbon atoms: 1s22s22p2: The two 2s electrons will occupy the same orbital, whereas the two 2p electrons will be in different orbital (and aligned the same direction) in accordance with Hund's rule.
Consider also the electron configuration of oxygen. Oxygen has 8 electrons. The electron configuration can be written as 1s22s22p4. To draw the orbital diagram, begin with the following observations: the first two electrons will pair up in the 1s orbital; the next two electrons will pair up in the 2s orbital. That leaves 4 electrons, which must be placed in the 2p orbitals. According to Hund’s rule, all orbitals will be singly occupied before any is doubly occupied. Therefore, two p orbital get one electron and one will have two electrons. Hund's rule also stipulates that all of the unpaired electrons must have the same spin. In keeping with convention, the unpaired electrons are drawn as "spin-up", which gives (Figure 1).
Purpose of Electron Configurations
When atoms come into contact with one another, it is the outermost electrons of these atoms, or valence shell, that will interact first. An atom is least stable (and therefore most reactive) when its valence shell is not full. The valence electrons are largely responsible for an element's chemical behavior. Elements that have the same number of valence electrons often have similar chemical properties.
Electron configurations can also predict stability. An atom is most stable (and therefore unreactive) when all its orbitals are full. The most stable configurations are the ones that have full energy levels. These configurations occur in the noble gases. The noble gases are very stable elements that do not react easily with any other elements. Electron configurations can assist in making predictions about the ways in which certain elements will react, and the chemical compounds or molecules that different elements will form.
Pauli Exclusion Principle
The Pauli Exclusion Principle states that, in an atom or molecule, no two electrons can have the same four electronic quantum numbers. As an orbital can contain a maximum of only two electrons, the two electrons must have opposing spins. This means if one electron is assigned as a spin up (+1/2) electron, the other electron must be spin-down (-1/2) electron.
Electrons in the same orbital have the same first three quantum numbers, e.g., \(n=1\), \(l=0\), \(m_l=0\) for the 1s subshell. Only two electrons can have these numbers, so that their spin moments must be either \(m_s = -1/2\) or \(m_s = +1/2\). If the 1s orbital contains only one electron, we have one \(m_s\) value and the electron configuration is written as 1s1 (corresponding to hydrogen). If it is fully occupied, we have two \(m_s\) values, and the electron configuration is 1s2 (corresponding to helium). Visually these two cases can be represented as
As you can see, the 1s and 2s subshells for beryllium atoms can hold only two electrons and when filled, the electrons must have opposite spins. Otherwise they will have the same four quantum numbers, in violation of the Pauli Exclusion Principle.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Electronic_Structure_of_Atoms_and_Molecules/Electronic_Configurations/Hund%27s_Rules.txt
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Spin pairing energy refers to the energy associated with paired electrons sharing one orbital and its effect on the molecules surrounding it. Electron pairing determining the direction of spin depends on several laws founded by chemists over the years such as Hund's law, the Aufbau principle, and Pauli's exclusion principle. An overview of the different types laws associated with the electron pairing rules.
Introduction
There are two different types of spin paring configurations for an atom or ion: paramagnetic or diamagnetic. Paramagnetic and diamagnetic configurations result from the amount of d electrons in a particular atom. The energy associated with the spin pairing of these configurations relies on a factor of three things, the atom (for its electronic configuration and number of d electrons), the Crystal Field Theory (field splitting of electrons), and the type of ligand field complex (tetrahedral or octahedral). Examples of these three factors affect on spin pairing are shown here;
Being diamagnetic means having all electrons paired and the individual magnetic effects cancel each other out.
Being paramagnetic means having unpaired electrons and the individual magnetic effects do not cancel each other out. The unpaired electrons carry a magnetic moment that gets stronger with the number of unpaired electrons causing the atom or ion to be attracted to an external magnetic field.
Spin Pairing Energy
According to Hund's Rule, it takes energy to pair electrons, therefore as electrons are added to an orbital, they do it in such a way that they minimize total energy; this causes the 2s orbital to be filled before the 2p orbital. When an electron can singly occupy a given orbital, in a paramagnetic state, that configuration results in high spin energy. However, when two electrons are forced to occupy the same orbital, they experience a interelectronic repulsion effect on each other which in turn increases the total energy of the orbital. The greater this repulsion effect, the greater the energy of the orbital. To calculate this repulsion effect Jorgensen and Slater founded that for any transition metal on the basis of first order perturbation theory can be solved by;
$E(S) = E(qd^n) + \left [S(S+1)- S(S+1) \right ] D$
where $E(qd^n)$ is the weighted mean energy of the configuration, $S$ is the spin quantum number, $S(S+1)$ is the average value of the total spin angular momentum and $D$ is the metal parameter. Therefore,
$\Delta E = E(S-1)-E(S) = 2SD$
It is also a general theory that spin pairing energy in the form of repulsion increase from P to D to S orbitals. An example of an element that does not follow this suit is Carbon, whose spin pairing energy increases in the opposite direction (S to D to P). For a clearer picture of how this formula works with the elements in the periodic table, see the attached table. C'dd refers to the repulsion associated with the 3dq occupation only, C', to the intracore repulsion, and C',d to the intershell repulsion between core and d electrons. For comparison, the first column shows D = E/2S, calculated from the frozen orbitals of the configuration average.
Spin Pairing Energy
In general:
• Electron spin pairing energy transition from ↑↑ (in two orbitals) to ↑↓ (in one orbital) is characterized by a decrease of the electronic repulsion.
• Atomic radii for transition metals decrease from left to right because added d electrons do not shield each other very well from the increasing nuclear charge (↑ $Z_{eff}$). Atomic radii toward the end of a period increase due to the greater electron-electron repulsion (↑ shielding) as electrons are added to occupied orbitals.
• Very closely associated with crystal field theory (repulsion between electrons of the ligands and the central metal ion) and bonding in complex ions such as octahedral, square-planar, and tetrahedral.
In the example above, the electrons can fill the d-orbitals in two different ways. The electrons can fill lower energy orbitals and pair with an existing electron there resulting in more stability (example on the right). Pairing energy is needed in order to force an electron to fill an orbital that is already occupied with an electron. The electrons can also fill higher energy orbitals and avoid the pairing energy (example on the left). This requires energy and reduces stability.
To figure out whether the electrons pair up or go into higher energy orbital depends on the crystal field splitting energy ($\Delta$). If the crystal field splitting energy (Δ) is greater than pairing energy, then greater stability would be obtained if the fourth and fifth electrons get paired with the ones in the lower level. If the crystal field splitting energy ($\Delta$) is less than the pairing energy, greater stability is obtained by keeping the electrons unpaired.
In this configuration, it is evident from previous information that the configuration on the left has a higher electronic pair spin than the configuration on the right due to the differing field splitting energy and max number of unpaired electrons. Weak ligands, such as $H_2$O and $F^-$, produce small crystal field splitting resulting in high-spin complexes and strongly paramagnetic. There are only two unpaired electrons in the configuration on the right, which is minimum amount of electrons known as low spin. Strong ligands, such as $NH_3$ and $CN^-$, produce large crystal field splitting, leading to low-spin complexes and weakly paramagnetic or sometimes even diamagnetic.
Table $1$: Example spin pairing energies for some first-row $d^4$, $d^5$, $d^6$, and $d^7$ complexes:
$d^n$ Ion P, kJ/mol (cm-1)
$d^4$
$Cr^{+2}$
$Mn^{+3}$
244.3 (20,425)
301.6 (25,215)
$d^5$
$Mn^{+2}$
$Fe^{+3}$
285.0 (23,825)
357.4 (29,875)
$d^6$ $Fe^{+2}$
$Co^{+3}$
229.1 (19,150)
282.6 (23,625)
$d^7$ $Co^{+2}$ 250 (20,800)
Reference
1. Wachters, A. J. H.; Nieuwpoort, W. C. Phys. Rev. B Solid State 1972, 5, 429 1.
2. Jorgensen, C. K. “Modern Aspects of Ligand Field Theory”; Elsevier: Amsterdam, New York, 1971.\
3. "The Pairing Energy of Co(III) + Co-ordination Chemistry." Chemical Forums (2007). Inorganic Chemistry Forum. Web. 24 May 2010.
4. Petrucci, Ralph H. General Chemistry Principles & Modern Applications, Tenth Edition. Upper Saddle River. New Jersey 2011.
5. pubs.acs.org/doi/abs/10.1021/ic00136a064
Problems
1. What state, paramagnetic or diamagnetic has a higher spin pairing repulsion?
2. What is the electron configuration of Mn?
3. What is the spin pairing configuration of Mn?
4. How does the spin pairing configuration of Mn affect the spin energy?
5. What is the electron spin energy differ between these two complexes:
A) [Fe(NO2)6]3−
B) [FeBr6]3−
Answers
1. Paramagnetic
2. [Ar] 3d3
3. Paramagnetic
4. Because Mn has a paramagnetic configuration, its spin energy is high
5. A) NO has a high crystal field splitting energy therefore causing the electrons to be forced together in lower state energy orbitals making most of them diamagnetic. This configuration causes this complex to have low spin energy.
B) Br has a very small crystal field splitting energy, causing the electrons to disperse among the orbitals freely. This allows a paramagnetic state, causing this complex to have high spin energy.
Spin Quantum Number
The Spin Quantum Number (\(m_s\)) describes the angular momentum of an electron. An electron spins around an axis and has both angular momentum and orbital angular momentum. Because angular momentum is a vector, the Spin Quantum Number (s) has both a magnitude (1/2) and direction (+ or -).
Each orbital can only hold two electrons. One electron will have a +1/2 spin and the other will have a -1/2 spin. Electrons like to fill orbitals before they start to pair up. Therefore the first electron in an orbital will have a spin of +1/2. After all the orbitals are half filled, the electrons start to pair up. This second electron in the orbital will have a spin of -1/2. If there are two electrons in the same orbital, it will spin in opposite directions.
Combinations of Quantum Numbers
• The three quantum numbers (n, l, and m) that describe an orbital are integers: 0, 1, 2, 3.
• The principal quantum number (n) cannot be zero. The allowed values of n are therefore 1, 2, 3, 4...
• The angular quantum number (l) can be any integer between 0 and n - 1.
• If n = 3, l can be either 0, 1, or 2.
• The magnetic quantum number (m) can be any integer between -l and +l.
• If l = 2, m can be -2, -1, 0, +1, or +2.
• Orbitals that have same value of principal quantum number form a Shell(n).
• Orbitals within the shells are divided into subshell (l)
• s:l = 0 p:l = 1 d:l = 2 f:l = 3
Exercise \(1\): Tungsten
What is the spin quantum number for Tungsten (symbol W)?
Answer
Tungsten has 4 electrons in the 5d orbital. Therefore 1 electron will go into each orbital (no pairing). The 4th electron will have a +1/2 spin.
Exercise \(2\): Gold
What is the spin quantum number for Gold (symbol Au)?
Answer
Gold has 9 electrons in the 5d orbital. Therefore the electrons will start to pair up, which means the 9th electron will pair up, giving it a -1/2 spin.
Exercise \(3\): Sulfur
What is the spin quantum number for Sulfur (symbol S)?
Answer
Sulfur has 4 electrons in the 3p orbitals. The 4th electron in this orbital will be the first one to pair up with another electron, therefore giving it a -1/2 spin.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Electronic_Structure_of_Atoms_and_Molecules/Electronic_Configurations/Spin_Pairing_Energy.txt
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The Aufbau model lets us take an atom and make predictions about its properties. All we need to know is how many protons it has (and how many electrons, which is the same as the number of protons for a neutral atom). We can predict the properties of the atom based on our vague idea of where its electrons are and, more importantly, the energy of those electrons.
How electrons fill in their positions around an atom is called the Aufbau Process (German: "building-up" process). The Aufbau Process is all about keeping electrons at their lowest possible energy and is the direct result of the Pauli Exclusion Principle. A corollary of Coulomb's law is that the energy of an electron is affected by attractive and repulsive forces. The closer an electron is to the nucleus, the lower its energy. The closer an electron is to another electron, the higher its energy.
Of course, a basic principle of thermodynamics is that a system will proceed to the lowest energy possible. That means, if an atom has only one electron, the electron will have quantum numbers that place it at the lowest possible energy. It will be as close as possible to the positive nucleus.
If an atom has a second electron, it will also be as close as possible to the nucleus. It could have the same quantum numbers as the first electron, except for spin. There is a trade-off, of course, because those two electrons will be close enough to repel each other. However, if it is a choice between that and taking a position much farther from the nucleus, the second electron will go ahead and pair up. These two electrons are often described as being "in the same orbital" and share their first three quantum numbers, so they are found somewhere in the same region of space. This first orbital, which has no directional restrictions, is called the 1s orbital.
There is only room for one orbital at this distance from the nucleus. A third electron has to occupy another orbital farther away, the 2s orbital. Again, this is a spherical orbital: the electron can be found in any direction. The 2 in 2s means the Principal Quantum Number is two (corresponding to the second general energy level). The s is a code for other quantum numbers; it means the electron can be found in any direction, just like the 1s electrons.
The second energy level is large enough to accommodate additional orbitals, but they are a little farther from the nucleus. These are called the 2p orbitals. They are regions of space along the x, y and z axes. There are three orbitals of this type, and they are just called px, py and pz to remind us that they are orthogonal to each other.
Expressed in a different way, an electron with Principal Quantum Number 2 can have four different combinations of its other quantum numbers. These combinations are denoted 2s, 2px, 2py and 2pz. The three 2p combinations are a little higher in energy than the 2s orbital.
A third electron will go into the 2s orbital. It is the lowest in energy. What about a fourth? Does it go into the 2s or a 2p? Once again the pairing energy is not quite as big as the energy jump up to the 2p orbital. The fourth electron pairs up in the 2s orbital.
A fifth electron goes into one of the 2p orbitals. It does not matter which one. We will say it is the px, arbitrarily. A sixth electron again could either pair up in the px, or it could go into the py. But the py level is really the same as the px, just in a different direction. The energy is the same. That means a sixth electron will go into the py rather than pair up in the px, where it would experience extra repulsion.
Note that the p orbitals are often drawn a little differently. For example, p orbitals are usually drawn in a way that shows that they have phase. Either the two lobes are colored differently to show that they are out of phase with each other, or they are shown with one lobe shaded and the other left blank.
This pattern is generally followed for all of the elements. The tally of how many electrons are found in each orbital is called the electron configuration. For example, hydrogen has only one electron. Its ground state configuration (that means, assuming the electron has not been excited to another orbital via addition of energy) is 1s1.
On the other hand, an atom with six electrons, such as the element carbon, has the configuration 1s22s22px12py1.
Periodicity
You may already know that electron configuration is the reason the periodic table works the way it does. Mendelev and others noticed certain elements had very similar properties, and that is because they have very similar electron configurations. Lithium has configuration 1s22s1 and its alkali sister, sodium, has configuration 1s22s22px22py22pz23s1. In both cases, the last electron added is an unpaired s electron. The last electron, or last few electrons, added to an atom generally play a strong role in how the atom behaves. This "frontier" electron is the one at the outer limits of the atom. If the atom is to interact with anything, the frontier electron will encounter the thing first. In contrast, the "core" electrons closer to the nucleus are more protected from the outside.
The order of electrons in an atom, from lowest to highest energy, is:
• 1s
• 2s
• 2p
There are some shortcuts we take with electron configurations. We tend to abbreviate "filled shells" (meaning all the possible orbitals with a given Principal Quantum Number are filled with electrons) and filled "sub-shells" (like 2s or 2p). First of all, in the case of p orbitals, if all the p orbitals are filled, we might just write 2p6 instead of 2px22py22pz2, because there is only one way all the orbitals could be filled. However, we would not necessarily write 2p2 instead of 2px12py1, because we may wish to make clear that the configuration does not involve two electrons in one p orbital at that point, as in 2px22py0.
Also, we dispense with orbital labels entirely to ignore core electrons in a filled shell. For example, instead of writing 1s22s1 for lithium, we can write [He]2s1. Instead of writing 1s22s22px22py22pz23s1 for sodium, we write [Ne]3s1. The [He] means all the electrons found in a helium atom, which is a noble gas. The [Ne] means all the electrons found in a neon atom. A noble gas is an unreactive element with a filled shell: helium, neon, argon, krypton, xenon or radon. The electrons beyond the noble gas shell are called the valence electrons.
Principal Quantum Number 3 actually allows a third set of orbitals. These are called the d orbitals that are a little like p orbitals, but they are two-dimensional rather than one-dimensional. A d electron, for example, might extend along the x axis and the y axis, but not in between the axes.
The d orbitals have five allowed orientations with:
• dxy extending along the x and y axes,
• dxz extending along the x and z axes,
• dyz extending along the y and z axes.
The other two orbitals extend between the axes instead, dx2-y2 rotated 45 degrees away from one of the other d orbitals and extends between the x and y axes. In the same way, you could imagine an orbital between the x and z and between the y and z axes, but that would make six different orientations. Quantum mechanical rules do not allow that. As a result, two of the possible combinations collapse into a mathematical sum, making just one orbital. We call this one the dz2 orbital.
However, just as the 2p orbitals are higher in energy than the 2s orbitals, the 3d orbitals are higher in energy than the 3p orbitals. The 3d level is very similar in energy to the 4s level. For that reason, calcium's last electrons go into a 4s orbital, not a 3d orbital. Calcium behaves much like magnesium as a result.
The order of electrons in an atom, from lowest to highest energy, is:
• 1s
• 2s
• 2p
• 3s
• 3p
• 4s
• 3d
• 4p
Reality is Less Simple
The atomic orbital energies do not properly follow this trend as shown pictorially below. For example, the relative energies of the 3d and 4s orbitals switch between calcium and scandium.
Exercise \(1\)
Write electron configurations for the following elements.
1. oxygen, O
2. sulfur, S
3. silicon, Si
4. nitrogen, N
5. argon, Ar
6. neon, Ne
Exercise \(2\)
Write abbreviated electron configurations for the following elements.
1. chlorine, Cl
2. calcium, Ca
3. aluminum, Al
4. phosphorus, P
Exercise \(3\)
Write abbreviated electron configurations for the following elements.
1. iron, Fe
2. copper, Cu
3. mercury, Hg
4. lead, Pb
5. arsenic, As
6. titanium, Ti
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Electronic_Structure_of_Atoms_and_Molecules/Electronic_Configurations/The_Aufbau_Process.txt
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The octet rule refers to the tendency of atoms to prefer to have eight electrons in the valence shell. When atoms have fewer than eight electrons, they tend to react and form more stable compounds. When discussing the octet rule, we do not consider d or f electrons. Only the s and p electrons are involved in the octet rule, making it useful for the main group elements (elements not in the transition metal or inner-transition metal blocks); an octet in these atoms corresponds to an electron configurations ending with \(s^2p^6\).
Introduction
In 1904, Richard Abegg formulated what is now known as Abegg's rule, which states that the difference between the maximum positive and negative valences of an element is frequently eight. This rule was used later in 1916 when Gilbert N. Lewis formulated the "octet rule" in his cubical atom theory. Atoms will react to get in the most stable state possible. A complete octet is very stable because all orbitals will be full. Atoms with greater stability have less energy, so a reaction that increases the stability of the atoms will release energy in the form of heat or light.
Octet Rule
A stable arrangement is attended when the atom is surrounded by eight electrons. This octet can be made up by own electrons and some electrons which are shared. Thus, an atom continues to form bonds until an octet of electrons is made.
1. Normally two electrons pairs up and forms a bond, e.g., \(H_2\)
2. For most atoms there will be a maximum of eight electrons in the valence shell (octet structure), e.g., \(CH_4\)
The other tendency of atoms is to maintain a neutral charge. Only the noble gases (the elements on the right-most column of the periodic table) have zero charge with filled valence octets. All of the other elements have a charge when they have eight electrons all to themselves. The result of these two guiding principles is the explanation for much of the reactivity and bonding that is observed within atoms: atoms seek to share electrons in a way that minimizes charge while fulfilling an octet in the valence shell.
Note
The noble gases rarely form compounds. They have the most stable configuration (full octet, no charge), so they have no reason to react and change their configuration. All other elements attempt to gain, lose, or share electrons to achieve a noble gas configuration.
Example 1: NaCl Salt
The formula for table salt is NaCl. It is the result of Na+ ions and Cl- ions bonding together. If sodium metal and chlorine gas mix under the right conditions, they will form salt. The sodium loses an electron, and the chlorine gains that electron. In the process, a great amount of light and heat is released. The resulting salt is mostly unreactive — it is stable. It will not undergo any explosive reactions, unlike the sodium and chlorine that it is made of. Why?
Solution
Referring to the octet rule, atoms attempt to get a noble gas electron configuration, which is eight valence electrons. Sodium has one valence electron, so giving it up would result in the same electron configuration as neon. Chlorine has seven valence electrons, so if it takes one it will have eight (an octet). Chlorine has the electron configuration of argon when it gains an electron.
The octet rule could have been satisfied if chlorine gave up all seven of its valence electrons and sodium took them. In that case, both would have the electron configurations of noble gasses, with a full valence shell. However, their charges would be much higher. It would be Na7- and Cl7+, which is much less stable than Na+ and Cl-. Atoms are more stable when they have no charge, or a small charge.
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This page looks at some of the problems with the usual way of explaining the electronic structures of the d-block elements based on the order of filling of the d and s orbitals. The way that the order of filling of orbitals is normally taught gives you an easy way of working out the electronic structures of elements. However, it does throw up problems when you come to explain various properties of the transition elements. This page takes a closer look at this, and offers a more accurate explanation which avoids the problems.
The Order of Filling Orbitals
The aufbau principle explains how electrons fill low energy orbitals (closer to the nucleus) before they fill higher energy ones. Where there is a choice between orbitals of equal energy, they fill the orbitals singly as far as possible (Hunds rules). The diagram (not to scale) summarizes the energies of the orbitals up to the 4p level.
The oddity is the position of the 3d orbitals, which are shown at a slightly higher level than the 4s. This means that the 4s orbital which will fill first, followed by all the 3d orbitals and then the 4p orbitals. Similar confusion occurs at higher levels, with so much overlap between the energy levels that the 4f orbitals do not fill until after the 6s, for example.
Everything is straightforward up to this point, but the 3-level orbitals are not all full - the 3d levels have not been used yet. But if you refer back to the energies of the orbitals, you will see that the next lowest energy orbital is the 4s - so that fills first.
K 1s22s22p63s23p64s1
Ca 1s22s22p63s23p64s2
d-block elements
d-block elements are thought of as elements in which the last electron to be added to the atom is in a d orbital (actually, that turns out not to be true! We will come back to that in detail later.) The electronic structures of the d-block elements are shown in the table below. Each additional electron usually goes into a 3d orbital. For convenience, [Ar] is used to represent 1s22s22p63s23p6.
Sc [Ar] 3d14s2
Ti [Ar] 3d24s2
V [Ar] 3d34s2
Cr [Ar] 3d54s1
Mn [Ar] 3d54s2
Fe [Ar] 3d6 4s2
Co [Ar] 3d74s2
Ni [Ar] 3d84s2
Cu [Ar] 3d104s1
Zn [Ar] 3d104s2
d-block ions
This is probably the most unsatisfactory thing about this approach to the electronic structures of the d-block elements. In all the chemistry of the transition elements, the 4s orbital behaves as the outermost, highest energy orbital. The reversed order of the 3d and 4s orbitals only seems to apply to building the atom up in the first place. In all other respects, the 4s electrons are always the electrons you need to think about first.
When d-block (first row) elements form ions, the 4s electrons are lost first.
Example \(1\): Iron
Consider the electronic structure of neutral iron and iron (III). To write the electronic structure for Fe3+:
• Fe: 1s22s22p63s23p63d64s2
• Fe3+: 1s22s22p63s23p63d5
The 4s electrons are lost first followed by one of the 3d electrons. This last bit about the formation of the ions is clearly unsatisfactory.
1. We say that the 4s orbitals have a lower energy than the 3d, and so the 4s orbitals are filled first.
2. We know that the 4s electrons are lost first during ionization. The electrons lost first will come from the highest energy level, furthest from the influence of the nucleus. So the 4s orbital must have a higher energy than the 3d orbitals.
Those statements are directly opposed to each other and cannot both be right.
When discussing ionization energies for these elements, you talk in terms of the 4s electrons as the outer electrons being shielded from the nucleus by the inner 3d levels. We say that the first ionization energies do not change much across the transition series, because each additional 3d electron more or less screens the 4s electrons from the extra proton in the nucleus.
The explanations around ionization energies are based on the 4s electrons having the higher energy, and so being removed first.
Where is the flaw in the logic?
The usual way of teaching this is an easy way of working out what the electronic structure of any atom is - with a few odd cases to learn like chromium or copper. The problems arise when you try to take it too literally. It is way of working out structures - no more than that. The flaw lies in the diagram we started with (Figure 1) and assuming that it applies to all atoms. In other words, we assume that the energies of the various levels are always going to be those we draw in this diagram. If you stop and think about it, that has got to be wrong.
As you move from element to element across the Periodic Table, protons are added to the nucleus and electrons surrounding the nucleus. The various attractions and repulsions in the atoms are bound to change as you do this - and it is those attractions and repulsions which govern the energies of the various orbitals. That means that student must rethink this on the basis that what we drew above is not likely to look the same for all elements.
The Solution
• The elements up to argon: There is no problem with these. The general pattern that we drew in the diagram above works well.
• Potassium and calcium: The pattern is still working here. The 4s orbital has a lower energy than the 3d, and so fills next. That entirely fits with the chemistry of potassium and calcium.
• The d-block elements: For reasons which are too complicated to go into at this level, once you get to scandium, the energy of the 3d orbitals becomes slightly less than that of the 4s, and that remains true across the rest of the transition series (hence, Figure 1 is incorrect as drawn).
So rather than working out the electronic structure of scandium by imagining that you just throw another electron into a calcium atom, with the electron going into a 3d orbital because the 4s is already full, you really need to look more carefully at it.
Remember that, in reality, for Sc through to Zn the 3d orbitals have the lower energy - not the 4s.
Example \(2\): Scandium
So why is not the electronic configuration of scandium [Ar] 3d3 rather than [Ar] 3d14s2?
Solution
Making Sc3+
Imagine you are building a scandium atom from boxes of protons, neutrons and electrons. You have built the nucleus from 21 protons and 24 neutrons, and are now adding electrons around the outside. So far you have added 18 electrons to fill all the levels up as far as 3p. Essentially you have made the ion Sc3+.
Making Sc2+
Now you are going to add the next electron to make Sc2+. Where will the electron go? The 3d orbitals at scandium have a lower energy than the 4s, and so the next electron will go into a 3d orbital. The structure is [Ar] 3d1.
Making Sc+
You might expect the next electron to go into a lower energy 3d orbital as well, to give [Ar] 3d2. But it doesn't. You have something else to think about here as well. If you add another electron to any atom, you are bound to increase the amount of repulsion. Repulsion raises the energy of the system, making it less energetically stable. It obviously helps if this effect can be kept to a minimum.
The 3d orbitals are quite compactly arranged around the nucleus. Introducing a second electron into a 3d orbital produces more repulsion than if the next electron went into the 4s orbital. There is not a very big gap between the energies of the 3d and 4s orbitals. The reduction in repulsion more than compensates for the energy needed to do this.
The energetically most stable structure for Sc+ is therefore [Ar] 3d14s1.
Making Sc:
Putting the final electron in, to make a neutral scandium atom, needs the same sort of discussion. In this case, the lowest energy solution is the one where the last electron also goes into the 4s level, to give the familiar [Ar] 3d14s2 structure.
Summary
In each of these cases we have looked at, the 3d orbitals have the lowest energy, but as we add electrons, repulsion can push some of them out into the higher energy 4s level.
• If you build up the scandium atom from scratch, the last electrons to go in are the two 4s electrons. These are the electrons in the highest energy level, and so it is logical that they will be removed first when the scandium forms ions. And that's what happens.
• The 4s electrons are also clearly the outermost electrons, and so will largely define the radius of the atom. The lower energy 3d orbitals are inside them, and will contribute to the screening. There is no longer any conflict between these properties and the order of orbital filling.
The difficulty with this approach is that you cannot use it to predict the structures of the rest of the elements in the transition series. In fact, what you have to do is to look at the actual electronic structure of a particular element and its ions, and then work out what must be happening in terms of the energy gap between the 3d and 4s orbitals and the repulsions between the electrons.
The common way of teaching this (based on the wrong order of filling of the 3d and 4s orbitals for transition metals) gives a method which lets you predict the electronic structure of an atom correctly most of the time. The better way of looking at it from a theoretical point of view no longer lets you do that. You can get around this, of course. If you want to work out a structure, use the old method. But remember that it is based on a false idea, and do not try to use it for anything else - like working out which electrons will be lost first from a transition element, for example.
Thinking about the other elements in the series in the same way as we did with scandium, in each case the 3d orbitals will take the first electron(s). Then at some point repulsion will push the next ones into the 4s orbital. When this happens varies from element to element.
Example \(3\): Vanadium
Vanadium has two more electrons than scandium, and two more protons as well, of course. Think about building up a vanadium atom in exactly the same way that we did scandium. We have the nucleus complete and now we are adding electrons. When we have added 18 electrons to give the argon structure, we have then built a V5+ ion.
Now look at what happens when you add the next 5 electrons.
V5+ [Ar]
V4+ [Ar]3d1
V3+ [Ar]3d2
V2+ [Ar]3d3
V+ [Ar]3d4
V [Ar]3d34s2
The energy gap between the 3d and 4s levels has widened. In this case, it is not energetically profitable to promote any electrons to the 4s level until the very end. In the ions, all the electrons have gone into the 3d orbitals. You couldn't predict this just by looking at it.
Example \(4\): Chromium
Why is the electronic structure of chromium [Ar]3d54s1 instead of [Ar]3d44s2?
Solution
Because that is the structure in which the balance of repulsions and the size of the energy gap between the 3d and 4s orbitals happens to produce the lowest energy for the system.
Many chemistry textbooks and teachers try to explain this by saying that the half-filled orbitals minimize repulsions, but that is a flawed, incomplete argument. You are not taking into account the size of the energy gap between the lower energy 3d orbitals and the higher energy 4s orbital.
Two rows directly underneath chromium in the Periodic Table is tungsten. Tungsten has exactly the same number of outer electrons as chromium, but its outer structure is 5d46s2, NOT 5d56s1.
In this case, the most energetically stable structure is not the one where the orbitals are half-full. You cannot make generalizations like this!
Conclusion
• The current method of teaching students to work out electronic structures is fine as long as you realize that that is all it is - a way of working out the overall electronic structures, but not the order of filling.
• You can say that for potassium and calcium, the 3d orbitals have a higher energy than the 4s, and so for these elements, the 4s levels fill before than the 3d. That, of course, is entirely true!
• Then you can say that, looking at the structures of the next 10 elements of the transition series, the 3d orbitals gradually fill with electrons (with some complications like chromium and copper). That is also true.
• What is not right is to imply that the 3d levels across these 10 elements have higher energies than the 4s. That is definitely not true, and causes the sort of problems we have been discussing.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Electronic_Structure_of_Atoms_and_Molecules/Electronic_Configurations/The_Order_of_Filling_3d_and_4s_Orbitals.txt
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To be able to use Crystal Field Theory (CFT) successfully, it is essential to first determine the electronic configuration of the central metal ion in any complex. This requires being able to recognize all the entities making up the complex and knowing whether the ligands are neutral or anionic, so that you can determine the oxidation number of the metal ion.
In many cases the oxidation number for first row transition metal ions will be either (II) or (III), but in any case you may find it easier to start with the M(II) from which you can easily add or subtract electrons to get the final electronic configuration.
Contributors and Attributions
The Department of Chemistry, University of the West Indies)
Evaluating Spin Multiplicity
Spin-multiplicity value and its corresponding spin state was first discovered by Friedrich Hund in 1925. The formula which is generally used for the prediction of spin multiplicity value is $(2S+1)$, where
$S = \underset{\text{spin quantum #}}{\sum m_s} \label{eq1}$
is time consuming. To keep the matter in mind a simple innovative method1,2,3 has to be introduced for calculation of spin-multiplicity value and thus its corresponding spin state, shown in Table-1, in the easiest way by ignoring the calculation of total spin quantum number (Equation \ref{eq1}). First of all we should classify the species (atoms, molecules, ions or complexes) for which spin multiplicity value should be evaluated into three types based on the nature of alignment of unpaired electrons present in them.
Species having unpaired electrons in upward alignment (↑)
In this case, spin multiplicity = (n+1); where n = number of unpaired electrons
Spin multiplicity = (n +1) = (1+1) = 2 (spin state = doublet); (2+1) = 3 (spin state = triplet) and (3 + 1) = 4 (spin state = quartet) respectively.
↑↓
↑↓
↑↓
Spin multiplicity = (n +1) = (2 + 1) = 3 (in this case ignore paired electrons) (spin state = triplet) and (1 + 1) = 2 (spin state = doublet)
↑↓
↑↓
↑↓
Spin multiplicity = (n +1) = (0 + 1) = 1 (spin state = singlet)
Species having unpaired electrons in downward alignment (↓)
In this case spin multiplicity = (-n+1). Here (-ve) sign indicate downward arrow.
Spin multiplicity = (-n +1) = (-1 + 1) = 0; (-2 + 1) = -1 and (-3 + 1) = -2 respectively.
↑↓
↑↓
↑↓
Spin multiplicity = (-n + 1) = (-2 + 1) = -1( ignore paired electrons) and (-1 + 1) = 0 respectively.
Species having unpaired electrons in both mixed alignment (↑)(↓)
In this case spin multiplicity = [(+n) + (-n) + 1], where, n = number of unpaired electrons in each alignment. Here, (+ve) sign and (–ve) sign indicate upward and downward alignment respectively.
Here total no of unpaired electrons = 2 in which one having upward direction (+1) and other having downward mode (-1).
Hence Spin multiplicity = [(+n) + (-n) +1] = [(+1) + (-1) + 1] = 1 (spin state = singlet)
Here the total no of unpaired electrons = 3 in which two unpaired electrons lie in upward (+2) and one unpaired electrons lie in downward (-1) . Hence Spin multiplicity = [(+n) + (-n) + 1] = [(+2) + (-1) + 1] = 2 (spin state = doublet)
Here the total no of unpaired electrons = 5 in which three unpaired electrons lie upward (+3) and two unpaired electrons lie downward (-2) .Hence Spin multiplicity = [(+n) + (-n) + 1] = [(+3) + (-2) +1] = 2 (spin state = doublet)
Table 1:Spin multiplicity value and its corresponding spin state
Spin multiplicity value
Spin state
1
Singlet
2
Doublet
3
Triplet
4
Quartet
5
Quintet
External Links
• communities.acs.org/docs/DOC-46667
• communities.acs.org/docs/<wbr/>DOC-45853
• www.drarijitdaschem.in/Spin%2...13%20issue.pdf
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Electronic_Structure_of_Atoms_and_Molecules/Electronic_Configurations/d-orbital_Occupation_and_Electronic_Configura.txt
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A hydrogen discharge tube is a slim tube containing hydrogen gas at low pressure with an electrode at each end. If a high voltage (5000 volts) is applied, the tube lights up with a bright pink glow. If the light is passed through a prism or diffraction grating, it is split into its various colors. This is a small part of the hydrogen emission spectrum. Most of the spectrum is invisible to the eye because it is either in the infrared or the ultraviolet region of the electromagnetic spectrum.
The photograph shows part of a hydrogen discharge tube on the left, and the three most apparent lines in the visible part of the spectrum on the right. (Ignore the "smearing," particularly to the left of the red line. This is caused by flaws in the way the photograph was taken. See note below.)
This photograph is by courtesy of Dr Rod Nave of the Department of Physics and Astronomy at Georgia State University, Atlanta.
Extending hydrogen's emission spectrum into the UV and IR
The hydrogen spectrum is complex, comprising more than the three lines visible to the naked eye. It is possible to detect patterns of lines in both the ultraviolet and infrared regions of the spectrum as well. These fall into a number of "series" of lines named after the person who discovered them. The diagram below shows three of these series, but there are others in the infrared to the left of the Paschen series shown in the diagram.
The diagram is quite complicated. Consider first at the Lyman series on the right of the diagram; this is the broadest series, and the easiest to decipher.
The frequency scale is marked in PHz—petaHertz. Peta means "1015 times". The value 3 PHz is equal to 3 × 1015 Hz. The quantity "hertz" indicates "cycles per second".
The Lyman series is a series of lines in the ultraviolet region. The lines grow closer and closer together as the frequency increases. Eventually, they are so close together that it becomes impossible to see them as anything other than a continuous spectrum. This is suggested by the shaded part on the right end of the series. At one particular point, known as the series limit, the series ends.
In Balmer series or the Paschen series, the pattern is the same, but the series are more compact. In the Balmer series, notice the position of the three visible lines from the photograph further up the page.
Frequency and Wavelength
The hydrogen spectrum is often drawn using wavelengths of light rather than frequencies. Unfortunately, because of the mathematical relationship between the frequency of light and its wavelength, two completely different views of the spectrum are obtained when it is plotted against frequency or against wavelength. The mathematical relationship between frequency and wavelength is the following:
Rearranging this gives equations for either wavelength or frequency:
$\lambda =\dfrac{c}{\nu}$
or
$\nu=\dfrac{c}{\lambda }$
There is an inverse relationship between the two variables—a high frequency means a low wavelength and vice versa.
Drawing the hydrogen spectrum in terms of wavelength
This is what the spectrum looks like plotted in terms of wavelength instead of frequency:
Compare this to the same spectrum in terms of frequency:
When juxtaposed, the two plots form a confusing picture. The remainder of the article employs the spectrum plotted against frequency, because in this spectrum it is much easier visualize what is occurring in the atom.
The Balmer and Rydberg Equations
In an amazing demonstration of mathematical insight, in 1885 Balmer came up with a simple formula for predicting the wavelength of any of the lines in what we now know as the Balmer series. Three years later, Rydberg generalized this so that it was possible to determine the wavelengths of any of the lines in the hydrogen emission spectrum. Rydberg's equation is as follows:
$\dfrac{1}{\lambda}=R_H \left( \dfrac{1}{n^2_1}-\dfrac{1}{n^2_2}\right)$
where
• $R_H$ is the Rydberg constant.
• $n_1$ and $n_2$ are integers (whole numbers). $n_2$ is always greater than $n_1$. In other words, if $n_1$ is, say, 2 then $n_2$ can be any whole number between 3 and infinity.
The various combinations of numbers that can be substituted into this formula allow the calculation the wavelength of any of the lines in the hydrogen emission spectrum; there is close agreement between the wavelengths generated by this formula and those observed in a real spectrum.
A modified version of the Rydberg equation can be used to calculate the frequency of each of the lines:
The origin of the hydrogen emission spectrum
The lines in the hydrogen emission spectrum form regular patterns and can be represented by a (relatively) simple equation. Each line can be calculated from a combination of simple whole numbers. Why does hydrogen emit light when excited by a high voltage and what is the significance of those whole numbers?
When unexcited, hydrogen's electron is in the first energy level—the level closest to the nucleus. But if energy is supplied to the atom, the electron is excited into a higher energy level, or even removed from the atom altogether. The high voltage in a discharge tube provides that energy. Hydrogen molecules are first broken up into hydrogen atoms (hence the atomic hydrogen emission spectrum) and electrons are then promoted into higher energy levels.
Suppose a particular electron is excited into the third energy level. It would tend to lose energy again by falling back down to a lower level. It can do this in two different ways. It could fall all the way back down to the first level again, or it could fall back to the second level and then, in a second jump, down to the first level.
Assigning particular electron jumps to individual lines in the spectrum
If an electron falls from the 3-level to the 2-level, it must lose an amount of energy exactly equal to the energy difference between those two levels. That energy which the electron loses is emitted as light (which "light" includes UV and IR as well as visible radiation). Each frequency of light is associated with a particular energy by the equation:
The higher the frequency, the higher the energy of the light. If an electron falls from the 3-level to the 2-level, red light is seen. This is the origin of the red line in the hydrogen spectrum. From the frequency of the red light, its energy can be calculated. That energy must be exactly the same as the energy gap between the 3-level and the 2-level in the hydrogen atom.
The last equation can therefore be rewritten as a measure of the energy gap between two electron levels:
The greatest possible fall in energy will therefore produce the highest frequency line in the spectrum. The greatest fall will be from the infinity level to the 1-level. (The significance of the infinity level will be made clear later.)
The next few diagrams are in two parts, with the energy levels at the top and the spectrum at the bottom.
If an electron falls from the 6-level, the difference is slightly less than before, and so the frequency is slightly lower (because of the scale of the diagram, it is impossible to depict the levels beyond 7).
All other possible jumps to the first level make up the whole Lyman series. The spacings between the lines in the spectrum reflect the changes in spacings between the energy levels.
If the same is done for the 2-level, the Balmer series is shown. These energy gaps are all much smaller than in the Lyman series, and so the frequencies produced are also much lower.
The Paschen series is made up of the transitions to the 3-level, but they are omitted to avoid cluttering the diagram.
The significance of the numbers in the Rydberg equation
In the Rydberg equation, n1 and n2 represent the energy levels at either end of the jump that produces a particular line in the spectrum.
• In the Lyman series, $n_1 =1$, because electrons transition to the 1-level to produce lines in the Lyman series.
• In the Balmer series, $n_1 =2$, because electrons fall to the 2-level.
n2 is the level being jumped from. In the case before, in which a red line is produced by electrons falling from the 3-level to the 2-level, n2 is equal to 3.
The significance of the infinity level
The infinity level represents the highest possible energy an electron can have as a part of a hydrogen atom. If the electron exceeds that energy, it is no longer a part of the atom. The infinity level represents the point at which ionization of the atom occurs to form a positively charged ion.
Using the spectrum to find the ionization energy of hydrogen
When there is no additional energy supplied to it, hydrogen atom's electron is found at the 1-level. This is known as its ground state. If enough energy is supplied to move the electron up to the infinity level, the atom is ionized. The ionization energy per electron is therefore a measure of the difference in energy between the 1-level and the infinity level. In above diagrams, that particular energy jump produces the series limit of the Lyman series.
The frequency of the Lyman series limit can be used to calculate the energy required to promote the electron in one atom from the 1-level to the point of ionization. This energy can then be used to calculate the ionization energy per mole of atoms. A problem with this approach is that the frequency of a series limit is quite difficult to find accurately from a spectrum because the lines are so close together in that region that the spectrum looks continuous.
Finding the frequency of the series limit graphically
The following is a list of the frequencies of the seven most widely spaced lines in the Lyman series, together with the increase in frequency between successive lines.
As the lines become closer together, the increase in frequency is lessened. At the series limit, the gap between the lines is zero. Consequently, if the increase in frequency is plotted against the actual frequency, the curve can be extrapolated to the point at which the increase becomes zero, the frequency of the series limit.
In fact, two graphs can be plotted from the data in the table above. The frequency difference is related to two frequencies. For example, the figure of 0.457 is found by subtracting 2.467 from 2.924. Which of the two values should be plotted against 0.457 does not matter, as long as consistency is maintained—the difference must always be plotted against either the higher or the lower figure. At the limit, the two frequency numbers are the same.
As illustrated in the graph below, plotting both of the possible curves on the same graph makes it easier to decide exactly how to extrapolate the curves. Because these are curves, they are much more difficult to extrapolate than straight lines.
Both lines indicate a series limit at about 3.28 x 1015 Hz.
With this information, it is possible calculate the energy needed to remove a single electron from a hydrogen atom. Recall the equation above:
The energy gap between the ground state and the point at which the electron leaves the atom can be determined by substituting the frequency and looking up the value of Planck's constant from a data book.
$\begin{eqnarray} \Delta E &=& h\nu \ &=& (6.626 \times 10^{-34})(3.28 \times 10^{15}) \ &=& 2.173 \times 10^{-18}\ J \end{eqnarray}$
This is the ionization energy for a single atom. To find the normally quoted ionization energy, this value is multiplied by the number of atoms in a mole of hydrogen atoms (the Avogadro constant) and then dividing by 1000 to convert joules to kilojoules.
$\begin{eqnarray} Ionization\ energy &=& (2.173 \times 10^{-18})( 6.022 \times 10^{23})( \frac{1}{1000}) \ &=& 1310\ kJ\ mol^{-1} \end{eqnarray}$
This compares well with the normally quoted value for hydrogen's ionization energy of 1312 kJ mol-1.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Electronic_Structure_of_Atoms_and_Molecules/Hydrogen%27s_Atomic_Emission_Spectrum.txt
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Generally magnetic properties of diatomic molecules or ions whose total number electrons lie in the range (1-20) can be evaluated with the help of Molecular orbital theory (MO theory)1,2. The present study involves3-5 three (03) new formulae by just manipulating the number of unpaired electrons (n) for determination of magnetic properties without MO theory using mod function (based on Applied Mathematics) and by means of these n values one can easily stumble the magnetic moment values in Bohr-Magneton using spin only formula
$\mu_s = \sqrt{n(n+2)} \mu_B$
where
• $\mu_B$ is the Bohr Magneton (unit of magnetic moment) and
• $n$ is the number of unpaired electrons.
Classification
First of all we classify the molecules or ions depending on the total number of electrons present in them in the following three (03) sets.
• Set 1: Molecules or ions with (1-3), (3-5), (5-7), (7-10), or (13-16) electrons
• Set 2: Molecules or ions with (10-13) or (16-19) electrons
• Set 3: Molecules or ions with 20 electrons
Then for different set we have to use three different formulae to calculate the number of unpaired electrons which have been presented in Table 1 and thus magnetic moment ($\mu_s$) can be evaluated in the following way:
SET 1: Species with (1-3), (3-5), (5-7), (7-10), or (13-16) Electrons
For the prediction of number of unpaired electrons (n) of molecules or ions having total number of electrons (1-3),(3-5),(5-7),(7-10) and (13-16)electrons:
In this case, the number of unpaired electrons n = [ I (ND - total electrons) I ]
Here, ND = next digit i.e. digit next to minimum digit and ‘I I’ indicates mod function.
Eg:Molecules or ions having (1-3)electrons, in this case ND = 2 because here minimum digit is 1.
+
He2+ (3electrons), the total number of electrons will be 3, ND = 2, Hence, unpaired electron n = I (ND - total electrons) I = I (2-3) I = 1. Hence, Magnetic Moment μs = √n(n+2) $\mu_B$ = √ 1(1+2) BM = √3 BM = 1.73BM.
For the molecules or ions containing (3-5)electrons, (5-7)electrons, (7-10)electrons, and (13-16)electrons the ND value will be 4, 6, 8 and 14 respectively. Hence, the value of n = [ I (4-total electrons) I ]; [ I (6- total electrons) I ] [ I (8- total electrons) I ] and [ I (14- total electrons) I ] respectively.
SET 2: Species with (10-13) or (16-19) Electrons
For the prediction of number of unpaired electrons (n) of molecules or ions having total number of electrons (10-13) and (16-19):
In this case, the number of unpaired electrons n = [ I (PD - total electrons) I ]
Here, PD = Penultimate electron digit (i.e. before last electron).
-
The $C_2^-$ diatomic ion has 13 electrons, so PD = 12. Hence, unpaired electron n = I (12 - total electrons) I = I (12-13) I = 1
Hence, Magnetic Moment μs = √n(n+2) $\mu_B$ = √ 1(1+2) BM = √3 BM = 1.73BM
2
The $F_2$ diatomic molecules has 18 electrons, the total number of electrons will be 18, PD = 18. Hence, unpaired electron n = I (18 - total electrons) I = I (18-18) I = 0
Hence, Magnetic Moment μs = √n(n+2) $\mu_B$ = √ 0(0+2) BM = 0 BM = Diamagnetic in nature.
SET 3: Species with 20 Electrons
For the prediction of number of unpaired electrons (n) of molecules or ions having total number of electrons 20:
In this case, the number of unpaired electrons n = [ (20 - total electrons) ]
2
The $Ne_2$ diatomic molecules has 20 electrons, the total number of electrons will be 20. Hence, unpaired electron n = (20 - total electrons) = (20-20) = 0
Hence, Magnetic Moment μs = √n(n+2) $\mu_B$ = √ 0(0+2) BM = 0 BM = Diamagnetic in nature.
Table 1: Magnetic moments of homonulcear and heteronuclear diatomic species
Species (Molecules or ions)
Total Number of electrons
Number of unpaired electrons (n)
Magnetic moment (μs) in Bohr Magneton ($\mu_B$)
Magnetic Behavior
H2+
1
1
1.73
Paramagnetic
H2, He22+
2
0
0
Diamagnetic
H2-,He2+
3
1
1.73
Paramagnetic
He2,
4
0
0
Diamagnetic
Li2+,He2-
5
1
1.73
Paramagnetic
Li2, He22-, Be22+
6
0
0
Diamagnetic
Be2+,Li2-
7
1
1.73
Paramagnetic
Be2,Li22-
8
0
0
Diamagnetic
Be2-,B2+
9
1
1.73
Paramagnetic
B2, Be22-, HF
10
2
2.82
Paramagnetic
B2-,C2+
11
1
1.73
Paramagnetic
C2,B22-,N22+, CN+
12
0
0
Diamagnetic
C2-,N2+
13
1
1.73
Paramagnetic
N2,CO,NO+,C22-,CN-,O22+
14
0
0
Diamagnetic
N2-,NO,O2+
15
1
1.73
Paramagnetic
NO-,O2
16
2
2.82
Paramagnetic
O2-
17
1
1.73
Paramagnetic
F2,O22-,HCl
18
0
0
Diamagnetic
F2-
19
1
1.73
Paramagnetic
Ne2
20
0
0
Diamagnetic
External Links
• communities.acs.org/docs/DOC-46667
• communities.acs.org/docs/DOC-45853
• www.drarijitdaschem.in/Innova...Views%20in.pdf
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Electronic_Structure_of_Atoms_and_Molecules/Magnetic_Behavior_of_Diatomic_Species.txt
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Bond-order usually predicted from the Molecular Orbital Theory1,2. Molecular Orbital Theory (MOT) was first proposed by Friedrich Hund and Robert Mulliken in 1933. They developed an approach to covalent bond formation which is based upon the effects of the various electron fields upon each other and which employs molecular orbital rather than atomic orbital. Each such orbital characterizing the molecule as a whole is described by a definite combination of quantum numbers and possesses relative energy value.
In this article text based learning approaches have been highlighted by innovative and time economic way3-5 to enhance interest of students’ who belong to paranoia zone in chemical bonding. In this pedagogical survey, I have tried to hub one time economic pedagogy by including four (04) new formulae in the field of chemical education. This article explores the results and gives implications for context based teaching, learning and assessment in a time economic way.
Classification
First of all we classify the molecules or ions into the following four sets based on total number of electrons present in them.
SET 1: Molecules and ions having total no of electrons within the range (1-2):
In such case Bond order = n/2; [Where n = Total no of electrons]
Eg. H2 (Total electrons = 2), Therefore B.O. = n/2 = 2/2 = 1
SET 2: Molecules and ions having total no of electrons within the range (2-6):
In such case Bond order = I 4- n I / 2 ;
where n = Total no of electrons, ‘I I’ indicates Mod function i.e. the value of bond order is always positive]
Eg. Li2+(5electrons) Therefore B.O. = I 4-5 I / 2 = 1/2 = 0.5.
SET 3: Molecules and ions having total no of electrons within the range (6-14):
In such case Bond order = I 8-n I / 2
Eg: CO (Total electrons = 6+8=14), Therefore B.O.= I 8-14 I / 2 = 3
SET 4: Molecules and ions having total no of electrons within the range (14-20):
In such case Bond order = (20-n) / 2 ; [Where n = Total no of electrons]
Eg. NO (Total electrons = 15), Therefore B.O. = 20-15/2 = 2.5
Table 1: Bond order values for homonuclear and heteronuclear diatomic species having (1-20)electrons
Species (Molecules or ions)
Total Number of electrons (n)
Bond-Order (B.O.)
Bond-Order Values for the species having (1-2) electrons ; Bond order = n/2
H2+
H2, He22+
1
2
0.5
1
Bond-Order Values for the species having (2-6) electrons ; Bond order = I 4- n I / 2
H2-,He2+
He2,
Li2+,He2-
Li2, He22-, Be22+
3
4
5
6
0.5
0
0.5
1
Bond-Order Values for the species having (6-14) electrons ; Bond order = I 8- n I / 2
Be2+, Li2-
Be2, Li22-
Be2-, B2+
B2, Be22-, HF
B2-, C2+
C2, B22-, N22+, CN+
C2-, N2+
N2, CO, NO+, C22-, CN-,O22+
7
8
9
10
11
12
13
14
0.5
0
0.5
1
1.5
2
2.5
3
Bond-Order Values for the species having (14-20) electrons; Bond order = (20-n) / 2
N2-, NO, O2+
NO-, O2
O2-
F2, O22-, HCl
F2-
Ne2
15
16
17
18
19
20
2.5
2
1.5
1
0.5
0
Graphical Presentation of Bond-Order
The graphical representation presented in Fig. 1 shows that bond-order gradually increases to 1 in the range (0-2) electrons then falls to zero in the range (2-4) electrons then it further rises to 1 for (4-6) electrons and once again falls to zero for (6-8) electrons then again rises to 3 in the range (8-14) electrons and then finally falls to zero for (14-20) electrons. For total no of electrons 2, 6 and 14, we may use multiple formulae, because they fall in the overlapping region in which they intersect with each other.
External Links
1. communities.acs.org/docs/DOC-46667
2. communities.acs.org/docs/<wbr/>DOC-45853
3. www.drarijitdaschem.in/Innova...Views%20in.pdf
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Innovative study for the prediction of bond order in case of oxide based acid radicals have been discussed here1-3.
In case of oxide based acid radicals
Bond Order (B.O.) = Valency of the peripheral atom + (Charge on Acid Radical / Total number of peripheral atoms)
Eg.:
• ClO4- : (Valency of one Peripheral atom Oxygen = 2, Charge on acid radical = -1, Total Number of Peripheral atoms = 04), Therefore B.O. = 2 + (-1/4) = 1.75
• ClO3- : (Valency of one Peripheral atom Oxygen = 2, Charge on acid radical = -1, Total Number of Peripheral atoms = 03), Therefore B.O. = 2 + (-1/3) = 1.66
• ClO2- : (Valency of one Peripheral atom Oxygen = 2, Charge on acid radical = -1, Total Number of Peripheral atoms = 02), Therefore B.O. = 2 + (-1/2) = 1.5
• AsO43- : (Valency of one Peripheral atom Oxygen = 2, Charge on acid radical = -3, Total Number of Peripheral atoms = 04), Therefore B.O. = 2 + (-3/4) = 1.25
• AsO33-: (Valency of one Peripheral atom Oxygen = 2, Charge on acid radical = -3, Total Number of Peripheral atoms = 03), Therefore B.O. = 2 + (-3/3) = 1.0
• SO42- : (Valency of Peripheral atom Oxygen = 2, Charge on acid radical = -2, Number of Peripheral atoms = 04), Therefore B.O. = 2 + (-2/4) = 1.5
• SO32- : (Valency of Peripheral atom Oxygen = 2, Charge on acid radical = -2, Number of Peripheral atoms = 03), Therefore B.O. = 2 + (-2/3) = 1.33
• PO43- ; (Valency of Peripheral atom Oxygen = 2, Charge on acid radical = -3, Number of Peripheral atoms = 04), Therefore B.O. = 2 + (-3/4) = 1.25
• BO33- ; (Valency of Peripheral atom Oxygen = 2, Charge on acid radical = -3, Number of Peripheral atoms = 03), Therefore B.O. = 2 + (-3/3) = 1
• CO32- ;(Valency of Peripheral atom Oxygen = 2,Charge on acid radical = -2, Number of Peripheral atoms = 03), Therefore B.O. = 2 + (-2/3) = 1.33
• SiO44-:(Valency of Peripheral atom Oxygen = 2,Charge on acid radical = - 4, Number of Peripheral atoms = 04), Therefore B.O. = 2 + (- 4/4) = 1
Relation (Bond order vs. Bond length, Bond Strength, Bond energy, Thermal stability and Reactivity)
B.O. α 1 / Bond length or Bond distance;
B.O. α Bond strength;
B.O. α Bond Energy;
B.O. α Thermal Stability; B.O. α 1 / Reactivity
Correlation (Literature values of bond-distances of some oxide based acid radicals with their predicted bond order values)
Literature values of the Cl-O average bond lengths in ClO4-,ClO3- and ClO2- ; As-O bond lengths in AsO43- and AsO33- with respect to their bond order values suggest that with increasing bond-order M-O bond length (Where M = Cl, As etc.) decreases which is shown in Table-1.
Table 1:Bond-distances and their predicted bond order values
Oxide Based Acid Radicals
Bond-Order Values
Avg. M-O Bond-Distances
As per Literature (Å)
Remarks
ClO4-
1.75
1.50
Increasing Bond-Order decreases Bond Length
ClO3-
1.6
1.57
ClO2-
1.5
1.64
AsO43-
1.25
1.75
AsO33-
1.0
1.77
External Links
• communities.<wbr/>acs.org/docs/DOC-46667
• communities.acs.org/docs/DOC-45853
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Electronic_Structure_of_Atoms_and_Molecules/Predicting_the_Bond-Order_of_Oxides_based_Acid_Radicals.txt
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Prof. Linus Pauling (1931) first developed the Hybridization state theory in order to explain the structure of molecules such as methane (CH4).1This concept was developed for simple chemical systems but this one applied more widely later on and from today’s point of view it is considered an operative empirical for excusing the structures of organic and inorganic compounds along with their related problems. An innovative method proposed for the determination of hybridization state on time economic ground 2,3,4.
Prediction of sp, sp2, sp3 Hybridization state
We Know, hybridization is nothing but the mixing of orbital’s in different ratio to form some newly synthesized orbitals called hybrid orbitals. The mixing pattern is as follows:
s + p (1:1) - sp hybrid orbital; s + p (1:2) - sp2 hybrid orbital ; s + p (1:3) - sp3 hybrid orbital
Formula used for the determination of sp, sp2 and sp3 hybridization state:
Power on the Hybridization state of the central atom = (Total no of σ bonds around each central atom -1)
All single (-) bonds are σ bond, in double bond (=) there is one σ and 1π, in triple bond (≡) there is one σ and 2π. In addition to these each lone pair (LP) and Co-ordinate bond can be treated as one σ bond subsequently.
Eg.:
a. In NH3: central atom N is surrounded by three N-H single bonds i.e. three sigma (σ) bonds and one lone pair (LP) i.e. one additional σ bond. So, in NH3 there is a total of four σ bonds [3 bond pairs (BPs) + 1 lone pair (LP)] around central atom N. Therefore, in this case power of the hybridization state of N = 4-1 = 3 i.e. hybridization state = sp3.
b. In H2O: central atom O is surrounded by two O-H single bonds i.e. two sigma (σ) bonds and two lone pairs i.e. two additional σ bonds. So, altogether in H2O there are four σ bonds (2 bond pairs + 2 lone pairs) around central atom O, So, in this case power of the hybridization state of O = 4-1 =3 i.e. hybridization state of O in H2O = sp3.
c. In H3BO3:- B has 3σ bonds (3BPs but no LPs) and oxygen has 4σ bonds (2BPs & 2LPs) so, in this case power of the hybridization state of B = 3-1 = 2 i.e. B is sp2 hybridized in H3BO3. On the other hand, power of the hybridization state of O = 4-1= 3 i.e. hybridization state of O in H3BO3 is sp3.
d. In I-Cl: I and Cl both have 4σ bonds and 3LPs, so, in this case power of the hybridization state of both I and Cl = 4 - 1 = 3 i.e. hybridization state of I and Cl both are sp3.
e. In CH2=CH2: each carbon is attached with 2 C-H single bonds (2 σ bonds) and one C=C bond (1σ bond), so, altogether there are 3 sigma bonds. So, in this case, power of the hybridization state of both C = 3-1 = 2 i.e. hybridization state of both C’s are sp2.
Prediction of sp3d, sp3d2, and sp3d3 Hybridization States
In case of sp3d, sp3d2 and sp3d3 hybridization state there is a common term sp3 for which 4 sigma bonds are responsible. So, in addition to 4 sigma bonds, for each additional sigma, added one d orbital gradually as follows:-
5σ bonds = 4σ bonds + 1 additional σ bond = sp3d hybridization
6σ bonds = 4σ bonds + 2 additional σ bonds = sp3d2 hybridization
7σ bonds = 4σ bonds + 3 additional σ bonds = sp3d3 hybridization
Eg:-
a.IF4+: I has 7 e-s in its outermost shell, so, in this case, subtract one e- from 7 i.e. 7 – 1 = 6. So, out of 6 electrons, 4 electrons form 4 I-F bonds i.e. 4 sigma bonds and there is one LP. So, altogether there are 5 σ bonds. So, 5σ bonds = 4 σ bonds + 1 additional σ bond = sp3d hybridization
b.IF7: 7 I-F single bonds i.e. 7σ bonds = 4σ bonds + 3 additional σ bonds = sp3d3 hybridization.
c.ICl2-: I has 7 e-s in its outermost shell, so, in this case, add one e- with 7(overall charge on the compound) i.e. 07+1= 08. So, out of 08 electrons, 02 electrons form 02 I-Cl bonds i.e. 02 sigma bonds and there is 03 LPs. So, altogether there are 05σ bonds. So, 5σ bonds = 04 σ bonds + 01 additional σ bond = sp3d hybridization.
d. XeF4: Xe, an inert gas, consider 8 e-s in its outermost shell, 04 of which form 04 Xe-F sigma bonds and there is two LPs, i.e. altogether there is 06 σ bonds = 04 σ bonds + 02 additional σ bonds = sp3d2 hybridization.
In case of determination of the hybridization state by using the above method, one must have a clear idea about the outermost electrons of different family members in the periodic table as follows:
Family Outermost electrons
Nitrogen family 05
Oxygen family 06
Halogen family 07
Inert gas family 08
In case of cationic species you must remove requisite electron / electrons from the outermost orbit of the central atom and incase of anionic species you must add requisite electron with the outermost electrons of the central atom. Examples have been explored in Table 1.
Table 1: Total number of σ bonds and Hybridization State
Total number of sigma (σ) bonds
Nature of Hybridization State
Examples
2
sp
BeCl2, HgCl2,C2H2,CO2,CO,CdCl2, ZnCl2 etc.
3
sp2
BCl3, AlCl3,C2H4,C6H6,SO2,SO3,HNO3,
H2CO3,SnCl2, PbCl2 etc.
4
sp3
NH4+, BF4-, H2SO4, HClO4,PCl3, NCl3, AsCl3, HClO3,ICl2+,OF2,HClO2,SCl2,HClO, ICl, XeO3 etc.
5
sp3d
PCl5, SbCl5, SF4, ClF3, BrF3, XeF2, ICl2- etc.
6
sp3d2
SF6, AlF63-, SiF62-, PF6-, IF5, BrF5, XeOF4, XeF4, BrF4-, ICl4- etc.
7
sp3d3
IF7, XeF6 etc.
External Links
1. communities.<wbr/>acs.org/docs/DOC-46667
2. communities.acs.org/docs/<wbr/>DOC-45853
3. www.drarijitdaschem.in/Innova...Views%20in.pdf
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Electronic_Structure_of_Atoms_and_Molecules/Predicting_the_Hybridization_of_Simple_Molecules.txt
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In this article, formulae based mnemonics by classifying lone pair of electrons (localized or delocalized) have been highlighted in an innovative and time economic way to enhance interest of students’ on heterocyclic chemistry for determination of planarity by calculating Hybridization state of hetero atom and by prediction of Aromatic, Anti aromatic, non aromatic behavior of different heterocyclic compounds . Here, I have tried to hub three (03) time economic mnemonics by including two (02) formulae for the prediction of hybridization of hetero atom, aromatic and anti aromatic behavior of heterocyclic compounds. This article encourages students to solve multiple choice type questions (MCQs) on ‘Aromaticity of Heterocyclic compounds’ at different competitive examinations in a time economic way.
Introduction
The conventional methods1-7 for determination of hybridization state of hetero atom (planarity of molecule), prediction of aromatic and anti aromatic nature of heterocyclic compound is time consuming. Keeping this in mind, in this article, I have introduced three time economic innovative mnemonics by using two formulae for the prediction of hybridization state of hetero atom of heterocyclic compounds to determine its planarity and aromatic / anti aromatic / non aromatic nature of heterocyclic compounds containing one, two or more number of hetero atoms to make heterocyclic chemistry metabolic and interesting for students. This study also shows omission behavior of some heterocyclic compounds with respect to their aromatic/anti aromatic/non aromatic nature due to presence or absence of vacant d orbitals in DLP based hetero atoms and how lone pair electron discriminates prediction of hybridization state of hetero atom in heterocyclic compound with prediction of its Aromatic and Anti Aromatic nature.
Time Economic Innovative Mnemonics in Heterocyclic Chemistry
Classification of Lone Pair Electron
Lone Pair of electrons can be generally classified into two types as Delocalized lone pair of electron (DLP) and Localized lone pair of electron (LLP) as follows:
i)Delocalized lone pair of electron (DLP): When lone pair of electron of hetero atom undergo delocalization through conjugation then it is to be treated as delocalized lone pair of electron (DLP). Hetero atom (atom containing lone pair of electron) which is directly attached with single bonds only from all ends is to be considered as DLP containing hetero atom and its lone pair is to be treated as (DLP).
Example \(1\):
Eg. In Pyrrole lone pair of N atom is to be treated as DLP because it is directly attached with three single bonds only.
ii)Localized lone pair of electron (LLP): When lone pair of electron of hetero atom does not undergo delocalization through conjugation then it is to be treated as Localized lone pair of electron (LLP). Hetero atom (atom containing lone pair of electron) which is directly attached with single and double bonds with the ring system is to be considered as LLP containing hetero atom and its lone pair is to be treated as localized lone pair of electron (LLP).
Example \(2\):
Eg. In Pyridine lone pair of N atom is to be treated as LLP because it is directly attached with double and single bonds with the ring system.
Aromaticity
It was first devised by Hückel in 1931. Conventional method for prediction of Aromatic nature of organic compound:
1. Cyclic molecule,
2. Planer molecule in which all bonded atoms lie in same plane (having sp2 hybridized)
3. Conjugated molecule with conjugated π-electron system,
4. Contains (4n + 2) π electrons, where, n is a positive integer (n = 0,1,2,3 etc.)
Conventional method for Anti Aromatic nature of organic Compound:
Cyclic molecule,
Planer molecule in which all bonded atoms lie in same plane (having sp2 hybridized)
Conjugated molecule with conjugated π-electron system,
4nπ electrons, where, n is a positive integer (n = 0,1,2,3 etc.)
Conventional method for identification of Non Aromatic Nature of organic Compound:
If a compound violates any one of the above three conditions (1 or 2 or 3) then it is non aromatic in nature.
Planarity of Heterocyclic Compounds with the prediction of Hybridization State
Planarity of heterocyclic compounds depends on the nature of the hybridization state of carbon and hetero atoms present in it. When all atoms (carbon and hetero) in the heterocyclic compounds having sp2 hybridized then it is planar but when there is a mixing of sp2 and sp3 hybridization state then it is treated as non planar.
Hybridization state theory
Prof. Linus Pauling (1931) first developed the Hybridization state theory in order to explain the structure of molecules such as methane (CH4).This concept was developed for simple chemical systems but this one applied more widely later on and from today’s point of view it is considered an operative empirical for excusing the structures of organic and inorganic compounds along with their related problems.
Conventional method for prediction of hybridization state:
Hybridization state for a molecule can be calculated by the formula 0.5 (V+H−C+A), Where, V = Number of valance electrons in central atom, H = Number of surrounding monovalent atoms, C = Cationic charge, A = Anionic charge
Innovative Mnemonics for the prediction of hybridization state of hetero atom in the heterocyclic compounds with LLP:
Formula 1: Prediction of hybridization state of hetero atom
Power on the Hybridization state of the hetero atom =
(Total no of σ bonds around each hetero atom - 1)
This formula should be applicable up to 4 σ bonds.
If the power of the hybridization state will be 03, 02 and 01 then the hybridization state will be sp3, sp2 and sp respectively. All single (-) bonds are σ bond, in double bond (=) there is one σ and one π. In addition to these each localized lone pair of electron (LLP) can be treated as one σ bond. Hybridization State of Hetero atom with the help of LLP to find out the planarity in Heterocyclic Compounds are shown in Table-1 below.
Table-1 (Hybridization state of Hetero atom in Heterocyclic Compounds with the help of LLP)
Heterocyclic Compounds
(Planar/non planar)
Number of σ bonds around hetero atom
(from single and double bonds)
(A)
LLP
(localized Lone Pair of e-s)
(B)
Total Number of σ bonds around hetero atom
(A+B)
Power of the
Hybridization state
of the hetero atom
(Corresponding Hybridization state) = (A+B)-1
Pyrrole
(Planar)
03
0
(lone pair of electron undergo delocalization,DLP with the ring system)
03
02
(sp2 N)
Furan
(Planar)
02
01
(out of two lone pair of electrons, one undergo delocalization,DLP and other remain as LLP)
03
02
(sp2 O)
Thiophene
(Planar)
02
01
(out of two lone pair of electrons, one undergo delocalization, DLP and other remain as LLP)
03
02
(sp2 S)
Pyridine
(Planar)
02
01
03
02
(sp2 N)
Indole
(Planar)
03
0
03
02
(sp2 N)
Quinoline
(Planar)
02
01
03
02
(sp2 N)
(Planar)
02
01
03
02
(sp2 N)
(Planar)
03
(N1)
02
(N3)
0
(N1)
01
(N3)
03
03
02
(sp2 N1)
02
(sp2 N3)
(Planar)
02
(N1)
02
(N3)
01
(N1)
01
(N3)
03
03
02
(sp2 N1)
02
(sp2 N3)
(Planar)
02
(N1)
02
(N3)
02
(N7)
03
(N9)
01
(N1)
01
(N3)
01
(N7)
0
(N9)
03
03
03
03
02
(sp2 N1)
02
(sp2 N3)
02
(sp2 N7)
02
(sp2 N9)
(Planar)
02
(N)
02
(S)
01
(N)
01
(S)
(out of two lone pair of electrons on S, one undergo delocalization,DLP and other remain as LLP)
03
03
02
(sp2 N)
02
(sp2 S)
(Planar)
02
(N)
02
(S)
01
(N)
01
(S)
(out of two lone pair of electrons on S, one undergo delocalization,DLP and other remain as LLP)
03
03
02
(sp2 N)
02
(sp2 S)
(Planar)
02
(N1)
02
(N1)
01
(N1)
01
(N1)
03
03
02
(sp2 N1)
02
(sp2 N4)
(Planar)
02
(N1,N3 and N5)
01
(N1,N3 and N5)
03
02
(sp2 N1,N3,N5)
(Planar)
03
(N)
02
(S)
0
(N)
01
(S)
(out of two lone pair of electrons on S, one undergo delocalization, DLP and other is LLP)
03
03
02
(sp2 N)
02
(sp2 S)
(Planar)
02
(both N)
01
(both N)
03
02
(sp2 both N)
(Planar)
02
(N1,N2,N3,N4)
01
(N1,N2,N3,N4)
03
02
(sp2 All N)
(Planar)
02
01
03
02
(sp2 N)
(Planar)
02
01
03
02
(sp2 N)
(Non Planar)
03
01
04
03
(sp3 N)
(Non Planar)
02
02
04
03
(sp3 O)
Innovative Mnemonics for the Prediction of Aromatic, Anti Aromatic behavior of Heterocyclic Compounds with DLP:
The present study will be an innovative mnemonic involving calculation of ‘A’ value by just manipulating the no of π bonds within the ring system and delocalized lone pair of electron (DLP) with one (01).
The heterocyclic compound having cyclic, planar, conjugated (i.e. all the carbon atoms having same state of hybridization, sp2) with even number of ‘A’ value will be treated as aromatic in nature and with odd number of ‘A’ value will be treated as anti aromatic in nature.
Formula 2: Evaluation of A Value to predict Aromatic and Anti Aromatic Nature
A = πb+DLP+1(constant) = even no = Aromatic
A = πb+DLP+1(constant) = odd no = Anti Aromatic
where, πb = number of π bonds with in the ring system;
DLP = Delocalized lone pair of electron.
In case of a multi hetero atom based heterocyclic compound, containing both DLP and LLP hetero atoms,
Aromatic and Anti Aromatic behaviour should be predicted with respect to DLP based hetero atom only.
Example \(3\): Benzothiazole
Benzothiazole (Figure 1), is a multi hetero atom based heterocyclic compound, containing both DLP and LLP hetero atoms. Here, for N, DLP = 0 , LLP = 1 and for S, DLP = 1, LLP =1, so, in this case ‘A’ value should be calculated with respect to S only not N. Here, A = 4 + 1 + 1 = 6 (even no) = Aromatic.
But when heterocyclic compounds contain both LLP based hetero atoms then Aromaticity should be predicted with respect to that hetero atom which contains lowest possible position number as per IUPAC nomenclature or any one of the hetero atom.
Example \(3\): Imidazole
Imidazole (Figure 2) is a multi hetero atom based hetero cyclic compound in which, N1 is DLP based hetero atom and N3 is LLP based hetero atom. In this case Aromaticity should be predicted with respect to the DLP based hetero atom N1.
For N1, A = πb+DLP+1(constant) = 2+1+1 = 4 (even No) - Aromatic
Eg. Pyrimidine (Figure 3) is a multi hetero atom based hetero cyclic compound in which, both N1 & N3 are in same environment based hetero atoms (LLP based hetero atoms). In this case Aromaticity should be predicted with respect to N1 (lowest possible position number as per IUPAC nemenclature).
For N1, A = πb+DLP+1(constant) = 3+0+1 = 4 (even no) - Aromatic
Aromaticity of heterocyclic compounds have been illustrated in Table-2
Table-2 (Aromatic-Anti Aromatic and Non Aromatic behavior of heterocyclic compounds With the help of DLP)
Hetero Cyclic Compound
(Cyclic, Planar, Conjugated)
πb value
[πb =number of π bonds with in the ring system]
DLP
A value
[A = πb + DLP + 1(constant)]
(even No /odd No)
Remark on Nature
of compound
(Aromatic/Anti Aromatic)
Pyrrole
2
1
2 + 1 + 1 = 4
(even No)
Aromatic
Furan
2
1
( Here out of two lone pairs on O only one LP take part in delocalization)
2 + 1 + 1 = 4
(even No)
Aromatic
Thiophene
2
1
(Here out of two lone pairs on O only one LP take part in delocalization)
2 + 1 + 1 = 4
(even No)
Aromatic
Pyridine
3
0
3 + 0 + 1 = 4
(even No)
Aromatic
Indole
4
1
4 + 1 + 1 = 6
(even No)
Aromatic
Quinoline
5
0
5 + 0 + 1 = 6
(even No)
Aromatic
05
0
5 + 0 + 1 = 6
(even No)
Aromatic
02
01
(N1)
2 + 1 + 1 = 4
(even No)
Aromatic
(m-diazine)
03
0
(N1)
3 + 0 + 1 = 4
(even No)
Aromatic
04
01
(N9)
4 + 1 + 1 = 6
(even No)
Aromatic
02
01
(S)
2 + 1 + 1 = 4
(even No)
Aromatic
04
01
(S)
4 + 1 + 1 = 6
(even No)
Aromatic
03
0
3 + 0 + 1 = 4
(even No)
Aromatic
03
0
3 + 0 + 1 = 4
(even No)
Aromatic
07
0
7 + 0 + 1 = 8
(even No)
Aromatic
03
0
3 + 0 + 1 = 4
(even No)
Aromatic
04
0
4 + 0 + 1 = 5
(odd No)
Anti aromatic
02
0
2 + 0 + 1 = 3
(odd No)
Anti aromatic
Hetero Cyclic Compound
(Cyclic, non-planar)
πb value
[πb =number of π bonds with in the ring system]
DLP
A value
[A = πb + DLP + 1(constant)]
(even No/odd No)
Remark
on
Nature
of compound
-
-
-
Non Aromatic (non planar – sp3)
-
-
-
Non Aromatic (non planar – sp3)
Omission behavior of some heterocyclic compounds with respect to their Aromatic / Anti Aromatic and Non Aromatic nature :
Aromatic Behavior of some heterocyclic compounds containing different DLP based hetero atoms (one contains vacant d orbitals) :
In Phenothiazine (Figure 4), there is two DLP based hetero atoms N and S. In between N and S, since S
having vacant d orbitals, so, in this case ‘A’ value will be predicted with respect to DLP based S
hetero atom which contains vacant d orbitals only.
Here, A = πb + DLP + 1(constant) = 6 +1+1 = 8 (even no) = Aromatic.
Non Aromatic Behavior of some heterocyclic compounds containing same DLP based heteroatom having no d orbitals:
Omission behavior of some heterocyclic compounds will be observed (Figure 5 and 6),when there, is at least two hetero atoms (same or different) but both the hetero atoms do not have any d orbitals (such as O,N etc.) and they are in DLP based environment in the ring system.
These molecules have been studied with advanced molecular orbital techniques known as ‘ab initio calculations’. ‘Ab initio quantum chemistry methods’ are computational chemistry methods based on quantum chemistry8.
In the case of 1,2-dioxin, 1,4-dioxin and dibenzo-1,4-dioxin there is DLP based O atoms in all the molecules but still they will be non aromatic due to prevention of significant free electron delocalization (makes non conjugated). The π electrons from the carbon bonds and the lone pair electrons on the oxygen atoms do not overlap to a significant degree due to absence of vacant d orbitals in both O atoms in each case (pπ-dπ overlap is not possible here in conjugation). It makes these molecules non conjugated and thus allows the molecules to become non aromatic instead of aromatic (A value = even No).
In the heterocyclic compounds, where, there is two DLP based N atoms instead of two DLP based O atoms or there is one DLP N atom along with one DLP O atom, the same phenomena of non aromatic behavior will be observed. Because, both N and O atoms do not have any vacant d orbitals, and hence pπ-dπ overlap is not possible here in conjugation.
Anti Aromatic Behavior of some heterocyclic compounds containing same DLP based hetero atoms having vacant d orbitals:
These compounds (Figure 7) are anti aromatic, here both S atoms, having vacant d orbitals, contain one DLP and one LLP and here both DLP of both S atoms participate in the delocalization. Hence, for the prediction of ‘A’ value, consider both DLP (DLP = 2). Here, A = πb + DLP + 1 (Constant) = 2 + 2 + 1 = 5 (odd No) = Anti Aromatic.
Conclusions
It may be expected that these three time economic innovative mnemonics of heterocyclic chemistry will help the students of Undergraduate, Senior Undergraduate and Post-Graduate level to predict aromatic, anti aromatic and non aromatic character of heterocyclic compounds along with their omission behaviour. Experiment in vitro on 100 students showed that by using these two formulae students can save up to 5-10 minutes time in the examination hall to predict the aromatic, anti aromatic and non aromatic character of any heterocyclic compounds and their comparative study including omission behaviour with respect to the DLP and LLP based hetero atoms present on them. On the basis of this, I can strongly recommend to use these three time economic innovative mnemonics in the field of heterocyclic chemistry.
External Links: http://www.sciepub.com/WJCE/abstract/8888
Contributor
Dr. Arijit Das, Ph.D. (Inorganic Chemistry), MACS ( Invited,USA ), SFICS, MISC, MIAFS (India), Assistant Professor, Department of Chemistry, Ramthakur College, Agartala, Tripura(W), Tripura, India, Pin-799003.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Electronic_Structure_of_Atoms_and_Molecules/Prediction_of_Aromatic_Anti_Aromatic_and_Non_Aromatic_Character_of_Hete.txt
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Including: acid-base theories; strong and weak acids and bases; an introduction to \(pH\), \(K_w\) and \(pK_w\), \(K_a\) and \(pK_a\), \(K_b\) and \(pK_b\); buffer solutions; pH curves and indicators.
• 1. Theories of Acids and Bases
• 2. Strong and Weak Acids
This page explains the terms strong and weak as applied to acids. As a part of this it defines and explains what is meant by pH, Ka and pKa.
• 3. The Ionic Product for Water
This page explains what is meant by the ionic product for water. It looks at how the ionic product varies with temperature, and how that determines the pH of pure water at different temperatures.
• 4. Strong and Weak Bases
This page explains the terms strong and weak as applied to bases. As a part of this it defines and explains Kb and pKb. The usual way of comparing the strengths of bases is to see how readily they produce hydroxide ions in solution. This may be because they already contain hydroxide ions, or because they take hydrogen ions from water molecules to produce hydroxide ions.
• 6. Acid-Base Indicators
This page describes how simple acid-base indicators work, and how to choose the right one for a particular titration.
• 7. Buffer Solutions
This page describes simple acidic and alkaline buffer solutions and explains how they work.
• pH Titration Curves
Acid-Base Equilibria
This Module describes the Arrhenius, Brønsted-Lowry, and Lewis theories of acids and bases, and explains the relationships between them. It also explains the concept of a conjugate pair - an acid and its conjugate base, or a base and its conjugate acid.
The Arrhenius Theory of acids and bases
• Acids are substances that produce hydrogen ions in solution.
• Bases are substances that produce hydroxide ions in solution.
Neutralization happens because hydrogen ions and hydroxide ions react to produce water.
$H^+_{(aq)} + OH^-{(aq)} \rightarrow H_2O_{(l)}$
Hydrochloric acid is neutralized by both sodium hydroxide solution and ammonia solution. In both cases, you get a colorless solution which you can crystallize to get a white salt - either sodium chloride or ammonium chloride. These are clearly very similar reactions. The full equations are:
$NaOH_{(aq)} + HCl_{(aq)} \rightarrow NaCl_{(aq)} + H_2O_{(l)}$
$NH_{3(aq)} + HCl_{(aq)} \rightarrow NH_4Cl_{(aq)}$
In the sodium hydroxide case, hydrogen ions from the acid are reacting with hydroxide ions from the sodium hydroxide - in line with the Arrhenius theory. However, in the ammonia case, there do not appear to be any hydroxide ions! You can get around this by saying that the ammonia reacts with the water it is dissolved in to produce ammonium ions and hydroxide ions:
$NH_{3(aq)} + H_2O_{(l)} \rightleftharpoons NH^+_{4(aq)} + OH_{(aq)}^-$
This is a reversible reaction, and in a typical dilute ammonia solution, about 99% of the ammonia remains as ammonia molecules. Nevertheless, there are hydroxide ions there, and we can squeeze this into the Arrhenius theory. However, this same reaction also happens between ammonia gas and hydrogen chloride gas.
$NH_{3(g)} + HCl_{(g)} \rightarrow NH_4Cl_{(s)}$
In this case, there are no hydrogen ions or hydroxide ions in solution - because there is not any solution. The Arrhenius theory would not count this as an acid-base reaction, despite the fact that it is producing the same product as when the two substances were in solution.
The Brønsted-Lowry Theory of acids and bases
• An acid is a proton (hydrogen ion) donor.
• A base is a proton (hydrogen ion) acceptor.
The Brønsted-Lowry theory does not go against the Arrhenius theory in any way - it just adds to it. Hydroxide ions are still bases because they accept hydrogen ions from acids and form water. An acid produces hydrogen ions in solution because it reacts with the water molecules by giving a proton to them.
When hydrogen chloride gas dissolves in water to produce hydrochloric acid, the hydrogen chloride molecule gives a proton (a hydrogen ion) to a water molecule. A coordinate (dative covalent) bond is formed between one of the lone pairs on the oxygen and the hydrogen from the $HCl$. Hydronium ions, $H_3O^+_{(aq)}$, are pr oduced.
$H_2O + HCl \rightarrow H_3O^+ +Cl^-$
When an acid in solution reacts with a base, what is actually functioning as the acid is the hydronium ion. For example, a proton is transferred from a hydronium ion to a hydroxide ion to make water.
$H_3O^+_{(aq)} + OH^-_{(aq)} \rightarrow 2H_2O_{(l)}$
Showing the electrons, but leaving out the inner ones:
It is important to realize that whenever you talk about hydrogen ions in solution, $H^+_{(aq)}$, wha t you are actually talking about are hydronium ions.
The hydrogen chloride / ammonia problem
This is no longer a problem using the Brønsted-Lowry theory. Whether you are talking about the reaction in solution or in the gas state, ammonia is a base because it accepts a proton (a hydrogen ion). The hydrogen becomes attached to the lone pair on the nitrogen of the ammonia via a co-ordinate bond.
If it is in solution, the ammonia accepts a proton from a hydronium ion:
$NH_{3 (aq)} + H_3O_{(aq)}^+ \rightarrow NH^+_{4(aq)} + H_2O_{(l)}$
If the reaction occurs in the gas state, the ammonia accepts a proton directly from the hydrogen chloride:
$NH_{3 (g)} + HCl_{(g)} \rightarrow NH_4Cl_{(s)}$
Either way, the ammonia acts as a base by accepting a hydrogen ion from an acid.
Conjugate pairs
When hydrogen chloride dissolves in water, almost 100% of it reacts with the water to produce hydronium ions and chloride ions. Hydrogen chloride is a strong acid, and we tend to write this as a one-way reaction:
$H_2O + HCl \rightarrow H_3O^+ + Cl^-$
In fact, the reaction between HCl and water is reversible, but only to a very minor extent. To generalize, consider an acid $HA$, and think of the reaction as being reversible.
$HA + H_2O \rightleftharpoons H_3O^+ + A^-$
Thinking about the forward reaction:
• The $HA$ is an acid because it is donating a proton (hydrogen ion) to the water.
• The water is a base because it is accepting a proton from the $HA$.
However, there is also a back reaction between the hydronium ion and the $A^-$ i on:
• The $H_3O^+$ is an acid because it is donating a proton (hydrogen ion) to the $A^-$ ion.
• The $A^-$ ion is a base because it is accepting a proton from the $H_3O^+$.
The reversible reaction contains two acids and two bases. We think of them in pairs, called conjugate pairs.
When the acid, $HA$, loses a proton it forms a base, $A^-$, which can a ccept a proton back again to refom the acid, $HA$. These two are a conjugate pair. Members of a conjugate pair differ from each other by the presence or absence of the transferable hydrogen ion.
• If you are thinking about $HA$ as the acid, then $A^-$ is its c onjugate base.
• If you are thinking about $A^-$ as the base, then $HA$ is its con jugate acid.
The water and the hydronium ion are also a conjugate pair. Thinking of the water as a base, the hydronium ion is its conjugate acid because it has the extra hydrogen ion which it can give away again. Thinking about the hydronium ion as an acid, then water is its conjugate base. The water can accept a hydrogen ion back again to reform the hydronium ion.
A second example of conjugate pairs
This is the reaction between ammonia and water that we looked at earlier:
Think first about the forward reaction. Ammonia is a base because it is accepting hydrogen ions from the water. The ammonium ion is its conjugate acid - it can release that hydrogen ion again to reform the ammonia.
The water is acting as an acid, and its conjugate base is the hydroxide ion. The hydroxide ion can accept a hydrogen ion to reform the water. Looking at it from the other side, the ammonium ion is an acid, and ammonia is its conjugate base. The hydroxide ion is a base and water is its conjugate acid.
Amphoteric substances
You may possibly have noticed (although probably not!) that in one of the last two examples, water was acting as a base, whereas in the other one it was acting as an acid. A substance which can act as either an acid or a base is described as being amphoteric.
The Lewis Theory of acids and bases
This theory extends well beyond the things you normally think of as acids and bases.
• An acid is an electron pair acceptor.
• A base is an electron pair donor.
Lewis bases
It is easiest to see the relationship by looking at exactly what Brønsted-Lowry bases do when they accept hydrogen ions. Three Brønsted-Lowry bases we've looked at are hydroxide ions, ammonia and water, and they are typical of all the rest.
The Brønsted-Lowry theory says that they are acting as bases because they are combining with hydrogen ions. The reason they are combining with hydrogen ions is that they have lone pairs of electrons - which is what the Lewis theory says. The two are entirely consistent. So how does this extend the concept of a base? At the moment it doesn't - it just looks at it from a different angle.
But what about other similar reactions of ammonia or water, for example? On the Lewis theory, any reaction in which the ammonia or water used their lone pairs of electrons to form a co-ordinate bond would be counted as them acting as a base. Here is a reaction which you will find talked about on the page dealing with co-ordinate bonding. Ammonia reacts with BF3 by using its lone pair to form a co-ordinate bond with the empty orbital on the boron.
As far as the ammonia is concerned, it is behaving exactly the same as when it reacts with a hydrogen ion - it is using its lone pair to form a co-ordinate bond. If you are going to describe it as a base in one case, it makes sense to describe it as one in the other case as well.
Lewis acids
Lewis acids are electron pair acceptors. In the above example, the BF3 is acting as the Lewis acid by accepting the nitrogen's lone pair. On the Brønsted-Lowry theory, the BF3 has nothing remotely acidic about it. This is an extension of the term acid well beyond any common use.
What about more obviously acid-base reactions - like, for example, the reaction between ammonia and hydrogen chloride gas?
$NH_{3(g)} + HCl_{g)} \rightarrow NH^+_{4(s)} + Cl^-_{(s)}$
What exactly is accepting the lone pair of electrons on the nitrogen. Textbooks often write this as if the ammonia is donating its lone pair to a hydrogen ion - a simple proton with no electrons around it. That is misleading! You don't usually get free hydrogen ions in chemical systems. They are so reactive that they are always attached to something else. There aren't any uncombined hydrogen ions in HCl.
There isn't an empty orbital anywhere on the HCl which can accept a pair of electrons. Why, then, is the HCl a Lewis acid? Chlorine is more electronegative than hydrogen, and that means that the hydrogen chloride will be a polar molecule. The electrons in the hydrogen-chlorine bond will be attracted towards the chlorine end, leaving the hydrogen slightly positive and the chlorine slightly negative.
The lone pair on the nitrogen of an ammonia molecule is attracted to the slightly positive hydrogen atom in the HCl. As it approaches it, the electrons in the hydrogen-chlorine bond are repelled still further towards the chlorine. Eventually, a coordinate bond is formed between the nitrogen and the hydrogen, and the chlorine breaks away as a chloride ion.
This is best shown using the "curly arrow" notation commonly used in organic reaction mechanisms.
The whole HCl molecule is acting as a Lewis acid. It is accepting a pair of electrons from the ammonia, and in the process it breaks up. Lewis acids don't necessarily have to have an existing empty orbital.
Note
All you need to remember for Lewis acids and bases is:
• A Lewis acid is an electron pair acceptor.
• A Lewis base is an electron pair donor.
For all general purposes, stick with the Brønsted-Lowry theory.
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This page explains the terms strong and weak as applied to acids. As a part of this it defines and explains what is meant by pH, Ka and pKa.
Strong acids
We are going to use the Bronsted-Lowry definition of an acid. When an acid dissolves in water, a proton (hydrogen ion) is transferred to a water molecule to produce a hydroxonium ion and a negative ion depending on what acid you are starting from.
In the general case . . .
$HA + H_2O \rightleftharpoons H_3O^+ + A^- \tag{1}$
These reactions are all reversible, but in some cases, the acid is so good at giving away hydrogen ions that we can think of the reaction as being one-way. The acid is virtually 100% ionized.
For example, when hydrogen chloride dissolves in water to make hydrochloric acid, so little of the reverse reaction happens that we can write:
$H_2O_{(l)} + HCl_{(g)} \rightarrow H_3O^+_{(aq)} + Cl^-_{(aq)} \tag{2}$
At any one time, virtually 100% of the hydrogen chloride will have reacted to produce hydroxonium ions and chloride ions. Hydrogen chloride is described as a strong acid.
A strong acid is one which is virtually 100% ionized in solution.
Other common strong acids include sulphuric acid and nitric acid.
You may find the equation for the ionization written in a simplified form:
$HCl_{(aq)} \rightarrow H^+_{(aq)} + Cl^-_{(aq)} \tag{3}$
This shows the hydrogen chloride dissolved in the water splitting to give hydrogen ions in solution and chloride ions in solution.
This version is often used in this work just to make things look easier. If you use it, remember that the water is actually involved, and that when you write H+(aq) what you really mean is a hydroxonium ion, H3O+.
Strong acids and pH
pH is a measure of the concentration of hydrogen ions in a solution. Strong acids like hydrochloric acid at the sort of concentrations you normally use in the lab have a pH around 0 to 1. The lower the pH, the higher the concentration of hydrogen ions in the solution.
Defining pH
Example: Working out the pH of a strong acid
Suppose you had to work out the pH of 0.1 mol dm-3 hydrochloric acid. All you have to do is work out the concentration of the hydrogen ions in the solution, and then use your calculator to convert it to a pH.
With strong acids this is easy.
Hydrochloric acid is a strong acid - virtually 100% ionized. Each mole of HCl reacts with the water to give 1 mole of hydrogen ions and 1 mole of chloride ions
That means that if the concentration of the acid is 0.1 mol dm-3, then the concentration of hydrogen ions is also 0.1 mol dm-3.
Use your calculator to convert this into pH. My calculator wants me to enter 0.1, and then press the "log" button. Yours might want you to do it in a different order. You need to find out!
log10 [0.1] = -1
But pH = - log10 [0.1]
- (-1) = 1
The pH of this acid is 1.
Weak acids
A weak acid is one which doesn't ionize fully when it is dissolved in water. Ethanoic acid is a typical weak acid. It reacts with water to produce hydroxonium ions and ethanoate ions, but the back reaction is more successful than the forward one. The ions react very easily to reform the acid and the water.
$CH_3COOH + H_2O \rightleftharpoons CH_3COO^- + H_3O^+ \tag{4}$
At any one time, only about 1% of the ethanoic acid molecules have converted into ions. The rest remain as simple ethanoic acid molecules. Most organic acids are weak. Hydrogen fluoride (dissolving in water to produce hydrofluoric acid) is a weak inorganic acid that you may come across elsewhere.
Comparing the strengths of weak acids
The position of equilibrium of the reaction between the acid and water varies from one weak acid to another. The further to the left it lies, the weaker the acid is.
$HA + H_2O \rightleftharpoons H_3O^+ + A^- \tag{5}$
The acid dissociation constant, Ka
You can get a measure of the position of an equilibrium by writing an equilibrium constant for the reaction. The lower the value for the constant, the more the equilibrium lies to the left. The dissociation (ionization) of an acid is an example of a homogeneous reaction. Everything is present in the same phase - in this case, in solution in water. You can therefore write a simple expression for the equilibrium constant, Kc. Here is the equilibrium again:
$HA + H_2O \rightleftharpoons H_3O^+ + A^- \tag{5}$
You might expect the equilibrium constant to be written as:
$]K_c = \dfrac{[H_3O^+][A^-]}{[HA][H_2O]}$
However, if you think about this carefully, there is something odd about it.
At the bottom of the expression, you have a term for the concentration of the water in the solution. That's not a problem - except that the number is going to be very large compared with all the other numbers.
In 1 dm3 of solution, there are going to be about 55 moles of water.
If you had a weak acid with a concentration of about 1 mol dm-3, and only about 1% of it reacted with the water, the number of moles of water is only going to fall by about 0.01. In other words, if the acid is weak the concentration of the water is virtually constant. In that case, there isn't a lot of point in including it in the expression as if it were a variable. Instead, a new equilibrium constant is defined which leaves it out. This new equilibrium constant is called Ka.
You may find the Ka expression written differently if you work from the simplified version of the equilibrium reaction:
This may be written with or without state symbols.
It is actually exactly the same as the previous expression for Ka! Remember that although we often write H+ for hydrogen ions in solution, what we are actually talking about are hydroxonium ions. This second version of the Ka expression is not as precise as the first one.
To take a specific common example, the equilibrium for the dissociation of ethanoic acid is properly written as:
$CH_3COOH + H_2O \rightleftharpoons CH_3COO^- + H_3O^+ \tag{7}$
The Ka expression is:
If you are using the simpler version of the equilibrium . . .
$CH_3COOH \rightleftharpoons CH_3COO^- + H^+ \tag{8}$
. . . the Ka expression is:
The table shows some values of Ka for some simple acids:
acid Ka (mol dm-3)
hydrofluoric acid 5.6 x 10-4
methanoic acid 1.6 x 10-4
ethanoic acid 1.7 x 10-5
hydrogen sulphide 8.9 x 10-8
These are all weak acids because the values for Ka are very small. They are listed in order of decreasing acid strength - the Ka values get smaller as you go down the table. However, if you aren't very happy with numbers, that isn't immediately obvious. Because the numbers are in two parts, there is too much to think about quickly! To avoid this, the numbers are often converted into a new, easier form, called pKa.
An introduction to pKa
pKa bears exactly the same relationship to Ka as pH does to the hydrogen ion concentration:
$pK_a = -\log_{10} K_a \tag{9}$
If you use your calculator on all the Ka values in the table above and convert them into pKa values, you get:
acid Ka (mol dm-3) pKa
hydrofluoric acid 5.6 x 10-4 3.3
methanoic acid 1.6 x 10-4 3.8
ethanoic acid 1.7 x 10-5 4.8
hydrogen sulphide 8.9 x 10-8 7.1
Notice that the weaker the acid, the larger the value of pKa. It is now easy to see the trend towards weaker acids as you go down the table.
Note
• The lower the value for pKa, the stronger the acid.
• The higher the value for pKa, the weaker the acid.
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This page explains what is meant by the ionic product for water. It looks at how the ionic product varies with temperature, and how that determines the pH of pure water at different temperatures.
Kw and pKw
Water molecules can function as both acids and bases. One water molecule (acting as a base) can accept a hydrogen ion from a second one (acting as an acid). This will be happening anywhere there is even a trace of water - it doesn't have to be pure. A hydroxonium ion and a hydroxide ion are formed.
However, the hydroxonium ion is a very strong acid, and the hydroxide ion is a very strong base. As fast as they are formed, they react to poduce water again. The net effect is that an equilibrium is set up.
At any one time, there are incredibly small numbers of hydroxonium ions and hydroxide ions present. Further down this page, we shall calculate the concentration of hydroxonium ions present in pure water. It turns out to be 1.00 x 10-7 mol dm-3 at room temperature. You may well find this equilibrium written in a simplified form:
This is OK provided you remember that H+(aq) actually refers to a hydroxonium ion.
Defining the ionic product for water, Kw
Kw is essentially just an equilibrium constant for the reactions shown. You may meet it in two forms:
Based on the fully written equilibrium . . .
. . . or on the simplified equilibrium:
You may find them written with or without the state symbols. Whatever version you come across, they all mean exactly the same thing!
You may wonder why the water isn't written on the bottom of these equilibrium constant expressions. So little of the water is ionised at any one time, that its concentration remains virtually unchanged - a constant. Kw is defined to avoid making the expression unnecessarily complicated by including another constant in it.
The value of Kw
Like any other equilibrium constant, the value of Kw varies with temperature. Its value is usually taken to be 1.00 x 10-14 mol2 dm-6 at room temperature. In fact, this is its value at a bit less than 25°C.
pKw
The relationship between Kw and pKw is exactly the same as that between Ka and pKa, or [H+] and pH.
The Kw value of 1.00 x 10-14 mol2 dm-6 at room temperature gives you a pKw value of 14. Try it on your calculator! Notice that pKw doesn't have any units.
The pH of pure water
Why does pure water have a pH of 7?
That question is actually misleading! In fact, pure water only has a pH of 7 at a particular temperature - the temperature at which the Kw value is 1.00 x 10-14 mol2 dm-6.
This is how it comes about:
To find the pH you need first to find the hydrogen ion concentration (or hydroxonium ion concentration - it's the same thing). Then you convert it to pH. In pure water at room temperature the Kw value tells you that:
[H+] [OH-] = 1.00 x 10-14
But in pure water, the hydrogen ion (hydroxonium ion) concentration must be equal to the hydroxide ion concentration. For every hydrogen ion formed, there is a hydroxide ion formed as well. That means that you can replace the [OH-] term in the Kw expression by another [H+].
[H+]2 = 1.00 x 10-14
Taking the square root of each side gives:
[H+] = 1.00 x 10-7 mol dm-3
Converting that into pH:
pH = - log10 [H+]
pH = 7
That's where the familiar value of 7 comes from.
The variation of the pH of pure water with temperature
The formation of hydrogen ions (hydroxonium ions) and hydroxide ions from water is an endothermic process. Using the simpler version of the equilibrium:
The forward reaction absorbs heat.
According to Le Chatelier's Principle, if you make a change to the conditions of a reaction in dynamic equilibrium, the position of equilibrium moves to counter the change you have made. According to Le Chatelier, if you increase the temperature of the water, the equilibrium will move to lower the temperature again. It will do that by absorbing the extra heat.
That means that the forward reaction will be favored, and more hydrogen ions and hydroxide ions will be formed. The effect of that is to increase the value of Kw as temperature increases. The table below shows the effect of temperature on Kw. For each value of Kw, a new pH has been calculated using the same method as above. It might be useful if you were to check these pH values yourself.
T (°C) Kw (mol2 dm-6) pH
0 0.114 x 10-14 7.47
10 0.293 x 10-14 7.27
20 0.681 x 10-14 7.08
25 1.008 x 10-14 7.00
30 1.471 x 10-14 6.92
40 2.916 x 10-14 6.77
50 5.476 x 10-14 6.63
100 51.3 x 10-14 6.14
You can see that the pH of pure water falls as the temperature increases.
Caution
If the pH falls as temperature increases, does this mean that water becomes more acidic at higher temperatures? NO!
A solution is acidic if there is an excess of hydrogen ions over hydroxide ions. In the case of pure water, there are always the same number of hydrogen ions and hydroxide ions. That means that the water remains neutral - even if its pH changes.
The problem is that we are all so familiar with 7 being the pH of pure water, that anything else feels really strange. Remember that you calculate the neutral value of pH from Kw. If that changes, then the neutral value for pH changes as well.
At 100°C, the pH of pure water is 6.14. That is the neutral point on the pH scale at this higher temperature. A solution with a pH of 7 at this temperature is slightly alkaline because its pH is a bit higher than the neutral value of 6.14. Similarly, you can argue that a solution with a pH of 7 at 0°C is slightly acidic, because its pH is a bit lower than the neutral value of 7.47 at this temperature.
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This page explains the terms strong and weak as applied to bases. As a part of this it defines and explains Kb and pKb. The usual way of comparing the strengths of bases is to see how readily they produce hydroxide ions in solution. This may be because they already contain hydroxide ions, or because they take hydrogen ions from water molecules to produce hydroxide ions.
Strong bases
A strong base is something like sodium hydroxide or potassium hydroxide which is fully ionic. You can think of the compound as being 100% split up into metal ions and hydroxide ions in solution.
Each mole of sodium hydroxide dissolves to give a mole of hydroxide ions in solution.
Some strong bases like calcium hydroxide aren't very soluble in water. That doesn't matter - what does dissolve is still 100% ionised into calcium ions and hydroxide ions. Calcium hydroxide still counts as a strong base because of that 100% ionisation.
Working out the pH of a strong base
Remember that:
\[pH = -\log_{10} [H^+]\]
Since pH is a measure of hydrogen ion concentration, how can a solution which contains hydroxide ions have a pH? To understand this, you need to know about the ionic product for water. Wherever there is water, an equilibrium is set up. Using the simplified version of this equilibrium:
\[ H_2O_{(l)} \rightleftharpoons H^+_{(aq)} + OH^-_{(aq)}\]
In the presence of extra hydroxide ions from, say, sodium hydroxide, the equilibrium is still there, but the position of equilibrium has been shifted well to the left according to Le Chatelier's Principle. There will be far fewer hydrogen ions than there are in pure water, but there will still be hydrogen ions present. The pH is a measure of the concentration of these.
An outline of the method of working out the pH of a strong base
• Work out the concentration of the hydroxide ions.
• Use Kw to work out the hydrogen ion concentration.
• Convert the hydrogen ion concentration to a pH.
Example 1
What is the pH of 0.500 mol dm-3 sodium hydroxide solution:?
Solution
Because the sodium hydroxide is fully ionic, each mole of it gives that same number of moles of hydroxide ions in solution.
[OH-] = 0.500 mol dm-3
Now you use the value of Kw at the temperature of your solution. You normally take this as 1.00 x 10-14 mol2 dm-6.
[H+] [OH-] = 1.00 x 10-14
This is true whether the water is pure or not. In this case we have a value for the hydroxide ion concentration. Substituting that gives:
[H+] x 0.500 = 1.00 x 10-14
If you solve that for [H+], and then convert it into pH, you get a pH of 13.7.
Weak bases
Ammonia is a typical weak base. Ammonia itself obviously doesn't contain hydroxide ions, but it reacts with water to produce ammonium ions and hydroxide ions.
\[ NH_{3(aq)} + H_2O_{(l)} \rightleftharpoons NH^+_{4(aq)} + OH^-_{(aq)}\]
However, the reaction is reversible, and at any one time about 99% of the ammonia is still present as ammonia molecules. Only about 1% has actually produced hydroxide ions. A weak base is one which doesn't convert fully into hydroxide ions in solution. When a weak base reacts with water, the position of equilibrium varies from base to base. The further to the left it is, the weaker the base.
You can get a measure of the position of an equilibrium by writing an equilibrium constant for the reaction. The lower the value for the constant, the more the equilibrium lies to the left. In this case the equilibrium constant is called Kb. This is defined as:
\[ K_b = \dfrac{[B:H^+][OH^-]}{[B:]}\]
The relationship between Kb and pKb is exactly the same as all the other "p" terms in this topic:
\[pK_b = -\log_{10} K_b\]
The table shows some values for \(K_b\) and \(pK_b\) for some weak bases.
base Kb (mol dm-3) pKb
C6H5NH2 4.17 x 10-10 9.38
NH3 1.78 x 10-5 4.75
CH3NH2 4.37 x 10-4 3.36
CH3CH2NH2 5.37 x 10-4 3.27
As you go down the table, the value of Kb is increasing. That means that the bases are getting stronger. As Kb gets bigger, pKb gets smaller. The lower the value of pKb, the stronger the base. This is exactly in line with the corresponding term for acids, pKa - the smaller the value, the stronger the acid.
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This page describes how simple acid-base indicators work, and how to choose the right one for a particular titration.
Indicators as weak acids
Litmus
Litmus is a weak acid and is one of the oldest forms of a pH indicator and is used to test materials for acidity.
Chemical structure of 7-hydroxyphenoxazone, the chromophore of litmus components.
It has a seriously complicated molecule which we will simplify to HLit. The "H" is the proton which can be given away to something else. The "Lit" is the rest of the weak acid molecule. There will be an equilibrium established when this acid dissolves in water. Taking the simplified version of this equilibrium:
The un-ionised litmus is red, whereas the ion is blue. Now use Le Chatelier's Principle to work out what would happen if you added hydroxide ions or some more hydrogen ions to this equilibrium.
Adding hydroxide ions:
Adding hydrogen ions:
If the concentrations of HLit and Lit - are equal: At some point during the movement of the position of equilibrium, the concentrations of the two colors will become equal. The color you see will be a mixture of the two.
The reason for the inverted commas around "neutral" is that there is no reason why the two concentrations should become equal at pH 7. For litmus, it so happens that the 50 / 50 color does occur at close to pH 7 - that's why litmus is commonly used to test for acids and alkalis. As you will see below, that isn't true for other indicators.
Methyl orange
Methyl orange is one of the indicators commonly used in titrations. In an alkaline solution, methyl orange is yellow and the structure is:
Now, you might think that when you add an acid, the hydrogen ion would be picked up by the negatively charged oxygen. That's the obvious place for it to go. Not so!
In fact, the hydrogen ion attaches to one of the nitrogens in the nitrogen-nitrogen double bond to give a structure which might be drawn like this:
You have the same sort of equilibrium between the two forms of methyl orange as in the litmus case - but the colors are different.
You should be able to work out for yourself why the color changes when you add an acid or an alkali. The explanation is identical to the litmus case - all that differs are the colors.
In the methyl orange case, the half-way stage where the mixture of red and yellow produces an orange color happens at pH 3.7 - nowhere near neutral. This will be explored further down this page.
Phenolphthalein
Phenolphthalein is another commonly used indicator for titrations, and is another weak acid.
In this case, the weak acid is colorless and its ion is bright pink. Adding extra hydrogen ions shifts the position of equilibrium to the left, and turns the indicator colorless. Adding hydroxide ions removes the hydrogen ions from the equilibrium which tips to the right to replace them - turning the indicator pink.
The half-way stage happens at pH 9.3. Since a mixture of pink and colorless is simply a paler pink, this is difficult to detect with any accuracy!
The pH range of indicators
The importance of pKind
Think about a general indicator, HInd - where "Ind" is all the rest of the indicator apart from the hydrogen ion which is given away:
$HInd_{(aq)} \rightleftharpoons H^+_{(aq)} + Ind^-_{(aq)} \tag{1.6.1}$
Because this is just like any other weak acid, you can write an expression for $K_a$ for it. We will call it $K_{ind}$ to stress that we are talking about the indicator.
$K_{ind} = \dfrac{[H^+][Ind^-]}{[HInd]} \tag{1.6.2}$
Think of what happens half-way through the color change. At this point the concentrations of the acid and its ion are equal.
$[Ind^-] = [HInd] \tag{1.6.3}$
In that case, they will cancel out of the Kind expression.
$K_{ind} = \dfrac{[H^+]\cancel{[Ind^-]}}{\cancel{[HInd]}} \tag{1.6.4}$
$K_{ind} = [H^+] \tag{1.6.5}$
You can use this to work out what the pH is at this half-way point. If you re-arrange the last equation so that the hydrogen ion concentration is on the left-hand side, and then convert to pH and pKind, you get:
$[H^+] = K_{ind} \tag{1.6.5}$
$pH = -\log_{10} [H^+] = -\log_{10} K_{ind} = pK_{ind} \tag{1.6.6}$
That means that the end point for the indicator depends entirely on what its pKind value is. For the indicators we've looked at above, these are:
indicator pKind
litmus 6.5
methyl orange 3.7
phenolphthalein 9.3
The pH range of indicators
Indicators don't change color sharply at one particular pH (given by their pKind). Instead, they change over a narrow range of pH. Assume the equilibrium is firmly to one side, but now you add something to start to shift it. As the equilibrium shifts, you will start to get more and more of the second color formed, and at some point the eye will start to detect it. For example, suppose you had methyl orange in an alkaline solution so that the dominant color was yellow. Now start to add acid so that the equilibrium begins to shift. At some point there will be enough of the red form of the methyl orange present that the solution will begin to take on an orange tint. As you go on adding more acid, the red will eventually become so dominant that you can no longer see any yellow.
There is a gradual smooth change from one color to the other, taking place over a range of pH. As a rough "rule of thumb", the visible change takes place about 1 pH unit either side of the pKind value. The exact values for the three indicators we've looked at are:
indicator pKind pH range
litmus 6.5 5 - 8
methyl orange 3.7 3.1 - 4.4
phenolphthalein 9.3 8.3 - 10.0
The litmus color change happens over an unusually wide range, but it is useful for detecting acids and alkalis in the lab because it changes color around pH 7. Methyl orange or phenolphthalein would be less useful.
This is more easily seen diagramatically.
For example, methyl orange would be yellow in any solution with a pH greater than 4.4. It couldn't distinguish between a weak acid with a pH of 5 or a strong alkali with a pH of 14.
Choosing indicators for titrations
Remember that the equivalence point of a titration is where you have mixed the two substances in exactly equation proportions. You obviously need to choose an indicator which changes color as close as possible to that equivalence point. That varies from titration to titration.
Strong acid vs. strong base
The next diagram shows the pH curve for adding a strong acid to a strong base. Superimposed on it are the pH ranges for methyl orange and phenolphthalein.
You can see that neither indicator changes color at the equivalence point. However, the graph is so steep at that point that there will be virtually no difference in the volume of acid added whichever indicator you choose. However, it would make sense to titrate to the best possible color with each indicator. If you use phenolphthalein, you would titrate until it just becomes colorless (at pH 8.3) because that is as close as you can get to the equivalence point. On the other hand, using methyl orange, you would titrate until there is the very first trace of orange in the solution. If the solution becomes red, you are getting further from the equivalence point.
Strong acid vs. weak base
This time it is obvious that phenolphthalein would be completely useless. However, methyl orange starts to change from yellow towards orange very close to the equivalence point. You have to choose an indicator which changes color on the steep bit of the curve.
Weak acid vs. strong base
This time, the methyl orange is hopeless! However, the phenolphthalein changes color exactly where you want it to.
Weak acid vs. weak base
The curve is for a case where the acid and base are both equally weak - for example, ethanoic acid and ammonia solution. In other cases, the equivalence point will be at some other pH.
You can see that neither indicator is any use. Phenolphthalein will have finished changing well before the equivalence point, and methyl orange falls off the graph altogether. It may be possible to find an indicator which starts to change or finishes changing at the equivalence point, but because the pH of the equivalence point will be different from case to case, you can't generalise. On the whole, you would never titrate a weak acid and a weak base in the presence of an indicator.
Sodium carbonate solution and dilute hydrochloric acid
This is an interesting special case. If you use phenolphthalein or methyl orange, both will give a valid titration result - but the value with phenolphthalein will be exactly half the methyl orange one.
It so happens that the phenolphthalein has finished its color change at exactly the pH of the equivalence point of the first half of the reaction in which sodium hydrogencarbonate is produced.
$Na_2CO_{3(aq)} + HCl_{(aq)} \rightarrow NaCl_{(aq)} + NaHCO_{3(aq)} \tag{1.6.1}$
The methyl orange changes color at exactly the pH of the equivalence point of the second stage of the reaction.
$NaHCO_{3(aq)} + HCl_{(aq)} \rightarrow NaCl_{(aq)} + CO_{2(g)} + H_2O_{(l)} \tag{1.6.2}$
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This page describes simple acidic and alkaline buffer solutions and explains how they work.
What is a buffer solution?
A buffer solution is one which resists changes in pH when small quantities of an acid or an alkali are added to it.
• Acidic buffer solutions: An acidic buffer solution is simply one which has a pH less than 7. Acidic buffer solutions are commonly made from a weak acid and one of its salts - often a sodium salt. A common example would be a mixture of ethanoic acid and sodium ethanoate in solution. In this case, if the solution contained equal molar concentrations of both the acid and the salt, it would have a pH of 4.76. It wouldn't matter what the concentrations were, as long as they were the same. You can change the pH of the buffer solution by changing the ratio of acid to salt, or by choosing a different acid and one of its salts.
• Alkaline buffer solutions: An alkaline buffer solution has a pH greater than 7. Alkaline buffer solutions are commonly made from a weak base and one of its salts. A frequently used example is a mixture of ammonia solution and ammonium chloride solution. If these were mixed in equal molar proportions, the solution would have a pH of 9.25. Again, it doesn't matter what concentrations you choose as long as they are the same.
How do buffer solutions work?
A buffer solution has to contain things which will remove any hydrogen ions or hydroxide ions that you might add to it - otherwise the pH will change. Acidic and alkaline buffer solutions achieve this in different ways.
Acidic buffer solutions
We'll take a mixture of ethanoic acid and sodium ethanoate as typical. Ethanoic acid is a weak acid, and the position of this equilibrium will be well to the left:
Adding sodium ethanoate to this adds lots of extra ethanoate ions. According to Le Chatelier's Principle, that will tip the position of the equilibrium even further to the left.
The solution will therefore contain these important things:
• lots of un-ionized ethanoic acid;
• lots of ethanoate ions from the sodium ethanoate;
• enough hydrogen ions to make the solution acidic.
Other things (like water and sodium ions) which are present aren't important to the argument.
Adding an acid to this buffer solution
The buffer solution must remove most of the new hydrogen ions otherwise the pH would drop markedly. Hydrogen ions combine with the ethanoate ions to make ethanoic acid. Although the reaction is reversible, since the ethanoic acid is a weak acid, most of the new hydrogen ions are removed in this way.
Since most of the new hydrogen ions are removed, the pH won't change very much - but because of the equilibria involved, it will fall a little bit.
Adding an alkali to this buffer solution
Alkaline solutions contain hydroxide ions and the buffer solution removes most of these. This time the situation is a bit more complicated because there are two processes which can remove hydroxide ions. Removal by reacting with ethanoic acid
The most likely acidic substance which a hydroxide ion is going to collide with is an ethanoic acid molecule. They will react to form ethanoate ions and water.
Because most of the new hydroxide ions are removed, the pH doesn't increase very much. Removal of the hydroxide ions by reacting with hydrogen ions Remember that there are some hydrogen ions present from the ionization of the ethanoic acid.
Hydroxide ions can combine with these to make water. As soon as this happens, the equilibrium tips to replace them. This keeps on happening until most of the hydroxide ions are removed.
Again, because you have equilibria involved, not all of the hydroxide ions are removed - just most of them. The water formed re-ionizes to a very small extent to give a few hydrogen ions and hydroxide ions.
Alkaline buffer solutions
We'll take a mixture of ammonia and ammonium chloride solutions as typical. Ammonia is a weak base, and the position of this equilibrium will be well to the left:
Adding ammonium chloride to this adds lots of extra ammonium ions. According to Le Chatelier's Principle, that will tip the position of the equilibrium even further to the left. The solution will therefore contain these important things:
• lots of unreacted ammonia;
• lots of ammonium ions from the ammonium chloride;
• enough hydroxide ions to make the solution alkaline.
• Other things (like water and chloride ions) which are present aren't important to the argument.
Adding an acid to this buffer solution
There are two processes which can remove the hydrogen ions that you are adding.
Removal by reacting with ammonia
The most likely basic substance which a hydrogen ion is going to collide with is an ammonia molecule. They will react to form ammonium ions.
Most, but not all, of the hydrogen ions will be removed. The ammonium ion is weakly acidic, and so some of the hydrogen ions will be released again.
Removal of the hydrogen ions by reacting with hydroxide ions
Remember that there are some hydroxide ions present from the reaction between the ammonia and the water.
Hydrogen ions can combine with these hydroxide ions to make water. As soon as this happens, the equilibrium tips to replace the hydroxide ions. This keeps on happening until most of the hydrogen ions are removed.
Again, because you have equilibria involved, not all of the hydrogen ions are removed - just most of them.
Adding an alkali to this buffer solution
The hydroxide ions from the alkali are removed by a simple reaction with ammonium ions.
Because the ammonia formed is a weak base, it can react with the water - and so the reaction is slightly reversible. That means that, again, most (but not all) of the the hydroxide ions are removed from the solution.
Calculations involving buffer solutions
Acidic buffer solutions
This is easier to see with a specific example. Remember that an acid buffer can be made from a weak acid and one of its salts.
Let's suppose that you had a buffer solution containing 0.10 mol dm-3 of ethanoic acid and 0.20 mol dm-3 of sodium ethanoate. How do you calculate its pH? In any solution containing a weak acid, there is an equilibrium between the un-ionized acid and its ions. So for ethanoic acid, you have the equilibrium:
The presence of the ethanoate ions from the sodium ethanoate will have moved the equilibrium to the left, but the equilibrium still exists.
That means that you can write the equilibrium constant, Ka, for it:
Where you have done calculations using this equation previously with a weak acid, you will have assumed that the concentrations of the hydrogen ions and ethanoate ions were the same. Every molecule of ethanoic acid that splits up gives one of each sort of ion.
That's no longer true for a buffer solution:
If the equilibrium has been pushed even further to the left, the number of ethanoate ions coming from the ethanoic acid will be completely negligible compared to those from the sodium ethanoate. We therefore assume that the ethanoate ion concentration is the same as the concentration of the sodium ethanoate - in this case, 0.20 mol dm-3.
In a weak acid calculation, we normally assume that so little of the acid has ionized that the concentration of the acid at equilibrium is the same as the concentration of the acid we used. That is even more true now that the equilibrium has been moved even further to the left.
So the assumptions we make for a buffer solution are:
Now, if we know the value for Ka, we can calculate the hydrogen ion concentration and therefore the pH.
Ka for ethanoic acid is 1.74 x 10-5 mol dm-3.
Remember that we want to calculate the pH of a buffer solution containing 0.10 mol dm-3 of ethanoic acid and 0.20 mol dm-3 of sodium ethanoate.
Then all you have to do is to find the pH using the expression
pH = -log10 [H+]
You will still have the value for the hydrogen ion concentration on your calculator, so press the log button and ignore the negative sign (to allow for the minus sign in the pH expression).
You should get an answer of 5.1 to two significant figures. You can't be more accurate than this, because your concentrations were only given to two figures.
You could, of course, be asked to reverse this and calculate in what proportions you would have to mix ethanoic acid and sodium ethanoate to get a buffer solution of some desired pH. It is no more difficult than the calculation we have just looked at.
Suppose you wanted a buffer with a pH of 4.46. If you un-log this to find the hydrogen ion concentration you need, you will find it is 3.47 x 10-5 mol dm-3.
Feed that into the Ka expression.
All this means is that to get a solution of pH 4.46, the concentration of the ethanoate ions (from the sodium ethanoate) in the solution has to be 0.5 times that of the concentration of the acid. All that matters is that ratio.
In other words, the concentration of the ethanoate has to be half that of the ethanoic acid.
One way of getting this, for example, would be to mix together 10 cm3 of 1.0 mol dm-3 sodium ethanoate solution with 20 cm3 of 1.0 mol dm-3 ethanoic acid. Or 10 cm3 of 1.0 mol dm-3 sodium ethanoate solution with 10 cm3 of 2.0 mol dm-3 ethanoic acid. And there are all sorts of other possibilities.
Alkaline buffer solutions
We are talking here about a mixture of a weak base and one of its salts - for example, a solution containing ammonia and ammonium chloride. The modern, and easy, way of doing these calculations is to re-think them from the point of view of the ammonium ion rather than of the ammonia solution. Once you have taken this slightly different view-point, everything becomes much the same as before.
So how would you find the pH of a solution containing 0.100 mol dm-3 of ammonia and 0.0500 mol dm-3 of ammonium chloride? The mixture will contain lots of unreacted ammonia molecules and lots of ammonium ions as the essential ingredients. The ammonium ions are weakly acidic, and this equilibrium is set up whenever they are in solution in water:
\[ NH^+_{4 (aq)} \rightleftharpoons NH_{3(aq)} + H^+_{(aq)}\]
You can write a Ka expression for the ammonium ion, and make the same sort of assumptions as we did in the previous case:
The presence of the ammonia in the mixture forces the equilibrium far to the left. That means that you can assume that the ammonium ion concentration is what you started off with in the ammonium chloride, and that the ammonia concentration is all due to the added ammonia solution.
The value for Ka for the ammonium ion is 5.62 x 10-10 mol dm-3. Remember that we want to calculate the pH of a buffer solution containing 0.100 mol dm-3 of ammonia and 0.0500 mol dm-3 of ammonium chloride. Just put all these numbers in the Ka expression, and do the sum:
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This page describes how pH changes during various acid-base titrations.
Sorting out some confusing terms
When you carry out a simple acid-base titration, you use an indicator to tell you when you have the acid and alkali mixed in exactly the right proportions to "neutralise" each other. When the indicator changes color, this is often described as the end point of the titration. In an ideal world, the color change would happen when you mix the two solutions together in exactly equation proportions. That particular mixture is known as the equivalence point.
For example, if you were titrating sodium hydroxide solution with hydrochloric acid, both with a concentration of 1 mol dm-3, 25 cm3 of sodium hydroxide solution would need exactly the same volume of the acid - because they react 1 : 1 according to the equation.
In this particular instance, this would also be the neutral point of the titration, because sodium chloride solution has a pH of 7.
But that isn't necessarily true of all the salts you might get formed. For example, if you titrate ammonia solution with hydrochloric acid, you would get ammonium chloride formed. The ammonium ion is slightly acidic, and so pure ammonium chloride has a slightly acidic pH. That means that at the equivalence point (where you had mixed the solutions in the correct proportions according to the equation), the solution wouldn't actually be neutral. To use the term "neutral point" in this context would be misleading. Similarly, if you titrate sodium hydroxide solution with ethanoic acid, at the equivalence point the pure sodium ethanoate formed has a slightly alkaline pH because the ethanoate ion is slightly basic.
To summarize:
• The term "neutral point" is best avoided.
• The term "equivalence point" means that the solutions have been mixed in exactly the right proportions according to the equation.
• The term "end point" is where the indicator changes color. As you will see on the page about indicators, that isn't necessarily exactly the same as the equivalence point.
Monoprotic Titration curves
All the following titration curves are based on both acid and alkali having a concentration of 1 mol dm-3. In each case, you start with 25 cm3 of one of the solutions in the flask, and the other one in a burette. Although you normally run the acid from a burette into the alkali in a flask, you may need to know about the titration curve for adding it the other way around as well. Alternative versions of the curves have been described in most cases.
Titration curves for strong acid vs. strong base
We'll take hydrochloric acid and sodium hydroxide as typical of a strong acid and a strong base.
Running acid into the alkali
You can see that the pH only falls a very small amount until quite near the equivalence point. Then there is a really steep plunge. If you calculate the values, the pH falls all the way from 11.3 when you have added 24.9 cm3 to 2.7 when you have added 25.1 cm3.
Running alkali into the acid
This is very similar to the previous curve except, of course, that the pH starts off low and increases as you add more sodium hydroxide solution.
Again, the pH doesn't change very much until you get close to the equivalence point. Then it surges upwards very steeply.
Titration curves for strong acid vs. weak base
This time we are going to use hydrochloric acid as the strong acid and ammonia solution as the weak base.
Running acid into the alkali
Because you have got a weak base, the beginning of the curve is obviously going to be different. However, once you have got an excess of acid, the curve is essentially the same as before.
At the very beginning of the curve, the pH starts by falling quite quickly as the acid is added, but the curve very soon gets less steep. This is because a buffer solution is being set up - composed of the excess ammonia and the ammonium chloride being formed.
Notice that the equivalence point is now somewhat acidic ( a bit less than pH 5), because pure ammonium chloride isn't neutral. However, the equivalence point still falls on the steepest bit of the curve. That will turn out to be important in choosing a suitable indicator for the titration.
Running alkali into the acid
At the beginning of this titration, you have an excess of hydrochloric acid. The shape of the curve will be the same as when you had an excess of acid at the start of a titration running sodium hydroxide solution into the acid.
It is only after the equivalence point that things become different.
A buffer solution is formed containing excess ammonia and ammonium chloride. This resists any large increase in pH - not that you would expect a very large increase anyway, because ammonia is only a weak base.
Titration curves for weak acid v strong base
We'll take ethanoic acid and sodium hydroxide as typical of a weak acid and a strong base.
Running acid into the alkali
For the first part of the graph, you have an excess of sodium hydroxide. The curve will be exactly the same as when you add hydrochloric acid to sodium hydroxide. Once the acid is in excess, there will be a difference.
Past the equivalence point you have a buffer solution containing sodium ethanoate and ethanoic acid. This resists any large fall in pH.
Running alkali into the acid
The start of the graph shows a relatively rapid rise in pH but this slows down as a buffer solution containing ethanoic acid and sodium ethanoate is produced. Beyond the equivalence point (when the sodium hydroxide is in excess) the curve is just the same as that end of the HCl - NaOH graph.
Titration curves for weak acid v weak base
The common example of this would be ethanoic acid and ammonia.
It so happens that these two are both about equally weak - in that case, the equivalence point is approximately pH 7.
Running acid into the alkali
This is really just a combination of graphs you have already seen. Up to the equivalence point it is similar to the ammonia - HCl case. After the equivalence point it is like the end of the ethanoic acid - NaOH curve.
Notice that there isn't any steep bit on this graph. Instead, there is just what is known as a "point of inflexion". That lack of a steep bit means that it is difficult to do a titration of a weak acid against a weak base.
A summary of the important curves
The way you normally carry out a titration involves adding the acid to the alkali. Here are reduced versions of the graphs described above so that you can see them all together.
Polyprotic Titration Curves
Adding hydrochloric acid to sodium carbonate solution
The overall equation for the reaction between sodium carbonate solution and dilute hydrochloric acid is:
If you had the two solutions of the same concentration, you would have to use twice the volume of hydrochloric acid to reach the equivalence point - because of the 1 : 2 ratio in the equation.
Suppose you start with 25 cm3 of sodium carbonate solution, and that both solutions have the same concentration of 1 mol dm-3. That means that you would expect the steep drop in the titration curve to come after you had added 50 cm3 of acid. The actual graph looks like this:
The graph is more complicated than you might think - and curious things happen during the titration. You expect carbonates to produce carbon dioxide when you add acids to them, but in the early stages of this titration, no carbon dioxide is given off at all. Then - as soon as you get past the half-way point in the titration - lots of carbon dioxide is suddenly released.
The graph is showing two end points - one at a pH of 8.3 (little more than a point of inflection), and a second at about pH 3.7. The reaction is obviously happening in two distinct parts.
In the first part, complete at A in the diagram, the sodium carbonate is reacting with the acid to produce sodium hydrogencarbonate:
You can see that the reaction doesn't produce any carbon dioxide. In the second part, the sodium hydrogencarbonate produced goes on to react with more acid - giving off lots of CO2.
That reaction is finished at B on the graph. It is possible to pick up both of these end points by careful choice of indicator. That is explained on the separate page on indicators.
Adding sodium hydroxide solution to dilute ethanedioic acid
Ethanedioic acid was previously known as oxalic acid. It is a diprotic acid, which means that it can give away 2 protons (hydrogen ions) to a base. Something which can only give away one (like HCl) is known as a monoprotic acid.
The reaction with sodium hydroxide takes place in two stages because one of the hydrogens is easier to remove than the other. The two successive reactions are:
If you run sodium hydroxide solution into ethanedioic acid solution, the pH curve shows the end points for both of these reactions.
The curve is for the reaction between sodium hydroxide and ethanedioic acid solutions of equal concentrations.
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• Balanced Equations and Equilibrium Constants
In a balanced chemical equation, the total number of atoms of each element present is the same on both sides of the equation. Stoichiometric coefficients are the coefficients required to balance a chemical equation. The coefficients relate to the equilibrium constants because they are used to calculate them.
• Calculating an Equilibrium Concentration
To calculate an equilibrium concentration from an equilibrium constant, an understanding of the concept of equilibrium and how to write an equilibrium constant is required. Equilibrium is a state of dynamic balance where the ratio of the product and reactant concentrations is constant.
• Calculating An Equilibrium Concentrations
• Calculating an Equilibrium Constant, Kp, with Partial Pressures
• Determining the Equilibrium Constant
• Difference Between K And Q
Sometimes it is necessary to determine in which direction a reaction will progress based on initial activities or concentrations. In these situations, the relationship between the reaction quotient, Qc , and the equilibrium constant, Kc , is essential in solving for the net change. With this relationship, the direction in which a reaction will shift to achieve chemical equilibrium, whether to the left or the right, can be easily calculated.
• Dissociation Constant
Page notifications Off Share Table of contents The dissociation constant specifies the tendency of a substance AxBy to reversibly dissociate (separate) in a solution into smaller components A and B.
• Effect of Pressure on Gas-Phase Equilibria
Le Chatelier's Principle states that a system at equilibrium will adjust to relieve stress when there are changes in the concentration of a reactant or product, the partial pressures of components, the volume of the system, and the temperature of reaction.
• Equilibrium Calculations
• Kc
This page defines the equilibrium constant and introduces the equilibrium constant expressed in terms of concentrations, Kc. It assumes familiarity with the concept of dynamic equilibrium, as well as the terms "homogeneous" and "heterogeneous" as applied to chemical reactions. The two types of dynamic equilibria (homogeneous and heterogeneous) are discussed separately below, because the equilibrium constants are defined differently.
• Kp
This page explains equilibrium constants expressed in terms of partial pressures of gases, Kp. It covers an explanation of the terms mole fraction and partial pressure, and looks at Kp for both homogeneous and heterogeneous reactions involving gases. The page assumes that you are already familiar with the concept of an equilibrium constant, and that you know about Kc - an equilibrium constant expressed in terms of concentrations
• Law of Mass Action
• Mass Action Law
• The Equilibrium Constant
The equilibrium constant, K, expresses the relationship between products and reactants of a reaction at equilibrium with respect to a specific unit.This article explains how to write equilibrium constant expressions, and introduces the calculations involved with both the concentration and the partial pressure equilibrium constant.
• The Reaction Quotient
The reaction quotient ( Q ) measures the relative amounts of products and reactants present during a reaction at a particular point in time. The reaction quotient aids in figuring out which direction a reaction is likely to proceed, given either the pressures or the concentrations of the reactants and the products. The Q value can be compared to the Equilibrium Constant, K , to determine the direction of the reaction that is taking place.
Chemical Equilibria
In a balanced chemical equation, the total number of atoms of each element present is the same on both sides of the equation. Stoichiometric coefficients are the coefficients required to balance a chemical equation. These are important because they relate the amounts of reactants used and products formed. The coefficients relate to the equilibrium constants because they are used to calculate them. For this reason, it is important to understand how to balance an equation before using the equation to calculate equilibrium constants.
Introduction
There are several important rules for balancing an equation:
1. An equation can be balanced only by adjusting the coefficients.
2. The equation must include only the reactants and products that participate in the reaction.
3. Never change the equation in order to balance it.
4. If an element occurs in only one compound on each side of the equation, try balancing this element first.
5. When one element exists as a free element, balance this element last.
Example $1$:
$H_2\; (g) + O_2 \; (g) \rightleftharpoons H_2O \; (l) \nonumber$
Because both reactants are in their elemental forms, they can be balanced in either order. Consider oxygen first. There are two atoms on the left and one on the right. Multiply the right by 2 $H_2(g) + O_2(g) \rightleftharpoons 2H_2O(l) \nonumber$
Next, balance hydrogen. There are 4 atoms on the right, and only 2 atoms on the left. Multiply the hydrogen on left by 2 $2H_2(g) + O_2(g) \rightleftharpoons 2H_2O(l)\nonumber$
Check the stoichiometry. Hydrogen: on the left, 2 x 2 = 4; on right 2 x 2= 4. Oxygen: on the left: 1 x 2 = 2; on the right 2 x 1 = 2 . All atoms balance, so the equation is properly balanced.
$2H_2(g) + O_2(g) \rightleftharpoons 2H_2O(l)\nonumber$
Example $2$:
$Al \; (s) + MnSO_4 \; (aq) \rightleftharpoons Al_2(SO_4)_3 + Mn ; (s) \nonumber$
First, consider the SO42- ions. There is one on the left side of the equation, and three on the right side. Add a coefficient of three to the left side. $Al(s) + 3MnSO_4(aq) \rightleftharpoons Al_2(SO_4)_3 + Mn(s) \nonumber$
Next, check the Mn atoms. There is one on the right side, but now there are three on the left side from the previous adjustment. Add a coefficient of three on the right side. $Al(s) + 3MnSO_4(aq) \rightleftharpoons Al_2(SO_4)_3 + 3Mn(s)\nonumber$
Consider Al. There is one atom on the left side and two on the right side. Add a coefficient of two on the left side. Make sure there are equal numbers of each atom on each side. $2Al(s) + 3MnSO_4(aq) \rightleftharpoons Al_2(SO_4)_3 + 3 Mn(s)\nonumber$
Example $3$:
$P_4S_3 + KClO_3 \rightleftharpoons P_2O_5 + KCl + SO_2 \nonumber$
This problem is more difficult. First, look at the P atoms. There are four on the reactant side and two on the product side. Add a coefficient of two to the product side. $P_4S_3 + KClO_3 \rightleftharpoons 2P_2O_5 + KCl + SO_2\nonumber$
Next, consider the sulfur atoms. There are three on the left and one on the right. Add a coefficient of three to the right side. $P_4S_3 + KClO_3 \rightleftharpoons 2P_2O_5 + KCl + 3SO_2\nonumber$
Now look at the oxygen atoms. There are three on the left and 16 on the right. Adding a coefficient of 16 to the KClO3 on the left and the KCl on the right preserves equal numbers of K and Cl atoms, but increases the oxygen. $P_4S_3 + 16KClO_3 \rightleftharpoons 2P_2O_5 + 16KCl + 3 SO_2\nonumber$
Tripling the other three species (P4S3, P2O5, and SO2) balances the rest of the atoms. $3P_4S_3 + 16 KClO_3 \rightleftharpoons 2(3)P_2O_5 + 16KCl + 3(3)SO_2\nonumber$
Simplify and check. $3P_4S_3 + 16KClO_3 \rightleftharpoons 6P_2O_5 + 16KCl + 9SO_2\nonumber$
Chemical Equilibrium
Balanced chemical equations can now be applied to the concept of chemical equilibrium, the state in which the reactants and products experience no net change over time. This occurs when the forward and reverse reactions occur at equal rates. The equilibrium constant is used to determine the amount of each compound that present at equilibrium. Consider a chemical reaction of the following form:
$aA + bB \rightleftharpoons cC + dD\nonumber$
For this equation, the equilibrium constant is defined as:
$K_c = \dfrac{[C]^c [D]^d}{[A]^a [B]^b} \nonumber$
The activities of the products are in the numerator, and those of the reactants are in the denominator. For Kc, the activities are defined as the molar concentrations of the reactants and products ([A], [B] etc.). The lower case letters are the stoichiometric coefficients that balance the equation.
An important aspect of this equation is that pure liquids and solids are not included. This is because their activities are defined as one, so plugging them into the equation has no impact. This is due to the fact that pure liquids and solids have no effect on the physical equilibrium; no matter how much is added, the system can only dissolve as much as the solubility allows. For example, if more sugar is added to a solution after the equilibrium has been reached, the extra sugar will not dissolve (assuming the solution is not heated, which would increase the solubility). Because adding more does not change the equilibrium, it is not accounted for in the expression.
K is related to to the Balanced Chemical Reaction
The following are concepts that apply when adjusting K in response to changes to the corresponding balanced equation:
• When the equation is reversed, the value of K is inverted.
• When the coefficients in a balanced equation are multiplied by a common factor, the equilibrium constant is raised to the power of the corresponding factor.
• When the coefficients in a balanced equation are divided by a common factor, the corresponding root of the equilibrium constant is taken.
• When individual equations are combined, their equilibrium constants are multiplied to obtain the equilibrium constant for the overall reaction.
A balanced equation is very important in using the constant because the coefficients become the powers of the concentrations of products and reactants. If the equation is not balanced, then the constant is incorrect.
K IS ALSO RELATED TO THE BALANCED CHEMICAL EQUATION OF GASES
For gas-phase equilibria, the equation is a function of the reactants' and products' partial pressures. The equilibrium constant is expressed as follows:
$K_p = \dfrac{P_C^c P_D^d}{P_A^a P_B^b} \nonumber$
P represents partial pressure, usually in atmospheres. As before, pure solids and liquids are not accounted for in the equation. Kc and Kp are related by the following equation:
$K_p = K_c(RT)^{\Delta n} \nonumber$
where
$\Delta n = (c+d) - (a+b) \nonumber$
This represents the change in gas molecules. a,b,c and d are the stoichiometric coefficients of the gas molecules found in the balanced equation.
Neither Kc nor Kp have units. This is due to their formal definitions in terms of activities. Their units cancel in the calculation, preventing problems with units in further calculations.
c
$PbI_2 \rightleftharpoons Pb \; (aq) + I \; (aq) \nonumber$
First, balance the equation.
Check the Pb atoms. There is one on each side, so lead can be left alone for now. Next check the I atoms. There are two on the left side and one on the right side. To fix this, add a coefficient of two to the right side. $PbI_2 \rightleftharpoons Pb^{2+}(aq) + 2I^-(aq)\nonumber$
Check to make ensure the numbers are equal. $PbI_2 \rightleftharpoons Pb^{2+}(aq) + 2I^-(aq)\nonumber$
Next, calculate find Kc. Use these concentrations: Pb- 0.3 mol/L, I- 0.2 mol/L, PbI2- 0.5 mol/L
$K_c = \dfrac{(0.3) * (0.2)^2}{(0.5)} \nonumber$
$K_c= 0.024\nonumber$
Note: If the equation had not been balanced when the equilibrium constant was calculated, the concentration of I- would not have been squared. This would have given an incorrect answer.
utomatic number to work, you need to add the "AutoNum" template (preferably at the end) to the page.
Example $5$
$SO_2 \; (g) + O_2 \; (g) \rightleftharpoons SO_3 \; (g) \nonumber$
First, make sure the equation is balanced.
Check to make sure S is equal on both sides. There is one on each side. Next look at the O. There are four on the left side and three on the right. Adding a coefficient to the O2 on the left is ineffective, as the S on right must also be increased. Instead, add a coefficient to the SO2 on the left and the SO3 on the right. $2SO_2 + O_2 \rightleftharpoons 2SO_3\nonumber$
The equation is now balanced. $2SO_2 + O_2 \rightleftharpoons 2SO_3\nonumber$
Calculate Kp. The partial pressures are as follows: SO2- 0.25 atm, O2- 0.45 atm, SO3- 0.3 atm
$K_p = \dfrac{(0.3)^2}{(0.25)^2 \times (0.45)}$
$K_p= 3.2$
Contributors and Attributions
• Charlotte Hutton, Sarah Reno, Curtis Kortemeier
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Chemical_Equilibria/Balanced_Equations_and_Equilibrium_Constants.txt
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The Equilibrium Constant is useful in finding an unknown concentration in a given reaction if the other concentrations are given. This is special, because no matter what concentrations are given the reaction will always have the same ratio of product concentration to reactant concentration.
Calculating An Equilibrium Concentrations
1. Contributors and Attributions
Contributors and Attributions
• Charlotte Hutton, Wen Yu,Mahtab Danai (UCD)
Calculating an Equilibrium Constant Using
The equilibrium constant is known as $K_{eq}$. A common example of $K_{eq}$ is with the reaction:
$aA + bB \rightleftharpoons cC + dD$
$K_{eq} = \dfrac{[C]^c[D]^d}{[A]^a[B]^b}$
where:
• At equilibrium, [A], [B], [C], and [D] are either the molar concentrations or partial pressures.
• Products are in the numerator. Reactants are in the denominator.
• The exponents are the coefficients (a,b,c,d) in the balanced equation.
• Solids and pure liquids are omitted. This is because the activities of pure liquids and solids are equal to one, therefore the numerical value of equilibrium constant is the same with and without the values for pure solids and liquids.
• $K_{eq}$ does not have units. This is because when calculating activity for a specific reactant or product, the units cancel. So when calculating $K_{eq}$, one is working with activity values with no units, which will bring about a $K_{eq}$ value with no units.
Various $K_{eq}$
All the equilibrium constants tell the relative amounts of products and reactants at equilibrium. For any reversible reaction, there can be constructed an equilibrium constant to describe the equilibrium conditions for that reaction. Since there are many different types of reversible reactions, there are many different types of equilibrium constants:
• $K_{c}$: constant for molar concentrations
• $K_{p}$: constant for partial pressures
• $K_{sp}$: solubility product
• $K_{a}$: acid dissociation constant for weak acids
• $K_{b}$: base dissociation constant for weak bases
• $K_{w}$: describes the ionization of water ($K_{w} = 1 \times 10^{-14}$)
Calculating Kp
Referring to equation:
$aA + bB \rightleftharpoons cC + dD$
$K_p = \dfrac{(P_C)^c(P_D)^d}{(P_A)^a(P_B)^b}$
Partial Pressures: In a mixture of gases, it is the pressure an individual gas exerts. The partial pressure is independent of other gases that may be present in a mixture. According to the ideal gas law, partial pressure is inversely proportional to volume. It is also directly proportional to moles and temperature.
Example $1$
At equilibrium in the following reaction at room temperature, the partial pressures of the gases are found to be $P_{N_2}$ = 0.094 atm, $P_{H_2}$ = 0.039 atm, and $P_{NH_3}$ = 0.003 atm.
$\ce{N_2 (g) + 3 H_2 (g) \rightleftharpoons 2 NH_3 (g)} \nonumber$
What is the $K_p$ for the reaction?
Solution
First, write $K_{eq}$ (equilibrium constant expression) in terms of activities.
$K = \dfrac{(a_{NH_3})^2}{(a_{N_2})(a_{H_2})^3} \nonumber$
Then, replace the activities with the partial pressures in the equilibrium constant expression.
$K_p = \dfrac{(P_{NH_3})^2}{(P_{N_2})(P_{H_2})^3} \nonumber$
Finally, substitute the given partial pressures into the equation.
$K_p = \dfrac{(0.003)^2}{(0.094)(0.039)^3} = 1.61 \nonumber$
Example $2$
At equilibrium in the following reaction at 303 K, the total pressure is 0.016 atm while the partial pressure of $P_{H_2}$ is found to be 0.013 atm.
$\ce{3 Fe_2O_3 (s) + H_2 (g) \rightleftharpoons 2 Fe_3O_4 (s) + H_2O (g)} \nonumber$
What is the $K_p$ for the reaction?
Solution
First, calculate the partial pressure for $\ce{H2O}$ by subtracting the partial pressure of $\ce{H2}$ from the total pressure.
\begin{align*} P_{H_2O} &= {P_{total}-P_{H_2}} \[4pt] &= (0.016-0.013) \; atm \[4pt] &= 0.003 \; atm \end{align*}
Then, write K (equilibrium constant expression) in terms of activities. Remember that solids and pure liquids are ignored.
$K = \dfrac{(a_{H_2O})}{(a_{H_2})}\nonumber$
Then, replace the activities with the partial pressures in the equilibrium constant expression.
$K_p = \dfrac{(P_{H_2O})}{(P_{H_2})}\nonumber$
Finally, substitute the given partial pressures into the equation.
$K_p = \dfrac{(0.003)}{(0.013)} = 0.23 \nonumber$
Example $3$
A flask initially contained hydrogen sulfide at a pressure of 5.00 atm at 313 K. When the reaction reached equilibrium, the partial pressure of sulfur vapor was found to be 0.15 atm.
$\ce{2 H_2S (g) \rightleftharpoons 2 H_2 (g) + S_2 (g) } \nonumber$
What is the $K_p$ for the reaction?
Solution
For this kind of problem, ICE Tables are used.
$\ce{2H2S (g)}$ $\rightleftharpoons$ $\ce{2H2(g)}$ + $\ce{S2(g)}$
Initial Amounts 5.00 atm 0 atm 0 atm
Change in Amounts -0.3 atm +0.3 atm +0.15 atm
Equilibrium Amounts 4.7 atm 0.3 atm 0.15 atm
Now, set up the equilibrium constant expression, $K_p$.
$K_p = \dfrac{(P_{H_2})^2(P_{S_2})}{(P_{H_2S})^2} \nonumber$
Finally, substitute the calculated partial pressures into the equation.
\begin{align*} K_p &= \dfrac{(0.3)^2(0.15)}{(4.7)^2} \[4pt] &= 6.11 \times 10^{-4} \end{align*}
• Bianca Yau
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Chemical_Equilibria/Calculating_An_Equilibrium_Concentrations/Balanced_Equations_And_Equilibrium_Constan.txt
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Le Chatlier's Principle states that when a system at equilibrium undergoes a change in temperature, volume, or amount of moles present in a reactant or product, the system will respond in order to reach equilibrium. Think of a system at equilibrium as a balanced scale (equal weights on both sides) and when one side gains more weight, the scale will have to adjust the other side in order to reach equilibrium.
Introduction
In terms of volume changes within a system at equilibrium, the following applies:
• When there is a decrease in volume, the equilibrium will shift to favor the direction that produces fewer moles of gas.
• When there is an increase in volume, the equilibrium will shift to favor the direction that produces more moles of gas.
It's important to remember that these rules only apply to equations in which gases are involved. If only solids and aqueous solutions are present, volume changes will have no effect on the equilibrium.
Conceptual Questions
1. What happens when you increase the volume in a system at equilibrium that has equal moles of reactants and products and you increase the volume, in which direction will a net change occur in order to restore equilibrium? Decrease?
2. What happens when you increase the volume in an equilibrium system in which there are more moles of reactants than products?
3. What happens when you decrease the volume in an equilibrium system in which there are more moles of products than reactants?
4. What happens when you increase the volume in an equilibrium system in which there are more moles of products than reactants?
5. What happens when you decrease the volume in an equilibrium system in which there are more moles of reactants than products?
Practice Problems
How could one increase the amount of reactants produced in the following equation:
\(N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)\)
a) increase volume b) decrease volume
Explain what would happen in each of the following equations if there is a DECREASE in volume:
\(2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)\)
\(H_2(g) + I_2(g) \rightleftharpoons 2HI(g)\)
\(CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)\)
\(AgCl(s) + \rightleftharpoons Ag^{+1}(aq) + Cl^-(aq)\)
Summary
Le Chatlier's Principle applies to changes in the following:
1. pressure changes
2. concentration changes
3. volume changes
When any system at equilibrium is disturbed, it will attempt to reach a new equilibrium.
Answers
1. Because there is an equal number of moles on both sides of the reaction, an increase in volume will have no effect on the equilibrium and thus there is no shift in the direction. Similarly, when you decrease the volume there is no effect on the equilibrium.
2. Because there are more moles of reactants, an increase in volume will shift the equilibrium to the left in order to favor the reactants.
3. When there is a decrease in volume, the equilibrium will shift towards the side of the reaction with fewer moles. In this case, there are fewer moles of reactants and so the equilibrium will favor the reactants and shift to the left.
4. An increase in volume always favors the direction that produces more moles of gas and because in this case there are more moles of products, the reaction will shift to the right and produce more moles products.
5. Because a decrease in volume always favors the direction that produces fewer moles, this system will shift to right and produce more moles of products.
Answers
1. In order to increase the amount of reactants formed, increase the volume so that it will shift to the right (towards the side with more moles of gas).
2. Effects of a Decrease in Volume
• The products are favored; shift to right.
• Same number of moles on each side of equation; no effect.
• The reactants are favored; shift to left.
• Because no gases are present there is no change.
Outside Links
1. www.chm.davidson.edu/ChemistryApplets/equilibria/volume.html
2. 209.85.173.104/search?q=cache...&client=safari
• Leah Hughes
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Chemical_Equilibria/Calculating_An_Equilibrium_Concentrations/Effect_Of_Volume_Changes_On_Gas-phase_Equi.txt
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When calculating equilibrium constants of solutions, the concentrations of each component in the solution are used to calculate K, which is the equilibrium constant. When working with concentrations, the equilibrium constant is designated as Kc. However, when working with a mixture of gases, concentrations are not used, but instead the gases' partial pressures. By first calculating the equilibrium constant in terms of pressure, designated Kp, Kc can then be calculated by using a simple formula.
How do you begin writing the equilibrium constant for a mixture of gases?
To explain this clearly, let us look at a simple example, the reaction:
2NH3(g) <---> N2(g) + 3H2(g).
1) First, each component of the mixture in the equilibrium constant is written in terms of their activities. The activity of each component is the partial pressure of each component divided by their partial pressure "reference-states." Writing the equilibrium constant in terms of partial pressures requires it to be referred to as Kp.
K = ( (aN2)(aH2)3 / (aNH3)2) eq
aN2 = PN2/Po
aH2 = PH2/Po
aNH3 = PNH3/Po
2) Next, take the activities, showed in terms of pressure, of the products raised to the power of their coefficients, and divided by the activities of the reactants. The activities of the reactants are also raised to the power of their coefficients. To simplify the expression, the partial pressure "reference-state" of each component can be divided out.
Kp = ( (PN2/Po)(PH2/Po)3 / (PNH3/Po)2 ) eq
Po ( (PN2)(PH2)3/(PNH3)2) eq
3) To relate the pressures of this expression to concentrations, the ideal gas law is used. The ideal gas law is PV=nRT, where P is the pressure, V is the volume, n is the number of moles of the substnace, R is the constant 0.08206 L atm/K mol, and T is the temperature in Kelvins. By inverting the ideal gas law equation and solving for the concentration, n/V, or moles per liter, the concentrations of the gases are expressed by their partial pressures divided by RT.
{N2} = n/v = PN2/RT
{H2} = n/V = PH2/RT
{NH3} = n/V = PNH3/RT
4) Now, since we have solved for the concentrations of each gas component, we write them in terms of concentration activities. This is the concentrations of each component divided by the concentration reference state.
aN2 = {N2}/co = (PN2/RT)/co
aH2 = {H2}/co = (PH2/RT)/co
aNH3 = {NH3}/co = (PNH3/RT)/co
The last and final step is entering these activities into the expression we earlier derived in step 2 and dividing out RT.
Kp = Po( ({N2}RT)({H2}RT)3/({NH3}RT)2 ) eq
Kp = (Po/RT)({N2}{H2}3/{NH3}2) eq
This gives the equilibrium constant expression for a mixture of gases (Kp).
Kp and Kc can be related by setting the two equations equal to each other and divided out the reference states. This gives the simple formula:
Kp = Kc(RT)difference in coefficients of gas components only
Please Note:
Po is the partial pressure reference-state and it equals about 1 atm
co is the concentration reference-state and it equals about 1mol/L
eq is the term used to indicate that the components are at equilibrium
{x} is the concentration of the component x where x is a variable used to indicate a substance
Equilibrium constant expression do not include those components in a reaction that are pure solids or liquids (please refer to the corresponding internal link below)
Practice Problems
1) Gas A and Gas B react to form Gas C. The reaction performed can be written as A(g)+B(g)
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```
C(g). At equilibrium, the partial pressures of A, B, and C are 1 atm, 0.50 atm, and 0.75 atm respectively. Find Kp for the reaction.
Solution:
First, write out the equilibrium constant expression for Kp. Kp= Po(
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```
)eq
Remember that Po= 1 bar which is essentially equal to 1 atm. Next, plug in the partial pressures for the corresponding gases.
Kp=
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```
)
=1.5
2) A(g), at 750 mmHg, reacts with B(g), at 760 mmHg, to form C(g). Kp for the reaction is 1.5 X 10-5 and the reaction at equilibrium can be written A(g)+B(g)
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```
2C(g). Find the partial pressure of C.
Solution
First set up the equilibrium constant expression for Kp. Kp= Po(
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```
)eq
C is raised to the power of 2 since 2 is the coefficient of C. Since we know Kp we can subsitute that into the equation. Also, convert the partial pressure of A and B to atm. PA= 750 mmHg X
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```
=0.98684 & PB= 1 atm
1.5 X 105 =
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```
(1.5 X10-5)(0.98684)= C2
sqrt(1.48026X10-5)=C
C= 3.8474147X10-3 or C= 3.9X10-3
3) For the following reaction at equilibrium 2SO2(g)+ O2
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```
2SO3(g) Kp is 3.4 at 1000K. What is Kc for the reaction?
Solution:
We know that Kp=Kc(RT)(Difference in coefficients of gaseous products and reactants)
We also know Kp= 3.4, R is a constant which is 0.08206L atm mol-1K-1, T=1000K
(RT) is raised to the power of (Difference in coefficients of gaseous products and reactants) so since 2SO2(g)+ 1O2
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```
2SO3(g) (Difference in coefficients of gaseous products and reactants)= (2)-(2+1)= -1
Rewrite the equation, Kp=Kc(RT)(Difference in coefficients of gaseous products and reactants), to solve for Kc and subsitute in the values. ^Kc=
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```
Kc= 2.79X102
4) For the following reaction at equilibrium, N2(g)+ 3H2(g)
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```
2NH3(g), Kp is 2.25X10-6. If the values of the partial pressures of N2 and NH3 are 3.5X10-3 and 2.0X10-5 respectively, what is the partial pressure of H2?
Solution
Set up the equilibrium constant expression for Kp. Kp= Kp= Po
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```
eq
Substitute the values in to the correct places.
2.25X10-6=
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```
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```
=
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```
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```
=0.0507936508
Take the cubed root of the number to find the Partial pressure of H2
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```
=0.37 atm
5) 1 mol of A and 1 mol of B are placed in a 2.0 L flask. The following reaction at equilibrium is established at 500K. A(g)+B(g)
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```
C(g) Kp= 15.0 For the equilibrium established, find the partial pressure of C.
Solution
First, we must find the partial pressures of A and B. We know that by using P=
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```
, we can say PA=PB=
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```
=
20.515
Now create an ICE table.
A(g) + B(g)
UndefinedNameError: reference to undefined name 'math' (click for details)
```Callstack:
at (Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Chemical_Equilibria/Calculating_An_Equilibrium_Concentrations/Writing_Equilibrium_Constant_Expressions_Involving_Gases), /content/body/div[2]/p[35]/span/span/span/span/span/span/span, line 1, column 1
```
C(g)
I 20.515 20.515 0
C -x -x +x
E (20.515-x) (20.515-x) (x)
Now set up the equilibrium constant expression for Kp= 15.0=
UndefinedNameError: reference to undefined name 'math' (click for details)
```Callstack:
at (Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Chemical_Equilibria/Calculating_An_Equilibrium_Concentrations/Writing_Equilibrium_Constant_Expressions_Involving_Gases), /content/body/div[2]/p[39]/span/span, line 1, column 1
```
15(20.515-x)2= x Use FOIL
15(420.865225-41.03x+x2)= x distribute & multiply by 15
15x2-615.45x+6312.978375=x subtract the x to make the equation equal to 0
15x2-616.45x+6312.978375=0 Use the quadratic formula
UndefinedNameError: reference to undefined name 'math' (click for details)
```Callstack:
at (Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Chemical_Equilibria/Calculating_An_Equilibrium_Concentrations/Writing_Equilibrium_Constant_Expressions_Involving_Gases), /content/body/div[2]/p[44]/span/span/span[1], line 1, column 2
```
and
UndefinedNameError: reference to undefined name 'math' (click for details)
```Callstack:
at (Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Chemical_Equilibria/Calculating_An_Equilibrium_Concentrations/Writing_Equilibrium_Constant_Expressions_Involving_Gases), /content/body/div[2]/p[44]/span/span/span[2], line 1, column 2
```
a=15 b=-616.45 c= 6312.978375
UndefinedNameError: reference to undefined name 'math' (click for details)
```Callstack:
at (Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Chemical_Equilibria/Calculating_An_Equilibrium_Concentrations/Writing_Equilibrium_Constant_Expressions_Involving_Gases), /content/body/div[2]/p[46]/span/span/span[1], line 1, column 2
```
and
UndefinedNameError: reference to undefined name 'math' (click for details)
```Callstack:
at (Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Chemical_Equilibria/Calculating_An_Equilibrium_Concentrations/Writing_Equilibrium_Constant_Expressions_Involving_Gases), /content/body/div[2]/p[46]/span/span/span[2], line 1, column 2
```
We find that there are two solutions. PC= 19.38 or 21.7
19.38 is the correct solution since 21.7 is too large.
Contributors and Attributions
• Kathryn Rashel
• Lisa Peterson
• Kyle Catabay (UCD)
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Chemical_Equilibria/Calculating_An_Equilibrium_Concentrations/Writing_Equilibrium_Constant_Expressions_I.txt
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To calculate an equilibrium concentration from an equilibrium constant, an understanding of the concept of equilibrium and how to write an equilibrium constant is required. Equilibrium is a state of dynamic balance where the ratio of the product and reactant concentrations is constant.
Calculating an Equilibrium Constant Kp with Partial Pressures
\(K_p\) is the equilibrium constant calculated from the partial pressures of a reaction equation. It is used to express the relationship between product pressures and reactant pressures. It is a unitless number, although it relates the pressures.
Contributors and Attributions
• Sarah Reno (UC Davis)
Determining the Equilibrium Constant 2
"An equilibrium constant expression describes the relationship among the concentrations (or partial pressures) of the substances present in a system at equilibrium" (General Chemistry).
For a given chemical reaction:
$aA + bB \rightleftharpoons cC + dD$
the equilibrium constant expression is as follows:
$K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}$
To calculate the equilibrium constant (also known as the dissociation constant), the concentrations of each species in the reaction at equilibrium must be measured. Consider the general acid dissociation equation:
$HA + H_2O \rightleftharpoons A^- + H_3O^+$
Where HA is the acid, H2O is water, A- is the conjugate base of the acid, and H3O+ is the hydronium ion, a protonated water molecule.
Any equilibrium problem given provides some needed information, including a variety of numbers and/or units. Either Ka or Kb, the dissociation constants for the acid or the base in the equation, respectively, will be provided. For this example, Ka = 2.2 x 10-4 and the initial concentration of HA, the acid in solution is 2 mol L-1.
The equilibrium concentrations are calculated using an ICE Table. An ICE Table is a way to keep all your information organized throughout the process. ICE stands for 'Initial', 'Concentration', and 'Equilibrium'.
$HA + H_2O \rightleftharpoons A^- + H_3O^+$
HA + H2O A- + H3O+
I 2 mols/L n/a 0 0
C -x n/a +x +x
E 2-x n/a x x
In this case,
$K_a = \frac{[A^-][H_3O^+]}{[HA]}$
H2O is ignored in the ICE Table and when calculating the dissociation constant because it is a pure liquid. Any hypothetical solids would also be ignored.
Substituting the values from the "E" row of the table,
$\begin{eqnarray} K_a &=& \frac{(x)(x)}{2-x} \ &=& \frac{x^2}{2-x} \end{eqnarray}$
A simple way to solve this equation is to first estimate the value for x. Assume x is very small, so the quantity 2-x is essentially 2. Keep in mind that the value for Ka is known. The equation simplifies to the following:
$x^2 = 2K_a$
The calculated value for x is very close to 0. It can be plugged into the 2-x term in the original equation to find a more accurate value for x. A few iterations of this process will converge on the correct concentration of H3O+ and, equivalently, A- at equilibrium in this specific case. In this example, the concentration of H3O+ and A- is 0.02086 or 2.09 x 10-2 mol L-1
The relationship between Ka and K b is the following:
$K_W = K_a \times K_b$
Kw is equal to 10-14. If the problem provides a value for Ka but the reaction involves a base, Kb can be calculated by dividing Kw by Ka. The same holds true if Kb is given for an acid dissociation problem.
• Gregory Arch
Determining the Equilibrium Constant
The Equilibrium Constant is unitless. It simply signifies the ratio between the forward and reversal rates of a chemical reaction at equilibrium.
Introduction
There are a few different Equilibrium Constants such as:
• \(K_c\) (dealing with concentration)
• \(K_p\) (dealing with partial pressure)
For the sake of brevity, let us stick to the former, \(K_c\) to follow this Module.
Generally, we follow this equation when dealing with the calculation of the equilibrium constant, \(K_c\):
\(K_c = \dfrac{Products}{Reactants}\)
• Where the concentrations of the Products are multiplied in the numerator (top) and the concentrations of the Reactants are multiplied in the denominator (bottom).
• We separate and denote each chemicals' concentration inside of closed brackets.
Important: We only include the concentrations of aqueous and gaseous substances. Do not include those of solids or liquids!
The Calculation of K
1. Determine whether the corresponding chemical (i.e. HCl, NaCl, etc…) is on the products or reactants side of the chemical equation. Be sure to only include those in the (aq) or (g) states.
2. Observe your ICE table and look at the values in the E row of the table.
3. Based on your observations from step 2, place those values in the general equation stated above (products divided by reactants).
4. Multiply, then divide your answers in order to obtain the equilbrium constant.
This guide assumes you know how to setup and ICE table. If you do not, please refer to this excellent guide: Using ICE Tables. Caution: Remember that we are obtaining the equilibrium constant K, NOT the Q. Q deals with using the initial values from the I row of the ICE table.
That's All? Be Careful…
• If your balanced chemical equation contains a coefficient (i.e., 2, 3, 4, ect…) for a specific chemical (i.e. 2H3O+) then you MUST raise it's concentration to the power of that coefficient. (i.e. [H3O+]2)
Example
Take this general, easy to follow example, where A/B/C are different chemicals and M is molarity (mols/liter).
Question: Calculate the equilibrium constant for a reaction between .100 M A and .300 M B, given that the product, C has a concentration of .200 M and that these values are those when the reaction is in a state of equilbrium. Here is the chemical equation and assume it is balanced.
\(A_{(aq)} + B_{(aq)} \rightleftharpoons 3C_{(aq)}\)
Solution:
Given that the concentrations at equilibrium, we can assume that the ICE table (omit IC as they are not needed here) ends up looking like:
A(aq) + B(aq) \(\rightleftharpoons \) 3C(aq)
0.100 0.300 0.200
Based on the steps mentioned above:
1. A and B are reactants, while C is a product. They are all in the aqueous (aq) state, so their concentrations are important to us.
2. According to the E row of the ICE table (shown), we see the concentrations at equilibrium of A, B and C.
3. Recall:
\(K_c = \dfrac{Products}{Reactants}\)
Thus:
\(K_c = \dfrac{(0.200)^3}{(0.100)(0.300)}\)
Notice that the concentration of C (0.200) is raised to the power of 3, since the number 3 is the coefficient of C.
4. Following basic order of operations of mathematics, first calculating exponential values then multiplying out the products and the reactants, and finally dividing them, we obtain our answer as follows:
\(K_c = 0.267\)
Therefore, we can state that the ratio of the forward and reversal rates of this reaction is 0.267
Extra Info: When Balancing a Chemical Equation, K is Manipulated…
• If you reverse the equation, then invert K. In other words, 1/K.
• If you multiply the entire equation by a number, then raise K to the power of that number you multiplied by.
• If you divide the entire equation by a number, then root K according to that number (i.e. dividing by 2 will cause a square root of K, or dividing by 3 would cause a cube root of K).
External Links
• Senior Chemistry video tutorials are an excellent visual (and comical) tool for students, in my opinion.
• Here is an example of determining the value of K with Senior Chemistry. Enjoy and learn. www.youtube.com/watch?v=9aYe5_Xa2-I
• The General Chemistry text book comes with an access code for it's online material. Use the media resources provided by Pearson Education! They often include excerpts of the text book on their website of important equations and information. wps.prenhall.com/wps/media/access/Pearson_Default/2905/2974919/login.html
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Chemical_Equilibria/Calculating_an_Equilibrium_Concentration.txt
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Sometimes it is necessary to determine in which direction a reaction will progress based on initial activities or concentrations. In these situations, the relationship between the reaction quotient, $Q_c$, and the equilibrium constant, $K_c$, is essential in solving for the net change. With this relationship, the direction in which a reaction will shift to achieve chemical equilibrium, whether to the left or the right, can be easily calculated.
Introduction
$K_c$ can be used calculate the final concentrations at equilibrium for a reaction using an ICE table and the natural progression of the reaction, from left to right or from right to left. However, what if you do not know which way the reaction will progress? A simple relationship between $K_c$ and the reaction quotient, known as $Q_c$, can help. The reaction quotient, $Q$, expresses the relative ratio of products to reactants at a given instant. Using either the initial concentrations or initial activities of all the components of the reaction, the progression of an reaction can easily be determined.
Given the general chemical reaction
$aA +bB \rightleftharpoons gG +hH$
$Q$ may be expressed as the following equations:
$Q= \dfrac{a_{init}^ga_{init}^h}{a_{init}^a a_{init}^b}$
or
$Q_{c}= \dfrac{[G]_{init}^g[H]_{init}^h}{[A]_{init}^a[B]_{init}^b}$
Remember that the concentrations of liquids and solids do not change, so they are excluded from the expression. As shown above, the value of $Q$ can be found by raising the products to the power of their coefficients, or stoichiometric factors, divided by the reactants raised to their coefficients. If the concentration of products in the numerator is much larger than that of reactants in the denominator, $Q$ will be a large value. On the other hand, a small amount of products (small numerator) divided by a large value for the concentration of reactants (large denominator) would result in a small value for Q. The expressions for Q are very similar to those for $K$:
$K= \dfrac{a_G^g a_{H}^h}{a_{A}^a a_{B}^b}$
or
$K_c = \dfrac{[G]^g [H]^h}{[A]^a [B]^b}$
To determine which direction a reaction will go towards, simply compare $Q_c$, the initial concentration ratio, to $K_c$, the equilibrium constant, and evaluate the results.
Q vs. K: What Does It Mean?
When you set $Q$ against $K$, there are five possible relationships:
• $Q=K$
• $Q=0$
• $Q<K$
• $Q= \infty$ and
• $Q>K$.
To properly predict which way a reaction will progress, you must know these relationships.
Situation 1: Q = K
When Q=K, the system is at equilibrium and there is no shift to either the left or the right.
Take, for example, the reversible reaction shown below:
$CO_{(g)}+2H_{2 \; (g)} \rightleftharpoons CH_{3}OH_{(g)}$
The value of Kc at 483 K is 14.5. If Q=14.5, the reaction is in equilibrium and will be no evolution of the reaction either forward or backwards.
Situation 2: Q < K
When Q<K, there are more reactants than products. As a result, some of the reactants will become products, causing the reaction to shift to the right.
Consider again:
$CO_{(g)}+2H_{2 \; (g)} \rightleftharpoons CH_{3}OH_{(g)}$
For Q<K:
$CO_{(g)}+2H_{2 \; (g)} \longrightarrow CH_{3}OH_{(g)}$
so that equilibrium may be established.
Q Equals Zero
If Q=0, then Q is less than K. Therefore, when Q=0, the reaction shifts to the right (forward). An easy way to remember this relationship is by thinking, “once you have nothing, the only thing left to do is to move forward.” If Q equals to zero, the reaction will shift forward (to the right):
$CO_{(g)}+2H_{2 \; (g)} \longrightarrow CH_{3}OH_{(g)}$
Situation 3: Q > K
When Q>K, there are more products than reactants. To decrease the amount of products, the reaction will shift to the left and produce more reactants. For Q>K:
$CO_{(g)}+2H_{2 \; (g)} \longleftarrow CH_{3}OH_{(g)}$
Q Equals Infinity
When Q=∞, the reaction shifts to the left (backwards). This is a variation of when Q>>>K.
$CO_{(g)}+2H_{2 \; (g)} \longleftarrow CH_{3}OH_{(g)}$
Remembering the Relationship Between K and Q
An easy way to remember these relationships is by thinking of the > or < as the mouth of an alligator. The alligator will "eat" in the direction that the reaction shifts as long as $Q$ is written before $K$.
Handy Chart Outlining the Relationships of Q and K
Remembering these simple relationships will aid you to solving for the progression of a reaction. A chart outlining them can be found below.
Predicting the Shift of a Reaction Without Calculations
Depending on what a problem asks of you, sometimes it is unnecessary to make any calculations at all. Take, for example, the now familiar reversible reaction listed below:
$CO_{(g)}+2H_{2 \; (g)} \rightleftharpoons CH_{3}OH_{(g)}$
What do you think will happen if more of the product, methanol (CH3OH), is added? Equilibrium will be disrupted, and the increase in products mean that Q>K. In order to re-establish equilibrium, the reaction will progress to the left, towards the reactants. This means some of the added methanol will break down into carbon monoxide and hydrogen gas.
$CO_{(g)}+2H_{2 \; (g)} \longleftarrow CH_{3}OH_{(g)}$
Now, what if more of the reactants, carbon monoxide and hydrogen gas, are a? You should realize that this would upset the equilibrium. Q<K, because the value for the amount of reactants, or the denominator of the Q expression, has increased. To establish equilibrium again, the reaction will favor the product, so the reaction will progress to the right.
$CO_{(g)}+2H_{2 \; (g)} \longrightarrow CH_{3}OH_{(g)}$
The ideas illustrated above show Le Chatelier's Principle whereby when an equilibrated system is subjected to a change in temperature, pressure, or concentration of a species in the reaction, the system responds by achieving a new equilibrium that partially offsets the impact of the change. Predicting which way a reaction will go can be the easiest thing that you will ever do in chemistry!
Example: Putting It All Together
To properly use the relationship between Q and K, you must know how to set it up. Take, for example, the reaction below:
If you start with 4.00M CH4, 2.00M C2H2, and 3.00M H2, which direction will the reaction progress to reach equilibrium?
Problems
1) Consider this reaction:
$2NOBr_{(g)} \leftrightharpoons 2NO_{(g)}+Br_{2}$
If Kc= 0.0142 and the initial concentrations are 1.0 M NOBr, 0.2M NO, and 0.8M Br2, which way will the reaction progress to reach equilibrium?
2) What is Q and its purpose?
3) Consider the reaction following in equilibrium:
$N_{2}O_{4 \; (g)} \leftrightharpoons 2NO_{2 \; (g)}$
If more N2O4 is added, which way will the reaction proceed?
4) Consider the following reaction:
$CO_{(g)}+Cl_{2 \; (g)} \leftrightharpoons COCl_{2 \; (g)}$
With a Kc of 1.2 x 103 at 668 K, is the reaction in equilibrium when there are 5.00 mol CO(g), 2.00 mol Cl2(g), and 6.00 mol of COCl2(g) in a 3.00L flask? If not, which direction will the reaction progress to reach equilibrium?
5) Consider the following reaction:
$H_{2 \; (g)}+I_{2 \; (g)} \leftrightharpoons 2HI_{(g)}$
If Kc=50.2 at 718 K and the initial concentrations are 0.5 M H2, 0.15M I2, and 0.05M HI, which way will the reaction progress?
6) Consider the following reaction:
$2COF_{2 \; (g)} \leftrightharpoons CO_{2 \; (g)}+CF_{4 \; (g)}$
If Kc= 2.00 at 473 K and the initial concentrations are 2.0 M CO2, 4.0 M CF4, and 0.5 M COF2, which way will the reaction progress?
7) Consider the following reaction:
$2SO_{2 \; (g)}+O_{2 \: (g)} \leftrightharpoons 2SO_{3 \; (g)}$
Kc=100. With the initial masses of 20 g SO2, 13 g O2, and 25 g SO3 in a 5.0 L container, which way will the reaction progress.
Solutions
1) The reactions shifts to the left, towards the reactants.
NOBr= 1M, NO= 0.2M, Br2= 0.8M
$Q_c= \dfrac {[0.2]^2 [0.8]}{[1]^2}$
$Q_c= 0.032$
Therefore, Qc> Kc and the reactions shifts towards the reactants.
2) Q is a reaction quotient, which helps determine if a reaction will shift forward or backwards. As a system approaches towards equilibrium, Q approaches towards K.
3) The reaction will proceed to the right.
4) No, it is not at equilibrium. Since Q<K, the reaction will shift to the right to reach equilibrium.
5) Q = 0.033, so Q<K. The reaction will shift to the right.
6) Q = 32.0, so Q>K. The reaction will shift to the left.
7) Q = 12, so Q<K. The reaction will shift to the right.
Contributors and Attributions
• Rubi Medrano (UCD), Irene Ly (UCD)
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Chemical_Equilibria/Difference_Between_K_And_Q.txt
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The dissociation constant specifies the tendency of a substance MxNy to reversibly dissociate (separate) in a solution (often aqueous) into smaller components M and N:
$M_xN_{y(aq)} + H_2O_{(l)} \rightleftharpoons xM_{(aq)} + yN_{(aq)} \tag{1}$
The dissociation constant is denoted Kd and is calculated by
$K_d =\dfrac{a_M^x·a_N^y}{a_{M_xN_y}·a_{H_2O}} \approx \dfrac{[M]^x[N]^y}{[M_xN_y](1)} \tag{2}$
where A represents the activity of a species, and [M], [N], and [MxNy] are the molar concentrations of the entities M, N, and MxNy. Because water is the solvent, and the solution is assumed to be dilute, the water is assumed to be pure, and the activity of pure water is defined as 1. The activities of the solutes are approximated with molarities. The dissociation constant is an immediate consequence of the law of mass action which describes equilibria in a more general way. The dissociation constant is also sometimes called ionization constant when applied to salts. The inverse of the dissociation constant is called association constant.
Dissociation constant of water
Formally, the dissociation (autoprotolysis) of water follows the following equation:
$H_2O_{(l)} \rightleftharpoons H_{(aq)}^+ + OH_{(aq)}^- \tag{3}$
or
$H_2O_{(l)} + H_2O_{(l)} \rightleftharpoons H_3O_{(aq)}^+ + OH_{(aq)}^- \tag{4}$
These two reaction equations are thermodynamically equivalent, thus the law of mass action for the two equations must be equivalent. Their equivalency is shown by
$K_d = \dfrac{a_{H^+}·a_{OH^-}}{a_{H_2O}} = \dfrac{a_{H_3O^+}·a_{OH^-}}{a_{H_2O}^2}\approx\dfrac{[H^+][OH^-]}{1} =\dfrac{[H_3O^+][OH^-]}{[1]^2} = [H^+][OH^-]=[H_3O^+][OH^-] =1.00\times 10^{-14}=K_w\tag{5}$
In these reactions (and equations), the activity of water, as the solvent in a dilute solution, is approximated as the activity of pure water, which has a defined value of 1. The activity of the ions, as solutes, can be approximated as the molarity of the ions. Thus, equation 3 and equation 4 both have the same law of mass action and the same $K_d$, which is commonly written as $K_w$.
The value of Kw changes considerably with temperature. Consequently this variation must be taken into account when making precise measurements (i.e. when determining the pH).
Water Temperature
[°C]
Kw
[10-14]
pKw
0 0.1 14.92
10 0.3 14.52
18 0.7 14.16
25 1.2 13.92
30 1.8 13.75
50 8.0 13.10
60 12.6 12.90
70 21.2 12.67
80 35 12.46
90 53 12.28
100 73 12.14
Acid base reactions
The dissociation constant can also be applied to the reaction of acids with water, sometimes called the deprotonation of acids. These reactions can be written in two, equivalent forms, depending on whether water is shown as the solvent, or not:
$HA_{(aq)} + H_2O_{(l)} \rightleftharpoons H_3O^+_{(aq)} + A^-_{(aq)} \tag{6}$
or
$HA_{(aq)} \rightleftharpoons H^+_{(aq)} + A^-_{(aq)} \tag{7}$
In either case, the dissociation constant is denoted as Ka, and because the activity of water has a value of 1, there is no net difference in whether water is formally included in the reaction or the law of mass action equation. The greater the dissociation constant of an acid the stronger the acid. Polyprotic acids (e.g. carbonic acid or phosphoric acid) show several dissociation constants, because more than one proton can be separated (one after the other):
H3A $\rightleftharpoons$H+ + H2A- Ka1 = [H+][H2A-]/[H3A] pKa1 = -lg(Ka1)
H2A- $\rightleftharpoons$H+ + HA2- Ka2 = [H+][HA2-]/[H2A-] pKa2 = -lg(Ka2)
HA2- $\rightleftharpoons$H+ + A3- Ka3 = [H+][A3-]/[HA2-] pKa3 = -lg(Ka3)
A list of acid dissociation constants can be found here.
Other applications
The concept of the dissociation constant is applied in various fields of chemistry and pharmacology. In protein-ligand binding the dissociation constant describes the affinity between a protein and a ligand. A small dissociation constant indicates a more tightly bound ligand. In the case of antibody-antigen binding the inverted dissociation constant is used and is called affinity constant, which, confusingly, is also written as $K_a$.
Contributors and Attributions
{{template.ContribHans()}}
Modified by Tom Neils (Grand Rapids Community College)
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Chemical_Equilibria/Dissociation_Constant.txt
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Le Chatelier's Principle states that a system at equilibrium will adjust to relieve stress when there are changes in the concentration of a reactant or product, the partial pressures of components, the volume of the system, and the temperature of reaction. There are three ways to change the pressure of a constant-temperature reaction system involving gaseous components:
1. Add or remove a gaseous reactant or product: Adding or remove a gaseous reactant or product changes the concentrations. If the concentration of reactant or product is increased, the system will shift away from the side in which concentration was increased (i.e. if the concentration of reactants is increased, the system will shift toward the products. If more products are added, the system will shift to form more reactants). Conversely, if the concentration of reactant or product is decreased, the system will shift toward the side in which concentration was decreased (i.e. If reactants are removed, the system will shift to form more reactants. If the concentration of products is decreased, the equilibrium will shift toward the products).
2. Add an inert gas (one that is not involved in the reaction) to the constant-volume reaction mixture: This will increase the total pressure of the system, but will have no effect on the equilibrium condition. That is, there will be no effect on the concentrations or the partial pressures of reactants or products.
3. Change the volume of the system: When the volume is changed, the concentrations and the partial pressures of both reactants and products are changed. If the volume is decreased, the reaction will shift towards the side of the reaction that has fewer gaseous particles. If the volume is increased, the reaction will shift towards the side of the reaction that has more gaseous particles.
When a system at equilibrium undergoes a change in pressure, the equilibrium of the system will shift to offset the change and establish a new equilibrium. The system can shift in one of two ways:
• Toward the reactants (i.e. in favor of the reverse reaction)
• Toward the products (i.e. in favor of the forward reaction)
The effects of changes in pressure can be described as follows (this only applies to reactions involving gases):
• When there is an increase in pressure, the equilibrium will shift towards the side of the reaction with fewer moles of gas.
• When there is a decrease in pressure, the equilibrium will shift towards the side of the reaction with more moles of gas.
Pressure is inversely related to volume. Therefore, the effects of changes in pressure are opposite of the effects of changes in volume. Additionally, this does not apply to a change in the pressure in the system due to the addition of an inert gas.
Problems
1. Consider the decomposition of NOCl: \[2NOCl_{(g)} \rightleftharpoons 2NO_{(g)} + Cl_{2 (g)}\]
In which direction will the reaction shift if the overall pressure is decreased? Does this favor the forward reaction or the reverse reaction?
2. Consider the decomposition of HBr: \[2HBr_{(g)} \rightleftharpoons H_{2 (g)} + Br_{2 (g)}\]
In which direction will the reaction shift when overall pressure is increased? Which direction will it shift when overall pressure is decreased?
3. Consider the reaction: \[C_{(s)} + 2H_{2 (g)} \rightleftharpoons CH_{4 (g)}\]
What will happen to the equilibrium if the overall pressure is increased? (In which direction will the reaction shift? Does it favor reactants or products? Does this favor the formation of CH4? Is the rate of the forward reaction greater than the rate of the reverse reaction?)
4. Consider the decomposition of MgCO3: \[MgCO_{3(s)} \rightleftharpoons MgO_{(s)} + CO_{2 (g)}\]
Will the formation of CaCO3 or the decomposition of CaCO3 occur faster if the overall pressure is increased?
5. Consider the reaction in a closed container: \[2SO_{2 (g)} + O_{2 (g)} \rightleftharpoons 2SO_{3 (g)}\]
You want the reaction to favor the formation of SO3. You have two options: decrease the overall volume of the container, or increase the overall volume. Which should you choose?
Solutions
1. There are 2 moles of gas particles on the side of the reactants, and 3 moles of gas particles on the side of the products. (Note: 2 moles on the reactant's side come from 2 moles NOCl; 3 moles on the product's comes from 2 moles NO + 1 mole Cl2.). A decrease in pressure favors the side with more particles.
The reaction will shift towards the products, and will favor the forward reaction.
2. There are 2 moles of gas particles on the side of the reactants, and 2 moles of gas particles on the side of the products. Increasing pressure favors the side with fewer particles, and decreasing pressure favors the side with more particles. However, because there is an equal number of particles on both sides, change in pressure will have no effect on the system.
No effect - the reaction will not shift in either direction regardless of pressure changes.
3. There are 2 moles of gas particles on the side of the reactants, and 1 mole of gas particles on the side of the products. Increasing pressure favors the side with fewer particles.
∴ The reaction will shift towards the products. This means that the reaction will favor the forward reaction, which means that it favors the formation of CH4, and that the rate of the forward reaction is greater than the rate of the reverse reaction.
4. There are 0 moles of gas particles on the side of the reactants, and 1 mole of gas particles on the side of the products.
Increasing the pressure favors the side with fewer particles, so the reaction favors the reactants. The products are the decomposition of MgCO3, while the reactants are the formation of MgCO3.
The formation of MgCO3 is favored, meaning that the formation of MgCO3 will occur faster than its decomposition.
5. There are 3 moles of gas particles on the side of the reactants, and 2 moles of gas particles on the side of the products.
Because you want the reaction to favor the formation SO3, you want the reaction to favor the forward reaction/shift to the right, which is the side with fewer moles of gas particles. For a system to shift towards the side of a reaction with fewer moles of gas, you need to increase the overall pressure. Recall that pressure and volume are inversely related, so in order to increase the overall pressure, you need to decrease the overall volume.
∴ You should decrease the overall volume.
Contributors and Attributions
• Christina Chen (UC Davis)
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Chemical_Equilibria/Effect_of_Pressure_on_Gas-Phase_Equilibria.txt
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Strategy for Equilibrium Calculations
• Reinterpret the problem.
• Make required assumptions.
• Equate quantities according to chemical relationship.
• Solve the equations, and make approximations if necessary.
• Check for validity of the answers.
Kc
This page defines the equilibrium constant and introduces the equilibrium constant expressed in terms of concentrations, Kc. It assumes familiarity with the concept of dynamic equilibrium, as well as the terms "homogeneous" and "heterogeneous" as applied to chemical reactions. The two types of dynamic equilibria (homogeneous and heterogeneous) are discussed separately below, because the equilibrium constants are defined differently.
• A homogeneous equilibrium is one in which all species are present in the same phase. Common examples include gas-phase or solution reactions.
• A heterogeneous equilibrium is one in which species exist in more than one phase. Common examples include reactions involving solids and gases, or solids and liquids.
Homogeneous equilibria
This is the more straightforward case. It applies where everything in the equilibrium mixture is in the same phase. An example of a gaseous homogeneous equilibrium is the conversion of sulfur dioxide to sulfur trioxide at the heart of the Contact Process:
\[2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)} \tag{1}\]
A commonly used liquid example is the esterification reaction between an organic acid and an alcohol, as in the example below:
\[ CH_3COOH_{(l)} + CH_3CH_2OH_{(l)} \rightleftharpoons CH_3COOCH_2CH_{3(l)} + H_2O_{(l)}\tag{2}\]
Writing an expression for \(K_c\)
Consider the general equilibrium reaction:
\[ aA + bB \rightleftharpoons cC + dD\tag{3}\]
No state symbols are specified, but the reaction is assumed to be homogeneous here. The measured equilibrium concentrations of the reactants and products are used to calculate the equilibrium constant, as shown in the figure below. The equilibrium constant always has the same value (provided the temperature does not change), irrespective of the initial concentrations of A, B, C and D. It is also unaffected by a change in pressure or the presence (or absence) of a catalyst.
Compare this with the chemical equation for the equilibrium. The convention is that the substances on the right side of the equation are written in the numerator of the Kc expression, and those on the left side in the denominator. The exponents are the coefficients of the reactants and products from the equation.
Example 1: Esterification
A typical esterification reaction is given below:
\[ CH_3COOCH_2CH_{3(l)} + H_2O_{(l)} \rightleftharpoons CH_3COOH_{(l)} + CH_3CH_2OH_{(l)}\]
There is only one molecule of each reactant or product involved in the reaction, so all the exponents in the equilibrium constant expression are "1," and need not be included in the \(K_c\) ex pression, shown below:
As long as the temperature is held constant, \(K_c\) has a constant value regardless of the proportions of acid and alcohol used. At room temperature, this value is approximately 4 for this reaction.
Example 2: Hydrolysis of Esters
This is the reverse of the previous esterification reaction:
\[ CH_3COOCH_2CH_{3(l)} + H_2O_{(l)} \rightleftharpoons CH_3COOH_{(l)} + CH_3CH_2OH_{(l)}\]
The \(K_c\) ex pression is written as follows:
\[ K_c = \dfrac{[CH_3COOH][CH_3CH_2OH]}{[CH_3COOCH_2CH_3][H_2O]}\]
Compared with the previous \(K_c\), this expressio n is simply inverted. Its value at room temperature is approximately 1/4 (0.25).
It is helpful to write down the chemical equation for a equilibrium reaction whenever discussing an equilibrium constant, to ensure the expression is written correctly (products over reactants).
The Contact Process Equilibrium
Recall that the equation for this process is as follows:
\[ 2SO_{2(g)} + O_2 {(g)} \rightleftharpoons 2SO_{3(g)} \tag{4}\]
In this case, the \(K_c\) expressio n includes some nontrivial exponents:
\[ K_c = \dfrac{[SO_3]^2}{[SO_2]^2[O_2]} \tag{5}\]
Although everything is present as a gas, because \(K_c\) is used he re, the activities are measured in mol dm-3. There is another equilibrium constant, \(K_p\), that i s more frequently used for gases.
The Haber process equilibrium
The chemical equation for the Haber process is given below:
\[ N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)} \tag{6}\]
The \(K_c\) expression is as follows:
\[ K_c = \dfrac{[NH_3]^2}{[N_2][H_2]^3} \tag{7}\]
\(K_c\) in heterogeneous Equilibria
An examples of a heterogeneous equilibrium is the equilibrium established if steam is in contact with red hot carbon. In this reaction, a gas reacts with a solid:
\[ H_2O_{(g)} + C_{(s)} \rightleftharpoons H_{2(g)} + CO_{(g)} \tag{8}\]
Another heterogeneous equilibrium involves shaking copper with silver nitrate solution; this reaction involves solids and aqueous ions:
\[ Cu_{(s)} + 2Ag^+_{(aq)} \rightleftharpoons Cu^{2+}_{(aq)} + 2Ag_{(s)} \tag{9}\]
The important consideration for a heterogeneous equilibrium is that there are no solid terms in an equilibrium constant expression; therefore, not every reactant or product will be accounted in \(K_c\).
Example 3: Carbon with Steam
\[ H_2O_{(g)} + C_{(s)} \rightleftharpoons H_{2(g)} + CO_{(g)}\]
The equilibrium constant expression is written the same way as in previous examples, omitting the solid carbon term:
\[ K_c = \dfrac{[H_2][CO]}{[H_2O]}\]
Example 4: Copper wtih Silver Ions
Consider the redox reaction between solid copper and silvir ions in solution:
\[ Cu_(s)+ + 2Ag^+_{(aq)} \rightleftharpoons Cu^{2+}_{(aq)} + 2Ag_{(s)}\]
Both the copper on the left-hand side and the silver on the right are solids. Both are left out of the equilibrium constant expression:
\[ K_c = \dfrac{[Cu_{2+}]}{[Ag^+]^2}\]
Example 5: Heating \(CaCO_3\)
This equilibrium is only established if calcium carbonate is heated in a closed system, preventing carbon dioxide from escaping:
\[CaCO_{3(s)} \rightleftharpoons CaO_{(s)} + CO_{2(g)} \]
The only nonsolid species in this system is carbon dioxide, so it is the only term in the equilibrium constant expression:
\[ K_c = [CO_2]\]
Contributors and Attributions
Jim Clark (Chemguide.co.uk)
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Chemical_Equilibria/Equilibrium_Calculations.txt
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This page explains equilibrium constants expressed in terms of partial pressures of gases, Kp. It covers an explanation of the terms mole fraction and partial pressure, and looks at \(K_p\) for both homogeneous and heterogeneous reactions involving gases. The page assumes that you are already familiar with the concept of an equilibrium constant, and that you know about \(K_c\) - an equilibrium constant expressed in terms of concentrations
Before we can go any further, there are two terms relating to mixtures of gases that you need to be familiar with.
Mole fraction
If you have a mixture of gases (A, B, C, etc), then the mole fraction of gas A is worked out by dividing the number of moles of A by the total number of moles of gas.
The mole fraction of gas A is often given the symbol xA. The mole fraction of gas B would be xB - and so on.
Pretty obvious really! For example, in a mixture of 1 mole of nitrogen and 3 moles of hydrogen, there are a total of 4 moles of gas. The mole fraction of nitrogen is 1/4 (0.25) and of hydrogen is 3/4 (0.75).
Partial pressure
The partial pressure of one of the gases in a mixture is the pressure which it would exert if it alone occupied the whole container. The partial pressure of gas A is often given the symbol PA. The partial pressure of gas B would be PB - and so on.
There are two important relationships involving partial pressures. The first is again fairly obvious. The total pressure of a mixture of gases is equal to the sum of the partial pressures.
It is easy to see this visually:
Gas A is creating a pressure (its partial pressure) when its molecules hit the walls of its container. Gas B does the same. When you mix them up, they just go on doing what they were doing before. The total pressure is due to both molecules hitting the walls - in other words, the sum of the partial pressures. The more important relationship is the second one:
Learn it! That means that if you had a mixture made up of 20 moles of nitrogen, 60 moles of hydrogen and 20 moles of ammonia (a total of 100 moles of gases) at 200 atmospheres pressure, the partial pressures would be calculated like this:
gas mole fraction partial pressure
nitrogen 20/100 = 0.2
0.2 x 200 = 40 atm
hydrogen 60/100 = 0.6 0.6 x 200 = 120 atm
ammonia 20/100 = 0.2 0.2 x 200 = 40 atm
Partial pressures can be quoted in any normal pressure units. The common ones are atmospheres or pascals (Pa). Pascals are exactly the same as N m-2 (newtons per square meter).
\(K_p\) in homogeneous gaseous equilibria
A homogeneous equilibrium is one in which everything in the equilibrium mixture is present in the same phase. In this case, to use Kp, everything must be a gas. A good example of a gaseous homogeneous equilibrium is the conversion of sulfur dioxide to sulfur trioxide at the heart of the Contact Process:
Writing an expression for Kp
We are going to start by looking at a general case with the equation:
If you allow this reaction to reach equilibrium and then measure (or work out) the equilibrium partial pressures of everything, you can combine these into the equilibrium constant, Kp.
Just like Kc, \(K_p\) always has the same value (provided you don't change the temperature), irrespective of the amounts of A, B, C and D you started with.
Kp has exactly the same format as Kc, except that partial pressures are used instead of concentrations. The gases on the right-hand side of the chemical equation are at the top of the expression, and those on the left at the bottom.
The Contact Process equilibrium
You will remember that the equation for this is:
Kp is given by:
The Haber Process equilibrium
The equation for this is:
. . . and the \(K_p\) expression is:
Kp in heterogeneous equilibria
A typical example of a heterogeneous equilibrium will involve gases in contact with solids.
Writing an expression for \(K_p\) for a heterogeneous equilibrium
Exactly as happens with Kc, you don't include any term for a solid in the equilibrium expression.
The next two examples have already appeared on the \(K_c\) page.
The equilibrium produced on heating carbon with steam
Everything is exactly the same as before in the expression for Kp, except that you leave out the solid carbon.
The equilibrium produced on heating calcium carbonate
This equilibrium is only established if the calcium carbonate is heated in a closed system, preventing the carbon dioxide from escaping.
The only thing in this equilibrium which isn't a solid is the carbon dioxide. That is all that is left in the equilibrium constant expression.
Contributors and Attributions
Jim Clark (Chemguide.co.uk)
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Chemical_Equilibria/Kp.txt
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The law of mass action only rigorously applies to elementary reactions (in that otherwise the exponents do not match stoichiometric coefficients). These are reactions whose equations match which particles are colliding on the microscopic scale. Consider the reaction
$r_1 R_1 + r_2 R_2 + \ldots + r_n R_n <=> p_1 P_1 + p_2 P_2 + \ldots + p_m P_m \label {1}$
Reaction $(1)$ is elementary if and only if it proceeds in a single step. At some time $t = t_c$ all reactant particles $R_1, R_2, \ldots, R_n$ collide. Quantum mechanically this collision is not too well defined, so view it to be Newtonian in essence. Not to mention: the probability $P$ of the simultaneous collision goes to zero as $n \to \infty$. This is why in actuality most elementary steps involve at most three individual constituents.
We therefore sacrifice some generality and instead consider the easier to handle
$R_1(g) + R_2(g) <=> P_1(g) + P_2(g) \label {1'}$
The reactant molecules must collide in order for a reaction to take place. Therefore we must know the collision frequency in order to determine the reaction rate. This rate must be the product of some probability and the collision frequency. Collision Theory gives a way to calculate this collision frequency. In doing so Maxwell-Boltzmann distributions are used to calculate the ratio of atoms that can react within a given concentration. Most of this 'stuff' gets swept up in the reaction rate constant.
Energies
Collision theory postulates that particles only initiate a transformation when they have enough energy. This is called activation energy and denoted by $E_a$. Maxwell-Boltzmann distribution
$\frac{\mathrm{d}N_v}{N} = 4\pi\left(\frac{m}{2\pi k_bT}\right)^{3/2}v^2\exp\left(-\frac{mv^2}{2k_bT}\right)\mathrm{d}v\label{2}$
allows us to estimate the fraction of particles that have the required amount of movement. Note we are technically assuming total energy = kinetic energy. Similarly relativistic effects are ignored, and no cap is placed on maximum velocity. With these simplifications, the fraction of interesting particles is
$f = \frac{N^*}{N} = \frac{1}{k_bT}\int_{E_a}^\infty\exp\left(-\frac{E}{k_bT}\right)\mathrm{d}E\label{3}$
which evaluates to
$f = \exp\left(-\frac{E_a}{k_bT}\right)\label{3'}.$
Number of interesting collisions
Say $Z^*$ is the number of collisions that matter. We will look for a function $Z$ such that
$Z^* = Z\cdot f. \label {4}$
The Maxwell-Boltzmann distribution (Equation $\ref{2}$) gives for a single kind of particle (say two $\ce{R1}$'s) that there are
$Z = 2N^2 \left( \sigma_{\ce{R1}} \right)^2 \sqrt{\dfrac{\pi RT}{M_{\ce{R1}}}} \label{5}$
collisions in $1$ second per $1\ \mathrm{cm^3}$ if $N$ is the number of particles per square centimeter. (This is the traditional starting point unit.) Hence the rate of such a reaction would be
$v_r = \frac{2Z^*}{N_A} \cdot 10^3\ \ \ \left(\frac{\mathrm{mol}}{\mathrm{s\cdot dm^3}}\right). \label{6}$
Reaction rate $v_r$
If you like, you are now able to substitute $f$ from Equation $\ref{3'}$ and $Z$ via Equation $\ref{5}$ into Equation $\ref{4}$. Then insert Equation $\ref{4}$ into Equation $\ref{6}$. After necessary manipulation, this yields Equation $\ref{7}$.
$v_r = \underbrace{4 \times 10^{-3} N_A \left(\sigma_{\ce{R1}} \right)^2\sqrt{\frac{\pi RT}{M_{\ce{R1}}}}\exp\left(-\frac{E_a}{k_bT}\right) }_k \cdot \overbrace{\left(\frac{N}{N_A}\cdot 10^3\right)^2}^{c^2}. \label{7}$
Coming back to our actual reaction (Equation $\ref{1'}$) involves more relative quantities. For example, $\sigma_{\ce{R1}}$ becomes
$\sigma_{\text{aver}} = 0.5\left(d_{\ce{R1}} + d_{\ce{R2}} \right).$
So it is a bit more difficult. But the result is analogous to Equation $\ref{7}$.
$v_r = \underbrace{2\sqrt{2} \cdot 10^{-3} N_A\left(\sigma_\text{aver}\right)^2\sqrt{\pi RT\left(\frac{1}{M_{\ce{R1}}} + \frac{1}{M_{\ce{R2}}}\right)}\exp\left(-\frac{E_a}{k_bT}\right)}_k \cdot c_{\ce{R1}} \cdot c_{\ce{R2}}$
More compactly, if $\ce{R1} = A$ and $\ce{R2} = B$
$v_r = k[A][B]\label{8}$
which is what we set out to prove.
• This derivation assumes that every active collision leads to a reaction. When theoretically computed $k$ were compared to experimental, an extra factor was introduced, $P$. This is called the steric factor, and in classical collision theory $P$ remains empirical in essence.
Collision theory is for elementary steps
As your quote suggests, collision theory is a first theoretical explanation for the proportionality to products of concentrations. It does not generally explain various exponentiation. It does not have to either. The higher (or non-integer or negative) exponents usually derive from the mechanism itself. In other words, the fact that common reactions are not elementary comes into play.
For instance, the transition
$\ce{2Br- + H2O2 + 2H+ -> Br2 + H2O}$
is experimentally found to follow
$v_r = k\ce{[H2O2][H+][Br^{-}]}\label {a}.$
Mathematically, we can verify that one possible mechanism is
$\ce{H+ + H2O2 <=>[K] H2O+-OH,} \tag{fast equilibrium}$
$\ce{H2O+-OH + Br- ->[k_2] HOBr + H2O,}\tag{slow}$
$\ce{HOBr + H+ + Br- ->[k_3] Br2 + H2O.}\tag{fast}$
Indeed,
$K_c = \frac{\ce{[H2O+-OH]}}{\ce{[H+]}\ce{[H2O2]}} \implies \ce{[H2O+-OH]} = K_c\ce{[H+]}\ce{[H2O2]}\label {b}.$
Applying the method of stationary concentration gives
$v\left(\ce{HOBr}\right) = 0 \implies k_2\ce{[H2O+-OH]}\ce{[Br-]} = k_3\ce{[H2O2]}\ce{[H+]}\ce{[Br-]}\label {c}.$
The overall rate of the reaction is characterised by the rate of formation of bromine $\ce{Br2}$. So,
$v_r = v\left(\ce{Br2}\right) = k_3\ce{[H2O2]}\ce{[H+]}\ce{[Br-]} \overset{(c)}{=} k_2\ce{[H2O+-OH]}\ce{[Br-]} \overset{(b)}{=} k_2K_c\ce{[H+][H2O2]}\ce{[Br-]}$
or more briefly using $k_2K_c = k$
$v_r = k\ce{[H2O2][H+]}\ce{[Br-]}\label {d}.$
This only shows that it might be a valid mechanism, not that the reaction actually follows such a pathway. Therefore, while we can use collision theory to derive the law of mass action as a first approximation, the exponents for non-elementary reactions are determined via experiment.
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Learning Objectives
• Identify a system, an open system, a closed system and the environment of the system.
• Define a state of equilibrium.
• Describe the mass action law.
• Apply the mass action law to write expressions for equilibrium constants.
• Write the equilibrium constant expression for any reaction equation.
Heat is energy flowing from a high temperature object to a low temperature object. When the two objects are at the same temperature, there is no net flow of energy or heat. That is why a covered cup of coffee will not be colder than or warmer than the room temperature after it has been in there for a few hours. This phenomenon is known as equilibrium. In this example, we deal with the flow of energy.
Equilibria happen in phase transitions. For example, if the temperature in a system containing a mixture of ice and water is uniformly 273.15 K, the net amount of ice formed and melted will be zero. The amount of liquid water will also remain constant, if no vapor escapes from the system. In this case, three phases, ice (solid) water (liquid), and vapor (gas) are in equilibrium with one another. Similarly, equilibrium can also be established between the vapor phase and the liquid at a particular temperature. Equilibrium conditions also exist between solid phase and vapor phases. These are phase equilibria.
Chemical reactions may not be as complete as we have assumed in Stoichiometry calculations. For example, the following reactions are far short of completion.
$\ce{2 NO2 \rightleftharpoons N2O4}$
$\ce{3 H2 + N2 \rightleftharpoons 2 NH3}$
$\ce{H2O + CO \rightleftharpoons H2 + CO2}$
Let us consider only the first reaction in this case. At room temperature, it is impossible to have pure $\ce{NO2}$ or $\ce{N2O4}$. However, in a sealed tube (closed system), the ratio
$\ce{\dfrac{[N2O4]}{[NO2]^2}}$
is a constant. This phenomenon is known as chemical equilibrium. Such a law of nature is called the law of mass action or mass action law.
Of course, when conditions, such as pressure and temperature, change, a period of time is required for the system to establish an equilibrium. Before we introduce the mass action law, it is important for us to identify a system or a closed system in our discussion. The law provides an expression for a constant for all reversible reactions. For systems that are not at equilibrium yet, the ratio calculated from the mass action law is called a reaction quotient Q. The Q values of a closed system have a tendency to reach a limiting value called equilibrium constant K over time. A system has a tendency to reach an equilibrium state.
A Closed System for the Equilibrium State
To discuss equilibrium, we must define a system, which may be a cup of water, a balloon, a laboratory, a planet or a universe. Thus, for discussion purposes, we define an isolated portion of the universe under consideration as a system, and anything outside of the system is called environment. When the system under consideration is isolated from its environment in such a way that there is no energy or mass transferred into or out of the system, the system is said to be an isolated system. In a isolated system, changes continue, but eventually there is no NET change over time; such a state is called an equilibrium state.
For example, a glass containing water is an open system. Evaporation lets water molecules escape into the air by absorbing energy from the environment until the glass is empty. When covered and insulated it is a closed system. Water vapor in the space above water eventually reaches an equilibrium vapor pressure. In fact, measuring of temperature itself requires the thermometer to be at the same state as the system it measures. We read the temperature of the thermometer when heat transfer between the thermometer and the system stops (at equilibrium). Equilibrium states are reached for physical as well as chemical reactions. Equilibrium is dynamic in the sense that changes continue, but the net change is zero.
Reversible Chemical Reactions
Heat transfer, vaporization, melting, and other phase changes are physical changes. These changes are reversible and you have already experienced them. Many chemical reactions are also reversible. For example
$\underset{colorless}{\ce{N2O4}} \rightleftharpoons \underset{brown}{\ce{2 NO2}}$
and
$\ce{N2 + 3 H2 \rightleftharpoons 2 NH3}$
are reversible chemical reactions.
The Law of Mass Action
The law of mass action is universal, applicable under any circumstance. However, for reactions that are complete, the result may not be very useful. We introduce the mass action law by using a general chemical reaction equation in which reactants $\ce{A}$ and $\ce{B}$ react to give products $\ce{C}$ and $\ce{D}$.
$a\, \ce A + b\, \ce B \rightarrow c\, \ce C + d\, \ce D$
where a, b, c, d are the coefficients for a balanced chemical equation. The mass action law states that if the system is at equilibrium at a given temperature, then the following ratio is a constant:
$\dfrac{[\ce C]^c [\ce D]^d}{[\ce A]^a [\ce B]^b} = K_{\ce{eq}}$
The square brackets "[ ]" around the chemical species represent their concentrations. This is the ideal law of chemical equilibrium or law of mass action.
The Reaction Quotient Q vs. the Equilibrium Constants K
If the system is NOT at equilibrium, the ratio is different from the equilibrium constant. In such cases, the ratio is called a reaction quotient which is designated as Q.
$\dfrac{[\ce C]^c [\ce D]^d}{[\ce A]^a [\ce B]^b} = Q$
A system not at equilibrium tends to become at equilibrium, and the changes will cause changes in Q so that its value approaches the equilibrium constant, K:
$Q \rightarrow K_{\ce{eq}}$
The mass action law gives us a general method to write the expression for the equilibrium constant of any reaction. At this stage, you should be able to write the equilibrium expression for any reaction equation. If you are not sure from the above general theory, here are some examples. It is more important for you to understand WHY the equilibrium constants are expressed this way than what the equilibrium expression is.
Example $1$
Write the the equilibrium constant expression for the reaction equation:
$\ce{NH3 + HOAc \rightleftharpoons NH4+ + OAc-} \nonumber$
Hint
$\ce{\dfrac{[NH4+] [OAc- ]}{[NH3] [HOAc]}} = K\: (\textrm{unitless constant}) \nonumber$
Example $2$
For the ionization of an acid,
$\ce{H2SO4 \rightleftharpoons 2 H+ + SO4^2-} \nonumber$
what is the equilibrium constant expression?
Hint
The equilibrium constant is
$\ce{\dfrac{[H+]^2 [SO4^2- ]}{[H2SO4]}} = K \ce M^2 \nonumber$
where M = mol/L. Note the unit for K.
Example $3$
For the reaction equation:
$\ce{Cu^2+ + 6 NH3 \rightleftharpoons Cu(NH3)6^2+} \nonumber$
what is the equilibrium constant expression?
Hint
The expression is
$\ce{\dfrac{[Cu(NH3)6^2+]}{[Cu^2+] [NH3]^6}} = K \ce M^{-6} \nonumber$
The equilibrium constant depends on the written equation.
Example $1$
Discuss the ionization of oxalic acid, $\ce{H2C2O4}$, in two stages.
Solution
Experimentally, it has been shown that
$\ce{H2C2O4 \rightleftharpoons H+ + HC2O4- }\tag{1}$
$\ce{\dfrac{[HC2O4- ] [H+]}{[H2C2O4]}} = K_1 = \mathrm{0.059\: M} \nonumber$
The second ionization constant is much smaller:
$\ce{HC2O4- \rightleftharpoons H+ + C2O4^2- }\tag{2}$
$\ce{\dfrac{[H+] [C2O4^2- ]}{[HC2O4- ]}} = K_2 = \mathrm{0.000064\: M} \nonumber$
The overall ionization can be obtained by adding (1) and (2) to give (3).
$\ce{H2C2O4 \rightleftharpoons 2 H+ + C2O4^2- }\tag{3} \nonumber$
and the equilibrium constant is
$\ce{\dfrac{[H+]^2 [C2O4^2- ]}{[H2C2O4]}} = K_{3}\: \ce M^2 \nonumber$
It is obvious that
\begin{align*} K_{3} &= K_1 \times K_2\ &= 3.8\times10^{-6}\: \ce M^2 \end{align*}
Please confirm the above obvious relationship to satisfy yourself.
Example $5$
At some temperature, the equilibrium constant is 4.0 for the reaction
$\ce{CO_{\large{(g)}} + 3 H_{2\large{(g)}} \rightleftharpoons CH_{4\large{(g)}} + H2O_{\large{(g)}}} \nonumber$
What is the equilibrium constant expression and value for the reaction,
$\ce{CH_{4\large{(g)}} + H2O_{\large{(g)}} \rightleftharpoons CO_{\large{(g)}} + 3 H_{2\large{(g)}}}? \nonumber$
Hint
$\ce{\dfrac{[CO] [H2]^3}{[CH4] [H2O]}} = \dfrac{1}{4.0} = 0.25$
Example $6$
At some temperature, the equilibrium constant is 4.0 for the reaction
$\ce{CO_{\large{(g)}} + 3 H_{2\large{(g)}} \rightleftharpoons CH_{4\large{(g)}} + H2O_{\large{(g)}}}$.
What is the equilibrium constant expression and value for the reaction,
$\ce{\dfrac{1}{3} CO_{\large{(g)}} + H_{2\large{(g)}} \rightleftharpoons \dfrac{1}{3} CH_{4\large{(g)}} + \dfrac{1}{3} H2O_{\large{(g)}}}$?
Hint
$\mathrm{\dfrac{[CH_4]^{1/3} [H_2O]^{1/3}}{[CO]^{1/3} [H_2]} = 4.0^{1/3} = 1.59}$
The application of the mass action law leads to the method to write the expression for the equilibrium constant. The law is given in a general form, and these examples should help you grasp the method.
Questions
1. A closed container has $\ce{N2O4}$ and $\ce{NO2}$ gases in it, and it has been placed in the lab for many days. What would you consider the container and the gases to be?
1. an open system
2. a closed system
3. not a system
2. A closed container has $\ce{N2O4}$ and $\ce{NO2}$ gases in it, and it has been placed in the lab for many days. Is the system at an equilibrium state for the following chemical reaction?
$\ce{N2O4 \rightleftharpoons 2 NO2}$
3. A closed container has $\ce{N2O4}$ and $\ce{NO2}$ gases in it, and it has just been placed in an ice/water bath. Is the system at an equilibrium state for the following chemical reaction?
$\ce{N2O4 \rightleftharpoons 2 NO2}$
4. $\ce{HCOOH + CN- \rightleftharpoons HCN + HCOO-}$ is 5E5, what is the equilibrium constant for the reaction equation
$\ce{HCN + HCOO- \rightleftharpoons HCOOH + CN-}$?
5. For reactions taking place in the gas phase, the equilibrium constant is normally expressed in terms of partial pressures of the reactants and products. If C represents the concentration, and other symbols of the ideal gas equation are used, which of the following is correct?
1. $C = \dfrac{RT}{PV}$
2. $C = \dfrac{RT}{P}$
3. $C = \dfrac{RT}{V}$
4. $C = \dfrac{RT}{PM}$
5. $C = \dfrac{P}{RT}$
6. $C = \dfrac{PV}{RT}$
6. If Kc represents the equilibrium constant in terms of concentration, and Kp represents that in terms of partial pressure, which of the following is correct?
1. $K_c = K_p$
2. $K_c \propto K_p$
3. $K_c = \dfrac{1}{K_p}$
4. $K_c \propto \dfrac{1}{K_p}$
5. $K_c \times K_p = RT$
7. For which of the following gas-phase reaction equations are the equilibrium constants unitless quantities?
1. $\ce{2 H2 + O2 \rightleftharpoons 2 H2O}$
2. $\ce{2 NO \rightleftharpoons N2 + O2}$
3. $\ce{COCl2 \rightleftharpoons CO + Cl2}$
4. $\ce{CO + H2O \rightleftharpoons CO2 + H2}$
(Give the letter(s) of your choice(s), e.g., ad).
Solutions
1. b
Discussion...
Since it has been in the lab, the temperature of the system is the same as its environment.
2. yes
Discussion...
Only a few minutes are required for the reaction to reach an equilibrium state. $\ce{NO2}$ is a brown gas
3. no
Discussion...
Heat will be extracted from the container, and that causes the equilibrium to shift. At 0 deg C, the equilibrium is shifted to have lots of
$\ce{N2O4}$.
4. 1/5E5 = 2E-6
Discussion...
Since the reaction equation is reversed, use the relationship
$K_{reverse} = \dfrac{1}{K_{forward}}$.
If you reverse the reaction equation, take the reciprocal of the previous equilibrium constant.
5. e
Discussion...
By definition,
$C = \dfrac{n}{V}$. Derive it from $n R T = P V$.
6. b
Discussion...
Statements (c), (d), and (e) may be true for special cases, but they are not generally true. In this list, "Kc is proportional to Kp" is true, but a more quantative relationship will be derived.
7. bd
Discussion...
Both reaction equations in (b) and (d) have equal numbers of reactants and products.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Chemical_Equilibria/Mass_Action_Law.txt
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Consider the following gas-phase chemical reaction:
\(H_2 + I_2 \rightleftharpoons 2HI\)
The apparatus shown belows contains 0.02 mol/L I2 gas in the left syringe and 0.01 mol/L H2 gas in the right syringe. When the two syringes are depressed, the two gases are mixed and the above reaction occurs. The graph at the right shows the variations of the H2, I2, and HI concentrations as a function of time. Run the experiment and examine the concentration-time plots.
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<applet align="middle" archive="stoppedFlow.jar,DataGraph4_.jar,STools4.jar" code="dataGraph.DataGraph.class" codebase="/deki/vce/" height="400" hspace="5" name="dataGraph1" title="Java(TM)" vspace="5" width="400"><param name="Function"/><param name="DataFile"/><param name="ShowControls" value="false"/></applet>
As the reaction progresses, the concentrations of iodine and hydrogen decrease as they are consumed while the concentration of hydrogen iodide increases as it is formed. Eventually, however, the concentrations of all three species reach constant values. This behavior is a result of the reverse in which hydrogen iodide reacts to form iodine and hydrogen. Initially there is no hydrogen iodide, so the reverse reaction cannot occur. As hydrogen iodide accumulates, the reverse reaction can progress and becomes significant.
Eventually the rate at which iodine is being consumed by the forward reaction is perfectly balanced by the rate at which it is being produced by the back reaction. The same is true for hydrogen and hydrogen iodide. When the rates of forward and reverse reactions are perfectly balanced in this way, the reaction is said to be at equilibrium.
Contributors and Attributions
• Dr. David Blauch (Davidson College)
Dynamic equilibrium
At dynamic equilibrium, the reaction rate of the forward reaction is equal to the reaction rate of the backward reaction.
Contributors and Attributions
• Esther Lee (UCD), Jiaxu Wang, Jonathan Wang
Equilibrium Constant
Concepts
For the hypothetical chemical reaction:
\[aA + bB \rightleftharpoons cC + dD\]
the equilibrium constant is defined as:
\[K_C = \dfrac{[C]^c[D]^d}{[A]^a[B]^b}\]
The notation [A] signifies the molar concentration of species A. An alternative expression for the equilibrium constant involves partial pressures:
\[K_P = \dfrac{P_C^c P_D^d}{P_A^aP_B^b}\]
Note that the expression for the equilibrium constant includes only solutes and gases; pure solids and liquids do not appear in the expression. For example, the equilibrium expression for the reaction
\[CaH_2 (s) + 2H_2O (g) \rightleftharpoons Ca(OH)_2 (s) + 2H_2 (g)\]
is the following:
\[K_C = \dfrac{[H_2]^2}{[H_2O]^2}\]
Observe that the gas-phase species H2O and H2 appear in the expression but the solids CaH2 and Ca(OH)2 do not appear.
The equilibrium constant is most readily determined by allowing a reaction to reach equilibrium, measuring the concentrations of the various solution-phase or gas-phase reactants and products, and substituting these values into the Law of Mass Action.
Experiment
Objective: Determine the equilibrium constant for a chemical reaction.
Part 1.
In this part of the experiment, the following reaction is studied:
\[NaHCO_3 (s) \rightleftharpoons NaOH (s) + CO_2 (g) \]
The steps in the experiment are those that would be performed in practice in the laboratory:
1. A 1 gram portion of sodium hydrogen carbonate powder is placed in the glass bulb
2. The bulb is evacuated (that is, all gases are removed from the flask)
3. The bulb is heated to 800 K and the reaction is allowed to reach equilibrium
At room temperature, the reaction does not proceed at a measurable rate. At 800 K, however, the rate of reaction is relatively fast and equilibrium is quickly achieved.
A manometer is attached to the bulb to measure the partial pressure of carbon dioxide in the bulb. Assume the sodium hydrogen carbonate occupies a negligible fraction of the volume of the 500 mL bulb. Also assume that the connections to the manometer have a negligible volume. (The stopcock to the bulb is always open, because the bulb is always open to the manometer so that the pressure may be continually monitored. Both air and carbon dioxide are colorless gases.)
Use the experimental data to calculate KP and KC for this reaction. In this part of the experiment, the value KP and KC are provided for verification.
Dr. David Blauch (Davidson College)
Kinetically vs Thermodynamically Stable
Kinetics and thermodynamics are related to each other in ways that can be explained by using chemical reactions. A discussion of kinetics and thermodynamics requires an explanation of the underlying relationships between the two, through application to chemical reactions and several examples from natural processes.
Contributors and Attributions
• Gal Ozery (UCD)
Characteristics
Chemical equilibrium is a state in which the rate of the forward reaction equals the rate of the backward reaction. In other words, there is no net change in concentrations of reactants and products. This kind of equilibrium is also called dynamic equilibrium.
Contributors and Attributions
• Vanessa Chan (UCD)
• Lorraine Alborzfar (UCD
Principles of Chemical Equilibrium
The chemical equilibrium state describes concentrations of reactants and products in a reaction taking place in a closed system,which no longer change with time. In other words, the rate of the forward reaction equals the rate of the reverse reaction, such that the concentrations of reactants and products remain fairly stable, in a chemical reaction. Equilibrium is denoted in a chemical equation by the ⇌ symbol.
Contributors and Attributions
• Preya Sheth (UCD)
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Chemical_Equilibria/Principles_of_Chemical_Equilibria/Basic_Concepts.txt
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The equilibrium constant, K, expresses the relationship between products and reactants of a reaction at equilibrium with respect to a specific unit.This article explains how to write equilibrium constant expressions, and introduces the calculations involved with both the concentration and the partial pressure equilibrium constant.
Homogeneous Reactions
A homogeneous reaction is one where the states of matter of the products and reactions are all the same (the word "homo" means "same"). In most cases, the solvent determines the state of matter for the overall reaction. For example, the synthesis of methanol from a carbon monoxide-hydrogen mixture is a gaseous homogeneous mixture, which contains two or more substances:
$CO (g)+ 2H_2 (g) \rightleftharpoons CH_3OH (g)$
At equilibrium, the rate of the forward and reverse reaction are equal, which is demonstrated by the arrows. The equilibrium constant, however, gives the ratio of the units (pressure or concentration) of the products to the reactants when the reaction is at equilibrium.
The synthesis of ammonia is another example of a gaseous homogeneous mixture:
$N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$
Heterogeneous Reactions
A heterogeneous reaction is one in which one or more states within the reaction differ (the Greek word "heteros" means "different"). For example, the formation of an aqueous solution of lead(II) iodide creates a heterogeneous mixture dealing with particles in both the solid and aqueous states:
$PbI_{2 (s)} \rightleftharpoons Pb^{+2}_{(aq)} + 2I^-_{(aq)}$
The decomposition of sodium hydrogen carbonate (baking soda) at high elevations is another example of a heterogeneous mixture, this reaction deals with molecules in both the solid and gaseous states:
$2NaHCO_{3 (s)} \rightleftharpoons Na_2CO_{3 (s)} + H_2O_{ (g)} + CO_{2 (g)}$
$C_{(s)} + O_{2 (g)} \rightleftharpoons CO_{2 (g)}$
This difference between homogeneous and heterogeneous reactions is emphasized so that students remember that solids, pure liquids, and solvents are treated differently than gases and solutes when approximating the activities of the substances in equilibrium constant expressions.
Writing Equilibrium Constant Expressions
The numerical value of an equilibrium constant is obtained by letting a single reaction proceed to equilibrium and then measuring the concentrations of each substance involved in that reaction. The ratio of the product concentrations to reactant concentrations is calculated. Because the concentrations are measured at equilibrium, the equilibrium constant remains the same for a given reaction independent of initial concentrations. This knowledge allowed scientists to derive a model expression that can serve as a "template" for any reaction. This basic "template" form of an equilibrium constant expression is examined here.
Equilibrium Constant of Activities
The thermodynamically correct equilibrium constant expression relates the activities of all of the species present in the reaction. Although the concept of activity is too advanced for a typical General Chemistry course, it is essential that the explanation of the derivation of the equilibrium constant expression starts with activities so that no misconceptions occur. For the hypothetical reaction:
$bB + cC \rightleftharpoons dD + eE$
the equilibrium constant expression is written as
$K = \dfrac{a_D^d ·a_E^e}{a_B^b · a_C^c}$
*The lower case letters in the balanced equation represent the number of moles of each substance, the upper case letters represent the substance itself.
• If $K > 1$ then equilibrium favors products
• If $K < 1$ then equilibrium favors the reactants
Equilibrium Constant of Concentration
To avoid the use of activities, and to simplify experimental measurements, the equilibrium constant of concentration approximates the activities of solutes and gases in dilute solutions with their respective molarities. However, the activities of solids, pure liquids, and solvents are not approximated with their molarities. Instead these activities are defined to have a value equal to 1 (one).The equilibrium constant expression is written as $K_c$, as in the expression for the reaction:
$HF_{(aq)} + H_2O_{(l)} \rightleftharpoons H_3O^+_{(aq)} + F^-_{(aq)}$
$K_c = \dfrac{a_{H_3O^+}· a_{F^-}}{a_{HF} · a_{H_2O}} ≈ \dfrac{[H_3O^+][F^-]}{[HF](1)} = \dfrac{[H_3O^+][F^-]}{[HF]}$
Here, the letters inside the brackets represent the concentration (in molarity) of each substance. Notice the mathematical product of the chemical products raised to the powers of their respective coefficients is the numerator of the ratio and the mathematical product of the reactants raised to the powers of their respective coefficients is the denominator. This is the case for every equilibrium constant. A ratio of molarities of products over reactants is usually used when most of the species involved are dissolved in water. A ratio of concentrations can also be used for reactions involving gases if the volume of the container is known. .
Equilibrium Constant of Pressure
Gaseous reaction equilibria are often expressed in terms of partial pressures. The equilibrium constant of pressure gives the ratio of pressure of products over reactants for a reaction that is at equilibrium (again, the pressures of all species are raised to the powers of their respective coefficients). The equilibrium constant is written as $K_p$, as shown for the reaction:
$aA_{(g)} + bB_{(g)} \rightleftharpoons gG_{(g)} + hH_{(g)}$
$K_p= \dfrac{p^g_G \, p^h_H}{ p^a_A \,p^b_B}$
• Where $p$ can have units of pressure (e.g., atm or bar).
Conversion of Kc to Kp
To convert Kc to Kp, the following equation is used:
$K_p = K_c(RT)^{\Delta{n_{gas}}}$
where:
• R=0.0820575 L atm mol-1 K-1 or 8.31447 J mol-1 K-1
• T= Temperature in Kelvin
• Δngas= Moles of gas (product) - Moles of Gas (Reactant)
Reaction Quotient
Another quantity of interest is the reaction quotient, $Q$, which is the numerical value of the ratio of products to reactants at any point in the reaction. The reaction quotient is calculated the same way as is $K$, but is not necessarily equal to $K$. It is used to determine which way the reaction will proceed at any given point in time.
$Q = \dfrac{[G]^g[H]^h}{[A]^a[B]^b}$
• If $Q > K$, then the reactions shifts to the left to reach equilibrium
• If $Q < K$, then the reactions shifts to the right to reach equilibrium
• If $Q = K$ then the reaction is at equilibrium
The same process is employed whether calculating $Q_c$ or $Q_p$.
Heterogeneous Mixture
The most important consideration for a heterogeneous mixture is that solids and pure liquids and solvents have an activity that has a fixed value of 1. From a mathematical perspective, with the activities of solids and liquids and solvents equal one, these substances do not affect the overall K or Q value. This convention is extremely important to remember, especially in dealing with heterogeneous solutions.
Example $1$
In a hypothetical reaction:
$aA_{(s)} + bB_{(l)} \rightleftharpoons gG_{(aq)} + hH_{(aq)}$
The equilibrium constant expression is written as follows:
$K_c = \dfrac{[G]^g[H]^h}{1 \times 1} = [G]^g[H]^h$
In this case, since solids and liquids have a fixed value of 1, the numerical value of the expression is independent of the amounts of A and B. If the product of the reaction is a solvent, the numerator equals one, which is illustrated in the following reaction:
$H^+_{(aq)} + OH^–_{(aq)} \rightarrow H_2O_{ (l)}$
The equilibrium constant expression would be:
$K_c= \dfrac{1}{ [H^+][OH^-]}$
which is the reciprocal of the autoionization constant of water ($K_w$)
$K_c = \dfrac{1}{K_w}=1 \times 10^{14}$
Manipulation of Constants
The equilibrium constant expression must be manipulated if a reaction is reversed or split into elementary steps. When the reaction is reversed, the equilibrium constant expression is inverted. The new expression would be written as:
$K'= \dfrac{1}{\dfrac{[G]^g[H]^h}{[A]^a[B]^b}} = \dfrac{[A]^a[B]^b}{[G]^g[H]^h}$
When there are multiple steps in the reaction, each with its own K (in a scenario similar to Hess's law problems), then the successive K values for each step are multiplied together to calculate the overall K.
Activities
Because the concentration of reactants and products are not dimensionless (i.e. they have units) in a reaction, the actual quantities used in an equilibrium constant expression are activities. Activity is expressed by the dimensionless ratio $\frac{[X]}{c^{\circ}}$ where $[X]$ signifies the molarity of the molecule and c is the chosen reference state:
$a_b=\dfrac{[B]}{c^{\circ}}$
For gases that do not follow the ideal gas laws, using activities will accurately determine the equilibrium constant that changes when concentration or pressure varies. Thus, the units are canceled and $K$ becomes unitless.
Practice Problems
1. Write the equilibrium constant expression for each reaction.
1. $2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}$
2. $N_2O_{ (g)} + \dfrac{1}{2} O_{2(g)} \rightleftharpoons 2NO_{(g)}$
3. $Cu_{(s)} + 2Ag^+_{(aq)} \rightleftharpoons Cu^{+2}_{(aq)} + 2Ag_{(s)}$
4. $CaCO_{3 (g)} \rightleftharpoons CaCO_{(s)} + CO_{2 (g)}$
5. $2NaHCO_{3 (s)} \rightleftharpoons Na_2CO_{3 (s)} + CO_{2 (g)} + H_2O_{ (g) }$
2. What is the $K_c$ of the following reaction? $2SO_{2 (g)} + O_{2 (g)} \rightleftharpoons 2SO_{3 (g)}$ with concentration $SO_{2(g)} = 0.2 M O_{2 (g)} = 0.5 M SO_{3 (g)} = 0.7 \;M$ Also, What is the $K_p$ of this reaction? At room temperature?
3. For the same reaction, the differing concentrations: $SO_{2 (g)} = 0.1\; M O_{2(g)} = 0.3\; M \;SO_{3 (g)} = 0.5\; M$ Would this go towards to product or reactant?
4. Write the Partial Pressure Equilibrium: $C_{(s)} + O_{2 (g)} \rightarrow CO_{2 (g)}$
5. Write the chemicl reaction for the following equilibrium constant: $K_p= \dfrac{P^2_{HI}}{P_{H_2} \times P_{I_2}}$
Outside Links
• For more information on equilibrium constant expressions please visit the Wikipedia site: http://en.Wikipedia.org/wiki/Equilibrium_constant
• The image below can be found here: image.tutorvista.com/content/chemical-equilibrium/reaction-rate-time-graph.gif
Answers to Practice Problems
1. $K_c = \dfrac{ [SO_3]^2}{[O_2][SO_2]^2}$
2. $Kc = \dfrac{ [NO]^2}{[O_2]^{0.5}[N_2O]}$
3. $Kc = \dfrac{ [Cu^{+2}]}{[Ag^+]^2 }$
4. $Kc = \dfrac{ [CO_2]}{[CaCO_3]}$
5. $K_c = [H_2O][CO_2]$
What is $K_c$ for the Reaction
1) Kc: 24.5
Kp: 1.002 Atm
2) Qc= 83.33 > Kc therefore the reaction shifts to the left
1. $K_p= \dfrac{P_{CO_2}} {P_{O_2}}$
2. $H_2 (g)+ I_2 (g) \rightarrow 2HI(g)$
Contributors and Attributions
• Heather Voigt
• Modified by Tom Neils (Grand Rapids Community College)
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The reaction quotient ($Q$) measures the relative amounts of products and reactants present during a reaction at a particular point in time. The reaction quotient aids in figuring out which direction a reaction is likely to proceed, given either the pressures or the concentrations of the reactants and the products. The $Q$ value can be compared to the Equilibrium Constant, $K$, to determine the direction of the reaction that is taking place.
K vs. Q
The main difference between $K$ and $Q$ is that $K$ describes a reaction that is at equilibrium, whereas $Q$ describes a reaction that is not at equilibrium. To determine $Q$, the concentrations of the reactants and products must be known. For a given general chemical equation:
$aA + bB \rightleftharpoons cC + dD \tag{1}\nonumber$
the Q equation is written by multiplying the activities (which are approximated by concentrations) for the species of the products and dividing by the activities of the reactants. If any component in the reaction has a coefficient, indicated above with lower case letters, the concentration is raised to the power of the coefficient. $Q$ for the above equation is therefore:
$Q_c = \dfrac{[C]^c[D]^d}{[A]^a[B]^b} \tag{2}\nonumber$
Note
This equation only shows components in the gaseous or aqueous states. Each pure liquid or solid has an activity of one and can be functionally omitted. Equilibrium constants really contain a ratio of concentrations (actual concentration divided by the reference concentration that defines the standard state). Because the standard state for concentrations is usually chosen to be 1 mol/L, it is not written out in practical applications. Hence, the ratio does not contain units.
A comparison of $Q$ with $K$ indicates which way the reaction shifts and which side of the reaction is favored:
• If $Q>K$, then the reaction favors the reactants. This means that in the $Q$ equation, the ratio of the numerator (the concentration or pressure of the products) to the denominator (the concentration or pressure of the reactants) is larger than that for $K$, indicating that more products are present than there would be at equilibrium. Because reactions always tend toward equilibrium (Le Châtelier's Principle), the reaction produces more reactants from the excess products, therefore causing the system to shift to the LEFT. This allows the system to reach equilibrium.
• If $Q<K$, then the reaction favors the products. The ratio of products to reactants is less than that for the system at equilibrium—the concentration or the pressure of the reactants is greater than the concentration or pressure of the products. Because the reaction tends toward reach equilibrium, the system shifts to the RIGHT to make more products.
• If $Q=K$, then the reaction is already at equilibrium. There is no tendency to form more reactants or more products at this point. No side is favored and no shift occurs.
Activity
Another important concept that is used in the calculation of the reaction quotient is called an activity. For example, consider the $Q$ equation for this acid/base reaction:
$\ce{CH_3CH_2CO_2H(aq) + H_2O(l) <=> H_3O^{+}(aq) + CH_3CH_2CO_2^{-}(aq)} \nonumber$
The $Q$ equation is written as the concentrations of the products divided by the concentrations of the reactants, but only including components in the gaseous or aqueous states and omitting pure liquid or solid states. The $Q$ equation for this example is the following:
$Q = \dfrac{[\ce{H3O^{+}(aq)}][\ce{CH3CH2CO2^{-}(aq)}]}{[\ce{CH3CH2CO2H(aq)}]} \nonumber$
Example 1
What is the $Q$ value for this equation? Which direction will the reaction shift if $K_c$ = 1.0?
$\ce{CO(g) + H_2O(g) \rightleftharpoons CO_2(g) + H_2(g)} \nonumber$
• [CO2(g)]= 2.0 M
• [H2(g)]= 2.0 M
• [CO(g)]= 1.0 M
• [H2O(g)]= 1.0 M
Solution
Step 1: Write the $Q$ formula:
$Q_c = \dfrac{[CO_2][H_2]}{[CO][H_2O]} \nonumber$
Step 2: Plug in given concentration values:
\begin{align*} Q_c &= \dfrac{(2.0)(2.0)}{(1.0)(1.0)} \4pt] &= 4.0 \end{align*} Step 3: Compare \(Q to K:
Because $4.0 > 1.0$, then $Q > K$ and the reaction shifts left toward the reactants.
Answer
Q= 4.0 and the reaction shifts left.
Example 2
Find the value of $Q$ and determine which side of the reaction is favored with $K=0.5$.
$\ce{HCl(g) + NaOH(aq) \rightleftharpoons NaCl(aq) + H_2O(l)} \nonumber$
with
• $[\ce{HCl}]= 3.2$
• $[\ce{NaOH}]= 4.3$
• $[\ce{NaCl}]=6$
Solution
Step 1: Write the $Q$ formula. Because the activity of a liquid is 1, we can omit the water component in the equation.
$Q_c = \dfrac{[NaCl{(aq)}]}{[HCl{(g)}][NaOH{(aq)}]}$
Step 2: Plug in given concentrations into the $Q$ formula:
$Q_c = \dfrac{[6]}{[3.2][4.3]}$
Step 3: Calculate using the given concentrations:
$Q = 0.436$
Step 4: Compare Q to K. The $Q$ value, 0.436, is less than the given $K$ value of 0.5, so $Q < K$.
Because $Q$ < K, the reaction is not at equilibrium and proceeds to the products side to reach dynamic equilibrium once again.
Answer: Q= 0.436 and the reaction favors the products.
Example 3
Given the equation with $K= 0.040$. Find $Q$ and determine which direction the reaction will shift to reach equilibrium.
$\ce{N_2(g) + 3H_2(aq) \rightleftharpoons 2NH_3(g)} \nonumber$
with
• $[\ce{N2}]= 0.04M$
• $[\ce{H2}]= 0.09M$
Solution
Step 1: Write the $Q$ formula:
$Q_c = \dfrac{[NH_3{(g)}]^2}{[N_2{(g)}][H_2{(g)}]^3}\nonumber$
Step 2: Plug in values. Because the concentrations for $N_2$ and $H_2$ were given, they can be inserted directly into the equation. However, no concentration value was given for NH3 so the concentration is assumed to be 0.
$Q_c = \dfrac{(0)^2}{(0.04)(0.09)^3}\nonumber$
Step 3: Solve for Q:
$Q=0\nonumber$
Step 4: Compare $Q$ to K. Because $K=0.04$ and $Q=0$, $K > Q$ and the reaction will shift right to regain equilibrium. Answer: $Q=0$, the reaction shifts right.
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Learning Objectives
• Describe chemical equilibria at the molecular level as a dynamic process.
Explaining Chemical Equilibria
This page looks at the basic ideas underpinning the idea of a chemical equilibrium. It talks about reversible reactions and how they behave if the system is closed. This leads to the idea of a dynamic equilibrium, and what the common term "position of equilibrium" means.
Reversible reactions
A reversible reaction is one which can be made to go in either direction depending on the conditions. If you pass steam over hot iron the steam reacts with the iron to produce a black, magnetic oxide of iron called triiron tetroxide, $Fe_3O_4$
$3 Fe (s) + 4H_2 O(g) \rightarrow Fe_3O_4(s) + 4H_2 (g)$
The hydrogen produced in the reaction is swept away by the stream of steam.
Under different conditions, the products of this reaction will also react together. Hydrogen passed over hot triiron tetroxide reduces it to iron. Steam is also produced.
$Fe_3O_4 (s) + 4H_2 (g) \rightarrow 3Fe(s) + 4H_2O (g)$
This time the steam produced in the reaction is swept away by the stream of hydrogen.
These reactions are reversible, but under the conditions normally used, they become one-way reactions. The products aren't left in contact with each other, so the reverse reaction can't happen.
Reversible reactions happening in a closed system
A Closed System is one in which no substances are either added to the system or lost from it. Energy can, however, be transferred in or out at will. In the example we've been looking at, you would have to imagine iron being heated in steam in a closed container. Heat is being added to the system, but none of the substances in the reaction can escape. The system is closed.
As the triiron tetroxide and hydrogen start to be formed, they will also react again to give the original iron and steam. So, if you analyzed the mixture after a while, what would you find? You would find that you had established what is known as a dynamic equilibrium. To explain what that means, we are going to use a much simpler example . . .
Dynamic equilibria
Imagine a substance which can exist in two forms - a blue form or an orange form - and that each form can react to give the other one. We are going to let them react in a closed system; neither form can escape. Assume that the blue form turns into the orange one much faster than the other way round. In any given time, these are the chances of the two changes happening:
You can simulate this very easily with some coloured paper cut up into small pieces (a different colour on each side), and a dice. The following are the real results of a "reaction". Started with 16 blue squares and looked at each one in turn and decided whether it should change colour by throwing a dice.
• A blue square was turned into an orange square (the bit of paper was turned over!) if I threw a 4, 5 or 6
• An orange square was turned into a blue square only if I threw a 6 while I was looking at that particular square.
Once I had looked at all 16 squares, I started the process all over again - but obviously with a different starting pattern. The diagrams show the results of doing this 11 times (plus the original 16 blue squares).
You can see that the "reaction" is continuing all the time. The exact pattern of orange and blue is constantly changing. However, the overall numbers of orange and of blue squares remain remarkably constant - most commonly, 12 orange ones to 4 blue ones.
Explaining the term "dynamic equilibrium"
The reaction has reached equilibrium in the sense that there is no further change in the numbers of blue and orange squares. However, the reaction is still continuing. For every orange square that turns blue, somewhere in the mixture it is replaced by a blue square turning orange. This is known as a dynamic equilibrium. The word dynamic shows that the reaction is still continuing. You can show dynamic equilibrium in an equation for a reaction by the use of special arrows. In the present case, you would write it as:
It is important to realize that this doesn't just mean that the reaction is reversible. It means that you have a reversible reaction in a state of dynamic equilibrium.
Note: The "forward reaction" and the "back reaction"
The change from left to right in the equation (in this case from blue to orange as it is written) is known as the forward reaction. The change from right to left is the back reaction.
In the example above, the equilibrium mixture contained more orange squares than blue ones. Position of equilibrium is a way of expressing this. You can say things like:
• "The position of equilibrium lies towards the orange."
• "The position of equilibrium lies towards the right-hand side."
If the conditions of the experiment change (by altering the relative chances of the forward and back reactions happening), the composition of the equilibrium mixture will also change.
For example, if changing the conditions produced more blue in the equilibrium mixture, you would say "The position of equilibrium has moved to the left" or "The position of equilibrium has moved towards the blue".
Reaching equilibrium from the other side
What happens if you started the reaction with orange squares rather than blue ones, but kept the chances of each change happening the same as in the first example? This is the result of my "reaction".
Once again, you can see that exactly the same position of equilibrium is being established as when we started with the blue squares. You get exactly the same equilibrium mixture irrespective of which side of the equation you start from - provided the conditions are the same in both cases.
A more formal look at dynamic equilibria
This is the equation for a general reaction which has reached dynamic equilibrium:
$A + B \rightleftharpoons C + D$
How did it get to that state? Let's assume that we started with $A$ and $B$. At the beginning of the reaction, the concentrations of A and $B$ were at their maximum. That means that the rate of the reaction was at its fastest. As $A$ and $B$ react, their concentrations fall. That means that they are less likely to collide and react, and so the rate of the forward reaction falls as time goes on.
In the beginning, there is not any $C$ and $D$, so there can't be any reaction between them. As time goes on, though, their concentrations in the mixture increase and they are more likely to collide and react. With time, the rate of the reaction between $C$ and $D$ increases:
Eventually, the rates of the two reactions will become equal. $A$ and $B$ will be converting into $C$ and $D$ at exactly the same rate as $C$ and $D$ convert back into $A$ and $B$ again.
At this point there will not be any further change in the amounts of $A$, $B$, $C$ and $D$ in the mixture. As fast as something is being removed, it is being replaced again by the reverse reaction. We have reached a position of dynamic equilibrium.
Summary
A dynamic equilibrium occurs when you have a reversible reaction in a closed system. Nothing can be added to the system or taken away from it apart from energy. At equilibrium, the quantities of everything present in the mixture remain constant, although the reactions are still continuing. This is because the rates of the forward and the back reactions are equal. If you change the conditions in a way which changes the relative rates of the forward and back reactions you will change the position of equilibrium - in other words, change the proportions of the various substances present in the equilibrium mixture. This is explored in detail on other pages in this equilibrium section.
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It is a common observation that most of the reactions when carried out in closed vessels do not go to completion, under a given set of conditions of temperature and pressure. In fact in all such cases, in the initial state, only the reactants are present but as the reaction proceeds, the concentration of reactants decreases and that of products increases. Finally a stage is reached when no further change in concentration of the reactants and products is observed.
Example 1
If a mixture of gaseous hydrogen and iodine vapors is made to react at 717 k in a closed vessel for about 2 - 3 hours, gaseous hydrogen iodide is produced according to the following equation:
$H_{2\; (g)} + I_{2\; (g)} \rightarrow 2HI_{(g)}$
But along with gaseous hydrogen iodide, there will be some amount of unreacted gaseous hydrogen and gaseous iodine left. On the other hand if gaseous hydrogen iodide is kept at 717K in a closed vessel for about 2 - 3 hours it decomposes to give gaseous hydrogen and gaseous iodine.
$2HI_{(g)} \rightarrow H_{2\; (g)} + I_{2\; (g)}$
In this case also some amount of gaseous hydrogen iodide will be left unreacted. This means that the products of certain reactions can be converted back to the reactants. These types of reactions are called reversible reactions.
Thus, in reversible reactions the products can react with one another under suitable conditions to give back the reactants. In other words, in reversible reactions the reaction takes place in both the forward and backward directions. The reversible reaction may be expressed as:
$H_{2\; (g)} + I_{2\; (g)} \rightleftharpoons 2HI_{(g)}$
These reversible reactions never go to completion if performed in a closed container. For a reversible chemical reaction, an equilibrium state is attained when the rate at which a chemical reaction is proceeding in forward direction equals the rate at which the reverse reaction is proceeding.
Contributors and Attributions
Binod Shrestha (University of Lorraine)
Reversible vs. Irreversible Reactions
Reversible and irreversible reactions are prevalent in nature and are responsible for reactions such as the breakdown of ammonia.
Introduction
It was believed that all chemical reactions were irreversible until 1803, when French chemist Claude Louis Berthollet introduced the concept of reversible reactions. Initially he observed that sodium carbonate and calcium chloride react to yield calcium carbonate and sodium chloride; however, after observing sodium carbonate formation around the edges of salt lakes, he realized that large amount of salts in the evaporating water reacted with calcium carbonate to form sodium carbonate, indicating that the reverse reaction was occurring.
Chemical reactions are represented by chemical equations. These equations typically have a unidirectional arrow ($\rightarrow$) to represent irreversible reactions. Other chemical equations may have a bidirectional harpoons ($\rightleftharpoons$) that represent reversible reactions (not to be confused with the double arrows $\leftrightarrow$ used to indicate resonance structures). To review the fundamentals of chemical reactions, click here: Chemical Reactions
Irreversible Reactions
A fundamental concept of chemistry is that chemical reactions occurred when reactants reacted with each other to form products. These unidirectional reactions are known as irreversible reactions, reactions in which the reactants convert to products and where the products cannot convert back to the reactants. These reactions are essentially like baking. The ingredients, acting as the reactants, are mixed and baked together to form a cake, which acts as the product. This cake cannot be converted back to the reactants (the eggs, flour, etc.), just as the products in an irreversible reaction cannot convert back into the reactants.
An example of an irreversible reaction is combustion. Combustion involves burning an organic compound—such as a hydrocarbon—and oxygen to produce carbon dioxide and water. Because water and carbon dioxide are stable, they do not react with each other to form the reactants. Combustion reactions take the following form:
$C_xH_y + O_2 \rightarrow CO_2 + H_2O$
Reversible Reactions
In reversible reactions, the reactants and products are never fully consumed; they are each constantly reacting and being produced. A reversible reaction can take the following summarized form:
$A + B \underset{k_{-1}} {\overset{k_1} {\rightleftharpoons}} C + D$
This reversible reaction can be broken into two reactions.
Reaction 1: $A + B \xrightarrow{k_1}C+D$
Reaction 2: $C + D \xrightarrow{k_{-1}}A+B$
These two reactions are occurring simultaneously, which means that the reactants are reacting to yield the products, as the products are reacting to produce the reactants. Collisions of the reacting molecules cause chemical reactions in a closed system. After products are formed, the bonds between these products are broken when the molecules collide with each other, producing sufficient energy needed to break the bonds of the product and reactant molecules.
Below is an example of the summarized form of a reversible reaction and a breakdown of the reversible reaction N2O4 ↔ 2NO2
Reaction 1 and Reaction 2 happen at the same time because they are in a closed system.
Blue: Nitrogen Red: Oxygen
Reaction 1 Reaction 2
Imagine a ballroom. Let reactant A be 10 girls and reactant B be 10 boys. As each girl and boy goes to the dance floor, they pair up to become a product. Once five girls and five boys are on the dance floor, one of the five pairs breaks up and moves to the sidelines, becoming reactants again. As this pair leaves the dance floor, another boy and girl on the sidelines pair up to form a product once more. This process continues over and over again, representing a reversible reaction.
Unlike irreversible reactions, reversible reactions lead to equilibrium: in reversible reactions, the reaction proceeds in both directions whereas in irreversible reactions the reaction proceeds in only one direction. To learn more about this phenomenon, click here: Chemical Equilibrium
If the reactants are formed at the same rate as the products, a dynamic equilibrium exists. For example, if a water tank is being filled with water at the same rate as water is leaving the tank (through a hypothetical hole), the amount of water remaining in the tank remains consistent.
Connection to Biology
There are four binding sites on a hemoglobin protein. Hemoglobin molecules can either bind to carbon dioxide or oxygen. As blood travels through the alveoli of the lungs, hemoglobin molecules pick up oxygen-rich molecules and bind to the oxygen. As the hemoglobin travels through the rest of the body, it drops off oxygen at the capillaries for the organ system to use oxygen. After expelling the oxygen, it picks up carbon dioxide. Because this process is constantly carried out through the body, there are always hemoglobin molecules picking or expelling oxygen and other hemoglobin molecules that are picking up or expelling carbon dioxide. Therefore, the hemoglobin molecules, oxygen, and carbon dioxide are reactants while the hemoglobin molecules with oxygen or carbon dioxide bound to them are the products. In this closed system, some reactants convert into products as some products are changing into reactants, making it similar to a reversible reaction.
Contributors and Attributions
• Heather Yee (UCD), Mandeep Sohal (UCD)
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A closed system containing at least two phases is called a heterogeneous system. Reactions (changes) between or among phases are driven by energy manifested in temperature of chemical potentials. When there is no net change in a closed system among the phases, the system is said to have reached an equilibrium condition. The application of the equilibrium principles to describe the changes between solids and their solutions is the focus in this section. For these salts, the concept of solubility product, Ksp, is used to described the formation of solutions and solids.
Heterogeneous Equilibria
Learning Objectives
• Recognize common ions from various salts, acids, and bases.
• Calculate concentrations involving common ions.
• Calculate ion concentrations involving chemical equilibrium.
• Apply the Newton's method to solve cubic equations.
The solubility products Ksp's are equilibrium constants in hetergeneous equilibria. If several salts are present in a system, they all ionize in the solution. If the salts contain a common cation or anion, these salts contribute to the concentration of the common ion. Contributions from all salts must be included in the calculation of concentration of the common ion. For example, a solution containing sodium chloride and potassium chloride will have the following relationship:
$\mathrm{[Na^+] + [K^+] = [Cl^-]}$
Consideration of charge balance or mass balance or both leads to the same conclusion.
Common Ions
When $\ce{NaCl}$ and $\ce{KCl}$ are dissolved in the same solution, the $\mathrm{ {\color{Green} Cl^-}}$ ions are common to both salts. In a system containing $\ce{NaCl}$ and $\ce{KCl}$, the $\mathrm{ {\color{Green} Cl^-}}$ ions are common ions.
$\mathrm{NaCl \rightleftharpoons Na^+ + {\color{Green} Cl^-}}$
$\mathrm{KCl \rightleftharpoons K^+ + {\color{Green} Cl^-}}$
$\mathrm{CaCl_2 \rightleftharpoons Ca^{2+} + {\color{Green} 2 Cl^-}}$
$\mathrm{AlCl_3 \rightleftharpoons Al^{3+} + {\color{Green} 3 Cl^-}}$
$\mathrm{AgCl \rightleftharpoons Ag^+ + {\color{Green} Cl^-}}$
For example, when $\ce{AgCl}$ is dissolved into a solution already containing $\ce{NaCl}$ (actually $\ce{Na+}$ and $\ce{Cl-}$ ions), the $\ce{Cl-}$ ions come from the ionization of both $\ce{AgCl}$ and $\ce{NaCl}$. Thus, $\ce{[Cl- ]}$ differs from $\ce{[Ag+]}$. The following examples show how the concentration of the common ion is calculated.
Example $1$
What are $\ce{[Na+]}$, $\ce{[Cl- ]}$, $\ce{[Ca^2+]}$, and $\ce{[H+]}$ in a solution containing 0.10 M each of $\ce{NaCl}$, $\ce{CaCl2}$, and $\ce{HCl}$?
Solution
Due to the conservation of ions, we have
$\mathrm{[Na^+] = [Ca^{2+}] = [H^+] = 0.10\: \ce M}$.
but
\begin{alignat}{3} &\ce{[Cl- ]} &&= && && \:\textrm{0.10 (due to NaCl)}\ & && && + &&\mathrm{\:0.20\: (due\: to\: CaCl_2)}\ & && && + &&\mathrm{\:0.10\: (due\: to\: HCl)}\ & &&= && &&\mathrm{\:0.40\: M} \end{alignat}
Exercise $1$
John poured 10.0 mL of 0.10 M $\ce{NaCl}$, 10.0 mL of 0.10 M $\ce{KOH}$, and 5.0 mL of 0.20 M $\ce{HCl}$ solutions together and then he made the total volume to be 100.0 mL. What is $\ce{[Cl- ]}$ in the final solution?
$\mathrm{[Cl^-] = \dfrac{0.1\: M\times 10\: mL+0.2\: M\times 5.0\: mL}{100.0\: mL} = 0.020\: M}$
Example $2$
What is the solubility (in moles per liter) of $\ce{AgCl}$ in a solution that is 0.05 M in $\ce{KCl}$? Ksp for $\ce{AgCl}$ is 1.0E-10.
Solution
Assume $x = \ce{[Ag+]}$ to be the molar solubility, then we can write the equilibrium concentration below the formula of the equilibrium.
$\begin{array}{cccccl} \ce{AgCl &\rightarrow &Ag+ &+ &Cl- &}\ &&x &&0.05+x &\leftarrow \ce{equilibrium\:\, concentration} \end{array}$
By definition of the Ksp, we have
$x (0.05 + x) = K_{\ce{sp}}$;
Thus,
$x = \dfrac{\textrm{1.0e-10}}{0.05} = \textrm{2e-9 M}$
Note that 0.05 + x is approximately 0.05, and you may use this approximation for the calculation.
Example $3$
What is the solubility of $\ce{Li2CO3}$ in a solution also containing 0.10 M of $\ce{K2CO3}$? The solubility product for $\ce{Li2CO3}$ is 0.0017.
Solution
In this case, $\ce{CO3^2-}$ is the common ion between the two salts.
$\ce{K2CO3 \rightleftharpoons 2 K+ + CO3^2-}$
$\ce{Li2CO3 \rightleftharpoons 2 Li+ + CO3^2-}$
This question implies that $\ce{K2CO3}$ is soluble. Thus, the only heterogeneous equilibrium to be considered is
$\ce{K2CO3 \rightleftharpoons 2 K+ + CO3^2-}$
Let $\ce{[Li+]} = 2 x$, then $\ce{[CO3^2- ]} = 0.10 + x$. Thus, we can write down the equilibrium equation again, and write the concentrations below the chemical formula:
$\begin{array}{ccccc} \ce{Li2CO3 &\rightleftharpoons &2 Li+ &+ &CO3^2-}\ &&2 x &&0.10+x \end{array}$
$\ce{[Li+]^2 [CO3^2- ]} = (2 x)^2 (0.10+x) = 0.0017$
Expanding the formula results in
$x^3 + 0.10 x^2 - 0.00043 = 0\.] A general method to solve this equation is to use the Newton's method. For which, we assume that \[y = x^3 + 0.10 x^2 - 0.00043 = 0.$
We assume the value of $x = 0.06$, and substitute in the equation to calculate y.
$y = 0.06^3 + 0.10\times 0.06^2 - 0.00043 = 0.000146$
We now assume $x = 0.05$, and substitute in the equation to evaluate y,
$y = 0.05^3 + 0.10\times 0.05^2 - 0.00043 = -0.000055$
Since the values for y calculated for x = 0.05 and 0.06 have different sign, the x value should lie between 0.05 and 0.06. We further assume the value for x = 0.053, and we obtain a y value
$y = 0.053^3 + 0.10\times 0.053^2 - 0.00043 = -2.2\times 10^{-7}$
Thus, the value of x should be greater than 0.053. We can increase x to 0.0531 or 0.0532, and evaluate y again:
$y = 0.0531^3 + 0.10\times 0.0531^2 - 0.00043 = 1.7\times 10^{-6}$
Thus, x value should lie between 0.0531 and 0.0530. By this method, we have evaluated x values to three significant figures. We only need two significant figures due to the nature of the data in the problem. Thus we have
$\ce{[Li+]} = 2 x = \textrm{0.106 \;M}$
DISCUSSION
The molar solubility of $\ce{Li2CO3}$ in a solution also containing 0.10 M of $\ce{K2CO3}$ is 0.053 mole per liter, but
$\ce{[Li+]} = 2 x = \textrm{0.106 M}$
This example illustrates the Newton's method for solving cubic equations. Note that if the Ksp is small, then x is a very small value. In this case, $0.10 + x \approx 0.10$. You do not need to use the Newton's method in this case.
Example $4$
What is the solubility of $\ce{Li2CO3}$ in a solution which is 0.20 M in $\ce{Na2CO3}$? The solubility product of $\ce{Li2CO3}$ is 1.7e-3 M3.
Solution
Since the solution contains 0.20 M $\ce{Na2CO3}$, $\ce{[CO3^2- ]} = \textrm{0.20 M}$. Assume the solubility to be x M of $\ce{Li2CO3}$, then we have
$\begin{array}{ccccc} \ce{Li2CO3 &\rightleftharpoons &2 Li+ &+ &CO3^2-}\ &&2 x &&x + 0.20\:\ce M \end{array}$
Thus,
\begin{align} (2 x)^2 (x+0.20) &= \textrm{1.7e-3 M}^3\ 4x^3 + 0.20 x^2 &= \textrm{1.7e-3 M}^3 \end{align}
$x^3 + 0.20 x^2 - 0.00043 = 0$
There is no definite way to solve this equation, but the Newton's method is very useful. In this method, we let
$y = x^3 + 0.20 x^2 - 0.00043$
We guesstimate x and evaluate y using the above expression. The solution is a value for x so that y=0. We refine x values between two x values that give positive and negative y values in progression as shown in the Table below.
y x Remarks
$2 \times 10^{4}$ 0.05 any small value for x
$5\times 10^{-4}$ 0.06 y larger, try x < 0.05
$-5\times 10^{-5}$ 0.04 0.04 < x < 0.05
$8\times 10^{-6}$ 0.0425 0.04 < x < 0.0425
$-1\times 10^{-5}$ 0.0415 0.0415 < x < 0.0425
$3\times 10^{-6}$ 0.0420 0.0415 < x < 0.0420
$-5\times 10^{-6}$ 0.0419 0.0415 < x < 0.0419
DISCUSSION
This is the same type of question as Example 3. As the carbonate concentration increases from 0.10 to 0.20 M, the solubility of lithium carbonate reduces to 0.042 M from 0.053 M. Note that the lithium ion concentration is 0.084 M in this case.
Note that when used to treat depression, lithium carbonate is usually called lithium in the health care profession.
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Learning Objectives
• Describe changes involving several phases (states) of materials.
• Apply equilibrium concept to interpret changes in system containing several phases.
The solubility products Ksp's are equilibrium constants in hetergeneous equilibria. If several salts are present in a system, they all ionize in the solution. If the salts contain a common cation or anion, these salts contribute to the concentration of the common ion. Contributions from all salts must be included in the calculation of concentration of the common ion. For example, a solution containing sodium chloride and potassium chloride will have the following relationship:
Such a tendency is called the ACTIVITY of the phase. As long as such a phase exists, its tendency or activity remains constant. However, the activities of substances in a gas phase are proportional to their partial pressures or concentrations. For a solution, their activities are proportional to their concentrations. Thus, their partial pressures or concentrations are indicators of their tendency to change.
At equilibrium, these tendencies of changes reach certain proportions such that the forward and reverse changes are balanced. Similar to the equilibrium conditions of homogeneous systems, heterogeneous systems also tend to reach equilibrium conditions. EQUILIBRIUM CONSTANTS can also be assigned to describe equilibrium conditions of heterogeneous systems.
We shall look at several types of heterogeneous systems to illustrate how we deal with their behavior or change.
Saturated Solutions
Saturated solutions are typical heterogeneous equilibria.
We all experience that when solid sugar is present in a sugar solution, putting more sugar in it will not increase its solubility. The sugar solution will not get any sweeter, that is if we really can taste the sweetness as proportional to the concentration.
The solution mentioned above is called a saturated solution. The main criterion for a saturated solution is that the amount in the solid phase remains constant, or the concentration remains constant.
A saturated solution is the result of equilibrium between the solute and its solution. For this type of equilibria, the equilibrium constant is the concentration of the saturated solution. We write the dissolution in an equation:
$\ce{C12H22O_{11\large{(s)}} \rightleftharpoons C12H22O_{11\large{(aq)}}}, \hspace{10px} K_{\ce c} = \ce{[C12H22O11]\: (saturated)}$
and $\ce{[C12H22O11]}$ represents the saturated concentration of the solution, $\ce{C12H22O_{11\large{(aq)}}}$. We simply treat the activity of the solid as 1 (unity).
When a salt dissolves, the solution contains ions rather than molecules. The equilibrium constant is the product of the ion concentrations. This is illustrated next.
Equilibrium of Salt Solutions
When a salt dissolves in water, ions rather than molecules are present in the solution. For example, when silver chloride dissolves in water, $\ce{Ag+}$ and $\ce{Cl-}$ or more precisely $\ce{Ag+(H2O)6}$ and $\ce{Cl- (H2O)6}$ are present in the solution. We write the dissolving process and the equilibrium constant this way:
$\ce{AgCl_{\large{(s)}} \rightleftharpoons Ag+ + Cl- }, \hspace{10px} K = \ce{[Ag+] [Cl- ]}$
Since the solubility of $\ce{AgCl}$ is small, the concentrations of $\ce{Ag+}$ and $\ce{Cl-}$ are very small. As we shall see later, the equilibrium constant of sparingly soluble salts is often designated as Ksp.
More examples of dissolved salts and equilibrium constants are:
$\ce{CaCO3 \rightleftharpoons Ca^2+ + CO3^2-}, \hspace{10px} K_{\ce{sp}} = \ce{[Ca^2+] [CO3^2- ]}$;
$\ce{Al3SO4 \rightleftharpoons 2 Al^3+ + 3 SO4^2-}, \hspace{10px} K_{\ce{sp}} = \ce{[Al^3+]^2 [SO4^2- ]^3}$
We shall deal with this type of equilibrium more extensively on other pages.
Henry's Law
Dissolving of a gas in a liquid involves changes of two phases. These types of changes are examples of heterogeneous equilibrium. For this type of equilibrium, the equilibrium constant is expressed by the partial pressure rather than by the ratio of pressure and concentration.
For example, the dissolution of oxygen in water and the equilibrium constant are usually written in this way:
$\ce{O_{3\large{(g)}} \rightleftharpoons O_{3\large{(aq)}}}, \hspace{10px} K_{\ce p} = \ce{\dfrac{1}{P(O3)}}$;
or
$\ce{O_{3\large{(aq)}} \rightleftharpoons O_{3\large{(g)}}}, \hspace{10px} K'_{\ce p} = \ce{P(O3)}$;
The concentration is not 1 (unity), but we chose to express the equilibrium constant in terms of the partial pressure of oxygen. Ideally, in a closed system, the partial pressure of oxygen changes as it dissolves in water, and eventually reaches an equilibrium. But due to the small solubility of $\ce{O3}$, the changes in partial pressure are not noticeable. Furthermore, this type of system has been investigated earlier by Henry, and he noticed that the solubility (concentration) of a gas in a liquid is proportional to the partial pressure. This is now known as Henry's law.
Heterogeneous Equilibria Involving Chemical Reactions
For heterogeneous equilibria, the equilibrium constants, K, should be expressed as a function of the concentrations of reactants and products of solution or gases. For convenience, the ACTIVITY of a SOLID or LIQUID is given as 1 (unity).
For example, when limestone or shell (of shell fish), $\ce{CaCO_{3\large{(s)}}}$, is heated, $\ce{CO3}$ gas is released leaving the $\ce{CaO}$ as a solid. We write the reaction and equilibrium constant in this form:
$\ce{CaCO_{3\large{(s)}} \rightleftharpoons CaO_{\large{(s)}} + CO_{3\large{(g)}}}, \hspace{10px} K_{\ce p} = \ce{P(CO3)}$
since the activities of the solids are considered unity.
The Kp is used to mean that the equilibrium constant is expressed in terms of partial pressures. In this example, Kp is the saturated partial pressure of $\ce{CO3}$ when $\ce{CaCO3}$ and $\ce{CaO}$ solids are present, and no net change will take place.
Phase Transition and Equilibrium
We are used to the idea that water vapor pressure is a constant at a definite temperature. We seldom think of it being an equilibrium constant, but it is. The reaction can be represented by,
$\ce{H2O_{\large{(l)}} \rightleftharpoons H2O_{\large{(g)}}}, \hspace{10px} K_{\ce p} = \ce{P(H2O)}$
$\ce{H2O_{\large{(s)}} \rightleftharpoons H2O_{\large{(g)}}}, \hspace{10px} K_{\ce p} = \ce{P(H2O)}$
On the phase diagram of water, the equilibrium conditions between various phases are marked by curves. The sublimation, evaporation, and melting curves show the dependence of equilibrium on pressure and temperature. A change in temperature will shift the equilibrium along the paths on these curves.
For your information, the vapor pressure of ice and water is listed in the data section. Here are some values:
T -10 -5 0 5 10 degree C
ice 1.95 3.01 4.579 mmHg
water 2.149 3.163 4.579 6.343 9.209 mmHg
What do you expect the vapor pressure of ice at 5 deg C is? Is it higher than or lower than 6.343 mmHg? Well, find out!
Questions
1. Does water vapor pressure increase or decrease as the temperature increases?
2. The vapor pressure over ice at the triple point (273.15 K) is 4.579 mmHg. What is the vapor pressure over water at the triple point?
3. At the same temperature, do you expect the vapor pressure of a salt solution to be higher or lower than that of pure water?
4. Aside from concentrations or activities, two other important factors affect equilibrium. What are they?
Solutions
1. Hint...
The higher the temp, the higher the vapor pressure.
Answer increase
Consider...
The equilibrium constant, $K = \ce{P(H2O)}$, varies with temperature.
2. Hint...
Since three phases coexist at the triple point, both solid and liquid have the same vapor pressure. Their equilibrium constants are the same.
Answer: 4.579 mmHg
Consider...
If the vapor pressures are different, the three phases cannot coexist.
3. Hint...
Vapor pressure of a solution usually is lower than that of a pure solvent.
Answer: lower
Consider...
The lowering of vapor pressure can be applied to explain the fact that salt solution freezes at lower temperature.
4. Consider...
Temperature and Pressure! Just a reminder: Le Chatelier's Principle deals with equilibrium factors.
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Learning Objectives
• Define Ksp, the solubility product.
• Explain solid/solution equilibria using $K_{sp}$ and Qsp.
• Calculate molarity of saturated solution from Ksp.
• Calculate $K_{sp}$ from molarity of saturated solution.
Phase transitions such as sublimation, deposition, melting, solidification, vaporization, and condensation are heterogeneous equilibria; so are the formation of crystals from a saturated solution, because a solid and its solution are separated phases. The picture shown here is the formation of solid from a gaseous solution. The equilibrium constants for saturated solution and solid formation (precipitate) are called solubility product, Ksp. For unsaturated and supersaturated solutions, the system is not at equilibrium, and ion products, Qsp, which have the same expression as Ksp, are used.
An oversaturated solution becomes a saturated solution by forming a solid to reduce the dissolved material. The crystals formed are called a precipitate. Often, however, a precipitate is formed when two clear solutions are mixed. For example, when a silver nitrate solution and sodium chloride solution are mixed, silver chloride crystals $\ce{AgCl_{\large{(s)}}}$ (a precipitate) are formed. $\ce{Na+}$and $\mathrm{\sideset{ }{_{3}^{-}}{NO}}$ are spectator ions.
$\mathrm{Ag^+_{\large{(aq)}} + \sideset{ }{_{\large{(aq)}}^{-}}{Cl} \rightarrow AgCl_{\large{(s)}}} \hspace{10px} \ce{(precipitate)}$
Silver chloride is one of the few chlorides that has a limited solubility. A precipitate is also formed when sodium carbonate is added to a sample of hard water,
$\mathrm{Ca^{2+}_{\large{(aq)}} + CO^{2-}_{3\large{(aq)}} \rightarrow CaCO_{3\large{(s)}}} \hspace{10px} \ce{(precipitate)}$
When a solid dissolves, we have the reverse reaction:
$\mathrm{AgCl_{\large{(s)}} \rightarrow Ag^+_{\large{(aq)}} + \sideset{ }{_{\large{(aq)}}^{-}}{Cl}} \hspace{10px} \ce{(dissolving)}$
$\ce{CaCO_{3\large{(s)}} \rightarrow Ca^2+_{\large{(aq)}} + CO^{2-}_{3\large{(aq)}}} \hspace{10px} \ce{(dissolving)}$
Solubility Products, Ksp, and Ion Products Qsp
Formations of precipitates are chemical equilibria phenomena, and we usually write these heterogeneous equilibrium in the following manner, and call the equilibrium constants solubility products, Ksp:
Equilibrium Expression for $K_{sp}$ and Qsp
$\ce{AgCl_{\large{(s)}} \rightleftharpoons Ag+_{\large{(aq)}} + Cl^{-}_{\large{(aq)}}}$ $\ce{[Ag+] [Cl- ]}$
$\ce{CaCO_{3\large{(s)}} \rightleftharpoons Ca^2+_{\large{(aq)}} + CO^{2-}_{3\large{(aq)}}}$ $\ce{[Ca^2+] [CO3- ]}$
$\ce{Li2CO_{3\large{(s)}} \rightleftharpoons 2 Li+_{\large{(aq)}} + CO^{2-}_{3\large{(aq)}}}$ $\ce{[Li+]^2 [CO3- ]}$
The solubility product, Ksp, is a special type of equilibrium constant given to a solution containing sparingly soluble salts. The symbols (aq) indicate that these ions are surrounded by water molecules; these ions are in the solution. Note the expression for the solubility product given above, please. These are special equilibrium constants, because the solid present has a constant tendency of being dissolved. Therefore, their role in $K_{sp}$ is a constant. They do not appear in the $K_{sp}$ expression. If the solution is not saturated, no precipitate will form. In this case, the product is called the ion product, Qsp, which is the product of the concentrations of the ions at any moment in time. and is equal to the solubility product for the salt when at equilibrium.
The Table of solubility product is given as Salt, $K_{sp}$ in the Handbook Section. In this table, the salts are divided into
• Carbonates, salts of $\ce{CO3^2-}$
• Chromates, salts of $\ce{CrO4^2-}$
• Halides, salts of $\ce{Cl-}$, $\ce{Br-}$, and $\ce{I-}$
• Hydroxides, salts of $\ce{OH-}$
• Oxalates, salts of $\ce{C2O4^2-}$
• Sulfates, salts of $\ce{SO4^2-}$
• Sulfides, salts of $\ce{S^2-}$
Qsp, $K_{sp}$ and Saturation
For some substances, formation of a solid or crystallization does not occur automatically whenever a solution is saturated. These substances have a tendency to form oversaturated solutions. For example, syrup and honey are oversaturated sugar solutions, containing other substances such as citric acids. For oversatureated solutions, Qsp is greater than Ksp. When a seed crystal is provided or formed, a precipitate will form immediately due to equilibrium of requiring Qsp to approach Ksp.
Sodium acetate trihydrate, $\ce{NaCH3COO\cdot 3H2O}$, when heated to 370 K will become a liquid. The sodium acetate is said to be dissolved in its own water of crystallization. The substance stays as a liquid when cooled to room temperature or even below 273 K. As soon as a seed crystal is present, crystallization occurs rapidly. In such a process, heat is released, and the liquid feels warm. Thus, the relationship among Qsp, $K_{sp}$ and saturation is given below:
Qsp < Ksp Unsaturated solution
Qsp = Ksp Saturated solution
Qsp > Ksp Oversaturated solution
Molar Solubilities and Solubility Products
Solubility products, Ksp, of salts are indirect indications of their solubilities expressed in mol/L (called molar solubility). However, the solubility products are more useful than molar solubility. The molar solubilities are affected when there are common ions present in the solution. We need to employ the solubility products to estimate the molar solubilities in these cases. When a salt is dissolved in pure water, solubility products and molar solubilities are related. This is illustrated using calcium carbonate. If x is the concentration of $\ce{Ca^2+}$ (= $\ce{[CO3^2-]}$) in the saturated solution, then
$K_{\ce{sp}} = x^2$
In this case, x is also called the molar solubility of $\ce{CaCO3}$. The following examples illustrate the relationship between solubility products, Ksp, and molar solubilities.
Example 1
The $K_{sp}$ for $\ce{AgCl}$ is 1.8e-10 M2. What is the molar solubility of $\ce{AgCl}$ in pure water?
Solution
Let x be the molar solubility, then
$\begin{array}{ccccc} \ce{AgCl &\rightleftharpoons &Ag+ &+ &Cl-}\ &&x &&x \end{array}$
\begin{align} x &= (\textrm{1.8e-10 M}^2)^{1/2}\ &= \textrm{1.3e-5 M} \end{align}
DISCUSSION
The solubility product, $K_{sp}$ is a better indicator than the usual solubility specification of g per 100 mL of solvent or moles per unit volume of solvent.
For the $\ce{AgCl}$ case, when the cation concentration is not the same as the anion concentration ($\ce{[Ag+] \neq [Cl- ]}$), solubility of $\ce{AgCl}$ can not be defined in terms of moles per L. In this case, the system can be divided into three zones. The condition $\ce{[Ag+] [Cl- ]} = K_{\ce{sp}}$, is represented by a line which divides the plane into two zones.
When $\ce{[Ag+] [Cl- ]} < K_{\ce{sp}}$, no precipitate will be formed.
When $\ce{[Ag+] [Cl- ]} > K_{\ce{sp}}$, a precipitate will be formed.
When $\ce{AgCl}$ and $\ce{NaCl}$ dissolve in a solution, both salts give $\ce{Cl-}$ ions. The effect of $\ce{[Cl- ]}$ on the solubility of $\ce{AgCl}$ is called the Common Ion Effect.
Example 2
The $K_{sp}$ for $\ce{Ag2CrO4}$ is 9e-12 M3. What is the molar solubility of $\ce{Ag2CrO4}$ in pure water?
Solution
Let x be the molar solubility of $\ce{Ag2CrO4}$, then
$\begin{array}{ccccc} \ce{Ag2CrO4 &\rightleftharpoons &2 Ag+ &+ &CrO4^2-}\ &&2 x &&x \end{array}$
$(2 x)^2 (x) = K_{\ce{sp}}$
\begin{align} x &= \left (\dfrac{\textrm{9e-12}}{4} \right )^{1/3}\ &= \textrm{1.3e-4 M} \end{align}
$\ce{[Ag+]} = \textrm{2.6e-4 M}$
and the molar solubility is 1.3e-4 M.
DISCUSSION
A similar diagram to the one given for $\ce{AgCl}$ can be drawn, but the shape of the curve representing the $K_{sp}$ is different.
Example 3
The $K_{sp}$ for $\ce{Cr(OH)3}$ is 1.2e-15 M4. What is the molar solubility of $\ce{Cr(OH)3}$ in pure water?
Solution
Let x be the molar solubility of $\ce{Cr(OH)3}$, then you have
$\begin{array}{ccccc} \ce{Cr(OH)3 &\rightleftharpoons &Cr^3+ &+ &3 OH-}\ &&x &&3 x \end{array}$
Thus,
\begin{align} x (3 x)^3 &= \textrm{1.2e-15}\ x &= \left (\dfrac{\textrm{1.2e-15}}{27} \right )^{1/4}\ &= \textrm{8.2e-5 M} \end{align}
Example 4
Very careful experiment indicates that the molar solubility of $\ce{Bi2S3}$ is 1.8e-15 M; what value of $K_{sp}$ does this compound have?
Solution
If the molar solubility of $\ce{Bi2S3}$ is 1.8e-15, then
$\begin{array}{ccccc} \ce{Bi2S3 &\rightleftharpoons &2 Bi^3+ &+ &3 S^2-}\ &&\textrm{3.6e-15} &&\textrm{5.4e-15} \end{array}$
\begin{align} K_{\ce{sp}} &= (\textrm{3.6e-15})^2 (\textrm{5.4e-15})^3\ &= \textrm{2.0e-72 M}^5 \end{align}
DISCUSSION
You should be able to calculate the $K_{sp}$ if you know the molar solubility.
Review of skills
1. Be able to write the $K_{sp}$ expression for the ionization of any salts, and calculate $K_{sp}$ from molar solubility.
2. Calculate $K_{sp}$ from molar solubility and vice versa. Pay attention to the stoichiometry of the salt. You may have to understand what the salt really is in order to know how they ionize in the solution. You have learned the method to calculate molar solubility from Ksp.
3. Perform calculations and be able to tell if a precipitate will form. The calculation is an important part of chemistry.
Questions
1. Experiment shows that the molar solubility of $\ce{CuCl}$ is $1.1 \times 10^{-3}$. What is its solubility product, Ksp?
2. The $K_{sp}$ for $\ce{Hg2I2}$ is $4.5 \times 10^{-29}$. What is its molar solubility in pure water? Remember $\ce{Hg2^2+}$ is a stable ion in solution.
3. What is the pH in a saturated solution of $\ce{Ca(OH)2}$?$K_{sp} = 5.5 \times 10^{-6}$ for $\ce{Ca(OH)2}$.
4. Chemical analysis gave $\ce{[Pb^2+]} = \textrm{0.012 M}$, and $\ce{[Br- ]} = \textrm{0.024 M}$ in a solution. From a table, you find $K_{sp}$ for $\ce{PbBr2}$ has a value of $4 \times 10^{-5}$. Is the solution saturated, oversaturated or unsaturated?
Solutions
1. Answer 1.2e-6
Hint...
The equilibrium equation and concentrations x are shown below
$\begin{array}{cccccc} \ce{CuCl &\rightleftharpoons &Cu+ &+ &Cl- &}\ &&x &&x &\hspace{10px}(x = \textrm{1.1e-3}) \end{array}$
$K_{\ce{sp}} = \ce{[Cu+][Cl- ]} = (\textrm{1.1e-3})^2 =\: ?\: \ce M^2$
If $\ce{CuCl2}$ is used, you would have $\ce{CuCl2 \rightleftharpoons Cu^2+ + 2 Cl-}$.
2. Answer 2.2e-10
Hint...
Let x be the molar solubility, then $\ce{[Hg2^2+]} = x$; and $\ce{[I- ]} = 2 x$;
$\begin{array}{ccccc} \ce{Hg2I2 &\rightleftharpoons &Hg2^2+ &+ &2 I-}\ x &&x &&2 x \end{array}$
Molar solubility = $\ce{[Hg2^2+]}$.
$\ce{[I- ]} = \ce{2 [Hg2^2+]}$.
$K_{\ce{sp}} = x\, (2 x)^2 = \textrm{4.5e-29}$; $x =\: ?$
3. Answer 12.35
Hint...
Let $\ce{[Ca^2+]} = x$, then $\ce{[OH- ]} = 2 x$. The equilibrium and concentration are represented below:
$\begin{array}{ccccc} \ce{Ca(OH)2 &\rightleftharpoons &Ca^2+ &+ &2 OH-}\ &&x &&2 x \end{array}$
$x = \left (\dfrac{\textrm{5.5e-6}}{4} \right )^{1/3}$
$\ce{[OH- ]} = 2 x$.
Note the stoichiometry of equilibrium.
4. Answer unsaturated
Hint...
[0.012] [0.024]2 = 6.9e-6 < 4e-5 (Ksp).
This question deals with the concept of ion product, Qsp.
If $Q_{\ce{sp}} = \ce{[Pb^2+][Br- ]^2} < K_{\ce{sp}}$ the solution is unsaturated.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Heterogeneous_Equilibria/Insoluble_Salts.txt
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Learning Objectives
• Calculate ion concentrations to maintain a heterogeneous equilibrium.
• Calculate pH required to precipitate a metal hydroxide.
• Design experiments to separate metal ions in a solution of mixtures of metals.
Chemical Separation by Precipitation
A mixture of metal ions in a solution can be separated by precipitation with anions such as $\ce{Cl-}$, $\ce{Br-}$, $\ce{SO4^2-}$, $\ce{CO3^2-}$, $\ce{S^2-}$, $\ce{Cr2O4^2-}$, $\ce{PO4^2-}$, $\ce{OH-}$ etc. When a metal ion or a group of metal ions form insoluble salts with a particular anion, they can be separated from others by precipitation. We can also separate the anions by precipitating them with appropriate metal ions.
There are no definite dividing lines between insoluble salts, sparingly soluble, and soluble salts, but concentrations of their saturated solutions are small, medium, and large. Solubility products are usually listed for insoluble and sparingly soluble salts, but they are not given for soluble salts. Solubility products for soluble salts are very large.
What type of salts are usually soluble, sparingly soluble and insoluble? The following are some general guidelines, but these are not precise laws.
• All nitrates are soluble. The singly charged large $\ce{NO3-}$ ions form salts with high solubilities. So do $\ce{ClO4-}$, $\ce{ClO3-}$, $\ce{NO2-}$, $\ce{HCOO-}$, and $\ce{CH3COO-}$.
• All chlorides, bromides, and iodides are soluble except those of $\ce{Ag+}$, $\ce{Hg2^2+}$, and $\ce{Pb^2+}$. $\ce{CaF2}$, $\ce{BaF2}$, and $\ce{PbF2}$ are also insoluble.
• All sulfates are soluble, except those of $\ce{Ba^2+}$, $\ce{Sr^2+}$, and $\ce{Pb^2+}$. The doubly charged sulfates are usually less soluble than halides and nitrates.
• Most singly charge cations $\ce{K+}$, $\ce{Na+}$, $\ce{NH4+}$ form soluble salts. However, $\ce{K3Co(NO2)6}$ and $\ce{(NH4)3Co(NO2)6}$ are insoluble.
These are handy rules for us to have if we deal with salts often. On the other hand, solubility is an important physical property of a substance, and these properties are listed in handbooks.
Chemical Separation of Metal Ions
Formation of crystals from a saturated solution is a heterogeneous equilibrium phenomenon, and it can be applied to separate various chemicals or ions in a solution. When solubilities of two metal salts are very different, they can be separated by precipitation. The Ksp values for various salts are valuable information, and some data are given in the Handbook of this website. On the Handbook Menu, clicking the item Salts Ksp will give the Ksp's of some salts. In the first two examples, we show how barium and strontium can be separated as chromate.
Example 1
The Ksp for strontium chromate is 3.6E-5 and the Ksp for barium chromate is 1.2E-10. What concentration of potassium chromate will precipitate the maximum amount of either the barium or the strontium chromate from an equimolar 0.30 M solution of barium and strontium ions without precipitating the other?
Solution
Since the Ksp for barium chromate is smaller, we know that $\ce{BaCrO4}$ will form a precipitate first as $\ce{[CrO4^2- ]}$ increases so that Qsp for $\ce{BaCrO4}$ also increases from zero to Ksp of $\ce{BaCrO4}$, at which point, $\ce{BaCrO4}$ precipitates. As $\ce{[CrO4^2- ]}$ increases, $\ce{[Ba^2+]}$ decreases. Further increase of $\ce{[CrO4^2- ]}$ till Qsp for $\ce{SrCrO4}$ increases to Ksp of $\ce{SrCrO4}$; it then precipitates.
Let us write the equilibrium equations and data down to help us think. Further, let x be the concentration of chromate to precipitate $\ce{Sr^2+}$, and y be that to precipitate $\ce{Ba^2+}$. According to the definition of Ksp we have,
$\begin{array}{cccccl} \ce{SrCrO4 &\rightarrow &Sr^2+ &+ &CrO4^2-}, &K_{\ce{sp}} = 3.6\times 10^{-5}\ &&0.30 &&x & \end{array}$
$x = \dfrac{\textrm{3.6e-5}}{0.30} = 1.2 \times 10^{-4} M$
$\begin{array}{cccccl} \ce{BaCrO4 &\rightarrow &Ba^2+ &+ &CrO4^2-}, &K_{\ce{sp}} = 1.2 \times 10^{-10}\ &&0.30 &&y & \end{array}$
$y = \dfrac{\textrm{1.2e-10}}{0.30} = 4.0 \times 10^{-10} \;M$
The Ksp's for the two salts indicate $\ce{BaCrO4}$ to be much less soluble, and it will precipitate before any $\ce{SrCrO4}$ precipitates. If chromate concentration is maintained a little less than 1.2e-4 M, $\ce{Sr^2+}$ ions will remain in the solution.
DISCUSSION
In reality, to control the increase of $\ce{[CrO4^2- ]}$ is very difficult.
Example 2
The Ksp for strontium chromate is $3.6\times 10^{-5}$ and the Ksp for barium chromate is $1.2\times 10^{-10}$. Potassium chromate is added a small amount at a time to first precipitate $\ce{BaCrO4}$. Calculate $\ce{[Ba^2+]}$ when the first trace of $\ce{SrCrO4}$ precipitate starts to form in a solution that contains 0.30 M each of $\ce{Ba^2+}$ and $\ce{Sr^2+}$ ions.
Solution
From the solution given in Example 1, $\ce{[CrO4^2- ]} = 3.6\times 10^{-4}\; M$ when $\ce{SrCrO4}$ starts to form. At this concentration, the $\ce{[Ba^2+]}$ is estimated as follows.
$\ce{[Ba^2+]\,} \textrm{3.6\times 10^{-4} = 1.2\times 10^{-10}}$
The Ksp of $\ce{BaCrO4}$.
Thus,
$\ce{[Ba^2+]} = \textrm{3.33e-7 M}$
Very small indeed, compared to 0.30. In the fresh precipitate of $\ce{SrCrO4}$, the mole ratio of $\ce{SrCrO4}$ to $\ce{BaCrO4}$ is 0.30 / 3.33e-7 = 9.0e5. In other words, the amount of $\ce{Ba^2+}$ ion in the solid is only 1e-6 (1 ppm) of all metal ions, providing that all the solid was removed when $\ce{[CrO4^2- ]} = \textrm{3.6e-4 M}$.
DISCUSSION
The calculation shown here indicates that the separation of $\ce{Sr}$ and $\ce{Ba}$ is pretty good. In practice, an impurity level of 1 ppm is a very small value.
Example 3
What reagent should you use to separate silver and lead ions that are present in a solution? What data or information will be required for this task?
Solution
The Ksp's for salts of silver and lead are required. We list the Ksp's for chlorides and sulfates in a table here. These value are found in the Handbook Menu of our website as Salts Ksp.
Salt Ksp Salt Ksp
$\ce{AgCl}$ $1.8 \times 10^{-10}$ $\ce{Ag2SO4}$ $1.4\times 10^{-5}$
$\ce{Hg2Cl2}$ $1.3\times 10^{-18}$ $\ce{BaSO4}$ $1.1\times 10^{-10}$
$\ce{PbCl2}$ $1.7 \times 10^{-5}$ $\ce{CaSO4}$ $2.4\times 10^{-5}$
$\ce{PbSO4}$ $6.3\times 10^{-7}$
$\ce{SrSO4}$ $3.2\times 10^{-7}$
Because the Ksp's $\ce{AgCl}$ and $\ce{PbCl2}$ are very different, chloride, $\ce{Cl-}$, apppears a good choice of negative ions for their separation.
The literature also indicates that $\ce{PbCl2}$ is rather soluble in warm water, and by heating the solution to 350 K (80oC), you can keep $\ce{Pb^2+}$ ions in solution and precipitate $\ce{AgCl}$ as a solid. The solubility of $\ce{AgCl}$ is very small even at high temperatures.
DISCUSSION
Find more detailed information about the solubility of lead chloride as a function of temperature.
Can sulfate be used to separate silver and lead ions? Which one will form a precipitate first as the sulfate ion concentration increases? What is the $\ce{[Pb^2+]}$ when $\ce{Ag2SO4}$ begins to precipitate in a solution that contains 0.10 M $\ce{Ag+}$?
Questions
1. Iron(II) hydroxide is only sparingly soluble in water at 25oC; its Ksp is $7.9 \times 10^{-16}$. Calculate the solubility of iron(II) hydroxide in a buffer solution with pH = 7.00.
2. A solution contains 0.60 M $\ce{Ba^2+}$ and 0.30 M $\ce{Ca^2+}$; Ksp values for $\ce{BaCrO4}$ and $\ce{CaCrO4}$ are $1.2 \times 10^{-10}$ and $7.1 \times 10^{-4}$ respectively. What value of $\ce{[CrO4^2- ]}$ will result in a maximum separation of these two ions?
3. A solution contains 0.60 M $\ce{Ba^2+}$ and 0.30 M $\ce{Ca^2+}$; Ksp values for $\ce{BaCrO4}$ and $\ce{CaCrO4}$ are $1.2 \times 10^{-10}$ and $7.1 \times 10^{-4}$ respectively. Calculate the $\ce{[Ca^2+]/[Ba^2+]}$ ratio in the solution when $\ce{[CrO4^2- ]}$ is maintained at $1.2 \times 10^{-3}\; M$.
Solutions
1. Answer $\ce{[Fe^2+]} = \textrm{0.079 M}$
Consider...
$\ce{[OH- ]} = 10^{(-14+7)} = \textrm{1.00e-7 (buffer)}$.
$\ce{[Fe^2+]} (\textrm{1.00e-7})^2 = K_{\ce{sp}}$; $\ce{[Fe^2+]} =\: ?$
This $\ce{Fe^2+}$ concentration is low; it is not very soluble in a neutral solution (pH = 7).
What is $\ce{[Fe^2+]}$ in a solution whose pH = 6.00?
2. Answer $\ce{[CrO4^2- ]} = \textrm{2.37e-3 M}$
Consider...
Solid $\ce{BaCrO4}$ will form first as $\ce{[CrO4^2- ]}$ increases. The maximum $\ce{[CrO4^2- ]}$ to precipitate $\ce{CaCrO4}$ is estimated as follows.
$\ce{[CrO4^2- ]} = \dfrac{\textrm{7.1e-4}}{0.30} = \textrm{2.37e-3 M}$
Estimate $\ce{[Ba^2+]}$ when $\ce{[CrO4^2- ]} = \textrm{2.3e-3 M}$, slightly below the maximum concentration.
3. Answer $\ce{\dfrac{[Ca^2+]}{[Ba^2+]}} = \textrm{3e6}$
Consider...
$\ce{[Ba^2+]} = \dfrac{\textrm{1.2e-10}}{\textrm{1.2e-3}} = \textrm{1e-7}$;
$\ce{\dfrac{[Ca^2+]}{[Ba^2+]}} = \dfrac{0.3}{\textrm{1e-7}} =\: ?$ The ratio of three million is large!
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Heterogeneous_Equilibria/Ion_Separation.txt
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Learning Objectives
• Explain what complex ions or metal complexes are.
• Apply the concept of equilibrium in complex formation.
• Calculate concentrations by applying the formation constant.
• Derive the overall formation constant from stepwise formation constants.
• Derive the dissociation constant from formation constant.
Metals are Lewis acids because of their positive charge. When dissolved in water, they react with water to form hydrated compounds such as $\ce{Na(H2O)6+}$ and $\ce{Cu(H2O)6^2+}$. These are called metal complexes, or coordination compounds. These coordination reactions are Lewis acid-base reactions. The neutral molecules such as $\ce{H2O}$ and $\ce{NH3}$, and anions such as $\ce{CN-}$, $\ce{CH3COO-}$ are called ligands. The coordination reaction can be represented by
$\ce{CuSO4 + 6 H2O \rightleftharpoons} \ce{Cu(H2O)6^2+ + SO4^2-}$
Usually, copper sulfate solids are hydrated with 5 water molecules per CuSO_4 complex ( $\mathrm{ {\color{Periwinkle} CuSO_4\cdot(H_2O)_5}}$), and its color is light blue. When heated, it loses water of crystallization, and becomes $\ce{CuSO4}$, which is colorless.
Most people know that when ammonia is added to a $\mathrm{ {\color{Periwinkle} Cu(H_2O)_6^{2+}}}$ solution, it turns deep blue. This is due to the formation of complexes:
$\color{Periwinkle} {Cu(H_2O)_6^{2+} } + 4 NH_3 \: \rightleftharpoons {\color{Blue} Cu(H_2O)_2(NH_3)_4^{2+} + 4 H_2O}$
Actually, the above reaction takes place in steps. The $\ce{H2O}$ molecules are displaced one at a time as the ammonia concentration, $\ce{[NH3]}$, increases. As more $\ce{NH3}$ is bonded to $\ce{Cu^2+}$, the blue color deepens.
Ammonia forms complexes with many metals. It forms a very strong complex with $\ce{Ag+}$ such that $\ce{AgCl}$ solid will dissolve in ammonia solution.
$\ce{AgCl_{\large{(s)}} + 2 NH3 \rightleftharpoons Ag(NH3)2+ + 2 Cl-}$
The silver ammonia complex is colorless, however.
Other commonly encountered ligands are $\ce{CN-}$, $\ce{SCN-}$, $\ce{Cl-}$, ethylene diamine ($\ce{NH2CH3CH3NH2}$), and acetate ($\ce{CH3COO-}$). For example,
$\mathrm{ {\color{Tan} Fe(H_2O)_6^{3+}} + SCN^-}$ $\rightleftharpoons$ $\mathrm{ {\color{Red} Fe(H_2O)_5SCN^{2+}} + H_2O}$
brown blood red
Formation Constant of Complexes
Complex Kf
$\ce{Ag(NH3)2}$ 1.6e7
$\ce{Ag(S2O3)2^3-}$ 1.7e13
$\ce{Al(OH)4-}$ 7.7e33
$\ce{AlF6^3-}$ 6.7e19
$\ce{Zn(EDTA)^2-}$ 3.8e16
Formation of complexes is also a thermal dynamic phenomenon. For the equilibrium,
$\ce{Ag+ + 2 NH3 \rightleftharpoons Ag(NH3)2+}$,
the formation constant is very large,
$K_{\ce f} = \ce{\dfrac{[Ag(NH3)2+]}{[Ag+] [NH3]^2}} = \textrm{1.6e7 M}^{-2}$
because $\ce{[Ag+]}$ is very small in such an equilibrium.
The formation constants of some other complexes are given in a table form on the right. Although only three metal ions are involved, the complexes are formed by five ligands, and the overall formation constants range from 1.6e7 to 7.7e33. The EDTA is a complicated organic molecule, $\mathrm{( ^- OOCCH_2)_2}\ce{N-CH2-CH2-N(CH2COO- )2}$ with six sites ($\ce{4\: O}$ and $\ce{2\: N}$) embracing the zinc ion in the zinc complex.
Example 1
Calculate $\ce{[Ag+]}$ in a solution containing 0.10 M $\ce{AgNO3}$ and 1.0 M $\ce{NH3}$.
Solution
For simplicity of formulation, let $\mathrm{M = [Ag^+]}$. The equilibrium equation and concentration are:
$\begin{array}{cccccl}\ce{Ag+ &+ &2 NH3 &\rightleftharpoons &Ag(NH3)2+} &\hspace{10px}K_{\ce f} = \textrm{1.6e7 M}^{-2}\ x &&1.00-0.20+x &&0.10-x & \end{array}$
$\dfrac{0.10-x}{x (0.80+x)^2} = \textrm{1.6e7 M}^{-2}$
$x = \textrm{9.7e-9 M}$
DISCUSSION
If we assume $x \ce M = \ce{[Ag(NH3)2+]}$, the calculation will be very difficult. Try it and find out why.
Suppose we now introduce 0.10 M $\ce{Cl-}$ into the equilibrium; will a precipitate form? For $\ce{AgCl}$, Ksp = 1.8e-10.
Ans. Since 9.7e-9*0.10 = 9.7e-10 > Ksp, a precipitate will form.
Example 2
What is the solubility of $\ce{AgCl}$ in a solution which contains 1.0 M $\ce{NH3}$? For $\ce{Ag(NH3)2+}$, Kf = 1.6e7, and for $\ce{AgCl}$, Ksp = 1.8e-10.
Solution
First consider the equilibria:
$\ce{AgCl \rightleftharpoons Ag+ + Cl-} \hspace{15px} K_{\ce{sp}} = \textrm{1.8e-10 M}^2$
$\ce{Ag+ + 2 NH3 \rightleftharpoons Ag(NH3)2+} \hspace{15px} K_{\ce f} = \textrm{1.6e7 M}^{-2}$
Adding the two equations together results in the equilibrium equation below.
$\begin{array}{cccccccl}\ce{AgCl &+ &2 NH3 &\rightleftharpoons &Ag(NH3)2+ &+ &Cl-} &\:\:\: K = K_{\ce{sp}} K_{\ce f} = \textrm{2.9e-3}\ &&1.0-x &&x &&x &\:\:\: \Leftarrow \textrm{equilibrium concentrations} \end{array}$
where x is the molar solubility of $\ce{AgCl}$.
$\dfrac{x^2}{(1.0-x)^2} = \textrm{2.9e-3}$
$\dfrac{x}{(1.0-x)} = 0.054$
$x = \textrm{0.051 M}$
DISCUSSION
Answer the following questions and review the discussion in Example 1.
The solubility product of $\ce{AgBr}$ is 5.0e-13 M2. Estimate the molar solubility of $\ce{AgBr}$ in a 1.0 M $\ce{NH3}$ solution.
Ans. 2.8e-3 M
Stepwise Formation Constants and Overall Constants
As indicated earlier, the formation of a complex takes place in steps. The formation constants in these steps are called stepwise formation constants.
For the reaction,
$\ce{Ag+ + NH3 \rightleftharpoons Ag(NH3)+}$
$K_{\ce{\large f_{\Large 1}}} = \ce{\dfrac{[Ag(NH3)+]}{[Ag+] [NH3]}} = \textrm{2.2e3 M}$
And for the reaction,
$\ce{Ag(NH3)+ + NH3 \rightleftharpoons Ag(NH3)2+}$
$K_{\ce{\large f_{\Large 2}}} = \ce{\dfrac{[Ag(NH3)2+]}{[Ag(NH3)+] [NH3]}} = \textrm{7.2e3 M}$
And obviously, for the overall reaction,
$\ce{Ag+ + 2 NH3 \rightleftharpoons Ag(NH3)2+}$
$K_{\ce{\large f}} = \ce{\dfrac{[Ag(NH3)2+]}{[Ag+] [NH3]+}} = K_{\ce{\large f_{\Large 1}}} \times K_{\ce{\large f_{\Large 2}}} = \textrm{1.6e7 M}^2$
A generalized formula is
$K_{\ce{\large f}} = K_{\ce{\large f_{\Large 1}}} \times K_{\ce{\large f_{\Large 2}}} \times K_{\ce{\large f_{\Large 3}}} \times \: ...$
Dissociation Constants and the Formation Constants
The reverse reaction of the complex formation is called dissociation, and the equilibrium constant is called dissociation constant Kd.
\p\ce{Ag(NH3)2+ \rightleftharpoons Ag+ + 2 NH3}, \hspace{10px} K_{\ce d}.\]
Obviously, we have
\pK_{\ce d} = \dfrac{1}{K_{\ce f}}\]
Similar to stepwise formation constants, we can also apply the concept to give a stepwise dissociation constant.
Example 3
Calculate $\ce{[Ag+]}$ when equal volume of 0.10 M $\ce{AgNO3}$ and 0.10 M $\ce{Na2S2O3}$ solutions are mixed, giving Kf = 1.7e13 M-2 for $\ce{Ag(S2O3)2^3-}$.
Solution
When the solutions are mixed, $\ce{[Ag(S2O3)2^3- ]} = \textrm{0.05 M}$. Let $x \ce M = \ce{[Ag+]}$ at equilibrium. The dissociation equilibrium equation rather than the formation equilibrium equation is more convenient in this case.
$\begin{array}{ccccc} \ce{Ag(S2O3)2^3- &\rightleftharpoons &Ag+ &+ &2 S2O3^3-}\ 0.05-x &&x &&2 x \end{array}$
$\dfrac{x (2 x)^2}{0.05-x} = \dfrac{1}{\textrm{1.7e13}} = \textrm{5.9e-14 M}^2$
Since x is very small, 0.05-x ~ 0.050, and
$\ce{[Ag+]} = x = \left(\dfrac{\textrm{5.9e-14}}{4} \right)^{1/3} = \textrm{2.5e-5 M}$
The approximation is justified.
DISCUSSION
If you use the formation equilibrium equation and let $\ce{[Ag(S2O3)2^3- ]} = x$, the equation is very difficult to solve.
Questions
1. Which of the following is a complex ion?
$\ce{CH4}$, $\ce{H2O}$, $\ce{NH3}$, $\ce{Al(OH)4-}$, $\ce{CCl4}$, $\ce{CO3^2-}$, $\ce{NH4+}$
2. To a sample of 9.0 mL solution with $\ce{[NH3]} = \textrm{1.1 M}$, 1.0 mL of 0.1 M $\ce{AgNO3}$ solution is added. Calculate $\ce{[Ag+]}$. Assume the final volume to be 10.0 mL. The Kf for $\ce{Ag(NH3)2} = \textrm{1.6e7}$.
3. Calculate the dissociation constant Kd for $\ce{Ag(NH3)+}$, if Kf for $\ce{Ag(NH3)2} = \textrm{1.6e7}$.
4. The solubility product of $\ce{AgBr}$ is 5.0e-13 M2, and Kf = 1.6e7 M-2 for $\ce{Ag(NH3)2+}$. Estimate the molar solubility of $\ce{AgBr}$ in a 1.0 M $\ce{NH3}$ solution.
5. The solubility product of $\ce{AgI}$ is 8.3e-17 M2, and Kf = 1.6e7 M-2 for $\ce{Ag(NH3)2+}$. Estimate the molar solubility of $\ce{AgI}$ in a 1.0 M $\ce{NH3}$ solution.
6. The solubility product of $\ce{AgBr}$ is 5.0e-13 M2, and Kf = 1.7e13 M-2 for $\ce{Ag(S2O3)2^3-}$. Estimate the molar solubility of $\ce{AgBr}$ in a 1.0 M $\ce{Na2S2O3}$ solution.
Solutions
1. Answer $\ce{Al(OH)4-}$
Consider...
Explain what complex ions or metal complexes are.
2. Answer $\ce{[Ag+]} = \textrm{6.25e-10}$
Consider...
Assume $\ce{[Ag+]} = x$, then
$\begin{array}{cccccl} \ce{Ag+ &+ &2 NH3 &\rightleftharpoons &Ag(NH3)2+}, &K_{\ce f} = \textrm{1.6e7}\ x &&1.0 &&0.01 &\leftarrow \textrm{equilibrium concentration} \end{array}$
$\dfrac{0.01}{x 1.0^2} = \textrm{1.6e7}$; $x =\: ?$
Apply the concept of equilibrium in complex formation.
3. Answer = Kd = 6.25e-8
Consider...
Derive the dissociation constant from formation constant, and show that $K_{\ce d} = \dfrac{1}{K_{\ce f}}$.
4. Answer Molar solubility = 2.8e-3 M
Hint...
Review Example 2.
5. Answer Molar solubility = 3.6e-5 M
Discussion...
The solubilities for $\ce{AgCl}$, $\ce{AgBr}$, and $\ce{AgI}$ in 1.0 M $\ce{NH3}$ solution are 0.051, 0.0028, and 0.000036 M respectively.
6. Answer The molar solubility is 0.49 M
Discussion...
Use the method of Example 2, but no approximation can be made.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Heterogeneous_Equilibria/Metal_Coordination_Complexes.txt
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Learning Objectives
• Estimate the pH of the solution due to precipitate of a metal hydroxide.
• Calculate the maximum metal ion concentration when the pH is known.
• Explain behavior of amphoteric metal hydroxides.
Most metal hydroxides are insoluble; some such as $\ce{Ca(OH)2}$, $\ce{Mg(OH)2}$, $\ce{Fe(OH)2}$, $\ce{Al(OH)3}$ etc. are sparingly soluble. However, alkali metal hydroxides $\ce{CsOH}$, $\ce{KOH}$, and $\ce{NaOH}$ are very soluble, making them strong bases. When dissolved, these hydroxides are completely ionized. Since the hydroxide concentration, $\ce{[OH- ]}$, is an integrated property of the solution, the solubility of metal hydroxide depends on pH, pOH or $\ce{[OH- ]}$.
Alkali metal hydroxides $\ce{LiOH}$, $\ce{NaOH}$, $\ce{KOH}$, $\ce{CsOH}$ are soluble, and their solutions are basic. Hydroxides of alkali earth metals are much less soluble. For example, quicklime ($\ce{CaO}$) reacts with water to give slaked lime, which is slightly soluble.
$\begin{array}{ccccl} \ce{CaO &+ &H2O &\rightleftharpoons &\:Ca(OH)2}\ \textrm{quicklime} &&&&\textrm{slaked lime (slightly soluble)} \end{array}$
Milk of magnesia is $\ce{Mg(OH)2}$ (Ksp = 7e-12) suspension. In an acidic solution such as stomach juice, the following reaction takes place,
$\ce{Mg(OH)2 + H+ \rightleftharpoons Mg^2+ + 2 H2O}$
Thus, it can neutralize excess acid in the stomach.
Example 1
Calculate the maximum concentration of $\ce{Mg^2+}$ in a solution which contains a buffer so that pH = 3 at 298 K.
Solution
As usual, we write the equilibrium equation so that we can write the concentration below the formula. If we do not know the concentration, we assume it to be a variable x.
$\begin{array}{ccccc} \ce{Mg(OH)2 &\rightleftharpoons &Mg^2+ &+ &2 OH-}\ &&x && 1 \times 10^{-11} \end{array}$
$K_{\ce{sp}} = x (1 \times 10^{-11})^2 = 7 \times 10^{-12}$
Solving for x results in $x = 7 \times ^{10}$
DISCUSSION
This value certainly is too large, unrealistic.
Example 2
Calculate the pH of a saturated $\ce{Mg(OH)2}$ solution.
Solution
We assume the concentration to be x M of $\ce{Mg(OH)}$, and note that $\ce{[OH- ]} = 2 x$,
$\begin{array}{ccccc} \ce{Mg(OH)2 &\rightleftharpoons &Mg^2+ &+ &2 OH-}\ &&x &&2 x \end{array}$
$K_{\ce{sp}} = x (2 x)^2 = \textrm{7e-12}$
Solving for x; x = 1.2e-4
\begin{align} \ce{[OH]} &= \textrm{2.4e-4}\ \ce{pOH} &= 3.62 \end{align}
$\mathrm{pH = 14 - 3.62 = 10.38}$
DISCUSSION
The pH of a saturated lime ($\ce{Ca(OH)2}$) solution is about 10.0.
Amphoteric Hydroxides
Not all metal hydroxides behave the same way - that is precipitate as hydroxide solids. Metal hydroxides such as $\ce{Fe(OH)3}$ and $\ce{Al(OH)3}$ react with acids and bases, and they are called amphoteric hydroxide. In reality, $\ce{Al(OH)3}$ should be formulated as $\ce{Al(H2O)3(OH)3}$, and this neutral substance has a very low solubility. It reacts in the following way as $\ce{[H+]}$ increases.
\begin{align} \ce{Al(H2O)3(OH)3 + H3O+ &\rightleftharpoons Al(H2O)4(OH)2+ + HOH}\ \ce{Al(H2O)4(OH)2+ + H3O+ &\rightleftharpoons Al(H2O)5(OH)^2+ + H2O}\ \ce{Al(H2O)5(OH)^2+ + H3O+ &\rightleftharpoons Al(H2O)6^3+ + H2O} \end{align}
When the pH increases, the following reactions take place:
\begin{align} \ce{Al(H2O)3(OH)3 + OH- &\rightleftharpoons Al(H2O)2(OH)4- + H2O}\ \ce{Al(H2O)2(OH)4- + OH- &\rightleftharpoons Al(H2O)(OH)5^2- + H2O}\ \ce{Al(H2O)(OH)5^2- + OH- &\rightleftharpoons Al(OH)6^3- + H2O} \end{align}
The charged species are soluble in water. As a result, amphoteric hydroxides dissolve in acidic and basic solutions.
Questions
1. Assume the pH of gastric juice to be 2. Calculate the maximum $\ce{[Mg^2+]}$.
2. Calculate the pH of a 0.10 M $\ce{NH3}$ solution.
3. Calculate the maximum $\ce{[Fe^2+]}$ in a 0.10 M $\ce{NH3}$ solution. Give the value in M .
4. What are amphoteric metal hydroxides? (enter no more than one line)
Solutions
1. Answer $\mathrm{[Mg^{2+}] = 7e12\: M}$
Consider...
We assume the temperature to be 298 K, which is too low.
$\ce{[Mg^2+]} = \dfrac{\textrm{7e-12}}{(\textrm{1e-12})^2} =\: ?$
This value is unrealistically large. The result is correct, but meaningless.
2. Answer pH = 11.12
Consider...
$\ce{[OH- ]} = (0.10\times\textrm{1.8e-5})^{1/2} = \textrm{1.34e-3}$
$\ce{[H+]} = \dfrac{\textrm{1e-14}}{\textrm{1.34e-3}} = \textrm{7.5e-12}$, $\ce{pH} =\: ?$
This value is required for the calculation in next question. Better yet, remember that $\ce{[OH- ]} = \textrm{1.34e-4}$.
3. Answer 4.4e-10 M
Consider...
$\ce{[Fe^2+]} = \dfrac{\textrm{7.9e-16}}{(\textrm{1.34e-3})^2} =\: ?\:\ce M$
What is the value in g/L? Molar mass of $\ce{Fe}$ is 55.8 g/mol.
4. Answer Metal hydroxides such as $\ce{Fe(OH)3}$ and $\ce{Al(OH)3}$ that react with acids and bases are called amphoteric hydroxide.
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textbooks/chem/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Heterogeneous_Equilibria/Solubility_of_Metal_Hydroxides.txt
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