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Chapter 8: Problem 92

Consider the following bond lengths: \(\begin{array}{lllll}\mathrm{C}-\mathrm{O} & 143 \mathrm{pm} & \mathrm{C}=\mathrm{O} & 123 \mathrm{pm} & \mathrm{C} \equiv \mathrm{O} & 109 \mathrm{pm}\end{array}\) In the \(\mathrm{CO}_{3}^{2-}\) ion, all three \(\mathrm{C}-\mathrm{O}\) bonds have identical bond lengths of \(136 \mathrm{pm}\). Why?

Short Answer

Expert verified
The identical bond lengths of 136 pm for all three C-O bonds in the CO₃²⁻ ion can be explained by the resonance hybrid structure, in which the bond character is distributed uniformly across all three C-O bonds. The average bond length calculated from the given single and double bond lengths (along with equal contributions from the resonance structures) also equals 136 pm, thus validating that resonance leads to equal bond lengths in the carbonate ion.

Step by step solution

01

Draw the Lewis Structure of CO₃²⁻ Ion

(Write the content here) First, draw the Lewis structure for the carbonate ion (CO₃²⁻). In the carbonate ion, carbon has 4 valence electrons, and each oxygen atom has 6 valence electrons, totaling 24 valence electrons. Add two more electrons for the −2 charge. With a total of 26 valence electrons, the carbonate ion looks like this: \[ O=C=O^{-} \oplus O^{-}-C-O^{-} \oplus O^{-}-C-O^{-} \] It can be seen that there are three single-double bonded oxygen atoms in resonance with the central carbon atom.
02

Analyze the Resonance Structures

Resonance occurs when there is more than one possible Lewis structure with the same arrangement of atoms but different electron assignments to different positions (different bond orders). In the case of the carbonate ion, there are three resonance structures: \[ O = C - O^{-} \oplus O^{-} - C = O \oplus O^{-} - C - O^{-} \] In each resonance structure, there are two single bonds and one double bond. However, none of these individual resonance structures are an accurate representation of the true structure of the carbonate ion. Instead, the actual ion is a combination of the three resonance structures, meaning that the bond character is distributed uniformly across all three C-O bonds.
03

Calculate the Average Bond Length

To understand the identical bond lengths of 136pm in the carbonate ion, consider the resonance hybrid structure, which is an average of the bond lengths in the different resonance structures. Since there is one double bond and two single bonds in each resonance structure, their average bond lengths can be calculated as follows: \[ \frac{1 \times 123 \operatorname{pm} ( {C} = {O}) + 2 \times 143 \operatorname{pm}( {C} - {O})}{1 + 2} = \frac{123 + 286}{3} = 136 \operatorname{pm} \] Hence, the average bond length of the carbonate ion's C-O bonds is 136 pm.
04

Conclusion

The identical bond lengths of 136 pm for all three C-O bonds in the CO₃²⁻ ion can be explained by the resonance hybrid structure, in which the bond character is distributed uniformly across all three C-O bonds. The average bond length calculated from the given single and double bond lengths also equals 136 pm, thus validating that resonance leads to equal bond lengths in the carbonate ion.

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Most popular questions from this chapter

Think of forming an ionic compound as three steps (this is a simplification, as with all models): (1) removing an electron from the metal; (2) adding an electron to the nonmetal; and (3) allowing the metal cation and nonmetal anion to come together. a. What is the sign of the energy change for each of these three processes? b. In general, what is the sign of the sum of the first two processes? Use examples to support your answer. c. What must be the sign of the sum of the three processes? d. Given your answer to part \(\mathrm{c}\), why do ionic bonds occur? e. Given your above explanations, why is NaCl stable but not \(\mathrm{Na}_{2} \mathrm{Cl} ? \mathrm{NaCl}_{2} ?\) What about \(\mathrm{MgO}\) compared to \(\mathrm{MgO}_{2} ?\) \(\mathrm{Mg}_{2} \mathrm{O} ?\)

\(\mathrm{SF}_{6}, \mathrm{ClF}_{5}\), and \(\mathrm{XeF}_{4}\) are three compounds whose central atoms do not follow the octet rule. Draw Lewis structures for these compounds.

What do each of the following sets of compounds/ions have in common with each other? See your Lewis structures for Exercises 107 through 110 . a. \(\mathrm{XeCl}_{4}, \mathrm{XeCl}_{2}\) b. \(\mathrm{ICl}_{5}, \mathrm{TeF}_{4}, \mathrm{ICl}_{3}, \mathrm{PCl}_{3}, \mathrm{SCl}_{2}, \mathrm{SeO}_{2}\)

What is the electronegativity trend? Where does hydrogen fit into the electronegativity trend for the other elements in the periodic table?

Give an example of an ionic compound where both the anion and the cation are isoelectronic with each of the following noble gases. a. Ne c. \(\mathrm{Kr}\) b. \(\mathrm{Ar}\) d. Xe
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