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Chapter 11: Problem 96

Delocalized molecular orbitals are found in (a) \(\mathrm{H}_{2}\) (b) \(\mathrm{HS}^{-} ;\) (c) \(\mathrm{CH}_{4} ;\) (d) \(\mathrm{CO}_{3}^{2-}\) .

Short Answer

Expert verified
Among the given choices, only \( \mathrm{CO}_3^{2-} \) has delocalized molecular orbitals.

Step by step solution

01

Analyzing \( \mathrm{H}_2 \)

The molecular structure of \( \mathrm{H}_2 \) is simple, containing only two hydrogen atoms. The molecule has one bond, which is a sigma bond between the two hydrogen atoms. There aren't any pi electrons and hence, no delocalized electrons.
02

Analyzing \( \mathrm{HS}^- \)

The \( \mathrm{HS}^- \) molecule consists of a sigma bond between sulfur and hydrogen. The additional electron from the negative charge is localized on the sulfur atom. There are no pi electrons in this molecule, and hence, there are no delocalized electrons.
03

Analyzing \( \mathrm{CH}_4 \)

The \( \mathrm{CH}_4 \) molecule has a tetrahedral structure, with sigma bonds between carbon and the four hydrogen atoms. As there are only sigma bonds, there are no pi electrons, and as a result, there are no delocalized electrons.
04

Analyzing \( \mathrm{CO}_3^{2-} \)

In the \( \mathrm{CO}_3^{2-} \) ion, the carbon atom is bonded to three oxygen atoms. The bonding in \( \mathrm{CO}_3^{2-} \) can't be described accurately with a single Lewis structure. It has resonance structures and the extra electrons are delocalized over the three oxygen atoms. This means that there are delocalized molecular orbitals.

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Most popular questions from this chapter

In which of the following, \(\mathrm{CO}_{3}^{2-}, \mathrm{SO}_{2}, \mathrm{CCl}_{4}, \mathrm{CO}\) \(\mathrm{NO}_{2}^{-},\) would you expect to find \(s p^{2}\) hybridization of the central atom? Explain.

Match each of the following species with one of these hybridization schemes: \(s p, s p^{2}, s p^{3}, s p^{3} d, s p^{3} d^{2} .\) (a) \(\mathrm{PF}_{6}^{-}\) (b) \(\operatorname{COS} ;\) (c) \(\operatorname{SiCl}_{4} ;\) (d) \(\mathrm{NO}_{3}^{-}\);(e) AsF \(_{5}\)

Explain the essential difference in how the valencebond method and molecular orbital theory describe a covalent bond.

Lewis theory is satisfactory to explain bonding in the ionic compound \(\mathrm{K}_{2} \mathrm{O},\) but it does not readily explain formation of the ionic compounds potassium superoxide, \(\mathrm{KO}_{2}\), and potassium peroxide, \(\mathrm{K}_{2} \mathrm{O}_{2}\) (a) Show that molecular orbital theory can provide this explanation. (b) Write Lewis structures consistent with the molecular orbital explanation.

Consider the molecules \(\mathrm{CO}^{+}\) and \(\mathrm{CN}^{-}\) and use molecular orbital theory to answer the following: (a) Write the molecular orbital configuration of each ion (ignore the 1 s electrons). (b) Predict the bond order of each ion. (c) Which of these ions is paramagnetic? Which is diamagnetic? (d) Which of these ions do you think has the greater bond length? Explain.
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