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Chapter 11: Problem 103

Use the valence molecular orbital configuration to determine which of the following species is expected to have the lowest ionization energy: (a) \(\mathrm{C}_{2}^{+} ;\) (b) \(\mathrm{C}_{2}\) (c) \(\mathrm{C}_{2}^{-}\)

Short Answer

Expert verified
The species \(C_{2}^{-}\) is expected to have the lowest ionization energy.

Step by step solution

01

Understand the Molecular Ion Notations

The notation of a molecule in the form \(C_{2}^{+}\) or \(C_{2}^{-}\) represents a molecular ion where \(C_{2}^{+}\) has lost an electron and \(C_{2}^{-}\) has gained an electron. ⁠Subsequently, \(C_{2}\) is a neutral molecule with no charge.
02

Determine the Valence Molecular Orbital Configuration for Each Species

The valence molecular orbital configuration for each species is as follows: For \(C_{2}^{+}\), the highest energy electron resides in the \(\pi\) antibonding orbital; For \(C_{2}\), the highest energy electron is in the \(\pi\) orbital;For \(C_{2}^{-}\), the highest energy electron is in the \(\pi^{*}\) antibonding orbital. The asterisk indicates a higher energy antibonding orbital.
03

Compare Ionization Energies Based on Valence Molecular Orbital Configurations

Ionization energy increases with the energy of the most loosely bound electron. Hence, \(C_{2}^{-}\) has the lowest ionization energy because the highest energy electron lies in the \(\pi^{*}\) antibonding orbital, which is a higher energy level compared to the \(\pi\) orbital or \(\pi\) antibonding level. Removing an electron from the \(\pi^{*}\) antibonding orbital would require less energy than removing one from the \(\pi\) or \(\pi\) antibonding orbital.

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Most popular questions from this chapter

Propose a hybridization scheme to account for bonds formed by the central carbon atom in each of the following molecules: (a) hydrogen cyanide, HCN; (b) methyl alcohol, \(\mathrm{CH}_{3} \mathrm{OH} ;\) (c) acetone, \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CO}\) (d) carbamic acid,

Represent bonding in the carbon dioxide molecule, \(\mathrm{CO}_{2},\) by \((\mathrm{a})\) a Lewis structure and \((\mathrm{b})\) the valencebond method. Identify \(\sigma\) and \(\pi\) bonds, the necessary hybridization scheme, and orbital overlap.

Think of the reaction shown here as involving the transfer of a fluoride ion from \(\mathrm{ClF}_{3}\) to \(\mathrm{AsF}_{5}\) to form the ions \(\mathrm{ClF}_{2}^{+}\) and \(\mathrm{AsF}_{6}^{-}\). As a result, the hybridization scheme of each central atom must change. For each reactant molecule and product ion, indicate (a) its geometric structure and (b) the hybridization scheme for its central atom. $$ \mathrm{ClF}_{3}+\mathrm{AsF}_{5} \longrightarrow\left(\mathrm{ClF}_{2}^{+}\right)\left(\mathrm{AsF}_{6}^{-}\right) $$

Use the valence molecular orbital configuration to determine which of the following species is expected to have the greatest electron affinity: (a) \(\mathrm{C}_{2}^{+} ;\) (b) \(\mathrm{Be}_{2}\) (c) \(\mathrm{F}_{2} ;\) (d) \(\mathrm{B}_{2}^{+}\)

A conjugated hydrocarbon has an alternation of double and single bonds. Draw the molecular orbitals of the \(\pi\) system of 1,3,5 -hexatriene. If the energy required to excite an electron from the HOMO to the LUMO corresponds to a wavelength of \(256 \mathrm{nm},\) do you expect the wavelength for the corresponding excitation in 1,3,5,7 -octatetraene to be a longer or shorter wavelength?
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