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Chapter 9: Problem 74

(a) Sketch the molecular orbitals of the \(\mathrm{H}_{2}^{-}\) ion and draw its energy-level diagram. (b) Write the electron configuration of the ion in terms of its MOs. (c) Calculate the bond order in \(\mathrm{H}_{2}^{-}\). (d) Suppose that the ion is excited by light, so that an electron moves from a lower-energy to a higher- energy molecular orbital. Would you expect the excited-state \(\mathrm{H}_{2}^{-}\) ion to be stable? Explain.

Short Answer

Expert verified
The molecular orbitals of the 𝐻2− ion consist of a lower-energy bonding σ1s orbital and a higher-energy antibonding σ1s* orbital. The energy-level diagram displays these orbitals on the y-axis. The electron configuration for this ion is σ1s^2 σ1s*^1, and the bond order is 1/2. In the excited state, the electron configuration is σ1s^1 σ1s*^2, and the bond order is -1/2, meaning that the excited 𝐻2− ion is unstable and will dissociate into individual hydrogen atoms and an electron.

Step by step solution

01

Sketch Molecular Orbitals of the 𝐻2− ion

We have to construct the molecular orbitals of the 𝐻2− ion using atomic orbitals of hydrogen atoms. For each hydrogen atom, there is one 1s orbital. When the two 1s orbitals of the two hydrogen atoms combine, they form two new molecular orbitals, one constructive (lower-energy bonding σ1s orbital) and one destructive (higher-energy antibonding σ1s* orbital).
02

Draw the Energy-Level Diagram

Next, draw the energy-level diagram for the 𝐻2− ion. On the y-axis, you will have energy increasing upwards. Add the σ1s orbital at a lower energy level and the σ1s* orbital at a higher energy level. Label these orbitals and the individual atomic orbitals.
03

Write the Electron Configuration of the 𝐻2− ion

For a 𝐻2− ion, there are 3 electrons present. To write the electron configuration in terms of the MOs, we follow the Aufbau principle and fill the orbitals with the lowest energy first. The electron configuration will be as follows: σ1s^2 σ1s*^1
04

Calculate the Bond Order in 𝐻2− ion

The bond order can be calculated using the formula: Bond Order = (Bonding electrons - Antibonding electrons) / 2 For the 𝐻2− ion, there are 2 bonding electrons in σ1s and 1 antibonding electron in σ1s*. Bond Order = (2 - 1) / 2 = 1/2
05

Determine the Stability of the Excited-State 𝐻2− ion

In the excited state, an electron is promoted from a lower-energy molecular orbital to a higher-energy molecular orbital (from σ1s to σ1s* in this ion). The electron configuration of the excited state would be: σ1s^1 σ1s*^2 Now, we need to calculate the bond order for the excited state: Bond Order (excited) = (1 - 2) / 2 = -1/2 Since the bond order is negative for the excited state, this indicates that the excited 𝐻2− ion would be unstable. The repulsion between the nuclei will be greater than the bonding force resulting from the electrons, so the ion would dissociate into individual hydrogen atoms and an electron.

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Most popular questions from this chapter

(a) What is the difference between a localized \(\pi\) bond and a delocalized one? (b) How can you determine whether a molecule or ion will exhibit delocalized \(\pi\) bonding? (c) Is the \(\pi\) bond in \(\mathrm{NO}_{2}^{-}\) localized or delocalized?

In which of the following \(\mathrm{AF}_{n}\) molecules or ions is there more than one \(\mathrm{F}-\mathrm{A}-\mathrm{F}\) bond angle: \(\mathrm{SiF}_{4}, \mathrm{PF}_{5}, \mathrm{SF}_{4}, \mathrm{AsF}_{3} ?\)

Describe the bond angles to be found in each of the following molecular structures: (a) planar trigonal, (b) tetrahedral, (c) octahedral, (d) linear.

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Determine the electron configurations for \(\mathrm{CN}^{+}, \mathrm{CN},\) and \(\mathrm{CN}^{-}\). (a) Which species has the strongest \(\mathrm{C}-\mathrm{N}\) bond? (b) Which species, if any, has unpaired electrons?
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