(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 theion is excited by light, sothat 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? (e) Which of the following statements about part (d) is correct: (i) The light excites an electron from a bonding orbital to an antibonding orbital, (ii) The light excites an electron from an antibonding orbital to a bonding orbital, or (iii) In the excited state there are more bonding electrons than antibonding electrons?

Let's begin by drawing the molecular orbitals and energy level diagram of the \(\mathrm{H}_{2}^{-}\) ion. As it has two hydrogen atoms, we will have bonding (lower-energy) and antibonding (higher-energy) molecular orbitals formed by combining the 1s atomic orbitals of each hydrogen atom. The ion has a total of 3 electrons, with one in each hydrogen atom, and one extra. 

Now that we've sketched the molecular orbitals, it's time to write the electron configuration of \(\mathrm{H}_{2}^{-}\) in terms of its MOs. We have three electrons, and they will fill the available orbitals in order of increasing energy: 1. Two electrons will fill the \(\sigma_{1s}\) bonding molecular orbital. 2. One electron will fill the \(\sigma_{1s}^{*}\) antibonding molecular orbital. Thus, the electron configuration is: \(\sigma_{1s}^{2}\sigma_{1s}^{*1}\)

Next, let's calculate the bond order. Bond order is a measure of the strength of a chemical bond, and can be determined using the following formula: Bond Order = \(\frac{1}{2} \times (\text{number of bonding electrons} - \text{number of antibonding electrons})\) For our \(\mathrm{H}_{2}^{-}\) ion, we have: Bond Order = \(\frac{1}{2}\times(2 - 1) = \frac{1}{2}\) Hence, the bond order of the \(\mathrm{H}_{2}^{-}\) ion is 0.5.

Suppose the \(\mathrm{H}_{2}^{-}\) ion is excited by light, causing an electron to move from a lower-energy to a higher-energy molecular orbital. The electron will move from the \(\sigma_{1s}\) orbital to the \(\sigma_{1s}^{*}\) orbital. The excited state will have the following electron configuration: \(\sigma_{1s}^{1}\sigma_{1s}^{*2}\) Now, let's calculate the bond order for the excited-state configuration: Excited-State Bond Order = \(\frac{1}{2}\times(1-2)= - \frac{1}{2}\) The negative bond order indicates that the excited-state \(\mathrm{H}_{2}^{-}\) ion would not be stable.

Finally, let's determine which statement about part (d) is correct: (i) The light excites an electron from a bonding orbital to an antibonding orbital. Since we observed that the electron moved from the \(\sigma_{1s}\) orbital to the \(\sigma_{1s}^{*}\) orbital in the excited state, this statement is correct: The light excites an electron from a bonding orbital to an antibonding orbital. (ii) The light excites an electron from an antibonding orbital to a bonding orbital. This statement is incorrect as the electron moved from a lower-energy bonding orbital to a higher-energy antibonding orbital. (iii) In the excited state, there are more bonding electrons than antibonding electrons. This statement is also incorrect, as in the excited state, there is one bonding electron and two antibonding electrons. So, the correct statement is (i): The light excites an electron from a bonding orbital to an antibonding orbital.