In terms of the molecular orbital model, which species in each of the following two pairs will most likely be the one to gain an electron? Explain. a. CN or NO b. \(\mathrm{O}_{2}^{2+}\) or \(\mathrm{N}_{2}{ }^{2+}\)

For heteronuclear diatomic molecules like CN and NO, we'll be using the molecular orbital diagrams with the σ and π orbitals. Note that very often, these diagrams are very similar to molecular orbital diagrams of the most electronegative similar homonuclear diatomic molecules. The electronic configuration of \(\mathrm{O_{2}^{2+}}\) and \(\mathrm{N_{2}^{2+}}\) can be obtained by seeing how many valence electrons are filled into the orbitals following Hund's rule and the Aufbau principle on the molecular orbital diagrams of O2 and N2.

Determine the number of valence electrons in CN and NO molecules: - \(\mathrm{CN}\): C has 4 valence electrons, N has 5, so the total is 9. - \(\mathrm{NO}\): N has 5 valence electrons, O has 6, so the total is 11. To determine the molecular orbitals of CN and NO, let's begin with the closest homonuclear diatomic molecule, which is \(\mathrm{N}_{2}\). The electronic configuration for \(\mathrm{N}_{2}\): \(\sigma_{1s}^{2}\sigma_{1s}^{*2}\sigma_{2s}^{2}\sigma_{2s}^{*2}\pi_{2p}^{4}\sigma_{2p}^{2}\). Now, let's add or remove electrons to get the electron configurations for CN and NO: - \(\mathrm{CN}\): Has 9 valence electrons. Start with the N2 configuration and remove 1 electron: \(\sigma_{1s}^{2}\sigma_{1s}^{*2}\sigma_{2s}^{2}\sigma_{2s}^{*2}\pi_{2p}^{4}\sigma_{2p}^{1}\). - \(\mathrm{NO}\): Has 11 valence electrons. Start with the N2 configuration and add 1 electron: \(\sigma_{1s}^{2}\sigma_{1s}^{*2}\sigma_{2s}^{2}\sigma_{2s}^{*2}\pi_{2p}^{4}\sigma_{2p}^{2}\pi_{2p}^{*1}\).

Determine the molecular orbitals of the given ions: - \(\mathrm{O}_{2}^{2+}\): O2 has the configuration \(\sigma_{1s}^{2}\sigma_{1s}^{*2}\sigma_{2s}^{2}\sigma_{2s}^{*2}\pi_{2p}^{4}\sigma_{2p}^{2}\pi_{2p}^{*2}\). Remove 2 electrons to obtain \(\sigma_{1s}^{2}\sigma_{1s}^{*2}\sigma_{2s}^{2}\sigma_{2s}^{*2}\pi_{2p}^{4}\sigma_{2p}^{2}\pi_{2p}^{*0}\). - \(\mathrm{N}_{2}{ }^{2+}\): N2 has the configuration \(\sigma_{1s}^{2}\sigma_{1s}^{*2}\sigma_{2s}^{2}\sigma_{2s}^{*2}\pi_{2p}^{4}\sigma_{2p}^{2}\). Remove 2 electrons to obtain \(\sigma_{1s}^{2}\sigma_{1s}^{*2}\sigma_{2s}^{2}\sigma_{2s}^{*2}\pi_{2p}^{2}\sigma_{2p}^{2}\).

Now we will compare the electron affinities of each pair based on their electronic configurations: a. In the CN and NO pair: - \(\mathrm{CN}\) has a partially-filled σ2p orbital, while \(\mathrm{NO}\) has a partially-filled π2p* orbital. Given that the antibonding π* orbital is generally less stable (higher in energy) than the bonding σ orbital, CN is more likely to gain an electron. b. In the \(\mathrm{O}_{2}^{2+}\) and \(\mathrm{N}_{2}{ }^{2+}\) pair: - \(\mathrm{O}_{2}^{2+}\) has empty π2p* orbitals, while \(\mathrm{N}_{2}{ }^{2+}\) has partially-filled π2p orbitals. Since it's energetically more favorable for an electron to fill an empty antibonding π* orbital than a bonding π orbital, \(\mathrm{O}_{2}^{2+}\) is more likely to gain an electron. So, to conclude: a. CN is more likely to gain an electron than NO b. \(\mathrm{O}_{2}^{2+}\) is more likely to gain an electron than \(\mathrm{N}_{2}{ }^{2+}\)