(a) The nitric oxide molecule, NO, readily loses one electron to form the NO \(^{+}\) ion. Which of the following is the best explanation of why this happens: (i) Oxygen is more electronegative than nitrogen, (ii) The highest energy electron in NO lies in a \(\pi_{2 p}^{*}\) molecular orbital, or (iii) The \(\pi_{2 p}^{*}\) MO in NO is completely filled. (b) Predict the order of the \(\mathrm{N}-\mathrm{O}\) bond strengths in NO, NO^, and NO', and describe the magnetic properties of each. (c) With what neutral homonuclear diatomic molecules are the NO \(^{+}\) and \(\mathrm{NO}^{-}\) ions isoelectronic (same number of electrons)?

Let's assess each explanation from the problem statement: (i) Oxygen is more electronegative than nitrogen: Though it's true that oxygen is more electronegative than nitrogen, the statement alone doesn't fully explain why NO readily loses an electron. Electronegativity plays a role in determining the distribution of the electrons in the molecule but doesn't directly explain the ionization process. (ii) The highest energy electron in NO lies in a π₂p* molecular orbital: This statement suggests that the last electron fills an antibonding orbital (indicated by the *), which means the electron is more readily lost to form the ion NO⁺, with the loss of the electron resulting in an overall strengthening of the bond. (iii) The π₂p* MO in NO is completely filled: This statement is incorrect. If the π₂p* MO in NO were completely filled, losing an electron would destabilize the molecule. Instead, the highest energy molecular orbital is partially filled. Given these assessments, explanation (ii) is the best reason for why NO readily loses an electron.

To predict the bond strengths in NO, NO⁺, and NO⁻, we use molecular orbital theory. The order of bond strengths will be determined by the number of bonding electrons minus the number of antibonding electrons. For magnetic properties, we'll consider whether the molecules have unpaired electrons. NO: Nitrogen has 7 electrons, and oxygen has 8 electrons, so NO has 15 electrons. Using the molecular orbitals, we can determine the electron configuration. NO has 1 unpaired electron, making it paramagnetic. The bond order is (10-5)/2 = 2.5. NO⁺: When NO loses an electron, there will be 14 electrons in total. NO⁺ will have no unpaired electrons, which means it's diamagnetic. The bond order is (10-4)/2 = 3, so it has a stronger bond than NO. NO⁻: When NO gains an electron, there will be 16 electrons in total. NO⁻ will have 2 unpaired electrons, making it paramagnetic. The bond order is (10-6)/2 = 2, so it has a weaker bond than NO. Therefore, the order of bond strengths is NO⁺ > NO > NO⁻, and the magnetic properties are NO (paramagnetic), NO⁺ (diamagnetic), and NO⁻ (paramagnetic).

We now need to find the neutral homonuclear diatomic molecules that have the same number of electrons as NO⁺ and NO⁻. NO⁺ has 14 electrons. NO⁻ has 16 electrons. For the same number of electrons, we need to identify the neutral homonuclear diatomic molecules that have 14 or 16 electrons. Using the periodic table, we can identify these molecules: 14 electrons: Nitrogen has 7 electrons, so N₂ has 14 electrons in total. 16 electrons: Oxygen has 8 electrons, so O₂ has 16 electrons in total. Thus, the NO⁺ ion is isoelectronic with the N₂ molecule, and the NO⁻ ion is isoelectronic with the O₂ molecule.