Describe the bonding in \(\mathrm{NO}^{+}, \mathrm{NO}^{-},\) and \(\mathrm{NO}\) using both the localized electron and molecular orbital models. Account for any discrepancies between the two models.

In the localized electron model, we describe bonding in terms of hybridization of atomic orbitals and the overlap of these orbitals to form sigma and pi bonds. 1. \(\mathrm{NO}^+\): In this cation, nitrogen (N) has 4 valence electrons, and oxygen (O) has 6 valence electrons. Since there is a positive charge, we lose one electron. Thus, there are 9 valence electrons in total. We can describe nitrogen as having sp-hybridization, forming one sigma bond with oxygen. This leaves three unpaired electrons in nitrogen's 2p orbitals, and two unpaired electrons in oxygen's 2p orbitals. The unpaired electrons in the 2p orbitals can form two pi bonds, leading to a triple bond between nitrogen and oxygen. 2. \(\mathrm{NO}^-\): In this anion, nitrogen has 4 valence electrons, oxygen has 6 valence electrons, and we add one due to the negative charge. There are 11 valence electrons in total. Nitrogen can be described as having sp-hybridization, forming one sigma bond with oxygen. There are three unpaired electrons in nitrogen's 2p orbitals and three in oxygen's 2p orbitals. The unpaired electrons in the 2p orbitals can form two pi bonds, leading to a triple bond. The remaining electrons form a lone pair on oxygen. 3. \(\mathrm{NO}\): Nitrogen has 4 valence electrons, and oxygen has 6 valence electrons. There are 10 valence electrons in total. Nitrogen has sp-hybridization, forming one sigma bond with oxygen. There are three unpaired electrons in nitrogen's 2p orbitals and two in oxygen's 2p orbitals. The unpaired electrons in the 2p orbitals can form two pi bonds, leading to a triple bond. In all three cases, the localized electron model predicts a triple bond between nitrogen and oxygen.

In the molecular orbital model, we form molecular orbitals by combining atomic orbitals. The molecular orbitals are filled in the order of increasing energy, following the aufbau principle. First, we need to determine the molecular orbitals formed: - 1 \(2\sigma\) bonding orbital and 1 \(2\sigma^*\) antibonding orbital from the interaction of the 2s orbitals. - 1 \(2\sigma_g\) bonding orbital and 1 \(2\sigma_u^*\) antibonding orbital from the interaction of the 2p orbitals along the internuclear axis. - 4 bonding orbitals (\(2\pi_{x}\) and \(2\pi_{y}\)) and 4 antibonding orbitals (\(2\pi_{x}^*\) and \(2\pi_{y}^*\)) from the interaction of the 2p orbitals perpendicular to the internuclear axis. Next, we fill in the electrons: 1. \(\mathrm{NO}^+\): 9 valence electrons can be placed as follows: \[\sigma(2s)^2, \sigma^*(2s)^1, \sigma(2p)^2, \pi(2p_x)^2, \pi(2p_y)^2\] This leads to a bond order of \[\frac{1}{2}(6 - 3) = 1.5\] which indicates a bond between single and double bond character. 2. \(\mathrm{NO}^-\): 11 valence electrons result in: \[\sigma(2s)^2, \sigma^*(2s)^2, \sigma(2p)^2, \pi(2p_x)^2, \pi(2p_y)^2, \sigma^*(2p)^1\] The bond order is \[\frac{1}{2}(6 - 4) = 1\], which indicates a single bond character. 3. \(\mathrm{NO}\): With 10 valence electrons: \[\sigma(2s)^2, \sigma^*(2s)^2, \sigma(2p)^2, \pi(2p_x)^2, \pi(2p_y)^2\] This results in a bond order of \[\frac{1}{2}(6 - 2) = 2\], indicating a double bond character.

The localized electron model predicts triple bonds for all three molecular species. However, the molecular orbital model predicts different bond orders: 1.5 for \(\mathrm{NO}^+\), 1 for \(\mathrm{NO}^-\), and 2 for \(\mathrm{NO}\). The molecular orbital model provides a more accurate picture of the bonding, as it accounts for the differences in the bond strength and bond length of the three species. These discrepancies arise due to the difference in the ways electrons are distributed among orbitals in each model.