Which has the greater bond lengths: \(\mathrm{NO}_{2}^{-}\) or \(\mathrm{NO}_{3}^{-} ?\) Explain.

First, we will examine the resonance structures of \(\mathrm{NO}_{2}^{-}\) and \(\mathrm{NO}_{3}^{-}\) ions. This will help us understand the average bond order of nitrogen-oxygen bonds in these ions. In \(\mathrm{NO}_{2}^{-}\), the nitrogen has one single bond and one double bond with the two oxygen atoms. There are two resonance structures for this ion, where the double bond exists between the nitrogen and either of the oxygen atoms. In \(\mathrm{NO}_{3}^{-}\), the nitrogen atom has single bonds with all three oxygen atoms. The charges are distributed uniformly over all oxygen atoms. There are three resonance structures for this ion, where the single unpaired electron on one of the oxygens is delocalized.

The average bond order is the ratio of the total number of bonds between two atoms and the number of different resonance structures. To find the average bond order of nitrogen-oxygen bonds in the aforementioned ions, we need to divide the total number of bonds between nitrogen and oxygen atoms with the number of resonance structures. For \(\mathrm{NO}_{2}^{-}\): Average bond order = \( \frac{1 + 2}{2} = 1.5 \) For \(\mathrm{NO}_{3}^{-}\): Average bond order = \( \frac{1 + 1 + 1}{3} = 1 \)

With the bond order values calculated in the previous step, we can now make a comparison between the bond lengths of both ions. Typically, as bond order increases, the bond length decreases. Since the average bond order of nitrogen-oxygen bonds in \(\mathrm{NO}_{2}^{-} (1.5)\) is greater than that in \(\mathrm{NO}_{3}^{-} (1)\), the bond length in \(\mathrm{NO}_{2}^{-}\) is shorter than the bond length in \(\mathrm{NO}_{3}^{-}\). So, the ion with greater bond lengths is \(\mathrm{NO}_{3}^{-}\).