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

To determine the molecular structures, first, we will calculate the total number of valence electrons in each ion and draw their respective Lewis structures. For \(\mathrm{NO}_{2}^{-}\), the total number of valence electrons is 18 (5 from N, 12 from two O atoms, and 1 extra electron due to the negative charge). In the case of \(\mathrm{NO}_{3}^{-}\), the total number of valence electrons is 24 (5 from N, 18 from three O atoms, and 1 extra electron due to the negative charge). Now, let's draw the Lewis structures for the ions: \(\mathrm{NO}_{2}^{-}\): O - N = O N is bonded to two O atoms with one single bond and one double bond. The negative charge is placed on the O atom with the single bond. \(\mathrm{NO}_{3}^{-}\): O - N - O | O N is bonded to three O atoms with one single bond, and there are two double bonds between the O atoms. The negative charge will be distributed among the O atoms.

Both ions have resonance structures which have an effect on the distribution of electron pairs. In resonance structures, the electrons are shared among multiple bonds, resulting in an average of what these bonds might look like in reality. \(\mathrm{NO}_{2}^{-}\) resonance structures: O = N - O^(-) O^(-) - N = O \(\mathrm{NO}_{3}^{-}\) resonance structures: O = N - O | O^(-) O - N = O | O^(-) O^(-) - N - O | O

Resonance structures show that electrons are distributed among the bonds, and the bond length is an average of the localized single and double bonds in the different structures. In \(\mathrm{NO}_{2}^{-}\), the two N-O bonds alternate between single and double bonds in resonance structures, resulting in an average bond order of 1.5. In \(\mathrm{NO}_{3}^{-}\), the bonds between N and O atoms are always double in the different resonance structures, resulting in an average bond order of 2.

A higher bond order corresponds to a shorter bond length, and a lower bond order corresponds to a longer bond length. As the bond order of \(\mathrm{NO}_{2}^{-}\) is 1.5 (lower than the bond order of \(\mathrm{NO}_{3}^{-}\), which is 2), the bond length in \(\mathrm{NO}_{2}^{-}\) is greater than in \(\mathrm{NO}_{3}^{-}\). Therefore, the ion with the greater bond length is \(\mathrm{NO}_{2}^{-}\).