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

We begin by drawing the Lewis structures of both ions. Count the total number of valence electrons available, place the less electronegative atoms in the center, and add bonds or lone pairs on the peripheral atoms accordingly. For \(\mathrm{NO}_{2}^{-}\), there are 18 valence electrons (5 from N, 6 from each O, and 1 extra for the -1 charge). The O atoms are more electronegative, so N is placed in the center of the structure: O || N \ O For \(\mathrm{NO}_{3}^{-}\), there are 24 valence electrons (5 from N, 6 from each O, and 1 extra for the -1 charge). Similar to the previous structure, O atoms are placed around the central N atom: O \ N // \ O O

Now we calculate the formal charges for each atom in the structure to ensure that it is the most plausible structure, using the formula: \( Formal\ Charge = Group\ Number - (Number\ of\ Bonds + Number\ of\ Lone\ Pair\ Electrons) \) For \(\mathrm{NO}_{2}^{-}\), the formal charge on nitrogen is +1 and on one oxygen atom is -1. For \(\mathrm{NO}_{3}^{-}\), the formal charge on nitrogen is +1 and on one oxygen atom is -1, while the other two oxygen atoms have a formal charge of 0.

Now we identify any resonance structures, which are alternative arrangements of electrons that result in an identical overall distribution of electron density. Increasing the number of equivalent resonance structures results in a smaller bond length. For \(\mathrm{NO}_{2}^{-}\), there are two resonance structures with the double bond and the negative charge on different oxygen atoms. For \(\mathrm{NO}_{3}^{-}\), there are three equivalent resonance structures, with the double bond and the negative charge rotating around the three oxygen atoms.

Bond Length is inversely proportional to the bond order or number of resonance structures. Since \(\mathrm{NO}_{3}^{-}\) has more resonance structures (3) than \(\mathrm{NO}_{2}^{-}\) (2), the electrons are more evenly distributed, and the bonds in \(\mathrm{NO}_{3}^{-}\) will be shorter. So, \(\mathrm{NO}_{2}^{-}\) has greater bond lengths than \(\mathrm{NO}_{3}^{-}\).