Based on Lewis structures, predict the ordering, from shortest to longest, of \(\mathrm{N}-\mathrm{O}\) bond lengths in $\mathrm{NO}^{+}, \mathrm{NO}_{2}^{-},\( and \)\mathrm{NO}_{3}^{-}$.

(1) NO^+ ion: The total number of valence electrons for this ion is 10 (five from nitrogen and six from oxygen, minus one from the positive charge). The Lewis structure for NO^+ is: ``` O = N →→ ``` (2) NO2^− ion: The total number of valence electrons for this ion is 18 (five from nitrogen, six from each oxygen, and one from the negative charge). The Lewis structure for NO2^- is: ``` O ¬ N ¬ O ↓ ``` (3) NO3^− ion: The total number of valence electrons for this ion is 24 (five from nitrogen, six from each oxygen, and one from the negative charge). The Lewis structure for NO3^- is a resonance structure that represents an average of the three possible structures: ``` O ¬ N ¬ O ↓ ↓ O ```

(1) NO^+ ion: The N-O bond in NO^+ is a triple bond, so its bond order is 3. (2) NO2^− ion: One N-O bond in NO2^- is a double bond, and the other is a single bond. The average bond order for the two N-O bonds in NO2^- is \((1 + 2)/2 = 1.5\). (3) NO3^− ion: Each N-O bond in NO3^- is a double bond, and there are three equivalent resonance structures. The bond order for N-O in NO3^- is 2 (since each double bond has an order of 2).

Bond length is inversely proportional to bond order. A higher bond order corresponds to a shorter bond length. Considering the bond orders calculated in Step 2, we see that the order, from shortest to longest N-O bond length, is: Shortest: NO^+ (bond order = 3) < NO3^- (bond order = 2) < Longest: NO2^- (bond order = 1.5) Thus, the N-O bond lengths ordering is NO^+ < NO3^- < NO2^-.