Question

The correct order of $ N - O $ bond lengths in $NO, NO_2^-, NO_3^- $ and $ N_2O_4 $ is

• A. $N_2O_4 > NO_2^- > NO_3^- > NO$
• B. $ NO > NO_3^- > N_2O_4 > NO_2^-$
• C. $ NO_3^- > NO_2^- > N_2O_4 > NO$
• D. $ NO > N_2O_4 > NO_2^- > NO_3^- $

Answer 1

The Correct Option is C

 To determine the correct order of \(N - O\) bond lengths in \(NO, NO_2^-, NO_3^-\), and \(N_2O_4\), we need to understand the concepts of bond order and resonance. Bond length is inversely related to bond order: higher bond order means a shorter bond length and vice versa.

1. Nitric Oxide (\(\(NO\)\)):
   • In nitric oxide, there is an unpaired electron which results in a bond order of about 2.5 (higher bond order).
2. Nitrite ion (\(\(NO_2^-\)\):
   • In the nitrite ion, due to resonance, the bond order is 1.5.
3. Nitrate ion (\(\(NO_3^-\)\):
   • In the nitrate ion, resonance lowers the bond order to approximately 1.33 because it is averaged over three equivalent structures.
4. Dinitrogen Tetroxide (\(\(N_2O_4\)\):
   • Dimerizes from nitrogen dioxide \((\(NO_2\))\) and each \(\(NO_2\)\) unit contributes to the bond order through resonance.
   • When two \(\(NO_2\)\) molecules are joined, the average bond order between \(N\) and \(O\) in \(N_2O_4\) decreases from that in \(\(NO_2^-\)\)

Considering the bond order calculations as explained above:

• \(NO\) has the highest bond order and hence the shortest bond length.
• \(N_2O_4\), being the nitrosonium ketone in equilibrium, has a shorter bond length than \(NO_2^-\\), but longer than \(NO\).\)
• \(NO_2^-\\) has a longer bond length than \(N_2O_4\) due to lower bond order.
• \(NO_3^-\\) has the longest bond length due to the lowest bond order among these molecules.

Therefore, the correct order of \(N - O\) bond lengths is given by: \(NO_3^- > NO_2^- > N_2O_4 > NO\).