Question

During the formation of the $N_2{O}_4$ dimer from two molecules of $N{O}_2$, the odd electrons, one in each of the nitrogen atoms of the $N{O}_2$ molecules, get paired to form a

• A. weak $N-N$ bond, two $N-O$ bonds become equivalent and the other two $N-O$ bonds become nonequivalent
• B. weak $N-N$ bond and all the four $N-O$ bonds become equivalent
• C. strong $N-N$ bond and all the four $N-O$ bonds become non-equivalent
• D. strong $N-N$ bond and all the four $N-O$ bonds become equivalent.

Answer 1

Answer: (B)

The dimer $N_{2}S{O}_4$ has a planer structure, the odd electon on each of the $N$ atoms of the $N{O}_2$ molecules pairing to form a weak $N-N$ bond.
he $N-N$ bond is very long $(175pm)$, and is therefore weak.
It is much longer than the single bond $N-N$ distance of $147pm$ in hydrazing but there is no satiafactory explanation of why it is long.
All the four $N-O$ bonds are equivalent as the molecule is a resonance hybrid of the following forms: