In the formation of the dimer `N_(2)O_(4)`from two molecules of `NO_(2)`the odd electron on each of the `N` atoms of the `NO_(2)` molecule gets paired to form a

Options

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

Answer

2

Explanation

The dimer `N_(2)SO_(4)` has a planer structure, the odd electon on each of the `N` atoms of the `NO_(2)` molecules pairing to form a weak `N-N` bond.
he `N-N` bond is very long `(175" pm")`, and is therefore weak. It is much longer than the single bond `N-N` distance of `147` pm 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: