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:

weak N-N bond, two N-O bonds become equivalent and the other N-O bonds become nonequivalent

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weak N-N bond and all the four N-O bonds become equivalent

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strong N-N bond and all the four N-O bonds become Non-equivalent

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none of these

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Solution

The correct option is C weak N-N bond and all the four N-O bonds become equivalent
The $N - N$ single bond is weak and the $4$ $N - O$ bonds are equivalent.

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N - O - N

N_{2} O_{5}

O - N - O

N O_{2}

$N_{2}$ and $O_{2}$ are converted into monocations, $N_{2}^{+}$ and $O_{2}^{+}$ respectively. Which of the following is wrong ?

N_{2}

O_{2}

N_{2}^{+}

O_{2}^{+}

N O_{2}

N_{2} O_{4}

N O_{2}

N_{2} O

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