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(B) $1$. In $NO_2^+$,the hybridization of $N$ is $sp$,the geometry is linear,and the $O-N-O$ bond angle is $180^{\circ}$. The bond order is $2$. $2$.

Vedclass · Accepted Answer

Decrease of $O-N-O$ bond angle

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(B) $1$. In $NO_2^+$,the hybridization of $N$ is $sp$,the geometry is linear,and the $O-N-O$ bond angle is $180^{\circ}$. The bond order is $2$. $2$. In $NO_2$,the $N$ atom has one unpaired electron. The $O-N-O$ bond angle is approximately $134^{\circ}$. $3$. In $NO_2^-$,the hybridization of $N$ is $sp^2$,the geometry is bent,and the $O-N-O$ bond angle is approximately $115^{\circ}$. $4$. As we move from $NO_2^+$ to $NO_2^-$,the number of lone pairs on the central $N$ atom increases ($0$ to $1$),which causes greater repulsion,leading to a decrease in the $O-N-O$ bond angle. $5$. Simultaneously,the bond order decreases as electrons are added to antibonding orbitals,which leads to an increase in $N-O$ bond length. $6$. Therefore,the statement 'Decrease of $O-N-O$ bond angle' is correct.

Select the correct statement regarding the following change:
$NO_2^+ \to NO_2 \to NO_2^-$

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Select the correct statement regarding the following change:$NO_2^+ \to NO_2 \to NO_2^-$