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The correct order of increasing bond angles in the following triatomic species is :
A
$NO_2^+ \lt NO_2 \lt NO_2^-$
B
$NO_2^+ \lt NO_2^- \lt NO_2$
C
$NO_2^- \lt NO_2^+ \lt NO_2$
D
$NO_2^- \lt NO_2 \lt NO_2^+$
Detailed Solution
$NO_2^+$: the nitrogen has no lone pair and is $sp$ hybridised, so the ion is linear with a bond angle of $180^\circ$.
$NO_2$: the nitrogen is $sp^2$ hybridised and carries one odd (unpaired) electron. A single electron repels the bond pairs less than a lone pair does, so the bond angle is larger than $120^\circ$, about $132^\circ$ to $134^\circ$.
$NO_2^-$: the nitrogen is $sp^2$ hybridised and carries one lone pair. The lone pair-bond pair repulsion pushes the bonds closer, so the bond angle is less than $120^\circ$, about $115^\circ$.
Bond angles: $NO_2^+$ ($180^\circ$) > $NO_2$ (about $132^\circ$) > $NO_2^-$ ($115^\circ$)
Increasing order of bond angle: $NO_2^- \lt NO_2 \lt NO_2^+$
$NO_2$: the nitrogen is $sp^2$ hybridised and carries one odd (unpaired) electron. A single electron repels the bond pairs less than a lone pair does, so the bond angle is larger than $120^\circ$, about $132^\circ$ to $134^\circ$.
$NO_2^-$: the nitrogen is $sp^2$ hybridised and carries one lone pair. The lone pair-bond pair repulsion pushes the bonds closer, so the bond angle is less than $120^\circ$, about $115^\circ$.
Bond angles: $NO_2^+$ ($180^\circ$) > $NO_2$ (about $132^\circ$) > $NO_2^-$ ($115^\circ$)
Increasing order of bond angle: $NO_2^- \lt NO_2 \lt NO_2^+$