Which of the following is the correct decreas | vedclass

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(C) To determine the bond angle,we look at the hybridization and the number of lone pairs on the central nitrogen atom:$1$. $NO^{+}_2$: The nitrogen a

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$NO^{+}_2 > NO_2 > NO^{-}_2$

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(C) To determine the bond angle,we look at the hybridization and the number of lone pairs on the central nitrogen atom:$1$. $NO^{+}_2$: The nitrogen atom is $sp$ hybridized with $0$ lone pairs. The geometry is linear,so the bond angle is $180^{\circ}$.$2$. $NO_2$: The nitrogen atom is $sp^2$ hybridized with $1$ unpaired electron. The geometry is bent,and the bond angle is approximately $134^{\circ}$.$3$. $NO^{-}_2$: The nitrogen atom is $sp^2$ hybridized with $1$ lone pair. The geometry is bent,and the bond angle is approximately $115^{\circ}$.Therefore,the correct decreasing order of bond angles is $NO^{+}_2 > NO_2 > NO^{-}_2$.

Which of the following is the correct decreasing order of bond angles?

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