The correct order of increasing bond angles in the following...

Key Idea : As the number of lone pair of electrons increases, bond angle decreases. $\mathrm{NO}_2^{+}$ ion is isoelectronic with $\mathrm{CO}_2$ molecule. It is a linear ion and its central atom $\left(\mathrm{N}^{+}\right)$ undergoes $s p$-hybridisation, hence bond angle is $180^{{}^{\circ}}$. In $\mathrm{NO}_2^{-}$ ion, $\mathrm{N}$-atom undergoes $s p^2$ hybridisation. The angle between hybrid orbital should be $120^{{}^{\circ}}$ but one lone pair of electrons is lying on $\mathrm{N}$-atom, hence bond angle decreases to $115^{{}^{\circ}}$. In $\mathrm{NO}_2$ molecule, $\mathrm{N}$-atom has one unpaired electron in $s p^2$-hybrid orbital. The bond angle should be $120^{{}^{\circ}}$ but actually it is $132^{{}^{\circ}}$. It may be due to one unpaired electron in $s p^2-hybrid orbital. Therefore, the increasing order of$bond angles is : \underset{\left(115^{{}^{\circ}}\right)}{\mathrm{NO}_{2}^{-}} < \underset{\left(132^{{}^{\circ}}\right)}{\mathrm{NO}_2 } < \underset{\left(180^{{}^{\circ}}\right)}{\mathrm{NO}_{+}^{+}}