• # question_answer Arrange the following in order of decreasing bond angle, giving reason : $N{{O}_{2}},\,NO_{2}^{+},\,NO_{2}^{-}$

$NO_{2}^{+}>N{{O}_{2}}>NO_{2}^{-}$. This is because $NO_{2}^{+}$ has no lone pair of electrons (i.e. has only bond pairs on two sides) and hence it is linear. $N{{O}_{2}}$ has one unshared electron while $NO_{2}^{-}$ has one unshared electron pair. There are greater repulsions on $N-O$ bonds in case of $NO_{2}^{-}$ than in case of $N{{O}_{2}}$.