question_answer

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

Answer:

                \[NO_{2}^{+}>N{{O}_{2}}>NO_{2}^{-}\]. 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 \[N{{O}_{2}}\].