question_answer

The ONO angle is maximum in

A) \[\text{NO}_{3}^{-}\]   

B)         \[\text{NO}_{2}^{-}\]  

C)  \[\text{N}{{\text{O}}_{\text{2}}}\]        

D)         \[\text{NO}_{2}^{+}\]

Correct Answer: D

Solution :

In \[NO_{3}^{-},N\]atom undergoes \[s{{p}^{2}}-\]hybridisation. Thus, the ion has trigonal planar geometry with bond angle \[\text{120}{{\,}^{\text{o}}}\text{C}\text{.}\] \[\text{NO}_{2}^{-}\]has angular geometry (\[s{{p}^{2}}\]hybridisation) due to presence of one lone pair of electrons on N atom. Bond angle is \[\text{11}{{\text{6}}^{\text{o}}}\text{.}\] \[\text{N}{{\text{O}}_{\text{2}}}\]has angular geometry (\[\text{s}{{\text{p}}^{2}}\]hybridisation) with bond angle \[{{134}^{o}}\] \[\text{NO}_{\text{2}}^{\text{+}}\] (nitronium ion) has linear structure (sp hybridisation) with bond angle \[\text{18}{{\text{0}}^{o}}.\] \[\therefore \] The bond angle is maximum in \[\text{NO}_{\text{2}}^{\text{+}}\text{,ie,18}{{\text{0}}^{\text{o}}}\text{.}\]