question_answer

The ONO bond angle is maximum in:

A) \[{{\operatorname{NO}}_{3}}^{-}\]                 

B) \[{{\operatorname{NO}}_{2}}^{-}\]   

C) \[{{\operatorname{NO}}_{2}}\]                         

D) \[{{\operatorname{NO}}_{2}}^{+}\]

Correct Answer: D

Solution :

\[{{\operatorname{NO}}^{+}}_{2}\] \[{{\operatorname{NO}}^{-}}_{3}has s{{p}^{2}}\] hybridisation, resonating structure \[{{\operatorname{NO}}^{-}}_{2}\]has no unshared electron & has sp hybridisation shape is linear. \[(\underset{..}{\overset{..}{\mathop{o}}}\,\bar{-}N\bar{-}\underset{..}{\overset{..}{\mathop{o}}}\,) With bond angle = 180{}^\circ  N{{O}_{2}} has\]one unshared electron. Whereas\[N{{O}_{2}}^{-}\]has one unshared electron pair. Hence in\[N{{O}_{2}}^{-}\]repulsion on bond pair are more and angle is less