A) \[NO_{3}^{-}\]
B) \[NO_{2}^{-}\]
C) \[N{{O}_{2}}\]
D) \[NO_{2}^{+}\]
Correct Answer: D
Solution :
In \[NO_{3}^{-}\], N atom undergoes \[s{{p}^{2}}\]-hybridisation. Thus, the ion has trigonal planar geometry with bond angle \[{{120}^{o}}C\]. \[NO_{2}^{-}\] has angular geometry (\[s{{p}^{2}}\]hybridisation) due to presence of one lone pair of electrons on N atom. Bond angle is \[{{116}^{o}}\]. \[N{{O}_{2}}\] has angular geometry (\[s{{p}^{2}}\]hybridisation) with bond angle \[{{134}^{o}}\]. \[NO_{2}^{+}\] (nitronium ion) has linear structure (sp hybridisation) with bond angle \[{{180}^{o}}\]. \[\therefore \] The bond angle is maximum in\[NO_{2}^{+}\], ie,\[{{180}^{o}}\].You need to login to perform this action.
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