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 \[\text{NO}_{\text{2}}^{\text{+}}\text{,ie,18}{{\text{0}}^{\text{o}}}\text{.}\]You need to login to perform this action.
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