A) \[\text{NO}_{2}^{-}\text{and}\,\text{N}{{\text{H}}_{\text{3}}}\]
B) \[\text{BF}_{\text{3}}^{{}}\text{and}\,\text{NH}_{2}^{-}\]
C) \[\text{NH}_{\text{2}}^{\text{-}}\text{and}\,\text{H}_{2}^{{}}\text{O}\]
D) \[\text{BF}_{3}^{{}}\text{and}\,\text{NH}_{2}^{-}\]
Correct Answer: B
Solution :
Key Idea For \[s{{p}^{2}}\] hybridization, there must be \[3\sigma \]-bonds or \[2\sigma \]-bonds along with a lone pair of electrons. (i)\[NO_{2}^{-}\Rightarrow 2\sigma +1\,lp=3,ie,s{{p}^{2}}\]hybridization (ii)\[N{{H}_{3}}\Rightarrow 3\sigma +1\,lp=4,ie,s{{p}^{3}}\] hybridization (iii)\[B{{F}_{3}}\Rightarrow 3\sigma +0\,lp=3,ie,s{{p}^{2}}\] hybridization (iv)\[NH_{2}^{-}\Rightarrow 2\sigma +2\,lp=4,ie,s{{p}^{3}}\] hybridization (v)\[H_{2}^{{}}O\Rightarrow 2\sigma +2\,lp=4,ie,s{{p}^{3}}\] hybridization Thus, among the given pairs, only \[\text{B}{{\text{F}}_{\text{3}}}\] and \[\text{NO}_{2}^{-}\]have \[s{{p}^{2}}\] hybridization.
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