A) \[NO_{2}^{-}\] and \[N{{H}_{3}}\]
B) \[BF_{3}^{{}}\] and \[NH_{2}^{-}\]
C) \[NH_{2}^{-}\] and \[H_{2}^{{}}O\]
D)
\[BF_{3}^{{}}\] and 
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 \[B{{F}_{3}}\] and \[NO_{2}^{-}\] have \[s{{p}^{2}}\] hybridization. |
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