A) \[{{H}_{2}}O\]
B) \[B{{F}_{3}}\]
C) \[NH_{2}^{-}\]
D) \[PC{{l}_{3}}\]
Correct Answer: D
Solution :
[a] \[{{H}_{2}}O\Rightarrow \] |
[bp = bond pairs and Ip = lone pair] |
[b] |
[c] |
[d] |
Thus, in PCl3, the central atom P ha:, three bond pairs and one lone pair. |
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