A) \[ClOF_{4}^{-}\]
B) \[ClOF_{2}^{+}\]
C) \[BH_{4}^{-}\]
D) \[XeO{{F}_{2}}\]
Correct Answer: D
Solution :
\[ClO\text{F}_{4}^{-}\,\,=\,\,1\,\,l.p.\] \[ClO\text{F}_{2}^{+}\,\,=\,\,1\,\,l.p\] \[BH_{4}^{-}~=\,\,0\,\,l.p\] \[XeO{{F}_{2}}~=\,\,2\,\,l.p.\]
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