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Identify the reaction(s) in which the stereochemistry of underlined species has changed

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${XeF}_{2} + \underline{P} F_{5} \to [XeF]^{+} {[{PF}_{6}]}^{-}$

$\underline{Xe} F_{4} + {SbF}_{5} \to {[{XeF}_{3}]}^{+} {[{SbF}_{6}]}^{-}$

${XeF}_{4} + \underline{Sb} F_{5} \to {[{XeF}_{3}]}^{+} {[{SbF}_{6}]}^{-}$

$\underline{{XeF}_{6}} + 2 CsF \to {({Cs}^{+})}_{2} {[{XeF}_{8}]}^{2 -}$

answer is A, B, D, C.

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Detailed Solution

In the first reaction, Sb has sp³d² hybridization. After reaction it becomes sp³d.

Similarly, in third reaction, SbF5₂ has sp³d hybridization. It is changed after the reaction for Sb element.

Here, in all the reaction stereochemistry of underlined species has changed.

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