When SbF_5 reacts with XeF_4 to form an adduc | vedclass

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(B) In the reaction between $XeF_4$ and $SbF_5$,$XeF_4$ acts as a fluoride $(F^-)$ donor.The reaction is: $XeF_4 + SbF_5 \rightarrow [XeF_3]^+ + [SbF_

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$T$-shaped,octahedral

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(B) In the reaction between $XeF_4$ and $SbF_5$,$XeF_4$ acts as a fluoride $(F^-)$ donor.The reaction is: $XeF_4 + SbF_5 \rightarrow [XeF_3]^+ + [SbF_6]^-$.For the $[XeF_3]^+$ cation,the central $Xe$ atom has $7 + 4 - 1 = 10$ valence electrons,leading to $sp^3d$ hybridization with two lone pairs,resulting in a $T$-shaped structure.For the $[SbF_6]^-$ anion,the central $Sb$ atom has $5 + 6 + 1 = 12$ valence electrons,leading to $sp^3d^2$ hybridization with no lone pairs,resulting in an octahedral structure.

Column $I$	Column $II$
$(A). (CH_3)_2 SiCl_2$	$(p).$ Hydrogen halide formation
$(B). XeF_4$	$(q).$ Redox reaction
$(C). Cl_2$	$(r).$ Reacts with glass
$(D). VCl_5$	$(s).$ Polymerization
	$(t). O_2$ formation

When $SbF_5$ reacts with $XeF_4$ to form an adduct,the shapes of cation and anion in the adduct respectively are

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