Observe the following reaction: XeF_4 + SbF_5 | vedclass

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(C) $1$. The reaction is $XeF_4 + SbF_5 \rightarrow [XeF_3]^+ [SbF_6]^-$. $2$. The cationic part is $[XeF_3]^+$. The central atom $Xe$ has $8$ valence

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The anionic part of $[XeF]^+ [PF_6]^-$ is isostructural with the anionic part of the product.

Vedclass · Answer

(C) $1$. The reaction is $XeF_4 + SbF_5 \rightarrow [XeF_3]^+ [SbF_6]^-$. $2$. The cationic part is $[XeF_3]^+$. The central atom $Xe$ has $8$ valence electrons. It forms $3$ bonds with $F$ and has $2$ lone pairs. Total electron pairs = $5$,so hybridization is $sp^3d$. $3$. The anionic part is $[SbF_6]^-$. The central atom $Sb$ has $5$ valence electrons. It forms $6$ bonds with $F$ and has $0$ lone pairs. Total electron pairs = $6$,so hybridization is $sp^3d^2$. $4$. Option $A$: $XeOF_2$ has $sp^3d$ hybridization. $[SbF_6]^-$ has $sp^3d^2$. Incorrect. $5$. Option $B$: $I_3^-$ has $3$ lone pairs on the central $I$ atom. $[XeF_3]^+$ has $2$ lone pairs on the central $Xe$ atom. Incorrect. $6$. Option $C$: The anionic part of $[XeF]^+ [PF_6]^-$ is $[PF_6]^-$. $[PF_6]^-$ is octahedral $(sp^3d^2)$,and $[SbF_6]^-$ is also octahedral $(sp^3d^2)$. They are isostructural. Correct.

Observe the following reaction: $XeF_4 + SbF_5 \rightarrow [XeF_3]^+ [SbF_6]^-$. The correct statement$(s)$ regarding the product is/are:

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