The shape of XeF^{+}_3 is | vedclass

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Question

(C) The central atom $Xe$ has $8$ valence electrons.For $XeF^{+}_3$,the number of valence electrons is $8 + 3(7) - 1 = 28$.The number of electron pair

Vedclass · Accepted Answer

Bent $T$-shape

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(C) The central atom $Xe$ has $8$ valence electrons.For $XeF^{+}_3$,the number of valence electrons is $8 + 3(7) - 1 = 28$.The number of electron pairs around $Xe$ is calculated as:$Steric \ number = \frac{1}{2} (V + M - C + A) = \frac{1}{2} (8 + 3 - 1) = 5$.With $5$ electron pairs,the hybridization is $sp^3d$.There are $3$ bond pairs and $2$ lone pairs on the $Xe$ atom.According to $VSEPR$ theory,the $2$ lone pairs occupy the equatorial positions to minimize repulsion.This results in a bent $T$-shape for the $XeF^{+}_3$ ion.

The shape of $XeF^{+}_3$ is

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