The shape of XeF_{5}^{-} will be | vedclass

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(C) To determine the shape of $XeF_{5}^{-}$,we calculate the number of electron pairs around the central $Xe$ atom: Number of electron pairs $= \frac{

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planar

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(C) To determine the shape of $XeF_{5}^{-}$,we calculate the number of electron pairs around the central $Xe$ atom: Number of electron pairs $= \frac{1}{2} [V + M - C + A] = \frac{1}{2} [8 + 5 - 0 + 1] = 7$. Here,$V = 8$ (valence electrons of $Xe$),$M = 5$ (monovalent atoms),and $A = 1$ (anionic charge). With $7$ electron pairs,the electron geometry is pentagonal bipyramidal ($sp^{3}d^{3}$ hybridization). Out of these $7$ pairs,$5$ are bond pairs and $2$ are lone pairs. The two lone pairs occupy the axial positions to minimize repulsion,leaving the $5$ $Xe-F$ bonds in a single equatorial plane. Therefore,the molecular shape of $XeF_{5}^{-}$ is pentagonal planar.

The shape of $XeF_{5}^{-}$ will be

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