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(C) For $XeF_5^-$,the number of valence electrons on $Xe$ is $8$. The charge is $-1$,so total valence electrons $= 8 + 5 + 1 = 14$. Number of electron

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There are two non-bonding electron pairs,one above the plane and the other below the plane.

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(C) For $XeF_5^-$,the number of valence electrons on $Xe$ is $8$. The charge is $-1$,so total valence electrons $= 8 + 5 + 1 = 14$. Number of electron pairs $= 14 / 2 = 7$. This corresponds to $sp^3d^3$ hybridisation. The geometry is pentagonal planar with two non-bonding electron pairs,one above the plane and the other below the plane.

What is the correct statement with respect to $XeF_5^-$?

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