The number of lone pair of electrons present | vedclass

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(A) The valence shell of Xenon $(Xe)$ contains $8$ electrons. The number of lone pairs can be calculated using the formula: $	ext{Lone pairs} = \frac

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$1, 2, 3, 1$

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(A) The valence shell of Xenon $(Xe)$ contains $8$ electrons. The number of lone pairs can be calculated using the formula: $	ext{Lone pairs} = \frac{1}{2} (V - B)$,where $V$ is the number of valence electrons and $B$ is the number of bonding electrons.$1$. In $XeOF_4$: $Xe$ forms $4$ single bonds with $F$ and $1$ double bond with $O$,using $4 + 2 = 6$ electrons. Lone pairs = $\frac{1}{2} (8 - 6) = 1$.$2$. In $XeF_4$: $Xe$ forms $4$ single bonds with $F$,using $4$ electrons. Lone pairs = $\frac{1}{2} (8 - 4) = 2$.$3$. In $XeF_2$: $Xe$ forms $2$ single bonds with $F$,using $2$ electrons. Lone pairs = $\frac{1}{2} (8 - 2) = 3$.$4$. In $XeF_6$: $Xe$ forms $6$ single bonds with $F$,using $6$ electrons. Lone pairs = $\frac{1}{2} (8 - 6) = 1$.Thus,the number of lone pairs are $1, 2, 3, 1$ respectively.

Column-$I$	Column-$II$
$(a) XeF_{4}$	$(i) \text{Pyramidal}$
$(b) XeF_{6}$	$(ii) \text{Square planar}$
$(c) XeOF_{4}$	$(iii) \text{Distorted octahedral}$
$(d) XeO_{3}$	$(iv) \text{Square pyramidal}$

The number of lone pair of electrons present in the valence shell of Xenon $(Z=54)$ in $XeOF_4$,$XeF_4$,$XeF_2$,and $XeF_6$ are respectively:

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