In $\mathrm{XeF}_{2}, \mathrm{XeF}_{4}$ and $\mathrm{XeF}_{6}$, the number of lone pairs on Xe are…
In $\mathrm{XeF}_{2}, \mathrm{XeF}_{4}$ and $\mathrm{XeF}_{6}$, the number of lone pairs on Xe are respectively
$2,3,1$
$1,2,3$
$4,1,2$
$3,2,1$
Solution
In $\mathrm{XeF}_{2}$ Total number of valence electrons of $\mathrm{Xe}=8$, two electrons shared with $2 \mathrm{~F}$ atoms, 6 electrons left hence 3 lone pairs, in $\mathrm{XeF}_{4} 4$ shared with $4 \mathrm{~F}$ atoms 4 left hence 2 lone pairs; in $\mathrm{XeF}_{6} 6$ shared with $6 \mathrm{~F}$ atoms 2 left hence 1 lone pair.
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