In XeF_2, XeF_4, XeF_6,the number of lone pai | vedclass

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(D) The central atom $Xe$ has $8$ valence electrons.For $XeF_2$: $Xe$ forms $2$ bonds with $F$ atoms,using $2$ electrons. Remaining electrons = $8 - 2

Vedclass · Accepted Answer

$3, 2, 1$

Vedclass · Answer

(D) The central atom $Xe$ has $8$ valence electrons.For $XeF_2$: $Xe$ forms $2$ bonds with $F$ atoms,using $2$ electrons. Remaining electrons = $8 - 2 = 6$. Number of lone pairs = $6 / 2 = 3$.For $XeF_4$: $Xe$ forms $4$ bonds with $F$ atoms,using $4$ electrons. Remaining electrons = $8 - 4 = 4$. Number of lone pairs = $4 / 2 = 2$.For $XeF_6$: $Xe$ forms $6$ bonds with $F$ atoms,using $6$ electrons. Remaining electrons = $8 - 6 = 2$. Number of lone pairs = $2 / 2 = 1$.Thus,the number of lone pairs on $Xe$ in $XeF_2, XeF_4, XeF_6$ are $3, 2, 1$ respectively.

In $XeF_2, XeF_4, XeF_6$,the number of lone pairs on $Xe$ is respectively:

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