The type of hybridisation and number of lone | vedclass

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(A) To determine the hybridisation and lone pairs of $Xe$ in $XeOF_4$:$1$. The central atom is $Xe$,which has $8$ valence electrons.$2$. $Xe$ forms $4

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$sp^3d^2$ and $1$

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(A) To determine the hybridisation and lone pairs of $Xe$ in $XeOF_4$:$1$. The central atom is $Xe$,which has $8$ valence electrons.$2$. $Xe$ forms $4$ single bonds with $F$ atoms and $1$ double bond with the $O$ atom.$3$. Total number of electron pairs around $Xe$ = (Number of sigma bonds) + (Number of lone pairs).$4$. $Xe$ forms $5$ sigma bonds ($4$ with $F$ and $1$ with $O$) and $1$ pi bond with $O$.$5$. The number of lone pairs on $Xe$ = (Total valence electrons - Electrons used in bonding) / $2$ = $(8 - (4 	imes 1 + 1 	imes 2)) / 2 = (8 - 6) / 2 = 1$ lone pair.$6$. Steric number = (Number of sigma bonds) + (Number of lone pairs) = $5 + 1 = 6$.$7$. $A$ steric number of $6$ corresponds to $sp^3d^2$ hybridisation.$8$. Therefore,the hybridisation is $sp^3d^2$ and the number of lone pairs is $1$.

The type of hybridisation and number of lone pair$(s)$ of electrons of $Xe$ in $XeOF_4$ respectively are

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