According to Lewis theory,the total number of | vedclass

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(A) In $XeO_6^{4-}$ ion,the central atom is Xenon $(Xe)$.The oxidation state of $Xe$ is calculated as: $x + 6(-2) = -4$,which gives $x = +8$.$Xe$ has

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$6$

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(A) In $XeO_6^{4-}$ ion,the central atom is Xenon $(Xe)$.The oxidation state of $Xe$ is calculated as: $x + 6(-2) = -4$,which gives $x = +8$.$Xe$ has $8$ valence electrons. In the $XeO_6^{4-}$ ion,$Xe$ forms $6$ double bonds with $6$ oxygen atoms.Each double bond consists of $2$ electron pairs (one sigma and one pi bond),but in Lewis structure counting for geometry/$VSEPR$,we count the number of electron domains around the central atom.Here,$Xe$ is bonded to $6$ oxygen atoms,so there are $6$ bonding domains (bond pairs).Since all $8$ valence electrons of $Xe$ are used in bonding with $6$ oxygen atoms (each oxygen atom requires $2$ electrons to complete its octet,and the negative charge accounts for the extra electrons),there are no lone pairs remaining on the $Xe$ atom.Total number of electron pairs around the central atom = $6 	ext{ (bond pairs)} + 0 	ext{ (lone pairs)} = 6$.

According to Lewis theory,the total number of bond-pairs and lone pairs of electrons around the central atom of $XeO_6^{4-}$ ion is . . . . . . .

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