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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 $x = +8$.$Xe$ ha

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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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