VSEPR Theory Questions in English

(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 = $6 \text{ (bond pairs)} + 0 \text{ (lone pairs)} = 6$.

(B) In $\text{ClF}_3$,the central chlorine atom has $7$ valence electrons.
It forms $3$ covalent bonds with fluorine atoms,leaving $2$ lone pairs on the chlorine atom.
According to the $VSEPR$ theory,the steric number is $3 + 2 = 5$,which corresponds to $sp^3d$ hybridization.
The presence of $2$ lone pairs in the equatorial positions of the trigonal bipyramidal electron geometry results in a $T$-shaped molecular geometry.