Total Number of Sigma Bond Pairs and Lone Pairs in XeO6 4 Minus

To find the total number of $\sigma$ bond-pairs and lone pairs of electrons around the central Xenon ($\text{Xe}$) atom in the perxenate ion ($\text{XeO}_6^{4-}$), we analyze its Lewis structure and bonding:

1. Valence Electrons of the Central Atom: Xenon ($\text{Xe}$) belongs to Group 18 and possesses $8$ valence electrons in its ground state ($5s^2 5p^6$).

2. Oxidation State of Xenon in $\text{XeO}_6^{4-}$: Let $x$ be the oxidation state of Xenon: $x + 6(-2) = -4 \implies x = +8$ Since Xenon is in the $+8$ oxidation state, all $8$ of its valence electrons are involved in bonding.

3. Bonding Structure:

   • The central Xenon atom is bonded to six Oxygen atoms arranged octahedrally ($sp^3d^2$ hybridization).
   • To satisfy the overall $-4$ charge and octet requirements of Oxygen:
     • Four Oxygen atoms form single bonds ($\text{Xe}-\text{O}^-$), contributing $4$ $\sigma$ bonds.
     • Two Oxygen atoms form double bonds ($\text{Xe}=\text{O}$), contributing $2$ $\sigma$ bonds and $2$ $\pi$ bonds.

4. Counting $\sigma$ Bond-Pairs and Lone Pairs around $\text{Xe}$:

   • $\sigma$ bond-pairs: Each of the 6 Oxygen atoms forms one $\sigma$ bond with $\text{Xe}$. Therefore, the number of $\sigma$ bond-pairs $= 6$.
   • $\pi$ bond-pairs: $2$ $\pi$ bonds are formed using 2 valence electrons of $\text{Xe}$.
   • Electrons used by $\text{Xe}$: $6 \text{ (for } \sigma \text{ bonds)} + 2 \text{ (for } \pi \text{ bonds)} = 8$ valence electrons.
   • Lone pairs on $\text{Xe}$: Since all 8 valence electrons are engaged in bonding, the number of non-bonding lone pairs on $\text{Xe} = 0$.

5. Total Sum: $\text{Total} = (\text{Number of } \sigma \text{ bond-pairs}) + (\text{Number of lone pairs})$ $\text{Total} = 6 + 0 = 6$

Thus, the total number of $\sigma$ bond-pairs and lone pairs around the central atom is 6.