According to Lewis theory, the total number of $$\sigma$$ bond-pairs and lone pair of electrons around the central atom of $$\text{XeO}_6^{4-}$$ ion is _____.

The central atom is xenon. Xenon belongs to group 18, so it possesses $$8$$ valence electrons.

Step 1 - make the σ-bond framework.
$$\text{XeO}_6^{4-}$$ contains six Xe-O bonds. Formation of one σ-bond consumes one electron from Xe. Hence electrons of Xe used in σ-bonding = $$6$$.

Step 2 - electrons remaining on Xe.
Electrons initially present on Xe = $$8$$.
Electrons already used in σ-bonds = $$6$$.
Electrons still left on Xe = $$8-6=2$$. These two electrons stay together as one lone pair.

Step 3 - count σ-bond pairs and lone-pair(s).
Number of σ-bond pairs around Xe = $$6$$.
Number of lone pairs on Xe = $$1$$.

Therefore, the total of “σ bond-pairs + lone pair of electrons’’ around the central xenon atom is $$6 + 1 = 7$$.

Answer: 7