What is the hybridization of xecl4?

Of course, I can help with that.

The hybridization of the central xenon atom in XeCl₄ is sp³d².

Let's break down how we get there. Xenon, being a noble gas, starts with eight valence electrons. It forms four single bonds with the four chlorine atoms. This process uses up four of its valence electrons, one for each bond.

That leaves four electrons remaining on the central xenon atom. These four electrons pair up to form two lone pairs.

To determine the hybridization, we need to count the total number of electron domains around the xenon. This includes both the bonded atoms and the lone pairs. So, we have four bonds and two lone pairs, giving us a total of six electron domains.

To accommodate six electron domains, the atom needs six hybrid orbitals. This corresponds perfectly to sp³d² hybridization (one s + three p + two d orbitals = six orbitals). As a result of this arrangement, the molecule adopts a square planar shape to minimize repulsion between the lone pairs.

Hope that clears it up for you.