The type of hybridization and number of lone pair(s) of electrons of Xe in $$XeOF_4$$​ , respectively, are

Correct option is A

The correct answer is (A) sp²d² and 1

Explanation:

In XeOF₄ (Xenon oxytetrafluoride), we need to determine the hybridization and the number of lone pairs of electrons on the central atom (Xenon). Here's how we can solve it:

1. Electron Counting:

   • Xenon (Xe) has 8 valence electrons.

   • Oxygen (O) contributes 6 valence electrons.

   • Fluorine (F) contributes 7 valence electrons per atom, and there are 4 fluorine atoms.

   • Total valence electrons = 8 (from Xe) + 6 (from O) + 4 × 7 (from F) = 8 + 6 + 28 = 42 valence electrons.

2. Bonding and Hybridization:

   • Xenon will form 5 bonds: one with oxygen and four with fluorine.

   • Xenon uses its d-orbitals to accommodate the bonding, resulting in sp²d² hybridization. This means two p-orbitals and two d-orbitals are involved in bonding.

3. Lone Pairs:

   • Since there are 5 bonding pairs, this leaves 2 lone pairs of electrons on Xenon.

   • The electron distribution results in a square pyramidal geometry with 5 bonding pairs and 2 lone pairs.

Therefore, the correct answer should be sp²d² hybridization with 2 lone pairs.

Information Booster:

• sp²d² hybridization is characterized by 5 bonding regions, which are formed from a combination of s, p, and d orbitals.

• Xenon in XeOF₄ is in the + 6 oxidation state, as it bonds with four fluorine atoms and one oxygen atom.

• The square pyramidal geometry results from having 5 bonding pairs and 2 lone pairs, which are positioned to minimize electron repulsion.

• This molecule is an example of a noble gas compound, showcasing the ability of Xenon to form bonds with highly electronegative elements like oxygen and fluorine.