The correct geometry and hybridization of \(XeF_6\) are:

Distorted octahedral, \(sp^3d^3\)

The correct answer is option 4. Distorted octahedral, \(sp^3d^3\).

To determine the correct geometry and hybridization of \(XeF_6\), let us analyze the molecule step by step.

Xenon (Xe) is in Group 18 of the periodic table, so it has 8 valence electrons. Fluorine (F) has 7 valence electrons, and there are 6 fluorine atoms in \(XeF_6\), contributing \(6 \times 7 = 42\) valence electrons.  Total valence electrons = \(8 + 42 = 50\) electrons.

In \(XeF_6\), xenon forms 6 bonds with fluorine atoms. Each bond uses 2 electrons, so \(6 \times 2 = 12\) electrons are used in bonding. The remaining electrons (\(50 - 12 = 38\) electrons or 19 pairs) must be placed around the xenon and fluorine atoms as lone pairs. Each fluorine atom needs 3 lone pairs to complete its octet (\(6 \times 3 = 18\) pairs or 36 electrons). After satisfying the octet of all fluorine atoms, 2 electrons (or 1 lone pair) are left on the xenon atom.

Steric Number and Hybridization

Steric number: The steric number is calculated as the sum of the number of atoms bonded to the central atom (6 fluorine atoms) plus the number of lone pairs on the central atom (1 lone pair).

Steric number of \(Xe\) in \(XeF_6 = 6 + 1 = 7\).

Hybridization: A steric number of 7 corresponds to the hybridization \(sp^3d^3\).

For a steric number of 7, the ideal geometry is a pentagonal bipyramidal structure. Due to the presence of one lone pair, the molecule adopts a distorted octahedral geometry. The lone pair tends to occupy a position that causes the least repulsion, leading to distortion from the ideal structure.



The correct geometry and hybridization of \(XeF_6\) are:

Geometry: Distorted octahedral

Hybridization: \(sp^3d^3\)
Thus, the correct answer is option 4: Distorted octahedral, \(sp^3d^3\).