The correct geometry and hybridization of XeF_6 are:

The correct answer is Option (4) → Distorted octahedral, \(sp^3d^3\)

The shape follows from the number of electron pairs round the central atom. Xe has 8 valence electrons (5s²5p⁶); in \( \ce{XeF6} \) six of them form bonds with six F atoms and two remain as a lone pair:

    6 bond pairs + 1 lone pair = 7 electron pairs → sp³d³ hybridisation

(three 5p electrons are promoted to 5d orbitals, giving six unpaired electrons for the six Xe–F bonds). The lone pair pushes the six bonds out of a regular arrangement, so \( \ce{XeF6} \) has a distorted octahedral structure, as found in the gas phase.

Why the other options are incorrect

• Option (1) Distorted octahedral, \(sp^3d^2\) — Incorrect. sp³d² means six electron pairs, as in \( \ce{XeF4} \) (4 bond pairs + 2 lone pairs, square planar); \( \ce{XeF6} \) has seven.

• Option (2) Trigonal bipyramidal, \(sp^3d\) — Incorrect. sp³d means five electron pairs, as in \( \ce{XeF2} \) (2 bond pairs + 3 lone pairs, linear).

• Option (3) Octahedral, \(sp^3d^2\) — Incorrect. A regular octahedron fits six bond pairs and no lone pair (\( \ce{SF6} \)); it ignores the lone pair on Xe.

Note: This question is from the pre-rationalised NCERT book (2022-23). The topic is not there in the latest NCERT book (2026-27), and it is also out of the CUET 2026 syllabus.

NCERT (Pre-rationalised edition, 2022-23):
Xenon-Fluorine Compounds – “… XeF₆ has seven electron pairs (6 bonding pairs and one lone pair) and would, thus, have a distorted octahedral structure as found experimentally in the gas phase.”