The shape of \(XeF_6\) is :
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → Distorted octahedral
By VSEPR theory the shape depends on the number of bond pairs and lone pairs on the central atom. Xe has 8 valence electrons. In \( \ce{XeF6} \) six of them form six Xe–F bonds and the remaining two stay as one lone pair:
6 bond pairs + 1 lone pair = 7 electron pairs
Seven electron pairs cannot sit at the six corners of a regular octahedron; the lone pair distorts it.
Why the other options are incorrect
Option (1) Regular octahedral — Incorrect. Needs 6 bond pairs and no lone pair, as in \( \ce{SF6} \); \( \ce{XeF6} \) has one lone pair.
Option (2) Square planar — Incorrect. The shape of \( \ce{XeF4} \), with 4 bond pairs and 2 lone pairs.
Option (4) Square pyramidal — Incorrect. The shape of \( \ce{XeOF4} \), with 5 bonding positions and 1 lone pair.
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 – …
