The correct shape and hybridization of XeF_2 are:

The correct answer is Option (1) → linear; \(sp^3d\)

In \( \ce{XeF2} \) the Xe atom has 8 valence electrons; 2 are used in the two Xe–F bonds and 6 remain as three lone pairs. So there are 5 electron pairs (2 bond pairs + 3 lone pairs) around Xe, and the hybridisation is sp³d (trigonal bipyramidal arrangement). The three lone pairs occupy the equatorial positions to keep repulsion lowest, so the two F atoms lie on the axis and the molecule is linear (F–Xe–F = 180°).

Why the other options are incorrect

• Option (2) linear; sp³d² — Incorrect. sp³d² needs six electron pairs, as in \( \ce{XeF4} \); \( \ce{XeF2} \) has only five.

• Option (3) trigonal bipyramidal; sp³d — Incorrect. Trigonal bipyramidal is the arrangement of the electron pairs, but the shape of the molecule is decided by the atoms only, and that is linear.

• Option (4) trigonal bipyramidal; sp³d² — Incorrect. Both the shape and the hybridisation are wrong.

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 – “… The structures of the three xenon fluorides can be deduced from VSEPR … XeF₂ and XeF₄ have linear and square planar structures respectively.”