Structure of XeO_2F_2 as per VSEPR theory is:

The correct answer is Option (3) → See-Saw

By VSEPR theory the shape follows from the number of electron pairs (bond pairs and lone pairs) around the central atom; the pairs arrange themselves as far apart as possible, and the shape is named from the positions of the atoms only.

In \( \ce{XeO2F2} \), xenon has 8 valence electrons. Two are shared with the two F atoms and four go into the two Xe=O double bonds, leaving one lone pair:

    \( \displaystyle \text{lone pairs} = \frac{8 - 2(1) - 2(2)}{2} = 1 \)

So Xe has 4 bonded atoms + 1 lone pair = 5 electron domains, arranged as a trigonal bipyramid (sp³d). The lone pair takes an equatorial position, where it has fewer 90° repulsions; the two O atoms are equatorial and the two F atoms axial. Four atoms around Xe with one equatorial lone pair give the see-saw shape.

Why the other options are incorrect

• Option (1) Square planar — Incorrect. Needs 4 bond pairs and 2 lone pairs (six domains), as in \( \ce{XeF4} \).

• Option (2) Tetrahedral — Incorrect. Needs 4 bond pairs and no lone pair, as in \( \ce{XeO4} \).

• Option (4) Square pyramidal — Incorrect. Needs 5 bonded atoms and 1 lone pair, as in \( \ce{XeOF4} \).

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.