Match List - I with List - II.List - IList - II(A) XeO3(I) sp3d2(B) X…

The correct answer is Option (1) → (A)-(IV), (B)-(II), (C)-(I), (D)-(III)

The hybridisation of xenon is found from the number of electron pairs around Xe: the σ bonds (one per attached atom; a double bond Xe=O counts once) plus the lone pairs. Xenon has 8 valence electrons; each Xe–F bond uses 1 of them and each Xe=O bond uses 2.

    4 pairs → sp³; 5 pairs → sp³d; 6 pairs → sp³d²; 7 pairs → sp³d³

• (A) \( \ce{XeO3} \) — This matches (IV) sp³. Three Xe=O bonds use 6 electrons; 2 are left = 1 lone pair. 3 σ + 1 lone pair = 4 pairs: sp³, pyramidal.

• (B) \( \ce{XeF6} \) — This matches (II) sp³d³. Six Xe–F bonds use 6 electrons; 2 are left = 1 lone pair. 6 + 1 = 7 pairs: sp³d³, distorted octahedral.

• (C) \( \ce{XeOF4} \) — This matches (I) sp³d². Four Xe–F bonds and one Xe=O use 6 electrons; 1 lone pair is left. 5 + 1 = 6 pairs: sp³d², square pyramidal.

• (D) \( \ce{XeOF2} \) — This matches (III) sp³d. Two Xe–F bonds and one Xe=O use 4 electrons; 4 are left = 2 lone pairs. 3 + 2 = 5 pairs: sp³d, T-shaped.

Shortcut: The pair (A)–(IV) appears only in Option (1).

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.