JEE Advanced 2026 — Paper 2 — CHEM-II solution

JEE Advanced2026Paper 2CHEM-II

Correct statement(s) about the compounds $P$ , $Q$ and $R$ is(are)

A
(A) $P$ has two lone pairs of electrons on the central atom.
B
(B) $Q$ has a perfect octahedral geometry.
C
(C) $Q$ can act as a fluorinating agent.
D
(D) The molecular structure of $R$ is trigonal pyramidal.

Solution

Identify P, Q, R.

$P$ : Xe with excess F $_{2}$ (1:5 ratio) at 873 K and 7 bar gives XeF $_{4}$ .

$Q$ : XeF $_{4}$ + O $_{2}$ F $_{2}$ at 143 K is a known fluorination that pushes XeF $_{4}$ to XeF $_{6}$ , releasing O $_{2}$ . So Q = XeF $_{6}$ .

$R$ : Complete hydrolysis of XeF $_{6}$ gives XeO $_{3}$ and HF. So R = XeO $_{3}$ .

(A) P = XeF $_{4}$ . Total electron pairs around Xe: 4 bond pairs + 2 lone pairs (square planar geometry, AX $_{4}$ E $_{2}$ ). Two lone pairs on central atom. Correct.

(B) Q = XeF $_{6}$ . Total electron pairs: 6 bond pairs + 1 lone pair (AX $_{6}$ E). The single lone pair distorts the octahedron, so the geometry is distorted octahedral (often described as monocapped octahedral or pentagonal bipyramidal with vacancy), not perfect octahedral. Incorrect.

(C) Q = XeF $_{6}$ . It is a strong fluorinating agent, donating F to many substrates including silica. Correct.

(D) R = XeO $_{3}$ . Xe with three Xe=O bonds and one lone pair: AX $_{3}$ E gives trigonal pyramidal shape (analogous to NH $_{3}$ ). Correct.

$∴$ Correct options are (A), (C), (D).

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