JEE Advanced 2026 · Paper 2 · Chemistry Q6

ChemistryMultiple correct+4 / -1Question 42 of 54

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

$\ce{Xe(g) + F2(g)}$ (1 : 5 ratio) $\xrightarrow{873\ \mathrm K,\ 7\ \mathrm{bar}}$ P

P + $\ce{O2F2}$ $\xrightarrow{143\ \mathrm K}$ Q + $\ce{O2}$

Q + $\ce{H2O}$ $\xrightarrow{\text{complete hydrolysis}}$ R + HF

P has two lone pairs of electrons on the central atom.

Q has a perfect octahedral geometry.

Q can act as a fluorinating agent.

The molecular structure of R is trigonal pyramidal.

Answer and worked solution

Official answer: A, C, D

Checking the statements:

(A) $\ce{XeF4}$ is square planar with two lone pairs on Xe. True.
(B) $\ce{XeF6}$ has seven electron pairs (one lone pair), so it is a distorted octahedron. False.
(C) $\ce{XeF6}$ is a strong fluorinating agent. True.
(D) $\ce{XeO3}$ has three bond pairs and one lone pair: trigonal pyramidal. True.

JEE Advanced 2026 · Paper 2, Chemistry question 6