Match List I with List II
| List I | List II |
| (A) \(XeF_6\) | (I) Disproportionation reaction |
| (B) Liquid Helium | (II) Thermodynamically unstable |
| (C) \(HNO_2\) | (III) MRI |
| (D) \(O_3\) | (IV) Distorted octahedral |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A)-(IV), (B)-(III), (C)-(I), (D)-(II)
(A) \( \ce{XeF6} \) — This matches (IV) Distorted octahedral. Xe has six bond pairs and one lone pair (sp3d3); the lone pair distorts the octahedron.
(B) Liquid Helium — This matches (III) MRI. Liquid helium (b.p. 4.2 K) is the cryogenic agent that keeps the superconducting magnets of NMR spectrometers and Magnetic Resonance Imaging systems cold.
(C) \( \ce{HNO2} \) — This matches (I) Disproportionation reaction. Nitrogen in \( \ce{HNO2} \) is +3, between its higher and lower states, so it is both oxidised (to +5) and reduced (to +2).
\( \ce{3HNO2 -> HNO3 + H2O + 2NO} \)
(D) \( \ce{O3} \) — This matches (II) Thermodynamically unstable. The decomposition of ozone releases heat (ΔH negative) and increases entropy (ΔS positive), so ΔG is negative and ozone is unstable with respect to oxygen.
\( \ce{2O3(g) -> 3O2(g)} \), ΔH negative
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.