Match List - I with List - II and select CORRECT code for your answer :- List - I (Properties) List - II (Decreasing order) (P) Ionisation energy (1) Ne Na + (Q) Radii (2) S O (R) Electronegativit

Question

Match List - I with List - II and select CORRECT code for your answer :-

List - I (Properties)	List - II (Decreasing order)
(P) Ionisation energy	(1) $Ne > Na^{+}$
(Q) Radii	(2) $S > O$
(R) Electronegativity	(3) $Hf > Zr$
(S) Electron gain enthalpy (Magnitude only)	(4) $[He] 2 s^{2} 2 p^{4} > [He] 2 s^{2} 2 p^{3}$

Code :

PQRS

TestHub · Accepted Answer

P → 3: IE(Hf) > IE(Zr), mainly because lanthanide contraction raises Hf's effective nuclear charge. Q → 1: Ne and Na⁺ are isoelectronic; Na⁺ has a greater nuclear charge, so 𝑟 ( Ne ) > 𝑟 ( Na + ) . R → 4: Electronegativity of O ([He]2s² 2p⁴) is greater than N ([He]2s² 2p³). S → 2: The magnitude of electron-gain enthalpy is greater for S than O. Therefore, P, Q, R, S = 3, 1, 4, 2; option D.

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