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Match each periodic property in List-I with its correct order in List-II. List-I (Periodic property): (P) Ionisation energy, (Q) Atomic/ionic radii, (R) Electronegativity, (S) Electron affinity List-II (Order): (1) Sn < Pb, (2) N < C < O, (3) Li < Be < B, (4) O⁻ < S⁻ Select the correct matching code:

1. P-4, Q-1, R-3, S-2
2. P-1, Q-2, R-3, S-4
3. P-4, Q-3, R-2, S-1
4. P-1, Q-4, R-2, S-3

Correct answer: P-4, Q-1, R-3, S-2

Solution

P(IE) matches (4): O⁻ < S⁻ because the 2p orbital of O⁻ has higher electron density causing more repulsion, lowering IE compared to S⁻. Q(Radii) matches (1): Sn < Pb (atomic radius increases down group 14). R(EN) matches (3): Li < Be < B (EN increases across period 2). S(EA) matches (2): N < C < O (N has a stable half-filled 2p³ sub-shell so EA is anomalously low; C has EA lower than O).

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