Match the species in List I with the number of bond pairs and lone pairs on the central atom in List II. List I: (P) ClF3 (Q) XeF4 (R) XeF6 (S) ICl2-. List II: (1) 2 bond pairs, 3 lone pairs (2) 3 bond pairs, 2 lone pairs (3) 4 bond pairs, 2 lone pairs (4) 6 bond pairs, 1 lone pair.

JEE Advanced levelSingle correct optionDifficulty 3 of 3ph200

Match the species in List I with the number of bond pairs and lone pairs on the central atom in List II. List I: (P) $ClF X_{3}$ (Q) $XeF X_{4}$ (R) $XeF X_{6}$ (S) $ICl X_{2} X^{-}$ . List II: (1) 2 bond pairs, 3 lone pairs (2) 3 bond pairs, 2 lone pairs (3) 4 bond pairs, 2 lone pairs (4) 6 bond pairs, 1 lone pair.

Central-atom valence electrons (with charge) minus those used gives the lone pairs.

ClF X_{3}

7 - 3 = 4

electrons, 2 lone pairs.

XeF X_{4}

8 - 4 = 4

, 2 lone pairs.

XeF X_{6}

8 - 6 = 2

, 1 lone pair.

ICl X_{2} X^{-}

7 + 1 - 2 = 6

, 3 lone pairs. So P-2, Q-3, R-4, S-1.