JEE Advanced 2023 — Paper 1 — CHEM-IV solution

JEE Advanced2023Paper 1CHEM-IV

Match the electronic configurations in List-I with appropriate metal complex ions in List-II and choose the correct option.

[Atomic Number: Fe = 26, Mn = 25, Co = 27]

List-I	List-II
(P) $t_{2 g}^{6} e_{g}^{0}$	(1) [Fe(H $_{2}$ O) $_{6}$ ] $^{2 +}$
(Q) $t_{2 g}^{3} e_{g}^{2}$	(2) [Mn(H $_{2}$ O) $_{6}$ ] $^{2 +}$
(R) $e^{2} t_{2}^{3}$	(3) [Co(NH $_{3}$ ) $_{6}$ ] $^{3 +}$
(S) $t_{2 g}^{4} e_{g}^{2}$	(4) [FeCl $_{4}$ ] $^{-}$
	(5) [CoCl $_{4}$ ] $^{2 -}$

A
P $\to$ 1; Q $\to$ 4; R $\to$ 2; S $\to$ 3
B
P $\to$ 1; Q $\to$ 2; R $\to$ 4; S $\to$ 5
C
P $\to$ 3; Q $\to$ 2; R $\to$ 5; S $\to$ 1
D
P $\to$ 3; Q $\to$ 2; R $\to$ 4; S $\to$ 1

Solution

Determine the $d$ -electron count and field-splitting for each octahedral or tetrahedral complex.

[Fe(H $_{2}$ O) $_{6}$ ] $^{2 +}$ . Fe $^{2 +}$ is $d^{6}$ ; H $_{2}$ O is a weak-field ligand, so the high-spin octahedral configuration is $t_{2 g}^{4} e_{g}^{2}$ (matches S).

[Mn(H $_{2}$ O) $_{6}$ ] $^{2 +}$ . Mn $^{2 +}$ is $d^{5}$ ; with H $_{2}$ O (weak field) it is high spin: $t_{2 g}^{3} e_{g}^{2}$ (matches Q).

[Co(NH $_{3}$ ) $_{6}$ ] $^{3 +}$ . Co $^{3 +}$ is $d^{6}$ ; NH $_{3}$ is a strong-field ligand, so low spin in octahedral geometry gives $t_{2 g}^{6} e_{g}^{0}$ (matches P).

[FeCl $_{4}$ ] $^{-}$ . Fe $^{3 +}$ is $d^{5}$ in a tetrahedral field. The tetrahedral splitting puts the $e$ set below $t_{2}$ ; weak-field high spin gives $e^{2} t_{2}^{3}$ (matches R).

[CoCl $_{4}$ ] $^{2 -}$ . Co $^{2 +}$ is $d^{7}$ in tetrahedral field, configuration $e^{4} t_{2}^{3}$ . This does not appear in List-I.

Mapping: P $\to$ 3, Q $\to$ 2, R $\to$ 4, S $\to$ 1. Answer: (D).

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