JEE Advanced 2022 — Paper 1 — CHEM-III solution

JEE Advanced2022Paper 1CHEM-III

LIST-I contains metal species and LIST-II contains their properties.

LIST-I	LIST-II
(I) [Cr(CN) $_{6}$ ] $^{4 -}$	(P) $t_{2 g}$ orbitals contain 4 electrons
(II) [RuCl $_{6}$ ] $^{2 -}$	(Q) $μ$ (spin-only) = 4.9 BM
(III) [Cr(H $_{2}$ O) $_{6}$ ] $^{2 +}$	(R) low spin complex ion
(IV) [Fe(H $_{2}$ O) $_{6}$ ] $^{2 +}$	(S) metal ion in 4+ oxidation state
	(T) $d^{4}$ species

[Given: Atomic number of Cr = 24, Ru = 44, Fe = 26]

Match each metal species in LIST-I with their properties in LIST-II, and choose the correct option

A
I $\to$ R, T; II $\to$ P, S; III $\to$ Q, T; IV $\to$ P, Q
B
I $\to$ R, S; II $\to$ P, T; III $\to$ P, Q; IV $\to$ Q, T
C
I $\to$ P, R; II $\to$ R, S; III $\to$ R, T; IV $\to$ P, T
D
I $\to$ Q, T; II $\to$ S, T; III $\to$ P, T; IV $\to$ Q, R

Solution

(I) [Cr(CN) $_{6}$ ] $^{4 -}$ : Cr is in +2 oxidation state, so Cr $^{2 +}$ is $d^{4}$ . CN $^{-}$ is a strong-field ligand, so pairing occurs and the configuration is low-spin ( $t_{2 g}^{4} e_{g}^{0}$ ). Properties: low spin complex (R) and $d^{4}$ species (T).

(II) [RuCl $_{6}$ ] $^{2 -}$ : Charge balance gives Ru in +4 oxidation state. Ru $^{4 +}$ is $4 d^{4}$ . For $4 d$ metals, even Cl $^{-}$ enforces low-spin behaviour, so $t_{2 g}^{4}$ . Properties: $t_{2 g}$ contains 4 electrons (P) and metal in 4+ oxidation state (S).

(III) [Cr(H $_{2}$ O) $_{6}$ ] $^{2 +}$ : Cr $^{2 +}$ is $d^{4}$ . H $_{2}$ O is a weak-field ligand, so high-spin with 4 unpaired electrons. $μ = 4 (4 + 2) = 24 \approx 4.9$ BM. Properties: $μ$ = 4.9 BM (Q) and $d^{4}$ species (T).

(IV) [Fe(H $_{2}$ O) $_{6}$ ] $^{2 +}$ : Fe $^{2 +}$ is $d^{6}$ . H $_{2}$ O is weak field, so high-spin $t_{2 g}^{4} e_{g}^{2}$ with 4 unpaired electrons. $μ \approx 4.9$ BM, and $t_{2 g}$ contains 4 electrons. Properties: $t_{2 g}$ contains 4 electrons (P) and $μ$ = 4.9 BM (Q).

The correct matching is option (A).

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