JEE Advanced 2023 — Paper 2 — CHEM-II solution

JEE Advanced2023Paper 2CHEM-II

The complex(es), which can exhibit the type of isomerism shown by $[P t (N H_{3})_{2} B r_{2}]$ , is(are)

[en = $H_{2} N C H_{2} C H_{2} N H_{2}$ ]

A
$[P t (e n) (S C N)_{2}]$
B
$[Z n (N H_{3})_{2} C l_{2}]$
C
$[P t (N H_{3})_{2} C l_{4}]$
D
$[C r (e n)_{2} (H_{2} O) (S O_{4})]^{+}$

Solution

$[P t (N H_{3})_{2} B r_{2}]$ is a $d^{8}$ square-planar $P t (I I)$ complex of $M A_{2} B_{2}$ type, which shows cis-trans (geometrical) isomerism. We need other complexes that also display geometrical isomerism.

(A) $[P t (e n) (S C N)_{2}]$ : Square planar, but en is a bidentate ligand that must occupy two cis sites. The two remaining SCN ligands are forced to be cis as well, so no cis-trans isomerism. Incorrect.

(B) $[Z n (N H_{3})_{2} C l_{2}]$ : $Z n (I I)$ is $d^{10}$ and adopts a tetrahedral geometry. Tetrahedral $M A_{2} B_{2}$ complexes do not show geometrical isomerism. Incorrect.

(C) $[P t (N H_{3})_{2} C l_{4}]$ : $P t (I V)$ , $d^{6}$ , octahedral $M A_{2} B_{4}$ . The two $N H_{3}$ ligands can be cis or trans, giving geometrical isomers. Correct.

(D) $[C r (e n)_{2} (H_{2} O) (S O_{4})]^{+}$ : Octahedral with two en ligands plus $H_{2} O$ and $S O_{4}^{2 -}$ as the two monodentate ligands. The unidentate pair can be cis or trans (with the two en occupying the remaining four sites), so geometrical isomerism is possible. Correct.

Correct options: (C) and (D).

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