JEE Chemistry 2023 Question with Solution

Given: The complex $[\text{NiCl}_2\text{Br}_2]^{2-}$ is compared with $[\text{PtCl}_2\text{Br}_2]^{2-}$ after replacing $\text{Ni}^{2+}$ by $\text{Pt}^{2+}$.

Find: Which among A, B, C, D change.

For $[\text{NiCl}_2\text{Br}_2]^{2-}$, the metal ion is $\text{Ni}^{2+}$ with $d^8$ configuration. With weak field ligands $\text{Cl}^-$ and $\text{Br}^-$, the complex is tetrahedral.

A tetrahedral complex of type $\text{MA}_2\text{B}_2$ does not show geometrical isomerism and is optically inactive.

For $[\text{PtCl}_2\text{Br}_2]^{2-}$, the metal ion is $\text{Pt}^{2+}$ with $d^8$ configuration. As a $5d$ element, $\text{Pt}^{2+}$ forms a square planar complex.

A square planar complex of type $\text{MA}_2\text{B}_2$ can show cis and trans forms, so geometrical isomerism appears. Optical isomerism is still not observed here because of symmetry.

The change from tetrahedral geometry to square planar geometry changes the crystal field splitting pattern, so the magnetic properties also change.

Therefore, Geometry (A), Geometrical isomerism (B), and Magnetic properties (D) change. The correct option is D.

Given: Replace $\text{Ni}^{2+}$ by $\text{Pt}^{2+}$ in $[\text{NiCl}_2\text{Br}_2]^{2-}$.

Find: Which listed properties are altered.

1. Geometry:

• $[\text{NiCl}_2\text{Br}_2]^{2-}$ is tetrahedral.
• $[\text{PtCl}_2\text{Br}_2]^{2-}$ is square planar. So, A changes.

2. Geometrical isomerism:

• Tetrahedral $\text{MA}_2\text{B}_2$ complexes do not show geometrical isomerism.
• Square planar $\text{MA}_2\text{B}_2$ complexes show cis-trans isomerism. So, B changes.

3. Optical isomerism:

• The tetrahedral complex here is optically inactive.
• The square planar complex here is also optically inactive. So, C does not change.

4. Magnetic properties:

• The geometry change modifies crystal field splitting.
• Hence the magnetic behaviour changes. So, D changes.

Thus the changing properties are A, B and D, so the correct option is D.