JEE MainChemistry

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Changes in Coordination Geometry, Isomerism, and Magnetism from 3d to 5d Metal Centers

Q: Why is [PtCl_2Br_2]^(2-) square planar even though halide ligands are weak field ligands?

For 4d and 5d series transition metals (like Pt^(2+)), the crystal field splitting parameter Δ is very large (about 40-50% higher than 3d). Consequently, all ligands behave as strong field ligands and cause pairing, giving dsp^2 square planar geometry.

If Ni^(2+) is replaced by Pt^(2+) in the complex [NiCl_2 Br_2]^(2-), which of the following properties are expected to get changed? A. Geometry B. Geometrical isomerism C. Optical isomerism D. Magnetic properties

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AA, B and CBA, B and DCA and DDB and C

Step-by-step solutionView

Correct answer

Replacing 3d ${N i}^{2 +}$ with 5d ${P t}^{2 +}$ changes the complex from tetrahedral (paramagnetic, no geometrical isomerism) to square planar (diamagnetic, shows cis/trans geometrical isomerism), altering properties A, B, and D.

Option analysis

Why each option works or fails

A · A, B and C

Believing that square planar complexes of the form $[{M A}_{2} B_{2}]$ exhibit optical isomerism, or overlooking the change in magnetic behavior. Square planar complexes of the type $[{M A}_{2} B_{2}]$ have an in-plane mirror symmetry and are achiral; thus optical isomerism remains absent in both complexes.

B · A, B and D

This correctly identifies that geometry changes from tetrahedral to square planar, geometrical isomerism changes from none to cis/trans active, and magnetic properties change from paramagnetic ( $2$ unpaired electrons) to diamagnetic ( $0$ unpaired electrons). Identify that 5d metals have high crystal field splitting, forcing low-spin $d s p^{2}$ square planar geometry.

C · A and D

Overlooking that square planar ${[{P t C l}_{2} {B r}_{2}]}^{2 -}$ can exist as cis and trans geometrical isomers, whereas tetrahedral ${[{N i C l}_{2} {B r}_{2}]}^{2 -}$ cannot. Remember that a transition from tetrahedral to square planar introduces geometrical isomerism (cis and trans forms).

D · B and C

Ignoring the structural and magnetic change between 3d weak-field coordination and 5d strong crystal field splitting. Account for both geometry change ( $s p^{3}$ tetrahedral vs $d s p^{2}$ square planar) and spin pairing in 5d metal complexes.

Reviewed route

Solution

StepWorking

01Concept

For ${N i}^{2 +}$ ( $3 d^{8}$ ) with weak field halide ligands ( ${C l}^{-}, {B r}^{-}$ ), crystal field splitting $Δ_{t}$ is small, leading to $s p^{3}$ hybridization, tetrahedral geometry, paramagnetism (2 unpaired electrons), no geometrical isomerism, and no optical isomerism. For ${P t}^{2 +}$ ( $5 d^{8}$ ), $Δ_{o} / Δ_{s p}$ is very large due to high effective nuclear charge and $5 d$ radial expansion, forcing pairing to form $d s p^{2}$ square planar geometry, diamagnetism (0 unpaired electrons), cis/trans geometrical isomerism, and no optical isomerism.

02Option verdict

Neither $[{N i C l}_{2} {B r}_{2}]^{2 -}$ (tetrahedral $M a_{2} b_{2}$ has planes of symmetry) nor $[{P t C l}_{2} {B r}_{2}]^{2 -}$ (square planar $M a_{2} b_{2}$ has molecular plane of symmetry) exhibits optical isomerism, so optical isomerism (C) does not change.

03Option verdict

Geometry changes (tetrahedral to square planar: A), geometrical isomerism changes (0 isomers to 2 isomers [cis/trans]: B), and magnetic properties change (paramagnetic $μ \approx 2.83 BM$ to diamagnetic $μ = 0$ : D). Hence A, B, and D change.

04Option verdict

Overlooks that geometrical isomerism (B) also changes because square planar $M a_{2} b_{2}$ displays cis-trans isomerism whereas tetrahedral $M a_{2} b_{2}$ cannot.

05Option verdict

Omits geometry (A) and magnetic properties (D) which clearly change between $s p^{3}$ (paramagnetic) and $d s p^{2}$ (diamagnetic), and incorrectly includes optical isomerism (C).

✓Discriminator

Identify that 5d transition metal complexes with coordination number 4 are almost always square planar and low-spin, shifting geometry ( $T_{d} \to D_{4 h}$ ), spin state (paramagnetic $\to$ diamagnetic), and enabling cis/trans GI, while neither geometry allows optical isomerism.

Hints that build this answer step by step

What are the geometries of ${[{N i C l}_{2} {B r}_{2}]}^{2 -}$ and ${[{P t C l}_{2} {B r}_{2}]}^{2 -}$ , respectively?
${[{N i C l}_{2} {B r}_{2}]}^{2 -}$ is tetrahedral ( $s p^{3}$ ) and ${[{P t C l}_{2} {B r}_{2}]}^{2 -}$ is square planar ( $d s p^{2}$ ).
How do the isomerism and magnetic properties differ between the tetrahedral ${[{N i C l}_{2} {B r}_{2}]}^{2 -}$ and square planar ${[{P t C l}_{2} {B r}_{2}]}^{2 -}$ ?
Tetrahedral is paramagnetic with no geometrical isomerism; square planar is diamagnetic and exhibits cis/trans isomerism (neither shows optical isomerism).

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Chemistry
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 7 September 2026

Quick checks

Students also ask

Why is

[{P t C l}_{2} {B r}_{2}]^{2 -}

square planar even though halide ligands are weak field ligands?

For $4 d$ and $5 d$ series transition metals (like ${P t}^{2 +}$ ), the crystal field splitting parameter $Δ$ is very large (about 40-50% higher than $3 d$ ). Consequently, all ligands behave as strong field ligands and cause pairing, giving $d s p^{2}$ square planar geometry.

Answer

Replacing 3d ${N i}^{2 +}$ with 5d ${P t}^{2 +}$ changes the complex from tetrahedral (paramagnetic, no geometrical isomerism) to square planar (diamagnetic, shows cis/trans geometrical isomerism), altering properties A, B, and D.

Why each option works or fails

A: A, B and C - Believing that square planar complexes of the form $[{M A}_{2} B_{2}]$ exhibit optical isomerism, or overlooking the change in magnetic behavior. Square planar complexes of the type $[{M A}_{2} B_{2}]$ have an in-plane mirror symmetry and are achiral; thus optical isomerism remains absent in both complexes.
B · correct: A, B and D - This correctly identifies that geometry changes from tetrahedral to square planar, geometrical isomerism changes from none to cis/trans active, and magnetic properties change from paramagnetic ( $2$ unpaired electrons) to diamagnetic ( $0$ unpaired electrons). Identify that 5d metals have high crystal field splitting, forcing low-spin $d s p^{2}$ square planar geometry.
C: A and D - Overlooking that square planar ${[{P t C l}_{2} {B r}_{2}]}^{2 -}$ can exist as cis and trans geometrical isomers, whereas tetrahedral ${[{N i C l}_{2} {B r}_{2}]}^{2 -}$ cannot. Remember that a transition from tetrahedral to square planar introduces geometrical isomerism (cis and trans forms).
D: B and C - Ignoring the structural and magnetic change between 3d weak-field coordination and 5d strong crystal field splitting. Account for both geometry change ( $s p^{3}$ tetrahedral vs $d s p^{2}$ square planar) and spin pairing in 5d metal complexes.

Step-by-step solution

Concept: For ${N i}^{2 +}$ ( $3 d^{8}$ ) with weak field halide ligands ( ${C l}^{-}, {B r}^{-}$ ), crystal field splitting $Δ_{t}$ is small, leading to $s p^{3}$ hybridization, tetrahedral geometry, paramagnetism (2 unpaired electrons), no geometrical isomerism, and no optical isomerism. For ${P t}^{2 +}$ ( $5 d^{8}$ ), $Δ_{o} / Δ_{s p}$ is very large due to high effective nuclear charge and $5 d$ radial expansion, forcing pairing to form $d s p^{2}$ square planar geometry, diamagnetism (0 unpaired electrons), cis/trans geometrical isomerism, and no optical isomerism.
Option verdict: Neither $[{N i C l}_{2} {B r}_{2}]^{2 -}$ (tetrahedral $M a_{2} b_{2}$ has planes of symmetry) nor $[{P t C l}_{2} {B r}_{2}]^{2 -}$ (square planar $M a_{2} b_{2}$ has molecular plane of symmetry) exhibits optical isomerism, so optical isomerism (C) does not change.
Option verdict: Geometry changes (tetrahedral to square planar: A), geometrical isomerism changes (0 isomers to 2 isomers [cis/trans]: B), and magnetic properties change (paramagnetic $μ \approx 2.83 BM$ to diamagnetic $μ = 0$ : D). Hence A, B, and D change.
Option verdict: Overlooks that geometrical isomerism (B) also changes because square planar $M a_{2} b_{2}$ displays cis-trans isomerism whereas tetrahedral $M a_{2} b_{2}$ cannot.
Option verdict: Omits geometry (A) and magnetic properties (D) which clearly change between $s p^{3}$ (paramagnetic) and $d s p^{2}$ (diamagnetic), and incorrectly includes optical isomerism (C).
Discriminator: Identify that 5d transition metal complexes with coordination number 4 are almost always square planar and low-spin, shifting geometry ( $T_{d} \to D_{4 h}$ ), spin state (paramagnetic $\to$ diamagnetic), and enabling cis/trans GI, while neither geometry allows optical isomerism.

Changes in Coordination Geometry, Isomerism, and Magnetism from 3d to 5d Metal Centers

What feels right?

Replacing 3d ${N i}^{2 +}$ with 5d ${P t}^{2 +}$ changes the complex from tetrahedral (paramagnetic, no geometrical isomerism) to square planar (diamagnetic, shows cis/trans geometrical isomerism), altering properties A, B, and D.

Why each option works or fails

Solution

01Concept

02Option verdict

03Option verdict

04Option verdict

05Option verdict

✓Discriminator

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