JEE MainChemistry

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Coordination Compounds: JEE Main Chemistry Question with Solution

Q: Why is square planar even though halide ligands are weak field ligands?

For and series transition metals (like ), the crystal field splitting parameter is very large (about 40-50% higher than ). Consequently, all ligands behave as strong field ligands and cause pairing, giving square planar geometry.

\mathrm{Ni}^{2+}

is replaced by

\mathrm{Pt}^{2+}

in the complex

\left[\mathrm{NiCl}_2 \mathrm{Br}_2\right]^{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 $\mathrm{Ni}^{2+}$ with 5d $\mathrm{Pt}^{2+}$ 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 $\left[\mathrm{MA}_2\mathrm{B}_2\right]$ exhibit optical isomerism, or overlooking the change in magnetic behavior. Square planar complexes of the type $\left[\mathrm{MA}_2\mathrm{B}_2\right]$ have an in-plane mirror symmetry and are achiral; thus optical isomerism remains absent in both complexes.

B · A, B and D

None. 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 $dsp^2$ square planar geometry.

C · A and D

Overlooking that square planar $\left[\mathrm{PtCl}_2\mathrm{Br}_2\right]^{2-}$ can exist as cis and trans geometrical isomers, whereas tetrahedral $\left[\mathrm{NiCl}_2\mathrm{Br}_2\right]^{2-}$ 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 ( $sp^3$ tetrahedral vs $dsp^2$ square planar) and spin pairing in 5d metal complexes.

Reviewed route

Solution

StepWorking

01concept

For $\mathrm{Ni}^{2+}$ ( $3d^8$ ) with weak field halide ligands ( $\mathrm{Cl}^-, \mathrm{Br}^-$ ), crystal field splitting $\Delta_t$ is small, leading to $sp^3$ hybridization, tetrahedral geometry, paramagnetism (2 unpaired electrons), no geometrical isomerism, and no optical isomerism. For $\mathrm{Pt}^{2+}$ ( $5d^8$ ), $\Delta_o / \Delta_{sp}$ is very large due to high effective nuclear charge and $5d$ radial expansion, forcing pairing to form $dsp^2$ square planar geometry, diamagnetism (0 unpaired electrons), cis/trans geometrical isomerism, and no optical isomerism.

02option_verdict

Neither $[\mathrm{NiCl}_2\mathrm{Br}_2]^{2-}$ (tetrahedral $\mathrm{M}a_2b_2$ has planes of symmetry) nor $[\mathrm{PtCl}_2\mathrm{Br}_2]^{2-}$ (square planar $\mathrm{M}a_2b_2$ 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 $\mu \approx 2.83\text{ BM}$ to diamagnetic $\mu = 0$ : D). Hence A, B, and D change.

04option_verdict

Overlooks that geometrical isomerism (B) also changes because square planar $\mathrm{M}a_2b_2$ displays cis-trans isomerism whereas tetrahedral $\mathrm{M}a_2b_2$ cannot.

05option_verdict

Omits geometry (A) and magnetic properties (D) which clearly change between $sp^3$ (paramagnetic) and $dsp^2$ (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 ( $T_d \to D_{4h}$ ), 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 $\left[\mathrm{NiCl}_2\mathrm{Br}_2\right]^{2-}$ and $\left[\mathrm{PtCl}_2\mathrm{Br}_2\right]^{2-}$ , respectively?
$\left[\mathrm{NiCl}_2\mathrm{Br}_2\right]^{2-}$ is tetrahedral ( $sp^3$ ) and $\left[\mathrm{PtCl}_2\mathrm{Br}_2\right]^{2-}$ is square planar ( $dsp^2$ ).
How do the isomerism and magnetic properties differ between the tetrahedral $\left[\mathrm{NiCl}_2\mathrm{Br}_2\right]^{2-}$ and square planar $\left[\mathrm{PtCl}_2\mathrm{Br}_2\right]^{2-}$ ?
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
Concepts assessed: Chemistry
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 7 September 2026

Quick checks

Students also ask

Why is

[\mathrm{PtCl}_2\mathrm{Br}_2]^{2-}

square planar even though halide ligands are weak field ligands?

For $4d$ and $5d$ series transition metals (like $\mathrm{Pt}^{2+}$ ), the crystal field splitting parameter $\Delta$ 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.

Answer

Replacing 3d $\mathrm{Ni}^{2+}$ with 5d $\mathrm{Pt}^{2+}$ 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 $\left[\mathrm{MA}_2\mathrm{B}_2\right]$ exhibit optical isomerism, or overlooking the change in magnetic behavior. Square planar complexes of the type $\left[\mathrm{MA}_2\mathrm{B}_2\right]$ have an in-plane mirror symmetry and are achiral; thus optical isomerism remains absent in both complexes.
B · correct: A, B and D - None. 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 $dsp^2$ square planar geometry.
C: A and D - Overlooking that square planar $\left[\mathrm{PtCl}_2\mathrm{Br}_2\right]^{2-}$ can exist as cis and trans geometrical isomers, whereas tetrahedral $\left[\mathrm{NiCl}_2\mathrm{Br}_2\right]^{2-}$ 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 ( $sp^3$ tetrahedral vs $dsp^2$ square planar) and spin pairing in 5d metal complexes.

Step-by-step solution

concept: For $\mathrm{Ni}^{2+}$ ( $3d^8$ ) with weak field halide ligands ( $\mathrm{Cl}^-, \mathrm{Br}^-$ ), crystal field splitting $\Delta_t$ is small, leading to $sp^3$ hybridization, tetrahedral geometry, paramagnetism (2 unpaired electrons), no geometrical isomerism, and no optical isomerism. For $\mathrm{Pt}^{2+}$ ( $5d^8$ ), $\Delta_o / \Delta_{sp}$ is very large due to high effective nuclear charge and $5d$ radial expansion, forcing pairing to form $dsp^2$ square planar geometry, diamagnetism (0 unpaired electrons), cis/trans geometrical isomerism, and no optical isomerism.
option_verdict: Neither $[\mathrm{NiCl}_2\mathrm{Br}_2]^{2-}$ (tetrahedral $\mathrm{M}a_2b_2$ has planes of symmetry) nor $[\mathrm{PtCl}_2\mathrm{Br}_2]^{2-}$ (square planar $\mathrm{M}a_2b_2$ 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 $\mu \approx 2.83\text{ BM}$ to diamagnetic $\mu = 0$ : D). Hence A, B, and D change.
option_verdict: Overlooks that geometrical isomerism (B) also changes because square planar $\mathrm{M}a_2b_2$ displays cis-trans isomerism whereas tetrahedral $\mathrm{M}a_2b_2$ cannot.
option_verdict: Omits geometry (A) and magnetic properties (D) which clearly change between $sp^3$ (paramagnetic) and $dsp^2$ (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 ( $T_d \to D_{4h}$ ), spin state (paramagnetic $\to$ diamagnetic), and enabling cis/trans GI, while neither geometry allows optical isomerism.

Coordination Compounds: JEE Main Chemistry Question with Solution

What feels right?

Replacing 3d $\mathrm{Ni}^{2+}$ with 5d $\mathrm{Pt}^{2+}$ 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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