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JEE MainChemistry

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Electronic Configuration and Geometry of 4-Coordinate Nickel Complexes

Given below are two statements: Statement I:[Ni(CN) 4]^(2-) is square planar and diamagnetic complex. with dsp^2 hybridization for Ni but [Ni(CO)_4] is tetrahedral. paramagnetic and with sp^3 -hybridication for Ni. Statement II: [NiCl_4]^(2-) and [Ni(CO)_4] both have same d-electron configuration have same geometry and are paramagnetic. In light the above statements. choose the correct answer form the options given below:

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ABoth Statement I and Statement II are true.BBoth Statement I and Statement II are false.CStatement I is correct but statement II is false.DStatement I is incorrect but statement II is true.

Step-by-step solutionView

Correct answer

Both Statement I and Statement II are false because $[N i (C O)_{4}]$ is diamagnetic (not paramagnetic) and possesses a $3 d^{10}$ configuration, whereas $[{N i C l}_{4}]^{2 -}$ has a $3 d^{8}$ configuration.

Option analysis

Why each option works or fails

A · Both Statement I and Statement II are true.

Believing $[N i (C O)_{4}]$ is paramagnetic and that ${N i}^{0}$ in $[N i (C O)_{4}]$ has the same $d$ -electron count as ${N i}^{2 +}$ in $[{N i C l}_{4}]^{2 -}$ . Determine the oxidation state of nickel first: ${N i}^{0}$ in $[N i (C O)_{4}]$ has $4 s^{2} 3 d^{8}$ , which reorganizes under the strong-field $C O$ ligand into fully paired $3 d^{10}$ , making it diamagnetic.

B · Both Statement I and Statement II are false.

This option correctly identifies that both statements are false. Both statements are false: Statement I incorrectly labels $[N i (C O)_{4}]$ as paramagnetic, and Statement II falsely claims $[{N i C l}_{4}]^{2 -}$ ( $3 d^{8}$ ) and $[N i (C O)_{4}]$ ( $3 d^{10}$ ) have the same $d$ -electron configuration and both are paramagnetic.

C · Statement I is correct but statement II is false.

Overlooking the word 'paramagnetic' assigned to $[N i (C O)_{4}]$ in Statement I, assuming Statement I is completely true. Read every descriptor carefully: $[N i (C O)_{4}]$ has all electrons paired in the $3 d$ subshell ( $3 d^{10}$ ), meaning it is diamagnetic, rendering Statement I false.

D · Statement I is incorrect but statement II is true.

Assuming that because both $[{N i C l}_{4}]^{2 -}$ and $[N i (C O)_{4}]$ are tetrahedral complexes of nickel, they must share the same $d$ -electron configuration and magnetic behavior. Calculate the oxidation state of each metal center: $N i$ in $[{N i C l}_{4}]^{2 -}$ is $+ 2$ ( $3 d^{8}$ , paramagnetic), whereas $N i$ in $[N i (C O)_{4}]$ is $0$ ( $3 d^{10}$ , diamagnetic).

Reviewed route

Solution

StepWorking

01Concept

For 4-coordinate Ni complexes, first determine the oxidation state and d-electron count of Ni. Next, identify the ligand field strength (SFL vs WFL). Then, find the electron pairing, hybridization, geometry, and magnetic character.

02Option verdict

Statement I claims [Ni(CO)4] is paramagnetic. However, Ni is in the 0 oxidation state (3d8 4s2). The strong field CO ligand forces the 4s electrons into 3d. This gives a completely filled 3d10 system with no unpaired electrons (diamagnetic). Hence, Statement I is false. Statement II is also false. [NiCl4]2- is a d8 paramagnetic complex. In contrast, [Ni(CO)4] is a d10 diamagnetic complex.

03Option verdict

Both statements are false. Statement I is false because [Ni(CO)4] is diamagnetic, not paramagnetic. Statement II is also false. [NiCl4]2- has Ni2+ with a 3d8 configuration and is paramagnetic. In contrast, [Ni(CO)4] has Ni0 with a 3d10 configuration and is diamagnetic. Thus, they have different d-electron configurations and different magnetic properties.

04Option verdict

Statement I is incorrect because [Ni(CO)4] is diamagnetic, not paramagnetic.

05Option verdict

Statement II is incorrect because [NiCl4]2- (Ni2+, d8) and [Ni(CO)4] (Ni0, d10) do not have the same d-electron configuration, nor are they both paramagnetic.

✓Discriminator

[Ni(CO)4] has Ni(0) with 3d10 configuration due to strong field CO ligands, meaning it is strictly diamagnetic. This instantly invalidates both Statement I (calls it paramagnetic) and Statement II (calls both paramagnetic with same d-count).

Hints that build this answer step by step

What are the oxidation state, electron configuration, and magnetic property of nickel in $[N i (C O)_{4}]$ ?
$N i (0)$ with a $3 d^{10}$ configuration; all electrons are paired, so it is diamagnetic.
Based on the properties of $[N i (C O)_{4}]$ , is Statement I true or false?
Statement I is false because it claims $[N i (C O)_{4}]$ is paramagnetic.
Comparing $[{N i C l}_{4}]^{2 -}$ and $[N i (C O)_{4}]$ , do they have the same $d$ -electron configuration and magnetic behavior?
No; $[{N i C l}_{4}]^{2 -}$ has a $3 d^{8}$ configuration (paramagnetic) while $[N i (C O)_{4}]$ has a $3 d^{10}$ configuration (diamagnetic).

✓ 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: 22 September 2026

Quick checks

Students also ask

Why does [Ni(CO)4] become d10 instead of staying 3d8 4s2?

CO is a very strong field ligand. The crystal field stabilization gained is higher than the pairing energy. This causes the two 4s electrons to pair up into the 3d orbitals. It results in a fully filled 3d10 configuration and sp3 hybridization.

Answer

Both Statement I and Statement II are false because $[N i (C O)_{4}]$ is diamagnetic (not paramagnetic) and possesses a $3 d^{10}$ configuration, whereas $[{N i C l}_{4}]^{2 -}$ has a $3 d^{8}$ configuration.

Why each option works or fails

A: Both Statement I and Statement II are true. - Believing $[N i (C O)_{4}]$ is paramagnetic and that ${N i}^{0}$ in $[N i (C O)_{4}]$ has the same $d$ -electron count as ${N i}^{2 +}$ in $[{N i C l}_{4}]^{2 -}$ . Determine the oxidation state of nickel first: ${N i}^{0}$ in $[N i (C O)_{4}]$ has $4 s^{2} 3 d^{8}$ , which reorganizes under the strong-field $C O$ ligand into fully paired $3 d^{10}$ , making it diamagnetic.
B · correct: Both Statement I and Statement II are false. - This option correctly identifies that both statements are false. Both statements are false: Statement I incorrectly labels $[N i (C O)_{4}]$ as paramagnetic, and Statement II falsely claims $[{N i C l}_{4}]^{2 -}$ ( $3 d^{8}$ ) and $[N i (C O)_{4}]$ ( $3 d^{10}$ ) have the same $d$ -electron configuration and both are paramagnetic.
C: Statement I is correct but statement II is false. - Overlooking the word 'paramagnetic' assigned to $[N i (C O)_{4}]$ in Statement I, assuming Statement I is completely true. Read every descriptor carefully: $[N i (C O)_{4}]$ has all electrons paired in the $3 d$ subshell ( $3 d^{10}$ ), meaning it is diamagnetic, rendering Statement I false.
D: Statement I is incorrect but statement II is true. - Assuming that because both $[{N i C l}_{4}]^{2 -}$ and $[N i (C O)_{4}]$ are tetrahedral complexes of nickel, they must share the same $d$ -electron configuration and magnetic behavior. Calculate the oxidation state of each metal center: $N i$ in $[{N i C l}_{4}]^{2 -}$ is $+ 2$ ( $3 d^{8}$ , paramagnetic), whereas $N i$ in $[N i (C O)_{4}]$ is $0$ ( $3 d^{10}$ , diamagnetic).

Step-by-step solution

Concept: For 4-coordinate Ni complexes, first determine the oxidation state and d-electron count of Ni. Next, identify the ligand field strength (SFL vs WFL). Then, find the electron pairing, hybridization, geometry, and magnetic character.
Option verdict: Statement I claims [Ni(CO)4] is paramagnetic. However, Ni is in the 0 oxidation state (3d8 4s2). The strong field CO ligand forces the 4s electrons into 3d. This gives a completely filled 3d10 system with no unpaired electrons (diamagnetic). Hence, Statement I is false. Statement II is also false. [NiCl4]2- is a d8 paramagnetic complex. In contrast, [Ni(CO)4] is a d10 diamagnetic complex.
Option verdict: Both statements are false. Statement I is false because [Ni(CO)4] is diamagnetic, not paramagnetic. Statement II is also false. [NiCl4]2- has Ni2+ with a 3d8 configuration and is paramagnetic. In contrast, [Ni(CO)4] has Ni0 with a 3d10 configuration and is diamagnetic. Thus, they have different d-electron configurations and different magnetic properties.
Option verdict: Statement I is incorrect because [Ni(CO)4] is diamagnetic, not paramagnetic.
Option verdict: Statement II is incorrect because [NiCl4]2- (Ni2+, d8) and [Ni(CO)4] (Ni0, d10) do not have the same d-electron configuration, nor are they both paramagnetic.
Discriminator: [Ni(CO)4] has Ni(0) with 3d10 configuration due to strong field CO ligands, meaning it is strictly diamagnetic. This instantly invalidates both Statement I (calls it paramagnetic) and Statement II (calls both paramagnetic with same d-count).