JEE MainChemistry

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How do ligand field strength and metal d-electron configuration determine the geometry and magnetic moment of 4-coordinate complexes?

Q: Why is [Ni(CO)_4] tetrahedral when CO is a strong field ligand?

In [Ni(CO)_4], nickel is in the 0 oxidation state (3d^8 4s^2). Strong field CO forces the two 4s electrons into the 3d subshell, completely filling it (3d^10). With all 3d orbitals full, inner d -orbitals are unavailable for dsp^2, forcing sp^3 hybridization (tetrahedral).

Match the LIST-I with LIST-II [ LIST-I (Complex/Species) LIST-II (Shape & magnetic moment); [ A. [Ni(CO)_4]; B. [Ni(CN)_4]^(2-); C. [NiCl_4]^(2-); D. [MnBr_4]^(2-) ] [ I. Tetrahedral, 2.8BM; II. Square planar, 0BM; III. Tetrahedral, 0BM; IV. Tetrahedral, 5.9BM ]; ] Choose the correct answer from the options given below:

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AA-III, B-IV, C-II, D-IB Four paired arrows fill 3d, leaving one empty 3d box grouped with 4s and two empty 4p boxes under a bracket labeled dsp².

A-I, B-II, C-III, D-IVCA-III, B-II, C-I, D-IVDA-IV, B-I, C-III, D-II

Step-by-step solutionView

Correct answer

Determine the oxidation state and d-electron configuration of each central metal ion, pair electrons if strong-field ligands force pairing, and deduce geometry and unpaired electrons: [Ni(CO)4] is tetrahedral (0 BM, III), [Ni(CN)4]2- is square planar (0 BM, II), [NiCl4]2- is tetrahedral (2.8 BM, I), and [MnBr4]2- is tetrahedral (5.9 BM, IV).

Option analysis

Why each option works or fails

A · A-III, B-IV, C-II, D-I

Treating [Ni(CN)4]2- as tetrahedral with high spin or confusing the strong-field pairing behavior of cyanide with weak-field halide behavior. Recall that CN- is a strong-field ligand causing d-electron pairing in Ni(II) (d8), leaving one empty d-orbital for dsp2 square planar hybridization with 0 unpaired electrons.

B · A-I, B-II, C-III, D-IV

Failing to account for the zero oxidation state of Ni in [Ni(CO)4] and the migration of 4s electrons to the 3d subshell due to strong-field CO. Ni in [Ni(CO)4] is Ni(0) with configuration 3d8 4s2; strong CO ligands force pairing into 3d10, using 4s and 4p for sp3 tetrahedral hybridization (0 BM).

C · A-III, B-II, C-I, D-IV

This option correctly correlates the oxidation state, ligand field strength, hybridization, geometry, and spin-only magnetic moment for all four coordination complexes. Correctly matches: A with III ([Ni(CO)4] is sp3, tetrahedral, 0 BM), B with II ([Ni(CN)4]2- is dsp2, square planar, 0 BM), C with I ([NiCl4]2- is sp3, tetrahedral, 2 unpaired electrons, 2.8 BM), and D with IV ([MnBr4]2- is sp3, tetrahedral, 5 unpaired electrons, 5.9 BM).

D · A-IV, B-I, C-III, D-II

Assigning the maximum unpaired electrons (5.9 BM) to [Ni(CO)4] instead of [MnBr4]2-, ignoring the d-electron count of Mn(II) versus Ni(0). Mn(II) in [MnBr4]2- has a 3d5 configuration with weak-field Br- ligands, giving 5 unpaired electrons (sqrt(35) approx 5.9 BM).

Reviewed route

Solution

StepWorking

01Concept

For 4-coordinate complexes: Determine oxidation state, $d$ -electron configuration, and effect of ligand field strength (strong field ligands like $C O, C N^{-}$ cause pairing; weak field ligands like $C l^{-}, B r^{-}$ do not). Magnetic moment $μ = \sqrt{n (n + 2)} BM$ , where $n$ is the number of unpaired electrons.

02Option verdict

Matches B with IV and C with II, which wrongly assigns high-spin tetrahedral geometry with 5 unpaired electrons to $[N i (C N)_{4}]^{2 -}$ .

03Option verdict

Matches A with I, wrongly treating $[N i (C O)_{4}]$ as paramagnetic with 2 unpaired electrons instead of pairing $4 s^{2}$ into $3 d^{8}$ to form a $3 d^{10}$ diamagnetic complex.

04Option verdict

A: $N i (0) : 3 d^{8} 4 s^{2} \overset{C O}{\to} 3 d^{10} (s p^{3}) ⟹ Tetrahedral, 0 BM (III)$ . B: $N i (I I) : 3 d^{8} \overset{C N^{-}}{\to} 3 d^{8} paired (d s p^{2}) ⟹ Square planar, 0 BM (II)$ . C: $N i (I I) : 3 d^{8} \overset{C l^{-}}{\to} 3 d^{8} unpaired (n = 2, s p^{3}) ⟹ Tetrahedral, μ = \sqrt{8} \approx 2.8 BM (I)$ . D: $M n (I I) : 3 d^{5} \overset{B r^{-}}{\to} 3 d^{5} unpaired (n = 5, s p^{3}) ⟹ Tetrahedral, μ = \sqrt{35} \approx 5.9 BM (IV)$ . Thus, A-III, B-II, C-I, D-IV is correct.

05Option verdict

Matches A with IV, assigning 5 unpaired electrons to $[N i (C O)_{4}]$ which is a diamagnetic neutral complex.

✓Discriminator

Finding D with ${M n}^{2 +}$ ( $d^{5}$ , weak field $B r^{-}$ , $n = 5$ , $μ = 5.9 BM ⟹ IV$ ) and A with ${N i}^{0}$ ( $d^{10}$ , $n = 0, s p^{3} ⟹ Tetrahedral, 0 BM ⟹ III$ ) immediately identifies option C.

Hints that build this answer step by step

What are the oxidation states and d-electron configurations of the central metal ions in [Ni(CO)4] and [Ni(CN)4]2-?
[Ni(CO)4]: Ni(0) with 3d10 (after pairing from 3d8 4s2); [Ni(CN)4]2-: Ni(II) with 3d8
What are the shapes and magnetic moments of [Ni(CO)4] and [Ni(CN)4]2-?
[Ni(CO)4] is Tetrahedral (sp3, 0 BM); [Ni(CN)4]2- is Square planar (dsp2, 0 BM)
What are the shapes and magnetic moments of [NiCl4]2- and [MnBr4]2-?
[NiCl4]2- is Tetrahedral with 2 unpaired electrons (2.8 BM); [MnBr4]2- is Tetrahedral with 5 unpaired electrons (5.9 BM)

✓ Source and academic review

Question type: Match the following
Exam relevance: JEE Main · Chemistry
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 19 September 2026

Quick checks

Students also ask

Why is

[N i (C O)_{4}]

tetrahedral when

C O

is a strong field ligand?

In $[N i (C O)_{4}]$ , nickel is in the 0 oxidation state ( $3 d^{8} 4 s^{2}$ ). Strong field CO forces the two $4 s$ electrons into the $3 d$ subshell, completely filling it ( $3 d^{10}$ ). With all $3 d$ orbitals full, inner $d$ -orbitals are unavailable for $d s p^{2}$ , forcing $s p^{3}$ hybridization (tetrahedral).

Answer

Determine the oxidation state and d-electron configuration of each central metal ion, pair electrons if strong-field ligands force pairing, and deduce geometry and unpaired electrons: [Ni(CO)4] is tetrahedral (0 BM, III), [Ni(CN)4]2- is square planar (0 BM, II), [NiCl4]2- is tetrahedral (2.8 BM, I), and [MnBr4]2- is tetrahedral (5.9 BM, IV).

Why each option works or fails

A: A-III, B-IV, C-II, D-I - Treating [Ni(CN)4]2- as tetrahedral with high spin or confusing the strong-field pairing behavior of cyanide with weak-field halide behavior. Recall that CN- is a strong-field ligand causing d-electron pairing in Ni(II) (d8), leaving one empty d-orbital for dsp2 square planar hybridization with 0 unpaired electrons.
B: A-I, B-II, C-III, D-IV - Failing to account for the zero oxidation state of Ni in [Ni(CO)4] and the migration of 4s electrons to the 3d subshell due to strong-field CO. Ni in [Ni(CO)4] is Ni(0) with configuration 3d8 4s2; strong CO ligands force pairing into 3d10, using 4s and 4p for sp3 tetrahedral hybridization (0 BM).
C · correct: A-III, B-II, C-I, D-IV - This option correctly correlates the oxidation state, ligand field strength, hybridization, geometry, and spin-only magnetic moment for all four coordination complexes. Correctly matches: A with III ([Ni(CO)4] is sp3, tetrahedral, 0 BM), B with II ([Ni(CN)4]2- is dsp2, square planar, 0 BM), C with I ([NiCl4]2- is sp3, tetrahedral, 2 unpaired electrons, 2.8 BM), and D with IV ([MnBr4]2- is sp3, tetrahedral, 5 unpaired electrons, 5.9 BM).
D: A-IV, B-I, C-III, D-II - Assigning the maximum unpaired electrons (5.9 BM) to [Ni(CO)4] instead of [MnBr4]2-, ignoring the d-electron count of Mn(II) versus Ni(0). Mn(II) in [MnBr4]2- has a 3d5 configuration with weak-field Br- ligands, giving 5 unpaired electrons (sqrt(35) approx 5.9 BM).

Step-by-step solution

Concept: For 4-coordinate complexes: Determine oxidation state, $d$ -electron configuration, and effect of ligand field strength (strong field ligands like $C O, C N^{-}$ cause pairing; weak field ligands like $C l^{-}, B r^{-}$ do not). Magnetic moment $μ = \sqrt{n (n + 2)} BM$ , where $n$ is the number of unpaired electrons.
Option verdict: Matches B with IV and C with II, which wrongly assigns high-spin tetrahedral geometry with 5 unpaired electrons to $[N i (C N)_{4}]^{2 -}$ .
Option verdict: Matches A with I, wrongly treating $[N i (C O)_{4}]$ as paramagnetic with 2 unpaired electrons instead of pairing $4 s^{2}$ into $3 d^{8}$ to form a $3 d^{10}$ diamagnetic complex.
Option verdict: A: $N i (0) : 3 d^{8} 4 s^{2} \overset{C O}{\to} 3 d^{10} (s p^{3}) ⟹ Tetrahedral, 0 BM (III)$ . B: $N i (I I) : 3 d^{8} \overset{C N^{-}}{\to} 3 d^{8} paired (d s p^{2}) ⟹ Square planar, 0 BM (II)$ . C: $N i (I I) : 3 d^{8} \overset{C l^{-}}{\to} 3 d^{8} unpaired (n = 2, s p^{3}) ⟹ Tetrahedral, μ = \sqrt{8} \approx 2.8 BM (I)$ . D: $M n (I I) : 3 d^{5} \overset{B r^{-}}{\to} 3 d^{5} unpaired (n = 5, s p^{3}) ⟹ Tetrahedral, μ = \sqrt{35} \approx 5.9 BM (IV)$ . Thus, A-III, B-II, C-I, D-IV is correct.
Option verdict: Matches A with IV, assigning 5 unpaired electrons to $[N i (C O)_{4}]$ which is a diamagnetic neutral complex.
Discriminator: Finding D with ${M n}^{2 +}$ ( $d^{5}$ , weak field $B r^{-}$ , $n = 5$ , $μ = 5.9 BM ⟹ IV$ ) and A with ${N i}^{0}$ ( $d^{10}$ , $n = 0, s p^{3} ⟹ Tetrahedral, 0 BM ⟹ III$ ) immediately identifies option C.

How do ligand field strength and metal d-electron configuration determine the geometry and magnetic moment of 4-coordinate complexes?

What feels right?

Why each option works or fails

Solution

01Concept

02Option verdict

03Option verdict

04Option verdict

05Option verdict

✓Discriminator

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