![Q.18 The CORRECT statement(s) about [Ni(CN)4]2−, [Ni(CO)4] and [NiCl4]2− is(are) (Given: Atomic number of Ni: 28) (A) Both [Ni(CN)4]2− and [Ni(CO)4] are square planar complexes. (B) [Ni(CN)4]2− is diamagnetic and [NiCl4]2− is paramagnetic. (C) Both [Ni(CO)4] and [NiCl4]2− are paramagnetic. (D) [Ni(CN)4]2− is square planar and [NiCl4]2− is tetrahedral in shape.](data:image/svg+xml;base64,PHN2ZyB2aWV3Qm94PSIwIDAgMTkyMCAxMDgwIiB3aWR0aD0iMTkyMCIgaGVpZ2h0PSIxMDgwIiB4bWxucz0iaHR0cDovL3d3dy53My5vcmcvMjAwMC9zdmciPjwvc3ZnPg==)
Q.18 The CORRECT statement(s) about [Ni(CN)4]2−, [Ni(CO)4] and [NiCl4]2− is(are)
(Given: Atomic number of Ni: 28)
(A) Both [Ni(CN)4]2− and [Ni(CO)4] are square planar complexes.
(B) [Ni(CN)4]2− is diamagnetic and [NiCl4]2− is paramagnetic.
(C) Both [Ni(CO)4] and [NiCl4]2− are paramagnetic.
(D) [Ni(CN)4]2− is square planar and [NiCl4]2− is tetrahedral in shape.
Ni atomic number 28 governs the electron configuration and properties of these Ni(II) and Ni(0) complexes. The correct statements are (B) and (D), as determined by ligand field strength, hybridization, geometry, and electron pairing.
Nickel Electron Configuration
Nickel (atomic number 28) has the ground state configuration [Ar] 4s² 3d⁸. In [Ni(CN)₄]²⁻ and [NiCl₄]²⁻, Ni exists as Ni²⁺ ([Ar] 3d⁸) due to the 2− charge and four monoanionic ligands. In neutral [Ni(CO)₄], Ni is in the 0 oxidation state ([Ar] 4s² 3d⁸).
Option Analysis
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(A) Both [Ni(CN)₄]²⁻ and [Ni(CO)₄] are square planar: Incorrect. [Ni(CN)₄]²⁻ is square planar (dsp² hybridization from strong-field CN⁻ causing d⁸ pairing), but [Ni(CO)₄] is tetrahedral (sp³ hybridization).
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(B) [Ni(CN)₄]²⁻ is diamagnetic and [NiCl₄]²⁻ is paramagnetic: Correct. CN⁻ (strong field) pairs all 3d⁸ electrons (no unpaired electrons, diamagnetic); Cl⁻ (weak field) leaves two unpaired electrons in tetrahedral sp³ (paramagnetic).
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(C) Both [Ni(CO)₄] and [NiCl₄]²⁻ are paramagnetic: Incorrect. [Ni(CO)₄] is diamagnetic (CO strong field pairs all 10 d-electrons in tetrahedral sp³); [NiCl₄]²⁻ is paramagnetic.
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(D) [Ni(CN)₄]²⁻ is square planar and [NiCl₄]²⁻ is tetrahedral: Correct. Strong-field CN⁻ favors dsp² square planar; weak-field Cl⁻ favors sp³ tetrahedral.
Properties Summary
| Complex | Oxidation State | Hybridization | Geometry | Magnetic Property | Unpaired Electrons |
|---|---|---|---|---|---|
| [Ni(CN)₄]²⁻ | Ni²⁺ (3d⁸) | dsp² | Square planar | Diamagnetic | 0 |
| [Ni(CO)₄] | Ni⁰ (3d¹⁰ effective) | sp³ | Tetrahedral | Diamagnetic | 0 |
| [NiCl₄]²⁻ | Ni²⁺ (3d⁸) | sp³ | Tetrahedral | Paramagnetic | 2 |
In coordination chemistry for CSIR NET Life Sciences preparation, understanding [Ni(CN)4]2− [Ni(CO)4] [NiCl4]2− geometry magnetism is crucial, especially with Ni atomic number 28. These complexes showcase valence bond theory (VBT) principles where ligand field strength dictates hybridization, shape, and magnetic behavior.
Ligand Field Effects
CN⁻ and CO are strong-field ligands causing electron pairing, while Cl⁻ is weak-field. For Ni²⁺ (3d⁸), strong fields yield dsp² square planar [Ni(CN)4]2− (diamagnetic), weak fields yield sp³ tetrahedral [NiCl4]2− (paramagnetic with 2 unpaired e⁻). [Ni(CO)4] (Ni⁰) achieves 3d¹⁰ pairing for tetrahedral diamagnetism.
Exam Relevance
This multi-correct JEE/CSIR NET question tests VBT application. Options B and D are right: [Ni(CN)4]2− diamagnetic vs [NiCl4]2− paramagnetic, and square planar vs tetrahedral shapes. Practice similar problems on Ni atomic number 28 electron configuration for competitive success.
