Q.51 Chemistry ⭐ 360° Option Elimination Active

NEET 2025 Chemistry Question 51 Solution

Topic: Coordination Compounds · Official NTA Booklet Code 45

Question

Which of the following are paramagnetic? A. [NiCl₄]²⁻ B. Ni(CO)₄ C. [Ni(CN)₄]²⁻ D. [Ni(H₂O)₆]²⁺ E. Ni(PPh₃)₄ Choose the correct answer from the options given below:

✓

Verified Against NTA Final Answer Key

Option (3) : A and D only

📚 Coordination Compounds Medium

📋 Given Parameters & Problem Setup

Complex A [NiCl₄]²⁻ (Tetrachloridonickelate(II))

Complex B Ni(CO)₄ (Tetracarbonylnickel(0))

Complex C [Ni(CN)₄]²⁻ (Tetracyanidonickelate(II))

Complex D [Ni(H₂O)₆]²⁺ (Hexaaquanickel(II))

Complex E Ni(PPh₃)₄ (Tetrakis(triphenylphosphine)nickel(0))

⚡

The 30-Second Exam Hall Shortcut: ⚡ Nickel Coordination Electron Pairing Rule

⏱️ Save ~50 seconds

[NiCl₄]²⁻ has weak ligand Cl⁻ (sp³ tetrahedral, 2 unpaired e⁻). [Ni(H₂O)₆]²⁺ has d⁸ octahedral (t₂g⁶ eg², 2 unpaired e⁻). Both are PARAMAGNETIC! Ni(CO)₄ and Ni(PPh₃)₄ are d¹⁰ (diamagnetic). [Ni(CN)₄]²⁻ is dsp² square planar (diamagnetic)!

1 Complex A [NiCl₄]²⁻: Ni²⁺ (3d⁸), Cl⁻ is weak field ligand. Tetrahedral geometry (sp³), configuration is e⁴ t₂⁴ with 2 unpaired electrons ⟹ PARAMAGNETIC.

2 Complex B Ni(CO)₄: Ni⁰ (3d⁸ 4s²). CO is a very strong field ligand, forcing pairing into 3d¹⁰. Zero unpaired electrons ⟹ DIAMAGNETIC.

3 Complex C [Ni(CN)₄]²⁻: Ni²⁺ (3d⁸). CN⁻ is strong field, forces pairing in 3d into dsp² square planar. Zero unpaired electrons ⟹ DIAMAGNETIC.

4 Complex D [Ni(H₂O)₆]²⁺: Ni²⁺ (3d⁸) in octahedral crystal field. Configuration is always t₂g⁶ eg² with 2 unpaired electrons in eg ⟹ PARAMAGNETIC.

5 Complex E Ni(PPh₃)₄: Ni⁰ (3d⁸ 4s²), PPh₃ forces pairing into 3d¹⁰. Zero unpaired electrons ⟹ DIAMAGNETIC.

6 Paramagnetic complexes are A and D only!

7 Select Option (3) with complete certainty!

🛡️

Why You Can 100% Trust This Shortcut: Crystal Field Theory & Spin-Only Magnetic Moment (NCERT Chemistry · Coordination Compounds: μ = √(n(n+2)) BM).

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Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Electronic Configuration and Splitting of [NiCl₄]²⁻

In [NiCl₄]²⁻, nickel is in +2 oxidation state with electronic configuration [Ar] 3d⁸. Chloride (Cl⁻) is a weak field ligand, so the crystal field splitting Δ_t is less than the pairing energy P:

\text{Geometry: Tetrahedral } (sp^3), \quad \text{Configuration: } e^4 t_2^4

• There are 2 unpaired electrons (n = 2).
• Magnetic moment μ = √(2(2+2)) = √8 ≈ 2.83 BM.
• Therefore, [NiCl₄]²⁻ is PARAMAGNETIC.

Step 2

Electronic Configuration of Ni(CO)₄, [Ni(CN)₄]²⁻ and Ni(PPh₃)₄

Examine the low-spin and zero-valent complexes with strong π-acceptor ligands:

\text{Ni(CO)}_4 \text{ and Ni(PPh}_3)_4: \; \text{Ni}^0 (3d^8 4s^2) \xrightarrow{\text{pairing}} 3d^{10} 4s^0 \implies n = 0 \text{ (Diamagnetic)}

• [Ni(CN)₄]²⁻: Ni²⁺ (3d⁸). CN⁻ causes pairing into four d-orbitals, leaving one d-orbital empty for dsp² square planar hybridization (n = 0, Diamagnetic).

Step 3

Electronic Configuration and Splitting of [Ni(H₂O)₆]²⁺

In the octahedral complex [Ni(H₂O)₆]²⁺, nickel is in +2 oxidation state (3d⁸). In octahedral symmetry, the d⁸ configuration is uniquely determined regardless of ligand field strength:

\text{Configuration: } t_{2g}^6 e_g^2 \implies n = 2 \text{ unpaired electrons}

• Two unpaired electrons reside in the eg orbitals (dx²-y² and dz²).
• Therefore, [Ni(H₂O)₆]²⁺ is PARAMAGNETIC (μ ≈ 2.83 BM).
• Conclusion: A and D only are paramagnetic, matching Option (3).

⭐

360° Option Elimination Engine Key Feature

Diagnose every choice: Know exactly why the right option works and why distractors fail.

4-Option Deep Scan

1 A

✗ Incorrect

Verdict & Scientific Reason:

Incorrect. [Ni(CN)₄]²⁻ is square planar and diamagnetic (zero unpaired electrons).

⚠️

Student Fallacy (Why candidates pick this trap): Thinking all 4-coordinate Ni²⁺ complexes are tetrahedral and paramagnetic.

2 B

✗ Incorrect

Verdict & Scientific Reason:

Incorrect. Both Ni(CO)₄ and Ni(PPh₃)₄ are 3d¹⁰ diamagnetic complexes.

⚠️

Student Fallacy (Why candidates pick this trap): Assuming Ni(0) has unpaired electrons from 3d⁸.

3 C

✓ 100% Correct

Verdict & Scientific Reason:

100% correct. Both [NiCl₄]²⁻ (tetrahedral d⁸) and [Ni(H₂O)₆]²⁺ (octahedral d⁸) possess 2 unpaired electrons.

💡

Why this is airtight: None; perfect mastery of CFT.

4 D

✗ Incorrect

Verdict & Scientific Reason:

Incorrect. Complex E is diamagnetic.

⚠️

Student Fallacy (Why candidates pick this trap): Treating PPh₃ as a weak ligand.

⚠️

NTA Trap Alert: Square Planar [Ni(CN)₄]²⁻ vs Tetrahedral [NiCl₄]²⁻

Candidates often confuse [Ni(CN)₄]²⁻ with [NiCl₄]²⁻. Because CN⁻ is strong field, it pairs up all electrons in 3d⁸ to make a square planar diamagnetic complex, whereas Cl⁻ is weak field, yielding a tetrahedral paramagnetic complex!

🛡️ How to avoid the -1 mark: Remember: Ni²⁺ with CN⁻ is square planar & DIAMAGNETIC; with Cl⁻ it is tetrahedral & PARAMAGNETIC.

📖

Core NEET Syllabus Standard

NEET Chemistry Core Syllabus · Coordination Compounds Standard Benchmark

Core Scientific Principle Tested:

"d⁸ tetrahedral with weak ligands has 2 unpaired e⁻; d⁸ octahedral always has 2 unpaired e⁻ (t₂g⁶ eg²)."

Syllabus Focus: Magnetic Properties & Hybridization of Coordination Complexes

✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗

Cl⁻ leaves d⁸ unpaired (A), water in octahedron has 2 unpaired in eg (D) ⟹ A & D only!

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NEET 2025 Chemistry Question 51 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ Nickel Coordination Electron Pairing Rule

Step-by-Step Rigorous Workflow & Mathematical Proof

Electronic Configuration and Splitting of [NiCl₄]²⁻

Electronic Configuration of Ni(CO)₄, [Ni(CN)₄]²⁻ and Ni(PPh₃)₄

Electronic Configuration and Splitting of [Ni(H₂O)₆]²⁺

360° Option Elimination Engine Key Feature

NTA Trap Alert: Square Planar [Ni(CN)₄]²⁻ vs Tetrahedral [NiCl₄]²⁻

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

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