Q.66 Chemistry ⭐ 360° Option Elimination Active

NEET 2026 Chemistry Question 66 Solution

Topic: Coordination Compounds · Official NTA Booklet Code 50

Question

Among the species given below, the spin-only magnetic moment is highest for (Given: Atomic number of Ti = 22, Mn = 25, Fe = 26 and Co = 27)

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Verified Against NTA Final Answer Key

Option (1) : [Mn(CN)₆]³⁻

📚 Coordination Compounds Medium

📋 Given Parameters & Problem Setup

Atomic Numbers Ti = 22, Mn = 25, Fe = 26, Co = 27

Target Species with highest spin-only magnetic moment (μ_s = √[n(n+2)] BM)

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The 30-Second Exam Hall Shortcut: ⚡ Count Unpaired Electrons (n)

⏱️ Save ~60 seconds

Highest magnetic moment = greatest number of unpaired electrons (n)!

1 [Mn(CN)₆]³⁻: Mn³⁺ is d⁴. CN⁻ is strong field -> t_{2g}⁴ e_g⁰ -> 2 unpaired electrons (n = 2).

2 [Fe(CN)₆]³⁻: Fe³⁺ is d⁵. CN⁻ is strong field -> t_{2g}⁵ e_g⁰ -> 1 unpaired electron (n = 1).

3 [Co(NH₃)₆]³⁺: Co³⁺ is d⁶. NH₃ with Co³⁺ is strong field -> t_{2g}⁶ e_g⁰ -> 0 unpaired electrons (diamagnetic, n = 0).

4 [Ti(H₂O)₆]³⁺: Ti³⁺ is d¹ -> t_{2g}¹ e_g⁰ -> 1 unpaired electron (n = 1).

5 Highest n is n = 2 in [Mn(CN)₆]³⁻! Option (1) wins in 20 seconds!

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Why You Can 100% Trust This Shortcut: Spin-only magnetic moment formula μ = √[n(n+2)] BM is strictly monotonically increasing with the number of unpaired electrons n.

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Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Determine Oxidation State and d-Electron Count for Each Complex

Find the d-electron configuration of the central metal ion: 1. [Mn(CN)₆]³⁻: Mn + 6(-1) = -3 ⇒ Mn³⁺. Mn atom is [Ar] 3d⁵ 4s² ⇒ Mn³⁺ is 3d⁴. 2. [Fe(CN)₆]³⁻: Fe + 6(-1) = -3 ⇒ Fe³⁺. Fe atom is [Ar] 3d⁶ 4s² ⇒ Fe³⁺ is 3d⁵. 3. [Co(NH₃)₆]³⁺: Co + 6(0) = +3 ⇒ Co³⁺. Co atom is [Ar] 3d⁷ 4s² ⇒ Co³⁺ is 3d⁶. 4. [Ti(H₂O)₆]³⁺: Ti + 6(0) = +3 ⇒ Ti³⁺. Ti atom is [Ar] 3d² 4s² ⇒ Ti³⁺ is 3d¹.

\text{Mn}^{3+}: 3d^4, \quad \text{Fe}^{3+}: 3d^5, \quad \text{Co}^{3+}: 3d^6, \quad \text{Ti}^{3+}: 3d^1

• All four complexes possess octahedral geometry.

Step 2

Apply Crystal Field Theory (Ligand Field Splitting)

Examine ligand field strengths: - CN⁻ is a strong field ligand (Δ_o > P), causing electron pairing in t_{2g}. - For Mn³⁺ (d⁴): 4 electrons enter t_{2g} giving configuration t_{2g}⁴ e_g⁰. Number of unpaired electrons n = 2. - For Fe³⁺ (d⁵): 5 electrons enter t_{2g} giving configuration t_{2g}⁵ e_g⁰. Number of unpaired electrons n = 1. - For Co³⁺ (d⁶): NH₃ acts as a strong field ligand with Co³⁺ (high charge density), giving low-spin t_{2g}⁶ e_g⁰. Number of unpaired electrons n = 0 (diamagnetic). - For Ti³⁺ (d¹): configuration is t_{2g}¹ e_g⁰. Number of unpaired electrons n = 1.

n([\text{Mn}(\text{CN})_6]^{3-}) = 2, \quad n([\text{Fe}(\text{CN})_6]^{3-}) = 1, \quad n([\text{Co}(\text{NH}_3)_6]^{3+}) = 0, \quad n([\text{Ti}(\text{H}_2\text{O})_6]^{3+}) = 1

• Maximum unpaired electrons n = 2.

Step 3

Calculate Spin-Only Magnetic Moments

Using the spin-only formula μ = √[n(n+2)] BM: 1. [Mn(CN)₆]³⁻: μ = √[2(4)] = √8 ≈ 2.83 BM. 2. [Fe(CN)₆]³⁻: μ = √[1(3)] = √3 ≈ 1.73 BM. 3. [Co(NH₃)₆]³⁺: μ = 0 BM. 4. [Ti(H₂O)₆]³⁺: μ = √[1(3)] = √3 ≈ 1.73 BM.

\mu([\text{Mn}(\text{CN})_6]^{3-}) = 2.83\text{ BM} \quad (\text{Highest})

• Matches Option (1).

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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

✓ 100% Correct

Verdict & Scientific Reason:

Highest magnetic moment with n = 2 unpaired electrons (μ = 2.83 BM).

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Why this is airtight: None; accurate application of low-spin t_{2g} filling.

2 B

✗ Incorrect

Verdict & Scientific Reason:

Incorrect. Fe³⁺ (d⁵) low-spin has t_{2g}⁵, which leaves only 1 unpaired electron (μ = 1.73 BM).

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Student Fallacy (Why candidates pick this trap): Assuming d⁵ always has 5 unpaired electrons, forgetting CN⁻ forces pairing.

3 C

✗ Incorrect

Verdict & Scientific Reason:

Incorrect. Co³⁺ with NH₃ forms a diamagnetic low-spin complex with n = 0 unpaired electrons (μ = 0).

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Student Fallacy (Why candidates pick this trap): Treating NH₃ as a weak-field ligand for Co³⁺.

4 D

✗ Incorrect

Verdict & Scientific Reason:

Incorrect. Ti³⁺ (d¹) has only 1 unpaired electron (μ = 1.73 BM).

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Student Fallacy (Why candidates pick this trap): Thinking Ti³⁺ has more d-electrons.

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NTA Trap Alert: Assuming d⁵ Fe³⁺ Always Has the Highest Magnetic Moment

Students habitually think d⁵ (Fe³⁺) is always the most paramagnetic. With strong field CN⁻, pairing collapses d⁵ into t_{2g}⁵ (only ONE unpaired electron), whereas d⁴ has t_{2g}⁴ (TWO unpaired electrons)!

🛡️ How to avoid the -1 mark: Fill the three t_{2g} orbitals: 4 electrons leave 2 half-filled orbitals; 5 electrons leave only 1 half-filled orbital.

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Core NEET Syllabus Standard

NEET Chemistry Core Syllabus · Coordination Compounds Standard Benchmark

Core Scientific Principle Tested:

"In strong field octahedral environments (low spin), d⁴ has n=2 while d⁵ has n=1."

Syllabus Focus: Crystal field theory, electronic configurations of d-orbitals, magnetic properties

✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗

In strong field cage: d⁴ has two lonely souls, but d⁵ has only one!

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NEET 2026 Chemistry Question 66 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ Count Unpaired Electrons (n)

Step-by-Step Rigorous Workflow & Mathematical Proof

Determine Oxidation State and d-Electron Count for Each Complex

Apply Crystal Field Theory (Ligand Field Splitting)

Calculate Spin-Only Magnetic Moments

360° Option Elimination Engine Key Feature

NTA Trap Alert: Assuming d⁵ Fe³⁺ Always Has the Highest Magnetic Moment

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

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