Q.86 Chemistry ⭐ 360° Option Elimination Active

NEET 2021 Chemistry Question 86 Solution

Topic: Coordination Compounds · Official NTA Booklet Code M1

Question

Match List - I with List - II. List I: (a). [Fe(CN)₆]³⁻; (b). [Fe(H₂O)₆]³⁺; (c). [Fe(CN)₆]⁴⁻; (d). [Fe(H₂O)₆]²⁺. List II: (i). 5.92 BM; (ii). 0 BM; (iii). 4.90 BM; (iv). 1.73 BM. Choose the correct answer from the options given below.

✓

Verified Against NTA Final Answer Key

Option (4) : (a)-(iv), (b)-(i), (c)-(ii), (d)-(iii)

📚 Coordination Compounds Medium

📋 Given Parameters & Problem Setup

List I Complexes (a) [Fe(CN)6]^3-, (b) [Fe(H2O)6]^3+, (c) [Fe(CN)6]^4-, (d) [Fe(H2O)6]^2+

List II Magnetic Moments (i) 5.92 BM, (ii) 0 BM, (iii) 4.90 BM, (iv) 1.73 BM

⚡

The 30-Second Exam Hall Shortcut: ⚡ Unpaired Electrons to BM: n=1 $\to$ 1.73, n=5 $\to$ 5.92, n=0 $\to$ 0, n=4 $\to$ 4.90

⏱️ Save ~30 seconds

$[Fe(CN)_6]^{3-}$ : $Fe^{3+} (3d^5) + \text{strong } CN^- \implies t_{2g}^5 \implies n=1 \implies 1.73\text{ BM}$ (iv). $[Fe(H_2O)_6]^{3+}$ : $Fe^{3+} (3d^5) + \text{weak } H_2O \implies t_{2g}^3 e_g^2 \implies n=5 \implies 5.92\text{ BM}$ (i). $[Fe(CN)_6]^{4-}$ : $Fe^{2+} (3d^6) + \text{strong } CN^- \implies t_{2g}^6 \implies n=0 \implies 0\text{ BM}$ (ii). $[Fe(H_2O)_6]^{2+}$ : $Fe^{2+} (3d^6) + \text{weak } H_2O \implies t_{2g}^4 e_g^2 \implies n=4 \implies 4.90\text{ BM}$ (iii). Match (a)-(iv), (b)-(i), (c)-(ii), (d)-(iii) $\implies$ Option (4)!

1 Recall spin-only formula:

\mu_s = \sqrt{n(n+2)}\text{ BM}

2 Identify ligand strength:

CN^-

is strong field (pairs electrons);

H_2O

is weak field (high spin).

3 Complex (a)

[Fe(CN)_6]^{3-}

Fe^{3+} (d^5)

, low spin

\implies t_{2g}^5 \implies n = 1 \implies \mu = \sqrt{3} = 1.73\text{ BM}

-> (iv).

4 Complex (b)

[Fe(H_2O)_6]^{3-}

Fe^{3+} (d^5)

, high spin

\implies t_{2g}^3 e_g^2 \implies n = 5 \implies \mu = \sqrt{35} = 5.92\text{ BM}

-> (i).

5 Complex (c)

[Fe(CN)_6]^{4-}

Fe^{2+} (d^6)

, low spin

\implies t_{2g}^6 \implies n = 0 \implies \mu = 0\text{ BM}

-> (ii).

6 Complex (d)

[Fe(H_2O)_6]^{2-}

Fe^{2+} (d^6)

, high spin

\implies t_{2g}^4 e_g^2 \implies n = 4 \implies \mu = \sqrt{24} = 4.90\text{ BM}

-> (iii).

7 Match Option (4).

🛡️

Why You Can 100% Trust This Shortcut: Canonical Crystal Field Theory (CFT) electronic configurations and Gouy balance magnetic moments.

📐

Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Crystal Field Theory Principles for Octahedral Complexes

Under an octahedral ligand field, the five degenerate $d$ -orbitals split into lower $t_{2g}$ (triply degenerate) and higher $e_g$ (doubly degenerate) sets separated by crystal field splitting energy $\Delta_o$ . Strong field ligands ( $CN^-$ ) induce high $\Delta_o > P$ (pairing energy), producing low-spin paired configurations. Weak field ligands ( $H_2O$ ) have $\Delta_o < P$ , producing high-spin configurations obeying Hund's rule.

Step 2

Evaluating Complex (a): [Fe(CN)6]3-

Iron is in the $+3$ oxidation state ( $Fe^{3+}: [Ar]3d^5$ ). With strong-field cyanide ligands ( $CN^-$ ), pairing occurs: $t_{2g}^5 e_g^0$ . Number of unpaired electrons $n = 1$ : $\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73\text{ BM} \implies \mathbf{\text{(a)-(iv)}}$

Step 3

Evaluating Complex (b): [Fe(H2O)6]3+

Iron is in the $+3$ oxidation state ( $Fe^{3+}: [Ar]3d^5$ ). With weak-field aqua ligands ( $H_2O$ ), electrons occupy orbitals singly: $t_{2g}^3 e_g^2$ . Number of unpaired electrons $n = 5$ : $\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92\text{ BM} \implies \mathbf{\text{(b)-(i)}}$

Step 4

Evaluating Complex (c): [Fe(CN)6]4-

Iron is in the $+2$ oxidation state ( $Fe^{2+}: [Ar]3d^6$ ). With strong-field cyanide ligands ( $CN^-$ ), all 6 electrons completely pair in $t_{2g}$ : $t_{2g}^6 e_g^0$ . Number of unpaired electrons $n = 0$ (diamagnetic): $\mu = 0\text{ BM} \implies \mathbf{\text{(c)-(ii)}}$

Step 5

Evaluating Complex (d): [Fe(H2O)6]2+

Iron is in the $+2$ oxidation state ( $Fe^{2+}: [Ar]3d^6$ ). With weak-field aqua ligands ( $H_2O$ ), the configuration is $t_{2g}^4 e_g^2$ . Number of unpaired electrons $n = 4$ : $\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90\text{ BM} \implies \mathbf{\text{(d)-(iii)}}$

⭐

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 1

✗ Incorrect

Verdict & Scientific Reason:

Mismatches (b) with 0 BM and (c) with 5.92 BM.

⚠️

Student Fallacy (Why candidates pick this trap): Swapping high spin Fe(III) with low spin Fe(II).

2 2

✗ Incorrect

Verdict & Scientific Reason:

Assigns 0 BM to complex (a), which has an odd number of electrons ( $3d^5$ ) and cannot be diamagnetic.

⚠️

Student Fallacy (Why candidates pick this trap): Assuming complete pairing is possible with 5 electrons.

3 3

✗ Incorrect

Verdict & Scientific Reason:

Treats cyanide as a weak-field ligand and water as strong-field.

⚠️

Student Fallacy (Why candidates pick this trap): Inverting spectrochemical series.

4 4

✓ 100% Correct

Verdict & Scientific Reason:

Flawless 1-to-1 match of all four coordination complexes with their exact CFT magnetic moments.

💡

Why this is airtight: None; rigorous crystal field theory execution.

⚠️

NTA Trap Alert: Odd vs. Even Electron Diamagnetism Trap

Students frequently mix up $[Fe(CN)_6]^{3-}$ ( $Fe^{3+}, 3d^5$ ) and $[Fe(CN)_6]^{4-}$ ( $Fe^{2+}, 3d^6$ ). An odd-electron species ( $d^5$ ) can NEVER be diamagnetic ( $0\text{ BM}$ )! Only $d^6$ in a strong field can pair all 6 electrons to give $\mu = 0\text{ BM}$ .

🛡️ How to avoid the -1 mark: Check electron count:

Fe^{2+} (d^6) + CN^- \implies 0\text{ BM}

Fe^{3+} (d^5) + CN^- \implies 1\text{ unpaired } e^- \implies 1.73\text{ BM}

📖

Core NEET Syllabus Standard

NCERT Chemistry · Coordination Compounds Standard Benchmark

Core Scientific Principle Tested:

"Spin-only magnetic moment $\mu = \sqrt{n(n+2)}\text{ BM}$ is determined by the number of unpaired electrons governed by ligand field strength."

Syllabus Focus: Crystal Field Theory (Magnetic Properties of Complexes)

✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗

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NEET 2021 Chemistry Question 86 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ Unpaired Electrons to BM: n=1 →\to→ 1.73, n=5 →\to→ 5.92, n=0 →\to→ 0, n=4 →\to→ 4.90

Step-by-Step Rigorous Workflow & Mathematical Proof

Crystal Field Theory Principles for Octahedral Complexes

Evaluating Complex (a): [Fe(CN)6]3-

Evaluating Complex (b): [Fe(H2O)6]3+

Evaluating Complex (c): [Fe(CN)6]4-

Evaluating Complex (d): [Fe(H2O)6]2+

360° Option Elimination Engine Key Feature

NTA Trap Alert: Odd vs. Even Electron Diamagnetism Trap

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

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Solve 4 NEET Questions a Day — One per Subject

The 30-Second Exam Hall Shortcut: ⚡ Unpaired Electrons to BM: n=1 $\to$ 1.73, n=5 $\to$ 5.92, n=0 $\to$ 0, n=4 $\to$ 4.90