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 ↗

Ready to attempt under real exam pressure?

Test your speed and accuracy in the authentic 200-minute timed NTA simulator.

Launch NEET 2021 CBT Test →

← Previous (Q.85) All Chemistry Questions Grid Next (Q.87) →

Spotted a mistake in this solution? Report an error ([email protected])

⚖️ Legal & Fair Dealing Notice: Sourced from publicly released NEET 2021 question papers and NEET answer keys for private study, research, and non-commercial educational practice under Fair Dealing (Section 52(1)(a) of the Indian Copyright Act, 1957). Not affiliated with or endorsed by NTA or NMC.

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?

🔗 Related Questions — Coordination Compounds

Coordination Isomerism & IUPAC Nomenclature of Bimetallic Complexes

Ionic Dissociation & Ion Ratio in Potash Alum

Spectrochemical Series & Ligand Field Strength Ordering

Spin-Only Magnetic Moments of Octahedral 3d Complexes

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