Match the LIST-I (Complex/Species)A. [Ni(CO)₄]B. [Ni(CN)₄]²⁻C. [NiCl₄]²⁻D. [MnBr₄]...| Doubtify JEE

 

Match the List-I with List-II: Coordination Complexes, Shapes, and Magnetic Moments

Introduction

Coordination chemistry is a fascinating branch of inorganic chemistry that deals with the structure, bonding, and properties of complexes. In competitive exams like JEE, NEET, and GATE, questions often test your ability to relate complex formulas with their shapes and magnetic moments.

In this article, we will solve the following question in detail:

Question:

Match the LIST-I with LIST-II

LIST-I (Complex/Species)		LIST-II (Shape & Magnetic Moment)
A. [Ni(CO)₄]		I. Tetrahedral, 2.8 BM
B. [Ni(CN)₄]²⁻		II. Square planar, 0 BM
C. [NiCl₄]²⁻		III. Tetrahedral, 0 BM
D. [MnBr₄]²⁻		IV. Tetrahedral, 5.9 BM

We’ll go step-by-step to identify hybridization, geometry, and magnetic moment for each complex.

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Step 1: Understanding the Basics

Before matching, let’s recall some key concepts:

1️⃣ Magnetic Moment

• Magnetic moment (μ) in Bohr Magneton (BM) helps determine whether a complex is paramagnetic (unpaired electrons) or diamagnetic (all electrons paired).

• Formula:

  $$\mu =\sqrt{n(n+2)}$$

  where n = number of unpaired electrons.

2️⃣ Geometry & Hybridization

• Tetrahedral geometry → Usually sp³ hybridization.

• Square planar geometry → Usually dsp² hybridization.

• Strong field ligands (like CO, CN⁻) can cause pairing of electrons (low spin).

• Weak field ligands (like Cl⁻, Br⁻) often give high-spin complexes.

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Step 2: Analyzing Each Complex

A. [Ni(CO)₄]

• Oxidation state of Ni:
  CO is a neutral ligand → Ni oxidation state = 0.
  Electronic configuration of Ni: [Ar] 3d⁸ 4s².

• Effect of ligand:
  CO is a strong field ligand (from spectrochemical series) → causes pairing of electrons in 3d orbitals.

• Hybridization:
  After pairing, configuration becomes: 3d¹⁰ 4s⁰ 4p⁰ → sp³ hybridization.

• Geometry: Tetrahedral.

• Magnetic property: All electrons paired → diamagnetic (0 BM).

✅ Match: A → III (Tetrahedral, 0 BM).

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B. [Ni(CN)₄]²⁻

• Oxidation state of Ni:
  CN⁻ = -1 charge each → total -4. Overall charge = -2 → Ni oxidation state = +2.
  Ni²⁺: [Ar] 3d⁸.

• Effect of ligand:
  CN⁻ is a strong field ligand → causes pairing of electrons.

• Hybridization:
  After pairing, configuration becomes: 3d⁸ → inner orbital (dsp²) hybridization.

• Geometry: Square planar.

• Magnetic property: No unpaired electrons → diamagnetic (0 BM).

✅ Match: B → II (Square planar, 0 BM).

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C. [NiCl₄]²⁻

• Oxidation state of Ni:
  Cl⁻ = -1 each → total -4. Overall charge = -2 → Ni oxidation state = +2.
  Ni²⁺: [Ar] 3d⁸.

• Effect of ligand:
  Cl⁻ is a weak field ligand → no pairing of electrons.

• Hybridization:
  Uses outer orbital → sp³ hybridization.

• Geometry: Tetrahedral.

• Magnetic property: Two unpaired electrons →
  μ = √(2(2+2)) = √8 ≈ 2.8 BM.

✅ Match: C → I (Tetrahedral, 2.8 BM).

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D. [MnBr₄]²⁻

• Oxidation state of Mn:
  Br⁻ = -1 each → total -4. Overall charge = -2 → Mn oxidation state = +2.
  Mn²⁺: [Ar] 3d⁵.

• Effect of ligand:
  Br⁻ is a weak field ligand → no pairing of electrons.

• Hybridization:
  Outer orbital → sp³ hybridization.

• Geometry: Tetrahedral.

• Magnetic property: 5 unpaired electrons →
  μ = √(5(5+2)) = √35 ≈ 5.9 BM.

✅ Match: D → IV (Tetrahedral, 5.9 BM).

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Step 3: Final Matching Table

Complex/Species	Shape	Magnetic Moment	Hybridization	Match
[Ni(CO)₄]	Tetrahedral	0 BM (dia)	sp³	III
[Ni(CN)₄]²⁻	Square planar	0 BM (dia)	dsp²	II
[NiCl₄]²⁻	Tetrahedral	2.8 BM	sp³	I
[MnBr₄]²⁻	Tetrahedral	5.9 BM	sp³	IV

Final Answer:

A → III
B → II
C → I
D → IV

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Step 4: Why This Question is Important for JEE/NEET

This type of problem tests three concepts at once:

1. Crystal Field Theory (CFT) – to determine strong/weak ligands.

2. Magnetic moment calculation – to identify unpaired electrons.

3. Hybridization & geometry – to predict shapes.

Mastering these ensures you can quickly solve high-weightage inorganic chemistry questions.

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Step 5: Key Takeaways

• Strong field ligands (like CO, CN⁻) → low spin, pairing of electrons, diamagnetic complexes.

• Weak field ligands (like Cl⁻, Br⁻) → high spin, unpaired electrons, paramagnetic complexes.

• sp³ hybridization → tetrahedral; dsp² hybridization → square planar.

• Use μ = √(n(n+2)) for magnetic moment calculations.

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FAQ Section

Q1: Why is [Ni(CO)₄] tetrahedral and not square planar?
Because Ni is in the zero oxidation state and CO causes electron pairing in 3d, leading to sp³ hybridization.

Q2: How can I quickly identify strong and weak ligands?
Memorize the spectrochemical series:
CN⁻ > CO > NH₃ > H₂O > F⁻ > Cl⁻ > Br⁻ > I⁻.

Q3: What is the significance of magnetic moment in complexes?
It tells us about the number of unpaired electrons, which helps deduce the geometry and type of hybridization.

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Conclusion

By understanding ligand strength, oxidation states, and magnetic moments, you can quickly match complexes with their correct shapes and properties.
For this question, the correct matching is:

A → III, B → II, C → I, D → IV.

These concepts not only help in exams like JEE Main, JEE Advanced, NEET, but also build strong foundations in coordination chemistry.