Chemistry Inorganic - How to Solve: Coordination Compounds (IUPAC, Hybridisation, Isomerism, CFSE, Magnetic Moment, Colour) – NEET UG Guide

How to Solve: Coordination Compounds (IUPAC, Hybridisation, Isomerism, CFSE, Magnetic Moment, Colour) – NEET UG Guide

Introduction

Mastering coordination compounds unlocks 10-12 marks in NEET Chemistry—enough to push you from a 600 to a 650+ score. These compounds are in drugs, catalysts, and even hemoglobin, but on exam day, they test your ability to name, predict structure, and calculate magnetic properties under time pressure.

WHAT YOU NEED TO KNOW FIRST

1. Valency & Oxidation States – How to find the charge on a metal ion.
2. VSEPR Theory – Basic shapes of molecules (tetrahedral, square planar, octahedral).
3. Electronic Configuration – Filling of d-orbitals (Aufbau principle).

If you’re shaky on these, pause and review them now.

KEY TERMS & FORMULAS

1. IUPAC Nomenclature Rules (MEMORISE THIS)

• Order: [Ligand] → [Metal] → [Oxidation State in Roman numerals]
• Ligand Prefixes:
• 1 = mono- (usually omitted)
• 2 = di-
• 3 = tri-
• 4 = tetra-
• 5 = penta-
• 6 = hexa-
• Anionic Ligands End in "-o":
• Cl⁻ = chloro
• CN⁻ = cyano
• OH⁻ = hydroxo
• Neutral Ligands Keep Their Name:
• H₂O = aqua
• NH₃ = ammine
• CO = carbonyl
• Metal Naming:
• If complex is cationic, metal name stays the same.
• If complex is anionic, metal name ends in "-ate" (e.g., Fe → ferrate, Cu → cuprate).

2. Hybridisation & Geometry (MEMORISE THIS)

Note: - d²sp³ = Inner orbital complex (strong field ligands, low spin) - sp³d² = Outer orbital complex (weak field ligands, high spin)

3. Crystal Field Stabilisation Energy (CFSE) (MEMORISE THIS)

Formula: CFSE = (-0.4 × nₜ₂g + 0.6 × nₑg) × Δ₀ - nₜ₂g = Number of electrons in t₂g orbitals - nₑg = Number of electrons in eg orbitals - Δ₀ = Crystal field splitting energy (given or inferred from spectrochemical series)

Spectrochemical Series (MEMORISE ORDER): I⁻ < Br⁻ < Cl⁻ < F⁻ < OH⁻ < H₂O < NH₃ < en < NO₂⁻ < CN⁻ < CO (Weak field → Strong field)

4. Magnetic Moment (MEMORISE THIS)

Formula: μ = √[n(n+2)] BM - n = Number of unpaired electrons - BM = Bohr magnetons

5. Colour in Coordination Compounds

• Cause: d-d transitions (electrons absorb light and jump from t₂g → eg)
• Rule: The complementary colour of the absorbed light is observed.
• Absorbs blue → Appears orange
• Absorbs red → Appears green

STEP-BY-STEP METHOD

Step 1: Identify the Central Metal & Its Oxidation State

• Find the charge on the complex.
• Subtract the total charge of ligands from the complex charge to get the metal’s oxidation state.

Step 2: Name the Complex (IUPAC)

1. List ligands in alphabetical order (ignore prefixes like di-, tri-).
2. Name the metal (use "-ate" if complex is anionic).
3. Add oxidation state in Roman numerals in parentheses.

Step 3: Determine Hybridisation & Geometry

1. Count coordination number (number of ligands attached to metal).
2. Check if ligands are strong/weak field (use spectrochemical series).
3. Write electronic configuration of metal ion.
4. Pair electrons if strong field ligand (low spin).
5. Determine hybridisation based on coordination number and inner/outer orbitals.

Step 4: Find CFSE (If Required)

1. Split d-orbitals into t₂g and eg (for octahedral).
2. Fill electrons following Hund’s rule (pair only if strong field).
3. Apply CFSE formula (-0.4 for t₂g, +0.6 for eg).

Step 5: Calculate Magnetic Moment

1. Count unpaired electrons from electronic configuration.
2. Apply μ = √[n(n+2)] BM.

Step 6: Predict Colour (If Asked)

1. Find Δ₀ (from spectrochemical series or given data).
2. Determine absorbed wavelength (Δ₀ = hc/λ).
3. State observed colour (complementary to absorbed colour).

WORKED EXAMPLES

Example 1 – Basic: [Co(NH₃)₆]Cl₃

Question: Name the complex, find hybridisation, magnetic moment, and CFSE.

Step 1: Oxidation State - Complex charge = +3 (Cl₃⁻ balances it). - NH₃ is neutral → Co oxidation state = +3.

Step 2: IUPAC Name - Ligand: ammine (NH₃) - Metal: cobalt (cationic → no "-ate") - Name: Hexaamminecobalt(III) chloride

Step 3: Hybridisation & Geometry - Coordination number = 6 → Octahedral. - NH₃ is a strong field ligand → Low spin. - Co³⁺ electronic config: [Ar] 3d⁶ - Strong field → All 6 electrons pair in t₂g. - Hybridisation: d²sp³ (inner orbital, low spin).

Step 4: CFSE - t₂g⁶ eg⁰ → CFSE = (-0.4 × 6 + 0.6 × 0) × Δ₀ = -2.4 Δ₀

Step 5: Magnetic Moment - Unpaired electrons = 0 → μ = 0 BM (diamagnetic).

What we did and why: - We named it correctly by following IUPAC rules. - We used the spectrochemical series to decide low/high spin. - CFSE and magnetic moment depend on electron pairing, which changes with ligand strength.

Example 2 – Medium: [Ni(CN)₄]²⁻

Question: Name the complex, find hybridisation, geometry, and magnetic moment.

Step 1: Oxidation State - Complex charge = -2. - CN⁻ charge = -1 → 4 CN⁻ = -4. - Ni oxidation state = +2 (since -2 = Ni + (-4)).

Step 2: IUPAC Name - Ligand: cyano (CN⁻) - Metal: nickel (anionic → "-ate") - Name: Tetracyanonickelate(II) ion

Step 3: Hybridisation & Geometry - Coordination number = 4. - CN⁻ is a strong field ligand → Square planar (dsp²). - Ni²⁺ electronic config: [Ar] 3d⁸ - Strong field → All electrons pair → No unpaired electrons. - Hybridisation: dsp² (square planar).

Step 4: Magnetic Moment - Unpaired electrons = 0 → μ = 0 BM (diamagnetic).

What we did and why: - We recognised that CN⁻ forces square planar geometry for d⁸ metals. - Magnetic moment is zero because all electrons are paired in a strong field.

Example 3 – Exam-Style: [Fe(H₂O)₆]²⁺ vs [Fe(CN)₆]⁴⁻

Question: Compare their hybridisation, magnetic moment, and colour.

Step 1: Oxidation State (Both Fe²⁺) - [Fe(H₂O)₆]²⁺ → Fe²⁺ (H₂O neutral) - [Fe(CN)₆]⁴⁻ → Fe²⁺ (CN⁻ = -1 each → 6 × -1 = -6; complex = -4 → Fe = +2)

Step 2: Hybridisation & Geometry - Both octahedral (CN = 6). - H₂O (weak field) → High spin (sp³d²) - Fe²⁺: [Ar] 3d⁶ → 4 unpaired electrons (t₂g⁴ eg²). - CN⁻ (strong field) → Low spin (d²sp³) - Fe²⁺: [Ar] 3d⁶ → 0 unpaired electrons (t₂g⁶ eg⁰).

Step 3: Magnetic Moment - [Fe(H₂O)₆]²⁺ → 4 unpaired → μ = √[4(4+2)] = 4.9 BM - [Fe(CN)₆]⁴⁻ → 0 unpaired → μ = 0 BM

Step 4: Colour - H₂O (weak field) → Small Δ₀ → Absorbs red → Appears green. - CN⁻ (strong field) → Large Δ₀ → Absorbs blue → Appears orange.

What we did and why: - We compared two complexes of the same metal but different ligands. - Ligand strength changes everything—hybridisation, magnetic moment, and colour. - Colour depends on Δ₀, which is larger for strong field ligands.

COMMON MISTAKES

EXAM TRAPS

1-MINUTE RECAP (Night Before Exam)

"Listen up—this is your 60-second cheat sheet for coordination compounds in NEET:

1. Naming: Ligands first (alphabetical), metal second, oxidation state in Roman numerals. Anionic complex? Metal ends in -ate.
2. Hybridisation: Count ligands (CN). 4 ligands? Could be sp³ (tetrahedral) or dsp² (square planar). 6 ligands? d²sp³ (low spin) or sp³d² (high spin).
3. Magnetic moment: Count unpaired electrons, plug into μ = √[n(n+2)]. No unpaired? Diamagnetic (μ = 0).
4. CFSE: Strong field? Electrons pair in t₂g. Weak field? Fill eg first. Formula: -0.4 × t₂g + 0.6 × eg.
5. Colour: Strong field → large Δ₀ → absorbs blue → looks orange. Weak field → small Δ₀ → absorbs red → looks green.

Last tip: If you see CN⁻ or CO, assume strong field. If you see H₂O or Cl⁻, assume weak field. Now go crush those 10 marks!