Chemistry Inorganic - How to Solve: d- and f-Block Elements (KMnO₄, K₂Cr₂O₇, Lanthanide Contraction, Colour, Magnetism) – NEET UG Guide

How to Solve: d- and f-Block Elements (KMnO₄, K₂Cr₂O₇, Lanthanide Contraction, Colour, Magnetism) – NEET UG Guide

Introduction

Mastering d- and f-block elements unlocks 5-7 direct questions in NEET Chemistry—worth 20-28 marks—and helps you predict colour, oxidation states, and magnetic properties in real-world applications like MRI contrast agents, catalysts, and pigments. If you skip this, you’re leaving easy marks on the table.

WHAT YOU NEED TO KNOW FIRST

1. Electronic configuration (Aufbau principle, Hund’s rule, Pauli exclusion principle).
2. Oxidation states (common +2, +3, +6, +7 for d-block; +3 for f-block).
3. Magnetic properties (paramagnetism vs. diamagnetism, spin-only formula).

KEY TERMS & FORMULAS

1. Oxidation States

• KMnO₄ (Potassium permanganate):
• Mn oxidation state = +7 (MEMORISE THIS).
• Used in titrations (acidic medium: MnO₄⁻ → Mn²⁺).
• K₂Cr₂O₇ (Potassium dichromate):
• Cr oxidation state = +6 (MEMORISE THIS).
• Used in oxidation reactions (acidic medium: Cr₂O₇²⁻ → Cr³⁺).

2. Lanthanide Contraction

• Definition: Gradual decrease in atomic/ionic radii across the lanthanide series (La to Lu) due to poor shielding of 4f electrons.
• Effect:
• Similar sizes of 2nd and 3rd transition series (e.g., Zr ≈ Hf, Nb ≈ Ta).
• Higher density in 3rd series (e.g., Os > Ru).

3. Colour in Transition Metals

• Cause: d-d transitions (absorption of visible light when electrons jump between split d-orbitals).
• Formula (Crystal Field Splitting Energy, Δ):
• Δ = hc/λ (MEMORISE THIS)
  • h = Planck’s constant (6.626 × 10⁻³⁴ J s)
  • c = speed of light (3 × 10⁸ m/s)
  • λ = wavelength of absorbed light (nm)
• Complementary colour rule:
• If a complex absorbs blue (450 nm), it appears orange (MEMORISE THIS).

4. Magnetic Properties

• Paramagnetic: Unpaired electrons → attracted to magnetic field.
• Diamagnetic: All electrons paired → repelled by magnetic field.
• Spin-only magnetic moment (μ):
• μ = √[n(n+2)] BM (MEMORISE THIS)
  • n = number of unpaired electrons
  • BM = Bohr magneton (unit)

STEP-BY-STEP METHOD

Step 1: Identify the Element & Its Block

• d-block: Groups 3-12 (Sc to Zn, Y to Cd, La to Hg).
• f-block: Lanthanides (Ce to Lu) + Actinides (Th to Lr).

Step 2: Write Electronic Configuration

• d-block: [Noble gas] (n-1)d¹⁻¹⁰ ns¹⁻².
• Example: Cr (Z=24) → [Ar] 3d⁵ 4s¹ (exception: half-filled stability).
• f-block: [Xe] 4f¹⁻¹⁴ 5d⁰⁻¹ 6s² (lanthanides) or [Rn] 5f¹⁻¹⁴ 6d⁰⁻¹ 7s² (actinides).

Step 3: Determine Oxidation States

• Rule: Sum of oxidation states = charge on complex.
• Example (KMnO₄):
• K = +1, O = -2 (×4 = -8).
• Let Mn = x.
• +1 + x + (-8) = 0 → x = +7.

Step 4: Predict Colour

• Check d-electrons: If d⁰ or d¹⁰, colourless (no d-d transitions).
• If d¹-d⁹: Use spectrochemical series (strong field ligands → larger Δ → higher energy absorption).
• Example: [Co(H₂O)₆]²⁺ (pink) vs. [Co(NH₃)₆]³⁺ (yellow-orange).

Step 5: Calculate Magnetic Moment

• Count unpaired electrons (n):
• Example: Fe²⁺ (d⁶) → 4 unpaired electrons (high-spin).
• Apply spin-only formula:
• μ = √[4(4+2)] = √24 ≈ 4.9 BM.

Step 6: Apply Lanthanide Contraction (if asked)

• Compare sizes: 2nd vs. 3rd transition series.
• Example: Zr (160 pm) ≈ Hf (159 pm) (due to lanthanide contraction).

WORKED EXAMPLES

Example 1 – Basic: Oxidation State of Cr in K₂Cr₂O₇

Question: What is the oxidation state of Cr in K₂Cr₂O₇? Steps:
1. K = +1 (×2 = +2).
2. O = -2 (×7 = -14).
3. Let Cr = x (×2 = 2x).
4. +2 + 2x - 14 = 0 → 2x = +12 → x = +6. Answer: +6. What we did and why: Used the sum of oxidation states = charge on compound rule. K and O have fixed oxidation states, so we solved for Cr.

Example 2 – Medium: Colour of [Ti(H₂O)₆]³⁺

Question: Why is [Ti(H₂O)₆]³⁺ violet? Steps:
1. Ti³⁺ = d¹ (1 unpaired electron).
2. H₂O is a weak field ligand → small Δ.
3. Absorbs yellow-green light (500-550 nm).
4. Complementary colour = violet (MEMORISE THIS). Answer: Violet (due to d-d transition absorbing yellow-green light). What we did and why: Linked d-electron count → ligand field strength → absorbed wavelength → complementary colour.

Example 3 – Exam-Style: Magnetic Moment of Co³⁺ in [CoF₆]³⁻

Question: Calculate the magnetic moment of Co³⁺ in [CoF₆]³⁻ (F⁻ is a weak field ligand). Steps:
1. Co³⁺ = d⁶.
2. F⁻ is weak field → high-spin complex (4 unpaired electrons).
3. μ = √[n(n+2)] = √[4(6)] = √24 ≈ 4.9 BM. Answer: 4.9 BM. What we did and why: Determined spin state first (weak field → high-spin), then applied the spin-only formula.

COMMON MISTAKES

EXAM TRAPS

1-MINUTE RECAP (NIGHT BEFORE EXAM)

"Listen up—this is your 60-second cheat sheet for d- and f-block elements in NEET:

1. Oxidation states: KMnO₄ = +7, K₂Cr₂O₇ = +6. Memorise these.
2. Colour: d⁰/d¹⁰ = colourless. Others absorb light—complementary colour appears.
3. Magnetism: Count unpaired electrons → μ = √[n(n+2)] BM.
4. Lanthanide contraction: 2nd and 3rd series have similar sizes (Zr ≈ Hf).
5. Ligands matter: Strong field (CN⁻, CO) → low-spin, weak field (F⁻, H₂O) → high-spin.

Now go crush those 5-7 questions tomorrow!