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The d- and f-Block Elements

Chemistry · Class 12NCERT pp. 89–1142 sets × 25 questionsFree

Key facts and formulas for The d- and f-Block Elements, three worked sample questions, and two full practice sets. Every question has a hint, a reason for every option and a step-by-step solution, with a calculator or animation where one helps. English and Hinglish.

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Formula sheet for this chapter

IdeaRule or fact
Spin-only momentμ = √(n(n+2)) BM; n = 1,2,3,4,5 gives 1.73, 2.83, 3.87, 4.90, 5.92
Colour of ionsNeeds partly filled d (or f); d⁰ and d¹⁰ (Sc³⁺, Zn²⁺) and La³⁺, Lu³⁺ are colourless
Electrons per oxidantMnO₄⁻ (acid) 5 e⁻; MnO₄⁻ (neutral) 3 e⁻; Cr₂O₇²⁻ 6 e⁻
Equivalent weightEq. wt = molar mass ÷ electrons gained or lost
Titration molesmmol = mL × M; moles of reductant = moles of oxidant × n(oxidant) ÷ n(reductant)
Chromate and dichromate2CrO₄²⁻ + 2H⁺ ⇌ Cr₂O₇²⁻ + H₂O; Cr stays +6, yellow in base, orange in acid
ExceptionsCr 3d⁵4s¹, Cu 3d¹⁰4s¹, Gd 4f⁷5d¹6s²; Zn, Cd, Hg have low m.p. (d¹⁰)
Lanthanoid contractionPoor 4f shielding: size falls La to Lu; Zr and Hf alike; basic strength of Ln(OH)₃ falls
Oxides and Cu⁺Higher oxidation state gives more acidic oxide; Cu⁺ disproportionates: 2Cu⁺ → Cu²⁺ + Cu
ActinoidsMax states Th +4, Pa +5, U +6, Np +7; 5f, 6d, 7s close in energy

Three sample questions

Q1. Which of the following 3d ions has a spin-only magnetic moment of 2.83 BM (high spin, aqueous)?
  1. Cu²⁺
  2. Ni²⁺
  3. Fe²⁺
  4. Mn²⁺
Hint and solution

Hint: Use μ = √(n(n+2)) backwards: which n gives 2.83? Then count unpaired d electrons in each ion.

  1. μ = √(n(n+2)) = 2.83 BM — The spin-only formula links the moment to the number of unpaired electrons n.
  2. n = 2 gives √(2×4) = √8 = 2.83 — Trying n = 2 gives exactly 2.83 BM, so we need an ion with two unpaired electrons.
  3. Cu²⁺ d⁹:1, Ni²⁺ d⁸:2, Fe²⁺ d⁶:4, Mn²⁺ d⁵:5 — Count unpaired electrons in each d configuration. Only Ni²⁺ has two.
  4. Answer: Ni²⁺ — Ni²⁺ has 2 unpaired electrons and 2.83 BM. Option B.
Answer: B. 2.83 BM means n = 2: √(2×4) = √8. Ni²⁺ is d⁸ with 2 unpaired electrons. Cu²⁺ is 1, Fe²⁺ is 4 and Mn²⁺ is 5.
Q2. Which of the following has the highest spin-only magnetic moment?
  1. Fe²⁺
  2. Co²⁺
  3. Ni²⁺
  4. Mn²⁺
Hint and solution

Hint: Magnetic moment rises with the number of unpaired electrons. Write the d configuration of each ion.

  1. Fe²⁺ d⁶, Co²⁺ d⁷, Ni²⁺ d⁸, Mn²⁺ d⁵ — Write the d configuration of every ion from its atom after losing the 4s electrons.
  2. Unpaired: 4, 3, 2, 5 — Fill the five d orbitals singly first. d⁵ has one electron in each orbital.
  3. μ = √(n(n+2)): 4.90, 3.87, 2.83, 5.92 — Put each n into the formula. The largest n gives the largest moment.
  4. Mn²⁺ with 5.92 BM is highest — Mn²⁺ has the most unpaired electrons, so it wins. Option D.
Answer: D. Mn²⁺ is d⁵ (5 unpaired): 5.92 BM. Fe²⁺ 4.90, Co²⁺ 3.87, Ni²⁺ 2.83.
Q3. The magnetic moment of a transition metal ion is 3.87 BM. The number of unpaired electrons is
  1. 3
  2. 2
  3. 4
  4. 5
Hint and solution

Hint: Square 3.87 to get n(n+2), then find the whole number n that fits.

  1. μ² = n(n+2) = (3.87)² ≈ 15 — Square both sides of the spin-only formula to remove the root.
  2. n² + 2n − 15 = 0 — Rearrange into a quadratic equation in n.
  3. (n + 5)(n − 3) = 0, n = 3 — Only the positive root is physical, so n = 3.
  4. Check: √(3×5) = √15 = 3.87 — The check matches the given moment. Option A.
Answer: A. √(n(n+2)) = 3.87 gives n(n+2) = 15, so n = 3.

How to use this chapter

Read the NCERT pages first, then attempt Set 1 without the hints. Note every wrong answer in a mistake list, read the reason for each option, and retry Set 2 two days later. A topic is solid when you can say why each wrong option is wrong.

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