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CBSE · Class 12 · Chemistry

The d- and f-Block Elements

Standard difficulty · 20 questions · 40 marks

Name: Date: Score: ____ / 40

Show your working for Sections B and C. Answers are on the last page — try every question first!

Section A — Choose the correct answer8 × 1 = 8 marks
1.
Transition elements have partly filled:
  1. p orbitals
  2. f orbitals only
  3. s orbitals only
  4. d orbitals in the atom or a common ion
2.
The ground-state electronic configuration of chromium (Cr, Z = 24) is:
  1. [Ar] 3d5 4s1
  2. [Ar] 3d6
  3. [Ar] 4s2 4p4
  4. [Ar] 3d3 4s2
3.
The electronic configuration of Ni2+ (Z of Ni = 28) is:
  1. [Ar] 3d7 4s1
  2. [Ar] 3d9
  3. [Ar] 3d8
  4. [Ar] 3d6 4s2
4.
The spin-only magnetic moment of Ti4+ (Z of Ti = 22) is:
  1. √8 ≈ 2.83 BM
  2. √3 ≈ 1.73 BM
  3. 0 BM (diamagnetic)
  4. √35 ≈ 5.92 BM
5.
The oxidation state of Mn in MnO4− is:
  1. +7
  2. +6
  3. +8
  4. +5
6.
Which of these ions is colourless in aqueous solution?
  1. Cr3+
  2. Ti4+
  3. Mn2+
  4. Cr2+
7.
Which has the larger spin-only magnetic moment, Fe2+ or Ti3+?
  1. Fe2+
  2. neither; both are diamagnetic
  3. Ti3+
  4. both have the same moment
8.
Assertion (A): Aqueous solutions of V2+ salts are coloured.
Reason (R): All transition-metal ions are coloured, whatever their d-electron count.
  1. Both A and R are true, and R is the correct explanation of A.
  2. Both A and R are true, but R is not the correct explanation of A.
  3. A is true, but R is false.
  4. A is false, but R is true.
Section B — Short answer8 × 2 = 16 marks
9.
Why are transition metal ions usually coloured?
10.
Write the electronic configuration of Sc3+ (Z of Sc = 21), and find the number of unpaired electrons and the spin-only magnetic moment.
11.
Find the oxidation state of Fe in FeO42−, and the number of 3d electrons the Fe atom has in that state.
12.
How many moles of oxalic acid are oxidised by 0.015 mol of KMnO4 in acid solution?
13.
A metal ion has a spin-only magnetic moment of 4.90 BM. How many unpaired electrons does it have? Name one ion that fits.
14.
How many 4f electrons does Lu3+ (Z = 71) have, and how many of them are unpaired?
15.
45 mL of 0.16 M KMnO4 solution reacts completely with H2O2 in acid. How many moles of H2O2 react?
16.
Compare Ti4+ and Co3+: give the d-electron configuration and spin-only magnetic moment of each, and say which would be coloured.
Section C — Word problems4 × 4 = 16 marks
17.
Describe the preparation of potassium dichromate from chromite ore, with equations.
18.
For each of Cu2+, Cu+, Co2+: write the 3d configuration, the number of unpaired electrons, the spin-only magnetic moment, and say whether its aqueous solution is coloured.
19.
25 mL of a oxalic acid solution needs 37 mL of 0.07 M KMnO4 for complete reaction in acid. Write the ionic equation and find the molarity of the oxalic acid solution.
20.
Compare Lu3+ (Z = 71) and Gd3+ (Z = 64): give the number of 4f electrons and unpaired electrons in each, and say which ion is smaller and why.
AmbifulWorksheets
Practice that never repeats
CBSE · Class 12 · Chemistry

Answer key

The d- and f-Block Elements · AW-12C-DFB-2026-V1 · For parents — fold back or tear off before handing the sheet to your child.

QAnswerWorking
1(d) d orbitals in the atom or a common ion—
2(a) [Ar] 3d5 4s1a half-filled or filled 3d set is extra stable
3(c) [Ar] 3d84s electrons are removed before 3d
4(c) 0 BM (diamagnetic)3d0: 0 unpaired electrons
5(a) +7O is −2, K is +1; the sum equals the charge
6(b) Ti4+Ti4+ is 3d0: no d–d transition is possible
7(a) Fe2+Fe2+: 4 unpaired; Ti3+: 1 unpaired
8(c) A is true, but R is false.A is true: V2+ is 3d3. R is false.
9Their partly filled d orbitals split in the presence of ligands, and an electron absorbs visible light to jump to a higher d level (d–d transition); the colour seen is the complementary colour.—
10[Ar] 3d0; 0 unpaired; 0 BM (diamagnetic)μ = √(n(n + 2)) BM
11+6; 2 (3d2)Z of Fe = 26; d electrons = Z − 18 − oxidation state
120.0375 mol2MnO4− + 5C2O42− + 16H+ → 2Mn2+ + 10CO2 + 8H2O
134 unpaired; e.g. Cr2+ or Mn3+ or Fe2+ or Co3+√(n(n + 2)) = 4.90 ⇒ n = 4
144f14: 14 electrons; 0 unpairedLn3+ has [Xe] 4fⁿ with n = Z − 57
150.018 molmoles of KMnO4 = 0.16 × 45/1000; 2MnO4− + 5H2O2 + 6H+ → 2Mn2+ + 5O2 + 8H2O
16Ti4+: 3d0, 0 BM (diamagnetic), colourless; Co3+: 3d6, √24 ≈ 4.90 BM, colouredd0 and d10 ions are colourless
17Chromite is fused with sodium carbonate in air: 4FeCr2O4 + 8Na2CO3 + 7O2 → 8Na2CrO4 + 2Fe2O3 + 8CO2. The yellow sodium chromate is acidified: 2Na2CrO4 + 2H+ → Na2Cr2O7 + 2Na+ + H2O. Adding KCl gives the less soluble orange K2Cr2O7: Na2Cr2O7 + 2KCl → K2Cr2O7 + 2NaCl.—
18Cu2+: 3d9, 1 unpaired, √3 ≈ 1.73 BM, coloured; Cu+: 3d10, 0 unpaired, 0 BM (diamagnetic), colourless; Co2+: 3d7, 3 unpaired, √15 ≈ 3.87 BM, colouredZ: Cu 29, Cu 29, Co 27
192MnO4− + 5C2O42− + 16H+ → 2Mn2+ + 10CO2 + 8H2O; 0.259 Mmoles of oxalic acid = 2.5 × moles of KMnO4
20Lu3+: 4f14, 0 unpaired; Gd3+: 4f7, 7 unpaired; Lu3+ is smaller because of the lanthanoid contraction (poor shielding by 4f electrons as Z rises)Ln3+: 4fⁿ, n = Z − 57