d- and f-Block Elements

13 NEET-UG practice questions on d- and f-Block Elements, part of Chemistry. The first 13 are below; the first 5 with answers and explanations (tap Show answer after you have chosen), the rest in Learn with a free account.

1. Question 1 · Hard

   Pyrolusite is fused with KOH in air to give a green compound X. On acidification, X gives a purple compound Y along with MnO₂. Identify X and Y.

   1. AX = K₂MnO₄; Y = Mn₂O₇
   2. BX = K₂MnO₄; Y = KMnO₄
   3. CX = KMnO₄; Y = K₂MnO₄
   4. DX = Mn₂O₇; Y = KMnO₄

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   Answer: B · X = K₂MnO₄; Y = KMnO₄

   2MnO₂ + 4KOH + O₂ → 2K₂MnO₄ + 2H₂O gives green manganate, with Mn at +6. In acid it disproportionates: 3MnO₄²⁻ + 4H⁺ → 2MnO₄⁻ + MnO₂ + 2H₂O, giving purple permanganate at +7 and MnO₂ at +4.

   • A: Mn₂O₇ is an oily covalent oxide, not the purple salt formed here.
   • C: Swaps the colours; permanganate is purple and manganate is green.
   • D: Mn₂O₇ is not formed by fusing with KOH.

   Remember this

   Manganate is green and stable in alkali; in neutral or acidic solution it disproportionates to permanganate and MnO₂.

2. Question 2 · Medium

   In neutral or faintly alkaline solution, potassium permanganate oxidises iodide. The manganese and iodine products of this reaction are

   1. AMnO₄²⁻ and I₂
   2. BMnO₂ and I₂
   3. CMn²⁺ and I₂
   4. DMnO₂ and IO₃⁻

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   Answer: D · MnO₂ and IO₃⁻

   2MnO₄⁻ + H₂O + I⁻ → 2MnO₂ + 2OH⁻ + IO₃⁻; the charge is −3 on each side. Mn goes from +7 to +4 and I from −1 to +5.

   • A: Manganate needs strongly alkaline medium, and it does not give I₂ here.
   • B: Right manganese product, but iodide is oxidised further, to iodate.
   • C: These are the products in acidic solution.

   Remember this

   Permanganate gives Mn²⁺ in acid and MnO₂ in neutral or faintly alkaline medium; here iodide becomes iodate.

3. Question 3 · Medium

   Arrange these lanthanoid ions in increasing order of ionic radius. A. Nd³⁺ B. Lu³⁺ C. La³⁺ D. Gd³⁺

   1. AD, B, A, C
   2. BB, D, A, C
   3. CB, A, D, C
   4. DC, A, D, B

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   Answer: B · B, D, A, C

   The atomic numbers rise from La (57) through Nd (60) and Gd (64) to Lu (71). Lanthanoid contraction makes the M³⁺ radius fall steadily along this sequence.

   • A: Places Gd³⁺ below Lu³⁺, ignoring where each lies in the series.
   • C: Places Nd³⁺ below Gd³⁺, though it comes earlier in the series.
   • D: Gives the decreasing order, the reverse of what is asked.

   Remember this

   M³⁺ radius falls smoothly across the lanthanoids, so La³⁺ is the largest and Lu³⁺ the smallest.

4. Question 4 · Medium

   Assertion A: Aqueous Cu²⁺ salts are coloured. Reason R: A neutral copper atom has the configuration [Ar] 3d¹⁰ 4s¹. Choose the correct answer.

   1. AA is correct but R is not correct
   2. BBoth A and R are correct and R is the correct explanation of A
   3. CA is not correct but R is correct
   4. DBoth A and R are correct but R is not the correct explanation of A

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   Answer: D · Both A and R are correct but R is not the correct explanation of A

   Both are true, but R describes the neutral atom with a full 3d subshell. The colour comes from Cu²⁺, which is d⁹ and has an unpaired electron available for d-d transitions.

   • A: R is true; it is the standard configuration of copper.
   • B: R describes the neutral atom; the colour comes from the d⁹ ion.
   • C: A is true; Cu²⁺ salts are blue in solution.

   Remember this

   Colour depends on the d count of the ion, so check the ion and not the neutral atom.

5. Question 5 · Easy

   Lanthanoid contraction is caused by

   1. Athe strong shielding by the 5d electrons
   2. Bimperfect shielding of one 4f electron by another
   3. Cthe rise in the number of 6s electrons along the series
   4. Dthe decrease in nuclear charge along the series

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   Answer: B · imperfect shielding of one 4f electron by another

   Along the series, nuclear charge rises by one with each step while the added 4f electron shields poorly. The effective nuclear charge therefore rises and the size falls.

   • A: It is the poor shielding by 4f electrons that matters, not strong 5d shielding.
   • C: The 6s electrons remain two throughout the series.
   • D: Nuclear charge rises, not falls, along the series.

   Remember this

   The poor shielding of 4f electrons also makes radii of the 4d and 5d series nearly equal.

6. Question 6 · Medium

   Match List-I with List-II. List-I (ion; Z of Ti = 22, V = 23, Cr = 24, Fe = 26) A. Ti³⁺ B. V³⁺ C. Cr³⁺ D. Fe²⁺ List-II (spin-only moment in BM) I. 3.87 II. 4.90 III. 1.73 IV. 2.83

   1. AA-III, B-IV, C-II, D-I
   2. BA-III, B-I, C-IV, D-II
   3. CA-III, B-IV, C-I, D-II
   4. DA-IV, B-III, C-I, D-II

   Answer and explanation in LearnFree account

7. Question 7 · Hard

   In acid solution, 30 mL of 0.02 M KMnO₄ just oxidises the Fe²⁺ in a sample. What volume of 0.01 M K₂Cr₂O₇ would be needed for an identical sample in acid solution?

   1. A100 mL
   2. B60 mL
   3. C30 mL
   4. D50 mL

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8. Question 8 · Hard

   Given below are two statements: Statement I: Because of lanthanoid contraction, the atomic radii of Zr and Hf differ widely. Statement II: Eu²⁺ acts as a strong oxidising agent because it changes readily to the common +3 state. Choose the correct answer.

   1. ABoth Statement I and Statement II are incorrect
   2. BStatement I is incorrect but Statement II is correct
   3. CBoth Statement I and Statement II are correct
   4. DStatement I is correct but Statement II is incorrect

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9. Question 9 · Medium

   Which ion is a strong oxidising agent in aqueous solution because its reduction gives a stable half-filled d⁵ configuration?

   1. ACr³⁺
   2. BV²⁺
   3. CMn³⁺
   4. DFe²⁺

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10. Question 10 · Medium

    Given below are two statements: Statement I: On acidifying a yellow chromate solution, orange dichromate forms with no change in the oxidation state of chromium. Statement II: In the Cr₂O₇²⁻ ion, two tetrahedra share one corner through a Cr–O–Cr link. Choose the correct answer.

    1. ABoth Statement I and Statement II are incorrect
    2. BStatement I is correct but Statement II is incorrect
    3. CBoth Statement I and Statement II are correct
    4. DStatement I is incorrect but Statement II is correct

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11. Question 11 · Easy

    Calculate the spin-only magnetic moment of Fe³⁺. (Z of Fe = 26.)

    1. A4.90 BM
    2. B3.87 BM
    3. C1.73 BM
    4. D5.92 BM

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12. Question 12 · Easy

    Atoms of two 3d-series elements hold only one 4s electron in the ground state, unlike the usual 4s² pattern. These two elements are

    1. ASc and Zn
    2. BMn and Fe
    3. CV and Ni
    4. DCr and Cu

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13. Question 13 · Medium

    Compared with lanthanoids, actinoids show a wider range of oxidation states mainly because

    1. Aactinoids have much smaller atomic radii than lanthanoids
    2. Bthe 5f, 6d and 7s levels have comparable energies
    3. Cactinoids have no electrons in their 5f subshell
    4. Dthe 4f electrons are lost more easily than the 6s electrons

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