NEET The d- and f-Block Elements PYQs & Practice Questions

10 free NEET Chemistry questions on The d- and f-Block Elements, starting with previous year questions (2016–2026). Tap “Show answer” to check each one. No login needed.

Topics: Electronic Configuration of d-Block Elements · Oxidation States of Transition Metals · Magnetic Properties of Transition Metal Ions · Interstitial Compounds · Catalytic Properties of Transition Metals · Potassium Dichromate · Lanthanoid Contraction · Colour of Transition Metal Compounds · General Properties of d-Block Elements · Potassium Permanganate · Electronic Configuration of d-block elements · Magnetic Properties

1. Q1NEET 2026Thermal Decomposition of KMnO4

   The green paramagnetic species formed by heating KMnO4 at 513 K is:

   • A. Mn3O4
   • B. MnO
   • C. KO2
   • D. K2MnO4
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   Answer: C — KO2

   On heating, KMnO4 decomposes: 2KMnO4 → K2MnO4 + MnO2 + O2. The product K2MnO4 (potassium manganate, containing Mn in +6 oxidation state with one unpaired electron) is green in colour and paramagnetic.

2. Q2NEET 2026Electronic Configuration of Lanthanide Ions - Unpaired Electrons

   The lanthanide ion having four unpaired electrons is: (Given: Atomic numbers of Ce = 58, Nd = 60, Tb = 65 and Ho = 67)

   • A. Ce3+
   • B. Tb3+
   • C. Ho3+
   • D. Nd3+
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   Answer: B — Tb3+

   Ce3+: [Xe]4f1, 1 unpaired electron. Tb3+: [Xe]4f8, which has 6 unpaired electrons (7 orbitals, 8 electrons: 7 fill singly, 1 pairs up, leaving 6 unpaired). Ho3+: [Xe]4f10, which has 4 unpaired electrons (7 orbitals, 10 electrons: 7 fill singly, then 3 more pair up in 3 of those orbitals, leaving 7-3=4 unpaired). Nd3+: [Xe]4f3, 3 unpaired electrons. So Ho3+ has 4 unpaired electrons.

3. Q3NEET 2025Industrial Catalysts

   Match List - I with List - II. List-I: A. Haber process, B. Wacker oxidation, C. Wilkinson catalyst, D. Ziegler catalyst. List-II: I. Fe catalyst, II. PbCl2, III. [(PPh3)3RhCl], IV. TiCl4 with Al(CH3)3. Choose the correct answer from the options given below :

   • A. A-I, B-II, C-IV, D-III
   • B. A-II, B-III, C-I, D-IV
   • C. A-I, B-II, C-III, D-IV
   • D. A-I, B-IV, C-III, D-II
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   Answer: B — A-II, B-III, C-I, D-IV

   Haber process uses Fe catalyst (I). Wacker oxidation uses PdCl2 (listed as PbCl2 here, I believe a typo for PdCl2) (II). Wilkinson catalyst is [(PPh3)3RhCl] (III). Ziegler-Natta catalyst is TiCl4 with Al(CH3)3/AlEt3 (IV). So A-I, B-II, C-III, D-IV.

4. Q4NEET 2023Stability of Oxidation States in Aqueous Solution

   The stability of Cu2+ is more than Cu+ salts in aqueous solution due to:

   • A. enthalpy of atomization.
   • B. hydration energy.
   • C. second ionisation enthalpy.
   • D. first ionisation enthalpy.
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   Answer: A — enthalpy of atomization.

   In aqueous solution, the relative stability of Cu2+ over Cu+ is determined by hydration energy, which is proportional to the charge on the ion and inversely proportional to ionic size. Cu2+, having a higher charge, has a much higher hydration energy than Cu+, which compensates for the higher ionisation enthalpy required to remove the second electron, making Cu2+ more stable in aqueous solution.

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5. Q5NEET 2023Oxidation States and Properties of Transition Metal Oxides

   Which of the following statements are INCORRECT? A. All the transition metals except scandium form MO oxides which are ionic. B. The highest oxidation number corresponding to the group number in transition metal oxides is attained in Sc2O3 to Mn2O7. C. Basic character increases from V2O3 to V2O4 to V2O5. D. V2O4 dissolves in acids to give VO4^3- salts. E. CrO is basic but Cr2O3 is amphoteric. Choose the correct answer from the options given below.

   • A. B and D only
   • B. C and D only
   • C. B and C only
   • D. A and E only
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   Answer: A — B and D only

   Statement C is incorrect — basic character actually DECREASES from V2O3 to V2O4 to V2O5 (as the oxidation state and positive charge on V increases, effective nuclear charge and electronegativity increase, decreasing basic character). Statement D is incorrect — V2O4 dissolves in acids to give VO2+ salts, not VO4^3- salts. Statements A, B, and E are correct. So C and D only are incorrect.

6. Q6NEET 2022Ionization Enthalpy

   Gadolinium has a low value of third ionization enthalpy because of:

   • A. high exchange enthalpy
   • B. high electronegativity
   • C. high basic character
   • D. small size
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   Answer: —

   Gadolinium has configuration 4f7 5d1 6s2. After losing 2 electrons, Gd2+ has 4f7 5d1 configuration which has high exchange enthalpy due to the half-filled 4f7 stability, making the third ionization relatively easier (low third IE).

7. Q7NEET 2022Ionisation Enthalpy of Lanthanides

   Gadolinium has a low value of third ionisation enthalpy because of:

   • A. small size
   • B. high exchange enthalpy
   • C. high electronegativity
   • D. high basic character
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   Answer: A — small size

   Gadolinium has configuration [Xe]4f7 5d1 6s2. Removing the third electron gives Gd3+ with [Xe]4f7, a stable half-filled f-subshell configuration, which has high exchange energy, making the third ionisation comparatively easier (lower ionisation enthalpy).

8. Q8NEET 2022Ores and Composition

   Match List-I with List-II. List-I (Ores): (a) Haematite (b) Magnetite (c) Calamine (d) Kaolinite List-II (Composition): (i) Fe3O4 (ii) ZnCO3 (iii) Fe2O3 (iv) [Al2(OH)4Si2O5] Choose the correct answer from the options given below:

   • A. (a)-(i), (b)-(ii), (c)-(iii), (d)-(iv)
   • B. (a)-(iii), (b)-(i), (c)-(ii), (d)-(iv)
   • C. (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)
   • D. (a)-(i), (b)-(iii), (c)-(ii), (d)-(iv)
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   Answer: A — (a)-(i), (b)-(ii), (c)-(iii), (d)-(iv)

   Haematite is Fe2O3(iii), Magnetite is Fe3O4(i), Calamine is ZnCO3(ii), Kaolinite is [Al2(OH)4Si2O5](iv). So (a)-(iii),(b)-(i),(c)-(ii),(d)-(iv).

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9. Q9NEET 2021Lanthanoid Contraction

   Zr(Z=40) and Hf(Z=72) have similar atomic and ionic radii because of:

   • A. having similar chemical properties
   • B. belonging to same group
   • C. diagonal relationship
   • D. lanthanoid contraction
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   Answer: C — diagonal relationship

   Due to lanthanoid contraction, the sizes of 4d and 5d series elements of the same group become approximately equal, explaining Zr and Hf's similar radii.

10. Q10NEET 2020Properties of transition metals and interstitial compounds

    Identify the incorrect statement.

    • A. Cr2+ (d4) is a stronger reducing agent than Fe2+ (d6) in water.
    • B. The transition metals and their compounds are known for their catalytic activity due to their ability to adopt multiple oxidation states and to form complexes.
    • C. Interstitial compounds are those that are formed when small atoms like H, C or N are trapped inside the crystal lattices of metals.
    • D. The oxidation states of chromium in CrO4^2- and Cr2O7^2- are not the same.
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    Answer: C — Interstitial compounds are those that are formed when small atoms like H, C or N are trapped inside the crystal lattices of metals.

    Chromium is in the +6 oxidation state in both CrO4^2- (chromate) and Cr2O7^2- (dichromate), so the statement that their oxidation states are not the same is incorrect.