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NEET d- and f-Block Elements MCQs with answers

499 questions on d- and f-Block Elements for NEET 2027. Below are 15 to try; tap a question to see its answer and explanation.

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15 sample questions

easy

1. Which of the following is an example of an interstitial compound explicitly mentioned in the NCERT textbook under d-block elements?

  1. Fe₄[Fe(CN)₆]₃
  2. Fe₃C
  3. FeSO₄·7H₂O
  4. Fe₂(SO₄)₃
Show answer and explanation

Answer: (B) Fe₃C

According to NCERT, transition metals form interstitial compounds such as TiC, Mn₄N, Fe₃C, and VH₀.₅₆ when small atoms like C, H, or N are trapped inside their crystal lattices.

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2. Given below are two statements: Statement I: The maximum oxidation state shown by any 3d transition metal in its compounds is +7. Statement II: Manganese is able to exhibit the +7 oxidation state because it has seven electrons available in its valence shell (3d⁵ 4s²). In the light of the above statements, choose the most appropriate answer:

  1. Both Statement I and Statement II are correct
  2. Statement I is correct but Statement II is incorrect
  3. Statement I is incorrect but Statement II is correct
  4. Both Statement I and Statement II are incorrect
Show answer and explanation

Answer: (A) Both Statement I and Statement II are correct

In the 3d series, the highest oxidation state shown is +7, observed in manganese compounds such as Mn₂O₇ and KMnO₄. This occurs because all seven valence electrons (five in 3d and two in 4s) can participate in bonding.

hard

3. An interstitial alloy of iron and nitrogen is formed by nitrogen atoms occupying 12.5% of the octahedral voids of an fcc lattice of iron. What is the empirical formula of this interstitial nitride?

  1. Fe₈N
  2. Fe₄N
  3. Fe₆N
  4. Fe₂N
Show answer and explanation

Answer: (A) Fe₈N

In an fcc lattice, there are 4 iron atoms per unit cell and 4 octahedral interstitial voids. If 12.5% (which is 1/8) of the octahedral voids are occupied by nitrogen atoms, the number of nitrogen atoms per unit cell = 4 × (1/8) = 0.5. The ratio of Fe to N is 4 : 0.5 = 8 : 1, giving the empirical formula Fe₈N.

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4. Potassium permanganate is prepared from the mineral:

  1. Cinnabar
  2. Pyrolusite (MnO₂)
  3. Chromite
  4. Magnetite
Show answer and explanation

Answer: (B) Pyrolusite (MnO₂)

MnO₂ is fused with KOH and an oxidising agent like KNO₃ to give dark green K₂MnO₄, which disproportionates in neutral or acidic solution to permanganate.

NCERT Class 12

easy

5. What is the maximum oxidation state exhibited by actinoids, and in which pair of elements is it observed?

  1. +6, exhibited by U and Th
  2. +7, exhibited by Pa and U
  3. +8, exhibited by Pu and Am
  4. +7, exhibited by Np and Pu
Show answer and explanation

Answer: (D) +7, exhibited by Np and Pu

According to NCERT, neptunium (Np) and plutonium (Pu) exhibit the maximum oxidation state of +7 among the actinoids. After these elements, the maximum oxidation state decreases towards the end of the series.

hard

6. Although the standard reduction potential E° for Ce⁴⁺/Ce³⁺ is +1.74 V (favouring the reduction of Ce⁴⁺ to Ce³⁺), Ce⁴⁺ solutions can persist in aqueous media because:

  1. The rate of oxidation of water by Ce⁴⁺ is kinetically extremely slow.
  2. The reaction of Ce⁴⁺ with water is thermodynamically non-spontaneous.
  3. Ce⁴⁺ forms an insoluble precipitate of Ce(OH)₄ immediately.
  4. The E° value for water oxidation is more positive than +1.74 V.
Show answer and explanation

Answer: (A) The rate of oxidation of water by Ce⁴⁺ is kinetically extremely slow.

Even though Ce⁴⁺ is a powerful oxidising agent capable of oxidising water to O₂ (E° = +1.23 V), the reaction rate is kinetically very slow. Consequently, Ce(IV) solutions are sufficiently stable to be used as standard analytical reagents.

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7. In an acidic medium, how many moles of hydrogen sulfide (H₂S) are oxidized to elemental sulfur by 0.5 mole of potassium dichromate (K₂Cr₂O₇)?

  1. 0.5 mole
  2. 1.5 moles
  3. 1.0 mole
  4. 3.0 moles
Show answer and explanation

Answer: (B) 1.5 moles

The balanced redox reaction is: Cr₂O₇²⁻ + 14H⁺ + 3H₂S → 2Cr³⁺ + 3S + 7H₂O. From the stoichiometry, 1 mole of Cr₂O₇²⁻ oxidizes 3 moles of H₂S. Therefore, 0.5 mole of Cr₂O₇²⁻ oxidizes 0.5 × 3 = 1.5 moles of H₂S.

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8. What is the spin-only magnetic moment of the neptunyl ion, [NpO₂]²⁺ (Np, Z = 93)?

  1. 1.73 BM
  2. 0 BM
  3. 2.83 BM
  4. 3.87 BM
Show answer and explanation

Answer: (A) 1.73 BM

In [NpO₂]²⁺, neptunium is in the +6 oxidation state. For Np (Z = 93, [Rn] 5f⁴ 6d¹ 7s²), Np⁶⁺ has the electronic configuration [Rn] 5f¹. With n = 1 unpaired electron, the spin-only magnetic moment is μ = √(1(1 + 2)) = √3 ≈ 1.73 BM.

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9. The correct order of increasing magnetic moment (spin-only) for the following high-spin divalent ions is:

  1. Fe²⁺ < Co²⁺ < Mn²⁺ < Ni²⁺
  2. Co²⁺ < Ni²⁺ < Fe²⁺ < Mn²⁺
  3. Ni²⁺ < Co²⁺ < Fe²⁺ < Mn²⁺
  4. Mn²⁺ < Fe²⁺ < Co²⁺ < Ni²⁺
Show answer and explanation

Answer: (C) Ni²⁺ < Co²⁺ < Fe²⁺ < Mn²⁺

The number of unpaired electrons in these high-spin M²⁺ ions are: Ni²⁺ (3d⁸) has 2 unpaired electrons, Co²⁺ (3d⁷) has 3 unpaired electrons, Fe²⁺ (3d⁶) has 4 unpaired electrons, and Mn²⁺ (3d⁵) has 5 unpaired electrons. Since magnetic moment increases with the number of unpaired electrons, the correct order is Ni²⁺ < Co²⁺ < Fe²⁺ < Mn²⁺.

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10. According to the NCERT oxidation states table for actinoids, which of the following elements exhibits all oxidation states from +3 to +7?

  1. Thorium (Th)
  2. Plutonium (Pu)
  3. Uranium (U)
  4. Curium (Cm)
Show answer and explanation

Answer: (B) Plutonium (Pu)

Both Neptunium (Np) and Plutonium (Pu) exhibit oxidation states of +3, +4, +5, +6, and +7. Thorium shows predominantly +4, Uranium exhibits up to +6, and Curium exhibits +3 and +4.

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11. Consider the following statements regarding interstitial compounds of transition metals: Statement I: They are typically non-stoichiometric and cannot be assigned standard oxidation states to the metal atoms. Statement II: The entrapment of small non-metal atoms in the crystal lattice prevents the sliding of metal layers, significantly increasing hardness. Choose the correct option:

  1. Both Statement I and Statement II are correct
  2. Statement I is correct but Statement II is incorrect
  3. Statement I is incorrect but Statement II is correct
  4. Both Statement I and Statement II are incorrect
Show answer and explanation

Answer: (A) Both Statement I and Statement II are correct

Both statements are correct. Interstitial compounds are non-stoichiometric (e.g., TiH₁.₇, VH₀.₅₆) and lack traditional ionic or covalent bonding, which prevents assigning standard oxidation states. The presence of interstitial atoms hinders dislocation movement (sliding of metal planes), making them considerably harder than pure metals.

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12. Given below are two statements: Statement I: Like lanthanoids, actinoids show a regular decrease in ionic radii with increasing atomic number in a given oxidation state. Statement II: Unlike lanthanoids, the unevenness in the metallic radii of actinoids is far greater than that of lanthanoids due to irregularities in electronic configurations. In the light of the above statements, choose the most appropriate answer:

  1. Both Statement I and Statement II are correct.
  2. Statement I is correct but Statement II is incorrect.
  3. Statement I is incorrect but Statement II is correct.
  4. Both Statement I and Statement II are incorrect.
Show answer and explanation

Answer: (A) Both Statement I and Statement II are correct.

NCERT explicitly notes that the actinoid contraction causes a smooth decrease across M³⁺ and M⁴⁺ series, just like lanthanoids. However, the metallic radii show far greater irregularities than lanthanoids, largely due to variations in their electronic configurations and involvement of 5f electrons in bonding.

easy

13. Match List-I (3d Element) with List-II (Ground state valence shell electronic configuration) and choose the correct option: List-I: (A) Scandium (B) Chromium (C) Cobalt (D) Copper List-II: (I) 3d¹⁰ 4s¹ (II) 3d⁵ 4s¹ (III) 3d¹ 4s² (IV) 3d⁷ 4s²

  1. (A)-(III), (B)-(I), (C)-(IV), (D)-(II)
  2. (A)-(IV), (B)-(II), (C)-(III), (D)-(I)
  3. (A)-(III), (B)-(II), (C)-(IV), (D)-(I)
  4. (A)-(III), (B)-(IV), (C)-(II), (D)-(I)
Show answer and explanation

Answer: (C) (A)-(III), (B)-(II), (C)-(IV), (D)-(I)

The ground state valence configurations are: Sc (Z = 21) is 3d¹ 4s², Cr (Z = 24) is 3d⁵ 4s¹, Co (Z = 27) is 3d⁷ 4s², and Cu (Z = 29) is 3d¹⁰ 4s¹.

easy

14. Which of the following lanthanoid atoms has completely filled 4f orbitals in its ground-state neutral electronic configuration?

  1. Lutetium (Lu, Z = 71) only
  2. Ytterbium (Yb, Z = 70) only
  3. Both Ytterbium (Yb, Z = 70) and Lutetium (Lu, Z = 71)
  4. Terbium (Tb, Z = 65) and Thulium (Tm, Z = 69)
Show answer and explanation

Answer: (C) Both Ytterbium (Yb, Z = 70) and Lutetium (Lu, Z = 71)

The ground-state electronic configuration of Yb (Z = 70) is [Xe] 4f¹⁴ 6s² and that of Lu (Z = 71) is [Xe] 4f¹⁴ 5d¹ 6s². Hence, both Yb and Lu neutral atoms possess completely filled 4f¹⁴ subshells in their ground states.

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15. The ability of transition metals to form an exceptionally wide variety of complex compounds is primarily due to: (i) Comparatively smaller sizes of metal ions (ii) High ionic charges (iii) Availability of vacant d-orbitals of appropriate energy (iv) Completely filled d-subshells in their elemental states Select the correct combination of factors:

  1. (i), (ii) and (iv) only
  2. (i), (ii) and (iii) only
  3. (ii), (iii) and (iv) only
  4. (i) and (iii) only
Show answer and explanation

Answer: (B) (i), (ii) and (iii) only

NCERT states that transition metals form a large number of complex compounds due to: (1) comparatively smaller sizes of the metal ions, (2) high ionic charges, and (3) the availability of vacant d-orbitals for the accommodation of lone pairs of electrons donated by the ligands. Statement (iv) is incorrect since transition metals have incompletely filled d-subshells.

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