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Actinoids: configuration and oxidation states: NEET MCQs with answers

89 questions on this topic of d- and f-Block Elements (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. Which of the following actinoids exhibits a stable +2 oxidation state due to a completely filled 5f¹⁴ electronic configuration?

  1. Fermium (Fm, Z = 100)
  2. Nobelium (No, Z = 102)
  3. Mendelevium (Md, Z = 101)
  4. Lawrencium (Lr, Z = 103)
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Answer: (B) Nobelium (No, Z = 102)

Nobelium (Z = 102) has the ground-state configuration [Rn] 5f¹⁴ 7s². Loss of two 7s electrons yields No²⁺ with a fully-filled, exceptionally stable [Rn] 5f¹⁴ configuration.

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2. The actinoid contraction is greater than the lanthanoid contraction because:

  1. Actinoids have fewer electrons
  2. 4f orbitals are larger
  3. Actinoids are lighter
  4. 5f electrons shield more poorly than 4f electrons
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Answer: (D) 5f electrons shield more poorly than 4f electrons

5f electrons provide poorer shielding than 4f electrons, so the contraction from element to element is greater in the actinoid series.

NCERT Class 12

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3. Given below are two statements: Statement I: Americium (Am, Z = 95) exhibits oxidation states ranging from +2 to +6. Statement II: Californium (Cf, Z = 98) shows +2, +3, and +4 oxidation states in its compounds. In light of the above statements, choose the correct 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
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Answer: (A) Both Statement I and Statement II are correct

According to NCERT Table 8.11, Americium exhibits +2, +3, +4, +5, and +6 oxidation states, with +3 being the most stable. Californium exhibits +2, +3, and +4 oxidation states, with +3 being the principal state.

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4. Match List-I with List-II to identify the correct ground-state electronic configurations: List-I (Element): (A) Protactinium (Pa, Z = 91) (B) Uranium (U, Z = 92) (C) Neptunium (Np, Z = 93) (D) Plutonium (Pu, Z = 94) List-II (Outer configuration): (I) 5f⁴ 6d¹ 7s² (II) 5f² 6d¹ 7s² (III) 5f⁶ 7s² (IV) 5f³ 6d¹ 7s² Choose the correct answer:

  1. A-II, B-IV, C-III, D-I
  2. A-II, B-I, C-IV, D-III
  3. A-IV, B-II, C-I, D-III
  4. A-II, B-IV, C-I, D-III
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Answer: (D) A-II, B-IV, C-I, D-III

Protactinium (Z = 91) is [Rn] 5f² 6d¹ 7s² (II). Uranium (Z = 92) is [Rn] 5f³ 6d¹ 7s² (IV). Neptunium (Z = 93) is [Rn] 5f⁴ 6d¹ 7s² (I). Plutonium (Z = 94) is [Rn] 5f⁶ 7s² (III). Thus, the correct sequence is A-II, B-IV, C-I, D-III.

easy

5. What is the ground-state electronic configuration of the protactinium atom (Pa, Z = 91)?

  1. [Rn] 5f³ 6d² 7s⁰
  2. [Rn] 5f¹ 6d² 7s²
  3. [Rn] 5f³ 7s²
  4. [Rn] 5f² 6d¹ 7s²
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Answer: (D) [Rn] 5f² 6d¹ 7s²

According to NCERT, protactinium (Pa, Z = 91) has the ground-state electronic configuration [Rn] 5f² 6d¹ 7s². The 5f and 6d orbitals are close in energy in early actinoids, leading to occupancy in both subshells.

hard

6. Given below are two statements: Statement I: In the actinoid series, the 5f electrons are more effectively shielded from nuclear charge than the 4f electrons in lanthanoids. Statement II: 5f orbitals extend further in space beyond the 6s and 6p orbitals than 4f orbitals do beyond 5s and 5p orbitals. In light of the above statements, choose the correct 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
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Answer: (C) Statement I is incorrect but Statement II is correct

Statement I is incorrect because 5f electrons have poorer shielding power than 4f electrons. Statement II is correct because 5f orbitals have a larger radial extent and extend further in space, enabling participation in bonding.

easy

7. What is the electronic configuration of the tripositive curium ion, Cm³⁺ (Z = 96)?

  1. [Rn] 5f⁶ 6d¹
  2. [Rn] 5f⁷
  3. [Rn] 5f⁸
  4. [Rn] 5f⁵ 6d²
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Answer: (B) [Rn] 5f⁷

Neutral curium (Z = 96) has the ground-state configuration [Rn] 5f⁷ 6d¹ 7s². Ionization to form Cm³⁺ involves the loss of two 7s electrons and one 6d electron, yielding the half-filled stable configuration [Rn] 5f⁷.

easy

8. Which actinoid exhibits strictly only the +3 oxidation state, showing no other oxidation states in its compounds?

  1. Thorium
  2. Protactinium
  3. Lawrencium
  4. Americium
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Answer: (C) Lawrencium

Lawrencium (Lr, Z = 103) is at the very end of the actinoid series where the 5f subshell is fully filled (5f¹⁴). It exclusively exhibits the +3 oxidation state, similar to lutetium in the lanthanoid series.

hard

9. Which of the following elements has its maximum oxidation state numerically equal to its total number of 5f, 6d, and 7s valence electrons outside the radon core?

  1. Americium (Am)
  2. Neptunium (Np)
  3. Curium (Cm)
  4. Californium (Cf)
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Answer: (B) Neptunium (Np)

Np (Z = 93) has the outer electronic configuration 5f⁴ 6d¹ 7s², giving 4 + 1 + 2 = 7 valence electrons. It exhibits a maximum oxidation state of +7, where all 7 valence electrons participate in bonding.

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10. Which of the following actinoid ions has a completely filled 5f subshell and exhibits a stable +2 oxidation state in solution?

  1. Md²⁺ (Mendelevium, Z = 101)
  2. No²⁺ (Nobelium, Z = 102)
  3. Lr²⁺ (Lawrencium, Z = 103)
  4. Fm²⁺ (Fermium, Z = 100)
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Answer: (B) No²⁺ (Nobelium, Z = 102)

Nobelium (Z = 102) has the ground-state configuration [Rn] 5f¹⁴ 7s². The loss of two 7s electrons yields the No²⁺ ion with the configuration [Rn] 5f¹⁴, which possesses an exceptionally stable completely filled 5f subshell. Hence, No²⁺ is readily formed and stable.

hard

11. Match List-I (Actinoid Ion) with List-II (Number of 5f electrons): List-I: (A) U⁴⁺ (B) Np³⁺ (C) Cm³⁺ (D) Am³⁺ List-II: (I) 5f⁴ (II) 5f² (III) 5f⁶ (IV) 5f⁷ Choose the correct match from the options given below:

  1. (A)-(II), (B)-(III), (C)-(IV), (D)-(I)
  2. (A)-(I), (B)-(II), (C)-(III), (D)-(IV)
  3. (A)-(II), (B)-(I), (C)-(IV), (D)-(III)
  4. (A)-(IV), (B)-(I), (C)-(II), (D)-(III)
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Answer: (C) (A)-(II), (B)-(I), (C)-(IV), (D)-(III)

For U (Z=92, [Rn] 5f³ 6d¹ 7s²), U⁴⁺ is 5f². For Np (Z=93, [Rn] 5f⁴ 6d¹ 7s²), Np³⁺ is 5f⁴. For Cm (Z=96, [Rn] 5f⁷ 6d¹ 7s²), Cm³⁺ is 5f⁷. For Am (Z=95, [Rn] 5f⁷ 7s²), Am³⁺ is 5f⁶. Therefore, the matching is (A)-(II), (B)-(I), (C)-(IV), (D)-(III).

easy

12. Given below are two statements: Statement I: The electronic configurations of actinoids show irregularities because the 5f, 6d, and 7s levels are of comparable energies. Statement II: The 5f electrons can take part in bonding to a far greater extent than the 4f electrons because 5f orbitals extend further into space. In the light of the above statements, choose the correct 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.
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Answer: (A) Both Statement I and Statement II are correct.

Both statements are direct NCERT facts: 5f, 6d, and 7s orbitals are of comparable energy in actinoids causing configuration irregularities, and 5f orbitals extend spatially farther than 4f orbitals, allowing 5f electrons to participate more extensively in bonding.

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