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s-, p-, d- and f-block classification: NEET MCQs with answers

50 questions on this topic of Classification of Elements and Periodicity in Properties (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. Identify the element that does NOT belong to the same block as the other three:

  1. Thallium (Z = 81)
  2. Bismuth (Z = 83)
  3. Thorium (Z = 90)
  4. Polonium (Z = 84)
Show answer and explanation

Answer: (C) Thorium (Z = 90)

Thallium (Group 13), Bismuth (Group 15), and Polonium (Group 16) all belong to the p-block. Thorium (Z = 90) is an actinoid belonging to the f-block.

medium

2. An element with configuration [Ar] 3d¹⁰ 4s² 4p⁵ is placed in:

  1. p-block, group 15
  2. p-block, group 17
  3. s-block, group 2
  4. d-block, group 12
Show answer and explanation

Answer: (B) p-block, group 17

The last electron enters a 4p orbital with 7 valence electrons (4s² 4p⁵), so it is a p-block element in group 17 (bromine).

NCERT Class 11

hard

3. Assertion (A): All elements belonging to the f-block are placed formally in Group 3 of the modern periodic table. Reason (R): Lanthanoids (Z = 57 to 71) and actinoids (Z = 89 to 103) exhibit chemical properties closely resembling group 3 elements Sc and Y.

  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
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Answer: (C) (A) is true but (R) is false

Assertion is true; f-block elements fall conceptually in Group 3 between groups 2 and 4 to preserve periodic table width. However, Reason is false: Lanthanoids span Z = 58 to 71 (not 57 to 71) and Actinoids span Z = 90 to 103 (not 89 to 103), as La (Z = 57) and Ac (Z = 89) are d-block elements.

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4. An element has the ground state configuration [Rn] 5f¹⁴ 6d⁴ 7s². The period, group, and block of this element are respectively:

  1. Period 7, Group 6, f-block
  2. Period 7, Group 4, d-block
  3. Period 7, Group 6, d-block
  4. Period 6, Group 6, d-block
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Answer: (C) Period 7, Group 6, d-block

The principal quantum number of the outermost shell is n = 7, so it belongs to Period 7. The 5f subshell is completely filled (5f¹⁴) and the last differentiating electron enters the 6d subshell (6d⁴), placing it in the d-block (specifically, Seaborgium, Z = 106). Group number for d-block = ns + (n-1)d = 2 + 4 = 6.

easy

5. The general outer electronic configuration of p-block elements is:

  1. ns¹⁻²
  2. (n-2)f¹⁻¹⁴ (n-1)d⁰⁻¹ ns²
  3. ns² np¹⁻⁶
  4. (n-1)d¹⁻¹⁰ ns⁰⁻²
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Answer: (C) ns² np¹⁻⁶

p-block elements (groups 13 to 18) have ns² np¹⁻⁶ outer configurations, with the last electron entering a p orbital.

NCERT Class 11

hard

6. The total number of s-block, p-block, d-block, and f-block elements present in the sixth period (Period 6) of the periodic table are, respectively:

  1. 2, 6, 10, 14
  2. 2, 8, 10, 12
  3. 2, 6, 8, 16
  4. 2, 6, 14, 10
Show answer and explanation

Answer: (A) 2, 6, 10, 14

Period 6 fills the 6s, 4f, 5d, and 6p subshells. The capacities are 2 electrons for 6s (s-block: Cs, Ba), 14 electrons for 4f (f-block: Ce to Lu), 10 electrons for 5d (d-block: La, Hf to Hg), and 6 electrons for 6p (p-block: Tl to Rn). Thus, the counts are 2, 6, 10, and 14.

easy

7. An element belongs to the 5th period and the 15th group of the modern periodic table. What is the total number of unpaired electrons in the ground state of a neutral atom of this element?

  1. 1
  2. 3
  3. 2
  4. 5
Show answer and explanation

Answer: (B) 3

Group 15 elements have the general valence shell configuration ns² np³. For period 5, the valence shell configuration is 5s² 5p³. By Hund's rule of maximum multiplicity, the three electrons in the 5p subshell occupy the three 5p orbitals individually with parallel spins, resulting in exactly 3 unpaired electrons.

easy

8. An element has atomic number Z = 56. To which group and block does this element belong in the modern periodic table?

  1. Group 2, s-block
  2. Group 14, p-block
  3. Group 6, d-block
  4. Group 3, f-block
Show answer and explanation

Answer: (A) Group 2, s-block

Barium has Z = 56, with the electronic configuration [Xe] 6s². Since the differentiating electron enters the 6s orbital and the valence shell has 2 electrons, it belongs to Group 2 and the s-block.

easy

9. The correct IUPAC temporary symbol and group number for the undiscovered transactinide element with atomic number 120 are respectively:

  1. Ubn and Group 2
  2. Ubn and Group 12
  3. Ubn and Group 1
  4. Ube and Group 2
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Answer: (A) Ubn and Group 2

For Z = 120, the IUPAC name is unbinilium (symbol Ubn). Following the filling order after Oganesson (Z = 118, completes period 7, group 18), Z = 119 will occupy 8s¹ (group 1) and Z = 120 will occupy 8s² (group 2, s-block) of period 8.

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10. The general valence shell electronic configuration of an f-block element is:

  1. (n-2)f⁰⁻¹⁴ (n-1)d⁰⁻¹ ns²
  2. (n-2)f¹⁻¹⁴ (n-1)d⁰⁻² ns²
  3. (n-2)f¹⁻¹⁴ (n-1)d⁰⁻¹ ns²
  4. (n-2)f¹⁻¹⁴ (n-1)d¹⁻² ns²
Show answer and explanation

Answer: (C) (n-2)f¹⁻¹⁴ (n-1)d⁰⁻¹ ns²

According to NCERT, the outer electronic configuration of f-block elements (inner-transition elements) is generally represented as (n-2)f¹⁻¹⁴ (n-1)d⁰⁻¹ ns², where the penultimate d-orbital has either 0 or 1 electron and the valence s-orbital is filled.

easy

11. Consider the following elements with their atomic numbers: P: Z = 37, Q: Z = 52, R: Z = 72, S: Z = 98 Which of the following correctly matches each element to its respective block?

  1. P - s-block, Q - p-block, R - f-block, S - d-block
  2. P - p-block, Q - s-block, R - d-block, S - f-block
  3. P - s-block, Q - d-block, R - p-block, S - f-block
  4. P - s-block, Q - p-block, R - d-block, S - f-block
Show answer and explanation

Answer: (D) P - s-block, Q - p-block, R - d-block, S - f-block

P (Z = 37, Rb) is [Kr] 5s¹ (s-block). Q (Z = 52, Te) is [Kr] 4d¹⁰ 5s² 5p⁴ (p-block). R (Z = 72, Hf) is [Xe] 4f¹⁴ 5d² 6s² (d-block, group 4). S (Z = 98, Cf) is an actinoid with [Rn] 5f¹⁰ 7s² (f-block).

hard

12. An element with ground-state electron configuration [Kr] 4d¹⁰ 5s⁰ belongs to which group and block of the modern periodic table?

  1. Group 10, d-block
  2. Group 12, d-block
  3. Group 8, d-block
  4. Group 18, p-block
Show answer and explanation

Answer: (A) Group 10, d-block

The configuration [Kr] 4d¹⁰ 5s⁰ corresponds to Palladium (Z = 46). Since the last filled shell is 4d, it belongs to the d-block. The group number for d-block elements is (number of (n-1)d electrons + number of ns electrons) = 10 + 0 = 10.

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