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NEET Classification of Elements and Periodicity in Properties MCQs with answers

219 questions on Classification of Elements and Periodicity in Properties 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. The IUPAC temporary symbol for an element with atomic number 119 is:

  1. Uen
  2. Uun
  3. Une
  4. Uue
Show answer and explanation

Answer: (D) Uue

According to IUPAC rules, the root for 1 is 'un' and for 9 is 'enn'. Therefore, 119 corresponds to un + un + enn + ium = Ununennium, and its three-letter symbol is Uue.

medium

2. Given below are two statements: Statement I: The maximum oxidation state shown by any p-block element is equal to the total number of its valence electrons (group number minus 10). Statement II: The stability of this maximum oxidation state increases as we move down any p-block group. 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
Show answer and explanation

Answer: (B) Statement I is correct but Statement II is incorrect

Statement I is correct as the group oxidation state corresponds to total valence electrons (ns + np). Statement II is incorrect because due to the inert pair effect, the stability of the highest group oxidation state decreases down the group.

hard

3. Which of the following pairs of elements consists of an f-block element and a d-block element, respectively?

  1. Uranium (Z = 92) and Tantalum (Z = 73)
  2. Cerium (Z = 58) and Lutetium (Z = 71)
  3. Lanthanum (Z = 57) and Thorium (Z = 90)
  4. Gadolinium (Z = 64) and Promethium (Z = 61)
Show answer and explanation

Answer: (A) Uranium (Z = 92) and Tantalum (Z = 73)

Uranium (Z = 92) is an actinoid belonging to the f-block (5f series), while Tantalum (Z = 73) has the electronic configuration [Xe] 4f¹⁴ 5d³ 6s², which makes it a d-block transition element. In contrast, both Ce and Lu or Gd and Pm are f-block elements, while Lanthanum is conventionally grouped in the d-block and Thorium in the f-block.

medium

4. Consider the following statements regarding the blocks of the periodic table: (a) All elements in the f-block are metals. (b) The p-block contains metals, non-metals, and metalloids. (c) No non-metal is present in the d-block. (d) All elements of group 12 are representative elements. Which of the above statements are correct?

  1. (a), (b), and (c) only
  2. (b) and (d) only
  3. (a) and (c) only
  4. (a), (b), (c), and (d)
Show answer and explanation

Answer: (A) (a), (b), and (c) only

Statements (a), (b), and (c) are correct: f-block and d-block elements are all metals, and the p-block contains metals, metalloids, and non-metals. Statement (d) is incorrect because representative elements are elements of s- and p-blocks only (Groups 1, 2, and 13-18); Group 12 belongs to the d-block.

easy

5. Fluorine displays only a -1 oxidation state in its compounds, whereas other halogens display oxidation states of -1, +1, +3, +5, and +7. This is primarily because fluorine:

  1. Has the lowest electron gain enthalpy among halogens
  2. Has a higher bond dissociation enthalpy than chlorine
  3. Lacks vacant d-orbitals in its valence shell
  4. Shows inert pair effect
Show answer and explanation

Answer: (C) Lacks vacant d-orbitals in its valence shell

Fluorine belongs to the second period with the valence configuration 2s² 2p⁵. Because it lacks vacant d-orbitals, it cannot unpair its electrons to expand its octet and exhibit positive oxidation states.

hard

6. Consider the following four elements with atomic numbers 19, 23, 35, and 60. How many of these are classified as representative (main group) elements?

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

Answer: (A) 2

Representative or main group elements are elements of the s-block and p-block (excluding group 18 sometimes, though generally s and p). Z = 19 (Potassium, s-block) and Z = 35 (Bromine, p-block) are representative elements. Z = 23 (Vanadium) is a transition (d-block) element, and Z = 60 (Neodymium) is an inner-transition (f-block) element. Hence, exactly 2 elements are representative.

easy

7. Which of the following processes is strictly endothermic in nature?

  1. S(g) + e⁻ → S⁻(g)
  2. Cl(g) + e⁻ → Cl⁻(g)
  3. F(g) + e⁻ → F⁻(g)
  4. O⁻(g) + e⁻ → O²⁻(g)
Show answer and explanation

Answer: (D) O⁻(g) + e⁻ → O²⁻(g)

The addition of a second electron to a mononegative anion like O⁻(g) experiences strong electrostatic repulsion from the existing negative charge. Energy must be supplied to overcome this repulsion, making all second electron gain enthalpies endothermic.

easy

8. The oxidation state of an element in a particular compound reflects the formal charge acquired by its atom based on electronegativity differences. What is the oxidation state of oxygen in OF₂ and O₂F₂ respectively?

  1. +2 and +1
  2. -2 and -1
  3. -1 and +1
  4. +1 and +2
Show answer and explanation

Answer: (A) +2 and +1

Fluorine is the most electronegative element and always exhibits an oxidation state of -1 in its compounds. Therefore, in OF₂, oxygen has an oxidation state of +2, while in O₂F₂, each oxygen atom has an oxidation state of +1.

easy

9. Lanthanoids and actinoids together constitute the:

  1. f-block
  2. d-block
  3. s-block
  4. p-block
Show answer and explanation

Answer: (A) f-block

In lanthanoids (4f) and actinoids (5f) the differentiating electron enters (n-2)f orbitals, so they form the f-block (inner transition elements).

NCERT Class 11

easy

10. The usual valence of group 1 elements is:

  1. 3
  2. 7
  3. 1
  4. 2
Show answer and explanation

Answer: (C) 1

Group 1 elements have one valence electron (ns¹) and typically show a valence of 1, forming M⁺ ions.

NCERT Class 11

hard

11. Two compounds of phosphorus are PCl₅ and PCl₃. Which of the following statements is correct regarding their oxidation states and Lewis acid-base behavior?

  1. Phosphorus in PCl₅ has an oxidation state of +5 and can act only as a Lewis base
  2. Phosphorus in PCl₅ has an expanded octet and cannot react further with chloride ions
  3. Phosphorus in PCl₃ is in +3 state and cannot act as a Lewis base
  4. Phosphorus in PCl₅ has an oxidation state of +5 and acts as a Lewis acid by accepting an electron pair into its vacant d-orbitals
Show answer and explanation

Answer: (D) Phosphorus in PCl₅ has an oxidation state of +5 and acts as a Lewis acid by accepting an electron pair into its vacant d-orbitals

In PCl₅, phosphorus is in its maximum oxidation state (+5). Because it has vacant 3d orbitals, it can accept an electron pair from a chloride ion (Cl⁻) to form [PCl₆]⁻, thus acting as a Lewis acid.

medium

12. An element has the ground state outer electronic configuration 5d¹ 6s². To which group and period does it belong?

  1. Group 3, Period 5
  2. Group 2, Period 5
  3. Group 4, Period 6
  4. Group 3, Period 6
Show answer and explanation

Answer: (D) Group 3, Period 6

The valence shell principal quantum number is n = 6, so it belongs to Period 6. The configuration 5d¹ 6s² corresponds to Lanthanum (Z = 57). The group number for d-block elements is given by (electrons in (n-1)d + electrons in ns) = 1 + 2 = 3.

medium

13. An element with atomic number Z = 120 has not yet been synthesized. Based on the Modern Periodic Table, which of the following predictions about this element is INCORRECT?

  1. Its temporary IUPAC symbol is Ubn
  2. It would belong to Period 8 and Group 2
  3. Its outermost valence electronic configuration would be 7s²
  4. It would be an alkaline earth metal
Show answer and explanation

Answer: (C) Its outermost valence electronic configuration would be 7s²

Element 120 lies in the 8th period directly below Radium (Z = 88). Its temporary IUPAC name is Unbinilium (symbol Ubn), and it belongs to Group 2 (alkaline earth metals). Its outermost electronic configuration would be [Og] 8s², not 7s².

medium

14. Given below are two statements: Statement I: The first ionization enthalpy of zinc is higher than that of copper. Statement II: The electron is removed from the 4s-orbital in zinc, which is completely filled (4s²), whereas in copper it is removed from a half-filled 4s-orbital (4s¹). 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
Show answer and explanation

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

Zn has the configuration [Ar] 3d¹⁰ 4s², and Cu has [Ar] 3d¹⁰ 4s¹. Removing an electron from the fully filled, stable 4s² subshell of Zn requires more energy (906 kJ/mol) than removing an electron from the singly occupied 4s¹ orbital of Cu (745 kJ/mol). Thus, both statements are correct.

medium

15. An element with ground state electronic configuration [Ar] 3d⁵ 4s¹ exhibits a wide variety of oxidation states. What is the highest oxidation state this element can show?

  1. +5
  2. +7
  3. +6
  4. +4
Show answer and explanation

Answer: (C) +6

The configuration [Ar] 3d⁵ 4s¹ belongs to Chromium (Z = 24). It can utilize all six valence electrons (one 4s and five 3d) to exhibit a maximum oxidation state of +6 (e.g., in Cr₂O₇²⁻ and CrO₃).

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