Are these the official NEET questions on Classification of Elements and Periodicity in Properties?

Yes. They are the questions NTA set in each paper, in its wording. The solutions are written by BrainIt.

Classification of Elements and Periodicity in Properties NEET PYQs with Solutions (2020 to 2026) PDF

The questions come from NEET 2026, Re-NEET 2026 and NEET 2020. Each shows its answer and solution when you open it, so you can attempt it first. For what the chapter covers and how NEET tests it, see the Classification of Elements and Periodicity in Properties guide.

Practise on the CBT screen6 questions, 6 minutes, marked +4 / −1 Download PDF3 pages, 0.7 MB, with answers and solutions

NEET 2026 · the full paper

NEET 2026 · Q64
The correct order of increasing metallic character of Na, Be, P, Mg and Si is:
1. (1)
  $P < Si < Be < Mg < Na$
2. (2)
  $P < Si < Na < Mg < Be$
3. (3)
  $P < Mg < Be < Si < Na$
4. (4)
  $Be < Si < P < Mg < Na$
Answer and solution
Answer: (1)
Metallic character decreases across a period (left to right) and increases down a group, and it is inversely related to electronegativity. Using Pauling electronegativities (Na 0.9 < Mg 1.2 < Be 1.5 < Si 1.8 < P 2.1), metallic character increases in the reverse order: P < Si < Be < Mg < Na.
NEET 2026 · Q88
Identify the incorrect statement from the following:
1. (1)
  The largest and the smallest species among Mg, $\mathrm{Mg}^{2+}$ , Al and $\mathrm{Al}^{3+}$ are Al and $\mathrm{Mg}^{2+}$ respectively.
2. (2)
  The IUPAC name of the element with atomic number 107 is Unnilseptium.
3. (3)
  The similarity in behaviour of Li with Mg is referred to as 'diagonal relationship'.
4. (4)
  The oxidation state and covalency of Al in $[\mathrm{AlCl}(\mathrm{H}_{2}\mathrm{O})_{5}]^{2+}$ are 3 and 6, respectively.
Answer and solution
Answer: (1)
Cationic size decreases with increasing positive charge and across a period, so among Mg, $\mathrm{Mg}^{2+}$ , Al and $\mathrm{Al}^{3+}$ the largest species is neutral Mg and the smallest is $\mathrm{Al}^{3+}$ , not Al and $\mathrm{Mg}^{2+}$ ; hence statement (1) is incorrect and is the answer. The others are correct: element 107 is Unnilseptium, Li and Mg show a diagonal relationship, and Al in the given complex has oxidation state +3 with covalency 6.

Re-NEET 2026 · the full paper

Re-NEET 2026 · Q53
Among the following options, the correct trend in the electron gain enthalpy is
1. (1)
  F > Cl > Br > I
2. (2)
  Br > Cl > F > I
3. (3)
  Cl > F > Br > I
4. (4)
  I > Br > Cl > F
Answer and solution
Answer: (3)
Chlorine, not fluorine, releases the most energy on gaining an electron. The 2p subshell of fluorine is so compact that the incoming electron meets strong inter-electronic repulsion, which offsets the favourable nuclear attraction; chlorine's larger 3p shell has no such penalty. Down the rest of the group the value falls with increasing size, giving Cl > F > Br > I. NCERT Class 11, Classification of Elements and Periodicity in Properties.
Re-NEET 2026 · Q70
Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R.

Assertion (A): The first ionization enthalpy of O is lower than that of N and F.

Reason (R): The loss of an electron from O leads to stable half-filled p orbital

In light of the above statements, choose the most appropriate answer from the options given below:
1. (1)
  Both A and R are correct and R is the correct explanation of A
2. (2)
  Both A and R are correct and R is NOT the correct explanation of A
3. (3)
  A is correct but R is not correct.
4. (4)
  A is not correct but R is correct
Answer and solution
Answer: (1)
Oxygen is $[\mathrm{He}]2\mathrm{s}^2 2\mathrm{p}^4$ ; the fourth 2p electron must pair up in an already-occupied orbital and suffers extra repulsion, so it is comparatively easy to remove. Losing it gives $\mathrm{O}^{+}$ with the $2\mathrm{p}^3$ half-filled configuration, which is extra stable through exchange energy. That is precisely why oxygen's first ionization enthalpy dips below nitrogen's despite nitrogen having a smaller nuclear charge, and stays below fluorine's. R therefore explains A. NCERT Class 11, Classification of Elements and Periodicity in Properties.

NEET 2020 · the full paper

NEET 2020 · Q114
Identify the incorrect match.
List I
List II
P) Unnilunium
I) Mendelevium
Q) Unniltrium
II) Lawrencium
R) Unnilhexium
III) Seaborgium
S) Unununnium
IV) Darmstadtium
1. (1)
  Q-II
2. (2)
  R-III
3. (3)
  S-IV
4. (4)
  P-I
Answer and solution
Answer: (3)
Unnilunium, unniltrium and unnilhexium correctly correspond to elements 101, 103 and 106. The printed 'Unununnium' denotes element 111, whose official name is roentgenium, not darmstadtium; therefore S-IV is the incorrect match. Chapter: Classification of Elements and Periodicity in Properties.
NEET 2020 · Q116
Match each oxide with its nature and select the correct option.
List I
List II
P) CO
I) Basic
Q) BaO
II) Neutral
R) Al $_{2}$ O $_{3}$
III) Acidic
S) Cl $_{2}$ O $_{7}$
IV) Amphoteric
1. (1)
  P-II, Q-I, R-IV, S-III
2. (2)
  P-III, Q-IV, R-I, S-II
3. (3)
  P-IV, Q-III, R-II, S-I
4. (4)
  P-I, Q-II, R-III, S-IV
Answer and solution
Answer: (1)
CO is neutral, BaO basic, Al $_{2}$ O $_{3}$ amphoteric and Cl $_{2}$ O $_{7}$ acidic. Thus P-II, Q-I, R-IV, S-III. Chapter: Classification of Elements and Periodicity in Properties.

List I	List II
P) Unnilunium	I) Mendelevium
Q) Unniltrium	II) Lawrencium
R) Unnilhexium	III) Seaborgium
S) Unununnium	IV) Darmstadtium

List I	List II
P) CO	I) Basic
Q) BaO	II) Neutral
R) Al $_{2}$ O $_{3}$	III) Acidic
S) Cl $_{2}$ O $_{7}$	IV) Amphoteric

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Classification of Elements and Periodicity in Properties: NEET previous year questions

NEET 2026 · the full paper

Re-NEET 2026 · the full paper

NEET 2020 · the full paper

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