Modem periodic law and present form of the periodic table, s, p. d and f block elements, periodic trends in properties of elements atomic and ionic radii, ionization enthalpy, electron gain enthalpy, valence, oxidation states, and chemical reactivity.: NEET MCQs with Answers

15 free questions on Modem periodic law and present form of the periodic table, s, p. d and f block elements, periodic trends in properties of elements atomic and ionic radii, ionization enthalpy, electron gain enthalpy, valence, oxidation states, and chemical reactivity. for NEET (UG), starting with previous-year NEET and AIPMT questions. Try each one, then tap Show answer for the answer and a short explanation.

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Question 1

Previous-year question
Which sequence correctly lists a neutral, an acidic, a basic and an amphoteric oxide, in that order? [NEET (Manipur) 2023]
  1. NO, ZnO, CO₂, CaO
  2. ZnO, NO, CaO, CO₂
  3. NO, CO₂, ZnO, CaO
  4. NO, CO₂, CaO, ZnO
Show answer
Answer: (D) NO, CO₂, CaO, ZnO
Explanation: NO is a neutral oxide, CO₂ is acidic, CaO is basic and ZnO is amphoteric. Hence the sequence NO, CO₂, CaO, ZnO is correct.

Question 2

Previous-year question
Which one of the following is correctly matched? [NEET (Manipur) 2023]
  1. Basic oxides ⇒ In₂O₃, K₂O, SnO₂
  2. Neutral oxides ⇒ CO, NO₂, N₂O
  3. Acidic oxides ⇒ Mn₂O₇, SO₂, TeO₃
  4. Amphoteric oxides ⇒ BeO, Ga₂O₃, GeO
Show answer
Answer: (C) Acidic oxides ⇒ Mn₂O₇, SO₂, TeO₃
Explanation: Mn₂O₇ (Mn in +7 state), SO₂ and TeO₃ are all acidic oxides, so this set is correctly matched. The other sets each contain a misfit: SnO₂ is amphoteric, not basic; NO₂ is an acidic oxide, not neutral; and GeO is not amphoteric, since NCERT describes it as distinctly acidic.

Question 3

Previous-year question
Assertion: Ionisation enthalpy increases along each series of the transition elements from left to right, although small variations occur. [NEET (Manipur) 2023]
Reason: There is a corresponding increase in nuclear charge which accompanies the filling of electrons in the inner d-orbitals.
  1. (A) is correct but (R) is not correct.
  2. (A) is not correct but (R) is correct.
  3. Both (A) and (R) are correct and (R) is the correct explanation of (A).
  4. Both (A) and (R) are correct but (R) is not the correct explanation of (A).
Show answer
Answer: (C) Both (A) and (R) are correct and (R) is the correct explanation of (A).
Explanation: Both are true and R explains A: across each transition series the nuclear charge increases while the added electrons go into the inner (n−1)d orbitals, which shield poorly, so ionisation enthalpy increases from left to right. Small irregularities occur because of changes in the relative energies of 4s and 3d orbitals and the extra stability of d⁵ and d¹⁰ configurations.

Question 4

Previous-year question
Which of the following statements are true? [NEET 2025]
A. Unlike Ga that has a very high melting point, Cs has a very low melting point.
B. On Pauling scale, the electronegativity values of N and Cl are not the same.
C. Ar, K⁺, Cl⁻, Ca²⁺ and S²⁻ are all isoelectronic species.
D. The correct order of the first ionisation enthalpies of Na, Mg, Al and Si is Si > Al > Mg > Na.
E. The atomic radius of Cs is greater than that of Li and Rb.
  1. A, B and E only
  2. C and E only
  3. C and D only
  4. A, C and E only
Show answer
Answer: (B) C and E only
Explanation: C is true: Ar, K⁺, Cl⁻, Ca²⁺ and S²⁻ all have 18 electrons. E is true: atomic radius increases down group 1, so Cs is larger than Li and Rb. A is false because Ga itself has a very low melting point (≈303 K); B is false because N and Cl both have electronegativity 3.0 on the Pauling scale; D is false because the correct order of first ionisation enthalpy is Si > Mg > Al > Na.

Question 5

Previous-year question
Which of the following statements is incorrect? [NEET (Sawaimadhopur) 2024]
  1. Na₂O is a basic oxide and Cl₂O₇ is an acidic oxide.
  2. The following four species are called isoelectronic species: O²⁻, F⁻, Na⁺ and Mg²⁺.
  3. Among the four species Mg, Al, Mg²⁺ and Al³⁺, the smallest one is Al.
  4. The oxidation number and coordination number (or covalency) of Al in [AlCl(H₂O)₅]²⁺ are +3 and 6, respectively.
Show answer
Answer: (C) Among the four species Mg, Al, Mg²⁺ and Al³⁺, the smallest one is Al.
Explanation: Cations are much smaller than their parent atoms, and of the isoelectronic ions Mg²⁺ and Al³⁺ the one with higher nuclear charge is smaller. So the smallest of Mg, Al, Mg²⁺ and Al³⁺ is Al³⁺, not Al, and the statement naming Al is the incorrect one. The other statements are true: Na₂O is basic and Cl₂O₇ is acidic; O²⁻, F⁻, Na⁺ and Mg²⁺ all have 10 electrons; and in [AlCl(H₂O)₅]²⁺, x − 1 = +2 gives +3, while one Cl and five H₂O give coordination number 6.

Question 6

Previous-year question
Match List-I (elements) with List-II (atomic radii in pm) and choose the correct answer. [NEET (Overseas) 2022]
List-I (Elements):
(a) O
(b) C
(c) B
(d) N
List-II (Atomic radii, pm):
(i) 88
(ii) 74
(iii) 66
(iv) 77
  1. (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)
  2. (a)-(i), (b)-(iv), (c)-(iii), (d)-(ii)
  3. (a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)
  4. (a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)
Show answer
Answer: (C) (a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)
Explanation: Atomic radius decreases across period 2: B (88 pm) > C (77 pm) > N (74 pm) > O (66 pm). Hence O-(iii) 66, C-(iv) 77, B-(i) 88, N-(ii) 74, i.e. (a)-(iii), (b)-(iv), (c)-(i), (d)-(ii).

Question 7

Previous-year question
Which of the following electronic configurations belong to main group (representative) elements? [NEET 2025]
A. [Ne]3s¹
B. [Ar]3d³4s²
C. [Kr]4d¹⁰5s²5p⁵
D. [Ar]3d¹⁰4s¹
E. [Rn]5f⁰6d²7s²
  1. B and E only
  2. A and C only
  3. D and E only
  4. A, C and D only
Show answer
Answer: (B) A and C only
Explanation: Main group elements are the s- and p-block elements. [Ne]3s¹ (Na, s-block) and [Kr]4d¹⁰5s²5p⁵ (I, p-block) are main group. [Ar]3d³4s² (V) and [Ar]3d¹⁰4s¹ (Cu) are d-block transition elements, and [Rn]5f⁰6d²7s² (Th) is an f-block element.
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Question 8

Previous-year question
Match List-I (IUPAC official name) with List-II (IUPAC symbol) and choose the correct answer. [NEET (Sawaimadhopur) 2024]
List-I (IUPAC official name):
A. Mendelevium
B. Meitnerium
C. Moscovium
D. Nobelium
List-II (IUPAC symbol):
I. Mt
II. Mc
III. No
IV. Md
  1. A-IV, B-II, C-I, D-III
  2. A-IV, B-I, C-II, D-III
  3. A-I, B-IV, C-II, D-III
  4. A-III, B-I, C-II, D-IV
Show answer
Answer: (B) A-IV, B-I, C-II, D-III
Explanation: Mendelevium = Md, Meitnerium = Mt, Moscovium = Mc and Nobelium = No. Hence A-IV, B-I, C-II, D-III.

Question 9

Previous-year question
Match each element in Column I with its category in Column II and choose the correct match. [NEET (COVID-19) 2020]
Column I:
(a) Copper
(b) Fluorine
(c) Silicon
(d) Cerium
Column II:
(i) Non-metal
(ii) Transition metal
(iii) Lanthanoid
(iv) Metalloid
  1. (a)-(ii), (b)-(iv), (c)-(i), (d)-(iii)
  2. (a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)
  3. (a)-(iv), (b)-(iii), (c)-(i), (d)-(ii)
  4. (a)-(i), (b)-(ii), (c)-(iii), (d)-(iv)
Show answer
Answer: (B) (a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)
Explanation: Copper is a transition metal (d-block), fluorine is a non-metal, silicon is a metalloid and cerium (Z = 58) is a lanthanoid. Hence (a)-(ii), (b)-(i), (c)-(iv), (d)-(iii).

Question 10

Previous-year question
Match the following oxides with their nature and choose the correct option. [NEET 2020] Oxide: (A) CO (B) BaO (C) Al₂O₃ (D) Cl₂O₇ Nature: (i) Basic (ii) Neutral (iii) Acidic (iv) Amphoteric
  1. A–(ii), B–(i), C–(iv), D–(iii)
  2. A–(iii), B–(iv), C–(i), D–(ii)
  3. A–(iv), B–(iii), C–(ii), D–(i)
  4. A–(i), B–(ii), C–(iii), D–(iv)
Show answer
Answer: (A) A–(ii), B–(i), C–(iv), D–(iii)
Explanation: Carbon monoxide reacts with neither acids nor bases and is neutral; barium oxide is a strongly basic metal oxide; aluminium oxide dissolves in both acids and alkalis and is amphoteric; and dichlorine heptoxide is the anhydride of perchloric acid and is strongly acidic.

Question 11

Previous-year question
For the second period elements the correct increasing order of first ionisation enthalpy is [NEET 2019]
  1. Li < Be < B < C < N < O < F < Ne
  2. Li < B < Be < C < O < N < F < Ne
  3. Li < B < Be < C < N < O < F < Ne
  4. Li < Be < B < C < O < N < F < Ne
Show answer
Answer: (B) Li < B < Be < C < O < N < F < Ne
Explanation: The general rise across period 2 is broken twice: beryllium's filled 2s² makes its value higher than boron's, and nitrogen's half-filled 2p³ makes its value higher than oxygen's. Allowing for both anomalies gives Li < B < Be < C < O < N < F < Ne.

Question 12

Previous-year question
The formation of the oxide ion O²⁻(g) from an oxygen atom requires first an exothermic and then an endothermic step as shown below. Thus the process of formation of O²⁻ in the gas phase is unfavourable even though O²⁻ is isoelectronic with neon. It is due to the fact that [AIPMT 2015 RS]
  1. electron repulsion outweighs the stability gained by achieving a noble gas configuration
  2. O⁻ ion has a comparatively smaller size than an oxygen atom
  3. oxygen is more electronegative
  4. addition of an electron in oxygen results in a larger size of the ion
Show answer
Answer: (A) electron repulsion outweighs the stability gained by achieving a noble gas configuration
Explanation: The second electron must be forced on to a species that is already negatively charged, and the electrostatic repulsion involved requires 780 kJ mol⁻¹. This far outweighs the stabilisation gained from attaining the neon configuration.

Question 13

Previous-year question
The species Ar, K⁺ and Ca²⁺ contain the same number of electrons. In which order do their radii increase? [AIPMT 2015]
  1. Ca²⁺ < Ar < K⁺
  2. Ca²⁺ < K⁺ < Ar
  3. K⁺ < Ar < Ca²⁺
  4. Ar < K⁺ < Ca²⁺
Show answer
Answer: (B) Ca²⁺ < K⁺ < Ar
Explanation: All three are isoelectronic with 18 electrons, so radius decreases as nuclear charge increases: Ca (20) > K (19) > Ar (18). The radii therefore increase in the order Ca²⁺ < K⁺ < Ar.

Question 14

Previous-year question
Which of the following orders of ionic radii is correctly represented? [AIPMT 2014]
  1. H⁻ > H⁺ > H
  2. Na⁺ < F⁻ < O²⁻
  3. F⁻ > O²⁻ > Na⁺
  4. Al³⁺ > Mg²⁺ > N³⁻
Show answer
Answer: (B) Na⁺ < F⁻ < O²⁻
Explanation: Na⁺, F⁻ and O²⁻ are isoelectronic (10 electrons); size increases as the nuclear charge decreases, so Na⁺ < F⁻ < O²⁻. The correct order for hydrogen species is H⁻ > H > H⁺, and for the last option N³⁻ > Mg²⁺ > Al³⁺.

Question 15

Previous-year question
Identify the incorrect match. [NEET 2020] Name / IUPAC Official Name: (A) Unnilunium – (i) Mendelevium; (B) Unniltrium – (ii) Lawrencium; (C) Unnilhexium – (iii) Seaborgium; (D) Unununnium – (iv) Darmstadtium
  1. (B), (ii)
  2. (C), (iii)
  3. (D), (iv)
  4. (A), (i)
Show answer
Answer: (C) (D), (iv)
Explanation: Unununnium is element 111, which is officially Roentgenium (Rg); Darmstadtium is element 110. The other three matches are correct: 101 is Mendelevium, 103 is Lawrencium and 106 is Seaborgium.

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