Classification of Elements and Periodicity in Properties NEET PYQs (2005–2026) with Solutions

10 solved Classification of Elements and Periodicity in Properties PYQs

Real NEET and AIPMT questions from this chapter, newest first. Try each one, then open the answer and the step-by-step explanation.

1. Question 1 (NEET 2026)

   NEET 2026EasyElectron gain enthalpy of halogens

   Among the following options, the correct trend in the electron gain enthalpy is

   1. AOption A: $\mathrm{F > Cl > Br > I}$
   2. BOption B: $\mathrm{Br > Cl > F > I}$
   3. COption C: $\mathrm{Cl > F > Br > I}$
   4. DOption D: $\mathrm{I > Br > Cl > F}$

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   Correct answer: (C) $\mathrm{Cl > F > Br > I}$

   Explanation

   Down a group, electron gain enthalpy usually becomes less negative because the atom gets bigger and the added electron is farther from the nucleus. Fluorine is the exception: its compact $2p$ subshell is so small that the incoming electron feels strong repulsion from the electrons already there. So F has a less negative electron gain enthalpy than Cl. NCERT values (in $\mathrm{kJ\ mol^{-1}}$): Cl $-349$, F $-328$, Br $-325$, I $-295$. In magnitude, the trend is $\mathrm{Cl > F > Br > I}$, option (C).

2. Question 2 (NEET 2026)

   NEET 2026EasyMetallic character and periodic trends

   The correct order of increasing metallic character of Na, Be, P, Mg and Si is:

   1. AOption A: P < Si < Be < Mg < Na
   2. BOption B: P < Si < Na < Mg < Be
   3. COption C: P < Mg < Be < Si < Na
   4. DOption D: Be < Si < P < Mg < Na

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   Correct answer: (A) P < Si < Be < Mg < Na

   Explanation

   Metallic character decreases across a period and increases down a group. Therefore increasing metallic character: P < Si < Be < Mg < Na. Hence option (A) is correct.

3. Question 3 (NEET 2026)

   NEET 2026EasyIonization enthalpy anomaly of oxygen

   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. AOption A: Both A and R are correct and R is the correct explanation of A
   2. BOption B: Both A and R are correct and R is NOT the correct explanation of A
   3. COption C: A is correct but R is not correct.
   4. DOption D: A is not correct but R is correct

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   Correct answer: (A) Both A and R are correct and R is the correct explanation of A

   Explanation

   First ionization enthalpies: N ($2p^3$) and F (higher nuclear charge) are both higher than O. So Assertion A is correct. Oxygen is $1s^2 2s^2 2p^4$. Two of its four $2p$ electrons share one orbital, and their mutual repulsion makes one of them easy to remove. Removing it gives $\mathrm{O^+}$ ($2p^3$), a stable half-filled $p$ subshell. Nitrogen already has the stable half-filled $2p^3$ configuration, so removing an electron from N is harder. So R is correct and explains A: option (A).

4. Question 4 (NEET 2026)

   NEET 2026MediumAnomalous behaviour of second period elements

   The correct statement is

   1. AOption A: Boron has a maximum covalency of four.
   2. BOption B: Beryllium has three valence orbitals.
   3. COption C: Magnesium has a maximum covalency of four.
   4. DOption D: Aluminium has five valence orbitals.

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   Correct answer: (A) Boron has a maximum covalency of four.

   Explanation

   Second-period elements have only four valence orbitals ($2s$ and three $2p$) and no $d$ orbitals. So their maximum covalency is four. (A) is correct: boron can form at most four bonds, e.g. $\mathrm{[BF_4]^-}$. (B) is wrong: beryllium has four valence orbitals ($2s$, $2p_x$, $2p_y$, $2p_z$), not three. (C) is wrong: magnesium is a third-period element with $3s$, $3p$ and $3d$ orbitals, so its covalency can go beyond four, e.g. six in $\mathrm{[Mg(H_2O)_6]^{2+}}$. (D) is wrong: aluminium has nine valence orbitals ($3s$, $3p$, $3d$), so it can expand its covalency to six, e.g. $\mathrm{[AlF_6]^{3-}}$. So the answer is option (A).

5. Question 5 (NEET 2026)

   NEET 2026MediumPeriodic trends, diagonal relationship and coordination compounds

   Identify the incorrect statement from the following:

   1. AOption A: The largest and the smallest species among Mg, Mg²⁺, Al and Al³⁺ are Al and Mg²⁺ respectively
   2. BOption B: The IUPAC name of the element with atomic number 107 is Unnilseptium
   3. COption C: The similarity in behaviour of Li with Mg is referred to as diagonal relationship
   4. DOption D: The oxidation state and covalency of Al in [AlCl(H₂O)₅]²⁺ are 3 and 6, respectively

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   Correct answer: (A) The largest and the smallest species among Mg, Mg²⁺, Al and Al³⁺ are Al and Mg²⁺ respectively

   Explanation

   Among Mg, Mg²⁺, Al and Al³⁺, neutral Mg has the largest size while Al³⁺ has the smallest size due to high positive charge causing greater contraction. Therefore statement (1) is incorrect.

6. Question 6 (NEET 2025)

   NEET 2025MediumPeriodic Trends in Properties of Elements

   Which of the following statements are true? 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. $\mathrm{Ar}$, $\mathrm{K^+}$, $\mathrm{Cl^-}$, $\mathrm{Ca^{2+}}$, and $\mathrm{S^{2-}}$ are all isoelectronic species. D. The correct order of the first ionization enthalpies of Na, Mg, Al, and Si is: $$\mathrm{Si > Al > Mg > Na}$$ E. The atomic radius of Cs is greater than that of Li and Rb. Choose the correct answer from the options given below:

   1. AOption A: A, B, and E only
   2. BOption B: C and E only
   3. COption C: C and D only
   4. DOption D: A, C, and E only

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   Correct answer: (B) C and E only

   Explanation

   Statement C is correct because $\mathrm{Ar}$, $\mathrm{K^+}$, $\mathrm{Cl^-}$, $\mathrm{Ca^{2+}}$, and $\mathrm{S^{2-}}$ each contain 18 electrons and are therefore isoelectronic species. Statement E is also correct because atomic radius increases down Group 1, hence: $$\mathrm{Cs > Rb > Li}$$ Statement A is incorrect because gallium does not have a very high melting point. Statement B is incorrect because on the Pauling scale, nitrogen and chlorine have nearly equal electronegativity values. Statement D is incorrect because the correct order of first ionization enthalpy is: $$\mathrm{Si > Mg > Al > Na}$$ Therefore, the correct answer is: $$\mathrm{C\ and\ E\ only}$$

7. Question 7 (NEET 2024)

   NEET 2024MediumPeriodic Trends

   Arrange the following elements in increasing order of first ionization enthalpy: Li, Be, B, C, N

   1. AOption A: Li < Be < C < B < N
   2. BOption B: Li < Be < N < B < C
   3. COption C: Li < Be < B < C < N
   4. DOption D: Li < B < Be < C < N

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   Correct answer: (D) Li < B < Be < C < N

   Explanation

   Ionization enthalpy generally increases across a period. However, Be has higher ionization enthalpy than B due to completely filled 2s orbital. Thus the increasing order is: $$Li < B < Be < C < N$$ Hence, option (D) is correct.

8. Question 8 (NEET 2023)

   NEET 2023EasyIonization Enthalpy and Electron Gain

   The element expected to form largest ion to achieve the nearest noble gas configuration is:

   1. AOption A: O
   2. BOption B: F
   3. COption C: N
   4. DOption D: Na

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   Correct answer: (C) N

   Explanation

   Nitrogen gains three electrons to achieve the nearest noble gas configuration (Ne), forming N³⁻, which has the largest ionic radius among the given ions.

9. Question 9 (NEET 2022)

   NEET 2022EasyPeriodic Classification

   The IUPAC name of an element with atomic number 119 is:

   1. AOption A: ununoctium
   2. BOption B: ununennium
   3. COption C: unnilennium
   4. DOption D: unununium

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   Correct answer: (B) ununennium

   Explanation

   The temporary systematic IUPAC name for element 119 is formed from digits 1-1-9: un + un + enn + ium = ununennium Hence option (B) is correct.

10. Question 10 (NEET 2020)

    NEET 2020EasyAcidic, Basic and Amphoteric Oxides

    Match the following:

    1. AOption A: (a) (i) (b) (ii) (c) (iii) (d) (iv)
    2. BOption B: (a) (ii) (b) (i) (c) (iv) (d) (iii)
    3. COption C: (a) (iii) (b) (iv) (c) (i) (d) (ii)
    4. DOption D: (a) (iv) (b) (iii) (c) (ii) (d) (i)

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    Correct answer: (B) (a) (ii) (b) (i) (c) (iv) (d) (iii)

    Explanation

    CO is a neutral oxide, BaO is a basic oxide, Al₂O₃ is amphoteric and Cl₂O₇ is an acidic oxide. Therefore the correct sequence is (ii), (i), (iv), (iii).