Classification of Elements and Periodicity in Properties JEE Main previous year questions with solutions

5 solved JEE Main questions on Classification of Elements and Periodicity in Properties, free to read — no sign-in needed. The full chapter has 190 questions; sign in to attempt the remaining 185 in the exam simulator.

1. Q1JEE Main 2026 (08 Apr, Shift 2)Periodic properties
   Match List-I with List-II.List-I Electronic configuration of neutral atom (where $n=2$)List-II $1^{st}$ Ionization Energy (kJ mol$^{-1}$)A. $ns^2$I. $2080$B. $ns^2np^1$II. $899$C. $ns^2np^3$III. $800$D. $ns^2np^6$IV. $1402$Choose the correct answer from the options given below:
   1. A.A-II, B-III, C-IV, D-I
   2. B.A-IV, B-III, C-II, D-I
   3. C.A-III, B-II, C-IV, D-I
   4. D.A-III, B-II, C-I, D-IV
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   Answer: (A)

   Given $n=2$, the electronic configurations correspond to elements of the second period. A. $ns^2 \Rightarrow 2s^2$ is Beryllium (Be). B. $ns^2np^1 \Rightarrow 2s^2 2p^1$ is Boron (B). C. $ns^2np^3 \Rightarrow 2s^2 2p^3$ is Nitrogen (N). D. $ns^2np^6 \Rightarrow 2s^2 2p^6$ is Neon (Ne). The first ionization energy generally increases across a period from left to right. However, Beryllium has a higher ionization energy than Boron due to the stable fully-filled $2s$ subshell. Nitrogen has a high ionization energy due to its stable half-filled $2p$ subshell. Neon, being a noble gas, has the highest ionization energy. The order of first ionization energy is B $\lt$ Be $\lt$ N $\lt$ Ne. Arranging the given values in increasing order: $800 \lt 899 \lt 1402 \lt 2080$ kJ mol$^{-1}$. Thus, the matching is: A. Be $\rightarrow 899$ (II) B. B $\rightarrow 800$ (III) C. N $\rightarrow 1402$ (IV) D. Ne $\rightarrow 2080$ (I) This corresponds to A-II, B-III, C-IV, D-I. Answer: A-II, B-III, C-IV, D-I
2. Q2JEE Main 2025 (03 Apr, Shift 2)Earlier attempts of periodic classification
   Match the LIST-I with LIST-II. $\begin{array}{|l|l|l|l|}\hline & \begin{array}{c} \text{LIST-I} \\ \text{(Family)} \end{array} & & \begin{array}{c} \text{LIST-II} \\ \text{(Symbol of Element)} \end{array} \\ \hline \text{A.} & \begin{array}{l} \text{Pnicogen} \\ \text{(group 15)} \end{array} & \text{I.} & \text{Ts} \\ \hline \text{B.} & \text{Chalcogen} & \text{II.} & \text{Og} \\ \hline \text{C.} & \text{Halogen} & \text{III.} & \text{Lv} \\ \hline \text{D.} & \text{Noble gas} & \text{IV.} & \text{Mc} \\ \hline\end{array}$ Choose the correct answer from the options given below :
   1. A.A-IV, B-I, C-II, D-III
   2. B.A-IV, B-III, C-I, D-II
   3. C.A-III, B-I, C-IV, D-II
   4. D.A-II, B-III, C-IV, D-I
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   Answer: (B)

   (A) Pnictogen $\Rightarrow$ Mc (Moscovium), Atomic No. $=115$ (B) Chalcogen $\Rightarrow \mathrm{Lv}$ (Livermorium), Atomic No. $=116$ (C) Halogen $\Rightarrow \mathrm{Ts}$ (Tennessine), Atomic No. $=117$ (D) Noble gas $\Rightarrow \mathrm{Og}$ (Oganesson), Atomic No. $=118$
3. Q3JEE Main 2024 (08 Apr, Shift 2)Modern periodic table
   Identify the correct statements about p-block elements and their compounds. (A) Non metals have higher electronegativity than metals. (B) Non metals have lower ionisation enthalpy than metals. (C) Compounds formed between highly reactive nonmetals and highly reactive metals are generally ionic. (D) The non-metal oxides are generally basic in nature. (E) The metal oxides are generally acidic or neutral in nature. Choose the correct answer from the options given below :
   1. A.(B) and (D) only
   2. B.(A) and (C) only
   3. C.(D) and (E) only
   4. D.(B) and (E) only
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   Answer: (B)

   As electronegativity increases non-metallic nature increases. Along the period ionisation energy increases. High electronegativity difference results in ionic bond formation. Oxides of metals are generally basic and that of non-metals are acidic in nature.
4. Q4JEE Main 2026 (06 Apr, Shift 2)Periodic properties
   In a period, the first ionisation enthalpy of the element at extreme left and the negative electron gain enthalpy of the extreme right element, except noble gases, are respectively.
   1. A.lowest and lowest
   2. B.highest and lowest
   3. C.lowest and highest
   4. D.highest and highest
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   Answer: (C)

   In a period, as we move from left to right, the atomic size decreases and the effective nuclear charge increases. Due to the large atomic size and low effective nuclear charge, the element at the extreme left (alkali metal) has the lowest first ionisation enthalpy in the period. As we move to the extreme right (excluding noble gases), the elements (halogens) have the smallest atomic size and highest effective nuclear charge in the period. They require only one electron to complete their octet. Thus, they have the highest negative electron gain enthalpy. Therefore, the first ionisation enthalpy of the extreme left element is the lowest, and the negative electron gain enthalpy of the extreme right element is the highest. Answer: lowest and highest
5. Q5JEE Main 2026 (06 Apr, Shift 1)Periodic properties
   Given below are two statements: Statement I: Aluminium is more electropositive than thallium as the standard electrode potential value of $E^{\circ}_{\text{Al}^{3+}/\text{Al}}$ is negative and $E^{\circ}_{\text{Tl}^{3+}/\text{Tl}}$ is positive. Statement II: The sum of first three ionization enthalpies of boron is very high when compared to that of aluminium. Due to this reason boron forms covalent compounds only and aluminium forms $\text{Al}^{3+}$ ion. In the light of the above statements, choose the correct answer from the options given below
   1. A.Both Statement I and Statement II are true
   2. B.Both Statement I and Statement II are false
   3. C.Statement I is true but Statement II is false
   4. D.Statement I is false but Statement II is true
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   Answer: (A)

   Statement I: The standard reduction potential $E^{\circ}_{\text{Al}^{3+}/\text{Al}}$ is $-1.66$ V, which is highly negative, indicating that aluminium has a strong tendency to lose electrons and is highly electropositive. The standard reduction potential $E^{\circ}_{\text{Tl}^{3+}/\text{Tl}}$ is $+1.26$ V, which is positive, indicating that thallium is less electropositive than aluminium. Thus, Statement I is true. Statement II: Boron has a very small atomic size, resulting in a very high sum of its first three ionization enthalpies. This high energy requirement prevents boron from losing three electrons to form $\text{B}^{3+}$ ions, so it forms only covalent compounds. In contrast, aluminium has a larger size and a significantly lower sum of its first three ionization enthalpies, allowing it to easily form $\text{Al}^{3+}$ ions in its compounds. Thus, Statement II is true. Both Statement I and Statement II are true. Answer: Both Statement I and Statement II are true