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 192 questions; sign in to attempt the remaining 187 in the exam simulator.

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1. Q1JEE Main 2026 (04 Apr, Shift 1)Atomic Mass and Atomic Size
   A monoatomic anion $(A^-)$ has $45$ neutrons and $36$ electrons. Atomic mass, group in the periodic table and physical state at room temperature of the element $(A)$ respectively are
   1. A.$80, 17$, liquid
   2. B.$81, 16$, solid
   3. C.$80, 16$, gas
   4. D.$81, 15$, gas
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   Answer: (A)

   Number of electrons in $A^{-}$ is $36$. Number of electrons in neutral atom $A = 36 - 1 = 35$. Atomic number of $A = 35$. Number of protons in $A = 35$. Atomic mass of $A = \text{Number of protons} + \text{Number of neutrons} = 35 + 45 = 80$. The element with atomic number $35$ is Bromine ($\text{Br}$). Bromine belongs to Group $17$ of the periodic table. The physical state of Bromine at room temperature is liquid. Answer: $80, 17$, liquid
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 2023 (13 Apr, Shift 2)Periodic properties
   . Identify the correct order of standard enthalpy of formation of sodium halides.
   1. A.$NaI\lt NaBr\lt NaF\lt NaCl$
   2. B.$NaI\lt NaBr\lt NaCl\lt NaF$
   3. C.$NaF\lt NaCl\lt NaBr\lt NaI$
   4. D.$NaCl\lt NaF\lt NaBr\lt NaI$
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   Answer: (B)

   The lattice energy of an ionic compound is the energy required to separate one mole of the compound into its constituent ions in the gas phase. Lattice energy$\propto \dfrac{1}{\left({r}_{+}+{r}_{-}\right)}$ The lattice energy is inversely proportional to the sum of the ionic radii $\left({r}_{+}+{r}_{-}\right)$ of the ions involved. When the size of the ions increases, the distance between them also increases, leading to a decrease in the attractive forces between the ions and a decrease in the lattice energy. Therefore, if the size of the ions increases, the lattice energy generally decreases. So Order : $NaF\gt NaCl\gt NaBr\gt NaI$
5. Q5JEE Main 2021 (20 Jul, Shift 2)Electronic configuration
   Outermost electronic configuration of a group $13$ element, $E,$ is $4{s}^{2},4{p}^{1}.$ The electronic configuration of an element of p-block period-five placed diagonally to element, $E$ is:
   1. A.$[Kr]3{d}^{10}4{s}^{2}4{p}^{2}$
   2. B.$[Ar]3{d}^{10}4{s}^{2}4{p}^{2}$
   3. C.$[Xe]5{d}^{10}6{s}^{2}6{p}^{2}$
   4. D.$[Kr]4{d}^{10}5{s}^{2}5{p}^{2}$
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   Answer: (D)

   The element $E$ is $Ga$ and the diagonal element of ${5}^{\text{th}}$ period is $Sn50$ having outer electronic configuration will be $[Kr]5{s}^{2}4{d}^{10}5{p}^{2}.$