JEE Main Chemistry: Periodic Table And Periodicity High-Yield Practice Questions, Formulas & PYQ Breakdown (2026-27)

Correct Answer: (a)

Explanation: Statement I: "Nitrogen has more ionisation enthalpy and electronegativity than beryllium." **Ionisation Enthalpy:** Nitrogen, being a non-metal and located in group 15, has a higher ionisation energy than beryllium, a metal in group 2. For example, the first ionisation energy of nitrogen is around $1402 \, \text{kJ/mol}$, while for beryllium it is around $900 \, \text{kJ/mol}$. **Electronegativity:** Nitrogen is significantly more electronegative than beryllium. This reflects in their chemical behavior, as non-metals tend to attract electrons more strongly than metals. **Conclusion for Statement I:** This statement is correct. Statement II: "$\mathrm{CrO}_3, \mathrm{~B}_2 \mathrm{O}_3$ are acidic oxide." **Chromium Trioxide ($\mathrm{CrO}_3$):** This oxide is acidic. It reacts with water to form chromic acid: $\mathrm{CrO}_3 + \mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{H}_2\mathrm{CrO}_4.$ **Diboron Trioxide ($\mathrm{B}_2 \mathrm{O}_3$):** This oxide is also acidic. It reacts with water to yield boric acid: $\mathrm{B}_2 \mathrm{O}_3 + 3 \, \mathrm{H}_2 \mathrm{O} \rightarrow 2 \, \mathrm{H}_3 \mathrm{BO}_3.$ **Conclusion for Statement II:** This statement is also correct. Since both statements are correct, the answer is: Option A Both Statement I and Statement II are correct.

Correct Answer: (c)

Explanation: Amphoteric oxide: $\mathrm{SnO}, \mathrm{PbO}, \mathrm{SnO}_2, \mathrm{PbO}_2, \mathrm{Ga}_2 \mathrm{O}_3$ Acidic oxide : $\mathrm{CO}_2, \mathrm{~B}_2 \mathrm{O}_3$ Neutral : CO

Correct Answer: (d)

Explanation: SO$_3$ - Acidic, Na$_2$O - Basic, N$_2$O - Neutral, Al$_2$O$_3$ - Amphoteric

Correct Answer: (d)

Explanation: Option D is correct: A - II: **Ra (Radium)** belongs to the **s-block** as it is an alkaline earth metal. B - I: **Uuq (Ununquadium)** is now known as **Fl (Flerovium)** and belongs to the **p-block**, as it is a part of the post-transition metals. C - IV: **Ds (Darmstadtium)** belongs to the **d-block**, as it is a transition metal. D - III: **Fm (Fermium)** belongs to the **f-block**, as it is an actinide element. These elements are classified based on their electron configurations, which determine their position in the periodic table's blocks.

Correct Answer: (d)

Explanation: Tetrahalide, $\mathrm{PbI}_4$ does not exist because iodine is a good reducing agent and it reduces $\mathrm{Pb}^{4+}$ to $\mathrm{Pb}^{2+}$. So, the compound formed is $\mathrm{PbI}_2$ not $\mathrm{PbI}_4$.