Classification of Elements and Periodicity — JEE Advanced · 2026 · 17 May 2026 · Paper 2 Question with Solution

The correct statement(s) regarding the periodic properties of elements is (are)

(A)Second ionization enthalpy of carbon atom is less than that of boron atom✓ Correct
(B)Increasing order of ionic radii: $Al^{3 +} < Mg^{2 +} < Na^{+}$ ✓ Correct
(C)Under identical conditions, in solid state, the density of potassium metal is more than density of sodium metal
(D)The H-H bond is weaker than F-F bond.

Answer: (A), (B)

Solution

(A) Correct: Second ionization enthalpy order is B > C. For boron, the second electron is removed from a stable 2s² configuration, while for carbon it's from 2s² (after first ionization from 2p). The removal from stable configuration requires more energy.

(B) Correct: For isoelectronic species (all have 10 electrons), as nuclear charge increases, ionic radius decreases. Thus:

Na^{+} (Z = 11) > Mg^{2 +} (Z = 12) > Al^{3 +} (Z = 13)

(C) Incorrect: Density of Na > density of K. Although K is larger, its mass increase is not proportional to volume increase, making it less dense than Na.

(D) Incorrect: H-H bond dissociation energy (435.8 kJ/mol) is much greater than F-F bond (158.8 kJ/mol). The F-F bond is weak due to strong lone pair-lone pair repulsion.

Therefore, options (A) and (B) are correct.

JEE Advanced · 2026 · 17 May 2026 · Paper 2 · One or more correct options

JEE Advanced · 2026 · 17 May 2026 · Paper 2 · Classification of Elements and Periodicity

Solution

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