JEE Advanced Chemistry Question (2026) — Solution

Step-by-step solution

For (A): The electronic configuration of B ^+ is 1s^2 2s^2 and that of C ^+ is 1s^2 2s^2 2p^1. Removing an electron from the fully filled, more penetrating 2s orbital of B ^+ requires more energy than removing a 2p electron from C ^+. Thus, the second ionization enthalpy of carbon is less than that of boron. Statement (A) is correct. For (B): Al ^ 3+ , Mg ^ 2+ , and Na ^+ are isoelectronic species with 10 electrons each. The nuclear charge increases in the order Na ^+ (11) For (C): The density of potassium (0.86 g/cm ^3) is less than that of sodium (0.97 g/cm ^3) due to an abnormal increase in atomic volume caused by the presence of empty 3d orbitals in potassium. Statement (C) is incorrect. For (D): The bond dissociation enthalpy of H - H (436 kJ/mol ) is much higher than that of F - F (159 kJ/mol ) due to strong interelectronic repulsions between the lone pairs on the small fluorine atoms. Statement (D) is incorrect. Answer: Second ionization enthalpy of carbon atom is less than that of boron atom.; Increasing order of ionic radii: Al ^ 3+ < Mg ^ 2+ < Na ^+