Periodic Table & Periodicity

Concept:Periodic trends in ionization enthalpy, ionic radii, density, and bond energy are evaluated for given statements.Explanation:For option A:Carbon has configuration $1s^2 2s^2 2p^2$.Its singly charged ion $\mathrm{C}^{+}$ has $1s^2 2s^2 2p^1$.Boron has $1s^2 2s^2 2p^1$.Its $\mathrm{B}^{+}$ has $1s^2 2s^2$ (filled $2s$ subshell).Removing an electron from $\mathrm{B}^{+}$ requires more energy due to the stable filled subshell.Thus second ionization enthalpy of carbon is less than that of boron.Option A is correct.For option B:$\mathrm{Al}^{3+}$, $\mathrm{Mg}^{2+}$, $\mathrm{Na}^{+}$ are isoelectronic (10 electrons each).Ionic radius decreases as nuclear charge increases.Proton numbers: Al = 13, Mg = 12, Na = 11.Hence size order: $\mathrm{Al}^{3+} < \mathrm{Mg}^{2+} < \mathrm{Na}^{+}$.Option B is correct.For option C:Actual densities: Na = $0.97\ \mathrm{g/cm^3}$, K = $0.86\ \mathrm{g/cm^3}$.Sodium is denser than potassium, so the given statement is incorrect.For option D:Bond energies: H–H = $435.88\ \mathrm{kJ/mol}$, F–F = $155\ \mathrm{kJ/mol}$.The F–F bond is weaker due to lone pair repulsion on small fluorine atoms.Thus the statement "H–H bond is weaker than F–F bond" is incorrect.Answer:Options A and B are correct.