Periodic Properties Enthalpy Ionic Radii and Density Comparison

To determine the correct statement(s) regarding the periodic properties of elements, we analyze each option individually:

Analysis of Option (A):

• The electronic configuration of a neutral Carbon atom ($Z = 6$) is $1s^2 2s^2 2p^2$.

• After the first ionization, the monovalent carbon cation $\text{C}^+$ has the configuration: $\text{C}^+: 1s^2 2s^2 2p^1$

• The second ionization enthalpy ($\text{IE}_2$) of carbon corresponds to the removal of an electron from the $2p$ subshell of $\text{C}^+$: $\text{C}^+ (1s^2 2s^2 2p^1) \rightarrow \text{C}^{2+} (1s^2 2s^2) + e^-$

• The electronic configuration of a neutral Boron atom ($Z = 5$) is $1s^2 2s^2 2p^1$.

• After the first ionization, the monovalent boron cation $\text{B}^+$ has the configuration: $\text{B}^+: 1s^2 2s^2$

• The second ionization enthalpy ($\text{IE}_2$) of boron corresponds to the removal of an electron from the fully filled, stable $2s$ subshell of $\text{B}^+$: $\text{B}^+ (1s^2 2s^2) \rightarrow \text{B}^{2+} (1s^2 2s^1) + e^-$

• Removing an electron from a fully-filled $2s$ orbital ($\text{B}^+$) requires significantly higher energy than removing an electron from a $2p$ orbital ($\text{C}^+$).

• Therefore, $\text{IE}_2(\text{Carbon}) < \text{IE}_2(\text{Boron})$.

Thus, Option (A) is correct.

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Analysis of Option (B):

• The species $\text{Al}^{3+}$, $\text{Mg}^{2+}$, and $\text{Na}^+$ are all isoelectronic species containing $10$ electrons each with the electronic configuration $1s^2 2s^2 2p^6$.
• For isoelectronic ions, ionic radius is inversely proportional to the nuclear charge ($Z$): $\text{Ionic Radius} \propto \frac{1}{Z}$
• The respective nuclear charges (atomic numbers) are: $Z(\text{Al}^{3+}) = 13, \quad Z(\text{Mg}^{2+}) = 12, \quad Z(\text{Na}^+) = 11$
• Higher nuclear charge pulls the electron cloud closer to the nucleus, resulting in a smaller ionic radius. Therefore: $\text{Al}^{3+} < \text{Mg}^{2+} < \text{Na}^+$

Thus, Option (B) is correct.

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Analysis of Option (C):

• Down Group 1 (alkali metals), both atomic mass and atomic volume increase.
• Potassium ($\text{K}$) exhibits an anomaly due to a unusually large increase in atomic volume relative to its atomic mass (attributed to the presence of vacant $3d$ orbitals in the electronic shell shell structure).
• Consequently, the density of potassium ($\approx 0.86 \text{ g/cm}^3$) is less than the density of sodium ($\approx 0.97 \text{ g/cm}^3$): $\text{Density}(\text{K}) < \text{Density}(\text{Na})$

Thus, Option (C) is incorrect.

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Analysis of Option (D):

• The $\text{H}-\text{H}$ bond energy is approximately $436 \text{ kJ mol}^{-1}$.
• The $\text{F}-\text{F}$ bond energy is significantly lower, approximately $158.8 \text{ kJ mol}^{-1}$, due to strong inter-electronic lone pair-lone pair repulsions between the non-bonding valence electrons on the small fluorine atoms.
• Therefore, the $\text{H}-\text{H}$ bond is much stronger than the $\text{F}-\text{F}$ bond.

Thus, Option (D) is incorrect.

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Conclusion: The correct statements are (A) and (B).