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

Correct Options:
1. Second ionization enthalpy of carbon atom is less than that of boron atom.
2. Increasing order of ionic radii: Al³⁺ < Mg²⁺ < Na⁺

Step-by-Step Explanation:

Let us analyze each statement given in the options to determine its correctness:

Statement 1: Second ionization enthalpy of carbon atom is less than that of boron atom.
First, write down the ground-state electronic configurations of neutral Carbon (C, Z = 6) and Boron (B, Z = 5):

$\text{C}:1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^2$

$\text{B}:1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^1$

After removing the first electron (first ionization):

${\text{C}}^{+}:1{\text{s}}^{2}2{\text{s}}^{2}2{\text{p}}^1$

${\text{B}}^{+}:1{\text{s}}^{2}2{\text{s}}^2$

The second ionization enthalpy (${\text{IE}}_2$) corresponds to removing an electron from ${\text{C}}^{+}$ and ${\text{B}}^{+}$ respectively:
- For ${\text{C}}^{+}$, the electron is removed from a 2p orbital ($2{\text{p}}^1\to 2{\text{p}}^0$).
- For ${\text{B}}^{+}$, the electron must be removed from a fully filled, stable 2s orbital ($2{\text{s}}^2\to 2{\text{s}}^1$).
Because a fully filled 2s subshell has higher stability and greater penetration effect than a 2p subshell, removing an electron from ${\text{B}}^{+}$ requires more energy than removing an electron from ${\text{C}}^{+}$.
Therefore, the second ionization enthalpy of carbon is less than that of boron. This statement is correct.

Statement 2: Increasing order of ionic radii: Al³⁺ < Mg²⁺ < Na⁺
The given species, ${\text{Al}}^{3+}$, ${\text{Mg}}^{2+}$, and ${\text{Na}}^{+}$, are isoelectronic species because each contains 10 electrons.
For isoelectronic species, as the atomic number (nuclear charge $\text{Z}$) increases, the electrostatic attraction of the nucleus for the electrons increases, which pulls the electron cloud closer and decreases the ionic radius.

Nuclear charges ($\text{Z}$):
- ${\text{Al}}^{3+}:\text{Z}=13$
- ${\text{Mg}}^{2+}:\text{Z}=12$
- ${\text{Na}}^{+}$

Since nuclear charge follows the order $\text{Al}>\text{Mg}>\text{Na}$, the ionic radii follow the opposite order:
${\text{Al}}^{3+}<{\text{Mg}}^{2+}<{\text{Na}}^{+}$.
This statement is correct.

Statement 3: Under identical conditions, in solid state, the density of potassium metal is more than density of sodium metal.
Generally, density increases down a group in alkali metals. However, potassium (K) is an exception. Due to an unusually large increase in atomic volume (caused by the presence of vacant 3d orbitals and poor packing efficiency in its crystal lattice), potassium is actually less dense than sodium (density of Na ≈ 0.97 g/cm³, K ≈ 0.86 g/cm³). Therefore, this statement is incorrect.

Statement 4: The H–H bond is weaker than F–F bond.
The H–H bond dissociation enthalpy is approximately 435.8 kJ/mol, whereas the F–F bond dissociation enthalpy is unusually low (around 158.8 kJ/mol) due to strong inter-electronic repulsions between the non-bonding lone pairs on adjacent small fluorine atoms. Thus, the H–H bond is significantly stronger than the F–F bond. Therefore, this statement is incorrect.

Conclusion:
The correct statements are the 1st and 2nd statements.