Animated Solution for Chemistry - Periodicity in Properties: The correct statement(s) regarding the periodic properties of elements is (are)

The beauty of Chemistry lies not just in its rules, but in its fascinating exceptions. This problem is a masterclass in periodic properties, testing your ability to look beyond the obvious trends and dig into the quantum mechanical realities of atoms. Let's decode each statement step by step.

Statement (A) asks us to compare the second ionization enthalpy ($I{E}_2$) of Carbon and Boron. To understand this, we must look at the electronic configurations of their unipositive ions, because $I{E}_2$ is the energy required to remove an electron from the $+1$ state.

For Carbon ($Z=6$), the neutral atom is $1s^{2}2s^{2}2p^2$. After losing one electron, ${\text{C}}^{+}$ becomes $1s^{2}2s^{2}2p^1$.

For Boron ($Z=5$), the neutral atom is $1s^{2}2s^{2}2p^1$. After losing one electron, ${\text{B}}^{+}$ becomes $1s^{2}2s^2$.

Notice the configuration of ${\text{B}}^{+}$? It has a fully filled $2s$ subshell. Fully filled subshells are exceptionally stable due to their symmetrical electron distribution. Removing an electron from this stable $2s^2$ core requires a massive amount of energy. On the other hand, removing the single $2p^1$ electron from ${\text{C}}^{+}$ is relatively easier. Therefore, the second ionization enthalpy of Boron is significantly greater than that of Carbon. Statement (A) is absolutely correct.

Statement (B) presents three ions: ${\text{Al}}^{3+}$, ${\text{Mg}}^{2+}$, and ${\text{Na}}^{+}$. What do they have in common? Let's count their electrons.

Aluminum ($Z=13$) loses 3 electrons to have 10. Magnesium ($Z=12$) loses 2 to have 10. Sodium ($Z=11$) loses 1 to have 10. They are isoelectronic species—they all possess exactly 10 electrons.

When the number of electrons is constant, the size of the ion is dictated entirely by the nuclear charge (the number of protons). As the number of protons increases, the effective nuclear charge ($Z_{eff}$) increases. A stronger nucleus pulls the same 10 electrons much closer to itself.

Since Aluminum has the highest nuclear charge ($13$ protons), it exerts the strongest pull, making ${\text{Al}}^{3+}$ the smallest ion. Sodium has the weakest pull ($11$ protons), making ${\text{Na}}^{+}$ the largest. Thus, the increasing order of ionic radii is indeed ${\text{Al}}^{3+}<{\text{Mg}}^{2+}<{\text{Na}}^{+}$. Statement (B) is correct.

Statement (C) claims that Potassium is denser than Sodium. Let's investigate the general trend. As we move down a group, both atomic mass and atomic volume increase. Usually, the increase in mass outpaces the increase in volume, leading to a higher density.

However, Potassium is a notorious exception. When we transition from Sodium (Period 3) to Potassium (Period 4), there is a sudden availability of vacant $3d$ orbitals. This causes an abnormal and massive expansion in the atomic volume of Potassium. Because density is mass divided by volume, this huge spike in the denominator causes the overall density of Potassium ($0.86{\text{ g/cm}}^3$) to drop below that of Sodium ($0.97{\text{ g/cm}}^3$). Therefore, Sodium is denser than Potassium, making Statement (C) incorrect.

Finally, Statement (D) compares the bond dissociation energies of ${\text{H}}_2$ and ${\text{F}}_2$. Fluorine is the most electronegative element, so one might intuitively think it forms the strongest bonds. But intuition can be misleading in the quantum world.

Fluorine atoms are extremely small, and each atom carries three lone pairs of electrons. When two Fluorine atoms come close to form a covalent bond, these lone pairs are forced into a very tight space. The resulting interelectronic repulsion is fierce, acting like a compressed spring trying to push the atoms apart. This significantly weakens the $\text{F-F}$ bond ($158.8\text{ kJ/mol}$).

Hydrogen, conversely, has no lone pairs. Two Hydrogen atoms can get very close, forming a highly stable, short, and strong bond ($435.8\text{ kJ/mol}$). Thus, the $\text{H-H}$ bond is much stronger than the $\text{F-F}$ bond. Statement (D) is incorrect.

In conclusion, only statements (A) and (B) stand true against the rigorous laws of periodic properties.