Animated Solution for Chemistry - s and p-Block Elements: Among the statements (I-IV), the correct ones are : (I) Be has smaller atomic radius compared to Mg. (II) Be has higher ionisation enthalpy than Al. (III) Charge/radius ratio of Be is greater than that of Al. (IV) Both Be and Al form mainly covalent compounds.

The periodic table is not just a static grid of elements; it is a dynamic map that reveals the hidden behavioral patterns of atoms. This question takes us on a thrilling journey through some of the most fundamental periodic trends: atomic radius, ionization enthalpy, and the fascinating phenomenon known as the diagonal relationship. Let's break down each statement to uncover the chemical truths hidden within.

The Tug-of-War of Atomic Radius

Our first stop is Group 2, the alkaline earth metals. When we compare Beryllium ($\mathrm{Be}$) and Magnesium ($\mathrm{Mg}$), we must look at their positions. Beryllium sits at the top of the group in Period 2, while Magnesium is right below it in Period 3.

As we descend any group in the periodic table, a new principal quantum shell is added to the atoms. Even though the nuclear charge (the number of protons) increases, the addition of a whole new layer of electrons drastically increases the distance between the nucleus and the outermost shell. Furthermore, the inner electrons shield the outer electrons from the nucleus's pull. Consequently, the atomic radius increases down the group. Therefore, it is an undeniable fact that the atomic radius of Beryllium is smaller than that of Magnesium ($r_{\mathrm{Be}}<r_{\mathrm{Mg}}$). Statement I is correct.

The Fortress of Ionisation Enthalpy

Next, we compare the ionization enthalpy of Beryllium with Aluminum ($\mathrm{Al}$). Ionization enthalpy is the energy required to rip an electron away from an isolated gaseous atom. This is where electronic configuration plays a starring role.

Beryllium has an atomic number of 4, giving it a configuration of $1s^{2}2s^2$. Notice that its outermost subshell, the $2s$ orbital, is fully filled. This symmetry provides extra stability, making the atom highly reluctant to give up an electron.

Aluminum, on the other hand, has a configuration of $[\mathrm{Ne}]3s^{2}3p^1$. Its outermost electron resides alone in a $3p$ orbital. This single $p$-electron is further from the nucleus and is heavily shielded by the inner $s$ and $p$ electrons. Because it is less tightly bound, it requires significantly less energy to remove. Thus, despite Aluminum having a higher nuclear charge, Beryllium boasts a higher ionization enthalpy ($I{E}_{\mathrm{Be}}>I{E}_{\mathrm{Al}}$). Statement II is correct.

The Magic of the Diagonal Relationship

Statement III introduces a beautiful anomaly: the diagonal relationship. In the periodic table, elements situated diagonally across from each other (moving one group right and one period down) often exhibit strikingly similar properties. Beryllium (Group 2, Period 2) and Aluminum (Group 13, Period 3) are the classic poster children for this phenomenon.

Why does this happen? As you move across a period from left to right, the ionic charge increases and the ionic size decreases. Conversely, as you move down a group, the ionic size increases. When you move diagonally, these two opposing trends effectively cancel each other out!

The result is that the charge-to-radius ratio (also known as ionic potential, $\varphi =\frac{q}{r}$) for ${\mathrm{Be}}^{2+}$ and ${\mathrm{Al}}^{3+}$ becomes almost identical. Because their ratios are approximately equal, stating that Beryllium's ratio is strictly greater than Aluminum's is factually incorrect. Statement III is the imposter.

Fajans' Rules and Covalent Character

Finally, we arrive at the nature of the bonds these elements form. Because both Beryllium and Aluminum possess a very high charge-to-radius ratio, they wield immense polarizing power.

According to Fajans' Rules, when a small, highly charged cation approaches a larger anion, it exerts a massive pull on the anion's electron cloud. It distorts the cloud, dragging the electron density into the space between the two nuclei. This sharing of electron density is the very definition of a covalent bond. Therefore, despite being metals, both Beryllium and Aluminum predominantly form covalent compounds (like ${\mathrm{BeCl}}_2$ and ${\mathrm{AlCl}}_3$). Statement IV is correct.

By systematically applying our knowledge of periodic trends and Fajans' rules, we have deduced that Statements I, II, and IV are correct, while Statement III falls short due to the diagonal relationship. This leads us confidently to option (b).