Check statement (I): Atomic radius of Be vs Mg

Be and Mg belong to Group 2. Atomic radius increases down a group due to addition of a new shell.

$\text{Be} < \text{Mg}$

So, Be has smaller atomic radius compared to Mg.

Hence, (I) is correct.

Check statement (II): Ionization enthalpy of Be vs Al

Be has electronic configuration: $1s^2 2s^2$ Al has electronic configuration: $[Ne] 3s^2 3p^1$

Be has a completely filled $2s$ subshell, which is relatively stable. Also, its valence electrons are in the second shell and are held more strongly than the $3p$ electron of Al.

Therefore, $\text{IE}_1(\text{Be}) > \text{IE}_1(\text{Al})$

Hence, (II) is correct.

Check statement (III): Charge/radius ratio of Be vs Al

We compare the cations usually formed:

• Be forms $\text{Be}^{2+}$
• Al forms $\text{Al}^{3+}$

Charge/radius ratio means: $\frac{z}{r}$

Although $\text{Be}^{2+}$ is very small, $\text{Al}^{3+}$ has higher charge and also very small size, giving it a larger charge/radius ratio than Be.

Thus, $\frac{2}{r_{\text{Be}}} < \frac{3}{r_{\text{Al}}} \quad \text{(effectively, Al has greater polarising power)}$

Hence, (III) is incorrect.

Check statement (IV): Nature of compounds of Be and Al

Both Be and Al form mainly covalent compounds because their small size and relatively high charge density give them strong polarising power.

Examples:

• $\text{BeCl}_2$ is covalent
• $\text{AlCl}_3$ is largely covalent

Hence, (IV) is correct.

Conclusion

Correct statements are: $\text{(I), (II), and (IV)}$

Therefore, the correct option is: $\boxed{\text{A}}$