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Step-by-Step Explanation
Step 1: Compare the atomic radii of Be and Mg
Beryllium (Be) lies before Magnesium (Mg) in the same group (Group 2) in the periodic table. Moving down a group, the atomic radius generally increases due to an increase in the principal quantum number (additional shells). Hence, $\text{Be}$ has a smaller atomic radius compared to $\text{Mg}$. This makes Statement (I) correct.
Step 2: Compare the first ionization enthalpies of Be and Al
Beryllium (Be) has the electronic configuration $1s^2 2s^2$, which is relatively stable (fully filled $2s$ orbital). Aluminum (Al) has the configuration $[Ne] 3s^2 3p^1$, and the presence of a $3p$ electron is generally held less tightly than the $2s$ electron in Be. Therefore, it takes more energy to remove an electron from Be than from Al. Consequently, the first ionization enthalpy of Be is higher than that of Al, making Statement (II) correct.
Step 3: Compare the charge/radius ratios of Be and Al
The ratio of charge (nuclear charge) to radius depends on both the nuclear charge (atomic number) and the size of the ion/atom. Aluminum (Al), having a higher atomic number than Be, can have a higher effective nuclear charge when comparing ions or atoms under similar conditions. Generally, $ \frac{\text{Charge}}{\text{Radius}} $ for Al is greater than or at least comparable to that of Be, contradicting Statement (III), which claims Be's ratio is greater. Hence, (III) is incorrect.
Step 4: Nature of compounds formed by Be and Al
Beryllium (Be) and Aluminum (Al), given their relatively high charge-to-size ratios among their respective groups, often show a tendency to form covalent bonds rather than purely ionic bonds. This makes Statement (IV) that both Be and Al form mainly covalent compounds correct.
Conclusion
From the analysis above:
Statement (I) is correct.
Statement (II) is correct.
Statement (III) is incorrect.
Statement (IV) is correct.
Hence, the correct statements are (I), (II), and (IV).
