Question

Statement-I: Al is more electropositive than Tl because \( E^\circ_{\mathrm{Al^{3+}/Al}} \) is negative and \( E^\circ_{\mathrm{Tl^{3+}/Tl}} \) is positive.
Statement-II: For B atom, the sum of the first three ionization energies is very high, thus it forms covalent compounds.

• A. Both statement-I & statement-II are correct
• B. Both statement-I & statement-II are incorrect
• C. Statement-I is correct but statement-II is incorrect
• D. Statement-I is incorrect but statement-II is correct

Answer 1

The Correct Option is A

Step 1: Analyze Statement I.
Electropositivity refers to the ability of an element to donate electrons. The more negative the \( E_0 \) value of an element, the more easily it can lose electrons, making it more electropositive. For Al (Aluminum): \[ E_0_{\text{Al}^{3+}/\text{Al}} = -1.66 \, \text{V} \] This is negative, which means Al is more likely to lose electrons and is more electropositive. For Tl (Thallium): \[ E_0_{\text{Tl}^{3+}/\text{Tl}} = +0.74 \, \text{V} \] This positive value suggests that Tl is less likely to lose electrons compared to Al. Thus, Statement I is correct.

Step 2: Analyze Statement II.
Boron (B) has high ionization energies because it has a small atomic radius and a high effective nuclear charge. The sum of the first three ionization energies of boron is quite large, which reflects its tendency to form covalent compounds due to its reluctance to lose electrons easily. Thus, Statement II is also correct.

Step 3: Conclusion.
Both Statement I and Statement II are correct, so the correct answer is option (A).

Final Answer: (A) Both statement-I & statement-II are correct