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: Understanding the Question:
The question evaluates two independent statements about the group 13 elements: the relative electropositivity of Aluminum and Thallium, and the bonding nature of Boron based on its ionization energies.
Step 2: Detailed Explanation:
Statement-I: Aluminum has a standard reduction potential (\(E^\circ\)) of \(-1.66\) V, making it highly electropositive. In contrast, for Thallium, the \(+3\) oxidation state is unstable due to the "inert pair effect", and its \(E^\circ_{Tl^{3+}/Tl}\) is positive (\(+1.26\) V). This indicates that Tl$^{3+}$ is easily reduced to Tl, meaning Tl is much less electropositive than Al. Thus, Statement-I is correct.

Statement-II: Boron has a very small atomic size. The sum of its first three ionization energies (\(IE_1 + IE_2 + IE_3\)) is extremely high, preventing it from forming B$^{3+}$ ions easily. Consequently, Boron shares electrons to form covalent compounds rather than ionic ones. Thus, Statement-II is correct.
Step 3: Final Answer:
Since both statements are factually accurate, the answer is (1).