Question

Given below are two statements:

Statement I: Aluminium is more electropositive than thallium as the standard electrode potential value of $E^\circ_{Al^{3+}/Al}$ is negative and $E^\circ_{Tl^{3+}/Tl}$ is positive.
Statement II: The sum of first three ionization enthalpies of boron is very high when compared to that of aluminium. Due to this reason boron forms covalent compounds only and aluminium forms $Al^{3+}$ ion.

In the light of the above statements, choose the correct answer from the options given below

• A. Both Statement I and Statement II are true
• B. Both Statement I and Statement II are false
• C. Statement I is true but Statement II is false
• D. Statement I is false but Statement II is true

Hint:

Check the standard reduction potential values for $Al$ and $Tl$, and recall that Boron's small size leads to very high ionization energy, favoring covalent bonding.

Answer 1

The Correct Option is A

To determine the correctness of these statements, we must look at the periodic trends and electrochemical data of Group 13 elements.

Statement I addresses the concept of electropositivity. Electropositivity is the tendency of an element to lose electrons and form positive ions (cations). In aqueous solution, this tendency is quantitatively measured by the standard electrode potential ($E^\circ$). A more negative standard reduction potential indicates a greater tendency to lose electrons (oxidation) and hence higher electropositivity. For Aluminium, the reduction potential for the $Al^{3+}/Al$ couple is:

$$E^\circ_{Al^{3+}/Al} = -1.66\text{ V}$$

For Thallium, the value for the $Tl^{3+}/Tl$ couple is:

$$E^\circ_{Tl^{3+}/Tl} = +1.26\text{ V}$$

Since the value for Aluminium is highly negative and the value for Thallium is positive, Aluminium loses electrons much more easily than Thallium can lose three electrons to form $Tl^{3+}$. This confirms that Aluminium is indeed more electropositive than Thallium in this context. Thus, Statement I is true.

Statement II explains the bonding nature of Boron versus Aluminium based on ionization enthalpies. Boron has a very small atomic size, which means its valence electrons are very strongly held by the nucleus. The sum of its first three ionization enthalpies ($IE_1 + IE_2 + IE_3$) is extremely high (approximately $6887\text{ kJ/mol}$). This massive amount of energy is not easily compensated by the lattice energy in a crystal or hydration energy in a solution. As a result, Boron cannot easily form $B^{3+}$ ions and instead shares electrons to form covalent bonds. Aluminium, being larger, has a much lower sum of the first three ionization enthalpies (approximately $5137\text{ kJ/mol}$), allowing it to form $Al^{3+}$ ions in many of its compounds. Thus, Statement II is true.

Since both Statement I and Statement II are correct, the correct option is 1.