Question

Given below are two statements:
Statement I: The metallic radius of Al is less than that of Ga.
Statement II: The ionic radius of Al\(^{3+}\) is less than that of Ga\(^{3+}\).
In the light of the above statements, choose the most appropriate answer from the options given below:

• A. Both Statement I and Statement II are correct
• B. Statement I is Incorrect but Statement II is Correct
• C. Statement I is Correct but Statement II is Incorrect
• D. Both Statement I and Statement II are incorrect

Hint:

The metallic radius of an element generally increases down a group. The ionic radius decreases with an increase in positive charge on the ion.

Answer 1

The Correct Option is B

The problem presents two statements comparing the metallic (atomic) and ionic radii of Aluminum (Al) and Gallium (Ga). We need to evaluate the correctness of each statement to determine the most appropriate answer.

Concept Used:

The solution requires understanding periodic trends in atomic and ionic radii, along with a key exception known as the d-block contraction (or transition metal contraction).

1. Atomic/Metallic Radius Trend: Generally, the atomic radius increases as we move down a group in the periodic table. This is because each subsequent element has an additional principal energy level (electron shell), which increases the distance of the valence electrons from the nucleus.
2. d-Block Contraction: This is an exception to the general trend. For elements that come immediately after the first d-block series (like Gallium), the 10 electrons in the intervening d-orbitals (e.g., the 3d orbitals for Ga) are poor at shielding the outer valence electrons from the nuclear charge. The increase in nuclear charge is not effectively screened, leading to a higher effective nuclear charge (\(Z_{eff}\)) than expected. This pulls the valence electrons closer to the nucleus, causing the atom to be smaller than predicted by the general trend.
3. Ionic Radius Trend: The trend for ionic radii of ions with the same charge generally follows the atomic radius trend (increases down a group). This is because the ions of elements further down the group still have more electron shells.

Step-by-Step Solution:

Step 1: Evaluation of Statement I: "The metallic radius of Al is less than that of Ga."

Aluminum (Al) is in Period 3 and Gallium (Ga) is in Period 4, both belonging to Group 13. Based on the general periodic trend, one would expect the radius of Ga to be larger than that of Al because Ga has an additional electron shell (n=4 vs n=3).

However, we must consider the d-block contraction. Gallium (atomic number 31) has the electron configuration [Ar] 3d\(^{10}\) 4s\(^2\) 4p\(^1\). The 10 electrons in the 3d orbital are filled just before the 4p orbital. These 3d electrons provide very poor shielding for the outer 4s and 4p electrons.

Due to this poor shielding, the effective nuclear charge experienced by Gallium's valence electrons is unusually high, causing a contraction in its size. As a result, the metallic radius of Gallium is actually slightly smaller than that of Aluminum.

The experimentally determined metallic radii are approximately:

• Metallic radius of Al \(\approx\) 143 pm
• Metallic radius of Ga \(\approx\) 135 pm

Thus, the radius of Al is greater than the radius of Ga (Radius(Al) > Radius(Ga)). The statement claims that the radius of Al is less than that of Ga. Therefore, Statement I is incorrect.

Step 2: Evaluation of Statement II: "The ionic radius of Al\(^{3+}\) is less than that of Ga\(^{3+}\)."

Let's consider the formation of the ions Al\(^{3+}\) and Ga\(^{3+}\) by removing their three valence electrons.

• Al atom: [Ne] 3s\(^2\) 3p\(^1\) \(\rightarrow\) Al\(^{3+}\) ion: [Ne]
• Ga atom: [Ar] 3d\(^{10}\) 4s\(^2\) 4p\(^1\) \(\rightarrow\) Ga\(^{3+}\) ion: [Ar] 3d\(^{10}\)

After ionization, the outermost electrons in the Al\(^{3+}\) ion are in the n=2 principal energy level. For the Ga\(^{3+}\) ion, the outermost electrons are in the n=3 principal energy level (the 3d shell). Since Ga\(^{3+}\) has an additional electron shell compared to Al\(^{3+}\), its ionic radius is expected to be larger. The d-block contraction effect is less pronounced in the ions as the valence shells have been removed.

The experimentally determined ionic radii (for coordination number 6) are approximately:

• Ionic radius of Al\(^{3+}\) \(\approx\) 53.5 pm
• Ionic radius of Ga\(^{3+}\) \(\approx\) 62.0 pm

The radius of Al\(^{3+}\) is indeed less than the radius of Ga\(^{3+}\). Therefore, Statement II is correct.

Final Conclusion:

Based on the analysis, Statement I is incorrect and Statement II is correct. The correct option is the one that reflects this finding.