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

This problem involves two statements comparing the metallic (atomic) and ionic radii of Aluminum (Al) and Gallium (Ga). The goal is to assess the accuracy of each statement to identify the correct answer.

Concepts:

The solution relies on an understanding of periodic trends in atomic and ionic radii, with specific attention to the d-block contraction (or transition metal contraction).

1. Atomic/Metallic Radius Trend: Generally, atomic radius increases down a group in the periodic table due to the addition of electron shells, increasing the distance of valence electrons from the nucleus.
2. d-Block Contraction: This is an exception. For elements following the first d-block series (e.g., Gallium), the 10 electrons in intervening d-orbitals poorly shield outer valence electrons from nuclear charge. This leads to a higher effective nuclear charge (\(Z_{eff}\)), pulling valence electrons closer and resulting in a smaller atom than predicted by the general trend.
3. Ionic Radius Trend: For ions of the same charge, ionic radii generally follow the atomic radius trend (increasing down a group) due to the presence of more electron shells.

Solution Breakdown:

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

Al (Period 3, Group 13) and Ga (Period 4, Group 13) are in the same group. Conventionally, Ga should have a larger radius due to an additional electron shell. However, the d-block contraction applies to Ga. Gallium's electron configuration is [Ar] 3d\(^{10}\) 4s\(^2\) 4p\(^1\). The 3d electrons shield the outer electrons poorly. This results in a high effective nuclear charge for Ga, causing its size to contract. Consequently, Gallium's metallic radius is slightly smaller than Aluminum's.

Approximate metallic radii:

• Al \(\approx\) 143 pm
• Ga \(\approx\) 135 pm

Therefore, Radius(Al) > Radius(Ga). Statement I, claiming Radius(Al) < Radius(Ga), is incorrect.

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

Consider the ions Al\(^{3+}\) and Ga\(^{3+}\). They are formed by removing the three valence electrons.

• Al\(^{3+}\) ion: [Ne]
• Ga\(^{3+}\) ion: [Ar] 3d\(^{10}\)

The Al\(^{3+}\) ion's outermost electrons are in the n=2 shell. The Ga\(^{3+}\) ion's outermost electrons are in the n=3 shell. Ga\(^{3+}\) has an additional electron shell, making its ionic radius larger. The d-block contraction's effect is reduced in ions as valence shells are removed.

Approximate ionic radii (coordination number 6):

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

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

Conclusion:

Statement I is incorrect, and Statement II is correct. The correct option reflects this finding.