Question

Assertion (A) : \( E^{\circ}_{Sc^{3+}/Sc^{2+}} \) has low value.
Reason (R) : Because of the stability of \( Sc^{3+} \) ion which has a noble gas configuration.

• A. Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
• B. Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
• C. Assertion (A) is true, but Reason (R) is false.
• D. Assertion (A) is false, but Reason (R) is true.

Hint:

Scandium essentially only exists in the \( +3 \) oxidation state in its compounds due to this extreme stability of the \( d^0 \) configuration.
It is technically a transition element, but its chemistry is quite different from other transition metals because it doesn't show variable valency easily.

Answer 1

The Correct Option is A

Concept:

• Scandium (\( Sc \)) is the first element of the 3d transition series with atomic number 21.

• The electronic configuration of neutral \( Sc \) is \( [Ar] 3d^{1} 4s^{2} \).

• Stability in transition metal ions is often governed by the achievement of a noble gas configuration (\( d^0, d^5, \text{ or } d^{10} \)).

• Standard reduction potential (\( E^{\circ} \)) values indicate the tendency of a species to be reduced; a low (highly negative) value indicates high stability of the oxidized state.
Step 1: Determine the electronic configurations of the ions
For \( Sc^{3+} \): The ion is formed by losing all three valence electrons (two from \( 4s \) and one from \( 3d \)).
Configuration: \( [Ar] \) (equivalent to the noble gas Argon).
For \( Sc^{2+} \): The ion is formed by losing two electrons from the \( 4s \) orbital.
Configuration: \( [Ar] 3d^{1} \).

Step 2: Evaluate the stability of the states
The \( Sc^{3+} \) ion is exceptionally stable because it has a completely filled shell (noble gas configuration).
The \( Sc^{2+} \) ion is much less stable as it has one lone electron in the \( 3d \) orbital.
Therefore, \( Sc^{3+} \) has very little tendency to accept an electron to become \( Sc^{2+} \).

Step 3: Relate stability to \( E^{\circ} \) value
The reduction potential \( E^{\circ}_{Sc^{3+}/Sc^{2+}} \) refers to the process: \( Sc^{3+} + e^{-} \rightarrow Sc^{2+} \).
Because \( Sc^{3+} \) is so stable, this reduction is energetically unfavorable, resulting in a very low (highly negative) reduction potential. The final answer is (A).