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

Step 1: Work out the configurations involved.
Neutral scandium ($Z=21$) is $[Ar]3d^14s^2$. Removing three electrons to form $Sc^{3+}$ leaves just $[Ar]$, a complete noble gas shell, while $Sc^{2+}$ still carries one lone electron as $[Ar]3d^1$.
Step 2: Judge the relative stability of the two ions.
A filled noble gas configuration is unusually stable, so $Sc^{3+}$ sits in a comfortable, low energy state, whereas $Sc^{2+}$, with its single unpaired $d$ electron, is comparatively less settled.
Step 3: Connect this stability to the reduction potential.
The value $E^\circ_{Sc^{3+}/Sc^{2+}}$ describes how readily $Sc^{3+}$ accepts an electron to become $Sc^{2+}$. Since $Sc^{3+}$ is so content in its noble gas form, it strongly resists taking on another electron, making this reduction potential very low, strongly negative.
Step 4: Match statements to the option.
Both the Assertion and Reason are true, and the noble gas stability of $Sc^{3+}$ is exactly why the reduction potential is low. \[ \boxed{\text{Option (A)}} \]