Question

Assertion (A) : \( E^{\circ}_{\mathrm{Mn^{2+}/Mn}} \) value is highly negative.

Reason (R) : Because \( \mathrm{Mn^{2+}} \) is highly stable due to half-filled \( 3d^{5} \) configuration.

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

Hint:

Mn2+ has a stable half-filled 3d5 shell, so it is formed easily and E0 of Mn2+/Mn is highly negative. Both statements are true and R explains A.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The standard electrode potential \( E^{\circ}_{\mathrm{M^{2+}/M}} \) of the first transition series depends on three energies: atomisation, ionisation and hydration. A very stable \( \mathrm{M^{2+}} \) ion makes this value more negative.

Step 2: Check Assertion (A).
The value for \( \mathrm{Mn^{2+}/Mn} \) is -1.18 V. This is much more negative than the general trend of the series, so it is highly negative. (A) is TRUE.

Step 3: Check Reason (R).
The electronic configuration of Mn is \( [\mathrm{Ar}]\,3d^{5}4s^{2} \). After losing two electrons, \( \mathrm{Mn^{2+}} \) is \( 3d^{5} \), a half-filled and extra stable set.
So (R) is TRUE.

Step 4: Does R explain A?
Because \( \mathrm{Mn^{2+}} \) is so stable, the metal gives up electrons very easily, so the reduction \( \mathrm{Mn^{2+} + 2e^{-} \rightarrow Mn} \) is not favoured.
This gives the large negative value of \( E^{\circ} \). Hence R is the correct explanation of A.
So options (2), (3) and (4) are wrong.

Final Answer:
Both A and R are true, and R explains A, which is option (1). \[ \boxed{\text{Option (1)}} \]