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Assertion: E^(@) for Mn^(3+)//Mn^(2+) is...

Assertion: `E^(@)` for `Mn^(3+)//Mn^(2+)` is more positive than `Cr^(3+)//Cr^(2+)`
Reason: The third ionisation energy of `Mn` is larger than that of `Cr`.

A

if both Assertion and Reason are true and reason is the correct explanation of the assertion.

B

if both assertion and reason are true but reason is not the correct expalanation of the assertion

C

if assertion is true, reason is false

D

both assertion and reason are false statements.

Text Solution

AI Generated Solution

The correct Answer is:
To solve the question, we need to analyze both the assertion and the reason provided. ### Step-by-Step Solution: 1. **Understanding the Assertion**: - The assertion states that the standard electrode potential \( E^\circ \) for the half-reaction \( \text{Mn}^{3+} + e^- \rightarrow \text{Mn}^{2+} \) is more positive than that for \( \text{Cr}^{3+} + e^- \rightarrow \text{Cr}^{2+} \). - To verify this, we need to look at the standard electrode potentials of both half-reactions. Generally, \( E^\circ \) values become more negative as we move down the group in the periodic table. 2. **Electrochemical Series**: - In the electrochemical series, manganese (Mn) is positioned above chromium (Cr). This implies that the electrode potential for Mn is more positive than that of Cr. - Therefore, the assertion is correct: \( E^\circ (\text{Mn}^{3+}/\text{Mn}^{2+}) > E^\circ (\text{Cr}^{3+}/\text{Cr}^{2+}) \). 3. **Understanding the Reason**: - The reason states that the third ionization energy of manganese is larger than that of chromium. - The third ionization energy refers to the energy required to remove the third electron from an atom. A higher ionization energy indicates that it is more difficult to remove an electron, which is often due to a more stable electronic configuration. 4. **Electronic Configurations**: - Manganese has the electronic configuration \( [Ar] 3d^5 4s^2 \). When it loses two electrons to become \( \text{Mn}^{2+} \), it attains a half-filled \( 3d^5 \) configuration, which is particularly stable. - Chromium has the electronic configuration \( [Ar] 3d^5 4s^1 \). When it loses two electrons to become \( \text{Cr}^{2+} \), it goes to \( 3d^4 \), which is less stable compared to the half-filled configuration of manganese. 5. **Comparison of Ionization Energies**: - Since the half-filled \( 3d^5 \) configuration of \( \text{Mn}^{2+} \) is more stable than the \( 3d^4 \) configuration of \( \text{Cr}^{2+} \), it requires more energy to remove an electron from \( \text{Mn}^{2+} \) than from \( \text{Cr}^{2+} \). - Thus, the reason is also correct: the third ionization energy of manganese is indeed larger than that of chromium. 6. **Conclusion**: - Both the assertion and the reason are true, and the reason correctly explains the assertion. Therefore, the correct answer is that both the assertion and reason are true, and the reason is a correct explanation of the assertion. ### Final Answer: Both assertion and reason are true, and the reason is the correct explanation of the assertion.
To solve the question, we need to analyze both the assertion and the reason provided. ### Step-by-Step Solution: 1. **Understanding the Assertion**: - The assertion states that the standard electrode potential \( E^\circ \) for the half-reaction \( \text{Mn}^{3+} + e^- \rightarrow \text{Mn}^{2+} \) is more positive than that for \( \text{Cr}^{3+} + e^- \rightarrow \text{Cr}^{2+} \). - To verify this, we need to look at the standard electrode potentials of both half-reactions. Generally, \( E^\circ \) values become more negative as we move down the group in the periodic table. ...
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