The E° value for the Mn³⁺/Mn²⁺ couple is more positive than that of Cr³⁺/Cr²⁺ or Fe³⁺/Fe²⁺ due to change of

The correct answer is Option 2. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Chemistry Q64 – NEET-UG 2024

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Home Chemistry d- & f-Block ElementsQuestion #364

The

E^{\circ}

value for the

{Mn}^{3 +} / {Mn}^{2 +}

couple is more positive than that of

{Cr}^{3 +} / {Cr}^{2 +}

{Fe}^{3 +} / {Fe}^{2 +}

due to change of

Clue 1 of 3: Standard Electrode Potential

A more positive standard electrode potential ( $E^{\circ}$ ) for a reduction couple indicates a greater tendency for the reduction to occur, implying that the higher oxidation state is less stable than the lower oxidation state.

🧠 Full Solution Path

Step 1: Determine Electronic Configurations of Manganese Ions✦ Active

First, let's find the electronic configurations for the manganese ions involved in the couple:

- The atomic number of Manganese (Mn) is 25. Its ground state electronic configuration is $[Ar] 3 d^{5} 4 s^{2}$ .

- For ${Mn}^{2 +}$ , two electrons are removed from the $4 s$ orbital, resulting in the configuration $[Ar] 3 d^{5}$ . This is a highly stable half-filled d-orbital configuration.

- For ${Mn}^{3 +}$ , three electrons are removed (two from $4 s$ and one from $3 d$ ), resulting in the configuration $[Ar] 3 d^{4}$ .

💡 Teacher's Secret Hint

Remember to remove electrons from the outermost s-orbital first, then from the d-orbital, when forming cations of transition metals.

Step 2: Analyze the Reduction Process and Stability Change○ Expand

The ${Mn}^{3 +} / {Mn}^{2 +}$ couple represents the reduction of ${Mn}^{3 +}$ to ${Mn}^{2 +}$ :

{Mn}^{3 +} (3 d^{4}) + e^{-} \to {Mn}^{2 +} (3 d^{5})

In this reaction, ${Mn}^{3 +}$ (with a $3 d^{4}$ configuration) gains an electron to form ${Mn}^{2 +}$ (with a stable $3 d^{5}$ half-filled configuration). The strong tendency to achieve this highly stable $d^{5}$ configuration drives the reduction process.

💡 Teacher's Secret Hint

A more positive $E^{\circ}$ value indicates that the oxidized form (here, ${Mn}^{3 +}$ ) is relatively unstable and readily accepts an electron to form the more stable reduced form (here, ${Mn}^{2 +}$ ).

Step 3: Compare with Other Couples and Conclude○ Expand

The high stability gained by forming the $3 d^{5}$ configuration from $3 d^{4}$ makes the reduction of ${Mn}^{3 +}$ to ${Mn}^{2 +}$ very favorable, leading to a significantly positive $E^{\circ}$ value. This is why the $E^{\circ}$ for ${Mn}^{3 +} / {Mn}^{2 +}$ is more positive than for ${Cr}^{3 +} / {Cr}^{2 +}$ (where ${Cr}^{3 +}$ is $d^{3}$ , which is also stable due to $t_{2 g}^{3}$ configuration) or ${Fe}^{3 +} / {Fe}^{2 +}$ (where ${Fe}^{3 +}$ is $d^{5}$ , already stable).

The change in electronic configuration is from $d^{4}$ (for ${Mn}^{3 +}$ ) to $d^{5}$ (for ${Mn}^{2 +}$ ).

💡 Teacher's Secret Hint

The stability of $d^{5}$ (half-filled) and $d^{10}$ (fully-filled) configurations often plays a crucial role in determining the redox potentials of transition metals.

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Chemistry Question 64 – NEET-UG 2024

Clue 1 of 3: Standard Electrode Potential

🧠 Full Solution Path