Four successive members of first row transition element are listed below. Which one of them is expected to have highest `E_((M^(3+))/(M^(2+))^(ɵ)` value?

Options

A. Mn(Z = 25)
B. Fe(Z = 26)
C. Co(Z = 27)
D. Cr(Z = 24)

Answer

To determine which of the four successive members of the first-row transition elements has the highest \( E_{(M^{3+}/M^{2+})} \) value, we need to analyze the electronic configurations and the stability of the oxidation states of each element. The elements in question are Chromium (Cr), Manganese (Mn), Iron (Fe), and Cobalt (Co).

### Step-by-Step Solution:

1. **Identify the Elements and Their Atomic Numbers:**
- Chromium (Cr) - Atomic number 24
- Manganese (Mn) - Atomic number 25
- Iron (Fe) - Atomic number 26
- Cobalt (Co) - Atomic number 27

2. **Write the Electronic Configurations:**
- Chromium (Cr): \( [Ar] 3d^5 4s^1 \)
- Manganese (Mn): \( [Ar] 3d^5 4s^2 \)
- Iron (Fe): \( [Ar] 3d^6 4s^2 \)
- Cobalt (Co): \( [Ar] 3d^7 4s^2 \)

3. **Determine the Electronic Configurations of the Ions:**
- \( Cr^{3+} \): \( [Ar] 3d^3 \) (loses 1 electron from 4s and 2 from 3d)
- \( Mn^{3+} \): \( [Ar] 3d^4 \) (loses 1 electron from 4s and 1 from 3d)
- \( Fe^{3+} \): \( [Ar] 3d^5 \) (loses 2 electrons from 4s)
- \( Co^{3+} \): \( [Ar] 3d^6 \) (loses 2 electrons from 4s)

4. **Analyze the Stability of the Ions:**
- The stability of the \( M^{2+} \) state compared to \( M^{3+} \) is influenced by the electronic configuration.
- Half-filled and fully filled d-orbitals are generally more stable.
- \( Mn^{2+} \) has a half-filled \( 3d^5 \) configuration, which is particularly stable.
- \( Fe^{2+} \) has \( 3d^6 \), while \( Co^{2+} \) has \( 3d^7 \).

5. **Consider the Ionization Energies:**
- The reduction potential \( E_{(M^{3+}/M^{2+})} \) is influenced by the ionization energy required to remove an electron from \( M^{2+} \) to form \( M^{3+} \).
- As we move from Cr to Co, the effective nuclear charge increases, leading to higher ionization energies.

6. **Conclusion:**
- Among the four elements, Cobalt (Co) is expected to have the highest reduction potential \( E_{(M^{3+}/M^{2+})} \) because it has the highest ionization energy due to its increasing effective nuclear charge and the stability of its \( 3d^6 \) configuration.

### Final Answer:
**Cobalt (Co) is expected to have the highest \( E_{(M^{3+}/M^{2+})} \) value.**

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