For the four successive transition elements (Cr, Mn, Fe, and Co), the stability of `+2` oxidation state will be there in which of the following order ?
`(At. Nos. Cr = 24, Mn = 25, Fe = 26, Co = 27)`

Options

A. `Cr gt Mn gt Co gt Fe`
B. `Mn gt Fe gt Cr gt Co`
C. `Fe gt Mn gt Co gt Cr`
D. `Co gt Mn gt Fe gt Cr`

Explanation

To determine the stability of the +2 oxidation state for the transition elements chromium (Cr), manganese (Mn), iron (Fe), and cobalt (Co), we will analyze their electron configurations, reduction potentials, and the stability of their oxidation states step by step.

### Step 1: Determine Electron Configurations
- **Chromium (Cr)**: Atomic number 24
- Electron configuration: \( [Ar] 3d^5 4s^1 \)
- **Manganese (Mn)**: Atomic number 25
- Electron configuration: \( [Ar] 3d^5 4s^2 \)
- **Iron (Fe)**: Atomic number 26
- Electron configuration: \( [Ar] 3d^6 4s^2 \)
- **Cobalt (Co)**: Atomic number 27
- Electron configuration: \( [Ar] 3d^7 4s^2 \)

### Step 2: Analyze +2 Oxidation States
- For the +2 oxidation state, we remove electrons from the outermost shell (4s before 3d):
- **Cr**: \( Cr^{2+} \) → \( [Ar] 3d^4 \)
- **Mn**: \( Mn^{2+} \) → \( [Ar] 3d^5 \) (half-filled, stable)
- **Fe**: \( Fe^{2+} \) → \( [Ar] 3d^6 \)
- **Co**: \( Co^{2+} \) → \( [Ar] 3d^7 \)

### Step 3: Stability Based on Electron Configuration
- A half-filled \( d \) subshell (like in \( Mn^{2+} \)) is particularly stable.
- \( Cr^{2+} \) is not half-filled and is a strong reducing agent, which means it tends to oxidize to \( Cr^{3+} \).
- \( Fe^{2+} \) is more stable than \( Cr^{2+} \) because it does not easily oxidize to \( Fe^{3+} \).
- \( Co^{2+} \) is the least stable among these due to its higher tendency to be reduced.

### Step 4: Standard Reduction Potentials
- The stability of the +2 oxidation state can also be assessed using standard reduction potentials:
- \( E^\circ (Cr^{2+}/Cr) = -0.90 \, V \)
- \( E^\circ (Mn^{2+}/Mn) = -1.18 \, V \)
- \( E^\circ (Fe^{2+}/Fe) = -0.44 \, V \)
- \( E^\circ (Co^{2+}/Co) = -0.28 \, V \)

### Step 5: Order of Stability
- The more negative the reduction potential, the less stable the +2 oxidation state:
- **Most stable**: \( Mn^{2+} \) (lowest reduction potential)
- Next: \( Fe^{2+} \)
- Then: \( Cr^{2+} \)
- **Least stable**: \( Co^{2+} \)

### Final Order of Stability of +2 Oxidation State
The order of stability of the +2 oxidation state for the elements is:
1. \( Mn^{2+} \) (most stable)
2. \( Fe^{2+} \)
3. \( Cr^{2+} \)
4. \( Co^{2+} \) (least stable)

Thus, the final answer is:
**Mn > Fe > Cr > Co**