The d electron configuration of `Cr^(2+)`,` Mn^(2+)`, `Fe^(2+)` and `Ni^(2+)` aru `3d^(4)`, `3d^(5)` `3d^(6)`and `3d^(8)` respectively. Which one of the following aqua complexes will exhibit the maximum paramagnetic behaviour? (At. No. Cr= 24, Mn=25, Fe = 26, Ni = 28)

Options

A. `[Fe(H_(2)O)_(6)]^(2+)`
B. `[Ni(H_(2)O)_(6)]^(2+)`
C. `[Cr)H_(2)O)_(6)]^(2+0`
D. `[mn(H_(2)O)_(6)]^(2+0`

Answer

To determine which aqua complex exhibits the maximum paramagnetic behavior among `Cr^(2+)`, `Mn^(2+)`, `Fe^(2+)`, and `Ni^(2+)`, we need to analyze their d-electron configurations and the number of unpaired electrons present in each complex.

### Step-by-Step Solution:

1. **Identify the d-electron configurations:**
- `Cr^(2+)`: The electron configuration is `3d^4`.
- `Mn^(2+)`: The electron configuration is `3d^5`.
- `Fe^(2+)`: The electron configuration is `3d^6`.
- `Ni^(2+)`: The electron configuration is `3d^8`.

2. **Determine the number of unpaired electrons:**
- For `Cr^(2+)` (`3d^4`):
- The configuration can be represented as: ↑ ↑ ↑ ↑
- Number of unpaired electrons = 4.

- For `Mn^(2+)` (`3d^5`):
- The configuration can be represented as: ↑ ↑ ↑ ↑ ↑
- Number of unpaired electrons = 5.

- For `Fe^(2+)` (`3d^6`):
- The configuration can be represented as: ↑ ↑ ↑ ↑ ↓
- Number of unpaired electrons = 4 (since one pair is formed).

- For `Ni^(2+)` (`3d^8`):
- The configuration can be represented as: ↑ ↑ ↑ ↑ ↑ ↓ ↓
- Number of unpaired electrons = 2 (since four pairs are formed).

3. **Compare the number of unpaired electrons:**
- `Cr^(2+)`: 4 unpaired electrons.
- `Mn^(2+)`: 5 unpaired electrons.
- `Fe^(2+)`: 4 unpaired electrons.
- `Ni^(2+)`: 2 unpaired electrons.

4. **Conclusion:**
- The complex with the maximum number of unpaired electrons is `Mn^(2+)`, which has 5 unpaired electrons.
- Therefore, `Mn^(2+)` will exhibit the maximum paramagnetic behavior among the given aqua complexes.

### Final Answer:
`Mn^(2+)` exhibits the maximum paramagnetic behavior.