The d electron congfiguration of `Cr^(2+)` , `Mn^(2+)` , `Fe^(2+)` and `Ni^(2+)` are `3d^(4)` , `3d^(5)` , `3d^(6)` and `3d^(8)` respectively. Which one of the folowing aqua complexes will exhibit the minimum 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+)`
D. `[Mn(H_(2)O)_(6)`

Explanation

To determine which aqua complex exhibits the minimum paramagnetic behavior among `Cr^(2+)`, `Mn^(2+)`, `Fe^(2+)`, and `Ni^(2+)`, we need to analyze the d-electron configurations and count the number of unpaired electrons for each ion. The complex with the least number of unpaired electrons will exhibit the minimum paramagnetic behavior.

### Step-by-step Solution:

1. **Identify the d-electron configurations:**
- For `Cr^(2+)`: The atomic number of chromium (Cr) is 24. The electron configuration is `Ar 4s² 3d⁴`. For `Cr^(2+)`, we lose 2 electrons from the 4s orbital, resulting in `3d⁴`.
- For `Mn^(2+)`: The atomic number of manganese (Mn) is 25. The electron configuration is `Ar 4s² 3d⁵`. For `Mn^(2+)`, we lose 2 electrons from the 4s orbital, resulting in `3d⁵`.
- For `Fe^(2+)`: The atomic number of iron (Fe) is 26. The electron configuration is `Ar 4s² 3d⁶`. For `Fe^(2+)`, we lose 2 electrons from the 4s orbital, resulting in `3d⁶`.
- For `Ni^(2+)`: The atomic number of nickel (Ni) is 28. The electron configuration is `Ar 4s² 3d⁸`. For `Ni^(2+)`, we lose 2 electrons from the 4s orbital, resulting in `3d⁸`.

2. **Count the unpaired electrons:**
- For `Cr^(2+)` (`3d⁴`): The distribution of electrons in the d-orbitals is as follows:
```
↑ ↑ ↑ ↑
1 2 3 4 (4 unpaired electrons)
```
- For `Mn^(2+)` (`3d⁵`): The distribution is:
```
↑ ↑ ↑ ↑ ↑
1 2 3 4 5 (5 unpaired electrons)
```
- For `Fe^(2+)` (`3d⁶`): The distribution is:
```
↑ ↑ ↑ ↑ ↑
1 2 3 4 5 6 (4 unpaired electrons)
```
- For `Ni^(2+)` (`3d⁸`): The distribution is:
```
↑ ↑ ↑ ↑ ↑ ↑
1 2 3 4 5 6 7 8 (2 unpaired electrons)
```

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. **Determine the minimum paramagnetic behavior:**
The complex with the least number of unpaired electrons is `Ni^(2+)`, which has 2 unpaired electrons.

### Conclusion:
The aqua complex that exhibits the minimum paramagnetic behavior is `Ni^(2+)` with 2 unpaired electrons.