Write all possible oxidation states of an element having atomic number `25`.

Answer

To determine the possible oxidation states of the element with atomic number 25, we follow these steps:

### Step 1: Identify the Element
The element with atomic number 25 is manganese (Mn).

### Step 2: Write the Electronic Configuration
Manganese has the electronic configuration of:
\[ \text{Ar} \, 18 \, 3d^5 \, 4s^2 \]

### Step 3: Determine the Valence Electrons
From the electronic configuration, we can see that manganese has:
- 5 electrons in the 3d subshell
- 2 electrons in the 4s subshell

This totals to 7 valence electrons (5 from 3d and 2 from 4s).

### Step 4: Identify Possible Oxidation States
Manganese can lose electrons from both the 4s and 3d subshells. The possible oxidation states are determined by the number of electrons that can be lost:

1. **+2**: Losing 2 electrons from the 4s subshell.
2. **+3**: Losing 3 electrons (2 from 4s and 1 from 3d).
3. **+4**: Losing 4 electrons (2 from 4s and 2 from 3d).
4. **+5**: Losing 5 electrons (2 from 4s and 3 from 3d).
5. **+6**: Losing 6 electrons (2 from 4s and 4 from 3d).
6. **+7**: Losing all 7 electrons (2 from 4s and 5 from 3d).

### Conclusion
The possible oxidation states of manganese (atomic number 25) are:
- +2, +3, +4, +5, +6, and +7.