Highest `(+7)` oxidation state is shown by

Options

A. Co
B. Cr
C. V
D. Mn

Answer

To determine which element shows the highest oxidation state of +7 among the given options (cobalt, chromium, vanadium, magnesium), we can analyze the electronic configurations and possible oxidation states of each element.

### Step 1: Identify the electronic configurations of the elements.

1. **Cobalt (Co)**: Atomic number = 27
- Electronic configuration: \( [Ar] 4s^2 3d^7 \)

2. **Chromium (Cr)**: Atomic number = 24
- Electronic configuration: \( [Ar] 4s^1 3d^5 \)

3. **Vanadium (V)**: Atomic number = 23
- Electronic configuration: \( [Ar] 4s^2 3d^3 \)

4. **Magnesium (Mg)**: Atomic number = 12
- Electronic configuration: \( [Ne] 3s^2 \)

### Step 2: Determine the maximum oxidation states for each element.

- **Cobalt**: The highest oxidation state is +3 (it can lose 2 electrons from 4s and 1 from 3d).

- **Chromium**: The highest oxidation state is +6 (it can lose 1 electron from 4s and all 5 from 3d).

- **Vanadium**: The highest oxidation state is +5 (it can lose 2 electrons from 4s and 3 from 3d).

- **Magnesium**: The highest oxidation state is +2 (it can lose both electrons from 3s).

### Step 3: Identify the element that can achieve +7 oxidation state.

- **Manganese (Mn)**: Atomic number = 25
- Electronic configuration: \( [Ar] 4s^2 3d^5 \)
- Manganese can achieve +7 oxidation state by losing all 2 electrons from 4s and all 5 from 3d, resulting in \( Mn^{+7} \).

### Conclusion:
Among the given options, manganese (Mn) is the only element that can exhibit a +7 oxidation state. Therefore, the answer is:

**Manganese (Mn)** shows the highest oxidation state of +7.