Which of the following can show `+7` oxidation state?

Options

A. `Mn`
B. `F`
C. `In`
D. `N`

Answer

To determine which of the given elements can show a +7 oxidation state, we will analyze each option: Manganese (Mn), Fluorine (F), Indium (In), and Nitrogen (N).

### Step-by-Step Solution:

1. **Identify the Elements**: The elements we need to analyze are Manganese (Mn), Fluorine (F), Indium (In), and Nitrogen (N).

2. **Analyze Manganese (Mn)**:
- Manganese has an atomic number of 25.
- Its electronic configuration is:
\[ \text{Mn: } [\text{Ar}] 4s^2 3d^5 \]
- Manganese is a transition metal and can use both the 4s and 3d electrons for bonding.
- The maximum oxidation state of Mn can be calculated as:
\[ \text{Total electrons in } ns \text{ and } (n-1)d = 2 + 5 = 7 \]
- Therefore, Manganese can exhibit a +7 oxidation state.

3. **Analyze Fluorine (F)**:
- Fluorine has an atomic number of 9.
- Its electronic configuration is:
\[ \text{F: } [\text{He}] 2s^2 2p^5 \]
- Fluorine is a non-metal and is highly electronegative. It typically gains one electron to achieve a stable noble gas configuration.
- Thus, Fluorine can only exhibit a -1 oxidation state.

4. **Analyze Indium (In)**:
- Indium has an atomic number of 49.
- Its electronic configuration is:
\[ \text{In: } [\text{Kr}] 5s^2 4d^{10} 5p^1 \]
- Indium is a post-transition metal and typically exhibits oxidation states of +1 and +3.
- It does not have the capability to reach a +7 oxidation state.

5. **Analyze Nitrogen (N)**:
- Nitrogen has an atomic number of 7.
- Its electronic configuration is:
\[ \text{N: } [\text{He}] 2s^2 2p^3 \]
- Nitrogen can exhibit oxidation states of -3, +3, and +5, but it cannot reach +7 due to its limited number of available electrons.

6. **Conclusion**:
- Among the given elements, only Manganese (Mn) can show a +7 oxidation state. Therefore, the correct answer is:
\[ \text{Option 1: Manganese (Mn)} \]