Which one of the following exhibits the largest number of oxidation states?

Options

A. `Ti(22)`
B. `V(23)`
C. `Cr(24)`
D. `Mn(25)`

Answer

To determine which of the given elements exhibits the largest number of oxidation states, we will analyze the electronic configurations and the number of unpaired electrons for each element. The elements provided are titanium (Ti), vanadium (V), chromium (Cr), and manganese (Mn).

### Step-by-Step Solution:

1. **Identify the Atomic Numbers and Electronic Configurations:**
- Titanium (Ti): Atomic number 22
- Electronic configuration: [Ar] 3d² 4s²
- Vanadium (V): Atomic number 23
- Electronic configuration: [Ar] 3d³ 4s²
- Chromium (Cr): Atomic number 24
- Electronic configuration: [Ar] 3d⁵ 4s²
- Manganese (Mn): Atomic number 25
- Electronic configuration: [Ar] 3d⁵ 4s²

2. **Count the Number of Unpaired Electrons:**
- Titanium (Ti): 2 unpaired electrons (3d²)
- Vanadium (V): 3 unpaired electrons (3d³)
- Chromium (Cr): 6 unpaired electrons (3d⁵, due to half-filled stability)
- Manganese (Mn): 5 unpaired electrons (3d⁵)

3. **Determine the Possible Oxidation States:**
- Titanium (Ti): Can exhibit oxidation states of +2, +3, and +4 (total of 3 states).
- Vanadium (V): Can exhibit oxidation states of +2, +3, +4, +5 (total of 4 states).
- Chromium (Cr): Can exhibit oxidation states of +2, +3, +6 (total of 3 states).
- Manganese (Mn): Can exhibit oxidation states of +2, +3, +4, +6, +7 (total of 7 states).

4. **Conclusion:**
- Among the given elements, manganese (Mn) exhibits the largest number of oxidation states, which is 7.

### Final Answer:
**Manganese (Mn) exhibits the largest number of oxidation states.**

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