More stable compound among the following is

Options

A. `VF_(2)`
B. `VF_(3)`
C. `VF_(4)`
D. `VF_(5)`

Answer

To determine the more stable compound among the given options (Vf2, Vf3, Vf4, and Vf5), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Oxidation States**:
- The compounds are Vanadium Fluorides with Vanadium in different oxidation states: +2 (Vf2), +3 (Vf3), +4 (Vf4), and +5 (Vf5).

2. **Consider the Electronegativity of Fluorine**:
- Fluorine is the most electronegative element, which means it has a strong tendency to attract electrons. This property influences the stability of the compounds formed with transition metals.

3. **Analyze Stability Based on Oxidation States**:
- Generally, transition metals in higher oxidation states form more stable compounds with highly electronegative ligands like fluorine. This is because the electronegative ligand can stabilize the positive charge on the metal ion.

4. **Evaluate Each Compound**:
- Vf2 (Vanadium in +2 state) is less stable due to lower oxidation.
- Vf3 (Vanadium in +3 state) is more stable than Vf2 but less stable than Vf4 and Vf5.
- Vf4 (Vanadium in +4 state) is more stable than Vf2 and Vf3, but still less stable than Vf5.
- Vf5 (Vanadium in +5 state) is the highest oxidation state for Vanadium and is expected to be the most stable due to the strong electronegative effect of fluorine.

5. **Conclude the Most Stable Compound**:
- Based on the analysis, Vf5 is the most stable compound among Vf2, Vf3, Vf4, and Vf5.

### Final Answer:
The more stable compound among the following is **Vf5**.