In which compound does vanadium has an oxidation number of +4 ?

Options

A. `NH_4VO_2`
B. `VO_2+`
C. `VSO_4`
D. `VOSO_4`

Answer

To determine in which compound vanadium has an oxidation number of +4, we will analyze each of the given compounds step by step.

### Step 1: Analyze the first compound, \( \text{NH}_4\text{VO}_2 \)

1. **Break down the compound**: The compound can be separated into \( \text{NH}_4^+ \) and \( \text{VO}_2^- \).
2. **Assign oxidation states**:
- The oxidation number of nitrogen in \( \text{NH}_4^+ \) is -3.
- The oxidation number of oxygen is generally -2.
3. **Set up the equation**: Let the oxidation number of vanadium be \( X \).
- For \( \text{VO}_2^- \): \( X + 2(-2) = -1 \)
- This simplifies to \( X - 4 = -1 \) or \( X = +3 \).

### Step 2: Analyze the second compound, \( \text{VO}_2^+ \)

1. **Set up the equation**: Let the oxidation number of vanadium be \( X \).
- For \( \text{VO}_2^+ \): \( X + 2(-2) = +1 \)
- This simplifies to \( X - 4 = +1 \) or \( X = +5 \).

### Step 3: Analyze the third compound, \( \text{VSO}_4 \)

1. **Set up the equation**: Let the oxidation number of vanadium be \( X \).
- For \( \text{VSO}_4 \): \( X + \text{(oxidation state of SO}_4^{2-}) = 0 \)
- The oxidation state of \( \text{SO}_4^{2-} \) is -2.
- Thus, \( X - 2 = 0 \) or \( X = +2 \).

### Step 4: Analyze the fourth compound, \( \text{VOSO}_4 \)

1. **Set up the equation**: Let the oxidation number of vanadium be \( X \).
- For \( \text{VOSO}_4 \): \( X + 2(-2) + (-2) = 0 \)
- This simplifies to \( X - 4 = 0 \) or \( X = +4 \).

### Conclusion

From the analysis, we find that vanadium has an oxidation number of +4 in the compound \( \text{VOSO}_4 \).

### Final Answer
The compound in which vanadium has an oxidation number of +4 is \( \text{VOSO}_4 \).

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