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

Options

A. `NH_(4)VO^(2)`
B. `K_(4)[V(CN)_(6)]`
C. `VSO_(4)`
D. `VOSO_(4)`

Answer

To determine the oxidation state of vanadium in the given compounds, we will analyze each compound step by step.

### Step 1: Analyze the first compound, NH4VO2
1. **Write the formula**: NH4VO2
2. **Assign oxidation states**:
- NH4 (ammonium) has an oxidation state of +1.
- Oxygen (O) has an oxidation state of -2.
3. **Set up the equation**:
- Let the oxidation state of vanadium (V) be \( X \).
- The equation based on the charges will be:
\[ +1 + X + (-2 \times 2) = 0 \]
- Simplifying gives:
\[ +1 + X - 4 = 0 \implies X - 3 = 0 \implies X = +3 \]
4. **Conclusion**: In NH4VO2, vanadium has an oxidation state of +3.

### Step 2: Analyze the second compound, K4VN6
1. **Write the formula**: K4VN6
2. **Assign oxidation states**:
- Potassium (K) has an oxidation state of +1.
- Cyanide (CN) has an oxidation state of -1.
3. **Set up the equation**:
- Let the oxidation state of vanadium be \( X \).
- The equation based on the charges will be:
\[ (4 \times +1) + X + (-1 \times 6) = 0 \]
- Simplifying gives:
\[ 4 + X - 6 = 0 \implies X - 2 = 0 \implies X = +2 \]
4. **Conclusion**: In K4VN6, vanadium has an oxidation state of +2.

### Step 3: Analyze the third compound, VSO4
1. **Write the formula**: VSO4
2. **Assign oxidation states**:
- Sulfate (SO4) has an oxidation state of -2.
3. **Set up the equation**:
- Let the oxidation state of vanadium be \( X \).
- The equation based on the charges will be:
\[ X + (-2) = 0 \]
- Simplifying gives:
\[ X - 2 = 0 \implies X = +2 \]
4. **Conclusion**: In VSO4, vanadium has an oxidation state of +2.

### Step 4: Analyze the fourth compound, VOSO4
1. **Write the formula**: VOSO4
2. **Assign oxidation states**:
- Oxygen (O) has an oxidation state of -2.
- Sulfate (SO4) has an oxidation state of -2.
3. **Set up the equation**:
- Let the oxidation state of vanadium be \( X \).
- The equation based on the charges will be:
\[ X + (-2) + (-2) = 0 \]
- Simplifying gives:
\[ X - 4 = 0 \implies X = +4 \]
4. **Conclusion**: In VOSO4, vanadium has an oxidation state of +4.

### Final Answer
The compound in which vanadium has an oxidation state of +4 is **VOSO4**.