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 number of vanadium in various compounds, we will analyze each compound step by step.

### Step-by-Step Solution:

1. **Identify the Compounds**: The question asks for the oxidation state of vanadium in different compounds. Let's consider the compounds mentioned in the video transcript:
- NH₄VO₂
- K₄[VCN]₆
- VSO₄
- VOSO₄

2. **Analyze NH₄VO₂**:
- The ammonium ion (NH₄⁺) has a charge of +1.
- Oxygen (O) has an oxidation state of -2. Since there are 2 oxygen atoms, the total contribution is -4.
- Let the oxidation state of vanadium (V) be \( x \).
- The equation becomes:
\[ +1 + x - 4 = 0 \]
Solving for \( x \):
\[ x - 3 = 0 \implies x = +3 \]
- Thus, vanadium in NH₄VO₂ has an oxidation state of +3.

3. **Analyze K₄[VCN]₆**:
- Potassium (K) has a charge of +1. For 4 potassium ions, the total is +4.
- The cyanide ion (CN⁻) has a charge of -1. For 6 cyanide ions, the total is -6.
- Let the oxidation state of vanadium (V) be \( x \).
- The equation becomes:
\[ +4 + x - 6 = 0 \]
Solving for \( x \):
\[ x - 2 = 0 \implies x = +2 \]
- Thus, vanadium in K₄[VCN]₆ has an oxidation state of +2.

4. **Analyze VSO₄**:
- The sulfate ion (SO₄²⁻) has a charge of -2.
- Let the oxidation state of vanadium (V) be \( x \).
- The equation becomes:
\[ x - 2 = 0 \]
Solving for \( x \):
\[ x = +2 \]
- Thus, vanadium in VSO₄ has an oxidation state of +2.

5. **Analyze VOSO₄**:
- Oxygen (O) has an oxidation state of -2.
- The sulfate ion (SO₄²⁻) has a charge of -2.
- Let the oxidation state of vanadium (V) be \( x \).
- The equation becomes:
\[ x - 2 - 2 = 0 \]
Solving for \( x \):
\[ x = +4 \]
- Thus, vanadium in VOSO₄ has an oxidation state of +4.

### Conclusion:
The compound in which vanadium has an oxidation number of +4 is **VOSO₄**.