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 the oxidation states of vanadium in the given compounds step by step.

### Step 1: Analyze the first compound, NH4VO2
- In NH4VO2, we can assign oxidation states to find the oxidation state of vanadium (V).
- The ammonium ion (NH4+) has a charge of +1.
- The two oxygen atoms (O) each have an oxidation state of -2, contributing a total of -4.
- Let the oxidation state of vanadium be x.
- The equation can be set up as follows:
\[ +1 + x + (-4) = 0 \]
Simplifying this gives:
\[ x - 3 = 0 \implies x = +3 \]
- Thus, in NH4VO2, vanadium has an oxidation state of +3.

### Step 2: Analyze the second compound, VO2+
- In VO2+, we again assign oxidation states.
- The two oxygen atoms contribute -4 (2 × -2).
- Let the oxidation state of vanadium be x.
- The equation is:
\[ x + (-4) = +1 \]
Simplifying gives:
\[ x - 4 = 1 \implies x = +5 \]
- Thus, in VO2+, vanadium has an oxidation state of +5.

### Step 3: Analyze the third compound, VSO4
- In VSO4, we assign oxidation states.
- The sulfate ion (SO4) has a charge of -2, with sulfur having an oxidation state of +6.
- Let the oxidation state of vanadium be x.
- The equation is:
\[ x + (-2) = 0 \]
Simplifying gives:
\[ x - 2 = 0 \implies x = +2 \]
- Thus, in VSO4, vanadium has an oxidation state of +2.

### Step 4: Analyze the fourth compound, VOSO4
- In VOSO4, we assign oxidation states.
- The sulfate ion (SO4) has a charge of -2, with sulfur having an oxidation state of +6.
- Let the oxidation state of vanadium be x.
- The equation is:
\[ x + (-2) = 0 \]
This can be rearranged to:
\[ x - 2 = 0 \implies x = +4 \]
- Thus, in VOSO4, vanadium has an oxidation state of +4.

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

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