Each of the following ion contains vanadium the +5 oxidation state except

Options

A. `VO_(2)^(+)`
B. `V(OH)_(4)^(+)`
C. `VO^(2+)`
D. `[VO_(3)OH]^(2-)`

Answer

To determine which of the given ions contains vanadium in an oxidation state other than +5, we will analyze each option step by step.

### Step 1: Analyze VO2⁺
Let the oxidation state of vanadium be \( x \).
- The oxidation state of oxygen (O) is -2.
- The overall charge of the ion is +1.

The equation can be set up as follows:
\[ x + 2(-2) = +1 \]
\[ x - 4 = +1 \]
\[ x = +1 + 4 = +5 \]
**Conclusion:** The oxidation state of vanadium in VO2⁺ is +5.

### Step 2: Analyze VO(OH)4⁺
Let the oxidation state of vanadium be \( x \).
- The oxidation state of hydroxide (OH) is -1.
- There are four hydroxide ions, contributing a total of -4.
- The overall charge of the ion is +1.

The equation can be set up as follows:
\[ x + 4(-1) = +1 \]
\[ x - 4 = +1 \]
\[ x = +1 + 4 = +5 \]
**Conclusion:** The oxidation state of vanadium in VO(OH)4⁺ is +5.

### Step 3: Analyze VO2²⁺
Let the oxidation state of vanadium be \( x \).
- The oxidation state of oxygen (O) is -2.
- The overall charge of the ion is +2.

The equation can be set up as follows:
\[ x + 2(-2) = +2 \]
\[ x - 4 = +2 \]
\[ x = +2 + 4 = +6 \]
**Conclusion:** The oxidation state of vanadium in VO2²⁺ is +6.

### Step 4: Analyze VO3OH²⁻
Let the oxidation state of vanadium be \( x \).
- The oxidation state of oxygen (O) is -2.
- The oxidation state of hydroxide (OH) is -1.
- The overall charge of the ion is -2.

The equation can be set up as follows:
\[ x + 3(-2) + (-1) = -2 \]
\[ x - 6 - 1 = -2 \]
\[ x - 7 = -2 \]
\[ x = -2 + 7 = +5 \]
**Conclusion:** The oxidation state of vanadium in VO3OH²⁻ is +5.

### Final Conclusion:
From the analysis:
- VO2⁺: +5
- VO(OH)4⁺: +5
- VO2²⁺: +6 (not +5)
- VO3OH²⁻: +5

**The ion that does not contain vanadium in the +5 oxidation state is VO2²⁺.**