Oxidation state of "V" in `Rb_(4)K[HV_(10)O_(28)]` is .

Options

A. `+5`
B. `+6`
C. `+(7)/(5)`
D. `+4`

Answer

To find the oxidation state of vanadium (V) in the complex \( Rb_4K[HV_{10}O_{28}] \), we will follow these steps:

### Step 1: Identify the oxidation states of known elements
- Rubidium (Rb) is an alkali metal and has an oxidation state of +1.
- Potassium (K) also has an oxidation state of +1.
- Hydrogen (H) has an oxidation state of +1.
- Oxygen (O) has an oxidation state of -2.

### Step 2: Set up the equation for the oxidation state
Let the oxidation state of vanadium (V) be \( x \).

The complex can be represented as:
\[ 4 \times (+1) + 1 \times (+1) + 10 \times (x) + 28 \times (-2) = 0 \]

### Step 3: Substitute and simplify the equation
Substituting the known values into the equation:
\[ 4(1) + 1(1) + 10x + 28(-2) = 0 \]
This simplifies to:
\[ 4 + 1 + 10x - 56 = 0 \]
Combining the constants:
\[ 5 + 10x - 56 = 0 \]
This further simplifies to:
\[ 10x - 51 = 0 \]

### Step 4: Solve for \( x \)
Now, isolate \( x \):
\[ 10x = 51 \]
\[ x = \frac{51}{10} = 5.1 \]

### Conclusion
The oxidation state of vanadium (V) in \( Rb_4K[HV_{10}O_{28}] \) is +5.