Determine the oxidation number of the underlined atom in `Rb_4Na[HVul(10)O_(28)]`

Options

A. `+9`
B. `-4`
C. 0
D. `+5`

Answer

To determine the oxidation number of the underlined atom (Vanadium, V) in the compound \( Rb_4Na[HV_{10}O_{28}] \), we will follow these steps:

### Step 1: Identify the oxidation states of known elements
- **Rubidium (Rb)**: The oxidation number of rubidium is +1.
- **Sodium (Na)**: The oxidation number of sodium is +1.
- **Hydrogen (H)**: The oxidation number of hydrogen is +1.
- **Oxygen (O)**: The oxidation number of oxygen is -2.

### Step 2: Write the equation for the overall charge
The compound is neutral, so the sum of the oxidation numbers must equal zero. We can express this as:
\[ 4(\text{oxidation number of Rb}) + 1(\text{oxidation number of Na}) + 1(\text{oxidation number of H}) + 10(\text{oxidation number of V}) + 28(\text{oxidation number of O}) = 0 \]

### Step 3: Substitute the known oxidation states into the equation
Substituting the known values into the equation gives:
\[ 4(+1) + 1(+1) + 1(+1) + 10x + 28(-2) = 0 \]
Where \( x \) is the oxidation number of V.

### Step 4: Simplify the equation
Now, simplify the equation:
\[ 4 + 1 + 1 + 10x - 56 = 0 \]
Combine the constants:
\[ 6 + 10x - 56 = 0 \]
This simplifies to:
\[ 10x - 50 = 0 \]

### Step 5: Solve for the oxidation number of V
Now, isolate \( x \):
\[ 10x = 50 \\ x = 5 \]

### Conclusion
The oxidation number of Vanadium (V) in \( Rb_4Na[HV_{10}O_{28}] \) is **+5**.