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 in the compound \( Rb_4Na[HV_{10}O_{28}] \), we will follow these steps:

### Step 1: Identify the known oxidation states
- **Rubidium (Rb)**: As an alkali metal, it has an oxidation state of +1. Since there are 4 rubidium atoms, their total contribution is \( 4 \times (+1) = +4 \).
- **Sodium (Na)**: Also an alkali metal, it has an oxidation state of +1. Thus, for one sodium atom, the contribution is \( +1 \).
- **Hydrogen (H)**: In this compound, hydrogen typically has an oxidation state of +1. So, the contribution is \( +1 \).
- **Oxygen (O)**: Oxygen generally has an oxidation state of -2. Since there are 28 oxygen atoms, their total contribution is \( 28 \times (-2) = -56 \).

### Step 2: Set up the equation
Let the oxidation state of vanadium (V) be represented as \( x \). The total contribution from vanadium will be \( 10x \) (since there are 10 vanadium atoms).

Now, we can set up the equation based on the total charge of the compound, which is zero:

\[ (+4) + (+1) + (+1) + (10x) + (-56) = 0 \]

### Step 3: Simplify the equation
Combining the known values:

\[ 4 + 1 + 1 + 10x - 56 = 0 \]

This simplifies to:

\[ 10x - 50 = 0 \]

### Step 4: Solve for \( x \)
Now, we can solve for \( x \):

\[ 10x = 50 \]
\[ x = \frac{50}{10} = 5 \]

### Conclusion
The oxidation state of vanadium (V) in the compound \( Rb_4Na[HV_{10}O_{28}] \) is +5.

### Final Answer
The oxidation number of the underlined atom (V) is +5.

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