The oxidation state of tungsten in `Na_(2)W_(4)O_(13).10H_(2)O` is ?

Options

A. `+12`
B. `+6`
C. `0`
D. `+8`

Answer

To find the oxidation state of tungsten in the compound \( \text{Na}_2\text{W}_4\text{O}_{13} \cdot 10\text{H}_2\text{O} \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the oxidation states of known elements**:
- Sodium (Na) is an alkali metal and has an oxidation state of +1.
- Oxygen (O) typically has an oxidation state of -2.

2. **Write the formula for the compound**:
- The compound is \( \text{Na}_2\text{W}_4\text{O}_{13} \cdot 10\text{H}_2\text{O} \).
- We will focus on the \( \text{Na}_2\text{W}_4\text{O}_{13} \) part for calculating the oxidation state of tungsten.

3. **Set up the equation for the oxidation states**:
- Let the oxidation state of tungsten (W) be \( x \).
- The total oxidation state equation can be set up as follows:
\[ 2(\text{oxidation state of Na}) + 4(\text{oxidation state of W}) + 13(\text{oxidation state of O}) = 0 \]
- Substituting the known oxidation states:
\[ 2(+1) + 4(x) + 13(-2) = 0 \]

4. **Simplify the equation**:
- This simplifies to:
\[ 2 + 4x - 26 = 0 \]
- Combine like terms:
\[ 4x - 24 = 0 \]

5. **Solve for \( x \)**:
- Rearranging gives:
\[ 4x = 24 \]
- Dividing both sides by 4:
\[ x = 6 \]

6. **Conclusion**:
- The oxidation state of tungsten in \( \text{Na}_2\text{W}_4\text{O}_{13} \cdot 10\text{H}_2\text{O} \) is +6.

### Final Answer:
The oxidation state of tungsten (W) is +6.