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

Options

A. `+7`
B. `+6`
C. `+4`
D. `+4.5`

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 1: Identify the components of the compound
The compound consists of sodium (Na), tungsten (W), oxygen (O), and water (H2O). The water is in a hydrated form, which does not affect the oxidation state calculation of tungsten.

### Step 2: Write the formula without the water
We can simplify the problem by focusing on the anhydrous part of the compound:
\[ \text{Na}_2\text{W}_4\text{O}_{13} \]

### Step 3: Assign oxidation states to known elements
1. Sodium (Na) has an oxidation state of +1. Since there are 2 sodium atoms:
\[ \text{Total oxidation state from Na} = 2 \times (+1) = +2 \]

2. Oxygen (O) typically has an oxidation state of -2. Since there are 13 oxygen atoms:
\[ \text{Total oxidation state from O} = 13 \times (-2) = -26 \]

### Step 4: Set up the equation for the oxidation state of tungsten
Let the oxidation state of tungsten (W) be \( x \). Since there are 4 tungsten atoms, the total contribution from tungsten will be:
\[ \text{Total oxidation state from W} = 4x \]

### Step 5: Write the overall charge balance equation
The sum of the oxidation states in a neutral compound must equal zero:
\[ \text{Total oxidation state from Na} + \text{Total oxidation state from W} + \text{Total oxidation state from O} = 0 \]
Substituting the values we have:
\[ +2 + 4x - 26 = 0 \]

### Step 6: Solve for \( x \)
Rearranging the equation:
\[ 4x - 24 = 0 \]
\[ 4x = 24 \]
\[ x = \frac{24}{4} = 6 \]

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

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