The oxidation number of Cr in the product of alkaline oxidative fusion of `FeCr_2O_4` is

Explanation

To find the oxidation number of chromium (Cr) in the product of the alkaline oxidative fusion of `FeCr2O4`, we will follow these steps:

### Step 1: Write the reaction for alkaline oxidative fusion
The reaction for the alkaline oxidative fusion of `FeCr2O4` can be represented as follows:
\[ \text{FeCr}_2\text{O}_4 + \text{NaOH} + \text{O}_2 \rightarrow \text{Na}_2\text{CrO}_4 + \text{Fe}_2\text{O}_3 + \text{H}_2\text{O} \]

### Step 2: Balance the reaction
To balance the reaction, we need to ensure that the number of atoms of each element is the same on both sides. The balanced equation is:
\[ 4 \text{FeCr}_2\text{O}_4 + 16 \text{NaOH} + 7 \text{O}_2 \rightarrow 8 \text{Na}_2\text{CrO}_4 + 8 \text{Fe}_2\text{O}_3 + 8 \text{H}_2\text{O} \]

### Step 3: Identify the product
The product formed from the reaction is sodium chromate, which is represented as `Na2CrO4`.

### Step 4: Determine the oxidation state of Cr in `Na2CrO4`
Let the oxidation state of chromium (Cr) be \( x \). In sodium chromate (\( \text{Na}_2\text{CrO}_4 \)), we can set up the equation based on the known oxidation states of sodium and oxygen:
- Sodium (Na) has an oxidation state of +1.
- Oxygen (O) has an oxidation state of -2.

The equation can be set up as follows:
\[ 2(+1) + x + 4(-2) = 0 \]

### Step 5: Solve for \( x \)
Now, simplifying the equation:
\[ 2 + x - 8 = 0 \]
\[ x - 6 = 0 \]
\[ x = +6 \]

### Conclusion
The oxidation number of chromium (Cr) in the product \( \text{Na}_2\text{CrO}_4 \) is +6.