What is average oxidation state state of sulphur in product formed in given reaction?
`Na_(2)SO_(3)+Na_(2)S+I_(2)to . . . .+NaI`

Explanation

To find the average oxidation state of sulfur in the product formed in the given reaction, we will follow these steps:

### Step 1: Identify the product of the reaction
The reaction given is:
\[ \text{Na}_2\text{SO}_3 + \text{Na}_2\text{S} + \text{I}_2 \rightarrow \text{Na}_2\text{S}_2\text{O}_3 + 2\text{NaI} \]

From the reaction, we can see that the product formed is sodium thiosulfate, \(\text{Na}_2\text{S}_2\text{O}_3\).

### Step 2: Write the formula for the product
The formula for the product is:
\[ \text{Na}_2\text{S}_2\text{O}_3 \]

### Step 3: Assign oxidation states
In the compound \(\text{Na}_2\text{S}_2\text{O}_3\):
- Sodium (Na) has an oxidation state of +1.
- Oxygen (O) has an oxidation state of -2.

### Step 4: Set up the equation for oxidation states
Let the oxidation state of sulfur (S) be \(x\). In \(\text{Na}_2\text{S}_2\text{O}_3\), we have:
- 2 sodium atoms contribute: \(2 \times (+1) = +2\)
- 3 oxygen atoms contribute: \(3 \times (-2) = -6\)
- There are 2 sulfur atoms, contributing \(2x\).

The overall charge of the compound is neutral (0). Therefore, we can set up the equation:
\[ 2 + 2x - 6 = 0 \]

### Step 5: Solve for \(x\)
Rearranging the equation gives:
\[ 2x - 4 = 0 \]
\[ 2x = 4 \]
\[ x = 2 \]

### Step 6: Calculate the average oxidation state of sulfur
Since there are 2 sulfur atoms in \(\text{Na}_2\text{S}_2\text{O}_3\) and each has an oxidation state of +2, the average oxidation state of sulfur in the product is:
\[ \text{Average oxidation state} = \frac{2 \times 2}{2} = 2 \]

### Final Answer
The average oxidation state of sulfur in the product \(\text{Na}_2\text{S}_2\text{O}_3\) is +2.

---