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 from the reaction \( \text{Na}_2\text{SO}_3 + \text{Na}_2\text{S} + \text{I}_2 \), we will follow these steps:

### Step 1: Identify the Product
The reactants given are sodium sulfite (\( \text{Na}_2\text{SO}_3 \)) and sodium sulfide (\( \text{Na}_2\text{S} \)) in the presence of iodine (\( \text{I}_2 \)). The product formed from this reaction is sodium thiosulfate (\( \text{Na}_2\text{S}_2\text{O}_3 \)).

### Step 2: Determine the Structure of the Product
The structure of sodium thiosulfate (\( \text{S}_2\text{O}_3^{2-} \)) can be represented as:
- One sulfur atom is bonded to another sulfur atom (S-S bond).
- The other sulfur atom is bonded to three oxygen atoms (two with single bonds and one with a double bond).

### Step 3: Assign Oxidation States
To find the oxidation states of sulfur in \( \text{S}_2\text{O}_3^{2-} \):
1. Each oxygen typically has an oxidation state of -2 (for the double-bonded oxygen) and -1 (for the single-bonded oxygens).
2. Let the oxidation states of the two sulfur atoms be \( x \) and \( y \).

The charge balance equation for the thiosulfate ion can be set up as:
\[ x + y + 3(-2) = -2 \]
This simplifies to:
\[ x + y - 6 = -2 \implies x + y = 4 \]

### Step 4: Determine the Individual Oxidation States
From the structure, we can deduce:
- The sulfur atom that is double-bonded to oxygen typically has a higher oxidation state.
- The sulfur atom that is bonded to another sulfur atom will have an oxidation state of 0.

Assuming \( x = 2 \) (for the sulfur bonded to oxygen) and \( y = 0 \) (for the sulfur bonded to another sulfur), we can check:
\[ 2 + 0 = 4 \]
This is consistent with our earlier equation.

### Step 5: Calculate the Average Oxidation State
Now, we can calculate the average oxidation state of sulfur in \( \text{S}_2\text{O}_3^{2-} \):
\[ \text{Average oxidation state} = \frac{x + y}{\text{number of sulfur atoms}} = \frac{2 + 0}{2} = 1 \]

### Conclusion
Thus, the average oxidation state of sulfur in the product \( \text{Na}_2\text{S}_2\text{O}_3 \) is **1**.

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