In the reaction `2Na_(2)S_(2)O_(3)+I_(2) rarr Na_(2)S_(4)O_(6)+2NaI`, the oxidation state of S is -

Options

A. Increased
B. Decreased
C. Remain same
D. None

Answer

To determine the oxidation state of sulfur in the reaction:

\[ 2 \text{Na}_2\text{S}_2\text{O}_3 + \text{I}_2 \rightarrow \text{Na}_2\text{S}_4\text{O}_6 + 2 \text{NaI} \]

we can follow these steps:

### Step 1: Assign oxidation states to known elements
- Sodium (Na) is in Group 1 and has an oxidation state of +1.
- Oxygen (O) typically has an oxidation state of -2.

### Step 2: Write the oxidation state equation for the reactants
For the reactants \( \text{Na}_2\text{S}_2\text{O}_3 \):
- There are 2 sodium atoms: \( 2 \times (+1) = +2 \)
- There are 3 oxygen atoms: \( 3 \times (-2) = -6 \)
- Let the oxidation state of sulfur (S) be \( x \). There are 2 sulfur atoms, so we have \( 2x \).

The overall charge of the compound must equal 0:
\[ 2 + 2x - 6 = 0 \]
Simplifying this gives:
\[ 2x - 4 = 0 \implies 2x = 4 \implies x = +2 \]

### Step 3: Write the oxidation state equation for the products
For the products \( \text{Na}_2\text{S}_4\text{O}_6 \):
- Again, there are 2 sodium atoms: \( 2 \times (+1) = +2 \)
- There are 6 oxygen atoms: \( 6 \times (-2) = -12 \)
- Let the oxidation state of sulfur in this compound be \( y \). There are 4 sulfur atoms, so we have \( 4y \).

The overall charge of the compound must equal 0:
\[ 2 + 4y - 12 = 0 \]
Simplifying this gives:
\[ 4y - 10 = 0 \implies 4y = 10 \implies y = +2.5 \]

### Step 4: Conclusion
The oxidation state of sulfur changes from +2 in \( \text{Na}_2\text{S}_2\text{O}_3 \) to +2.5 in \( \text{Na}_2\text{S}_4\text{O}_6 \). Therefore, the oxidation state of sulfur in the reaction is:

**Answer: +2.5**