Sulphur shows a large number of oxidation states in its compounds such as `H_2S, SO_2, H_2SO_3,Na_2S_2O_3, H_2SO_4, H_2S_2O_7, H_2S_2O_4` and elemental S . In the reactions of sulphur compounds , the oxidation number of sulphur increases and decreases in different reactions and the compounds act as reducing and oxidising agents.
(i) `H_2S +HNO_3 rarr NO+S+H_2O`
(ii) `2Na_2S_2O_3+I_2rarr Na_2S_4O_6+2NaI`
(iii) `H_2SO_4+2HBr rarr 2H_2O+Br_2+SO_2`
(iv) `S_8+12OH^(-) rarr 4S^(2-) +2S_2O_3^(2-) +6H_2O`
What is the change in oxidation number of sulphur in reaction (ii)

Answer

To determine the change in oxidation number of sulfur in the reaction (ii) `2Na2S2O3 + I2 → Na2S4O6 + 2NaI`, we will follow these steps:

### Step 1: Identify the oxidation states of sulfur in the reactants.
In the compound `Na2S2O3` (sodium thiosulfate), we need to find the oxidation state of sulfur.

Using the X-method:
- Let the oxidation state of sulfur be \( x \).
- Sodium (Na) has an oxidation state of +1, and there are 2 sodium atoms, contributing +2.
- Oxygen (O) has an oxidation state of -2, and there are 3 oxygen atoms, contributing -6.

The equation becomes:
\[ 2(+1) + 2x + 3(-2) = 0 \]
\[ 2 + 2x - 6 = 0 \]
\[ 2x - 4 = 0 \]
\[ 2x = 4 \]
\[ x = +2 \]

Thus, the oxidation state of sulfur in `Na2S2O3` is +2.

### Step 2: Identify the oxidation states of sulfur in the products.
In the product `Na2S4O6` (sodium tetrathionate), we again use the X-method:
- Let the oxidation state of sulfur be \( y \).
- Sodium (Na) has an oxidation state of +1, and there are 2 sodium atoms, contributing +2.
- Oxygen (O) has an oxidation state of -2, and there are 6 oxygen atoms, contributing -12.

The equation becomes:
\[ 2(+1) + 4y + 6(-2) = 0 \]
\[ 2 + 4y - 12 = 0 \]
\[ 4y - 10 = 0 \]
\[ 4y = 10 \]
\[ y = +2.5 \]

Thus, the oxidation state of sulfur in `Na2S4O6` is +2.5.

### Step 3: Calculate the change in oxidation state.
Now, we can find the change in oxidation state of sulfur:
- Initial oxidation state in `Na2S2O3`: +2
- Final oxidation state in `Na2S4O6`: +2.5

The change in oxidation state is:
\[ \text{Change} = +2.5 - (+2) = +0.5 \]

### Final Answer:
The change in oxidation number of sulfur in reaction (ii) is +0.5.

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