Question

In which of the following reactions the change in oxidation statc of sulphur is fractional? (1) \(2 \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}+\mathrm{I}_{2} \longrightarrow \mathrm{Na}_{2} \mathrm{~S}_{4} \mathrm{O}_{6}+2 \mathrm{Nal}\) (2) \(\mathrm{Na}_{2} \mathrm{SO}_{3}+\mathrm{H}_{2} \mathrm{O}+\mathrm{J}_{2} \longrightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}+2 \mathrm{HI}\) (3) \(\mathrm{H}_{2} \mathrm{~S}+\mathrm{I}_{2} \longrightarrow 2 \mathrm{HI}\) (4) \(\mathrm{PbS}+4 \mathrm{H}_{2} \mathrm{O}_{2} \longrightarrow \mathrm{PbSO}_{4}+4 \mathrm{H}_{2} \mathrm{O}\)

Short answer

Reaction (1) has a fractional change in sulfur oxidation state.

Step-by-step solution

Understand the oxidation states of sulfur in reactants and products

Consider the reaction given and determine the oxidation states of sulfur in all the reactants and products in the equations.

Determine the oxidation states in each reaction

For each reaction, calculate the oxidation states of sulfur in both the reactants and products: 1. In the reaction: (1) \(2 \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3} + \mathrm{I}_{2} \longrightarrow \mathrm{Na}_{2} \mathrm{~S}_{4} \mathrm{O}_{6} + 2 \mathrm{Nal}\) - In \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\), sulfur has oxidation states of \(+2\) (+S) and \(+6+(-2)\times 3=+2\). - In \(\mathrm{Na}_{2} \mathrm{~S}_{4} \mathrm{O}_{6}\), the average oxidation state of sulfur is \(\frac{2 \times +2 + (4 \times -1)}{4} =\) fractional (1). 2. In the reaction: (2) \(\mathrm{Na}_{2} \mathrm{SO}_{3} + \mathrm{H}_{2} \mathrm{O} + \mathrm{I}_{2} \longrightarrow \mathrm{Na}_{2} \mathrm{SO}_{4} + 2 \mathrm{HI}\): - In \(\mathrm{Na}_{2} \mathrm{SO}_{3}\), sulfur has an oxidation state of \(+4\). - In \(\mathrm{Na}_{2} \mathrm{SO}_{4}\), sulfur has an oxidation state of \(+6\). 3. In the reaction: (3) \(\mathrm{H}_{2} \mathrm{~S} + \mathrm{I}_{2} \longrightarrow 2 \mathrm{HI}\): - In \(\mathrm{H}_{2} \mathrm{~S}\), sulfur has an oxidation state of \(-2\). - In \(\mathrm{HI}\), iodine changes but sulfur is not present as a product. 4. In the reaction: (4) \(\mathrm{PbS} + 4 \mathrm{H}_{2} \mathrm{O}_{2} \longrightarrow \mathrm{PbSO}_{4} + 4 \mathrm{H}_{2} \mathrm{O}\): - In \(\mathrm{PbS}\), sulfur has an oxidation state of \(-2\). - In \(\mathrm{PbSO}_{4}\), sulfur has an oxidation state of \(+6\).

Identify the reaction where sulfur changes to a fractional oxidation state

Based on the calculations: Only in reaction (1) \(2 \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3} + \mathrm{I}_{2} \longrightarrow \mathrm{Na}_{2} \mathrm{~S}_{4} \mathrm{O}_{6} + 2 \mathrm{Nal}\), sulfur has a fractional oxidation state of 2.5 in the product \(\mathrm{Na}_{2} \mathrm{~S}_{4} \mathrm{O}_{6}\).

Key concepts

fractional oxidation states

When dealing with oxidation states, we often come across whole number values. However, in some cases, the oxidation state of an element within a compound can be a fraction. This usually happens when there's a mix of different oxidation states of the same element within a compound.
For example, let's consider reaction (1) from the exercise: \(2 \mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3} + \mathrm{I}_{2} \rightarrow \mathrm{Na}_{2} \mathrm{S}_{4} \mathrm{O}_{6} + 2 \mathrm{NaI}\). Here, the compound \(\mathrm{Na}_{2} \mathrm{S}_{4} \mathrm{O}_{6}\) contains sulfur atoms with different oxidation states. The average oxidation state can be calculated:
\[\frac{2(2) + 4(-1)}{4} = 0.5\]
Note that 0.5 here is an average, implying a mix of oxidation states, which is referred to as a fractional oxidation state.
This concept helps in identifying specific reactions where elements feature fractional states.

redox reactions

Redox reactions involve the transfer of electrons between two species. Oxidation refers to the loss of electrons, while reduction refers to the gain. In the context of our problem:
Consider the reaction (2) \(\mathrm{Na}_{2} \mathrm{SO}_{3} + \mathrm{H}_{2} \mathrm{O} + \mathrm{I}_{2} \rightarrow \mathrm{Na}_{2} \mathrm{SO}_{4} + 2 \mathrm{HI}\).
Here, sodium sulfite (\(\mathrm{Na}_{2} \mathrm{SO}_{3}\)) is oxidized:
- Sulfur in \(\mathrm{Na}_{2} \mathrm{SO}_{3}\): +4
- Sulfur in \(\mathrm{Na}_{2} \mathrm{SO}_{4}\): +6
The sulfur atom increases its oxidation state from +4 to +6, indicating loss of electrons (oxidation). Meanwhile, iodine (\(\mathrm{I}_{2}\)) is reduced, as its oxidation state changes from 0 in \(\mathrm{I}_{2}\) to -1 in \(\mathrm{HI}\).
Understanding these changes helps identify and analyze redox reactions.

chemical reactions analysis

Analyzing chemical reactions requires understanding the changes in oxidation states of the elements involved. This involves calculating the oxidation states before and after the reaction to determine which elements are oxidized and which are reduced.
For example, in reaction (4) \(\mathrm{PbS} + 4 \mathrm{H}_{2} \mathrm{O}_{2} \rightarrow \mathrm{PbSO}_{4} + 4 \mathrm{H}_{2} \mathrm{O}\):

• Sulfur in \(\mathrm{PbS}\) has an oxidation state of -2.
• After the reaction, sulfur in \(\mathrm{PbSO}_{4}\) has an oxidation state of +6.

This significant increase in the oxidation number from -2 to +6 indicates sulfur has undergone oxidation.
By following these steps, students can systematically analyze any given chemical reaction, identify redox processes, and understand the behavior of different elements.