What are products in the following equation, `S+OH^(-) to `?

Options

A. `H_(2)S`
B. `S^(2-)`
C. `S_(2)O_(3)^(2-)`
D. `SO_(3)^(2-)`

Answer

To determine the products of the reaction between sulfur (S) and hydroxide ion (OH⁻), we can follow these steps:

### Step 1: Identify the Reactants
The reactants in the equation are sulfur (S) and hydroxide ion (OH⁻).

### Step 2: Analyze the Reaction
Sulfur can react with hydroxide ions in a basic medium. In this case, we expect sulfur to undergo a reaction where it can be both oxidized and reduced.

### Step 3: Determine Possible Products
When sulfur reacts with hydroxide ions, it can form:
- Sulfide ion (S²⁻)
- Thiosulfate ion (S₂O₃²⁻)

### Step 4: Write the Balanced Reaction
The balanced reaction can be represented as:
\[ S + 2OH⁻ \rightarrow S^{2-} + S_2O_3^{2-} + H_2O \]

### Step 5: Identify Oxidation States
- In elemental sulfur (S), the oxidation state is 0.
- In sulfide ion (S²⁻), the oxidation state is -2.
- In thiosulfate ion (S₂O₃²⁻), we can calculate the oxidation state of sulfur. Let the oxidation state of sulfur in thiosulfate be x:
\[ 2x + 3(-2) = -2 \quad \Rightarrow \quad 2x - 6 = -2 \quad \Rightarrow \quad 2x = 4 \quad \Rightarrow \quad x = +2 \]
So, in thiosulfate, sulfur has an oxidation state of +2.

### Step 6: Classify the Reaction
Since sulfur is both oxidized (from 0 to +2) and reduced (from 0 to -2), this reaction is classified as a disproportionation reaction.

### Step 7: Evaluate the Options
From the analysis, the products formed are:
- Sulfide ion (S²⁻) - Correct
- Thiosulfate ion (S₂O₃²⁻) - Correct
- Hydrogen sulfide (H₂S) - Incorrect
- Sulfite ion (SO₃²⁻) - Incorrect

### Conclusion
The products of the reaction are:
- Sulfide ion (S²⁻)
- Thiosulfate ion (S₂O₃²⁻)

### Final Answer
The correct options are B (Sulfide ion) and C (Thiosulfate ion).

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