Best reductant reduces chloroform to:

Options

A. `C_(2)H_(4)`
B. `CH_(4)`
C. `C_(2)H_(2)`
D. `C_(2)H_(6)`

Answer

To determine the best reductant that reduces chloroform (CHCl3) to its lowest oxidation state, we can follow these steps:

### Step 1: Determine the oxidation state of carbon in chloroform (CHCl3).
- The oxidation state of carbon (C) can be calculated using the formula:
\[ \text{Oxidation state of C} = x + 1 + 3(-1) = 0 \]
where \(x\) is the oxidation state of carbon. Simplifying this gives:
\[ x - 2 = 0 \implies x = +2 \]
- Therefore, the oxidation state of carbon in chloroform is +2.

### Step 2: Identify the oxidation states of the products in the options.
1. **C2H4 (Ethylene)**:
- The oxidation state of carbon in C2H4 is calculated as follows:
\[ 2x - 4 = 0 \implies 2x = 4 \implies x = +2 \]

2. **CH4 (Methane)**:
- The oxidation state of carbon in CH4 is:
\[ x + 4(-1) = 0 \implies x - 4 = 0 \implies x = -4 \]

3. **C2H2 (Acetylene)**:
- The oxidation state of carbon in C2H2 is:
\[ 2x - 2 = 0 \implies 2x = 2 \implies x = +1 \]

4. **C (Carbon)**:
- The oxidation state of carbon in elemental carbon is 0.

### Step 3: Compare the oxidation states.
- The oxidation states we found are:
- C2H4: +2
- CH4: -4
- C2H2: +1
- C: 0

### Step 4: Identify the lowest oxidation state.
- The lowest oxidation state among the products is -4, which corresponds to methane (CH4).

### Step 5: Conclusion.
- Therefore, the best reductant that reduces chloroform to its lowest oxidation state is **CH4** (methane).

### Final Answer:
The best reductant reduces chloroform to **CH4** (methane).

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