HCHO with conc. alkali forms two compounds. The change in oxidation number would be-

Options

A. (0 to –2) in both the compounds
B. (0 to +2) in both the compounds
C. (0 to +2) in one compound and (0 to –2) in the second compound
D. All are correct

Answer

To determine the change in oxidation number when HCHO (formaldehyde) reacts with concentrated alkali (like KOH), we can follow these steps:

### Step 1: Identify the Reactants and Products
The reactant is HCHO (formaldehyde) and the reagent is concentrated KOH. The reaction that occurs is known as the Cannizzaro reaction, which is a disproportionation reaction.

**Hint:** Look for the type of reaction that occurs when a carbonyl compound lacks alpha hydrogen atoms.

### Step 2: Understand the Cannizzaro Reaction
In the Cannizzaro reaction, one molecule of the aldehyde is oxidized to form a carboxylic acid salt, while another molecule is reduced to form an alcohol. In this case, formaldehyde will yield methanol (CH3OH) and potassium formate (HCOOK).

**Hint:** Remember that in disproportionation reactions, one species is oxidized while another is reduced.

### Step 3: Determine the Oxidation States
1. **For HCHO (formaldehyde):**
- The oxidation state of carbon (C) in HCHO is 0.

2. **For CH3OH (methanol):**
- The oxidation state of carbon in CH3OH can be calculated as follows:
- 3 Hydrogens contribute +3
- 1 Oxygen contributes -2
- Let the oxidation state of carbon be x.
- So, x + 3 - 2 = 0 → x = -2
- Therefore, the oxidation state of carbon in methanol is -2.

3. **For HCOOK (potassium formate):**
- The oxidation state of carbon in HCOOK can be calculated similarly:
- 1 Hydrogen contributes +1
- 2 Oxygens contribute -4 (2 × -2)
- Let the oxidation state of carbon be y.
- So, y + 1 - 4 + 1 = 0 → y = +2
- Therefore, the oxidation state of carbon in potassium formate is +2.

**Hint:** When calculating oxidation states, remember the charges contributed by hydrogen and oxygen.

### Step 4: Analyze the Changes in Oxidation States
- In the reaction:
- The carbon in HCHO changes from 0 to -2 in CH3OH (reduction).
- The carbon in HCHO changes from 0 to +2 in HCOOK (oxidation).

**Hint:** Identify which species undergoes oxidation and which undergoes reduction by looking at the changes in their oxidation states.

### Step 5: Conclusion
The overall change in oxidation number during the Cannizzaro reaction of HCHO with concentrated alkali results in:
- One molecule of formaldehyde is reduced (oxidation state decreases from 0 to -2).
- Another molecule of formaldehyde is oxidized (oxidation state increases from 0 to +2).

Thus, the change in oxidation number is both an increase and a decrease, confirming that the reaction involves both oxidation and reduction.

**Final Answer:** The change in oxidation number would be from 0 to +2 for one product and from 0 to -2 for the other product.