1 mole of `H_(3) PO_(2)` reacts completely to two moles of `I_(2)`. `I_(2)` converts to `I^(-)` , what will be the oxidation state of P in the product?

Answer

To determine the oxidation state of phosphorus (P) in the product formed when 1 mole of H₃PO₂ reacts completely with 2 moles of I₂, we can follow these steps:

### Step 1: Write the reaction
The reaction can be represented as:
\[ \text{H}_3\text{PO}_2 + 2 \text{I}_2 \rightarrow \text{Products} \]

### Step 2: Identify the oxidation states of reactants
In H₃PO₂ (phosphorous acid), we need to find the oxidation state of phosphorus (P). The oxidation states of hydrogen (H) is +1 and oxygen (O) is -2.

Let the oxidation state of phosphorus be \( x \). The overall charge of the molecule must be zero:
\[ 3(+1) + x + 2(-2) = 0 \]
\[ 3 + x - 4 = 0 \]
\[ x - 1 = 0 \]
\[ x = +1 \]

So, the oxidation state of phosphorus in H₃PO₂ is +1.

### Step 3: Determine the change in oxidation state
I₂ is reduced to I⁻. In this process, iodine (I) goes from an oxidation state of 0 in I₂ to -1 in I⁻.

Since 2 moles of I₂ are used, this indicates that 2 moles of iodine are reduced, meaning that the phosphorus must be oxidized to balance the reaction.

### Step 4: Calculate the oxidation state of phosphorus in the product
When phosphorus is oxidized, its oxidation state increases. In the product formed, we can assume that phosphorus is in a higher oxidation state.

Given that I₂ is reduced and phosphorus is oxidized, we can deduce that the oxidation state of phosphorus in the product will be +3. This is because it can be inferred that phosphorus in H₃PO₂ is oxidized to a higher oxidation state when it reacts with iodine.

### Conclusion
Thus, the oxidation state of phosphorus (P) in the product formed after the reaction is +3.