The equivalent weight of `H_3PO_2`,when it disproportionates into `PH_3` and `H_3PO_4`, is

Options

A. 82
B. 61.5
C. 33
D. 20.5

Answer

To find the equivalent weight of \( H_3PO_2 \) when it disproportionates into \( PH_3 \) and \( H_3PO_4 \), we can follow these steps:

### Step 1: Write the Disproportionation Reaction
The reaction for the disproportionation of \( H_3PO_2 \) can be written as:
\[ 2 H_3PO_2 \rightarrow PH_3 + H_3PO_4 \]

### Step 2: Determine the Changes in Oxidation States
In this reaction:
- The phosphorus in \( H_3PO_2 \) is oxidized to \( H_3PO_4 \) (oxidation state increases from +1 to +5).
- The phosphorus in \( H_3PO_2 \) is reduced to \( PH_3 \) (oxidation state decreases from +1 to -3).

### Step 3: Calculate the Number of Electrons Transferred
- For the oxidation to \( H_3PO_4 \):
- Change in oxidation state = \( +5 - (+1) = +4 \) (loss of 4 electrons).
- For the reduction to \( PH_3 \):
- Change in oxidation state = \( -3 - (+1) = -4 \) (gain of 4 electrons).

### Step 4: Determine the N-factor
The N-factor for the reaction can be calculated as the total number of electrons exchanged per molecule of \( H_3PO_2 \):
- In total, for \( 2 \) moles of \( H_3PO_2 \):
- \( 4 \) electrons are lost (oxidation) and \( 4 \) electrons are gained (reduction).
- Therefore, the N-factor for \( H_3PO_2 \) in this reaction is \( 4 \).

### Step 5: Calculate the Molar Mass of \( H_3PO_2 \)
The molar mass of \( H_3PO_2 \) can be calculated as follows:
- Molar mass = \( 3 \times 1 + 31 + 2 \times 16 = 3 + 31 + 32 = 66 \, g/mol \).

### Step 6: Calculate the Equivalent Weight
The equivalent weight is given by the formula:
\[ \text{Equivalent Weight} = \frac{\text{Molar Mass}}{\text{N-factor}} \]
Substituting the values we have:
\[ \text{Equivalent Weight} = \frac{66 \, g/mol}{4} = 16.5 \, g/equiv \]

### Final Answer
The equivalent weight of \( H_3PO_2 \) when it disproportionates into \( PH_3 \) and \( H_3PO_4 \) is \( 16.5 \, g/equiv \).

---