Equivalent weight of `H_(3)PO_(2)` when it disproportionates into `PH_(3)` and `H_(3)PO_(3)` is (mol.wt. of `H_(3)PO_(2)=M`)

Options

A. `M`
B. `(3M)/(4)`
C. `(M)/(2)`
D. `(M)/(4)`

Explanation

To find the equivalent weight of \( H_3PO_2 \) when it disproportionates into \( PH_3 \) and \( H_3PO_3 \), we will follow these steps:

### Step 1: Identify the oxidation states
In \( H_3PO_2 \), the oxidation state of phosphorus (P) is +1. In \( PH_3 \), the oxidation state of phosphorus is -3, and in \( H_3PO_3 \), the oxidation state of phosphorus is +3.

### Step 2: Determine the change in oxidation states
- When \( H_3PO_2 \) is converted to \( PH_3 \), the change in oxidation state is from +1 to -3. This is a change of 4 units (1 to -3).
- When \( H_3PO_2 \) is converted to \( H_3PO_3 \), the change in oxidation state is from +1 to +3. This is a change of 2 units (1 to 3).

### Step 3: Calculate the number of electrons transferred
- For the reaction forming \( PH_3 \), 4 electrons are gained (reduction).
- For the reaction forming \( H_3PO_3 \), 2 electrons are lost (oxidation).

### Step 4: Determine the total number of electrons transferred
In total, for each molecule of \( H_3PO_2 \) that disproportionates:
- 4 electrons are transferred to form \( PH_3 \).
- 2 electrons are transferred to form \( H_3PO_3 \).
Thus, the total electron transfer is \( 4 + 2 = 6 \) electrons.

### Step 5: Calculate the equivalent weight
The equivalent weight (EW) is given by the formula:
\[ \text{Equivalent Weight} = \frac{\text{Molar Weight}}{\text{Number of electrons transferred}} \]
In this case, the molar weight of \( H_3PO_2 \) is \( M \) and the total number of electrons transferred is 6.

Thus, the equivalent weight of \( H_3PO_2 \) is:
\[ \text{Equivalent Weight} = \frac{M}{6} \]

### Final Answer
The equivalent weight of \( H_3PO_2 \) when it disproportionates into \( PH_3 \) and \( H_3PO_3 \) is \( \frac{M}{6} \).

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