`P_(4)+3overset(ө)OH+3H_(2)Orarr3H_(2)Orarr3H_(2)PO_(2)^(ө)+PH_(3)`
Equivalent weight of `P_(4)` is

Options

A. `(M)/(4)`
B. `(M)/(12)`
C. `((M)/(4)+(M)/(12))`
D. `((M)/(2)+(M)/(6))`

Explanation

To find the equivalent weight of \( P_4 \) in the given reaction, we will follow these steps:

### Step 1: Write the Balanced Reaction
The balanced reaction is:
\[ P_4 + 3 \text{OH}^- + 3 \text{H}_2\text{O} \rightarrow 3 \text{H}_2\text{PO}_2^- + \text{PH}_3 \]

### Step 2: Determine the Oxidation States
- In \( P_4 \), the oxidation state of phosphorus (P) is \( 0 \).
- In \( H_2PO_2^- \), the oxidation state of phosphorus is \( -1 \).
- In \( PH_3 \), the oxidation state of phosphorus is \( +3 \).

### Step 3: Identify Oxidation and Reduction
- The phosphorus in \( P_4 \) is reduced to \( H_2PO_2^- \) (from \( 0 \) to \( -1 \)), which is a reduction process.
- The phosphorus in \( P_4 \) is oxidized to \( PH_3 \) (from \( 0 \) to \( +3 \)), which is an oxidation process.

### Step 4: Calculate the Change in Electrons
- For the reduction to \( H_2PO_2^- \):
- Change in oxidation number = \( 0 \) to \( -1 \) (1 electron per phosphorus).
- Since there are 3 phosphorus atoms in \( H_2PO_2^- \), total change = \( 3 \times 1 = 3 \) electrons.

- For the oxidation to \( PH_3 \):
- Change in oxidation number = \( 0 \) to \( +3 \) (3 electrons per phosphorus).
- Since there are 1 phosphorus atom in \( PH_3 \), total change = \( 1 \times 3 = 3 \) electrons.

### Step 5: Calculate the n-factor
- The total change in electrons for the reduction is \( 3 \) electrons (for \( H_2PO_2^- \)).
- The total change in electrons for the oxidation is \( 3 \) electrons (for \( PH_3 \)).
- Thus, the total n-factor for \( P_4 \) in this reaction is:
\[ n = 3 + 3 = 6 \]

### Step 6: Calculate the Equivalent Weight
The equivalent weight (EW) is given by the formula:
\[ \text{Equivalent Weight} = \frac{M}{n} \]
where \( M \) is the molar mass of \( P_4 \) and \( n \) is the n-factor.

The molar mass of \( P_4 \) is:
\[ M = 4 \times 31 = 124 \, \text{g/mol} \]

Thus, the equivalent weight of \( P_4 \) is:
\[ \text{Equivalent Weight} = \frac{124}{6} \approx 20.67 \, \text{g/equiv} \]

### Final Answer
The equivalent weight of \( P_4 \) is approximately \( 20.67 \, \text{g/equiv} \).

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