What is the equivalent weight of `P` in the following reaction?
`P_(4)+NaOH rarr NaH_(2)PO_(2)+PH_(3)`

Options

A. `(31)/(4)`
B. `(31)/(3)`
C. `(31)/(2)`
D. `31xx4//3`

Explanation

To find the equivalent weight of phosphorus (P) in the reaction:

\[ P_4 + NaOH \rightarrow NaH_2PO_2 + PH_3 \]

we will follow these steps:

### Step 1: Understand Equivalent Weight
The equivalent weight of an element in a reaction is defined as the molecular weight divided by the valence factor. The valence factor is determined by the number of electrons lost or gained by the element during the reaction.

### Step 2: Determine the Oxidation State Changes
In the reaction, phosphorus in \( P_4 \) undergoes both oxidation and reduction:
- In \( NaH_2PO_2 \), phosphorus has an oxidation state of +1.
- In \( PH_3 \), phosphorus has an oxidation state of -3.

### Step 3: Analyze Oxidation
For the oxidation half-reaction:
- \( P_4 \) (where P is 0) is converted to \( NaH_2PO_2 \) (where P is +1).
- Each phosphorus atom goes from 0 to +1, which means it loses 1 electron.
- Since there are 4 phosphorus atoms in \( P_4 \), the total loss of electrons is \( 4 \times 1 = 4 \) electrons.

**Valence Factor for Oxidation**: 4

### Step 4: Calculate Equivalent Weight for Oxidation
The molecular weight of phosphorus (P) is approximately 31 g/mol. Therefore, the equivalent weight for the oxidation process is calculated as:

\[ \text{Equivalent weight}_{\text{oxidation}} = \frac{4 \times 31}{4} = 31 \text{ g/equiv} \]

### Step 5: Analyze Reduction
For the reduction half-reaction:
- \( P_4 \) (where P is 0) is converted to \( PH_3 \) (where P is -3).
- Each phosphorus atom goes from 0 to -3, which means it gains 3 electrons.
- Since there are 4 phosphorus atoms in \( P_4 \), the total gain of electrons is \( 4 \times 3 = 12 \) electrons.

**Valence Factor for Reduction**: 12

### Step 6: Calculate Equivalent Weight for Reduction
Using the same molecular weight of phosphorus, the equivalent weight for the reduction process is calculated as:

\[ \text{Equivalent weight}_{\text{reduction}} = \frac{4 \times 31}{12} = \frac{124}{12} = \frac{31}{3} \text{ g/equiv} \]

### Step 7: Combine Equivalent Weights
The total equivalent weight of phosphorus in the reaction is the sum of the equivalent weights from oxidation and reduction:

\[ \text{Total Equivalent Weight} = \text{Equivalent weight}_{\text{oxidation}} + \text{Equivalent weight}_{\text{reduction}} = 31 + \frac{31}{3} \]

To combine these, we can express 31 as \( \frac{93}{3} \):

\[ \text{Total Equivalent Weight} = \frac{93}{3} + \frac{31}{3} = \frac{124}{3} \text{ g/equiv} \]

### Final Answer
Thus, the equivalent weight of phosphorus (P) in the given reaction is:

\[ \text{Equivalent Weight of } P = \frac{124}{3} \text{ g/equiv} \]