Number of moles of `K_(2)Cr_(2)O_(7)` in acidic medium required to oxidize one mole of `Cu_(3)P" to " CuSO_(4) " and " H_(3)PO_(4)` is :

Options

A. `11/6`
B. `6/11`
C. `3/5`
D. `5/3`

Explanation

To solve the problem of determining the number of moles of \( K_2Cr_2O_7 \) required to oxidize one mole of \( Cu_3P \) to \( CuSO_4 \) and \( H_3PO_4 \) in acidic medium, we can follow these steps:

### Step 1: Identify the oxidation states of the elements in \( Cu_3P \)
- Copper (Cu) in \( Cu_3P \) is in the +1 oxidation state.
- Phosphorus (P) in \( Cu_3P \) is in the -3 oxidation state.

### Step 2: Determine the oxidation states after the reaction
- In the products, copper is in the +2 oxidation state in \( CuSO_4 \).
- Phosphorus is in the +5 oxidation state in \( H_3PO_4 \).

### Step 3: Calculate the change in oxidation states
- For copper:
- Change from +1 to +2: \( +2 - (+1) = +1 \) for each copper atom.
- Since there are 3 copper atoms: \( 3 \times 1 = 3 \) electrons lost.

- For phosphorus:
- Change from -3 to +5: \( +5 - (-3) = +8 \) electrons gained.

### Step 4: Calculate the total number of electrons transferred
- Total change in electrons = Change in electrons from copper + Change in electrons from phosphorus
- Total = \( 3 + 8 = 11 \) electrons.

### Step 5: Determine the valency factor of \( Cu_3P \)
- The valency factor of \( Cu_3P \) is 11, as calculated from the total change in electrons.

### Step 6: Relate the equivalents of \( Cu_3P \) to \( K_2Cr_2O_7 \)
- In a redox reaction, the equivalents of the oxidizing agent (in this case, \( K_2Cr_2O_7 \)) must equal the equivalents of the reducing agent (in this case, \( Cu_3P \)).
- The formula for equivalents is:
\[ \text{Equivalents} = \text{Moles} \times \text{Valency Factor} \]
- For \( Cu_3P \):
\[ \text{Equivalents of } Cu_3P = 1 \text{ mole} \times 11 = 11 \]

### Step 7: Determine the valency factor of \( K_2Cr_2O_7 \) in acidic medium
- The valency factor of \( K_2Cr_2O_7 \) in acidic medium is 6.

### Step 8: Set up the equation for equivalence
- Let \( x \) be the number of moles of \( K_2Cr_2O_7 \) required.
- The equation based on equivalents is:
\[ 11 = x \times 6 \]

### Step 9: Solve for \( x \)
- Rearranging gives:
\[ x = \frac{11}{6} \]

### Conclusion
The number of moles of \( K_2Cr_2O_7 \) required to oxidize one mole of \( Cu_3P \) to \( CuSO_4 \) and \( H_3PO_4 \) in acidic medium is \( \frac{11}{6} \) moles.

---