A compound made of particles `A`, `B`, and `C` forms `ccp` lattice. Ions `A` are at lattice points, `B` occupy `TV_(s) C` occupy `OV_(s)`. If all the ions along one of the edge axis are removed, then formula of the compound is

Options

A. `A_(3.75)B_(8)C_(3.75)`
B. `A_(3.75)B_(4)C_(8)`
C. `A_(4)B_(8)C_(3.75)`
D. `A_(4)B_(3.75)C_(8)`

Explanation

To solve the problem step-by-step, we need to analyze the information given about the compound made of particles A, B, and C that forms a CCP (cubic close-packed) lattice.

### Step 1: Determine the number of A ions
In a CCP lattice:
- A ions are located at the lattice points, which include the corners and face centers.
- There are 8 corner atoms, each contributing \( \frac{1}{8} \) to the unit cell, and 6 face-centered atoms, each contributing \( \frac{1}{2} \).

Calculating the effective number of A ions:
\[ \text{Number of A ions} = \left(8 \times \frac{1}{8}\right) + \left(6 \times \frac{1}{2}\right) = 1 + 3 = 4 \]

### Step 2: Determine the number of B ions
B ions occupy the tetrahedral voids. In a CCP lattice, the number of tetrahedral voids is twice the number of octahedral voids. The number of octahedral voids is equal to the number of atoms in the unit cell (which is 4 for A).

Calculating the number of B ions:
\[ \text{Number of B ions} = 2 \times \text{Number of octahedral voids} = 2 \times 4 = 8 \]

### Step 3: Determine the number of C ions
C ions occupy the octahedral voids. As established, the number of octahedral voids is equal to the number of effective atoms in the unit cell (which is 4 for A).

Calculating the number of C ions:
\[ \text{Number of C ions} = \text{Number of octahedral voids} = 4 \]

### Step 4: Removing ions along one edge axis
When all ions along one edge axis are removed:
- From the corners, 2 A ions are removed (each corner contributes \( \frac{1}{8} \), so \( 2 \times \frac{1}{8} = \frac{1}{4} \)).
- From the octahedral voids, 1 C ion is removed (each octahedral void contributes \( \frac{1}{4} \), so \( 1 \times \frac{1}{4} = \frac{1}{4} \)).
- The number of B ions remains unchanged as they do not occupy the edge.

### Step 5: Calculate the remaining number of ions
After the removal:
- Remaining A ions:
\[ \text{Remaining A ions} = 4 - \frac{1}{4} = \frac{16}{4} - \frac{1}{4} = \frac{15}{4} \]
- Remaining B ions:
\[ \text{Remaining B ions} = 8 \]
- Remaining C ions:
\[ \text{Remaining C ions} = 4 - \frac{1}{4} = \frac{16}{4} - \frac{1}{4} = \frac{15}{4} \]

### Step 6: Write the formula of the compound
The formula of the compound can be expressed in terms of the remaining ions:
\[ \text{Formula} = A_{\frac{15}{4}} B_8 C_{\frac{15}{4}} \]

To express this in whole numbers, we can multiply through by 4:
\[ \text{Formula} = A_{15} B_{32} C_{15} \]

### Final Answer
The formula of the compound after removing the ions along one edge axis is:
\[ \text{Formula} = A_{15} B_{32} C_{15} \]