A compound made of particles `A`, `B`, and `C` forms `ccp` lattice. In the lattice, ions `A` occupy the lattice points and ions `B` and `C` occuphy the alternate `TV_(s)`. If all the ions along one of the body diagonals are removed, then formula of the compound is

Options

A. `A_(3.75)B_(3)C_(3)`
B. `A_(3.75)B_(3)C_(4)`
C. `A_(3)B_(3.75)C_(3)`
D. `A_(3)B_(3)C_(3.75)`

Explanation

To solve the problem, we need to analyze the structure of the compound formed by particles A, B, and C in a cubic close-packed (ccp) lattice. Here are the steps to derive the formula of the compound after removing the ions along one of the body diagonals.

### Step-by-Step Solution:

1. **Understanding the ccp Structure:**
- In a cubic close-packed (ccp) lattice, also known as face-centered cubic (FCC), the lattice points are occupied by ions A.
- The contribution of A ions from the lattice points:
- There are 8 corner atoms, each contributing \( \frac{1}{8} \) (total contribution from corners = \( 8 \times \frac{1}{8} = 1 \)).
- There are 6 face-centered atoms, each contributing \( \frac{1}{2} \) (total contribution from faces = \( 6 \times \frac{1}{2} = 3 \)).
- Therefore, total A ions = \( 1 + 3 = 4 \).

2. **Identifying Tetrahedral Voids:**
- In a ccp lattice, there are 8 tetrahedral voids.
- Since B and C occupy alternate tetrahedral voids, we can assume:
- Number of B ions = 4 (occupying half of the tetrahedral voids).
- Number of C ions = 4 (occupying the other half of the tetrahedral voids).

3. **Initial Composition:**
- Before removing any ions, the formula of the compound can be written as:
\[ A_4 B_4 C_4 \]

4. **Removing Ions Along the Body Diagonal:**
- When we remove all ions along one body diagonal, we need to identify how many of each type of ion is removed.
- The body diagonal of the cube passes through two corner A ions and one face-centered A ion, thus:
- Total A ions removed = 2 (corner contributions) + 1 (face-centered contribution) = 3 A ions.
- For B and C ions, since they occupy tetrahedral voids, the entire B and C ions along the body diagonal are also removed.
- Total B ions removed = 1 (entire B ion).
- Total C ions removed = 1 (entire C ion).

5. **Calculating Remaining Ions:**
- Remaining A ions = \( 4 - 3 = 1 \)
- Remaining B ions = \( 4 - 1 = 3 \)
- Remaining C ions = \( 4 - 1 = 3 \)

6. **Final Formula:**
- The final formula of the compound after removing the ions along the body diagonal is:
\[ A_1 B_3 C_3 \]

### Conclusion:
The formula of the compound after the removal of the ions along one of the body diagonals is \( A_1 B_3 C_3 \).