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 information given about the compound made of particles A, B, and C that forms a cubic close-packed (ccp) lattice. Let's break down the solution step by step.

### Step 1: Determine the number of A ions in the ccp lattice
In a cubic close-packed (ccp) lattice, also known as face-centered cubic (FCC), the number of atoms per unit cell is 4. This is because:
- There are 8 corner atoms, each contributing \( \frac{1}{8} \) to the unit cell.
- There are 6 face-centered atoms, each contributing \( \frac{1}{2} \) to the unit cell.

Calculating the total:
\[ \text{Total A atoms} = 8 \times \frac{1}{8} + 6 \times \frac{1}{2} = 1 + 3 = 4 \]
Thus, there are 4 A ions in the unit cell.

### Step 2: Determine the number of tetrahedral voids
In a ccp lattice, there are 8 tetrahedral voids per unit cell. However, since the problem states that ions B and C occupy alternate tetrahedral voids, we need to consider only half of these voids:
\[ \text{Number of tetrahedral voids occupied} = \frac{8}{2} = 4 \]
So, there are 4 tetrahedral voids occupied by B and C.

### Step 3: Analyze the removal of ions along the body diagonal
When all the ions along one body diagonal are removed, we need to consider how many A ions are affected. In a ccp lattice, the body diagonal passes through:
- 2 corner atoms (A)
- 1 face-centered atom (A)

Thus, the total number of A ions removed is:
\[ \text{Removed A ions} = 2 \text{ (from corners)} + 1 \text{ (from face center)} = 3 \]

### Step 4: Calculate the remaining A ions
Initially, there were 4 A ions. After removing 3 A ions:
\[ \text{Remaining A ions} = 4 - 3 = 1 \]

### Step 5: Calculate the remaining B and C ions
Since one tetrahedral void is removed when the body diagonal is removed, one of the ions (either B or C) occupying that void will also be removed. Initially, there were 4 ions of B and 4 ions of C:
- After removing one tetrahedral void, we have:
\[ \text{Remaining B ions} = 4 - 1 = 3 \]
\[ \text{Remaining C ions} = 4 \text{ (unchanged)} \]

### Step 6: Write the formula of the compound
Now we can write the formula of the compound based on the remaining ions:
\[ \text{Formula} = A_1 B_3 C_4 \]

### Final Answer
The final formula of the compound after the removal of the ions along the body diagonal is:
\[ \text{Formula} = A_1 B_3 C_4 \]