A crystal is made up of particles A. B and C. A forms fcc packing, B occupies all octahedral voids and C occupies all tetrahedral voids. If all the particles along one body diagonal are removed, then the formula of the crystal would be

Options

A. `ABC_2`
B. `A_2 B C_2`
C. `A_8 B_4 C_5`
D. `A_5 B_4 C_8`

Answer

To solve the problem, we need to determine the new formula of the crystal after removing the particles along one body diagonal. Let's break it down step by step.

### Step 1: Determine the number of particles A in the FCC structure.
- In a face-centered cubic (FCC) lattice, there are 4 atoms of A per unit cell.
- These atoms are located at the corners and face centers of the cube.

**Hint:** Remember that in an FCC structure, each corner atom contributes 1/8 to the unit cell and each face-centered atom contributes 1/2.

### Step 2: Identify the particles removed along the body diagonal.
- The body diagonal of a cube connects two opposite corners.
- In a cube, there are 8 corners, and along the body diagonal, we will remove 2 corner atoms.

**Hint:** Visualize the cube and identify the corners that lie on the body diagonal.

### Step 3: Calculate the contribution of the removed corner atoms.
- Each of the 2 corner atoms removed contributes 1/8 to the unit cell.
- Therefore, the total contribution of the removed corner atoms is:
\[ 2 \times \frac{1}{8} = \frac{2}{8} = \frac{1}{4} \]

### Step 4: Calculate the remaining number of A atoms.
- Initially, there were 4 A atoms in the FCC unit cell.
- After removing the contribution from the body diagonal:
\[ \text{Remaining A} = 4 - \frac{1}{4} = \frac{16}{4} - \frac{1}{4} = \frac{15}{4} \]

**Hint:** Always keep track of the contributions when atoms are removed from a crystal structure.

### Step 5: Determine the number of B atoms in the octahedral voids.
- In an FCC structure, there are 4 octahedral voids (OHV) per unit cell.
- Since we are removing atoms along the body diagonal, we need to consider how many octahedral voids are affected.
- After the removal, there will be 3 octahedral voids remaining.

**Hint:** Remember that the octahedral voids are located at the center of the cube and at the edge centers.

### Step 6: Determine the number of C atoms in the tetrahedral voids.
- In an FCC structure, there are 8 tetrahedral voids (THV) per unit cell.
- Each body diagonal has 2 tetrahedral voids that will be removed.
- Therefore, the remaining number of tetrahedral voids is:
\[ 8 - 2 = 6 \]

**Hint:** Tetrahedral voids are located at specific positions relative to the atoms in the FCC structure.

### Step 7: Write the new formula of the crystal.
- Now we have:
- A: \( \frac{15}{4} \)
- B: \( 3 \)
- C: \( 6 \)

- To simplify, we can multiply all coefficients by 4 to eliminate the fraction:
\[ A: 15, \quad B: 12, \quad C: 24 \]

- The simplified formula becomes:
\[ A_{15}B_{12}C_{24} \]

**Hint:** Always check if the coefficients can be simplified further.

### Final Answer:
The formula of the crystal after removing the particles along one body diagonal is \( A_{15}B_{12}C_{24} \).