In a unit cell, atoms A, B, C and D are present at corners, face - centres, body - centre and edge - centres respectively. If atoms touching one of the plane passing through two diagonally opposite edges are removed, then formula of compound is

Options

A. `ABCD_(2)`
B. `ABD_(2)`
C. `AB_(2)D_(2)`
D. `AB_(4)D_(5)`

Answer

To solve the problem, we need to analyze the arrangement of atoms in the unit cell and how the removal of certain atoms affects the overall formula of the compound. Here’s a step-by-step breakdown:

### Step 1: Identify the positions of the atoms
- Atoms A are located at the corners of the cube. There are 8 corners in a cube.
- Atom B is located at the face centers. There are 6 faces in a cube.
- Atom C is located at the body center. There is 1 body center in a cube.
- Atom D is located at the edge centers. There are 12 edges in a cube.

### Step 2: Calculate the contribution of each atom type
- **Contribution of A (corner atoms)**:
Each corner atom contributes \( \frac{1}{8} \) of an atom to the unit cell. Since there are 8 corners:
\[ \text{Total contribution of A} = 8 \times \frac{1}{8} = 1 \text{ atom} \]

- **Contribution of B (face-centered atoms)**:
Each face-centered atom contributes \( \frac{1}{2} \) of an atom to the unit cell. Since there are 6 faces:
\[ \text{Total contribution of B} = 6 \times \frac{1}{2} = 3 \text{ atoms} \]

- **Contribution of C (body-centered atom)**:
The body-centered atom contributes 1 atom to the unit cell:
\[ \text{Total contribution of C} = 1 \text{ atom} \]

- **Contribution of D (edge-centered atoms)**:
Each edge-centered atom contributes \( \frac{1}{4} \) of an atom to the unit cell. Since there are 12 edges:
\[ \text{Total contribution of D} = 12 \times \frac{1}{4} = 3 \text{ atoms} \]

### Step 3: Remove atoms touching the specified plane
The problem states that atoms touching a plane passing through two diagonally opposite edges are removed. This affects:
- **Corner atoms**: 4 corner atoms will be removed (2 from each edge).
- **Face-centered atoms**: No face-centered atoms are removed since they are not touching the specified plane.
- **Body-centered atom**: The body-centered atom is not removed.
- **Edge-centered atoms**: 4 edge-centered atoms will be removed (2 from each edge).

### Step 4: Calculate the remaining contributions
After removing the atoms:
- **Remaining A**:
\[ 1 - 4 \times \frac{1}{8} = 1 - \frac{1}{2} = \frac{1}{2} \text{ atom} \]

- **Remaining B**:
\[ 3 \text{ atoms (no removal)} \]

- **Remaining C**:
\[ 1 \text{ atom (no removal)} \]

- **Remaining D**:
\[ 3 - 4 \times \frac{1}{4} = 3 - 1 = 2 \text{ atoms} \]

### Step 5: Write the final formula
Now, we can write the formula based on the remaining contributions:
- A: \( \frac{1}{2} \)
- B: 3
- C: 1
- D: 2

To express the formula in whole numbers, we multiply by 2:
- A: 1
- B: 6
- C: 2
- D: 4

Thus, the final formula of the compound is:
\[ \text{A}_1\text{B}_6\text{C}_2\text{D}_4 \]