The correct statement (s) for cubic close packed (ccp) three dimensional structure is (are)

Options

A. The number of the neighbours of an atom present in the topmost layer is 12
B. The efficiency of atom packing is `74%`
C. The number of octahedral and tetrahedral voids per atom are 1 and 2 respectively
D. The unit cell edge length is `2sqrt(2)` times the radius of the atom

Explanation

To solve the question regarding the correct statements for the cubic close packed (CCP) three-dimensional structure, we will analyze each statement provided in the question.

### Step-by-Step Solution:

1. **Statement 1: The number of neighbors of the atom present in the topmost layer is 12.**
- **Analysis:** This statement is incorrect. The coordination number of an atom in the topmost layer of a cubic close-packed structure (also known as face-centered cubic, FCC) is not 12 because there are no atoms above it. The coordination number refers to the number of nearest neighbors surrounding an atom. In the case of an atom in the topmost layer, it is surrounded by 4 atoms in the same layer and 4 atoms in the layer below, giving a total of 8 neighbors.
- **Conclusion:** This statement is **false**.

2. **Statement 2: The efficiency of atom packing is 74%.**
- **Analysis:** This statement is correct. The packing efficiency of a cubic close-packed structure is indeed 74%. This can be calculated using the formula for packing fraction:
\[ \text{Packing Fraction} = \frac{Z \times \text{Volume of sphere}}{\text{Volume of cube}} \]
For FCC, \( Z = 4 \) (the number of atoms per unit cell), and the packing fraction comes out to be 74%.
- **Conclusion:** This statement is **true**.

3. **Statement 3: The number of octahedral and tetrahedral voids per atom is 1 and 2.**
- **Analysis:** This statement is also correct. In a cubic close-packed structure, there is 1 octahedral void for each atom and 2 tetrahedral voids for each atom. This is derived from the fact that the number of octahedral voids equals the number of atoms per unit cell (Z), and the number of tetrahedral voids is twice the number of atoms per unit cell.
- **Conclusion:** This statement is **true**.

4. **Statement 4: The unit cell edge length is \( 2\sqrt{2} \) times the radius of the atom.**
- **Analysis:** This statement is correct. In an FCC lattice, the relationship between the edge length (a) and the atomic radius (R) can be derived from the geometry of the face diagonal. The face diagonal is equal to \( a\sqrt{2} \) and is also equal to 4R (since it contains 3 atomic radii). Therefore, we can derive:
\[ a\sqrt{2} = 4R \implies a = \frac{4R}{\sqrt{2}} = 2\sqrt{2}R \]
- **Conclusion:** This statement is **true**.

### Final Conclusion:
The correct statements regarding the cubic close packed (CCP) structure are:
- Statement 2: True
- Statement 3: True
- Statement 4: True

### Summary of Correct Statements:
- Statement 2, Statement 3, and Statement 4 are correct.