Correct Statements Regarding Cubic Close Packed Crystal Structures

To determine the correct statements regarding a three-dimensional cubic close packed ($ccp$) structure, let us analyze each option individually:

1. Option (A): In a bulk $ccp$ (or $fcc$) structure, each sphere is surrounded by $12$ nearest neighbours ($6$ in its own layer, $3$ in the layer above, and $3$ in the layer below). However, for an atom located in the topmost layer, there is no layer above it. Thus, its coordination number is reduced to: $\text{Nearest neighbours} = 6 \text{ (same layer)} + 3 \text{ (layer below)} = 9$ Therefore, statement (A) is incorrect.

2. Option (B): In a $ccp$ unit cell, the number of effective atoms per unit cell is $Z = 4$. The relation between edge length $a$ and atomic radius $r$ along the face diagonal is: $\sqrt{2}a = 4r \implies a = 2\sqrt{2}r$ The packing efficiency is calculated as: $\text{Packing Efficiency} = \frac{\text{Volume occupied by } 4 \text{ atoms}}{\text{Volume of unit cell}} \times 100$ $\text{Packing Efficiency} = \frac{4 \times \frac{4}{3}\pi r^3}{a^3} \times 100 = \frac{\frac{16}{3}\pi r^3}{\left(2\sqrt{2}r\right)^3} \times 100 = \frac{\pi}{3\sqrt{2}} \times 100 \approx 74\%$ Therefore, statement (B) is correct.

3. Option (C): In a close-packed lattice containing $N$ constituent atoms:

   • The number of octahedral voids $= N$
   • The number of tetrahedral voids $= 2N$

   Therefore, per constituent atom: $\text{Number of octahedral voids per atom} = \frac{N}{N} = 1$ $\text{Number of tetrahedral voids per atom} = \frac{2N}{N} = 2$ Therefore, statement (C) is correct.

4. Option (D): In a $ccp$ structure, atoms touch along the face diagonal of the cubic unit cell. Thus: $\text{Length of face diagonal} = \sqrt{2}a = 4r$ $a = \frac{4r}{\sqrt{2}} = 2\sqrt{2}r$ Hence, the edge length of the unit cell is $2\sqrt{2}$ times the atomic radius. Therefore, statement (D) is correct.

Conclusion: The correct statements are (B), (C), and (D).