Question

In a AB unit cell (Rock salt type) assuming $A^{+}$ forming FCC:

• A. the nearest neighbour of $A^{+}$ is six $B^{-}$ ion
• B. the nearest neighbour of $B^{-}$ is six $A^{+}ion$
• C. the second neighbour of $A^{+}$ is twelve $A^{+}$
• D. the packing fraction of AB crystal is 0.79

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A the nearest neighbour of $A^{+}$ is six $B^{-}$ ion
B the nearest neighbour of $B^{-}$ is six $A^{+}ion$
C the second neighbour of $A^{+}$ is twelve $A^{+}$
D the packing fraction of AB crystal is 0.79

Based on FCC crystal, like NaCl lattice.
Packing fraction $=\frac{(4\times \frac{4}{3}\pi r_{A^{+}}^3)+(4\times \frac{4}{2}\pi r_{B^{-}}^3)}{a^3}=\frac{\left(4\frac{4}{3}\pi \frac{a^3}{(2\sqrt{2}{)}^3}\right)+(4\times \frac{4}{3}\pi \frac{a^3}{(2\sqrt{2}{)}^3})+(4\times \frac{4}{3}\pi \frac{a^3(0.414{)}^3}{(2\sqrt{2}{)}^3})}{a^3}=0.79$