Question

Select the correct statement(s) about an $AB$ unit cell (rock salt type), assuming that $A^{+}$ forms FCC lattice.

• 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 $0.74$

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 packing fraction of $AB$ crystal is $0.74$
D The second neighbour of $A^{+}$ is twelve $A^{+}$

Based on FCC crystal, like $NaCl$ lattice,
Packing fraction = $\frac{[4\times \frac{4}{3}\pi r_{A^{+}}^3]+[4\times \frac{4}{3}\pi r_{B^{-}}^3]}{a^3}$
$=\frac{[4\times \frac{4}{3}\pi \frac{a^3}{{\left(2\sqrt{2}\right)}^3}]+[4\times \frac{4}{3}\pi \frac{a^3{\left(0.414\right)}^3}{{\left(2\sqrt{2}\right)}^3}]}{a^3}=0.74$

Hence, the packing fraction is $0.74$.

Besides, the nearest neighbour of $A^{+}$ is $6$ $B^{-}$ ion. Similarly, for $B^{-}$, it is $6$ $A^{+}$ ion.

Also, the second neighbour of $A^{+}$ is $12$ $A^{+}$ ion.