Question

In a AB unit cell (Rock Salt type), assuming cations forming fcc lattice point :

• A. the nearest neighbours of $A^{+}$ are $6B^{-}$ ions
• B. a unit cell contains $8$ ions
• C. the second nearest neighbours of $A^{+}$ is ${12A}^{+}$
• D. the packing fraction of AB crystal $\frac{\sqrt{3}\pi }{8}$

Answer 1

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

The correct options are
A the nearest neighbours of $A^{+}$ are $6B^{-}$ ions
B a unit cell contains $8$ ions
C the second nearest neighbours of $A^{+}$ is ${12A}^{+}$

Each cation is surrounded by 6 anions. co-ordination no. of cation will be 6

No. of corner atoms = 8 atoms ,contribution to total no. of atoms will be 1/8

No. of face atoms= 6 atoms , contribution to total no. of atoms will be 1/2

No. of edge atoms= 12 atoms , contribution to total no. of atoms will be 1/4

No. of center atoms=1 atom , contribution to total no. of atoms will be 1

Total no. of atoms will be $(\frac{1}{8}\times 8)+(\frac{1}{2}\times 6)+(\frac{1}{4}\times 12)+(1\times 1)=8$

Atomic packing factor in crystal structure that is occupied by constituent particles. It is dimensionless and always less than unity.

$Atomicpackingfactor=\frac{no.ofatoms\times volumeofsphere}{volumeofunitcell}$

$APF=\frac{4\times 4/3\pi (0.102nm{)}^3+4\times 4/3\pi (0.181nm{)}^3}{(0.181nm{)}^3}$

Atomic packing factor - 0.65

Second nearest neighours of cation are 12 cations