Question

A compound is formed of two element ${}^{\prime }{A}^{\prime }$ and ${}^{\prime }{B}^{\prime }$. The atoms of element ${}^{\prime }{A}^{\prime }$ forms face centered cubic closed packing and atoms of ${}^{\prime }{B}^{\prime }$ occupies all the tetrahedral voids. What will be the formula of the compound?

• A. $A_{4}B$
• B. $A{B}_2$
• C. $A{B}_8$
• D. $A_{2}B$

Answer 1

The correct option is B $A{B}_2$

Atom 'A' at FCC lattice points means number of atom A in unit cell is:

$\frac{1}{8}\times 8+\frac{1}{2}\times 6=4$.

If in a close-packed structure (ccp or fcc) there are 'n' atoms or ions then the number of tetrahedral voids will be 2n. Thus the number of atom B is $=2\times 4=8$

Thus the compound is $A_4{B}_8$ or $A{B}_2$

Option B is correct