Question

In a fcc lattice, atom A forms the corner points and atom B forms the face centered point. If one atom B is missing from the face centered point, the formula of the ionic compound is

• A. $doesnotchange$
• B. $A{B}_2$
• C. $A_2{B}_5$
• D. $A_{2}B$

Answer 1

The correct option is C $A_2{B}_5$

A atom contribution = $\frac{1}{8}\times 8=1$
5 face centered B atom = $5\times \frac{1}{2}=2.5$
$\therefore$ Ratio of A : B = 1 : 2.5
= 2 : 5
$\therefore$ Formula $A_2{B}_5$