Question

For an FCC structure and a BCC structure having equal edge length ‘a’ and equal radius for anion, then the difference between cationic radius of BCC and FCC lattice is

• A. $\frac{a}{2}$
• B. $\frac{\sqrt{3}a}{2}$
• C. $\left(\frac{\sqrt{3}-\sqrt{2}}{2}\right)a$
• D. $\sqrt{3}a$

Answer 1

The correct option is C $\left(\frac{\sqrt{3}-\sqrt{2}}{2}\right)a$

$2r_{+b}+2r_a=\sqrt{3}a\Rightarrow r_{+b}+r_a=\frac{\sqrt{3}}{2}a$
$2r_{+F}+2r_a=\sqrt{2a}=r_{+F}+r_a=\frac{\sqrt{2}}{2}a$
$r_{+b}-r_{+F}=\left(\frac{\sqrt{3}-\sqrt{2}}{2}\right)a$