Question

In a unit cell, atoms $A,B,C$ and $D$ are present at corners, face-centres, body-centre and edge-centres respectively. If atoms touching one of the plane passing through two diagonally opposite edges are removed, then formula of compound is

• A. $ABC{D}_2$
• B. $AB{D}_2$
• C. $A{B}_2{D}_2$,
• D. $A{B}_4{D}_5$

Answer 1

Answer: (D)

Given atoms of one diagonal planes are to be removed.
$\therefore 4$ atoms from corner, $2$ atom from face centre and $1$ atom from body centre will be removed.
Total $A=4\times \frac{1}{8}=\frac{1}{2}$ [ $4$ atoms removed from corner]
Total $B=\frac{1}{2}\times 4=2$ [$2$ atoms are removed from face centre]
Total $C=0$ [$1$ atom removed from body centre]
Total $D=\frac{1}{4}\times 10=\frac{5}{2}$ [$2$ atom removed from edge]
$A_{\frac{1}{2}}{B}_2{C}_0{D}_{\frac{5}{2}}\to A_1{B}_4{D}_5$