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

The correct option is D $A{B}_4{D}_5$

Initially, atoms effectively in the lattice are as follows:

$A\to 8\times \frac{1}{8},B\to 6\times \frac{1}{2},C\to 1,D\to 12\times \frac{1}{4}$

After the removal of atoms,

atoms left at the corners$=8-4=4\Rightarrow A\to 4\times \frac{1}{8}=\frac{1}{2}$

atoms left at the face center$=6-2=4\Rightarrow B\to 4\times \frac{1}{2}=2$

atoms left at the body center$=0\Rightarrow C\to 0$

atoms left at edge center$=12-2=10\Rightarrow D\to 10\times \frac{1}{4}=2.5$

ratio of $A:B,C:D=0.5:2:0:2.5=1:4:0:5$

Hence, compound formula is $A{B}_4{D}_5$