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. $ABCD _{2}$
• B. $ABD _{2}$
• C. $AB _{2} D _{2}$,
• D. $AB _{4} D _{5}$

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]
$A _{\frac{1}{2}} B _{2} C _{0} D _{\frac{5}{2}} \rightarrow A _{1} B _{4} D _{5}$