Question

Find the lowest possible empirical formula in an arrangement of unit cell where A atoms are present at corners and alternate face centers, B atoms are present at alternate edge centres and C atoms are present at half of the mid of line joining opposite face centers. Assume: Any atom present in the inner locations of the unit cell should be considered completely within the unit cell structure.

• A. $A_{2}B{C}_6$
• B. $A_2{B}_{3}C$
• C. $A_{6}B{C}_6$
• D. $A_6{B}_3{C}_3$

Answer 1

The correct option is A $A_{2}B{C}_6$

At corner, an atom contributes $1/8^{th}$ part of it as it is shared by $8$ unit cell. As there are $8$ corners in a unit cell. So, Total combination of A present at corners

$\Rightarrow \frac{1}{8}\times 8=1$

Also present at alternate face centers. Therefore A will be present at any two faces opposite to each other. At faces contribution of the atom is $1/2$.

Hence, Total contribution of $A$ present at Alternate faces

$\Rightarrow \frac{1}{2}\times 2=1$

So, Total number of A atoms= $1+1=2$

B atom present at alternate edge center. There can be $4$ edge center on which B is present. Hence, the total contribution of B

$\Rightarrow \frac{1}{4}\times 4=1$

(Refer to Image 1)

C atom present on the half of mid of the line joining two face center i.e. on the line two atoms are present. As there are $6$ Faces hence $3$ lines joining the opposite face. One line contains $2$ atoms. So, the total number of the atom is $6$. As inside the unit, cell contribution is $1$. So, the total contribution of C atom is $1\times 6=6$

(Refer to Image 2)

So, the empirical formula is $A_{2}B{C}_6$