Question

In a unit cell, atoms A are present at all corner lattices, B atoms are present at alternate faces and all edge centers. Atoms C is present at face centers left from B and one at each body diagonal at a distance of $\frac{1}{4}$th of body diagonal from the corner.

The formula of the given solid is :

• A. $A_3{B}_8{C}_7$
• B. $A{B}_4{C}_6$
• C. $A_6{B}_4{C}_8$
• D. $A_2{B}_9{C}_{11}$

Answer 1

Answer: (D)

$A_2{B}_9{C}_{11}$

Total number of A atoms present in the unit cell $=\frac{1}{8}\times 8=1$

Total number of B atoms present in the unit cell $=\frac{1}{2}\times 3+\frac{1}{4}\times 12=1.5+3=4.5$

Total number of C atoms present in the unit cell $=\frac{1}{2}\times 3+4=5.5$

Hence, the formula of the given solid is $A{B}_{4.5}{C}_{5.5}$ or $A_2{B}_9{C}_{11}$.