Q13 / 150 · MHT CET 2026 (11 April Shift 2) · Chemistry
In a fcc arrangement of A and B atoms in which 'A' atoms are at the corners of the unit cell and 'B' atoms are at the face centers. One of the 'A' atoms is missing from one corner in unit cell. What is the simplest formula of a compound ?
- A$\text{A}_7\text{B}_8$
- B$\text{A}_7\text{B}_3$
- C$\text{AB}_3$
- Correct answerD$\text{A}_7\text{B}_{24}$
Solution
In an fcc unit cell, there are $8$ corners and $6$ face centers.
Since one 'A' atom is missing from a corner, the number of 'A' atoms at the corners is $7$.
Contribution of each corner atom to the unit cell is $\dfrac{1}{8}$.
Effective number of 'A' atoms per unit cell = $7 \times \dfrac{1}{8} = \dfrac{7}{8}$
The 'B' atoms are at the $6$ face centers.
Contribution of each face center atom to the unit cell is $\dfrac{1}{2}$.
Effective number of 'B' atoms per unit cell = $6 \times \dfrac{1}{2} = 3$
Ratio of atoms A to B = $\dfrac{7}{8} : 3 = 7 : 24$
The simplest formula of the compound is $\text{A}_7\text{B}_{24}$.
Answer: $\text{A}_7\text{B}_{24}$
Since one 'A' atom is missing from a corner, the number of 'A' atoms at the corners is $7$.
Contribution of each corner atom to the unit cell is $\dfrac{1}{8}$.
Effective number of 'A' atoms per unit cell = $7 \times \dfrac{1}{8} = \dfrac{7}{8}$
The 'B' atoms are at the $6$ face centers.
Contribution of each face center atom to the unit cell is $\dfrac{1}{2}$.
Effective number of 'B' atoms per unit cell = $6 \times \dfrac{1}{2} = 3$
Ratio of atoms A to B = $\dfrac{7}{8} : 3 = 7 : 24$
The simplest formula of the compound is $\text{A}_7\text{B}_{24}$.
Answer: $\text{A}_7\text{B}_{24}$
Hence, the correct answer is option: 'D'.