Question

If atoms along an axis connecting the opposite edge centres on a face are removed from $NaCl$ type solid AB then-new empirical formula of the remaining solid would be:
A. ${A_8}{B_5}$
B. $AB$
C.${A_3}{B_4}$
D. ${A_3}{B_8}$

Answer 1

Since the solid AB has $NaCl$ structure i.e., they have rock salt structure. Hence the A atoms occupy the FCC arrangement and atoms B occupy 12 edges as well as the centre of the unit cell.

Complete step by step answer: As the A atoms occupy the FCC arrangement that means A atom occupies the corner as well as the centre of each face.
Since the number of atoms present at the corner is 8 and the contribution of each atom present at the corner is 1/8.
Then the number of atom per unit cell can be calculated as: $8 \times \dfrac{1}{8} = 1$
Similarly, the number of atoms present on the center of each face is 6 and the contribution of each atom present at the face center is $\dfrac{1}{2}$ because each atom on the center of the face is shared by two unit cells. Hence the number of atom per unit cell can be calculated as: $6 \times \dfrac{1}{2} = 3$
Hence the total number of atoms A= $3 + 1 = 4$.

And the atom B is present at the center of the edge as well as the center of the unit cell. Since the number of atoms present at the center of edges is 12 and the contribution of each atom present at the edge center is $\dfrac{1}{4}$ because the atom present at the center of each edge is shared by 4 unit cells.
Hence the number of atoms present per unit cell for B is $12 \times \dfrac{1}{4} = 3$. And the number of atoms present at the center of the unit cell is 1.

Hence the total number of atoms B =$3 + 1 = 4$.

According to the question, there is the removal of two face-centered atoms along an axis hence the number of atoms present at the face center is 4. Again we know the contribution of each atom at the face center is $\dfrac{1}{2}$. Hence the total effective number of A atom is
$8 \times \dfrac{1}{8} + 4 \times \dfrac{1}{2} \\\ \Rightarrow 1 + 2 = 3 \\\$

Hence the new empirical formula for remaining solid represented as shown below
${\text{A:B}} \\\ {\text{3:4}} \\\ \Rightarrow {{\text{A}}_{\text{3}}}{{\text{B}}_{\text{4}}} \\\$
Hence the correct answer is option C.

Note: The number of particles present per unit cell is called unit cell constant (z). Thus the general expression for the calculation of the unit cell is:
$z = \dfrac{{{{\text{N}}_{\text{c}}}}}{{\text{8}}} + \dfrac{{{{\text{N}}_{\text{f}}}}}{2} + \dfrac{{{{\text{N}}_{\text{e}}}}}{{\text{4}}} + \dfrac{{{{\text{N}}_{\text{i}}}}}{{\text{1}}}$
Subscripts c, f, e, I represent corner, face center, edge center, and inside.