In a solid $$‘AB’$$ having the $$NaCl$$ structure, $$'A’$$ atoms occupy the corners of the cubic unit cell. If all the face-centered atoms along one of the axes are removed, then the resultant stoichiometry of the solid is
A.
$$A{B_2}$$
B.
$${A_2}B$$
C.
$${A_4}{B_3}$$
D.
$${A_3}{B_4}$$
Answer :
$${A_3}{B_4}$$
Solution :
Effective number of 'A’ atoms $$ = \left( {8 \times \frac{1}{8}} \right) + \left( {4 \times \frac{1}{2}} \right) = 3$$
Effective number of ‘B’ atoms $$ = \left( {12 \times \frac{1}{4}} \right) + 1 = 4$$
∴ Formula of the solid = $${A_3}{B_4}.$$
Releted MCQ Question on Physical Chemistry >> Solid State
Releted Question 1
$$CsBr$$ has $$bcc$$ structure with edge length 4.3. The shortest inter ionic distance in between $$C{s^ + }$$ and $$B{r^ - }$$ is :
In a solid $$‘AB’$$ having the $$NaCl$$ structure, $$'A’$$ atoms occupy the corners of the cubic unit cell. If all the face-centered atoms along one of the axes are removed, then the resultant stoichiometry of the solid is
A substance $${A_x}{B_y}$$ crystallizes in a face centred cubic $$(FCC)$$ lattice in which atoms $$'A'$$ occupy each corner of the cube and atoms $$'B'$$ occupy the centres of each face of the cube. Identify the correct composition of the substance $${A_x}{B_y}$$