In a solid A B having the NaCl structure, A atoms occupy the corners of the cubic unit cell. If all the face centred atoms along one of the axes are removed, then the resultant stoichiometry of the solid is

Question

In a solid A B having the NaCl structure, A atoms occupy the corners of the cubic unit cell. If all the face centred atoms along one of the axes are removed, then the resultant stoichiometry of the solid is (A)  AB2  (B)  A2 B  (C)  A4 B3  (D)  A3 B4

Tardigrade Admin · Accepted Answer

In NaCl , Na+ ' occupies body centre and edge centres while Cl - occupies corners and fuce centres, giving four Na +and four Cl - per unit cell. In the present case A represent Cl -and B represents Na +. Two face centres lies on one axis. ⇒ Number of A removed =2 × (1/2)=1 Number of B is removed because it is not present on face centres. ⇒ A reinaining =4-1=3, B remaining =4, Formula =A3 B4

Q. In a solid $A B$ having the $N a Cl$ structure, $A$ atoms occupy the corners of the cubic unit cell. If all the face centred atoms along one of the axes are removed, then the resultant stoichiometry of the solid is

$A B_{2}$

$A_{2} B$

$A_{4} B_{3}$

$A_{3} B_{4}$

Q. In a solid AB having the NaCl structure, A atoms occupy the corners of the cubic unit cell. If all the face centred atoms along one of the axes are removed, then the resultant stoichiometry of the solid is

AB2​

A2​B

A4​B3​

A3​B4​

Q. In a solid $A B$ having the $N a Cl$ structure, $A$ atoms occupy the corners of the cubic unit cell. If all the face centred atoms along one of the axes are removed, then the resultant stoichiometry of the solid is

$A B_{2}$

$A_{2} B$

$A_{4} B_{3}$

$A_{3} B_{4}$