In a crystal of an ionic compound, the ions B form the close packed lattice and the ions A occupy all the tetrahedral voids. The formula of the compound is

Question

In a crystal of an ionic compound, the ions B form the close packed lattice and the ions A occupy all the tetrahedral voids. The formula of the compound is (A) AB2 (B) A2B (C) AB (D) AB3

Tardigrade Admin · Accepted Answer

The number of tetrahedral voids is two per atom of the crystal in a close packed structure. Since tetrahedral voids are occupied by A ions, there are two A ions per B ion in the crystal. The formula of the compound is A2B.

Q. In a crystal of an ionic compound, the ions B form the close packed lattice and the ions A occupy all the tetrahedral voids. The formula of the compound is

$A B_{2}$

$A_{2} B$

AB

$A B_{3}$

The number of tetrahedral voids is two per atom of the crystal in a close packed structure. Since tetrahedral voids are occupied by $A$ ions, there are two $A$ ions per $B$ ion in the crystal. The formula of the compound is $A_{2} B$ .

Q. In a crystal of an ionic compound, the ions B form the close packed lattice and the ions A occupy all the tetrahedral voids. The formula of the compound is

AB2​

A2​B

AB

AB3​

The number of tetrahedral voids is two per atom of the crystal in a close packed structure. Since tetrahedral voids are occupied by A ions, there are two A ions per B ion in the crystal. The formula of the compound is A2​B.

$A B_{2}$

$A_{2} B$

$A B_{3}$

The number of tetrahedral voids is two per atom of the crystal in a close packed structure. Since tetrahedral voids are occupied by $A$ ions, there are two $A$ ions per $B$ ion in the crystal. The formula of the compound is $A_{2} B$ .