An ionic compound has a unit cell consisting of A ions at the corners of a cube and B ions on the centres of faces of the cube. The empirical formula of the compound would be

Question

An ionic compound has a unit cell consisting of A ions at the corners of a cube and B ions on the centres of faces of the cube. The empirical formula of the compound would be (A) AB (B)  A2B  (C)  AB3  (D)  A3B

Tardigrade Admin · Accepted Answer

Number of A ions at the corners of a cube (per unit cell) =8× (1/8)=1 Number of B ions at the face centres of a cube (per unit cell) =6× (1/2)=3 Hence, the empirical formula of the compound is AB3 .

Q. An ionic compound has a unit cell consisting of A ions at the corners of a cube and B ions on the centres of faces of the cube. The empirical formula of the compound would be

AB

$A_{2} B$

$A B_{3}$

$A_{3} B$

Number of A ions at the corners of a cube (per unit cell) $= 8 \times \frac{1}{8} = 1$ Number of B ions at the face centres of a cube (per unit cell) $= 6 \times \frac{1}{2} = 3$ Hence, the empirical formula of the compound is $A B_{3}$ .

Q. An ionic compound has a unit cell consisting of A ions at the corners of a cube and B ions on the centres of faces of the cube. The empirical formula of the compound would be

AB

A2​B

AB3​

A3​B

Number of A ions at the corners of a cube (per unit cell) =8×81​=1 Number of B ions at the face centres of a cube (per unit cell) =6×21​=3 Hence, the empirical formula of the compound is AB3​ .

$A_{2} B$

$A B_{3}$

$A_{3} B$

Number of A ions at the corners of a cube (per unit cell) $= 8 \times \frac{1}{8} = 1$ Number of B ions at the face centres of a cube (per unit cell) $= 6 \times \frac{1}{2} = 3$ Hence, the empirical formula of the compound is $A B_{3}$ .