Question

in a crystalline solid anions B are arranged in a cubic close packing. Cations A are equally distributed between octahedral and tetrahedral voids. If all the octahedral voids are occupied , what is the formula of the solid?

Answer 1

B is in ccp so B=4

A is on octa and tetra equally and octa is completely filled i.e in ccp octa=4 since octa and tetra are equal that means tetra=4 so this emplies that A=4tetra+4octa

A=8

Therefore A8B4=A2B

Hope this will help u

Answer 2

Answer:  A2B

Explanation:

So, we already know that B is arranged in cubic closed packing.

Thus, the number of atoms of B is 1/8*8 + 1/2*6 = 4.

Now, lets find the number of atome in A.

Its given that A is evenly distributed in octahedral and tetrahedreal voids, i.e A/2 atoms are there in octahedreal voids.

We already know that there are 4 atoms in octahedreal voids.

Thus, A= 4*2 which is 8.

Therefore, the formulua of the compound is A8B4.

Which can be simplified to A2B.