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Deriving Formula Of A Compound

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Deriving Formula Of A Compound - Lesson Summary

Arrangement of the cations and the anions the close packing:
The anions being larger in size form the basic crystal lattice and the cations being smaller in size occupy the voids. A smaller cation occupies the tetrahedral void while a relatively bigger cation occupies the octahedral void.

EX: sodium ions having size of 102 pm occupy the octahedral voids in a NaCl lattice and the zinc ions with size of 74 pm occupy the tetrahedral voids in a ZnS lattice.

The fraction of octahedral or tetrahedral voids occupied in a given lattice depends upon the chemical formula of the compound.

Problem:
In corundum, the oxide ions form the hexagonal close packing lattice and the Aluminum ions occupy 2/3rd of the octahedral voids. What is the formula of the compound corundum?

Answer:
The number of octahedral sites in a cubic close packing is the same as the number of constituent particles.

"N" oxide ions form the crystalline lattice, so the number of available octahedral sites = N.

2/3Noctahedral sites are occupied by the aluminium ions.


Number of tetrahedral & octahedral voids present in ccp or fcc unit cell:
One tetrahedral void is present in each of the small eight cubes of the ccp or fcc unit cell. 8 tetrahedral voids are present in ccp or fcc unit cell. In a ccp or fcc unit cell 4 atoms or ions are present per unit cell.

The number of tetrahedral voids =2*4 =8

The octahedral voids are also present at the body centre and at the centre of each edge in the ccp or fcc unit cell. The total number of octahedral voids present per unit cell is four.

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