Ionic structures in terms of anion packing

In the two closest packed sequences,. .. ABCABC. .. and. .. ABAB. .there are 2 A tetrahedral interstices and N octahedral interstices for every N anions. Structures are derived by placing cations into the interstices, making sure that the total positive charge on the cations is equal to the total negative charge on the anions. The formula of the structure can be found by counting up the numbers of ions of each sort present. 

Consider the structures that arise from a cubic close-packed array of X anions. If every octahedral position contains an M cation there are equal numbers of cations and anions in the structure. The formula of the compounds with this structure is MX, and the structure corresponds to the halite (NaCl, Bl) structure. In the case of halide anions, X, to maintain charge balance, each cation must have a charge of +1, and the alkali halide, MX, structure results. Should oxygen anions, 0, form the anion array, the cations must necessarily have a charge of 2- -, to ensure that the charges balance, and the oxides will have a formula MO, but still retain the halite structure. This structure is possessed by a number of oxides, including MgO, CaO, SrO, BaO, MnO, FeO, CoO, NiO and CoO. 

Should the anions adopt hexagonal close packing and all of the octahedral sites contain a cation, the hexagonal analogue of the halite structure is produced. In this case, the formula of the crystal is again MX. The structure is the nicolite (NiAs) 

Fraction of sites occupied Sequence of anion layers 

Structures containing cations in tetrahedral sites can be described in exactly the same way. In this case, there are twice as many tetrahedral sites as anions and so, if all sites are filled, the formula of the solid will be M2X. When half are filled this becomes MX, and so on. 

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