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A mineral having formula AB2 crystallises in ccp lattice with A atoms occupying the lattice points. The CN of A is 8 and of B is 4. What percentage of the tetrahedral sites is occupied by B atoms.
Dear sudhanshu,
Each host atom in a CCP lattice is surrounded by and touches 12 nearest neighbors, each at a distance of 2r:
When spheres of equal size (radius r) are packed together as closely as possible by stacking layers of hexagonally closest packed planes, the spheres themselves occupy 74% of the available space. The cavities of empty space ("holes" or "interstices") are found between layers.
For example, consider any two successive planes in a closest packed lattice. One atom in the A layer nestles in the triangular groove formed by three adjacent atoms in the B layer, and the four atoms touch along the edges of a regular tetrahedron. The edges of the tetrahedron are of length 2r, and the center of the tetrahedron is a cavity called the Tetrahedral hole (a common symbol for "tetrahedral" is Td). If a guest sphere in that cavity is in contact with all four host spheres, its radius is about 0.22r.
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