Q.In a solid atom M occupies ccp lattice and $\left(\dfrac{1}{3}\right)$ of tetrahedral voids are occupied by atom N. Find the formula of solid formed by M and N.

A tetrahedral void is the empty space enclosed by 4 spheres whose centres form a tetrahedron.

In a close-packed structure, when a second layer of spheres is placed over the triangular hollows of the first layer, each sphere of the second layer sits directly above the depression left by 3 touching spheres of the first layer. This arrangement of 4 spheres — 3 in one layer forming a triangular base, plus 1 sitting in the hollow above them — encloses a small void at the centre. Because the line joining the centres of these 4 spheres traces out a regular tetrahedron (with the void at its centroid), this empty space is called a tetrahedral void.

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A ccp (cubic close-packed / face-centred cubic) unit cell has 4 octahedral voids, the same as its atom count.

In a ccp/fcc unit cell:

• Atoms per unit cell = 8 corner atoms x 1/8 + 6 face-centre atoms x 1/2 = 1 + 3 = 4.
• Octahedral void positions: 1 at the body centre (fully inside, counts as 1) + 12 edge-centre voids x 1/4 share = 3.
• Total octahedral voids = 1 + 3 = 4. …

In a close-packed arrangement of N spheres, there are N octahedral voids and 2N tetrahedral voids.

In any close-packed structure (hcp or ccp/fcc) built from N spheres, each sphere contributes to voids in a fixed ratio: the number of octahedral voids = N and the number of tetrahedral voids = 2N (there are two tetrahedral voids for every sp …

With B forming a ccp lattice and A filling every octahedral void (equal in number to the close-packed atoms), the ratio A:B is 1:1.

In a cubic close-packed (ccp) arrangement, the number of octahedral voids equals the number of atoms forming the close-packed lattice. If B forms the ccp lattice, $Z_B = 4$ per unit cell, and the number of octahedr …