Q.The ratio of close packed atoms to tetrahedral hole in cubic packing is

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.

Diagram description (labelled): draw 3 spheres of equal size touching each other, arranged in a triangle, representing one layer; draw a 4th identical sphere resting in the depression at the centre of the triangle, directly above (or below) the first three. Label the small enclosed gap at the centre, where the 4 spheres nearly meet, as the "tetrahedral void"; an arrow/label pointing from the centre of the triangle of contact points to this gap makes the labelling clear. Each sphere in a close-packed lattice generates 2 tetrahedral voids per sphere.

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.

This illustrates the general rule: in any close-packed (ccp or hcp) structure, the number of octahedral voids equals the number of atoms (close-packed spheres) in the unit cell, while the number of tetrahedral voids is twice that.

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 sphere in the packing, located above and below each sphere along the body diagonal positions).

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 octahedral voids is also $4$. Since A occupies all the octahedral voids, $Z_A = 4$ too.

Ratio $A:B = 4:4 = 1:1$, giving formula $AB$.