Show that in a cubic packed structure, eight tetrahedral voids are present per unit cell.

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Solution

The cubic close-packed structure contains one atom at each corner of a unit cell and at the face centre of each unit cell. Each unit cell consists of 8 small cubes.

Each small cube contains 4 atoms at its alternate corner when these atoms are joined to each other lead to the creation of a tetrahedral void as shown below

since there are total 8 smaller cubes present at one unit cell and each smaller cube has one tetrahedral void hence, the total number of tetrahedral void present in each unit cell is equal to eight

As we know ccp structure has 4 atoms per unit cell. Thus, the total number of a tetrahedral void in one ccp unit cell is equal to 8.

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Similar questions

With the help of a labeled diagram show that there are four octahedral voids per unit cell in a cubic close-packed structure.

In the cubic close packing, the unit cell has

(a) 4 tetrahedral voids each of which is shared by four adjacent unit cell

(b) 4 tetrahedral voids within the unit cell

(d) 8 tetrahedral voids within the unit cells

Q. The number of octahedral voids in a unit cell of cubic close packed structure is:

The structure of a mixed oxide is cubic closed packed (ccp). The cubic unit cell of mixed oxide is composed of oxide ions. One fourth of the tetrahedral voids are occupied by divalent mtal A and the octahedral voids are occupied by a monovalent metal B. The formula of the oxide is:

In which of the following structures coordination number for cations and anions in the packed structure will be same ?

(a) $C l^{-}$ ions form fcc lattice and $N a^{+}$ ions occupy all octahedral voids of the unit cell

(b) $C a^{2 +}$ ions form fcc lattice and $F^{-}$ ions occupy all the eight tetrahedral voids of the unit cell

(d) $S^{2 -}$ ions form fcc lattice and $Z n^{2 +}$ ions go into alternate tetrahedral voids of the unit cell

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