LOCATING TETRAHEDRAL AND OCTAHEDRAL VOIDS
The close packed structures have both octahedral and tetrahedral voids. In a ccp structure, there is 1 octahedral void in the centre of the body and 12 octahedral void on the edges. Each one of which is common to four other unit cells. Thus, in cubic close packed structure.
Octahedral voids in the centre of the cube =1
Effective number of octahedral voids located at the 12 edge of = 12 × 1/4 = 3
∴ Total number of octahedral voids = 4
In ccp structure, there are 8 tetrahedral voids. In close packed structure, there are eight spheres in the corners of the unit cell and each sphere is in contact with three groups giving rise to eight tetrahedral voids
Circles labelled T represent the centers of the tetrahedral interstices in the ccp arrangement of anions. The unit cell "owns" 8 tetrahedral sites.
Circles labelled O represent centers of the octahedral interstices in the ccp arrangement of anions (fcc unit cell). The cell "owns" 4 octahedral sites.
Illustration 16. In a solid, oxide ions are arranged in ccp. Cations A occupy one – sixth of the tetrahedral voids and cations B occupy one third of the octahedral voids. What is the formula of the compound?
Solution: In ccp with each oxide there would be 2 tetrahedral voids and one octahedral voids 1/3rd octahedral voids is occupied by B and 1/6th tetrahedral void by A. Therefore the compound can be
Illustration 17. In a crystalline solid, having formula AB2O4, oxide ions are arranged in cubic close packed lattice while cations A are present in tetrahedral voids and cations B are present in octahedral voids
(i) What percentage of the tetrahedral voids is occupied by A?
(ii) What percentage of the octahedral voids is occupied by B?
Solution: In a cubic close packed lattice of oxide ions there would be two tetrahedral voids and one octahedral void for each oxide ion.
∴ For four oxide ions there would be 8 tetrahedral and four octahedral voids two are occupied by B.
Percentage of tetrahedral voids occupied by A = 1/8 × 100 = 12.5%
Percentage of tetrahedral voids occupied by B = 2/4 × 100 = 50%
Illustration 18. A binary solid has zinc blend structure with ions constituting the lattice and ions occupying 25% tetrahedral voids. The formula of solid is
(A) AB (B) A2B
(C) AB2 (D) AB4
Solution: (C) No. of ions in unit cell = 8 × 1/8 + 1/2 × 6 = 4
Now ion occupies 25% of tetrahedral voids
∴ No. of A+ = 8 × 25 / 100 = 2
Thus ratio of A+ to B– is 1:2
Hence formula is AB2
Illustration 19. A compound crystallises as follows:
Ions “A“ are at corners of a cubic unit cell and “B“ ions at face centres of a cubic unit cell and “C“ ions in 1/4th of the total tetrahedral void. Assuming if this is dissolved, only the ions in the tetrahedral voids are dissociated completely in water, which one of the following statements is true. (Assuming all are univalent ions) and also A and C are cations and B is an anions.
(A) Boiling point of same concentration of solution (100% dissociation) will be greater than that of this solution.
(B) Boiling point of same concentration of (100% dissociation) will be greater than that of this solution.
(C) Boiling point of same concentration of sucrose will be greater than this solution
(D) Data insufficient to predict.
Solution: (B) Compound is
So if C alone dissociates
AB3C2 ———> [AB3]2– + 2[C]+
Total 3 ions
(a) In a crystalline solid having molecular formula A2B anion B are arranged in cubic close packed lattice and cations A are equally distributed between octahedral and tetrahedral voids
(i) What percentage of octahedral voids is occupied?
(ii) What percentage of tetrahedral voids is occupied?
(b) In a compound, oxide ions have ccp arrangement cations A are present in 1/8th of the tetrahedral voids and cations B occupy half the octahedral voids. What is the simplest formula of the compound?
(i) What is the number of octahedral voids in case of H.C.P and F.C.C?
(ii) True or False
Number of tetrahedral voids is a whole number multiple of the number of octahedral void where n is any integer.