By knowing edge of a cubic crystal from x – ray differ action method and knowing type of crystal possessed by it. The density of crystal can be calculated.

For cubic crystal of ionic compound:-

Suppose the edge of unit cell = a pm
No. of atoms (ions) present per unit cell Z
Atomic [molecular] mass = M
Volume of unit cell = (a pm)3

Mass of the unit cell = No. of atoms in unit cell x Mass
of each atom = Z x m


1. A compound formed by elements A and B has cubic structure in which A atoms are at the corners of the cube and B atoms are at the face centres. Derive formula of compound?
Ans. As A atoms are present at the 8 corners of the cube, therefore number of atoms of A in the unit cell is = 1/8 x 8 = 1
As B atoms are present at the face centres of the C faces of the cube thus no. of atoms of B = 1/2 x 6 = 3
Ratio A:B = 1:3
So formula of compound is AB3

2. Atoms of element B form hcp lattices and element A occupy 2/3 of tetrahedral voids. What is formula of compound.
Ans. Let no. of atoms B in hcp = n
No. of tetrahedral voids = 2n
Atom A occupy 2/3 of tetrahedral voids
Thus no. of atom of A = 2/3 x 2n = 4n/3
Ratio A and B A:B = 4n/3:n = 4/3:1
=> 4:3
Hence formula of the compound is A4B3

3. Gold atomic mass 197.4 u, atomic radius = 0.144 nm crystalises in face centred unit cell calculate density.
Ans. For FCC unit cell 19

4. What is the coordination no. in the square three dimensional lattice?
Ans. The coordination no. would be 6 atoms
As each atom is surrounded by 6 atoms.

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