Question

A compound mx has fcc structure and a density 6.38g/ml. If the ionic radius of x2- is 1.08a, determine the ionic radius of of m2+ ion

Answer 1

Given:

• The compound MX has a FCC lattice.
• The density (d) = 6.38 g/mL
• The number of molecules of MX per unit cell (Z) = 4
• Avogadro's number (NA) = 6.023*10²³
• Let molar mass of MX be m g/mol
• The ionic radius of X²⁻ = 1.08A⁰

To find:

The ionic radius of M²⁺ ions.

Solution:

• Let the edge length of the unit cell of the lattice be a
• The formula  for density d = (Z*m)/(a³*NA) ⇒ a = ∛((Z*m)/(d*NA))
• Hence, the edge length a = ∛m A⁰
• Now, in FCC r(M²⁺) + r(X²⁻) = a/2 ⇒ r(M²⁺) = (0.5*∛m - 1.08) A⁰

Answer:

The ionic radius of M²⁺ ions = (0.5*∛m - 1.08) A⁰