Question

A compound MX has fcc structure and has a density 6.38g/ml.if the iconic radius of X2- is 1. 08A,determine the iconic radius 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) A

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