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
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