Calculate the ionic radius of a Cs+ ion, assuming that the cell edge length for CsCl is 0.4123 nm and that the ionic radius of a Cl- ion 0.181 nm.
CsCl has a bcc lattice. So, dbody =
Or dbody = 0.4123 nm = 0.7141 nm
The sum of the ionic radii of Cs+ and Cl- ions is half this distance ie
= 0.3572 nm
Ionic radius of Cs+ = 0.3571 – 0.181 = 0.1761
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