Question

Rubidium Chloride (RbCl) has NaCl like structure at normal pressures. If the radius of the Chloride ion is 1.54 Å, what is the unit cell edge length for RbCl? (Assuming anion-anion contact)

(a) 4.25 Å

(b) 4.78 Å

(c) 4.32 Å

(d) 5.14 Å

This question was posed to me in an interview.

Question is from Solid State in chapter Solid State of Chemistry – Class 12

Answer 1

The correct option is (c) 4.32 Å

The explanation: Given,

Radius of Chloride ion (r^–) = 0.154 nm

Distance between the centres of the Chloride ions = 2 x 0.154 = 0.308 nm

Let the edge length of cube = a

Distance between Rb^+ and Cl^– ions = a/2

Therefore, the distance between Cl^– ions = (2 x (a/2)^2)^1/2

     0.308 = (2 x (a/2)^2)^1/2

     0.094864 = 2 x (a/2)^2

     0.047432 = (a/2)^2

     0.218 = (a/2)

     a = 0.432 nm = 4.32 Å.