Question

If metallic atoms of mass 197 and radius 166 pm are arranged in ABCABC fashion then what is the surface area of each unit cell?

(a) 1.32 × 10^6 pm^2

(b) 1.32 × 10^-18pm^2

(c) 2.20 × 10^5 pm^2

(d) 2.20 × 10^-19 pm^2

The question was asked in an international level competition.

This is a very interesting question from Solid State in division Solid State of Chemistry – Class 12

Answer 1

The correct answer is (a) 1.32 × 10^6 pm^2

To explain: ABCABC arrangement is found in CCP.

In closed cubic packing, relation between edge length of unit cell, a, and radius of particle, r, is given as a=2\(\sqrt{2}\)r.

 Surface area (S.A.) = 6a^2

From the relationship,

a^2 = 8r^2

S.A. = 6a^2 = 48r^2

 When r = 166 pm, S.A. = 48(166pm) = 1.32 x 10^6 pm^2.