Question

Calculate the Coulomb attraction between the atoms of the following salts.

Cesium Chloride (CsCl)

Rutile (TiO₂)

This is all the information that was given. I wasn't sure how to find the radius. K should be a constant that equals 9.00*10^9 Nm^2 C^-2.

Answer 1

(1) Coulomb attraction between atoms of CsCl

F_A = - A/r²

A = A_o(Z₁e)(Z₂e)

e = charge on single electron = 1.602x10^‐19 C

Z1 = valance of charged ion, i.e., +1 for Cs and -1 for Cl

A_o = Propotionality constant = 9 x 10⁹ Nm^2 C^-2.

Substitute all the values

A = 9 x 10⁹ x (1 x 1.602x10^‐19) x (-1 x 1.602x10^‐19)

A = -23.097 x 10^-29

Here r is distance between centre of ions of Cs and Cl

Ionic radius of Cs⁺ is 174 pm and that of Cl^- is 181 pm.

Therefore distance between their centre will be = 174 + 181 = 355 pm = 355 x 10^-12 m.

r² = (3.55 x 10-10)² = 12.6025 x 10^-20

F_A = - A/r² = - (-23.097 x 10^-29) / 12.6025 x 10^-20 = 1.832 x 10^-9 N