Question

Calculate the Coulomb attraction between the atoms of the following salts. Cesium Chloride (CsCl) Rutile (TiO2)...

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.

Homework Answers

Answer #1

(1) Coulomb attraction between atoms of CsCl

FA = - A/r2

A = Ao(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

Ao = 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

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

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