Question

The atomic radii of Mg2+ and F- ions are 0.079 and 0.120 nm, respectively.

(a) Calculate the force of attraction between these two ions at their equilibrium inter-ionic separation (i.e., when the ions just touch each other).

(b) What is the force of repulsion at this same separation distance?

Answer 1

a) 1.165 × 10⁻⁸ N b)- 1.165 × 10⁻⁸ N

Step-by-step explanation:

Using Coulomb's law

F(attraction) =
`(Z1Z2qelectron)/(4piER^2)`

where

R = sum of the distance between the centers of charges = sum of ionic radii = 0.079 nm + 0.120nm = 0.199 nm = 0.199 × 10⁻⁹ m

Z₁ = valency of Mg²⁺ = 2

Z₂ = valency of F ⁻ = - 1

qelectron = charge on electron =1.062 × 10⁻19 C

E = permitivity of free space = 8.85 × 10 ⁻¹² C²/ Nm²

Fa= (1×2× (1.602 × 10⁻¹⁹)²) / (4× 3.142 × 8.85 × 10⁻¹² × (0.199 × 10⁻⁹)²) = 1.165 × 10⁻⁸ N

b) At equilibrium F of repulsion = - F of attraction = - 1.165 × 10⁻⁸ N