The atomic radii of a divalent cation and a monovalent anion are 0.46 nm and 0.131 nm, respectively.

(a) Calculate the force of attraction between these two ions at their equilibrium interionic separation (i. e., when the ions just touch one another).

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

1.32 * 10⁻⁹ N, - 1.32 * 10⁻⁹ N

Explanation:

distance between the charges r ₀ = 0.46 nm + 0.131 nm = 0.591 nm = 0.591 * 10⁻⁹ m

Using Coulomb's law,

Force of attraction = (Z₁) (Z₂) q² / (4 πε₀r₀²) where

Z₁ = valency of divalent = + 2

Z₂ = valency of monovalent = - 1

ε₀ = 8.85 * 10⁻¹² C / (Nm²)

Fa = (1 * 2 * (1.602 * 10⁻¹⁹ C) ²) / (4 π (8.85 * 10⁻¹² C/Nm²) (0.591 * 10⁻⁹ m) ²)

Fa = 1.32 * 10⁻⁹ N

b) Force of repulsion = force of attraction = - 1.32 * 10⁻⁹ N at equilibrium