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The atomic radii of a divalent cation and a monovalent anion are 0.97 nm and 0.135 nm, respectively.

Required:
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

Respuesta :

Answer:

A) F_a = 45.19 × 10^(-12) N

B) F_repulsive = -45.19 × 10^(-12)

Explanation:

We are given;

atomic radii of a divalent cation = 0.97 nm

atomic radii of monovalent anion = 0.135 nm

Distance between the 2 atomic centres; R = 0.97 + 0.135 = 1.105 nm = 1.105 × 10^(-9) m

A) Now, Force of attraction will be given by the formula;

F_a = [(z1 × e)•(z2 × e)]/(4πε_o•R²)

Where;

z1 is the valency of the divalent cation = 2

z2 is the valency of the monovalent anion = 1

e is electron charge = 1.602 × 10^(-19) C

ε_o is vacuum of permeability with a constant value of 8.85 × 10^(-12) C²/N.m²

Plugging in the relevant values;

F_a = [(2 × 1.602 × 10^(-19))•(1 × 1.602 × 10^(-19))]/(4π × 8.85 × 10^(-12) × (1.105 × 10^(-9))²)

F_a = (513.2808 × 10^(-40))/(11.35793096213 × 10^(-28))

F_a = 45.19 × 10^(-12) N

B) At same separation distance which is equilibrium;

F_repulsive = -F_a

F_repulsive = -45.19 × 10^(-12)