Question

The force of attraction between a divalent cation and a divalent anion is 1.64 x 10-8 N. If the ionic radius of the cation is 0.087 nm, what is the anion radius

Answer 1

The radius of the anion is 7.413 nm

How to calculate the force of attraction between charges

The force of attraction (F) is given by the formula:

• F = (1/4π∈r²)(Zc*e)(Za*e)

where:

∈ = permittivity of free space = 8.85*10⁻¹⁵ F/m

Zc = charge on the cation = +2

Zc = charge on the anion = -2

e = charge on an electron = 1.602 * 10⁻¹⁹ C

r = interionic distance

r = rc + ra

where rc and ra are the radius of the cation and anion respectively

F = 1.64 * 10⁻⁸ N

Therefore based on the equation of force of attraction:

1.64 *10⁻⁸ = [1/4π(8.85*10⁻¹⁵)r²](2 * 1.602*10⁻¹⁹)²

r² = 5.63 * 10⁻¹⁷

r = 7.50 nm

Since r = rc + ra

where rc = 0.087 nm

thus, ra = r - rc = 7.50 - 0.087

ra = 7.413 nm

Therefore, the radius of the anion is 7.413 nm

Learn more about ionic radius at: brainly.com/question/2279609