Question

The atomic radii of a divalent cation and a monovalent anion are 0.065 nm and 0.126 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)

Answer 1

Answer:

Explanation:

charge on divalent cation = 2 x 1.6 x 10⁻¹⁹ C

charge on monovalent anion = 1.6 x 10⁻¹⁹ C

distance between these charges = .065 + .126 nm

= .191 x 10⁻⁹ m

force of attraction

= k q₁q₂ / r²

= 9 x 10⁹ x 2 x 1.6 x 10⁻¹⁹  x 1.6 x 10⁻¹⁹ / ( .191 x 10⁻⁹ )²

= 1263.12 x 10⁻¹¹ N .