Question

Current Attempt in Progress The atomic radii of a divalent cation and a monovalent anion are 0.52 nm and 0.125 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). Enter your answer for part (a) in accordance to the question statement N (b) What is the force of repulsion at this same separation distance

Answer 1

Answer:

a)   F = 1.70 10⁻⁹N,   F = 1.47 10⁻⁸ N,

b) * the electronegative repulsion, from the repulsion by quantum effects

Explanation:

a) The atraicione force comes from the electric force given by Coulomb's law,

           F = $k \frac{ q_1 q_2}{r^2}$

divalent atoms

In this case q = 2q₀ where qo is the charge of the electron -1,6 10⁻¹⁹ C and the separation is given

           F = k q² / r²

           F = $2 \ 10^9 \ \frac{2 (1.6 \ 10^{-19} )^2}{ (0.52 10^{-9} )^2 }$

           F = 1.70 10⁻⁹N

monovalent atoms

in this case the load is q = q₀

           F = 2 \ 10^9 \  \frac{ (1.6 \  10^{-19} )^2}{ (0.125 10^{-9} )^2 }

           F = 1.47 10⁻⁸ N

b) repulsive forces come from various sources

* the electronegative repulsion of positive nuclei

* the electrostatic repulsion of the electrons when it comes to bringing the electron clouds closer together

* from the repulsion of electron clouds, by quantum effects