The force of attraction between a divalent cation and a divalent anion is 1.12 x 10-8 n. if the ionic radius of the cation is 0.071 nm, what is the anion radius?

= 0.2162 nm

Explanation:

The force of attraction can be written as:

Where,  

K is the Coulomb's constant having value 9×10⁹ N. m²/C²

 is the charge on the cation

 is the charge on the anion

e is electronic charge =

r is the distance between the cation and anion and in bond, it is the sum of the ionic radius of cation and anion

So,

Given, F =

Applying in the equation,

Also, 1 m =

So, r = 0.2872 nm

Also,

= 0.071 nm

Thus,

= 0.2162 nm

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