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The atomic radii of a divalent cation and a monovalent anion are 0.77 nm and 0.136 nm, respectively.1- Calculate the force of attraction between these two ions at their equilibrium interionic separation
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The atomic radii of a divalent cation and a monovalent anion are 0.77 nm and 0.136 nm, respectively.1- Calculate the force of attraction between these two ions at their equilibrium interionic separation (i.e., when the ions just touch one another).2-What is the force of repulsion at this same separation distance?

User Muffy
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1 Answer

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Answer:

A) attractive force between ions = 5.09x10^-19 N of attractive force

B) there will be no repulsion since both ions have opposite charges.

Step-by-step explanation:

Detailed explanation and calculation is shown in the image below.

The atomic radii of a divalent cation and a monovalent anion are 0.77 nm and 0.136 nm-example-1
User Aviad
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