Ion-dipole effects vs. atomic radius

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When dissolved in water, Li+ forms stronger ion-dipole bonds with water molecules compared to Na+. Although both ions have the same charge, Li+ has a smaller ionic radius, which allows water dipoles to be closer to the charge, resulting in a stronger Coulombic attraction. This increased interaction is attributed to Li+'s higher electronegativity and the effective use of the Born model, which describes solvation as a charged sphere within a medium characterized by its dielectric constant. The discussion highlights the importance of ionic size and electronegativity in determining the strength of ion-dipole interactions in aqueous solutions.
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When dissolved in water, which of the following ions will form stronger ion-dipole bonds with the water molecules? Li+ or Na+?

Both have roughly the same charge... Na has greater radius, but I don't see why or how that has any bearing on the problem.
 
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Radius changes distance between charges.
 
Borek said:
Radius changes distance between charges.

I'm sorry but could you please elaborate a little more on that?
Radius I guess does decrease the force between the metal ion and each water molecule (due to Coulomb's law), but it turns out that Li+ actually forms more bonds with water. Why is that?
 
Smaller ion means dipoles are closer to the charge, so the Coulomb force is larger.
 
I believe this has to do with the difference in electronegativity. Lithium has a higher electronegativity.
 
Solvation is often treated within a simple model, the Born model, which treats the atom as a charged sphere inside the medium assumed to be continuous and described by its dielectric constant. The model is still used a lot to describe the solvation of proteins even today.
Confer e.g.
http://pchemandyou.blogspot.com/2008/01/born-model-of-solvation.html
 
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