# Explain why these ions are more strongly hydrated in an aqueous solution?

I've already answered this question. All I want to know is if my reasoning is correct.

Case 1:

$$\ce{Rb+}$$

$$\ce{Na+} \leftarrow$$ This ion is more strongly favored

Case 2:

$$\ce{Mg^{2+}} \leftarrow$$ This ion is more strongly favored

$$\ce{Na+}$$

Reasoning:

For the first case, my guess would have to be because the electrons are much closer to the center of the nucleus (the atomic radius of the atom is very small) and makes it much easier to interact the positive dipole end of the water molecule (the hydrogen molecules).

For the second case, the reason why the magnesium ion is more favored is because even though both ions have roughly the same molar mass, one is more positively charged than the other and thus is more likely to attract the negative dipole end of the water molecule (the oxygen molecule).

Is my reasoning justified or is there something in particular that I'm not paying attention to?

• There's a problem with the reasoning for your first case: positive ions will much more strongly attract the oxygen atom than the hydrogen atoms in water. The oxygen atom in water has a locally high electron density (partial negative charge), while the hydrogen atoms has a locally low electron density (partial positive charge). The strongest attraction happens when oxygen atoms point towards the cations, and the hydrogen atoms point away. May 24, 2015 at 23:45

## 1 Answer

I believe your reasoning is correct (more or less). Solvation is influenced by charge to size ratios. Higher the charge to size ratio, the easier it is for the species to be solvated by water molecules.

Apply this idea to both your cases. (note: in the second case, I don't get your argument about molar mass. The idea is much simpler, magnesium's atomic radius is smaller than sodium's and it has a higher positive charge. What can we conclude from this?

Also, reconsider the first case: a positive ion would not be attracted to the positive end of a dipole.