Why is hydrated beryllium ion acidic?

Question

The beryllium ion, $\ce{Be^{2+}}$, forms the aquo complex $\ce{[Be(H2O)4]^{2+}}$. According to LibreTexts, this complex is acidic in solution: $$\ce{[Be(H2O)4]^{2+} + H2O -> [Be(H2O)3OH]+ + H3O+}$$

It is not clear to me why it should be acidic. I read here that "the hydrolysis happens because the $\ce{Be−O}$ bond is very strong and so in the hydrated ion this weakens the $\ce{O−H}$ bonds hence there is a tendency to lose protons."

I have two questions:

1. What is meant by a "very strong" $\ce{Be-O}$ bond? Is it stronger than those in other metal aquo complexes? If berylium's tendency to hold on to water ligands is unusually strong, is it due to its small ionic size?
2. Why does a strong $\ce{Be-O}$ make the $\ce{O-H}$ bond weaker? I know that oxygen forms a coordinate bond with metal cation so the electrons in the $\ce{O-H}$ bonds would get shifted towards oxygen. But as far as I can tell this only makes the $\ce{O-H}$ bonds more polar. Why does a polar bond have to be a weaker one?

Answer 1

What is meant by a "very strong" Be−O bond? If berylium's tendency to hold on to water ligands is unusually strong, is it due to its small ionic size?

You got that absolutely right. $\ce{Be\bond{-}O}$ is a strong bond because of the small size of Be. Smaller cation size means a stronger pull on the $\ce{O}$ electrons, thus reducing the bond length and hence increasing the bond strength.

[...]the electrons in the O−H bonds would get shifted towards oxygen. But as far as I can tell this only makes the $\ce{O\bond{−}H}$ bonds more polar.

You struck at the right point. Let's look at the original question. I'm sure you must know that a stronger acid has a higher tendency to release $\ce{H+}$ ions. This means that the electron pair should be drawn away from $\ce{H}$ atom in the $\ce{O\bond{−}H}$ bond. Thus, a polarized bond makes it easier for the H+ ion to 'escape' from the molecule.

Why does a polar bond have to be a weaker one?

It is not appropriate to answer this question in definitive terms. The strength of a bond is defined by its bond length, not its polarity. To address the confusion you're having, look back at the context of the question. $$\ce{[Be(H2O)4]^{2+} + H2O -> [Be(H2O)3OH]+ + H3O+}$$ We're talking about a compound with polar O-H bonds dissolved in water. Water has a high dielectric constant, and can easily break ionic/polar bonds since it liberates a high amount of solvation energy. This is because of its high dielectric constant.

So, are polar bonds weaker than non-polar bonds in water? Sure, I can see your line of thinking. But, always remember that ionic bonds can easily be broken by water, but that doesn't mean that ionic bonds are weaker than non-polar bonds.