Hydration of H⁺ ion

Question

I know that $\ce{H+}$ is not possible in water and it is present as $\ce{H3O+}$. But later on I come to know that even $\ce{H3O+}$ is not possible and that it is present as $\ce{H9O4+}$.

Why does this happen? What give that compound so much stability that is not present in $\ce{H3O+}$ or $\ce{H+}$?

Answer 1

The formation of bonds nearly always leads to a decrease in energy. This is desirable, since less energy $\implies$ stability, so wherever it is possible, bonds tend to form.

$\ce{H+}$ has an empty $s$ orbital. $\ce{H2O{:}}$ has a lone pair. These easily form something akin to a dative bond ($\ce{H2O{:}\bond{->}H+}$), giving $\ce{H3O+}$. Note that there is no single positively charged hydrogen--the charge is distributed over the entire molecule. Distribution of charge leads to more stability.

Now, $\ce{H9O4+}$ (called the "Eigen cation") is even stranger. You can imagine it like this: Three $\ce{H2O}$ molecules approach $\ce{H3O+}$ (remember, this cation is positively charged and water has a lone pair which is attracted to positive charge). Basically, charge-dipole attraction. Of course, the lone pair will "point" towards the hydronium.

Now, these charge-dipole attraction "bonds" resonate. The hydrogens of the central hydronium "detach" from the main molecule and attach to the peripheral ones. In reality, the electron cloud is delocalised and we have partial bonds.

Now, the charge is distributed even more. We can have other species like $\ce{H7O3+}$, $\ce{H15O7+}$, etc. Note that there are crowding issues as well - this is why $\ce{H_{103}O_{51}+}$ probably doesn't exist.

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Basically, the "original" $\ce{H+}$ is being solvated. The solvation delocalises the charge and forms extra bonds, reducing the net potential energy and increasing stability. That's basically it. It's similar to what happens when you dissolve any ionic compound in water.

Answer 2

I think that for this type of question a bit of context as to why you are asking the question, would be useful.

If for instance you are in an Intro Chem class I would say the following:

All chemical systems tend towards their most stable state (in the absence of external energy being applied).

$\ce{H^+}$ is just a single proton, a single positive charge without any negative charge to balance it out. Naked charges are very powerful forces, while it is possible for $\ce{H^+}$ to exist in isolation (even to have a number of them together in a system), the presence of other matter which might be used to balance out the naked positive charge will quickly lead to a reaction.

By forming $\ce{H_3O^+}$ the single proton $\ce{H^+}$ is able to spread its positive charge out over not just itself but the entirety of the bonded $\ce{H_2O}$ molecule. (Note that because Oxygen can be seen in resonance to distribute the negative charge associated with its loan pairs it is particularly good at stabilizing the extra positive charge).

Whats better than $\ce{H_3O^+}$? Getting even more water molecules to share the burden of the extra positive charge and producing $\ce{H_9O_4^+}$.

Thus the short answer I would give to your question would be:

There are two main factors:

1. Simply the distribution of that extra positive charge over a larger area/shared by more matter stabilizes the system.
2. The Oxygen in particular (being able to share the negative charge of its lone pair) can "handle" the extra positive charge better than the lone proton ($\ce{H^+}$)

Hope that helps. Because in my opinion, beyond this explanation the answer gets very complicated and as many have pointed out, broad.