The instability of hydrated carbon dioxide or "carbonic acid"

Question

Why is hydrated carbon dioxide - the predominant form of acid that one gets upon dissolution of carbon dioxide in solution - so unstable?

Is the below rationale valid?

1. Carbon in carbon dioxide has two (empty? no, but still vulnerable to attack) p-orbitals and bears a strong partial positive charge.
2. Oxygen's lone pair can attack an empty p-orbital and form a formal charge-separated complex with the carbon dioxide.
3. This is "hydrated" carbon dioxide or "carbonic acid." This form is extremely unstable and subject to disproportionation due to an unfavorable charge and the unfavorable nature of charge separation itself. The change in entropy also favors the products of disproportionation.
4. However, there exists a pathway to stability - that is - protonation of the oxygen with the negative formal charge by the oxygen bearing the positive formal charge.
5. This, however, is akin to a forbidden fruit; the ephemeral three-membered ring that would have to be formed exhibits "ring strain" (if you object to this term, can you please elaborate on your objection), and as a result, disproportionation is overwhelmingly favored, especially from an entropic standpoint (reconstitution of carbon dioxide gas is highly entropically favorable).

Answer 1

The situation is described by the following equilibrium $$\ce{H2O + CO2 <=> HO-CO-OH}$$ In a closed system, using Le Chatelier’s principle you can manipulate the equilibrium in either direction (e.g. pump in carbon dioxide and increase the concentration of carbonic acid, etc.). However in an open system, as carbon dioxide escapes, the amount of carbonic acid will decrease.

When all is said and done, water and carbon dioxide are two extremely stable molecules. Since $$\Delta G = -RT\ln K$$ Their stability is what drives the equilibrium to the left and why carbonic acid appears to be (relatively) so unstable.

Answer 2

This answer looks at the molecular structure characteristics behind the thermodynamics reported by Ron

We are dealing with the reaction

$\ce{CO2 +H2O <=> H2CO3}$

or more accurately

$\ce{O=C=O +H2O <=> (HO)2C=O}$

One might expect the equilibrium to favor the right side through the replacement of a pi bond with a sigma bond and the ability of the hydroxylated compounds to fit better with the hydrogen-bonded water structure. Such bonding features are common drivers of reactions between high oxidation-state nonmetaloxides and water. But the carbon dioxide molecule on the left has two particular features that disfavor such a reaction thermodynamically.

Carbon dioxide polarity: the whole truth

Many a chemistry textbook describes carbon dioxide as nonpolar, but it isn't. It is non-dipolar. Carbon dioxide has two strong opposing dipoles that cancel out as dipoles, but combine to make a powerful quadrupole. This quadrupole is well-sized to be solvated strongly with both the water dipole and the water quadrupole (look along the line from one hydrogen atom to the other, you see a quadrupole). Indeed the water quadrupole, whose dipole components are directed oppositely from the dipole components of the carbon dioxide quadrupole, fits like a hand in a glove when it comes to solvation. The resulting solvation capability is one reason carbon dioxide, despite its "nonpolarity", has fairly good solubility in water in the first place.

The revenge of 4n pi electrons

Carbon dioxide has an especially strong pi bonding stabilized by delocalization; it might be thought of as a linear-molecule counterpart to aromaticity. (In contrast with the usual form of aromaticity, the interaction favors $4n$ electrons; carbon dioxide with all atoms $sp$ hybridized gives $n=2$.) Reaction with water to break one of these pi bonds upsets this delocalization much like benzene resisting addition because of a similarly strong stabilization by delocalization.