Rate of O exchange with H2O
When you talk about the rate of exchange of oxygen with water it might be instructive to realise that actually any aldehyde or ketone in the presence of water is in equilibrium with the hydrate, as shown below.


Source: UC Davis

There are a few ways we know that this is indeed an equilibrium process, but the main method that was used to fully elucidate the mechanism was by using isotopically labelled water containing 18O (the most abundant isotope of oxygen is 16O).
By doing this the scientists could observe the 16O in the starting aldehyde being slowly replaced with 18O from the enriched water.
Given that an equilibrium is involved, there are actually two pertinent questions...

What is the position of the equilibrium?
How fast do we reach the equilibrium?

The table below might be of some use to you. On the left, you can see a variety of carbonyl compounds, and on the right, the equilibrium constants for the hydration.
Two of the compounds you listed (acetaldehyde and trifluoroacetaldehyde) are listed in the table, and you can see that the acetaldehyde is significantly less hydrated at equilibrium than the trifluoroacetaldehyde.


Source: J. Am. Chem. Soc. 1983, 105, 868 (American Chemical Society)

From the table, some trends can be picked out (excluding formaldehyde which is almost exclusively hydrated). There is here a relationship between kinetics and thermodynamics (the more favourable a compound is to form the hydrate, the faster it will form it)

Aldehydes are far more likely to be hydrated than ketones. This is largely though not exclusively a steric effect.

Electron withdawing/electronegative substituents favour hydration. On a very simple level this is due to the fact that they make the carbonyl more electron poor, thus promoting the addition of water via its lone pair.


The order should therefore be 4&gt;1&gt;&gt;2&gt;3   (fluorine is more electron withdrawing than chlorine)