There is a famous demonstration where, if you mix a $\ce{CoCl2}$-solution with the right amount of $\ce{HCl}$ you will end up with the pink $\ce{[Co(H2O)6]^2+}$ complex but when heating it to boiling temperatures it will change to blue as the $\ce{[CoCl4]^2-}$ complex is formed. This reaction is reversible as the mixture is cooled again. Now literature always describes that this works because that shift towards $\ce{[CoCl4]^2-}$ is endothermic, but why is this the case? 

What causes one to be more stable here than the other one but being substituted when heated? 

I have some ideas but I can't really compare them:

- First of all there should actually be an entropic advantage towards the chlorido-side, as 6 water ligands leave, meaning that the exothermic step here is against the natural direction of the entropy(?).

- Second, perhaps the chlorido ligand is the better leaving group. I know usually $\ce{H2O}$ is a good leaving group in organic molecules but then its usually a $\ce{H2O+}$. So perhaps chloride is the better leaving group here.

- Maybe there is a kinetic effect, that it is harder to displace 6 ligands than just four ligands

- Or, as much more water is present it just shifts towards the water side because of the huge excess of water?

- Chloride is a weaker ligand than water. But on the other hand the iron-fluorido complex is quite stable in water and towards other ligands. 

- So could it be due to the different ligand fields? The splitting in the tetrahedron is much smaller than in the octahedron. Although I don't know how this actually affects the stability. 

So those were the ideas I had but I can't really find a common thing among them besides the excess of water which often causes an equlibrium to shift.

By the way $\ce{[Cu(H2O)6]^2+}$ and $\ce{[CuCl4]^2-}$ do the exact same thing. 

Does anyone have an idea why this reaction is endothermic?