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?

  • 3
    $\begingroup$ Most of your ideas don’t matter at all. The thing about ‘leaving groups’ (in coordination chemistry one does not typically call them that) is not helpful. It is typically a property of the entire complex, most importantly of the LFSE, whether the complex tends to lose ligands easily or not. Most ligand exchanges also happen via a dissociative mechanism (one ligand leaves, another takes the free space) rather than the associative mechanism (one ligand joines in, this leads to another being kicked out) seen in organic nucleophilic substitutions. Therefore, I didn’t talk about those in the answer $\endgroup$
    – Jan
    Commented Sep 30, 2017 at 16:45
  • $\begingroup$ Anhydrous cobalt(II) chloride is blue in colour but on dissolving in water it changes to pink in colour with the change in coordination number to six. $\ce{Co(II)Cl2}$ has a blue colour in tetrahedral geometry. $$\ce{CoCl2·6H2O-> [Co(H2O)6]Cl2}$$ The anhydrous salt $\ce{Co(II)Cl2}$ has a blue colour in tetrahedral geometry and the hexahydrate $\ce{[Co(H2O)6]Cl2}$ is octahedrally coordinated complex. $\endgroup$
    – user100905
    Commented Nov 15, 2021 at 12:43

1 Answer 1


You may know the formula for Gibbs Free Energy $\Delta G$. This is taught as: $$\Delta G = \Delta H - T\Delta S\tag{1}$$

In this case, we would be observing $\Delta G$ for equilibrium $(2)$.

$$\ce{[Co(H2O)6]^2+ + 4 Cl- <=> [CoCl4]^2- + 6 H2O}\tag{2}$$

Within the temperature range between $\pu{25^\circ C}$ and $\pu{100^\circ C}$, we should not expect the enthalpic term $\Delta H$ to change significantly. In fact, it is usually correct at first or even second approximation to consider it constant. If we do, the only thing in $\Delta G$ that changes with temperature in equation $(1)$ is the $-T\Delta S$ term. It basically implies that a higher temperature will increase the entropic contribution.

You already noted that the product side is favoured entropically since five free-moving particles are turned into seven in the course of the reaction. Higher temperatures should thus favour the product side.

One more thing: since we have established that $\Delta S$ is positive, we can immediately see that $\Delta H$ must be positive, too; i.e. the reaction must be endothermic. This is because $T$ is positive by definition and therefore $-T\Delta S$ is always negative. However, the only way to tip an equilibrium to the opposite side is to change the sign of $\Delta G$ — which can only work if a positive $\Delta H$ can offset a negative $-T\Delta S$.


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