# Why chelate compounds are stable than nonchelated compounds?

Dec 3, 2016

Well, to compare them on equal footing, we would be considering the same bonds being made with each kind of compound.

Consider the following reaction:

["M"("NH"_3)_6]^(z+) + 3(en) rightleftharpoons ["M"(en)_3]^(z+) + 6"NH"_3

where $e n$ is ethylenediamine:

Visually, at first glance, the only difference between $e n$ and ${\text{NH}}_{3}$ is that $e n$ has a $- {\text{CH"_2-"CH}}_{2} -$ bridge connecting the two nitrogens, and that both "teeth" on $\boldsymbol{e n}$ bind to the same metal.

(They both have the same atoms that bind, and they both contribute zero charge.)

This will therefore have a net zero enthalpy for the breaking and formation of the six $\text{M"-"N}$ interactions, because each of the six ${\text{NH}}_{3}$'s would form one $\text{M"-"N}$ interaction, but each of the three $e n$'s would form two $\text{M"-"N}$ interactions.

All that means is that enthalpy is not a significant stabilizing factor.

The main difference we're looking for is that $e n$ is fixed to a cis configuration, so by binding via two "teeth", it is going to restrict its own mobility.

Entropy is smaller with more restricted motion.

Therefore, its entropy of bond formation is less positive, and the decrease in entropy due to the interaction corresponds to favoring $e n$ over ${\text{NH}}_{3}$ (since $\Delta {S}_{\text{rxn" = sum DeltaS_"bonds broken" - sum DeltaS_"bonds formed}}$).

We call this the chelate effect, where the binding of a chelating ligand like $e n$ in comparison to $N {H}_{3}$ is entropically favorable. Hence, a chelating ligand creates a more entropically stable compound.