Question

Why chelate compounds are stable than nonchelated compounds?

Answer 1

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 `en` is ethylenediamine:

![http://www.sigmaaldrich.com/]()

Visually, at first glance, the only difference between `en` and `"NH"_3` is that `en` has a `-"CH"_2-"CH"_2-` bridge connecting the two nitrogens, and that both "teeth" on `bb(en)` 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 `"M"-"N"` interactions, because each of the six `"NH"_3`'s would form one `"M"-"N"` interaction, but each of the three `en`'s would form two `"M"-"N"` interactions.

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

The main difference we're looking for is that `en` 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.

![https://figures.boundless-cdn.com/]()

Therefore, its entropy of bond formation is less positive, and the decrease in entropy due to the interaction corresponds to favoring `en` over `"NH"_3` (since `DeltaS_"rxn" = sum DeltaS_"bonds broken" - sum DeltaS_"bonds formed"`).

We call this the chelate effect, where the binding of a chelating ligand like `en` in comparison to `NH_3` is entropically favorable. Hence, a chelating ligand creates a more entropically stable compound.