# Question c2861

Nov 26, 2015

["Fe"("H"_2"O")_6]^(3+)

#### Explanation:

As you know, a Bronsted - Lowry acid is defined simply as a proton donor. A proton is also called a hydrogen ion, ${\text{H}}^{+}$.

This means that in order for a substance to be classified as a Bronsted - Lowry acid, it must be able to donate at least one hydrogen ion to a Bronsted - Lowry base, which is defined simply as a proton acceptor.

Out of all those four complex ions, the only one that can donate a proton (it can donate as many as three, actually) is the hexaaquairon(III) ion, ["Fe"("H"_2"O")_6]^(3+).

Now, when this complex ion is formed, the central iron atom will bind with six water molecules. More specifically, it will bind with a lone pair of electrons from each of those six water molecules.

As you would imagine, the positive charge of the cation will attract these now bonding electrons from the oxygen atoms.

This will in turn disturb the electron density of the $\text{O"-"H}$ bonds, which are already quite polar, since the bonding electrons that oxygen shares with hydrogen will spend even more time on oxygen.

Moreover, the $3 +$ charge will not remain concentrated on the iron cation, it will be distributed on the complex ion.

As a result, the partial positive hydrogen atoms will become even more partial positive, and thus easier to pick off by a base.

This means that in aqueous solution, one of the water molecules that are attached to the iron cation will donate one of those very partial positive hydrogen atoms to a free water molecule.

The structure of the complex ion will change, since it will now have five water molecules attached, plus an $\text{OH}$ group.

If you want, you can write the balanced chemical equation for the first ionization of the hexaaquairon(III) ion like this

["Fe"("H"_2"O")_6]_text((aq])^(3+) + "H"_2"O"_text((l]) rightleftharpoons ["Fe"("H"_2"O")_5("OH")]_text((aq])^(2+) + "H"_3"O"_text((aq])^(+)

None of the other complex ions have the capacity to donate a proton in aqueous solution, so your answer will be (3) "Fe"("H"_2"O")_6]^(3+)#, the hexaaquairon(III) ion.