What happens when you place #"Fe"_2"O"_3# in water?

Question

Owen Ambrose · Accepted Answer

Nothing; neither does it dissolve nor react with water.

The anion in #"Fe"_2"O"_3# is #"O"^(2-)#, which is a very, very strong #\mathbf(sigma)# donor. On the other hand, water is a much weaker #sigma# donor (its highest-occupied molecular orbital / HOMO is much lower in energy).

What it means is that the oxide anion makes a much stronger #sigma# interaction than water would, so water is incapable of breaking the #"Fe"^(3+)-"O"^(2-)# interaction and replacing it with an #"Fe"^(3+)-""^((delta^(-))) "OH"_2# interaction. It's not favorable in typical reaction conditions.

You can see other #sigma# donors such as #"NH"_3# and #"CO"# (which is also a #pi# acceptor) in the spectrochemical series:

The spectrochemical series of ligands is described in https://tutor.hix.ai

The ligands further to the left are not expected to capably displace the ligands further to the right (notice how #"O"^(2-)# is not the same as #"O"_2^(2-)#). A #sigma# donor with a #2-# charge is definitely a stronger lewis base than something that is neutral.

Therefore, neither does iron(III) oxide dissolve in nor react with water.

Jose Ayala · Answer

undefined When you place Fe2O3 (iron(III) oxide) in water, it undergoes a chemical reaction to form iron(III) hydroxide, also known as rust, and releases heat. The reaction equation is:

Fe2O3(s) + 6H2O(l) -> 2Fe(OH)3(s)

This reaction is an example of a hydration reaction.