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

Mar 6, 2016

Nothing. It doesn't dissolve in water, nor does it react with water.

The anion in ${\text{Fe"_2"O}}_{3}$ is ${\text{O}}^{2 -}$, which is a very, very strong $\setminus m a t h b f \left(\sigma\right)$ 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 ${\text{Fe"^(3+)-"O}}^{2 -}$ interaction and replacing it with an ${\text{Fe"^(3+)-""^((delta^(-))) "OH}}_{2}$ interaction. It's not favorable in typical reaction conditions.

You can see other $\sigma$ donors such as ${\text{NH}}_{3}$ and $\text{CO}$ (which is also a $\pi$ acceptor) in the spectrochemical series:

https://en.m.wikipedia.org/wiki/Spectrochemical_series#Spectrochemical_series_of_ligands

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

Hence, iron(III) oxide doesn't dissolve in water, nor does it react with water.