# Is hydronium an electrophile, nucleophile, both or neither?

Is hydronium an electrophile or a nucleophile? I expected $\ce{H3O+}$ to be a nucleophile due to the presence of electrons in the $\ce{2p}$ orbitals of $\ce{O}$. But the answer key of the test states that its none of them.

I can rationalise it this way: If $\ce{O}$ donates its 2 electrons then there would be 2 positive charges on $\ce{O}$ leading to instability.

But I am not too sure of my reasoning either. Isn't hydronium serving as a nucleophile in its reaction with $\ce{OH-}$ ? Why is it then neither an electrophile nor a nucleophile?

Google books results show that its mentioned as an electrophile in some books. While there's disagreement on other sides on the internet? What's the true classification?

• $\ce{H^{+}}$ (or equivalently $\ce{H3O^{+}}$) is an electrophile. If you react $\ce{C6H6}$ with $\ce{D3O^{+}}$ you will get $\ce{C6H5D}$ as the initial product, formed by the standard electrophilic aromatic substitution mechanism. – ron Mar 22 '18 at 1:07
• $\ce{H^{+}}$ is shorthand for $\ce{H3O^{+}}$ which is shorthand for the Zundel and Eigen cations. See here. – ron Mar 22 '18 at 3:11
• @ron I understand that $\ce{H+}$ is just shorthand for $\ce{H3O+}$. However, how do you explain that $\ce{H3O+}$ is an electrophile? We can explain that $\ce{H+}$ is an electrophile as it has an empty duplet, but we can't say the same for $\ce{H3O+}$. What other reasoning do we have then? – Gaurang Tandon Mar 22 '18 at 4:52
• @GaurangTandon Like I mentioned above, $\ce{H3O^{+}}$ doesn't really exist either. You actually just have a solvated proton that still needs 2 electrons. – ron Mar 22 '18 at 13:25
• @GaurangTandon I see no contradiction. – ron Mar 22 '18 at 13:42