I really don't know if it is even logical to ask that $\ce{H3O-}$ doesn't exist but someone asked me and I wasn't clear. Hope that some one will make me clear.

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    $\begingroup$ Well, it seems a valid question, while not particularly good. Your H3O- is H2O + H- and these react violently. $\endgroup$ – Mithoron Nov 6 '17 at 15:24
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    $\begingroup$ Someone should answer this with the MO scheme of water … $\endgroup$ – Jan Nov 7 '17 at 4:24
  • $\begingroup$ To me, this sounds like $\ce{H2}$ and $\ce{OH-}$. Both particles are fine, but have absolutely no intention to stay together. $\endgroup$ – Ivan Neretin Nov 7 '17 at 9:18
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    $\begingroup$ It does exist. It has been observed in the gas phase. $\endgroup$ – Raditz_35 Nov 7 '17 at 9:21
  • $\begingroup$ @Raditz_35 What kind of gas phase? Very low pressure and temperature, like in space? There many thoroughly unstable particles can exist. $\endgroup$ – Karl Nov 7 '17 at 11:00

I'm not a big fan of MOs when this can be explained using just AOs. For sure, MO theory gives a much better and profound explanation, and I'm surely waiting for someone to do that, but that's what I would answer if the kids asked me this.

Oxygen has 6 electrons on its outer shell, which has $\mathrm{2s}$ and $\mathrm{2p}$ orbitals. The arrangement of these electrons is such that $\mathrm{2s}$ sublevel is completed, and $\mathrm{2p}$ sublevel contains 4 electrons. Following Hund's rule, two electrons occupy one of three $\mathrm{2p}$ orbitals and other two come by one to each of $\mathrm{2p}$ orbitals left.

enter image description here

When water forms, oxygen shares two non-bonded electrons with hydrogen atoms, forming a covalent bond with them.

$\ce{H3O+}$ forms via donor-acceptor bonding, when oxygen supplies its lone pair, which it has two, to form a bond with $\ce{H+}$ ion.

To form $\ce{H3O-}$, you'll need an interaction between water molecule and $\ce{H-}$ ion. Even if we ignore the fact that hydride is a powerful base ($\mathrm pK_\mathrm a$ is around $42$) and will react violently with water, it would be impossible. In this theoretical case neither of the species has an orbital suitable to accept a lone pair from another species. Each has a complete outer shell.

Edit: it really exists though. Wow.


  1. Probing the transition state via photoelectron and photodetachment spectroscopy of H3O
  2. Calculation of the photodetachment spectrum for H3O
  • $\begingroup$ @Raditz_35 Yeap, never checked this. Edited my answer. Thanks. $\endgroup$ – wolphram Nov 7 '17 at 9:34
  • $\begingroup$ Thanks! I think the important part is why it is irrelevant for everyday chemistry = the reason for the question. $\endgroup$ – Raditz_35 Nov 7 '17 at 9:41
  • $\begingroup$ Please take your time and check out the mhchem package that this site has loaded for MathJax. Using the \ce command automatically gives the correct output (especially non-italicised element symbols!) for a number of intuitive inputs. See this post on meta for more information. $\endgroup$ – Jan Nov 7 '17 at 9:47

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