According to wikipedia $\ce{H^+3}$ and $\ce{H3}$ are possible and have been synthesized. It says that $\ce{H^+3}$ is stable too. Can anyone please tell me how is the bonding possible?

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Hydrogen has a valency of 1 but it still bonds like Ozone. How is this possible?


  • $\begingroup$ Because the cation is aromatic? (Bonding analogous to pi bonds in the cyclopropenyl cation.) $\endgroup$ Oct 21, 2020 at 0:11

1 Answer 1


Here are the vital parts from the links about $\ce{H3}$ in the question:

The molecule can only exist in an excited state [...] the electronic state for a trihydrogen cation with an electron delocalized around it is a Rydberg state.

and from second-level link

Rydberg states of the neutral triatomic hydrogen molecule can be viewed as an electron attached to a tightly bound $\ce{H3+}$ core

Basically, at least some states can be viewed as an $\ce{H3+}$ ion working as a nucleus in a bigger hydrogen-like atom. If said 'nucleus' catches the electron, the molecule immediately dissociates. So, let's focus for a bit on $\ce{H3+}$. In this ion a two-electron three-center bond is formed. Such bonding is actually pretty common, even if (almost never) mentioned in school. As long as orbitals of several atoms can overlap effectively and have fitting symmetry, they can form a single highly symmetrical bonding orbital, that can accept two electrons. In this case the bonding order between atoms is fractional, so in $D_\mathrm{3h}$ (triangular) $\ce{H3+}$, each of the three $\ce{H-H}$ bonds has bond order $1/3$. This kind of bonding plays a critical role in the chemistry of boranes (and Group 3 elements in general, maybe except thallium) and some families of compounds of d-block elements.

To obtain more intimate and correct understanding you should invest in molecular orbital theory. Your link for $\ce{H3+}$ contains a comprehensive description of bonding in terms of said theory.


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