I'm a high school student. I noticed $\ce{H+}$ ion is commonly present in my books while I didn't find any presence of $\ce{H-}$ ions in my books. However, I found on internet that $\ce{H-}$ also exists but it is less common. Because Hydrogen has just one electron, it can either receive one electron to be $\ce{H-}$ or omit an electron to be $\ce{H+}$. So, both should have the same possibility to exist. Then, why is $\ce{H+}$ more common than $\ce {H-}$?

The answer to the question might be obvious to most of the users here with their knowledge. But please share a detailed explanation that is suitable for a high school student.

  • 6
    $\begingroup$ The answer is actually not so obvious considering that the energy required to eject an electron out of an H atom is 13.60 eV, but when an additional electron is attached to an H atom, 0.75 eV are released. $\endgroup$
    – Loong
    Commented Aug 1, 2021 at 18:11
  • $\begingroup$ :/ Are you asking about actual separate entities existing in the world, or some letters in equations? As far as existence as chemical entities is concerned, both of them hardly exist at all. $\endgroup$
    – Mithoron
    Commented Aug 1, 2021 at 20:41
  • 1
    $\begingroup$ @Loong: as it was pointed out in the comments below, no actual H⁺ (having presumed internal energy 13.60 eV/ion) exists in solutions or solids. Ions denoted as H⁺ are actually acids whose conjugate bases are (neutral) molecules. That is, presumed hydron ions are actually protonated forms. These acids have much lower enthalpy. $\endgroup$ Commented Aug 2, 2021 at 7:46
  • 3
    $\begingroup$ It's not only for H the case. Removing an electron from Cu(I) to form Cu(II) requires a lot of energy as well, still we find mostly Cu(II) in solution because the energy of hydration is large enough to 'stabilize' or compensate for it. For hydrogen it requires about +1312 kJ/mol to ionize it but dissolving that H+ in water then gives you about -1168 kJ/mol of enthalpy back. $\endgroup$ Commented Aug 2, 2021 at 9:14

2 Answers 2


This is because we live in a world dominated by oxygen and water. In other words, it is an oxidized world. Most metals occur naturally in the form of oxides, silicates, halides, or other derivatives. Hydrogen occurs as $\ce{H+}$.

In a hypothetical world dominated by metals, all that could have turned out otherwise. Oxygen would be a scarcity, and would come in the form of metal oxides. Nitrogen would be found in nitrides, hydrogen in hydrides (so, a lot of $\ce{H-}$), and so on. There would be no free water or free oxygen.

In our world, it is the other way around. Water is ubiquitous (that is, found pretty much everywhere); oxygen is even more so. $\ce{H-}$ can't exist in their presence. It will quickly react with either and cease to be $\ce{H-}$. It can only exist in an artificial environment.

So it goes.


Actually, $\ce{H^-}$ is relatively common in organic chemistry. Carbocation rearrangement is often done through a hydride ion transfer within the ion, and hydride-ion transfers are featured in organic redox reactions such as sodium borohydride reduction and the Cannizzaro reaction.

  • 2
    $\begingroup$ It's important to point out that these are merely formal hydride transfers; at no point is there truly any hydride ion present. $\endgroup$ Commented Aug 1, 2021 at 19:48
  • 4
    $\begingroup$ Not that hydrogen cations are found as such in their reactions either ... . $\endgroup$ Commented Aug 1, 2021 at 19:50
  • 1
    $\begingroup$ True, but at least, one could argue that it is solvated $\ce{H+}$. $\ce{H3O+}$ is probably "closer" to $\ce{H+}$, than $\ce{BH4-}$ is to $\ce{H-}$. $\endgroup$ Commented Aug 1, 2021 at 19:54
  • 6
    $\begingroup$ None is really close. In both cases we have to accept molecular fragments or nothing at all. $\endgroup$ Commented Aug 1, 2021 at 19:59
  • 1
    $\begingroup$ @orthicrrson carbocation rearrangement is 3c-2e, no? $\endgroup$ Commented Aug 2, 2021 at 14:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.