0
$\begingroup$

A hydrogen anion has a bigger radius than neutral H . Why is this happening ? I mean both electrons exist at the same shell so this is not it. Both H- and H have 1 proton so this is not it as well . Both electrons of the hydrogen ion have the same energy level and they don't shield the charge of the core from one another. What is it then? Pauli repulsion?

$\endgroup$
8
  • 1
    $\begingroup$ Same shell all right, but they feel each other, don't they? $\endgroup$ Commented Aug 17, 2019 at 1:28
  • $\begingroup$ and if that was the case the H- couldnt exist since it would feel a total charge of 0 meaning it could be anywhere... $\endgroup$
    – Engineer
    Commented Aug 17, 2019 at 1:37
  • 1
    $\begingroup$ Not quite. Imagine the impossible situation where one electron was right next to the proton. Then the other electron would experience zero net attraction/repulsion. This would be like having a neutron and one electron. But the inverse square law implies that there is a difference between 1) when the 2 electrons are on the "same side" of the H atom and 2) being on "opposite sides" of the H atom. Make sense? $\endgroup$
    – Ed V
    Commented Aug 17, 2019 at 1:57
  • $\begingroup$ Ed V first of all electrons dont have a precise location until measured . $\endgroup$
    – Engineer
    Commented Aug 17, 2019 at 12:03
  • 1
    $\begingroup$ You have so many misconceptions, I don't think one question could cover them. Most important thing for this question is that you seem to think that data for ionic or atomic radii are for bare atom or ion which is not true. These are just calculated values for particular types of compounds. $\endgroup$
    – Mithoron
    Commented Aug 17, 2019 at 17:17

1 Answer 1

1
$\begingroup$

Electrostatic repulsion between electrons is the key to understanding $\ce{H-}$. The electrons might be delocalized, but they are also correlated - they avoid each other.

Both electrons of the hydrogen ion have the same energy level and they don't shield the charge of the core from one another.

The Aufbau and Pauli exclusion principles (and the assumption of hydrogenic orbitals) are insufficient to explain the structure and stability of $\ce{H-}$. In particular, it helps to drop the assumption that the electrons occupy the same spatial orbital and differ only according to their spin quantum number [1]:

As pointed out in the introduction Bethe [5] and Hylleraas [6] were the first authors to prove the stability of the negative hydrogen. But it was Chandrasekhar [1]who first introduced a clever wave function to describe the H− system which leads to a beautiful physical picture. The key concept introduced by Chandrasekhar was to break the symmetry between the two electrons, which is a way to introduce implicitly the electron correlation. [...]

We conclude that the negative hydrogen ion is a special atomic system whose stability depends completely on the electron correlation. [...] mean field theories and other approximations do not take into account correlation, preventing the possibility of predicting stability for the negative hydrogen ion.

Reference

[1]. Ignacio Lopez de Arbina de Frutos. Stability of the Negative Hydrogen Ion: Variational Approach with Electron Correlation.

$\endgroup$
4
  • 2
    $\begingroup$ Would have preferred to include a reference to a peer-reviewed article or textbook but the linked article does a good job of explaining the pertinent details. $\endgroup$
    – Buck Thorn
    Commented Aug 17, 2019 at 21:13
  • $\begingroup$ what about the attractive magnetic force? $\endgroup$
    – Engineer
    Commented Aug 18, 2019 at 0:44
  • $\begingroup$ due to the opposite spins? $\endgroup$
    – Engineer
    Commented Aug 18, 2019 at 0:44
  • 2
    $\begingroup$ @Engineer It's not clear what you are asking. Maybe you are talking about the exchange interaction? en.wikipedia.org/wiki/Exchange_interaction ? Quote: "Indeed, in general, the direct magnetic interaction between a pair of electrons (due to their electron magnetic moments) is negligibly small compared to this electric interaction." $\endgroup$
    – Buck Thorn
    Commented Aug 18, 2019 at 9:13

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