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There are no other electrons to collide, repel and kick Hydrogen's single electron to a distant nucleus. And that a single electron is tightly attracted to the nucleus by the electrostatic energy between them. So it seems to me, that Hydrogen does not require ionization energy at all.

But when I checked on the sources - the Hydrogen's ionization energy is relatively high.

Even Hydrogen makes $NaH$, which just doesn't stay for long and has extreme need for becoming $Na$ metal and Hydrogen gas.

So in this case why can't we say that Hydrogen does not require ionization energy?

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    $\begingroup$ I'm unfamiliar with the term ionic energy. By ionic energy, do you mean ionization energy? $\endgroup$
    – John Snow
    Commented Sep 11, 2014 at 22:37
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    $\begingroup$ I think that ionization-energy is a borderline case when it comes to tags, electrostatic-energy on the other hand is completely superfluous. $\endgroup$
    – Martin - マーチン ♦
    Commented Sep 12, 2014 at 16:57
  • $\begingroup$ @Martin I have no intention of earning 2reps on tag wiki. I just created them, because the context of my question refers to those. $\endgroup$
    – bonCodigo
    Commented Sep 12, 2014 at 17:04
  • $\begingroup$ That's alright, we'll see how they play out... $\endgroup$
    – Martin - マーチン ♦
    Commented Sep 12, 2014 at 17:09

2 Answers 2

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The ionization energy (IE) of an atom or molecule describes the minimum amount of energy required to remove an electron (to infinity) from the atom or molecule in the gaseous state.

With ionization energy, an electron is not "kicked out" by other electrons, but rather it is "the energy required for the electron to 'climb out' and leave the atom."

Since the electron "is drawn inwards by positive electrostatic potential," it would make sense to infer that the more "tightly drawn" the electron is to the nucleus, the more energy it would require to "climb out."

More of this can be found on the wikipedia page.

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    $\begingroup$ Also, Hydrogen does not have any other electrons to 'shield' the one electron from the nuclear charge, therefore the effective nuclear charge will be equivalent to the nuclear charge (not so in multi-electron atoms). $\endgroup$
    – LordStryker
    Commented Sep 12, 2014 at 11:28
  • $\begingroup$ I got the point about "tight, close" attraction between single electron and H nucleus. So how does this single electron acquire energy to fly/shift away from nucleus? $\endgroup$
    – bonCodigo
    Commented Sep 12, 2014 at 16:25
  • $\begingroup$ @bonCodigo I'm not sure what you mean. The ionization energy for a hydrogen atom is 13.6 eV (121 nm). If I take a sample of atomic hydrogen and hit it with UV light of 121 nm or higher energy (smaller wavelength), it will ionize. $\endgroup$
    – Geoff Hutchison
    Commented Sep 12, 2014 at 16:35
  • $\begingroup$ @GeoffHutchison So I guess that's why Hydrogen makes covalent bonds most of the time, apart from unstable NaH, LiH scenarios? $\endgroup$
    – bonCodigo
    Commented Sep 12, 2014 at 16:50
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    $\begingroup$ @bonCodigo It's a bit related. I'd generally revert to electronegativity arguments when discussing whether there are ionic or covalent bonds. The electronegativity of H is almost the same as C, so it does tend to form covalent bonds. NaH and LiH (any hydride) involves $H^-$. In this case, the extra electron in hydride anion is NOT tight or close, making it extra reactive. $\endgroup$
    – Geoff Hutchison
    Commented Sep 12, 2014 at 17:49
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The ionization energy of an atome is the minimum energy required to remove an elelectron from the atome in the gaseous state. In spite of the fact that there are no other electrons to collide, repel and kick Hydrogen's single electron to a distant nucleus. That single electron is tightly attracted to the nucleus by the electrostatic energy between them. So the electron would require energy to get rid of the electrostatic attractive force of the nucleus and escape from the atome.

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  • $\begingroup$ Nice- if you don't mind I continue the chat with @LordStryker. Feel free to join and chip in. $\endgroup$
    – bonCodigo
    Commented Sep 12, 2014 at 16:26

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