Is it possible to compare energy(enthalpy or similar) between an atom and its ion, possibly in the same state(gas)?

For example, $\ce{O}$ and $\ce{O^+}$(intentionally cation), $\ce{Na}$ and $\ce{Na^+}$.

At first I though of ionization energy, but its RHS(products) has electron, so I thought it can't be used to directly compare them.

  • 2
    $\begingroup$ The electron is assumed to have $0$ energy... $\endgroup$ – DHMO Sep 21 '16 at 15:53
  • $\begingroup$ @user34388 How about electron's kinetic energy? $\endgroup$ – user1448742 Sep 21 '16 at 15:55
  • $\begingroup$ Assume it's zero. $\endgroup$ – Zhe Dec 21 '16 at 14:43
  • $\begingroup$ Give us some context please. There are some good answers below, but I think you are aiming for something else. As a general rule, don't compare total energies of different compounds directly. I don't think it's really clear that you want the ionization energy but rather something more dubious $\endgroup$ – AMT Jan 20 '17 at 15:34

Ionization energy is, exactly, a comparison of the energy of (i) a given species and (ii) that same species with one electron removed.

Ionization energies can be thought of as a reaction like the following, using the oxygen atom as an example:

$$ \ce{O -> O+} + e^- $$

When evaluating quantities like the ionization energy, "external" energies (this is my term, not an official/formal one!) like kinetic and gravitational potential energy are disregarded. Further, the reference energy is generally defined to be zero for an isolated electron at "infinite separation".

Thus, the energy of the above reaction is:

$$ E_\mathrm{rxn} = \left(E_{\ce{O+}} + E_{e^-}\right) - E_{\ce O} $$

But, since the electron is taken as being in its reference state, this is simply:

$$ E_\mathrm{rxn} = E_{\ce{O+}} - E_{\ce O} \equiv IE $$

PLEASE NOTE that the above energies are only the internal/electronic energies of the $\ce O$ and $\ce{O+}$ species, and not their enthalpies $\left(H_{\ce{O+}},\,H_\ce{O}\right)$, free energies $\left(G_{\ce{O+}},\,G_\ce{O}\right)$, etc.


If you want to compare the energy between an atom and its parent ion you basically are indeed interested in the adiabatic ionization energy of your atom (i.e., the difference in energy between the atom and ion in their lowest energy state). This is the energy needed to remove the electron away from the ion core.

It might be instructive consider the spectrum of the hydrogen atom. You might know that the energy levels of the H atom w.r.t. its ground state are given by

$$E_n=E_\text{IP}-\frac{\mathcal{R}}{n^2}$$ where $E_n$ is the energy of the state with principal quantum number $n$, $E_\text{IP}$ is the adiabatic ionization energy, $\mathcal{R}$ is the Rydberg constant. In case of hydrogen, $E_\text{IP}=\mathcal{R}$. The hydrogen atom thus has an infinite amount of energy levels that become more closely spaced as $n$ increases and converges on the ionization energy. This corresponds to an electron that is so far away from the proton, that the system can be considered to consist of an isolated electron and ion. For Na and O, the formula for the energies is not so simple anymore, but you'll still find that the energy levels converge on the ionization energy of the atom.

Another way to look at it is by considering an low-energy collision between an ion and an electron.


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