I’d read that it takes energy to place electrons in pairs, just less than it would take to place two electrons in different subshells. I get that. What I don’t get is, then, why would electrons want to fill their orbitals with electron pairs at all? Wouldn’t it better to just fill them up with single electrons, as per Hund’s rule, and then leave them like that? Would that not be more energetically favorable? Why do atoms insist of having full valence shells, then?


2 Answers 2


Wouldn’t it better to just fill them up with single electrons, as per Hund’s rule, and then leave them like that? Would that not be more energetically favorable?

You have to consider two different scenarios here.

Case 1. Orbitals of different energies.

Consider a two electron system, say, a $\ce{He}$ atom, and let us concentrate on $\mathrm{1s}$ and $\mathrm{2s}$ orbitals exclusively. Naively, one could think of the total energy of the system being simply the sum of orbital energies. Then for $\mathrm{1s^2}$ configuration we get $E_{\mathrm{1s^2}} = \varepsilon_{\mathrm{1s}} + \varepsilon_{\mathrm{1s}}$, while for $\mathrm{1s^1 2s^1}$ configuration we get $E_{\mathrm{1s^1 2s^1}} = \varepsilon_{\mathrm{1s}} + \varepsilon_{\mathrm{2s}}$. Now clearly, since $\varepsilon_{\mathrm{1s}} < \varepsilon_{\mathrm{2s}}$, we could expect that $E_{\mathrm{1s^2}} < E_{\mathrm{1s^1 2s^1}}$, i.e. that $\mathrm{1s^2}$ configuration is indeed more energetically favourable than $\mathrm{1s^1 2s^1}$ one.

This is a bit naive way of looking at things since the total electronic energy is not equal to the sum of orbital energies. As you mentioned, pairing the electrons up takes some energy, so that $E_{\mathrm{1s^2}}$ is clearly not just the sum of $\varepsilon_{\mathrm{1s}}$ and $\varepsilon_{\mathrm{1s}}$, but this amount of energy is relatively small compared with the energy gain you get by having both electrons in the lower $\mathrm{1s}$ orbital. In other words, energy gain from placing both electrons in energetically more favourable $\mathrm{1s}$ orbital overcompensates for having to pair them up during the placement.

How do we know that? Well, we know it from the quantum theory, or, to be more precise, from its application to a chemical systems. Thus, a solid understanding of quantum mechanics & quantum chemistry is required for a quantitive discussion, but for now you can be certain that electrons fill the lowest available energy levels before filling higher levels. This is the principle known as the Aufbau principle which follows from quantum theory and generally holds.

Case 2. Orbitals of same energy.

Consider a $\ce{C}$ atom now and let us have a look at two electrons in $\mathrm{2p}$ orbitals. You could populate the same, say, $\mathrm{2p_x}$ orbital by a pair of electrons, or you could let the electrons populate two different orbitals, say, $\mathrm{2p_x}$ and $\mathrm{2p_y}$. Naively, we could expect the total electronic energy for both configurations to be equal to each other, since $\varepsilon_{\mathrm{2p_x}} = \varepsilon_{\mathrm{2p_y}}$, so that,

$$ E_{\mathrm{2p_x^2}} = \varepsilon_{\mathrm{2p_x}} + \varepsilon_{\mathrm{2p_x}} = \varepsilon_{\mathrm{2p_x}} + \varepsilon_{\mathrm{2p_y}} = E_{\mathrm{2p_x^1 2p_y^1}} \, , $$

but this is again "not quite right", since pairing the electrons up takes some energy. And this time there is no any gain to compensate for this effect since $\varepsilon_{\mathrm{2p_x}} = \varepsilon_{\mathrm{2p_y}}$, and as a result $\mathrm{2p_x^1 2p_y^1}$ configuration is energetically more favourable than $\mathrm{2p_x^2}$ one.


In terms of real world examples, here are two chemical species whose ground states have unpaired electrons in separate orbitals of the same energy: oxygen and cyclobutadiene. Oxygen, reactive as it is, is stable enough to constitute 20% of the atmosphere, whereas cyclobutadiene rapidly dimerizes or reacts with other species present.

These two examples conform to Hund's rule, and to case 2 in Wildcat's answer, above. Pairing the electrons in the same orbital would place them in closer proximity (hence higher energy) than placing them in two different orbitals where they remain unpaired. A further consequence of having unpaired electrons is that it makes the molecules paramagnetic. This state of having two unpaired electrons is called a "triplet" state since there are three possible spin states for the two electrons (up-up, up-down, and down-down) that can be observed spectroscopically in the presence of a magnetic field.

  • $\begingroup$ This seems more like a comment than an answer. Perhaps you could expand on why these molecules have unpaired electrons. $\endgroup$
    – bon
    Sep 21, 2015 at 17:37

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