I have faced several questions regarding the diamagnetic behaviour of certain organic compounds. I know that one has to check the electronic configuration of the atoms in the compound and see if there is an unpaired electron or not? Of all aromatic compounds (benzene, cyclopropene cation, cyclopentene anion) that I have come across one thing was common they were all diamagnetic. I have the following question in my mind

  • Are all aromatic compounds necessarily diamagnetic and if it is true what may be the proper reasoning for that?

  • Moreover, the compounds which are stabilised by resonance (hyper conjugation,+M effect) are necessarily diamagnetic. I have this doubt because by the electronic configuration tertiary butane cation should be paramagnetic ($2p^1$ configuration) but is still diamagnetic.

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The above is the structure of tertiary butyl cation. It is clear that if we consider the electronic configuration of the central carbon it is $1s^{2}2s^{2}2p^{1}$.So it must be paramagnetic. This is where my confusion is.

  • 1
    $\begingroup$ Nonsense, t-butyl cation has no electrons on p-orbital. $\endgroup$
    – Mithoron
    Dec 28, 2016 at 15:40
  • $\begingroup$ Mithoron is right. The central carbon atom is $\mathrm{sp^3}$ hybridized. $\endgroup$
    – DHMO
    Jan 13, 2017 at 14:58
  • 1
    $\begingroup$ @DHMO It isn't sp3, simply empty p orbital... $\endgroup$
    – Mithoron
    Jan 13, 2017 at 15:24
  • 1
    $\begingroup$ Sorry, I meant $\mathrm{sp^2}$ hybridized... $\endgroup$
    – DHMO
    Jan 13, 2017 at 15:25
  • $\begingroup$ As only a slight oversimplication, if you have an even number of electrons, the system is diamagnetic for simple organic molecules. There are enough exceptions: diradicals, diradicals created from filling degenerate antibonding orbitals, excited states. But as a rule for most applications, this statement is true. $\endgroup$
    – Zhe
    Jan 22, 2017 at 17:59

2 Answers 2


If all the electrons are spin paired then the molecule will be in a singlet state$^*$ and will be diamagnetic, if they are not so paired then it will be will be paramagnetic. The rules are the same whether or not the molecule is 'organic'.

Most 'normal' organic compounds, e.g. methane, benzene, etc. have paired electrons in each of the orbitals from the lowest one to the highest occupied one (HOMO), and thus are diamagnetic. The orbitals including the LUMO and up are unoccupied.

Radicals such as $\ce{CH3\cdot}$ have an unpaired electron (doublet state) and so are paramagnetic.

If a biradical is formed e.g. carbenes (see IUPAC Gold Book 'biradical' for an example structure) then it will depend upon whether the spins are paired in a singlet state (diamagnetic) or parallel as in a triplet state which will be paramagnetic. Often singlet and triplets are close in energy with the triplet being lowest. Similarly, dioxygen has two unpaired electrons and forms a triplet ground state so is paramagnetic. Nitric oxide has one unpaired electron and is paramagnetic.

If a molecule is put into an electronically excited state by absorbing a photon then both singlet and triplet states can be produced, however, theses are short lived and decay back to the ground state.

$^*$(singlet /doublet/triplet refer to spin multiplicity, spin quantum number $\pm 1/2$ and total spin e.g. $S=s_1+s_2$ gives multiplicity g as $g=2S+1$)

example of biradical from IUPAC Gold Book.


  • $\begingroup$ Thanks for the answer,but What about aromatic organic compounds?Are they necessarily diamagnetic? $\endgroup$
    – Pink
    Dec 28, 2016 at 12:54
  • $\begingroup$ What about tertiary butane cations.according to theory it must be paramagnetic but actually it is diamagnetic. $\endgroup$
    – Pink
    Dec 28, 2016 at 12:55
  • $\begingroup$ No, but almost all are. Have a look at the structure in the IUPAC Gold Book biradical page $\endgroup$
    – porphyrin
    Dec 28, 2016 at 12:56
  • $\begingroup$ @aniline. If experiment gives diamagnetic then thats what it is. It does not mean basic theory is wrong as this is supported by lots of data, but just that this particular calculation gives wrong answer possibly because singlet and triplet energy are close to one another and within error of calculation. $\endgroup$
    – porphyrin
    Dec 28, 2016 at 16:36
  • 3
    $\begingroup$ @aniline Go back to the drawing board with the tert-butyl cation. Remember that it is not a tert-butyl radical. $\endgroup$
    – Jan
    Dec 28, 2016 at 22:58

I see your point of confusion!

The above is the structure of tertiary butyl cation. It is clear that if we consider the electronic configuration of the central carbon it is $\mathrm{1s^2\,2s^2\,2p^1}$.

Well yes but no. Remember that you don’t have a lone atom in space but a carbon bound to three other atoms. These three σ bonds must originate from three orbitals pointing towards these three atoms. If we assume the electronic configuration to start off as (emphasis important) $\mathrm{2s^2\,2p^1}$, we cannot form three bonds, we only have one ‘ready’ orbital.

Therefore, this atom hybridises — or so our simplified view of the world. We mix the $\mathrm{2s}$ and two $\mathrm{2p}$ orbitals to give three $\mathrm{sp^2}$ hybrid orbitals. These now nicely align in $120^\circ$ angles and in one plane and are ready to bond to three other carbons.

There is a fourth p orbital on carbon. It remains there as a p orbital in the final structure. The three electrons we had are delivered into the three $\mathrm{sp^2}$ orbitals; thus the final $\mathrm{p}_z$ orbital is empty; completely electron-free.

Each of the electrons in the $\mathrm{sp^2}$ orbitals is paired up as a result of forming the σ bond. Therefore, there are no unpaired electrons in the tert-butyl cation and the species is diamagnetic.

The electronic configuration could be taken as $(\sigma_{\ce{C{1}-H}})^6\,(\sigma_{\ce{C{2}-H}})^6\,(\sigma_{\ce{C{3}-H}})^6\,(\sigma_{\ce{C{1}-C+}})^2\,(\sigma_{\ce{C{2}-C+}})^2\,(\sigma_{\ce{C{3}-C+}})^2\,\mathrm{2p^0}$. But nobody would ever do that.

You can add an electron into the $\mathrm{p}_z$ orbital to get a paramagnetic compound. Since you added an electron from the tert-butyl cation to arrive there, you no longer have a cation. Instead, you have created the tert-butyl radical. The electronic configuration from above would end in $\mathrm{2p^1}$.

Whenever you see the word radical, you can strongly assume that you have unpaired electrons. If you do not see it, you very likely have a diamagnetic organic molecule.

Note that NMR spectroscopy is intensely popular in organic chemistry. Because it bases on magnetic fields (and resonance therewith), it doesn’t really work with paramagnetic compounds. But more than $99.999~\%$ of known organic molecules have an NMR spectrum published.

  • $\begingroup$ Or a better answer that shows up in the next 3 days ;) His (and yours) will be tough to beat though, very thorough. $\endgroup$
    – airhuff
    Jan 17, 2017 at 3:52
  • $\begingroup$ @airhuff That too, of course; I am very open for that option ;) $\endgroup$
    – Jan
    Jan 17, 2017 at 22:53

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