This seems so simple, yet I just can't seem to manage. Apparently $\ce{S2N2}$ and $\ce{S4^2+}$ (which are isoelectronic) obey Huckel's rule (at least to some extent) and I was trying to draw the Lewis structures to show that, but I can't manage to draw them without having a diradical, which doesn't seem very likely to exist. And if there are radicals, it doesn't obey Huckel's rule anymore.

Can someone help me draw their Lewis structures that obey Huckel's rule?


1 Answer 1


Draw each ring with one double bond and the two atoms having no double bond with two electron pairs instead of one. Verify that the total electron count agrees with two positive charges on $\ce{S4}$ and zero on $\ce{S2N2}$. The double bond and the extra electron pairs on the two atoms are the six ring electrons in the Huckel rule.

  • 2
    $\begingroup$ You can use \ce for formatting chemistry expressions $\endgroup$ Apr 9, 2016 at 20:21
  • $\begingroup$ that works for $\ce{S_4^{2+}}$ but having one double bond doesn't work for $\ce{S_2N_2}$ unless it forms N-N and S-S bonds, rather than interchanging S-N bonds. The only possible structure I found is having two double bonds on one S atom to the two neighbouring S and N atoms $\endgroup$
    – name.disp
    Apr 10, 2016 at 17:23
  • $\begingroup$ One thing that you need to note with $S_4N_4$ is that we require the molecule to have zero charge. In many cases the individual atoms may still have positive and negative charges that cancel out. See for instance the Lewis structure for sulfur dioxide. With $S_4N_4$ you will have a positive charge on one sulfur atom, neutralized by a negative charge on nitrogen. $\endgroup$ Apr 10, 2016 at 17:46

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