P forms $\ce{P4}$ due to its inability to form double/triple bonds. So it makes a sort of tetrahedral structure. S however makes $\ce{S8}$ cycles w/ $sp^2$ hybridization. Why go all that way to connect the chain? Can't $\ce{S3}$ do the trick? It will still have $sp^2$. What makes $\ce{S8}$ more stable?


The ring you propose would have severe issues with ring strain, and would quickly 'pop-open' into a $\ce{S3}$ chain.

All other chains have similar benefits and penalties. The larger the ring, the greater the entropic cost for formation in a growing chain model. this is traded off with ring strain: too tight causes one kind of strian and too loose causes another.

$\ce{S8}$ is the 'sweet spot' for handling these competing factors.

As far as phosphorous goes, my understanding is that $\ce{P4}$ is the most reactive form of molecular phosphorous, and other allotropes are thermodynamically preferred.

  • $\begingroup$ White phosphorus $\ce{P4}$ is the only molecular phosphorous at standard temperature and pressure. All other modifications are polymer forms $\ce{P_{$n$}}$ with well knows structures, only red phosphorous is amorphous. $\endgroup$ – Martin - マーチン Mar 26 '15 at 11:16

Trisulfur does exist, but is not circular, it is ozone-like and unstable at common temperatures.

My guess is that while angle strain does exist in ring form, it is not that harsh. Unlike carbon, having natural bond angle ~109 degree, for bivalent sulfur it is closer to 90 degree (sulfur exhibits little to no hybridization). However, the S-S bond strength is only 40% of C-C bond and in trisulfur ring all S-S bonds would have dihedral angle equal to zero, which is unfavorable because of nonbonding pairs interaction. This issue is a lot more severe than in tetraphosphorus, and leads to ring opening.

For further information on sulfur allotropes see the Wikipedia article.

  • 1
    $\begingroup$ $$\scriptsize \underbrace{\left[\ce{S_{\alpha} <=>[95.6~^\circ\mathrm{C}] S_{\beta}}\right]}_{\normalsize\text{solid}} \ce{<=>[119.6~^\circ\mathrm{C}]} \underbrace{\left[\ce{S_{\lambda}<=>S_{\pi}<=>S_{\mu}}\right]}_{\normalsize\text{liquid}} \ce{<=>[44.46~^\circ\mathrm{C}]} \underbrace{\left[\ce{S8<=>S7<=>S6<=>S5<=>S4<=>S3<=>S2<=>S}\right]}_{\normalsize\text{gaseous}}$$ The liquid and gaseous phases are temperature dependent equilibria. Holleman Wiberg $\endgroup$ – Martin - マーチン Mar 26 '15 at 11:02
  • 2
    $\begingroup$ @Martin For the sake of completeness we should note that $\ce{S_{6+}}$ ring may be prepared and isolated at relatively low temperatures. Why exactly $\ce{S8}$ is the only one obtainable from common sources is indeed an interesting question without simple and clear answer. $\endgroup$ – permeakra Mar 26 '15 at 12:17
  • $\begingroup$ Absolutely right! I was typing up the formula for an answer myself, but then I noticed, you already stated what I had to say, so I just left the hard coding as a comment ;) $\endgroup$ – Martin - マーチン Mar 26 '15 at 12:19

enter image description here

Trisulfur or thiozone is significantly present in the gas phase of sulfur at high temperature and has been studied by rotational spectroscopy as reported in The Rotational Spectrum and Geometrical Structure of Thiozone J. Am. Chem. Soc., 2004, 126 (13), pp 4096–4097

The angle of $\ce{S3}$ is very close to isoelectronic $\ce{O3}$ (116.8 degrees), $\ce{SO2}$ (119.5 degrees), and $\ce{S2O}$ (118 degrees).

The above article states that the central sulfur is $sp^2$ hybridized.

Even though the ground state is bent, the cyclic form is only slightly higher in energy, about 5-10 kcal/mol higher.

There is evidence that the cyclic form can exist as a ligand of a transition metal complex, as reported in Transition Metal Complexes of Cyclic and Open Ozone and Thiozone J. Am. Chem. Soc., 2005, 127 (4), pp 1278–1285. According to this reference, early calculation incorrectly showed the cyclic form to be more stable for both ozone and thiozone "due to the fact that electron correlation plays an important role in these molecules, especially in the open form".


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.