Saturday, 23 April 2016

inorganic chemistry - Why does sulfur, but not oxygen, catenate?


Oxygen is a rather boring element. It has only two allotropes, dioxygen and ozone. Dioxygen has a double bond, and ozone has a delocalised cloud, giving rise to two "1.5 bonds".


On the other hand, sulfur has many stable allotropes, and a bunch of unstable ones as well. The variety of allotropes, is mainly due to the ability of sulfur to catenate.


But, sulfur does not have a stable diatomic allotrope at room temperature. I, personally would expect disulfur to be more stable than dioxygen, due to the possibility of pπ-dπ back-bonding.



So, why do sulfur and oxygen have such opposite properties with respect to their ability to catenate?



Answer



First, a note: while oxygen has fewer allotropes than sulfur, it sure has more than two! These include O, OX2, OX3, OX4, OX8, metallic O and four other solid phases. Many of these actually have a corresponding sulfur variant. However, you are right in a sense that sulfur has more tendency to catenate… let's try to see why!


Here are the values of the single and double bond enthalpies: BondDissociation energy / kJ mol1OO142SS268O=O499S=S352 This means that O=O is stronger than S=S, while OO is weaker than SS. So, in sulfur, single bonds are favoured and catenation is easier than in oxygen compounds.


It seems that the reason for the weaker S=S double bonds has its roots in the size of the atom: it's harder for the two atoms to come at a small enough distance, so that the 3p orbitals overlap is small and the π bond is weak. This is attested by looking down the periodic table: Se=Se has an even weaker bond enthalpy of 272kJ/mol. There is more in-depth discussion of the relative bond strengths in this question.


While not particularly stable, it's actually also possible for oxygen to form discrete molecules with the general formula HOXnH; water and hydrogen peroxide are the first two members of this class, but n goes up to at least 5. These "hydrogen polyoxides" are described further in this question.


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