Without any thermodynamic data, how would one compare the bond strength of one molecule to another?

I understand a major factor is bond order: triple bond strength > double bond strength > single bond strength. But what if the molecules have the same bond order.

Also, generally a higher electronegativity difference gives a stronger bond (Chemistry LibreTexts article). But what if the difference is insignificant?

For example, in an exam, I was asked to compare $\ce{O-O}$ vs $\ce{S-S}$, and $\ce{O=O}$ vs $\ce{S=S}$ bond strengths. How would I compare them?

  • 1
    $\begingroup$ Hmm, the second part seems to be a duplicate of Why is an S-S bond stronger than an O-O bond? Is it? $\endgroup$ – Gaurang Tandon Apr 28 '18 at 12:10
  • $\begingroup$ @GaurangTandon thanks for the link, it answers the example but I'm looking for a more general answer. $\endgroup$ – George Tian Apr 28 '18 at 13:13
  • $\begingroup$ While I understand that you've received an answer to the query of O−O vs S−S, and O=O vs S=S, I am not sure if there's a single (or even a proper exhaustive list of several) of catch-all rules for comparing the bond strength of any two given molecules. Hence, I'm voting to close as "too broad". Perhaps, if you could detail your question to focus on a specific subset of molecules? $\endgroup$ – Gaurang Tandon Apr 28 '18 at 14:57

You can compare the bond strengths of $\ce{O-O}$, $\ce{S-S}$, $\ce{S=S}$ and $\ce{O=O}$ by considering all possible effects that can alter the bond strengths. The effects which needs to be considered here are extent of π-overlap between the orbitals and inter electronic repulsions between the bonding atoms.

Firstly, it is obvious that double bonds will be relatively stronger than the single bonds. So, it is for sure that $\ce{O=O} $ and $\ce{S=S}$ will have higher bond strengths than $\ce{O-O}$ and $\ce{S-S}$. Now, we have to compare between two double bonds and two single bonds.

In $\ce{O-O}$, two similar $\ce{2p}$ orbitals of two oxygen atoms form head-on overlap (σ-bonding) with each other, whereas, in $\ce{S-S}$, two similar $\ce{3p}$ orbitals overlap. As it is a head on overlap, the extent of overlapping doesn't differ significantly, but for $\ce{2p}$ orbitals being smaller in size, two highly electronically dense $\ce{O}$ atoms comes much close to each other which hugely increase the inter electronic repulsions between the atoms. But in $\ce{S}$, the two bonding atoms remains much far from each other, and the electronic charge density is not also too much on each $\ce{S}$ as the $\ce{3p}$ orbitals are large in size, where the charge can easily diffuse. For this reason, $\ce{S-S}$ bonds become much stronger than $\ce{O-O}$ bonds. (The reason is similar for the bond dissociation energy order to decrease as in $\ce{Cl-Cl} > \ce{Br-Br} > \ce{F-F}$)

But if we compare $\ce{O=O}$ and $\ce{S=S}$, the charge density on oxygen is reduced as lone pairs on oxygen decreases, therefore that effect becomes negligible. In this case, the extent of π-bonding starts to dominate. Being small sized, $\ce{2p\pi-2p\pi}$ overlap in oxygen becomes much more strong and effective rather than $\ce{3p\pi-3p\pi}$ overlap (where extent of overlap is not so much) in $\ce{S=S}$. That's why the introduction of double bond becomes much more thermodynamically favourable for oxygen and therefore, $\ce{O=O}$ bond is stronger than $\ce{S=S}$ bond.

Thus the overall order of bond strength becomes, $$\ce{O=O > S=S > S-S > O-O}$$


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