How to qualitatively compare bond strength? [closed]








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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](https://chem.libretexts.org/Textbook_Maps/Inorganic_Chemistry_Textbook_Maps/Map%3A_Inorganic_Chemistry_(Wikibook)/Chapter_01%3A_Review_of_Chemical_Bonding/1.3%3A_Bond_polarity_and_bond_strength)). 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?




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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