# Predicting sigma bond overlap strengths of s-s, p-p, s-p, sp-sp etc

How can we qualitatively predict $\sigma-$bond strengths of overlap between:

• $\ce{s-s}$
• $\ce{s-p}$
• $\ce{p-p}$
• $\ce{sp-s}$
• $\ce{sp-p}$
• $\ce{sp-sp}$

etc.?

My school-book says $\ce{s-s}$ overlap bond strength is greater than $\ce{s-p}$ overlap which is greater than $\ce{p-p}$

Another book Inorganic Chemistry by JD Lee states this:

Are the facts stated in the above books correct or justified? Is there any order of orbital overlap strengths? If so can we predict it qualitatively?

I think the issue might be what orbitals are being referred to. Both books could be correct if they are referring to different $\mathrm{p-p}$ overlap. For example, suppose two atoms are bonding along the z-axis. $\mathrm{p_z}-\mathrm{p_z}$ overlap would be greater than $\mathrm{s-s}$ because the lobes of the p orbitals extend out to meet each other between the two molecules. If however we looked at $\mathrm{p_x}-\mathrm{p_x}$ overlap, the $\mathrm{s-s}$ overlap would be greater because the p lobes don't extend toward each other. Increasing hybridization increases overlap because it combines the directionality of p orbitals (that they extend out away from the atom) and the electronic density of s orbitals (larger space for overlap to occur).

• The schoolbook asked about p orbitals on internuclear axis only! I feel schoolbook may be wrong because as you said p orbitals are more directional and should form stronger bonds.. – jonsno Aug 20 '17 at 2:44
• @samjoe Well then for now, I would say trust the Inorganic chemistry book and that hybridization increases overlap. I'm not certain what the other textbook is referring to. – Tyberius Aug 20 '17 at 17:03
• @Tyberius According to this reasoning, shouldn't a p-p overlap be stronger than all of the overlap of hybridised orbitals? – FreakyLearner Mar 31 '18 at 9:47
• @Taufeeque I believe these measures are for the orbitals overlapping head on. The OP mentions the reasoning that the hybrids have larger directed lobes and so should have stronger bonding. – Tyberius Mar 31 '18 at 13:50
• @YUSUFHASAN so is your confusion why a hybridized orbital overlaps better than a pure p? It's really just a matter of hybridized orbitals having their density focused in one lobe, rather than equal lobes in the bonding and opposite direction. You could think of a chemical reaction changing the hybridization, but we can be more abstract and just think of the hybridization changing. Hybridization is just a mathematical construct and we can always choose an arbitrary mixing of the orbitals; We just happen to know that certain mixings due a better job at describing bonding. – Tyberius Oct 4 '18 at 4:02

Although the general trend $$\mathrm{s} < \mathrm{p} < \mathrm{sp}^n$$ makes sense, these magic numbers $$1.73$$, $$1.93$$, $$1.99$$, and $$2.00$$ seem to have just been pulled out of a hat. If these are really "approximate strengths of bonds", a good book would justify these by showing which bonds they use to come up with these numbers.

Just to show one counterexample, it is commonly known that the strength of C–H bonds increase going from alkanes to alkenes to alkynes. That is to say, the bond strengths increase in the order $$\ce{C_{sp^3}-H} < \ce{C_{sp^2}-H} < \ce{C_{sp}-H}$$. This clearly contradicts the supposed trend given in the book of $$\mathrm{sp}$$ orbitals forming slightly weaker bonds than $$\mathrm{sp^3}$$.

From the second table in this Wikipedia section:

Compound Carbon hybridisation C–H bond strength (kcal/mol)
Ethane sp3 101
Ethylene sp2 111
Benzene sp2 113
Acetylene sp 133

As far as I am aware, this trend is mainly rationalised using bond lengths.

Are the facts stated in the above books correct or justified?

Some parts of it (e.g. the general trend of $$\mathrm{s} < \mathrm{p} < \mathrm{sp}^n$$) might be correct, as already well addressed by Tyberius' answer, but other parts (especially these magic numbers) are highly debatable.