Why s-p orbital overlap is stronger than s-s overlap?

[tasnim rahman]

will the same amount of charge present at higher concentration in the binding region(between two nuclei), give a stronger bond, compared to the same amount of charge present at a lower concentration in the binding region? if so, could it be explained. is this the reason why s-p overlap is stronger than s-s overlap? need quick help
thank you

 

[jambaugh]

It is more a matter of getting the electrons of each atom as close as possible to the nuclei of the other atoms while avoiding other electrons and satisfying the Pauli exclusion principle.

I think (off the cuff here) that the stronger s-p overlap is due to the "sharpness" of the p orbitals which allow the electrons of e.g. the carbon atom's orbits to penetrate deeper toward the hydrogen nucleus in say Methane.

We must always remember that the orbitals are solutions to the quantum mechanics of a single electron in a central potential. Once atoms begin combining, and we consider many electrons these are only approximations. The orbitals are distorted and the ultimate answers to "why" lie in solving the (generally intractable) quantum many body problem.

 

[tasnim rahman]

Thanks very much for the answer Jambaugh. That must mean the closer the electrons are to each nucleus, the stronger the attraction and, so the stronger the bond, right? Thanks again.

 

[SpectraCat]

Thanks very much for the answer Jambaugh. That must mean the closer the electrons are to each nucleus, the stronger the attraction and, so the stronger the bond, right? Thanks again.

You have to be a little careful with this sort of statement. As jambaugh said, s & p orbitals are features of atoms, and lose their meanings in molecules. However, one can draw the following rationalizations about your question without getting into too much trouble (I think):

1) p-orbitals are more directional than s-orbitals (which are isotropic), therefore as jambaugh said, they are better suited to forming directional covalent bonds via orbital overlap.

2) p-orbitals have a node at the nucleus, so a p-electron "feels" the effect of it's own nucleus to a lesser extent than an s-electron. This means that a p-electron is more "available" to feel the effects of the nucleus on a neighboring atom than an s-electron.

3) These sort of arguments are only rationalizations, and should be confined to comparisons of systems with maximum similarity .. e.g. it would be hard to predict relative strengths of a 1-s orbital overlapping with a 2p orbital vs a 3s orbital based solely on these arguments, or even a 2s orbital on oxygen vs. a 2p orbital on carbon.

 

[tasnim rahman]

Thanks very much for the answer SpectraCat.But could it be explained in terms of, charge density, as in charge concentration; and also, amount of charge in the binding region? As much as I understood, the closer electrons are to each nuclei, getting more of their attractions, the stronger the bond, am I getting it right? I just want to have a qualitative idea, without the wave equations, that is.

 

[jambaugh]

Thanks very much for the answer SpectraCat.But could it be explained in terms of, charge density, as in charge concentration; and also, amount of charge in the binding region? As much as I understood, the closer electrons are to each nuclei, getting more of their attractions, the stronger the bond, am I getting it right? I just want to have a qualitative idea, without the wave equations, that is.

Thinking in terms of "charge clouds" you have to also view the electron's quantum orbits in terms of time. The "cloud" of charge density one visualizes is a time averaged density. This is important because you also must consider how well electrons can anti-correlate their positions over time so as to "avoid each other" while "getting cozy" with the protons in the nuclei. The electron's fermionic nature can't be totally ignored even in such qualitative musings as we are considering here.

 

[tasnim rahman]

Thanks very much Jambaugh.So, assuming similar spin interactions(spins pair when the bond forms) and similar electronic repulsions, for both s-s and s-p, the s-p bond will be stronger because, the p electron density can get comparatively closer to each nucleus, forming a stronger bond,is this right?

 

[ZealScience]

You mean s orbital and sp hibridization orbital or overlap, or simply overlap of s orbital and p orbital lying down (horizontal axis)?

I guess a dumbell p lying down overlap makes it closer between atoms than a spherical s orbital does.
And the sp orbital is deflected to the bonding site (where the resonnance happens) thus more resonnance stronger bond.

 

[tasnim rahman]

Thanks you very much ZealScience.But sorry i wasn't talking about hybridization or hybridization overlap or anything; just the simple s and p orbitals overlapping and increasing electron density in binding region, forming a bond. therefore as far as I understood I believe that s-p overlap is stronger than s-s overlap(assuming similar charge density in binding region in both cases), because it allows the charge density to get closer to each nucleus, forming a stronger bond.

 

[jambaugh]

Thanks very much Jambaugh.So, assuming similar spin interactions(spins pair when the bond forms) and similar electronic repulsions, for both s-s and s-p, the s-p bond will be stronger because, the p electron density can get comparatively closer to each nucleus, forming a stronger bond,is this right?

I think so...with all aforementioned qualifications.

 

[ZealScience]

Thanks you very much ZealScience.But sorry i wasn't talking about hybridization or hybridization overlap or anything; just the simple s and p orbitals overlapping and increasing electron density in binding region, forming a bond. therefore as far as I understood I believe that s-p overlap is stronger than s-s overlap(assuming similar charge density in binding region in both cases), because it allows the charge density to get closer to each nucleus, forming a stronger bond.

If you mean just s and p probably you can look at the wavefunction or probability distribution diagram of the p-orbital. You will find that comparing to s-orbital, p spend more time. s have more possibility near the nucleus, but less in the bonding site

 

[tasnim rahman]

If you mean just s and p probably you can look at the wavefunction or probability distribution diagram of the p-orbital. You will find that comparing to s-orbital, p spend more time. s have more possibility near the nucleus, but less in the bonding site

s orbitals spend more time close to their respective nuclei, and less in the binding region. p orbitals spend more time in the binding region; that is close both the nuclei; thus s-p overlap forming a stronger bond than s-s overlap. Thanks again for help, ZealScience.

 

[DrDu]

It is certainly not true in general that s-p overlap is stronger than s-s overlap and even where it is true, it does not automatically imply that the corresponding bonds are stronger. Usually, in first row elements (like C, N, O) s and p orbitals are of comparable size and also the overlaps are comparable, with the s-p overlap being slightly larger.
In contrast, in higher row elements (e. g. S, P, Si, Ge) the p orbitals are much more diffuse than the s orbitals. Hence s-p overlap is much reduced. That is also the main reason why elements of the higher rows do not form sp hybrid orbitals.

 

[tasnim rahman]

if we consider two pairs of hydrogen atoms. one pair of hydrogen atoms bond with overlap of their s-orbitals. The other pair bonds with the overlap of a s and a p orbital(I know its not possible, just assuming). Both bonds are formed through the process mentioned http://www.chemistry.mcmaster.ca/esam/Chapter_6/section_1.html" . Now which bond will be stronger? Is it the s-p overlap bond; due to the presence of a p-orbital allowing part of the charge density to get closer to each nuclei; compared to the s-s overlap? Thanks for the help Dr.Du

 

[DrDu]

if we consider two pairs of hydrogen atoms. one pair of hydrogen atoms bond with overlap of their s-orbitals. The other pair bonds with the overlap of a s and a p orbital(I know its not possible, just assuming). Both bonds are formed through the process mentioned http://www.chemistry.mcmaster.ca/esam/Chapter_6/section_1.html" . Now which bond will be stronger? Is it the s-p overlap bond; due to the presence of a p-orbital allowing part of the charge density to get closer to each nuclei; compared to the s-s overlap? Thanks for the help Dr.Du

As I tried to explain, this depends among other things on the size of the (hypothetical) p orbital as compared to the size of the s-orbital. You can easily see that the overlap of an s and a p orbital will go to 0 when the p orbital becomes bigger and bigger. On the other hand, the strength of the bond does not only depend on the overlap but also on the energy of the orbital.
So it would be better to restrict your argumentation to some reasonable class or real existing bonds e.g. C-C or C-H bonds or the like.