Timeline for Predicting sigma bond overlap strengths of s-s, p-p, s-p, sp-sp etc
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Oct 4, 2018 at 4:02 | comment | added | Tyberius♦ | @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. | |
| Oct 4, 2018 at 3:51 | comment | added | Yusuf Hasan | @Tyberius Again, didn't get you on the second part there. The sigma overlap of two sp2 hybridized atoms will always take place on the molecular axis itself, so where would the reorientation occur? Also,(unless you are somehow involving them in some kind of chemical reaction) how can atoms simply "go" from one hybridization to another? | |
| Oct 4, 2018 at 3:38 | comment | added | Tyberius♦ | @YUSUFHASAN what I was basically trying to convey is that internuclear bonding of p-orbitals (a sigma bond) should be stronger than noninternuclear bonding (like a pi bond). For the hybrids, we can always reorient the atoms such that the largest lobes are facing each other. So it two atoms went from sp to sp2 hybridization, we could always reorient them so their lobes are along their internuclear axis. | |
| Oct 4, 2018 at 2:38 | comment | added | Yusuf Hasan | @Tyberius I am unable to visualize how $\ce{pz}$ orbitals extend their lobes for bonding while $\ce{px}$ does not. I believe that the z-axis has been taken as the molecular axis by convention, but I can't really see how that helps. Also, if we are increasing the p- character in our hybridization(as an example, let it be $\ce{sp3}$), then isn't it possible that the overlap becomes weaker along any given axis(say y-axis) as compared to a bond formed by a pure $\ce{py}$ orbital, as percent y character in the former is surely less than a pure $\ce{py}$ orbital? | |
| Mar 31, 2018 at 13:50 | comment | added | Tyberius♦ | @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. | |
| Mar 31, 2018 at 9:47 | comment | added | FreakyLearner | @Tyberius According to this reasoning, shouldn't a p-p overlap be stronger than all of the overlap of hybridised orbitals? | |
| Aug 20, 2017 at 18:22 | vote | accept | jonsno | ||
| Aug 20, 2017 at 17:03 | comment | added | Tyberius♦ | @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. | |
| Aug 20, 2017 at 2:44 | comment | added | jonsno | 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.. | |
| Aug 19, 2017 at 22:12 | history | answered | Tyberius♦ | CC BY-SA 3.0 |