1
$\begingroup$

Our chemistry teacher told us that both VSEPR theory (which says that the electron pairs in the valence shell of an atom arrange themselves in such a way that repulsions among them are minimized and this arrangement of the electron pairs determines the shape of a particular molecule) and hybridization (which is the intermixing of a particular number of atomic orbitals to form equal number of new orbitals which have same shape and energy) can be used to determine the shape of a molecule and that they are independent of each other. I understand how the shape of a molecule can be predicted using the VSEPR theory but i cannot understand how hybridization helps in determining or rather predicting the shape of a particular molecule. I have searched the net but could not find any useful resource.

How does hybridization help in determining the shape of a molecule?

Improve this question
$\endgroup$
2
  • 5
    $\begingroup$ Classical hybridisation theory does not predict geometry but instead rationalises it. It's not Here I have some atoms connected by these bonds, hence the hybridisation is X, and from this geometry Y follows. It is more like Here I have some atoms connected by these bonds, they are arranged in geometry Y , and therefore the hybridisation is X. (Geometry here is found either by experiment, some semiquantitative theory or via computations.) $$$$ Things are likely different for modern hybridisation theory. You could also ask why hybridisation X mathematically rationalises geometry Y. $\endgroup$
    – Linear Christmas
    Sep 3, 2017 at 20:27
  • 2
    $\begingroup$ If you are in general Chemistry, I wouldn't worry about going about determining the geometry in this way. It is much simpler to work from a VSEPR approach. Determining the geometry entirely from hybridization requires understanding how to write linear combinations of wave functions. $\endgroup$
    – Tyberius ♦
    Sep 3, 2017 at 20:40

1 Answer 1

Reset to default
6
+50
$\begingroup$

The VSEPR theory can indeed be used to predict geometry a priori. You would count the number of substituents around a central atom, counting any lone pairs as substituents also, and then find an arrangement that maximises the distance between all substituents. This theory works well for elements of the first period and compounds in which the central element is in the highest oxidation state, but tends to fail very rapidly for all others, most notably because it cannot account for anything outside of the traditional 2-electron-2-centre bond and because it considers all lone pairs equal even though they are typically not.

The hybridisation theory can only be used to predict geometry in a very limited number of cases that involve carbon atoms, and it only works because carbon is extremely regular. However, using hybridisation to predict carbon geometry is basically circling back onto VSEPR: you check how many atoms are bound to a certain carbon and thereby deduce whether this carbon atom will be $\mathrm{sp,sp^2}$ or $\mathrm{sp^3}$. Carbocations and carbanions can also be accomodated for (by either counting the anionic lone pair or ignoring it — as in VSEPR).

In general, however, it is not possible to predict the geometry following a certain hybridisation. Rather, one determines the geometry by an independent mean (e.g. quantum chemical calculations, gas phase diffraction), deduces the bond angles (and bond lengths) from the geometry and then derives a hybridisation. In short: hybridisation follows geometry, not the other way around.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Then what is the purpose of hybridization? It provides no answers, makes no predictions. Is it utterly useless? $\endgroup$
    – Juan Perez
    Aug 3, 2021 at 19:51
  • $\begingroup$ @JuanPerez It simplifies things, especially for beginning learners. It is a good tool for a number of simple molecules often encountered and can be partially generalised in a way that is useful in most cases. Finally, it allows electronic predictions based on the geometry; this last one is probably the most powerful but also the one learners need the least. $\endgroup$
    – Jan
    Aug 4, 2021 at 12:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.