## Hybridized Orbitals

$sp, sp^{2}, sp^{3}, dsp^{3}, d^{2}sp^{3}$

Samantha Pedersen 2K
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### Hybridized Orbitals

In lecture today, Dr. Lavelle emphasized that the electrons in the hybridized 2sp^2 orbitals and unhybridized 2p orbital for ethene will all be unpaired. However, one of the four hybridized sp^3 orbitals for ammonia contained two paired electrons.

Will we only have paired electrons in hybridized orbitals when there is a lone pair? Otherwise, how do we know when we should and shouldn't have paired electrons in hybridized orbitals? Thank you!

shevanti_kumar_1E
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### Re: Hybridized Orbitals

The way to figure out the hybridization is to first look at the regions of electron density to determine the number of hybrid orbitals. Ammonia has four regions of electron density (3 single bonds N-H and one lone pair on the N). That means that the hybridization is sp^3. The sp^3 hybridization has four orbitals and they can be paired cause there isn't an energy difference (no need for an un-hybridized orbital). On the other hand for C2H4 which has 3 regions of electron density and sp^2 hybridization needs an unhybridized 2p orbital because it total the p and s orbitals have 4 orbitals. The sp2 only has 3 orbitals so one unhybridized 2p orbital is needed. Since the 2p unhybridized and 2sp^3 hybridized orbitals are different energy levels there is no spin pairs because the e repulsion is high.

Kailani_Dial_2K
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### Re: Hybridized Orbitals

Hi, so the central atom in ammonia is Nitrogen which has 5 valence electrons while the central atom in ethene is carbon which only has 4 valence electrons. When you do the hybrization of Ammonia, there are 4 regions of electron density created by the 3 bonds with Hydrogen and the lone pair. Since there are 4 regions of electron density, you should have 4 hybridized orbitals. Since there are 5 electrons and 4 orbitals one of the orbitals will have to have 2 electrons that have opposite spins. If you look at ethene there are 3 regions of electron density which means there will be three hybrid orbitals. However since you started with 4 atomic orbitals you must end up with 4 orbitals, so one of them becomes an unhybridized p-orbital. There are 4 electrons in carbon, so three of the 4 go into the three sp^2 orbitals. You now have one electron left correct? So, the way that you decide that it goes into the unhybridzed 2p orbital is based on energy. You want the atom to be in the lowest energy state possible. If you look at the Afuba Diagram before hybridization you will notice there is a large gap between the 2s and 2p orbital. If you look at the hybridized model, there is a smaller gap between the 2sp^2 orbital and the 2p orbital. Because the energy gap is so small, it is actually less energy to put the electron in the unhybridized orbital rather than to pair it in one of the hybridized orbitals.

What orbital you place th electrons in is not based on the pesence or absence of a lone pair, but is rather based on the energy conformation of the atom. If it is less energy to put it in the unhybrized orbital ( which it is) rather than paired with another electron in a hybridized orbital, then the electron will go into the unhybrized orbital.

Sorry this was so long, but I hope it helps