How are molecular orbitals determined? Who created the molecular orbital theory? How do you find hybridization orbitals? What are the steps associated with the process of constructing a hybrid orbital diagram? So ammonia before had the same thing, ammonia has 5 valence electrons so 2, 3, 4, 5, this should be the same I'm sorry they're kind of uneven, they should actually be the same in energy and we have the 5 electrons.
If we're going to hybridize all of them we need to have 1, 2, 3 of these are the same along with this fourth one so we need to have all 4 of these is the same, so we're goingto have again 4 equal in energy we're going to call it SP3, 1 from S, 3 from P 1, 2, 3, 4, 5 here's our lone pair and here's the hydrogens that are going to come in and bond with them all equal in energies so we have this new hybrid orbitals.
Okay about when we have multiple bonds? So in different cases we may have multiple bonds, double bonds and triple bonds. So what happens in those guys? Well one of those bonds within a multiple bond is called a sigma bond and again don't forget sigma bonds are hybridized so one of those bonds is going to be hybridized.
The rest of those bonds are called pi bonds, those pi bonds are just P orbitals overlapping each other, they're only P orbitals they're a little higher in energy, they actually are different in energy. So we're going actually keep then separated, so we have 1 sigma and 1 pi. So let's look at carbon dioxide here, we have a double bond alright of these is going to be a sigma bond and we're going to denote that with a sigma and one of the bonds is going to be a pi bond we'll denote it with pi.
These are only P orbitals, these are hybridized orbitals we're just talking about the carbon right now. Okay so carbon we already know looks like this we're going to save 2 of the P orbitals and I don't care which 2 I save it doesn't really matter, they're all the same in energy I don't care. I'm just going to save this just for practical purposes, these are going to be the ones to use in pi bonding, so I'm going to save those so they hybridize 1S and the other P.
Let's look at the oxygen, oxygen also has a sigma bond, a pi bond but notice it has lone pair. So the sigma bond and the lone pair are going to be hybridized but not the pi bond we're going to leave this lone. So we need 3 hybrid orbital, 1 from S and 2 from P, so it's going to be SP2. It doesn't matter that these guys have different types of orbitals, we just want to make sure that the orbitals within the atom itself are the same.
Donate Login Sign up Search for courses, skills, and videos. Worked examples: Finding the hybridization of atoms in organic molecules. Practice: Bond hybridization. Current timeTotal duration Google Classroom Facebook Twitter. Video transcript Voiceover: In this video, we're going to look at the SP three hybridization present in methane and ethane; let's start with methane. So that's CH four, if I want to draw a dot structure for methane, I would start with carbon, and its four valence electrons, and then we would put hydrogen around that; each hydrogen has one valence electron, so we go ahead and draw in our hydrogens with one valence electron, and that gives us the Lewis Dot structure.
Usually you see it drawn like this, with carbon with its four bonds to hydrogen around it, like that, and in methane, all of these bonds are equivalent, in terms of things like bond length and energy. And so the four valence electrons that carbon brought to the table over here, let me go ahead and highlight those four valence electrons, those should be equivalent, and if we look at the electron configuration for carbon, let's go ahead and do that right now.
It's one S two, so go ahead and put in two electrons in the one S orbital, two S two, go ahead and put in two electrons in the two S orbital, and then two P two, and so, I'm assuming you already know your electron configuration, so it would look something like that. If we look at those four valence electrons on our orbital notation here, that would be these four electrons here, the valence electrons in the outer shell. And if we look at this, this implies that carbon would only form two bonds, because I have these unpaired electrons right here, and everything's of different energies, and so, what we see from the dot structure and experimentally, doesn't quite match up with the electron configuration here, and so to explain this difference, Linus Pauling came up with the idea of hybridization.
And so, the first thing that he said was, you could go ahead and take out one of these electrons in the two S, and promote it up to the P orbital here, so let me go ahead and show that, so we've moved one of those electrons up to the two P orbital, so we're in the excited state now. And now we have the opportunity for carbon to form four bonds, however, those electrons are not of equivalent energy, and so Linus Pauling said, "Let's do something else here: "Let's go ahead and promote the two S orbital," so we're gonna take this S orbital, and we're gonna promote it in energy, and we're going to take these P orbitals and demote them in energy, so we're gonna lower those P orbitals, like that, so we have our P orbitals here.
Using the Lewis Structures , try to figure out the hybridization sp, sp 2 , sp 3 of the indicated atom and indicate the atom's shape.
The carbon has no lone pairs and is bonded to three hydrogens so we just need three hybrid orbitals, aka sp 2. Don't forget to take into account all the lone pairs. Every lone pair needs it own hybrid orbital. That makes three hybrid orbitals for lone pairs and the oxygen is bonded to one hydrogen which requires another sp 3 orbital. That makes 4 orbitals, aka sp 3. The carbon is bonded to two other atoms, that means it needs two hybrid orbitals, aka sp.
An easy way to figure out what hybridization an atom has is to just count the number of atoms bonded to it and the number of lone pairs. Double and triple bonds still count as being only bonded to one atom. Use this method to go over the above problems again and make sure you understand it.
It's a lot easier to figure out the hybridization this way. Introduction Carbon is a perfect example showing the value of hybrid orbitals.
Carbon's ground state configuration is: According to Valence Bond Theory , carbon should form two covalent bonds, resulting in a CH 2 , because it has two unpaired electrons in its electronic configuration. That would give us the following configuration: Now that carbon has four unpaired electrons it can have four equal energy bonds. Energy changes occurring in hybridization. Energy changes occurring in hybridization Hybridization of an s orbital with two p orbitals p x and p y results in three sp 2 hybrid orbitals that are oriented at o angle to each other Figure 3.
Example: sp 2 Hybridization in Aluminum Trihydride In aluminum trihydride, one 2s orbital and two 2p orbitals hybridize to form three sp 2 orbitals that align themselves in the trigonal planar structure. Example: sp 2 Hybridization in Ethene Similar hybridization occurs in each carbon of ethene.
Energy changes occurring in hybridization Figure 1: Notice how the energy of the electrons lowers when hybridized. Example: sp Hybridization in Magnesium Hydride In magnesium hydride, the 3s orbital and one of the 3p orbitals from magnesium hybridize to form two sp orbitals. Hybridization Example: sp Hybridization in Ethyne The hybridization in ethyne is similar to the hybridization in magnesium hydride.
References John Olmsted, Gregory M.
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