Which hybridization is associated with a steric number of 3

In the last video, we saw that SP three hybridized situation, we get four hybrid orbitals, and that's how many we need, the steric number tells us we need four hybridized orbitals, so we took one S orbital, and three P orbitals, and that gave us four, SP three hybrid orbitals, so this carbon must be SP three hybridized.

So let's go ahead, and draw that in here. So this carbon is SP three hybridized, and in the last video, we also drew everything out, so we drew in those four, SP three hybrid orbitals, for that carbon, and we had one valence electron in each of those four, SP three hybrid orbitals, and then hydrogen had one valence electron, in an un-hybridized S orbital, so we drew in our hydrogens, and the one valence electron, like that.

This head-on overlap; this is, of course, a sigma bond, so we talked about this in the last video. And so now that we have this picture of the methane molecule, we can think about these electron pairs, so these electron pairs are going to repel each other: like charges repel. And so, the idea of the VSEPR theory, tell us these electron pairs are going to repel, and try to get as far away from each other as they possibly can, in space.

And this means that the arrangement of those electron pairs, ends up being tetrahedral. So let's go ahead and write that. So we have a tetrahedral arrangement of electron pairs around our carbon, like that. When we think about the molecular geometry, so that's like electron group geometry, you wanna think about the geometry of the entire molecule.

I could think about drawing in those electrons, those bonding electrons, like that. So we have a wedge coming out at us in space, a dash going away from us in space, and then, these lines mean, "in the plane of the page. So the arrangement of the atoms turns out to also be tetrahedral, so let's go ahead and write that. So, tetrahedral. And, let's see if we can see that four-sided figure, so a tetrahedron is a four-sided figure, so we can think about this being one face, and then let's go ahead and draw a second face.

And if I draw a line back here, that gives us four faces to our tetrahedron. So our electron group geometry is tetrahedral, the molecular geometry of methane is tetrahedral, and then we also have a bond angle, let me go ahead and draw that in, so a bond angle, this hydrogen-carbon-hydrogen bond angle in here, is approximately point five degrees. All right, let's go ahead and do the same type of analysis for a different molecule, here. So let's do it for ammonia, next.

So we have NH three, if I want to find the steric number, the steric number is equal to the number of sigma bonds, so that's one, two, three; so three sigma bonds. Plus number of lone pairs of electrons, so I have one lone pair of electrons here, so three plus one gives me a steric number of four.

So I need four hybridized orbitals, and once again, when I need four hybridized orbitals, I know that this nitrogen must be SP three hybridized, because SP three hybridization gives us four hybrid orbitals, and so let's go ahead and draw those four hybrid orbitals.

So we would have nitrogen, and let's go ahead and draw in all four of those. So, one, two, three, and four; those are the four hybrid orbitals. When you're drawing the dot structure for nitrogen, you would have one electron, another electron, another electron, and then you'd have two in this one, like that.

And then you'd go ahead, and put in your hydrogens, so, once again, each hydrogen has one electron, in a hybridized S orbital, so we go ahead and draw in those hydrogens, so our overlap of orbitals, so here's a sigma bond, here's a sigma bond, and here's a sigma bond; so three sigma bonds in ammonia, and then we have this lone pair up here.

So the arrangement of these electron pairs, is just what we talked about before: So we have this tetrahedral arrangement of electron pairs, or electron groups, so the VSEPR theory tells us that's how they're going to repel. However, that's not the shape of the molecule, so if I go ahead and draw in another picture over here, to talk about the molecular geometry, and go ahead and draw in the bonding electrons, like that, and then I'll put in my non-bonding electrons, up here: this lone pair right here, housed in an SP three hybridized orbital.

So, the arrangement of the atoms turns out not to be tetrahedral, and that has to do with this lone pair of electrons up here, at the top. So, this lone pair of electrons is going to repel these bonding electrons more strongly than in our previous example, and because it's going to repel those electrons a little bit more strongly, you're not gonna get a bond angle of point five; it's going to decrease the bond angle.

So let me go ahead, and use the same color we used before, so this bond angle is not point five; it goes down a bit, because of the extra repulsion, so it turns out to be approximately degrees. And in terms of the shape of the molecule, we don't say "tetrahedral"; we say "trigonal-pyramidal. So, "trigonal" refers to the fact that nitrogen is bonded to three atoms here, so nitrogen is bonded to three hydrogens, so that takes care of the trigonal part.

The "pyramidal" part comes in, because when you're doing molecular geometry, you ignore lone pairs of electrons.

Concentrate on the electron pairs and other atoms linked directly to the concerned atom. This step is crucial and one can directly get the state of hybridization and shape by looking at the Lewis structure after practicing with few molecules. Note: When the concerned atom makes a dative bond with other atoms, it may acquire positive or negative charge depending on whether it is donating or accepting the lone pair while doing so respectively.

If it donates a lone pair, a positive charge is accumulated. If it receives a lone pair, a negative charge is acquired. Boron atom gets negative charge when it accepts a lone pair from hydride ion, H - in borohydride ion, BH 4 -. Now, based on the steric number, it is possible to get the type of hybridization of the atom. Consult the following table.

This case arises when there are no lone pairs on the given central atom. Note: The structure of a molecule includes both bond pairs and lone pairs. It is always arrived at from the steric number. Though the lone pairs affect the bond angles, their positions are not taken into account while doing so. The valency of carbon is 4 and hence it can form 4 sigma bonds with four hydrogen atoms.

Also remember that the valency of hydrogen is one. The bond angle is 19 o 28'. The valency of nitrogen is 3. Therefore it forms 3 bonds with three hydrogen atoms. There is also a lone pair on nitrogen. Nitrogen in ammonia is bonded to 3 hydrogen atoms.