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Hybridisation – Chemical Bonding

by Adam Le Gresley, PhD
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    00:00 draw your attention to the fact that the outer-phase lobes have been reduced in size.

    00:01 Okay. Now if it was that straightforward, then you'd have a relatively limited number of organic molecules that could be formed. However, things become a little bit more complicated in the case of carbon and, indeed, some other elements as well. And this is where you have a change in energy, formally, of atomic orbitals. Do you remember? We were looking at s orbitals in a single shell being of a lower energy to p orbitals in a slightly higher energy subshell. But this isn't the end of the story. If we look at methane, which has four covalent bonds evenly spaced in three dimensions following a tetrahedral structure with an angle of approximately 109 degrees between any of the two bonds, this suggests that each bond is somehow the same. This cannot be possible if you have one s orbital and three p orbitals with different geometries in different directions. And this is where the concept of hybridization comes in. So the way in which you can have sigma bonds formed which are equal in terms of their electron distribution, resulting in a uniform tetrahedral structure, is if the s and the p orbitals found in the element carbon are somehow hybridized—hybridization, in this context, meaning conversion of those 2s and 2p into one uniform atomic orbital type.

    01:43 So if we look at the current electron configuration for carbon, we have our 1s2, 2s2, and then we have two unpaired electrons in the p orbitals. If I bring it forward, it's possible to bring together those atomic orbitals to form a single orbital type. So we have the two p orbitals from carbon, and we have two electrons in the 2s orbital. If we look at the 2 p orbitals, you'll see that a tetrahedral structure for our methane is not possible, because you have a 90-degree angle and a 90-degree angle in the other direction for the different p orbitals: px, py, and pz. The reality is you have a mix-up of the s and the p orbitals, and this is very important, because in order for them to be uniform, they have to be uniform in terms not necessarily in terms of direction, but in terms of geometry and also in terms of energy. So let's see if we can explain this using electron configuration energy diagram. Because they are the same, we get a distribution, or methane-based structure, based on four identical (in terms of energy and shape) orbitals.

    03:04 So let's actually have a look at what that means. So on the left-hand side, we have the atomic orbitals for carbon that we are familiar with. Bear in mind we have the s orbital—the 2s, which has got two electrons in it from carbon—and we have the px, py, and pz. And there are two… one electron in, say, the px and one electron in the py. In hybridization for carbon, what happens is all four of those orbitals effectively come together and form four (as you can see, in terms of shape) identical hybrid orbitals. In this case, the degree of hybridization can be defined by the number of original atomic orbitals which is involved in that hybridization. So if you pay attention to the hybrid orbital that is formed, you'll see it says "4x sp3." This is telling us that there are four sp3 hybrid orbitals where, in this case, the s orbital and the three p orbitals all come together to create one orbital class, hence the name sp3 hybridization. What's happening from an energetic perspective, as you can see here, is that in our original case, we have two electrons in the 2s and two unpaired in the 2p. And in the process of hybridization, which occurs naturally, you end up with four unpaired electrons of an energy level which is somewhere in between the 2p and the 2s. Energy levels are the same. Shape is the same. Geometry depends on how they're interacting with other atoms. So if we take, for example, hydrogen and carbon, where car… hydrogen has its single s orbital, and we have an sp3 hybridized orbital from carbon, you can see that they form a sigma molecular orbital not unlike that one which I showed you for hydrogen and fluorine. Note you still have the sigma bond with the electron density distributed over that green area that you can see there, and the small lobe which is out of phase with the hydrogen atomic orbital. For methane, of course, in sp3 hybridized carbon, there are four of these hybrid atomic orbitals, and therefore, it's possible for four hydrogens to bind to a single carbon. Equally, it's possible for those four carbons to bind with a defined uniform tetrahedral structure with 109-degree bond angles between each of those sigma bonds formed. They're evenly spaced, as I've just said.

    05:59 But it's not just restricted to carbon. As you can see, oxygen can also be hybridized.

    06:05 If you recall, oxygen is in group 6; therefore, it has six electrons in its outer shell. That includes the 2s and also the 2p. Prior to hybridization, it has two electrons in the 2s, one paired set in the 2p, and then two unpaired electrons in the 2p. After hybridization, which is also uses the same nomenclature as we saw for carbon, all of those orbitals are brought together to form four sets of similar sp3 hybridized orbitals. And we see evidence of this in the structure of water. Note if you look at the structure of water, which is shown there on the board, we have 109-degree (or approximately 109-degree) bonding angle between the two hydrogens in H2O. Bear in mind the oxygen and the hydrogen are bound together by sigma molecular orbitals. Two of the sp3 orbitals will be filled with a pair of electrons, and two have one electron and are ready to form sigma bonds with the hydrogens.

    07:18 Hence, water has a bent shape, as you can see here. And knowledge of hybridization of orbitals isn't just something to talk about from an academic perspective; it's very important in terms of geometry. And as we'll see as we move along further in this course and we start looking at biological interactions of small molecules, we'll see that geometry is absolutely essential for the correct formation of a useful pharmacophore for medicinal chemistry.


    About the Lecture

    The lecture Hybridisation – Chemical Bonding by Adam Le Gresley, PhD is from the course Chemistry: Introduction.


    Included Quiz Questions

    1. …are identical in their shape and energy levels.
    2. …are identical in their spatial arrangement.
    3. …contain only anti-spin electrons.
    4. …contain only like-spin electrons.
    5. …do not take part in sigma bond formation by any means.
    1. S in SF4 ----- sp3 hybridization
    2. Be in BeCL2 ----- sp hybridization
    3. B in BH2 ----- sp2 hybridization
    4. C in CH4 ----- sp3 hybridization
    5. S in SF6 ----- sp3d2 hybridization
    1. …the number of original atomic orbitals participating in the hybridization process.
    2. …the number of original molecular orbitals formed in hybridization process.
    3. …the number of atoms forming sigma bonds with each other through their atomic orbitals.
    4. …the number of atoms forming sigma bonds with each other through their molecular orbitals.
    5. …the number of lone pairs present on the central atom of molecule which is participating in hybridization process.
    1. It enables the interaction of a molecule or ligand with its specific biological target to bring about a particular response with high efficiency.
    2. It neutralizes the negative charges present on the surface of ligand molecules.
    3. It acts as a reservoir for the extra lone pairs present on an atom during atomic orbital hybridization.
    4. It plays the role of a reservoir for the excess of nuclear charge present in the nucleus of an atom during the hybridization process.
    5. It neutralizes the positive charges present on the surface of ligand molecules.
    1. …given by formula: Number of molecular orbitals = Number of participating atomic orbitals.
    2. …given by formula: Number of molecular orbitals = (Number of participating atomic orbitals – 1).
    3. …given by formula: Number of molecular orbitals = (Number of participating atomic orbitals – 2/3).
    4. …given by formula: Number of molecular orbitals = (Number of participating atomic orbitals – 2).
    5. …given by formula: Number of molecular orbitals = (Number of participating atomic orbitals – 1/2).

    Author of lecture Hybridisation – Chemical Bonding

     Adam Le Gresley, PhD

    Adam Le Gresley, PhD


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