Why orbitals hybridize

To explain the bonding in methane, it is necessary to introduce the concept of hybridization and hybrid atomic orbitals. Hybridization is the mixing of the atomic orbitals in an atom to produce a set of hybrid orbitals. When hybridization occurs, it must do so as a result of the mixing of nonequivalent orbitals.

In other words, s and p orbitals can hybridize but p orbitals cannot hybridize with other p orbitals. Hybrid orbitals are the atomic orbitals obtained when two or more nonequivalent orbitals form the same atom combine in preparation for bond formation. In the current case of carbon, the single 2s orbital hybridizes with the three 2p orbitals to form a set of four hybrid orbitals, called sp 3 hybrids see Figure 3 below.

The sp 3 hybrids are all equivalent to one another. Spatially, the hybrid orbitals point towards the four corners of a tetrahedron. The process of sp 3 hybridization is the mixing of an s orbital with a set of three p orbitals to form a set of four sp 3 hybrid orbitals.

Each large lobe of the hybrid orbitals points to one corner of a tetrahedron. The four lobes of each of the sp 3 hybrid orbitals then overlap with the normal unhybridized 1s orbitals of each hydrogen atoms to form the tetrahedral methane molecule.

Use the link below to answer the following questions. Read only the sections on ammonia and water hybridization. Romeo and Juliet were two of the great lovers of all time. Their embrace allowed no other person to be a part of it — they only wanted to be with each other. It took outside intervention parents are like that! Paired electrons are similar to the lovers.

They do not bond covalently until they are unpaired. Then they can become a part of a larger chemical structure. The beryllium atom contains all paired electrons and so must also undergo hybridization. One of the 2s electrons is first promoted to the empty 2p x orbital see figure below. Now the hybridization takes place only with the occupied orbitals and the result is a pair of sp hybrid orbitals.

The two remaining p orbitals p y and p z do not hybridize and remain unoccupied see Figure 6 below. The geometry of the sp hybrid orbitals is linear, with the lobes of the orbitals pointing in opposite directions along one axis, arbitrarily defined as the x-axis see Figure 7.

Each can bond with a 1s orbital from a hydrogen atom to form the linear BeH 2 molecule. Post by » Fri Nov 23, am. Post by Parth Mungra » Fri Nov 23, am. Post by harperlacroix1a » Sat Nov 24, am. Post by Manas Jinka » Sun Nov 25, am. Post by Abhi4F » Sun Nov 25, am. Laurence Lavelle Skip to content. Okay so we know carbon has 4 electrons, we're going to denote them. Okay so we want to have 4 equal places where chlorine come in and bond with this carbon.

So we're going to hybridize all these orbitals to make 4 equal in energy orbitals. So we're going to have 4 new orbitals and we're going to call them the 1S and the 3 of them are P so we're going to call it SP3, 1 from S 3 from P. And we're going to spread these out just like the rule tells us to and we're going to say okay we have 4 electrons which gives us 4 equal places for chlorine to come in and actually bond with that carbon.

So chlorine is coming here, here, here and here and make 4 of those bonds like you see in the picture. Alright so what actually would get hybridized? What do we create to actually mix it so that these equal orbitals are necessary? So we know all single bonds are going to be hybridized because a single bond there's not one that's more energetic than the other. So all single bonds are going to be hybridized.

Because they're hybridized bonds we're going to now call single bonds sigma bonds, this is just the way they overlap, the way that orbitals overlap we're going to call them, denote them sigma bonds. And we also want to say that low in pairs are also going to be hybridized because they're not higher or lower in energy than those bonds either.

So let's look at ammonia as an example, ammonia if you look at nitrogen within ammonia it has these 2 lone pair of electrons. 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? It will be much easier to do this if you make a model. In the ethane molecule, the bonding picture according to valence orbital theory is very similar to that of methane.

Both carbons are sp 3 -hybridized, meaning that both have four bonds arranged with tetrahedral geometry. The carbon-carbon bond, with a bond length of 1. All of these are sigma bonds. Because they are formed from the end-on-end overlap of two orbitals, sigma bonds are free to rotate.

This means, in the case of ethane molecule, that the two methyl CH 3 groups can be pictured as two wheels on a hub, each one able to rotate freely with respect to the other.

The sp 3 bonding picture is also used to described the bonding in amines, including ammonia, the simplest amine. Just like the carbon atom in methane, the central nitrogen in ammonia is sp 3 -hybridized. With nitrogen, however, there are five rather than four valence electrons to account for, meaning that three of the four hybrid orbitals are half-filled and available for bonding, while the fourth is fully occupied by a non-bonding pair of electrons.

The bonding arrangement here is also tetrahedral: the three N-H bonds of ammonia can be pictured as forming the base of a trigonal pyramid, with the fourth orbital, containing the lone pair, forming the top of the pyramid. It would seem logical, then, to describe the bonding in water as occurring through the overlap of sp 3 -hybrid orbitals on oxygen with 1 s orbitals on the two hydrogen atoms.

In this model, the two nonbonding lone pairs on oxygen would be located in sp 3 orbitals. Some experimental evidence, however, suggests that the bonding orbitals on the oxygen are actually unhybridized 2 p orbitals rather than sp 3 hybrids. Both the hybrid orbital and the nonhybrid orbital models present reasonable explanations for the observed bonding arrangement in water, so we will not concern ourselves any further with the distinction.

The valence bond theory, along with the hybrid orbital concept, does a very good job of describing double-bonded compounds such as ethene. Three experimentally observable characteristics of the ethene molecule need to be accounted for by a bonding model:. Clearly, these characteristics are not consistent with an sp 3 hybrid bonding picture for the two carbon atoms.

Instead, the bonding in ethene is described by a model involving the participation of a different kind of hybrid orbital. Three atomic orbitals on each carbon — the 2 s , 2 p x and 2 p y orbitals — combine to form three sp 2 hybrids, leaving the 2 p z orbital unhybridized.