Now that we understand how to construct the atomic configuration for a particular element, we need to understand how this determines the bonds that are formed with other elements. For example, carbon (atomic number 6) has the atomic configuration 1s²2s²2p_x¹2p _y¹2p _z⁰ or 1s²2s²2p². How many bonds will carbon form with another element such as hydrogen (atomic number 1 and therefore 1s¹) and what shape will this molecule have?

If we look at the atomic configuration diagram for carbon, we note that there are two unpaired electrons. On this basis, we would predict that it would form two bonds with two hydrogens to form a CH₂ molecule and since the 2p orbitals are perpendicular to one another, that the CH₂ molecule would have an "L" shape. Unfortuntely, this is not correct.

Figure 1.

Understanding the solution to this dilemma requires mathematics, and once again, we will gloss over this. The solution, in simple terms, is to "mix" the 2s and three 2p orbitals together and form four new, hybrid orbitals called sp³ orbitals. That is, our "atomic configuration" for carbon is now shown in Figure 2.

Figure 2.

What do these sp³ orbitals look like? Again, the answer "falls out" of the mathematics but this orbital is essentially a distorted dumbell. Figure 3 shows an sp³ orbital in which one lobe of the dumbell is bigger than the other.

Figure 3. An sp³ orbital.

How are these four sp³ orbitals oriented in space? It turns out that they are oriented at 109^o to one another. If they were all drawn together, the large portion of the orbitals would point toward the corners of a tetrahedron. Figure 4 shows a tetrahedron (the dot is the center of the tetrahedron) and a second tetrahedron with just one of the sp³ orbtials.

Is this the only type of hybridization that we will see? It turns out that we need to discuss one other type of "mixing" called sp². As this name suggests, we will mix the 2s orbital with two, rather than three, 2p orbitals. This will leave behind one 2p orbital. An energy diagram is shown in Figure 5. You will note that I have not "loaded" the electrons according to our usual rule of filling the lowest energy sp² orbitals fully before filling the empty p orbital. For the moment, I will justify this simply by saying that the sp² orbitals and the 2p orbital are approximately the same energy, but the reason for this apparent "rule violation" will be clear in the next section on "molecular orbitals".

Figure 5. sp²-Hybridization

PROBLEMS:

1. Hybrid sp³ orbitals are formed from mixing one s orbital with three p orbitals. How would sp³ orbitals compare if we used 2s and three 2p orbitals versus 3s and three 3p orbitals?
2. Nitrogen has the atomic configuration 1s²2s²2p³. If nitrogen were sp³ hybridized, how would you fill the orbitals in Figure 2? What would Figure 2 look like for oxygen and fluorine?
3. Based on the answer in problem 2, we could display an sp³-hybridized nitrogen atom as shown below. What would oxygen and fluorine look like?

ANSWERS:

1. A mix of one 3s and three 3p orbitals would produce a larger sp³ orbital than 2s and three 2p orbitals, but the overall shape would be the same.

2.

3.