We'll use methane (CH4) as our exemplar to understand this process.
Let's think about the C atom at the center of methane. The electron configuration for a carbon atom is 1s2, 2s2, 2p2. So there are 4 valence electrons to make the 4 single bonds found in methane. That seems to make sense. However there is a problem. Two of those electrons are paired up in the 2s orbital and are not available to "partner" with an electron from a hydrogen atom.
A potential solution?
We could move one of the s electrons to the empty p orbital. That would give us 4 unpaired electrons.
Of course, there are several problems with this as a solution:
a) there is NO justification for this happening. (This is not a minor problem.)
b) It doesn't really solve the problem. Although we now have 4 electrons available for bonding, they aren't the same.
There are two aspects to the problem. Here they are in detail.
Problem 1:
Three of those electrons are in p orbitals and one is in an s orbital. That means that they will interact with other electrons differently. We don't even need to worry about HOW they will be different (stronger/weaker, shorter/longer). It only matters that they will be different. Methane has 4 bonds that are all IDENTICAL. That tells us that this solution is wrong.
Problem 2:
p orbitals exist at 90o angles to each other. The s orbital is spherically symmetric around the nucleus. So, if we use these orbitals for bonding, we should have three bonds at 90o to each other and one that has no defined bond angle and which, as a result can move around. This is definitely not what we find in the (perfectly symmetric) molecule of methane.
A Better solution - hybridization
Rather
than just shift an electron from one atomic orbital to another,
hybridization suggests that the orbitals of an atom are changed by
bonding. In other words the approach of 4 hydrogen nuclei and their
electrons causes changes in the electron structure of the carbon atom.
In simple terms, hybridization is a blender for orbitals and we are going to make an orbital smoothie. When you put yogurt, bananas and strawberries into a blender, what comes out is no longer any ONE of those things, but rather something that combines the properties of all of them.
When we take the 4 orbitals of carbon and put them in the hybridization blender, we pour out orbitals that are neither s nor p orbitals, but rather some blend of the two.
Since we put four orbitals (s, px, py and pz) into the blender, we will pour out four orbitals, but now all 4 are the same (same shape, same energy). That means that using Hund's rule, we can logically spread the 4 electrons out between them.
The label beneath the orbitals (sp3) is essentially the recipe. We put in a s orbital and 3 different p orbitals, so what came out of the blender were 4 sp3 orbitals.
In addition, since all four of these orbitals are identical, it makes sense for them to be spread evenly around the nucleus in 3-dimensional space - which just happens to be a tetrahedron.
Applying our "solution" beyond methane
Hybridization does an admirable job of explaining tetrahedrons in this way but, of course, there are other shapes and structures. If your goal is only to "get the right answer, go to the table at the bottom of the page, if you are curious where this goes...read on.Covalent bonds can be classified as either sigma (σ) bonds or pi (π) bonds.
σ bonds form as an "overlap" along the line between the two nuclei (what you normally think of as a bond. The first bond formed between any two atoms is ALWAYS a σ bond.
pi bonds form in two pieces - above and below the line between the nuclei. Second and third bonds formed between atoms are π bonds. So, if two atoms have a triple bond between them, they have formed one σ bond and two π bonds.
π bonds can only be formed from orbitals that have two parts. In other words,a π bond can only be made from p orbitals.
Because hybrid orbitals are so lopsided, they cannot form π bonds. That means that if an atom is making a π bond, it must have a p orbital that has NOT be hybridized. If the atom is making 2 pi bonds, it must have 2 un-hybridized orbitals.
Combining all of those ideas means that an atom making a double bond (perhaps oxygen, as seen here in this diagram from the earlier bonding pages)
each oxygen must keep 1 p orbital "out of the blender" to make the pi bond. That means that only 2 of the p orbitals went in the blender. The hybrids that are "poured" out of the blender are called sp2 orbitals. The resulting three orbitals will then arrange themselves into a trigonal planar structure at 120o.
A full list of the possible hybridizations and structures is in the table below.