Hybridization - Online Tutor, Practice Problems & Exam Prep

Hybridization is a process where atomic orbitals mix to form new hybrid orbitals, enabling elements to form more bonds and increase stability. This concept is tightly linked to the molecular shape of compounds, as the hybridization of the central atom correlates with the number of electron groups surrounding it. For example, a molecule with two electron groups has a linear geometry and an SP hybridization, indicating the mixing of one S orbital and one P orbital, leaving the other two P orbitals unhybridized. As the number of electron groups increases, so does the complexity of hybridization, from SP² in trigonal planar geometries to SP³D² in octahedral geometries, each reflecting the specific mix of S, P, and D orbitals. Understanding the relationship between electron geometry and hybridization is crucial for predicting molecular shapes and determining which orbitals are hybridized versus those left unhybridized.

Hybridization Concept 1

Hybridization represents the idea of balanced shell atomic orbitals mixing to form hybrid orbitals. And we're going to say in order to form more bonds and increase stability, elements must hybridize their atomic orbitals. The way we can think about it is we have our S orbital and our P orbitals. These are known as our atomic orbitals. They're going to mix together and doing this helps to create hybrid orbitals.

Now a hybrid orbital basically looks like this. It looks like almost like a P orbital, except one end is much smaller and the other end is much larger. So here we have S and P orbitals mixing together to give us this hybrid orbital. Now we're going to see how hybridization can connect to molecules and their molecular shapes.

Hybridization Concept 2

The hybridization of a central element can be connected to its number of electron groups and because of this there's a strong connection between hybridization and your electron geometry. So if we take a look here, when we have two electron groups, our electron geometry is linear. The hybridization is sp1. Now if we think about this sp1 there is a superscript here that we don't see, which is one in one. So 1 + 1 gives me two. This reflects the number of electron groups connected to the central element.

Now when we say sp1, that means we're hybridizing 1S orbital and one P orbit. So this is the portion that would be hybridized. But remember P orbitals? There are three P orbitals that exist. All we're doing is hybridizing the first one. What about the other two? The other two would be left unhybridized. So we'd say we have 2P orbitals that are left unhybridized.

If we look at 3 electron groups, our electron geometry is trigonal planar. Here the hybridization would be sp2. There is a superscript for S that we don't see, which is one. 1 + 2 gives ME3 which is the number of electron groups. So here what's hybridizing is 1S and 2P orbitals. So what's left unhybridized? Well, one of the P orbitals remaining is left unhybridized.

Next, 4 electron groups is tetrahedral. Its hybridization is sp3, so this would be 1 + 3 gives me 4/4 electron groups. Here my S and the three PS are all hybridized. That means I have zero unhybridized orbitals.

5 electron groups is trigonal bipyramidal. Here this would be sp3d. When we add this up, this is 1 + 3 + 1 which is five electron groups. So my S and my three PS are all hybridized and then one of my D orbitals is also hybridized. Remember D orbitals? There are five of them in total. Only one of them is being hybridized here. That leaves us with these four which are left unhybridized. So we'd say we have 4D orbitals that are left unhybridized.

And then finally here we're going to say we have 6 electron groups which is octahedral. Here the hybridization would be sp3d2, so that'd be 1 + 3 + 2, which is 6 electron groups. We're hybridizing the S, the three PS, and two of the DS. What's left unhybridized? Are these three remaining D orbitals. So here we'd have 3D orbitals that are left unhybridized.

OK, so just think of this, when you're looking at any type of molecular shape, try to connect it to its electron geometry. If you can do that, then you'll know the hybridization, you'll know what orbitals are being hybridized and which ones are left unhybridized.

Hybridization Example 1

Determine the hybridization of the sulfur atom within sulfur tetrabromide. So sulfur is in Group 6A so it has six valence electrons. And remember halogens when they're not in the center they only make single bonds. So our sulfur will single bond to the four bromines. Bromines are in Group 7A, so they have 7 valence electrons themselves. So there goes our 4 bromines.

And then remember sulfur is in Group 6A. It's only using four out of its six valence electrons, so two of them would exist as a lone pair. Now if we take a look here how many electron groups are around sulfur 1, 2, 3, 4, 5. Remember 5 electron groups would mean that were trigonal bipyramidal in terms of the electron geometry, but more importantly, that would mean that we're sp3d in terms of hybridization.

And remember, you can verify this by saying that S has a superscript that's invisible, which is 1 plus the superscript of three for P + 1 for D. Together they add up to five, which represents the five electron groups around the sulfur atom. So here our hybridization once again for the sulfur atom is sp3d.

How many of the following molecules have sp³d² hybridization on the central atom? 

SeCl₆                            XeCl₄                           IF₅                               AsCl₅

• A. 3
• B. 1
• C. 2
• D. 4

How many unhybridized orbitals does the beryllium atom possess in BeCl₂?

• A. 2
• B. 3
• C. 1
• D. 0

Draw and determine the hybridization and unhybridized orbitals for the following covalent compound. 

KrBr₄                                                                                       Hybridization:   

                                                                                               Unhybridized Orbitals:

Draw and determine the hybridization and unhybridized orbitals for the following covalent compound. So here we're dealing with Krypton Tetrabromide. Krypton is in group 8A, so it has 8 valence electrons. It's going to use 4 of them to connect to the 4 bromines. Bromines are in group 7A, so they have 7 valence electrons. So we're drawing out the bromines connected to one Krypton. Here we say that Krypton has used 4 out of its 8 total valence electrons, meaning that it has 4 remaining. They will be represented by 2 lone pairs, one here and one here. Now, if we look, how many groups, electron groups, are attached to the Krypton atom? We'd say 1, 2, 3, 4, 5, 6 electron groups. So remember, when you have 6 electron groups, that means your hybridization is s p3 d2 . So that will be our hybridization. Now remember, your hybridization also tells us how many are unhybridized as a result. We're going to say that when it comes to our s orbital, it's just one shape. Our p is 3 dumbbell shapes, and then our d is 5 shapes in total. Right? So we'd say that it'd be 4 four-leaf clovers, and one dumbbell with a ring around it. From the hybridization, it's telling us that the s is hybridized, all 3 p's are hybridized, and 2 d's are hybridized. So this would be hybridized. So what is left unhybridized? What's left unhybridized would be these 3 d orbitals. So those would be our final answers in terms of Krypton Tetrabromide.

Here’s what students ask on this topic:

What is hybridization?

Hybridization in chemistry refers to the concept where atomic orbitals mix to form new hybrid orbitals. This process is significant in explaining the geometry and bonding properties of molecules that cannot be adequately described by simple valence bond theory. Hybrid orbitals are a blend of s, p, and sometimes d orbitals, and they have different shapes and energies than the original atomic orbitals.

For example, in the methane (CH₄) molecule, carbon's ground state configuration does not explain its four equivalent bonds. Through hybridization, the one s and three p orbitals of carbon combine to form four sp³ hybrid orbitals, each of which forms a sigma bond with a hydrogen atom, resulting in a tetrahedral shape. Other types of hybridization include sp² and sp, which correspond to trigonal planar and linear geometries, respectively. The concept of hybridization helps us understand the actual arrangement of atoms in a molecule and the type of bonds formed.

How do you determine hybridization?

To determine the hybridization of an atom in a molecule, you'll need to follow a few steps:

1. Count the Number of Sigma Bonds: Look at the atom and count the number of single bonds it has. Each single bond is a sigma bond.
2. Count the Number of Lone Pairs: Identify the number of lone pairs of electrons on the atom.
3. Add Sigma Bonds and Lone Pairs: The sum of sigma bonds and lone pairs gives you the steric number, which is crucial for determining hybridization.
4. Determine Hybridization: Based on the steric number, you can determine the hybridization:

• Steric number 2 = sp hybridization (linear geometry, 180° bond angle)
• Steric number 3 = sp² hybridization (trigonal planar geometry, 120° bond angle)
• Steric number 4 = sp³ hybridization (tetrahedral geometry, 109.5° bond angle)
• Steric number 5 = sp³d hybridization (trigonal bipyramidal geometry)
• Steric number 6 = sp³d² hybridization (octahedral geometry)

Remember that double and triple bonds also count as one sigma bond for these purposes; the additional bonds are pi bonds, which do not affect.

How can the hybridization of the central atom be found?

To determine the hybridization of the central atom in a molecule, you can follow these steps:

1. Lewis Structure: Draw the Lewis structure of the molecule to determine the electron pairs around the central atom, including both bonding pairs (shared with other atoms) and lone pairs (non-bonding pairs).
2. Count Electron Pairs: Count the total number of electron pairs (both bonding and lone pairs) around the central atom.
3. Determine Hybridization: Use the number of electron pairs to determine the hybridization:

• 2 electron pairs: sp hybridization (linear geometry, 180° bond angle)
• 3 electron pairs: sp² hybridization (trigonal planar geometry, 120° bond angle)
• 4 electron pairs: sp³ hybridization (tetrahedral geometry, 109.5° bond angle)
• 5 electron pairs: sp³d hybridization (trigonal bipyramidal geometry, 90°, 120°, and 180° bond angles)
• 6 electron pairs: sp³d² hybridization (octahedral geometry, 90° and 180° bond angles)

Remember that double and triple bonds count as one electron pair when determining hybridization. The hybridization reflects the different types of orbitals that are mixing to

How do you determine hybrid orbitals?

To determine hybrid orbitals, you start by looking at the central atom in a molecule and identifying its electron configuration. The number of electron domains (regions of electron density such as bonds or lone pairs) around the central atom will help you determine the type of hybridization. Here's a step-by-step process:

1. Write the electron configuration: For the central atom, write out its ground state electron configuration.
2. Count electron domains: Count the number of bonds (single, double, or triple) and lone pairs on the central atom.
3. Determine hybridization: Use the number of electron domains to determine the hybridization:

• Two electron domains indicate sp hybridization (linear geometry, 180° bond angles).
• Three electron domains suggest sp² hybridization (trigonal planar geometry, 120° bond angles).
• Four electron domains mean sp³ hybridization (tetrahedral geometry, 109.5° bond angles).
• Five electron domains correspond to sp³d hybridization (trigonal bipyramidal geometry).
• Six electron domains are indicative of sp³d² hybridization (octahedral geometry).

Remember that the sum of the coefficients in the hybrid orbitals (s, p, d) will equal the number of hybrid orbitals formed, which corresponds to the number of electron domains.