Hybridization

Take one can of red paint and three cans of blue paint. Mix them all together and you get four identical cans of purple paint. The original colors are gone - you cannot scoop the red back out. You started with four cans, and you end with four cans, but now they are all the same new color. That is hybridization.

In atomic terms, a carbon atom has one 2s orbital and three 2p orbitals that look nothing alike - the s orbital is a sphere, and the p orbitals are dumbbells oriented along different axes. But when carbon forms four bonds (as in methane), those four different orbitals mix together to create four identical hybrid orbitals called sp3 hybrids. Each one is the same shape, the same energy, and points toward a corner of a tetrahedron. The original s and p orbitals no longer exist on that atom.

The Golden Rule: Orbitals In = Orbitals Out

The number of hybrid orbitals produced always equals the number of atomic orbitals that were mixed. Mix two orbitals, get two hybrids. Mix six orbitals, get six hybrids. No orbitals are created or destroyed - they are just reshaped.

This is why the name of the hybridization tells you everything. “sp3” means one s orbital + three p orbitals were mixed, producing four hybrid orbitals. “sp2” means one s + two p orbitals were mixed, producing three hybrids. The superscripts in the name are literally a recipe.

The Shortcut: Steric Number = Hybridization

Here is the fastest way to determine hybridization on the MCAT. Count the steric number - the total number of electron groups (bonds + lone pairs) around the central atom. That number maps directly to the hybridization.

| Steric Number | Orbitals Mixed | Hybridization | Geometry |
|:---:|:---|:---:|:---|
| 2 | 1 s + 1 p | sp | Linear |
| 3 | 1 s + 2 p | sp2 | Trigonal planar |
| 4 | 1 s + 3 p | sp3 | Tetrahedral |
| 5 | 1 s + 3 p + 1 d | sp3d | Trigonal bipyramidal |
| 6 | 1 s + 3 p + 2 d | sp3d2 | Octahedral |

That is the entire system. No exceptions, no special cases for the MCAT. Count groups, read off the hybridization.

sp3 Hybridization (4 Groups)

Mix one s orbital and three p orbitals to get four sp3 hybrid orbitals aimed at the corners of a tetrahedron. This is the most common hybridization in organic molecules.

CH4 (methane): Carbon has four bonding groups, zero lone pairs. Steric number = 4, so hybridization = sp3. The four C-H bonds point to the corners of a tetrahedron with 109.5-degree angles.

NH3 (ammonia): Nitrogen has three bonding groups plus one lone pair. Steric number = 4, so hybridization = sp3. The lone pair occupies one of the four sp3 orbitals. Even though the molecular geometry is trigonal pyramidal, the hybridization is still sp3 because hybridization depends on total electron groups, not just bonded atoms.

H2O (water): Oxygen has two bonding groups plus two lone pairs. Steric number = 4, hybridization = sp3. Two of the four sp3 orbitals hold lone pairs.

sp2 Hybridization (3 Groups)

Mix one s orbital and two p orbitals to get three sp2 hybrid orbitals in a trigonal planar arrangement (120 degrees apart). One p orbital remains unhybridized and is available to form a pi bond.

BF3 (boron trifluoride): Boron has three bonding groups, zero lone pairs. Steric number = 3, hybridization = sp2. The three B-F bonds are in a plane with 120-degree angles. Boron is electron-deficient here (only six electrons around it), which is why BF3 is a strong Lewis acid.

Ethylene (C2H4): Each carbon has three groups (two C-H bonds plus one C=C bond - remember, a double bond counts as one group). Steric number = 3, so each carbon is sp2 hybridized. The three sp2 orbitals form the trigonal planar framework, and the leftover unhybridized p orbital on each carbon overlaps sideways to form the pi bond of the double bond.

Carbonate ion (CO3 2-): The central carbon has three groups (three C-O bonds, some of which are double bonds in resonance structures). Steric number = 3, hybridization = sp2. The ion is flat and trigonal planar.

sp Hybridization (2 Groups)

Mix one s orbital and one p orbital to get two sp hybrid orbitals pointing in opposite directions (180 degrees apart). Two p orbitals remain unhybridized and can form up to two pi bonds.

CO2 (carbon dioxide): Carbon has two groups (two double bonds). Steric number = 2, hybridization = sp. The two C=O bonds point in opposite directions, making the molecule linear. The two unhybridized p orbitals on carbon each overlap with a p orbital on an oxygen to form two pi bonds.

BeCl2 (beryllium chloride): Beryllium has two bonding groups. Steric number = 2, hybridization = sp. Linear geometry with 180-degree bond angles.

HCN (hydrogen cyanide): Carbon has two groups (one C-H single bond and one C-N triple bond). Steric number = 2, hybridization = sp. The triple bond contains one sigma bond and two pi bonds formed by the two unhybridized p orbitals.

Expanded Octet Hybridization (5 and 6 Groups)

Elements in period 3 and below have access to d orbitals, which allows them to form more than four hybrid orbitals.

sp3d (5 groups): PCl5 is the classic example. Phosphorus has five bonding groups, so it mixes one s, three p, and one d orbital to create five sp3d hybrid orbitals in a trigonal bipyramidal arrangement.

sp3d2 (6 groups): SF6 is the textbook case. Sulfur has six bonding groups, so it mixes one s, three p, and two d orbitals to create six sp3d2 hybrid orbitals in an octahedral arrangement.

The Unhybridized Orbital Connection

Here is a detail that links this section directly to the next one: any p orbital that is not used in hybridization remains unhybridized and is available for pi bonding.

• sp3: All three p orbitals are used in hybridization. Zero unhybridized p orbitals. Zero pi bonds possible.
• sp2: Two p orbitals are hybridized, one remains. One unhybridized p orbital. One pi bond possible.
• sp: One p orbital is hybridized, two remain. Two unhybridized p orbitals. Two pi bonds possible.

This is why double bonds require sp2 hybridization (they need one unhybridized p orbital for the pi bond) and triple bonds require sp hybridization (they need two unhybridized p orbitals for two pi bonds). The connection between hybridization and bond type is not a coincidence - it is built into the system.

Common MCAT Traps

• Forgetting to count lone pairs. The nitrogen in NH3 is sp3 (four groups), not sp2 (three bonds). Always count lone pairs as electron groups.
• Counting double/triple bonds as multiple groups. A double bond is one group. A triple bond is one group. Only the number of positions matters.
• Assuming hybridization determines molecular geometry. Hybridization determines electron geometry. If lone pairs are present, the molecular geometry will differ from what the hybridization alone might suggest. NH3 is sp3 hybridized (tetrahedral electron geometry) but trigonal pyramidal in molecular geometry.