Hybridisation

Why pure orbitals are not enough

Carbon is $1s^2 2s^2 2p^2$, with two unpaired electrons. Yet it forms four bonds in methane. Even if one $2s$ electron moves up to $2p$, the four bonds would not be alike. Three $p$ orbitals lie at $90°$ to each other, so three bonds would be at $90°$, and the fourth would be different.

Methane has four identical C-H bonds, all at $109.5°$. Pure atomic orbitals cannot explain that. Pauling's idea of hybridisation can.

What hybridisation is

Hybridisation is the mixing of atomic orbitals of slightly different energies on one atom, to form new orbitals of equal energy and shape. These are hybrid orbitals.

• The number of hybrid orbitals equals the number of atomic orbitals that mix.
• All the hybrids of a set are alike in energy and shape.
• Hybrid orbitals point in definite directions, as far apart as possible.
• They make stronger bonds than pure orbitals, because one large lobe points at the other atom.

The conditions

• Only orbitals of the valence shell mix.
• The orbitals must have nearly the same energy.
• Promoting an electron is not needed for hybridisation.
• Half-filled orbitals and filled orbitals can both take part. A filled hybrid holds a lone pair.

sp hybridisation: beryllium chloride

Beryllium is $2s^2$. One electron is promoted to $2p$, then the $2s$ and one $2p$ mix into two sp hybrids. They point opposite ways, $180°$ apart. Each overlaps a $3p$ orbital of chlorine. So $\text{BeCl}_2$ is linear. Each sp hybrid is $50\%$ s and $50\%$ p.

sp2 hybridisation: boron trifluoride

Boron is $2s^2 2p^1$. After promotion, the $2s$ and two $2p$ orbitals mix into three sp2 hybrids. They lie in a plane, $120°$ apart. Each overlaps a $2p$ orbital of fluorine, so $\text{BF}_3$ is trigonal planar. Each hybrid is $33.3\%$ s.

sp3 hybridisation: methane, ammonia and water

In carbon, the $2s$ and all three $2p$ orbitals mix into four sp3 hybrids, pointing to the corners of a tetrahedron, $109.5°$ apart. Each overlaps a $1s$ orbital of hydrogen. Each hybrid is $25\%$ s.

Nitrogen in ammonia is also $sp^3$. Of its four hybrids, three hold one electron each and bond to hydrogen. The fourth holds a lone pair. The lone pair squeezes the angle to $107°$, and the molecule is trigonal pyramidal.

Oxygen in water is $sp^3$ too, with two lone pairs in two hybrids. The angle closes to $104.5°$, and the molecule is bent.

Carbon with double and triple bonds

Ethene, $\text{C}_2\text{H}_4$. Each carbon is $sp^2$. One sp2 hybrid of each carbon overlaps head on to form the C-C sigma bond. The other two bond to hydrogens. Each carbon keeps one unhybridised $2p$ orbital, upright to the plane. These two overlap sideways to form the pi bond.

Ethyne, $\text{C}_2\text{H}_2$. Each carbon is $sp$. One sp hybrid of each carbon forms the C-C sigma bond, and the other bonds to hydrogen. So the molecule is linear. Each carbon keeps two unhybridised $2p$ orbitals, at right angles to each other. They overlap in two pairs, to give two pi bonds.

Using d orbitals

Elements of the third period have empty $3d$ orbitals close in energy to $3s$ and $3p$. They can join in.

$sp^3d$: phosphorus pentachloride. Phosphorus is $3s^2 3p^3$. One $3s$ electron is promoted to $3d$. Then one $s$, three $p$ and one $d$ orbital mix into five sp3d hybrids, pointing to the corners of a trigonal bipyramid. Three equatorial P-Cl bonds lie in one plane, $120°$ apart. Two axial bonds are at $90°$ to that plane. The axial bonds suffer more repulsion and are slightly longer. They break more easily, which is why $\text{PCl}_5$ is very reactive.

$sp^3d^2$: sulphur hexafluoride. Sulphur is $3s^2 3p^4$. One $3s$ and one $3p$ electron move to $3d$. Then one $s$, three $p$ and two $d$ orbitals mix into six sp3d2 hybrids, pointing to the corners of an octahedron. Every angle between neighbours is $90°$.

A quick way to find the hybridisation

Count the electron pairs on the central atom, bonding and lone, as in VSEPR. Then:

The hybrids point the same way as the pairs in VSEPR: linear, trigonal planar, tetrahedral, trigonal bipyramidal and octahedral.

Practise with new numbers

Try it in Python

The program turns the number of electron pairs into a hybridisation, its s-character and its ideal angle, for six molecules.

Common mistakes

• Thinking hybrids hold only bonding pairs. In ammonia and water, some $sp^3$ hybrids hold lone pairs.
• Calling ethene's carbons $sp^3$. A carbon with a double bond keeps one $p$ orbital for the pi bond, so it is $sp^2$.
• Mixing orbitals of different shells. Only valence-shell orbitals of nearly equal energy mix.
• Mixing up shape and hybridisation. Water is $sp^3$ but bent. Hybridisation describes the arrangement of all the pairs.
• Giving all five bonds of $\text{PCl}_5$ the same length. The axial bonds are longer.

Remember these

• Hybridisation mixes orbitals of nearly equal energy into the same number of equal hybrids.
• $sp$: $180°$. $sp^2$: $120°$. $sp^3$: $109.5°$. $sp^3d$: trigonal bipyramidal. $sp^3d^2$: octahedral.
• Pairs on the central atom give the hybridisation: 2, 3, 4, 5, 6 for $sp$ to $sp^3d^2$.
• A double bond leaves a carbon $sp^2$, a triple bond leaves it $sp$. Unhybridised $p$ orbitals form the pi bonds.
• More s-character means the hybrid is held closer to the nucleus.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. What is the hybridisation of boron in $\text{BF}_3$, and what is the bond angle?

2. Why can pure $2s$ and $2p$ orbitals of carbon not explain methane?

3. How many unhybridised $p$ orbitals does each carbon in ethyne keep, and what do they do?

4. Which orbitals of sulphur mix in $\text{SF}_6$, and how many hybrids form?

5. Why are the axial bonds of $\text{PCl}_5$ longer than the equatorial ones?