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C. Structure and Properties of Matter · Structure and Properties of Matter

Valence Bond Theory: Hybrid Orbitals, Sigma and Pi Bonds

C3.2 C2.3About 65 minutes
Learning goals
  • I can explain why hybrid orbitals are needed to describe bonding in molecules such as methane.
  • I can predict sp, sp² and sp³ hybridization from the number of electron groups.
  • I can identify and count σ and π bonds in molecules.
  • I can relate π bonds to bond length, bond strength and restricted rotation.

Words to know

valence bond theory
A model in which covalent bonds form by the overlap of orbitals on two atoms, each sharing a pair of electrons.
hybrid orbital
An orbital formed by mixing atomic orbitals on one atom, such as sp³.
sigma (σ) bond
A bond formed by end-to-end overlap, with electron density along the axis between the nuclei.
pi (π) bond
A bond formed by side-by-side overlap of parallel p orbitals, with density above and below the axis.
delocalized electrons
Electrons spread over three or more atoms rather than shared by just two.

A puzzle about carbon

Carbon's configuration is 1s² 2s² 2p². Its two 2s electrons are paired, and only the two 2p electrons are unpaired. Using simple orbital overlap, carbon should make two bonds at 90°. But methane, CH₄, has four identical C–H bonds at 109.5°. VSEPR predicts the shape correctly, but how do the orbitals manage it?

The answer comes from valence bond theory, developed by Linus Pauling and others in the 1930s: a covalent bond forms when orbitals on two atoms overlap and share a pair of electrons, and atoms can mix (hybridize) their orbitals to make new ones pointing in the right directions.

Hybrid orbitals

When carbon bonds, its 2s and 2p orbitals combine into new hybrid orbitals. The number of hybrid orbitals always equals the number of atomic orbitals mixed.

  • sp³: one s + three p → four sp³ orbitals, pointing to the corners of a tetrahedron (109.5°). Used when the atom has 4 electron groups (CH₄, NH₃, H₂O, the carbons in ethane).
  • sp²: one s + two p → three sp² orbitals in a plane at 120°; one p orbital is left unhybridized, at right angles to the plane. Used for 3 electron groups (BF₃, each carbon in ethene, C₂H₄).
  • sp: one s + one p → two sp orbitals at 180°; two p orbitals are left over. Used for 2 electron groups (each carbon in ethyne, C₂H₂; the carbon in CO₂).

So the shortcut is: count the electron groups (as in VSEPR), and that tells you the hybridization. Lone pairs count as groups: the nitrogen in NH₃ is sp³.

(Many textbooks extend the idea to sp³d for five groups and sp³d² for six. Modern calculations suggest those molecules are better described in other ways, so this course uses hybridization mainly for 2, 3 and 4 groups.)

carbon atom: 2s² 2p²↑↓2s↑↑2pmix4 orbitals in,4 orbitals outfour equal sp³ hybrid orbitals↑↑↑↑sp³ sp³ sp³ sp³each holds 1 electron and forms one C–H bond:CH₄ has four identical bonds at 109.5°energy
Hybridization in methane (energy levels simplified). One 2s and three 2p orbitals combine into four equal sp³ hybrid orbitals, each with one electron, pointing to the corners of a tetrahedron.

Sigma and pi bonds

  • A sigma (σ) bond forms when orbitals overlap end to end, along the line between the two nuclei. Every single bond is a σ bond. σ bonds allow the atoms to rotate freely around the bond.
  • A pi (π) bond forms when two parallel, unhybridized p orbitals overlap side by side, putting electron density above and below the bond axis.

Double bond = 1 σ + 1 π. Triple bond = 1 σ + 2 π.

In ethene, each carbon is sp². The sp² orbitals form σ bonds to two H atoms and to the other carbon; the leftover p orbitals overlap side by side to form the π bond. Because twisting one end would break that side-by-side overlap, there is no free rotation around a C=C double bond. That is why molecules such as cis-but-2-ene and trans-but-2-ene are different compounds, as you see in the Organic Chemistry unit.

In ethyne (HC≡CH), each carbon is sp. The two leftover p orbitals on each carbon form two π bonds at right angles to each other.

σ bond: head-on overlapnucleusnucleuselectron density lies ONthe line between nucleiπ bond: side-by-side overlapdensity ABOVE and BELOW the axis;the solid line is the σ bond
A sigma (σ) bond forms by end-to-end overlap; a pi (π) bond forms when parallel p orbitals overlap side by side. A double bond is one σ plus one π bond; a triple bond is one σ plus two π bonds.
BondMake-upTypical lengthTypical bond energy
C–C (ethane)1 σ154 pmabout 350 kJ/mol
C=C (ethene)1 σ + 1 π134 pmabout 610 kJ/mol
C≡C (ethyne)1 σ + 2 π120 pmabout 840 kJ/mol
More shared electrons pull the nuclei closer and make the bond stronger. A π bond is weaker than a σ bond, so a double bond is less than twice as strong as a single bond. (Values are typical; tables differ slightly.)

Delocalized π electrons

In some molecules, p orbitals on several neighbouring atoms overlap, and the π electrons spread over all of them. These electrons are delocalized. This is the orbital picture of resonance: in the nitrate ion and in benzene (C₆H₆), the π electrons belong to the whole ring or ion, so all the bonds are identical. In graphite, the π electrons are delocalized over entire sheets of carbon atoms, which is why graphite conducts electricity (see the lesson on types of solids).

Worked example

Example 1: Hybridization from electron groups

State the hybridization of the central atom in (a) BF₃ (b) H₂O (c) CO₂ (d) NH₄⁺.

  1. (a) B has 3 electron groups → sp².
  2. (b) O has 2 bonds + 2 lone pairs = 4 groups → sp³.
  3. (c) C has 2 groups (two C=O) → sp.
  4. (d) N has 4 bonds → sp³.

BF₃ sp², H₂O sp³, CO₂ sp, NH₄⁺ sp³.

Worked example

Example 2: Counting σ and π bonds

How many σ and π bonds are in propene, CH₂=CH–CH₃? State the hybridization of each carbon.

  1. Draw it: C1 (=CH₂) has 2 C–H bonds and the C=C; C2 has 1 C–H, the C=C and a C–C; C3 (CH₃) has 3 C–H and the C–C.
  2. σ bonds: 6 C–H + 1 C–C + the σ part of C=C = 8.
  3. π bonds: 1 (in the C=C).
  4. C1 and C2 each have 3 electron groups → sp². C3 has 4 → sp³.

8 σ and 1 π; C1 and C2 are sp², C3 is sp³.

Worked example

Example 3: Hydrogen cyanide

Describe the bonding in H–C≡N using hybrid orbitals.

  1. C has 2 electron groups (C–H and C≡N) → sp. The two sp orbitals point 180° apart.
  2. One sp orbital overlaps the H 1s orbital: a C–H σ bond. The other overlaps an orbital on N: the C–N σ bond.
  3. The two leftover p orbitals on C overlap side by side with two p orbitals on N: two π bonds.
  4. Total: 2 σ + 2 π; the molecule is linear.

Carbon is sp; HCN has 2 σ bonds and 2 π bonds and is linear.

Just for fun

Chemists like to say a π bond is "a bond with a hole in the middle": its electron density sits above and below the line joining the nuclei, with none on the line itself. It may be the only bond whose most important feature is where it isn't.

Hands-on activity

Twist test: σ versus π bonds

You need: 2 large marshmallows or balls of modelling clay, 4 toothpicks, and 2 flat craft sticks or two strips of stiff cardboard. Grown-up note: toothpicks are sharp; supervise and keep them away from small children.

  1. Single bond. Join the two marshmallows with one toothpick through their centres (a σ bond). Hold one marshmallow still and twist the other. It turns freely.
  2. Double bond. Push the two flat sticks into the tops and bottoms of the marshmallows, parallel to the toothpick, one above and one below (these model the two lobes of the side-by-side π overlap).
  3. Try to twist one marshmallow now. You cannot do it without bending or breaking the sticks.
  4. Stick two toothpicks into each marshmallow as "C–H bonds", all in the plane at right angles to the sticks. In the double-bond model, notice that the sticks lock all four toothpicks in that one plane.

What to look for: a σ bond allows rotation; a π bond locks the atoms in place and keeps the molecule flat around a C=C bond. This is why ethene is planar and why cis and trans isomers of alkenes cannot turn into each other at room temperature.

Your turn

1

State the hybridization of carbon in CH₄, C₂H₄ and C₂H₂.

2

How many σ and π bonds are in a triple bond?

3

What is the hybridization of nitrogen in NH₃? Explain.

4

Count the σ and π bonds in ethene, C₂H₄.

5

Why is there no free rotation about a C=C double bond?

6

Why is a C=C double bond not twice as strong as a C–C single bond?

7

State the hybridization of the carbon in methanal, CH₂O, and count its σ and π bonds.

More practice: warm-up

1

How many hybrid orbitals form from one s and two p orbitals?

2

What angle is between sp hybrid orbitals?

3

Every single bond is what type of bond?

4

Which hybridization gives a tetrahedral arrangement?

5

How many π bonds are in a double bond?

More practice: on your own

1

Count the σ and π bonds in ethanoic acid, CH₃COOH.

2

State the hybridization of each carbon in propyne, HC≡C–CH₃.

3

What is the hybridization of the oxygen in water, and what is the shape?

4

Describe the π bonding in CO₂.

5

Which bond is shortest: C–C, C=C or C≡C? Why?

Challenge

1

Count all σ and π bonds in benzene, C₆H₆ (a ring of six carbons, each bonded to one H, drawn with alternating double bonds).

2

Acrylonitrile, CH₂=CH–C≡N, is used to make acrylic fibres. State the hybridization of each carbon and count σ and π bonds.

3

Allene, H₂C=C=CH₂, has a central carbon with two double bonds. Predict the hybridization of the central carbon and explain why the two CH₂ groups are at 90° to each other.

Check your understanding

Quiz

1.

What is the hybridization of carbon in methane?

2.

A double bond consists of:

3.

Which molecule contains an sp-hybridized carbon?

4.

How is a π bond formed?

5.

How many σ bonds are in ethane, C₂H₆?

6.

Why do cis and trans isomers of an alkene not convert into each other at room temperature?

For parents and teachers

Big idea (C3.2, C2.3): orbitals on an atom can mix (hybridize) so that bonds point in the directions VSEPR predicts. Single bonds are σ bonds; double and triple bonds add π bonds, which make bonds shorter and stronger and stop rotation.

How to help: ask your teen to count σ and π bonds in structures from food or medicine labels (draw the structure first). The marshmallow twist test makes restricted rotation memorable.

Common difficulties:

  • Forgetting to count lone pairs when deciding hybridization.
  • Counting a double bond as two σ bonds.
  • Thinking hybridization causes the shape; it is a way of describing bonding that agrees with the shape VSEPR predicts.