Sigma and Pi Bonds

Sigma and Pi Bonds

6 min read Updated Mar 26, 2026

Think about two ways to shake hands. In the normal handshake, you extend your arm straight out and grab the other person’s hand directly - palm to palm, along the line connecting your two bodies. Now imagine a different greeting: you and a friend stand side by side and link arms at the elbows, touching along the length of your forearms rather than at the tips. The first handshake is a sigma bond - head-on, direct overlap along the axis between two nuclei. The side-by-side arm link is a pi bond - lateral overlap above and below that axis.

Sigma Bonds: Head-On Overlap

A sigma (σ) bond forms when two orbitals overlap end-to-end, directly along the line (internuclear axis) connecting the two bonded nuclei. The electron density in a sigma bond is concentrated right between the nuclei, like a cylinder of electron cloud wrapping around the axis.

Hybridisation, sigma and pi, and what bond order does

Bonding
sp³ 4 groups one s + three p · no p left over C C tetrahedral · 109.5° 4 σ + 0 π ethane, CH₃–CH₃ bond order 1 free rotation sp² 3 groups one s + two p · one p left over C C trigonal planar · 120° 3 σ + 1 π ethene, CH₂=CH₂ bond order 2 rotation locked sp 2 groups one s + one p · two p left over C C linear · 180° 2 σ + 2 π ethyne, CH≡CH bond order 3 rotation locked Sigma head-on, pi side-on σ overlap on the axis π overlap above/below First bond: always σ. Second and third: π. π stops rotation, which is where cis and trans come from. Carbon-carbon bonds, measured length (pm) energy (kJ/mol) C–C 154 347 C=C 134 614 C≡C 120 839 Higher bond order, shorter bond, stronger bond, for the same pair of atoms.
1

Scroll sideways to see the whole map.

Bond order, length and strength move together. Adding a pi bond pulls the two carbons closer and makes the connection harder to break: 154 pm and 347 kJ/mol for a single bond, 120 pm and 839 kJ/mol for a triple. Shorter is always stronger, for the same pair of atoms.

Sigma bonds can form from many different orbital combinations:

  • s orbital + s orbital (as in H2)
  • s orbital + p orbital (as in HF)
  • p orbital + p orbital, overlapping head-on (as in F2)
  • Hybrid orbital + hybrid orbital (as in C-C bonds)
  • Hybrid orbital + s orbital (as in C-H bonds)

The common thread is always the same: the orbitals point directly at each other and overlap along the internuclear axis.

Pi Bonds: Lateral Overlap

A pi (π) bond forms when two unhybridized p orbitals line up parallel to each other and overlap sideways - above and below the internuclear axis. The electron density in a pi bond sits in two lobes, one above the plane and one below it. There is actually a node (zero electron density) right along the internuclear axis itself.

Pi bonds can only form from p orbitals overlapping laterally. They cannot form from s orbitals (which have no directional lobes to overlap sideways) and they do not form from hybrid orbitals (which are used for sigma bonding).

A pi bond is always the “second” or “third” bond between two atoms. You must have a sigma bond in place first - the head-on framework holds the atoms together at the correct distance for the p orbitals to overlap sideways.

The Counting Rule

This is one of the most tested concepts in general chemistry on the MCAT. Memorize this pattern:

  • Single bond = 1 sigma bond
  • Double bond = 1 sigma + 1 pi bond
  • Triple bond = 1 sigma + 2 pi bonds

Every bond between two atoms contains exactly one sigma bond. The sigma bond is always first. Any additional bonds beyond the first are pi bonds.

Rotation: Sigma Allows It, Pi Prevents It

Here is where sigma and pi bonds have a critical functional difference.

Sigma bonds allow free rotation. Because the electron density wraps symmetrically around the internuclear axis (like a cylinder), rotating one atom relative to the other does not disrupt the overlap. Imagine spinning a pencil that is stuck through the center of a donut - the donut does not care which way the pencil faces. Single bonds rotate freely, which is why molecules like ethane (C2H6) have rapidly interconverting conformations.

Pi bonds prevent rotation. The lateral overlap of p orbitals depends on those orbitals staying parallel. If you tried to rotate one atom 90 degrees, the p orbitals would become perpendicular to each other, the overlap would drop to zero, and the pi bond would break. This is why double bonds are rigid.

Worked Example: Ethylene (C2H4)

Ethylene has a carbon-carbon double bond with two hydrogens on each carbon.

Count the bonds:

  • 4 C-H bonds: each is a single bond = 4 sigma bonds
  • 1 C=C bond: the double bond = 1 sigma + 1 pi bond

Total: 5 sigma bonds + 1 pi bond

Each carbon is sp2 hybridized (3 groups: 2 H atoms + 1 C atom). The three sp2 orbitals form the three sigma bonds. The one unhybridized p orbital on each carbon overlaps laterally to form the single pi bond. The molecule is planar because sp2 hybridization creates a flat, 120-degree framework, and the pi bond locks everything in place.

Worked Example: Hydrogen Cyanide (HCN)

HCN has a single bond from H to C and a triple bond from C to N.

Count the bonds:

  • 1 H-C bond: single bond = 1 sigma bond
  • 1 C≡N bond: triple bond = 1 sigma + 2 pi bonds

Total: 2 sigma bonds + 2 pi bonds

Carbon is sp hybridized (2 groups: H and N). The two sp orbitals form the two sigma bonds (one to H, one to N). The two unhybridized p orbitals on carbon overlap laterally with two p orbitals on nitrogen to form the two pi bonds. The molecule is linear because sp hybridization produces a 180-degree geometry.

Worked Example: Acetic Acid (CH3COOH)

This slightly more complex molecule lets you practice on a real MCAT-style question.

Draw the structure: H3C - C(=O) - O - H. The molecule has:

  • 3 C-H bonds on the methyl group: 3 sigma bonds
  • 1 C-C bond: 1 sigma bond
  • 1 C=O bond: 1 sigma + 1 pi bond
  • 1 C-O bond: 1 sigma bond
  • 1 O-H bond: 1 sigma bond

Total: 7 sigma bonds + 1 pi bond

Notice that the methyl carbon (4 groups) is sp3 hybridized, while the carbonyl carbon (3 groups) is sp2 hybridized. Different carbons in the same molecule can have different hybridizations.

Bond Strength and Bond Length

Sigma bonds are generally stronger than pi bonds because head-on overlap is more effective than lateral overlap. However, a double bond (sigma + pi) is stronger overall than a single bond (sigma only), and a triple bond is stronger still.

Correspondingly, bond length decreases as bond order increases. A triple bond is shorter than a double bond, which is shorter than a single bond. More shared electrons pull the nuclei closer together.

| Bond Type | Bond Order | Relative Strength | Relative Length |
|:---|:---:|:---|:---|
| Single (sigma only) | 1 | Weakest | Longest |
| Double (sigma + pi) | 2 | Moderate | Moderate |
| Triple (sigma + 2 pi) | 3 | Strongest | Shortest |

These trends connect directly to the next section on bond energy, bond length, and bond order.

Common MCAT Traps

  • Saying pi bonds are “weaker” so double bonds are weak. Pi bonds individually are weaker than sigma bonds, yes. But a double bond (sigma + pi together) is stronger than a single bond (sigma only). The MCAT tests this distinction.
  • Forgetting that the first bond is always sigma. Even in a triple bond, one of the three bonds is a sigma bond. There is no such thing as a pure pi-bond-only connection between two atoms.
  • Miscounting bonds. Each hydrogen forms exactly one bond (one sigma). When counting sigma bonds in a molecule, do not forget the C-H, N-H, and O-H bonds - they add up quickly.
How many sigma and pi bonds are in a molecule of C2H2 (acetylene)?
Click to reveal answer

3 sigma bonds and 2 pi bonds. Acetylene has one H-C sigma bond on each end (2 total) and one C≡C triple bond (1 sigma + 2 pi). Grand total: 3 sigma + 2 pi. Each carbon is sp hybridized (2 groups), and the two unhybridized p orbitals on each carbon form the two pi bonds.

Why does a carbon-carbon double bond prevent rotation while a single bond allows it?
Click to reveal answer

The sigma bond allows rotation because its electron density is symmetrically distributed around the internuclear axis - rotating does not break the overlap. The pi bond, however, depends on lateral (side-by-side) overlap of parallel p orbitals. Rotating one carbon 90 degrees would make the p orbitals perpendicular, destroying the overlap and breaking the pi bond. Since breaking a bond requires significant energy, the double bond is effectively rigid under normal conditions.

A nitrogen atom in a molecule forms one double bond and one single bond. What is its hybridization, and how many unhybridized p orbitals does it have?
Click to reveal answer

Count the groups: 1 double bond (1 group) + 1 single bond (1 group) + 1 lone pair (likely, to satisfy the octet) = 3 groups. Hybridization = sp2. With sp2 hybridization, one of the three p orbitals remains unhybridized. That single unhybridized p orbital forms the pi bond component of the double bond. The three sp2 orbitals hold the two sigma bonds and the lone pair.