Sigma and pi bonds and bond parameters

Sigma bonds form by head-on overlap. Two atomic orbitals — $s$–$s$, $s$–$p$, $p$–$p$ along the internuclear axis, or overlaps involving hybrid orbitals — merge directly along the line joining the two nuclei. The electron density is concentrated symmetrically along that axis, which is why a lone $\sigma$ bond allows free rotation: spinning one atom about the bond axis does not disturb the overlap, because the overlap is cylindrically symmetric around that same axis.

Pi bonds form by sideways overlap. Two parallel, unhybridised $p$ orbitals, with their axes perpendicular to the internuclear axis, overlap laterally above and below (or in front of and behind) the bond line. This produces two separate lobes of electron density rather than one continuous cloud along the axis, and the overlap is inherently less effective than head-on overlap — which is the direct reason $\pi$ bonds are weaker than $\sigma$ bonds. Because the two lobes only line up when the $p$ orbitals stay parallel, rotating one end of the bond would tear the sideways overlap apart; a $\pi$ bond therefore locks the molecule against rotation about that bond, which is the structural origin of cis–trans (geometrical) isomerism in compounds with a C=C double bond.

The counting rule. In any multiple bond, exactly one component is always $\sigma$; every additional bond is $\pi$. A single bond is 1 $\sigma$; a double bond is 1 $\sigma$ + 1 $\pi$; a triple bond is 1 $\sigma$ + 2 $\pi$. A molecule can never have a bond made of two $\pi$ bonds and no $\sigma$ bond — the $\sigma$ framework is what holds the atoms together at all, and $\pi$ bonds only add on top of it.

The four bond parameters. Bond length is the equilibrium internuclear distance, in picometres (pm) or ångströms; it falls as bond order rises and as the bonded atoms shrink. Bond angle is the angle between two bonds sharing a central atom, fixed by the arrangement of electron pairs (the detailed geometry is VSEPR theory's job, not this article's). Bond enthalpy (bond dissociation energy) is the energy, in $\text{kJ mol}^{-1}$, needed to break one mole of a specific bond in the gas phase; it rises with bond order and falls as atoms get larger. Bond order ties these together as a single number:

where: $N_b$ is the number of electrons in bonding molecular orbitals and $N_a$ is the number of electrons in antibonding molecular orbitals. This is the formal molecular-orbital-theory definition. For the simple Lewis structures used at this level, bond order collapses to something much easier: just count the shared electron pairs between the two atoms — 1 for a single bond, 2 for a double bond, 3 for a triple bond, matching the $\sigma$-plus-$\pi$ count exactly.

Count the sigma and pi bonds in ethyne, $\text{HC}\equiv\text{CH}$, and use the result to rank its C–C bond against the C=C bond in ethene and the C–C bond in ethane.

Step 1. Identify every bond in the molecule. Each carbon is bonded to one hydrogen and to the other carbon by a triple bond. Each C–H bond is a single bond formed by head-on overlap, so it is one $\sigma$ bond; there are two C–H bonds, both $\sigma$.

Step 2. Break down the C≡C triple bond using the counting rule. It consists of exactly one $\sigma$ bond, formed by head-on overlap of $sp$ hybrid orbitals along the internuclear axis, plus two $\pi$ bonds, formed by sideways overlap of two perpendicular pairs of unhybridised $p$ orbitals.

Step 3. Total the bonds. The molecule has 3 $\sigma$ bonds (2 C–H + 1 C–C) and 2 $\pi$ bonds (both part of the C≡C triple bond), so the C≡C bond order is 3.

Step 4. Rank against the related bonds using the bond-order trend. Bond length falls and bond enthalpy rises as bond order rises, so the order-3 C≡C bond in ethyne (about $120\ \text{pm}$, $835\ \text{kJ mol}^{-1}$) is shorter and stronger than the order-2 C=C bond in ethene ($134\ \text{pm}$, $611\ \text{kJ mol}^{-1}$), which in turn is shorter and stronger than the order-1 C–C bond in ethane ($154\ \text{pm}$, $347\ \text{kJ mol}^{-1}$). Ethyne has 3 sigma bonds and 2 pi bonds, and its C≡C bond order of 3 makes it the shortest and strongest of the three carbon-carbon bonds compared.

Competitive stretch

Apply the same logic to $\text{N}_2$ versus hydrazine, $\text{N}_2\text{H}_4$. $:\text{N}\equiv\text{N}:$ gives bond order 3 (1 $\sigma$ + 2 $\pi$), while hydrazine's N–N link is a single $\sigma$ bond, bond order 1. Since bond enthalpy rises with bond order, $\text{N}\equiv\text{N}$ ($945\ \text{kJ mol}^{-1}$, $110\ \text{pm}$) is far stronger and shorter than N–N in hydrazine (about $160\ \text{kJ mol}^{-1}$, $145\ \text{pm}$) — why $\text{N}_2$ is famously unreactive while hydrazine, lacking $\pi$ character, serves as a rocket fuel.

• Treating a double bond as "two identical bonds" — it is always one strong, axial $\sigma$ bond plus one weaker, sideways $\pi$ bond, and the two are never interchangeable.
• Assuming free rotation is possible around any bond — rotation is unrestricted about a pure $\sigma$ bond but locked wherever $\pi$ character exists, which is exactly why C=C double bonds show cis–trans isomerism and C–C single bonds do not.
• Mixing up the Lewis-structure shared-pair count with the formal $(N_b - N_a)/2$ molecular-orbital formula — for ordinary non-resonance molecules the two give the same number, but they arrive there by different reasoning, and only the MO formula handles species with unpaired or antibonding electrons correctly.
• Assuming bond enthalpy scales exactly with bond order — a $\pi$ bond is always weaker than a $\sigma$ bond because sideways overlap is less effective than head-on overlap, so a triple bond is not simply three times as strong as a single bond.

Ask the tutor: Want to work through counting sigma and pi bonds in a bigger molecule like benzene, or compare bond parameters across a longer series of related compounds?