Sigma and pi bonds, intermolecular forces

This week we talk about bonds at two different scales. In the first half, the orbital structure of the bonds inside a molecule: σ and π. In the second half, the weak attractions between molecules: these determine which substance is a liquid and which a gas, and what dissolves in what. Do not mix the two: when you boil water the O–H bonds do not break, the molecules separate from each other.

Golden rules:

1. Every bond has exactly one σ. Anything more is π.
2. A π bond needs an unhybridized p orbital. An sp² carbon can make one π bond, an sp carbon two.
3. London forces exist in every molecule. Whether or not other forces exist.
4. For an H bond the H must be bonded to N, O or F.

σ and π bonds

In valence bond theory, a covalent bond forms by the overlap of half-filled orbitals of two atoms. The geometry of the overlap gives two types of bonds.

Sigma (σ) bond: the orbitals overlap end to end along the axis joining the two nuclei. The electron density is on the axis. s–s ($\mathrm{H_2}$), s–p (HCl), hybrid–hybrid or hybrid–s overlaps give σ bonds. Because a σ bond is symmetric about the axis, the atoms can rotate freely around the bond.

Pi (π) bond: two parallel p orbitals overlap side by side. The electron density is above and below the axis, in two lobes; there is none on the axis itself. π overlap is weaker than σ.

Ethylene: sp² and one π bond

In $\mathrm{C_2H_4}$ each C has three groups (two H and the other C): sp². The three sp² orbitals lie in a plane at 120° and make three σ bonds. On each C one $2p$ orbital that did not take part in hybridization stays perpendicular to the plane. These p orbitals of the two C overlap side by side to form the π bond.

ethylene  H2C=CH2
  σ: 4 C–H (sp2–1s) + 1 C–C (sp2–sp2) = 5
  π: 1 (2p–2p, perpendicular to the plane of the molecule)
  all atoms in the same plane, angles ~120°

For the π bond to stay intact the two p orbitals must be parallel; so the whole molecule is planar and rotation about C=C is prevented. This restriction is the source of cis-trans isomerism: cis-2-butene and trans-2-butene do not convert into each other at room temperature.

Acetylene: sp and two π bonds

In $\mathrm{C_2H_2}$ each C has two groups: sp, linear. The two sp orbitals make the σ bonds (one with H, one with the other C). The remaining two p orbitals are perpendicular to each other and to the axis; each makes a π bond with its counterpart on the opposite C.

acetylene HC≡CH
  σ: 2 C–H + 1 C–C = 3
  π: 2 (one py–py, one pz–pz)
  linear, 180°

The method for counting σ/π

1. Write the Lewis structure (show all bonds, expand abbreviations like $\mathrm{CH_3}$).
2. Number of σ = number of bonds between atom pairs (each group of lines is one σ).
3. Number of π = 1 for each double bond, 2 for each triple bond.

Example 1: acetic acid, $\mathrm{CH_3COOH}$.

        H   O
        |   ‖
    H - C - C - O - H
        |
        H
σ: 3 C–H + 1 C–C + 1 C=O (the σ part) + 1 C–O + 1 O–H = 7
π: 1 (C=O)
hybridization: the CH3 carbon is sp3, the carboxyl carbon is sp2

Example 2: $\mathrm{CO_2}$. C is sp (2 groups). Two C=O bonds: 2 σ + 2 π. The two π bonds are in mutually perpendicular planes.

PITFALL · making a π bond with a hybrid orbital. π bonds are always formed from unhybridized p orbitals. Hybrid orbitals carry only σ bonds and lone pairs. An sp³ atom cannot make a π bond; that is why every C with a double bond is at most sp².

Intermolecular forces

Intermolecular forces (IMF) are the electrostatic interactions that attract separate molecules to each other. They are much weaker than intramolecular bonds (hundreds of kJ/mol), usually 1-40 kJ/mol, but they determine the boiling point, vapor pressure, viscosity and solubility of a substance. Their collective name is van der Waals forces.

London dispersion forces

Even in a nonpolar molecule the electrons move constantly. If for a moment the electron cloud shifts to one side, an instantaneous dipole forms. This instantaneous dipole creates an induced dipole in the neighboring molecule, and the two attract each other.

• They exist in every molecule and atom (He, $\mathrm{CH_4}$, $\mathrm{I_2}$, $\mathrm{H_2O}$…).
• They depend on how easily the electron cloud is distorted, that is, on the polarizability. It increases as the number of electrons and the size of the molecule increase.
• Shape matters: long molecules touch with a wider surface. n-Pentane (36.1 °C) boils higher than the spherical neopentane (9.5 °C).

The halogens are the classic example: at room temperature $\mathrm{F_2}$ and $\mathrm{Cl_2}$ are gases, $\mathrm{Br_2}$ is a liquid, and $\mathrm{I_2}$ is a solid.

Dipole-dipole forces

Polar molecules have permanent dipoles. The $\delta^+$ end of one molecule attracts the $\delta^-$ end of its neighbor. Among molecules of similar molar mass, the polar one boils higher: $\mathrm{CH_3CHO}$ (44 g/mol, polar) 20 °C, propane (44 g/mol, nonpolar) −42 °C.

Hydrogen bond

A special and strong dipole-dipole interaction. The conditions:

1. The H atom must be covalently bonded to N, O or F (a very polar bond, the H being almost like a bare proton).
2. The neighboring molecule must have an N, O or F with a lone pair.

$\mathrm{O{-}H \cdots :O}$

Water, alcohols, carboxylic acids, amines, HF and DNA base pairs form hydrogen bonds. Its effect is clearly seen in the boiling points of group 14-17 hydrides:

After $\mathrm{H_2S}$ the trend that increases regularly with mass (London) is completely broken in water because of H bonding. In group 14 ($\mathrm{CH_4}$…$\mathrm{SnH_4}$) the trend is smooth because there is no H bonding.

Hydrogen bonding also explains why ice floats on water: in ice each molecule builds an open, spacious lattice with four H bonds. When it melts the lattice partly collapses, and liquid water is denser.

PITFALL · C–H does not form hydrogen bonds. $\mathrm{CH_3OCH_3}$ (dimethyl ether) and $\mathrm{C_2H_5OH}$ (ethanol) have the same formula ($\mathrm{C_2H_6O}$). In the ether all the H atoms are bonded to C: there is no H bonding among them, and it boils at −24 °C. Ethanol has an O–H: 78.4 °C.

Ion-dipole forces

The attraction between an ion and a polar molecule. The O ends of water gather around a cation, and the H ends around an anion (hydration). It allows ionic solids to dissolve in water and is the strongest intermolecular interaction in the list.

A comparison strategy

When comparing the boiling points of two substances, ask in order:

1. Is it ionic? Ionic solids melt and boil at very high temperatures.
2. Is there an H bond (N–H, O–H, F–H)? If so, it generally wins.
3. Is it polar? At similar mass, the polar one wins.
4. If all are of the same type: the larger molar mass / more electrons / longer shape wins.

TIP · do not underestimate London. If the molar mass difference is large, London forces can beat the H bond. $\mathrm{I_2}$ (nonpolar, 254 g/mol) is a solid at room temperature, water is a liquid. The rule applies “among molecules of similar size”.

Common mistakes