Physics · Atomic and quantum physics · Year 3

The molecule: bonding and spectra

How does a hydrogen molecule form, and why does CO₂ absorb infrared radiation?

The situation

In a laboratory an infrared spectrometer passes a beam through the gas in a cell: the nitrogen and oxygen in air hardly notice this beam, but carbon dioxide (CO₂) absorbs strongly around 4.26 and 15 µm — just as in the Earth's atmosphere. On the next bench a green laser lights up a liquid, and new colours appear in the scattered light that were not in the laser. Both have one cause: in a molecule the electrons bind the atoms together, while the nuclei vibrate and rotate. What joins two hydrogen atoms into one molecule, and what “fingerprint” does a molecule leave in light?

Topic

Molecules: chemical bonding and spectra. How a hydrogen molecule forms and why CO₂ absorbs infrared radiation

Key ideas

Covalent bonding
Vibrational and rotational spectra
Raman scattering

Formulas

E_{\pm} (R) = 2 E_{H} + \frac{e^{2}}{4 π ε_{0} R} + \frac{Q \pm A}{1 \pm S^{2}}

Heitler–London: $Q$ is the Coulomb integral, $A$ the exchange integral, $S$ the overlap of the orbitals. “+” is the singlet (opposite spins), the bonding state.

U (R) = D_{e} {(1 - e^{- a (R - R_{e})})}^{2} - D_{e}

Morse potential: $D_{e}$ is the depth of the well, $R_{e}$ the equilibrium distance (in H₂ 4.75 eV and 74 pm).

E_{v} = ℏ ω (v + \frac{1}{2}) - \frac{{[ℏ ω (v + \frac{1}{2})]}^{2}}{4 D_{e}}, D_{0} = D_{e} - E_{0}

Anharmonic vibrational levels and the dissociation energy (H₂: $D_{0} \approx 4.48$ eV).

E_{J} = h c B J (J + 1), {\tilde{ν}}_{R, P} \approx {\tilde{ν}}_{0} \pm 2 B m

The rigid rotor ( $B = h / 8 π^{2} c I$ ) and the R ( $m = J + 1$ ) and P ( $m = J$ ) branches of a vibration–rotation band.

{\tilde{ν}}_{S, a S} = {\tilde{ν}}_{0} \mp {\tilde{ν}}_{v i b}, \frac{I_{a S}}{I_{S}} \approx {(\frac{{\tilde{ν}}_{0} + {\tilde{ν}}_{v i b}}{{\tilde{ν}}_{0} - {\tilde{ν}}_{v i b}})}^{4} e^{- h c {\tilde{ν}}_{v i b} / k T}

Raman (inelastic) scattering: Stokes and anti-Stokes lines.

Theory

The adiabatic approximation and molecular forces

A proton is ≈ 1836 times heavier than an electron, so the electrons adjust to the motion of the nuclei almost instantly — this is the adiabatic (Born–Oppenheimer) approximation. First the nuclei are taken as fixed and the energy of the electrons is found for each distance $R$ ; this serves as the potential curve $U (R)$ for the nuclei, and the nuclei vibrate and rotate on this curve. So the energy of a molecule has three tiers: electronic (a few eV), vibrational (≈ 0.1–0.5 eV) and rotational (≈ $10^{- 4}$ – $10^{- 2}$ eV); the approximate ratio is $1 : \sqrt{m / M} : m / M$ . The molecular forces that hold the atoms are electrical in nature: attraction at long range, and strong repulsion when the electron shells overlap. When “Vibration” is switched on in the scene, the nuclei are shown slowed down (1 s ≈ 5 fs), while the electron cloud follows them at every moment.

H₂⁺ and H₂: the quantum nature of the chemical bond

The simplest molecule is the hydrogen molecular ion H₂⁺: one electron around two protons. If the wave function is taken as a sum of atomic orbitals $ψ_{a} \pm ψ_{b}$ , then in the “+” state the electron cloud thickens between the protons and pulls them together (bonding σg), while in the “−” state there is a nodal plane in the middle (antibonding σu*). For H₂, Heitler and London (1927) took the exchange of electrons into account: if the spins are opposite (singlet), the spatial function is symmetric and the energy falls — a covalent bond forms; if the spins are parallel (triplet), the Pauli principle squeezes the electrons out of the middle and the atoms repel. This attraction is a quantum effect of the exchange integral $A$ and has no classical analogue. Valency comes from this: H has one unpaired electron — it forms one bond; He has none — an He₂ molecule does not form. Model: Heitler–London gives $D_{e} \approx 3.16$ eV (87 pm), and LCAO gives 1.76 eV (132 pm) for H₂⁺; experiment gives 4.75 eV (74 pm) and 2.79 eV (106 pm) respectively — the dashed lines on the graph.

Types of bond and bond energy

Ionic bond: Na gives its electron to Cl — the “price” of this is $5.14 - 3.61 = 1.53$ eV, but the Coulomb attraction of the ions pays it back (the curve crosses the level of the neutral atoms at ≈ 940 pm); in a NaCl crystal the lattice energy per ion pair is about 8 eV. A covalent bond is a shared electron pair; it is directional (in diamond each atom has 4 neighbours). In a metallic bond the valence electrons belong to the whole crystal as an “electron gas” and glue the positive ions together (Cu: 3.49 eV/atom). Van der Waals forces come from the attraction of fluctuating dipoles (energy $\propto - 1 / r^{6}$ ) and are very weak: ≈ 10 meV for a pair of Ar atoms, which is why argon melts at −189 °C. A hydrogen bond is an H atom between two electronegative atoms (O–H···O), ≈ 0.2 eV: the properties of water and ice and the DNA double helix come from it. The greater the bond energy, the harder the crystal and the higher its melting point.

Vibrational, rotational and electronic spectra

In a Morse well the vibrational levels crowd closer together going up (anharmonicity). Even the lowest level is not at the bottom of the well: because of the zero-point energy, to break H₂ takes $D_{0} = D_{e} - E_{0} \approx 4.48$ eV (at 0 K); the 4.52 eV (436 kJ/mol) in books is the bond enthalpy at room temperature. For a molecule to absorb an IR photon, the dipole moment must change during the vibration; the selection rules are $Δ v = \pm 1$ , $Δ J = \pm 1$ , so the HCl band splits into P and R branches with no line in the middle. The spacing of neighbouring lines is ≈ $2 B$ — the bond length is found from it. CO, with a permanent dipole, gives a pure rotational (microwave) spectrum: the J = 0 → 1 transition is at 115.27 GHz — radio astronomers map molecular clouds with this line. N₂, O₂, H₂ and CO₂ have no permanent dipole — they have no pure rotational absorption spectrum either (though they do have a rotational Raman spectrum). In electronic spectra the transition is vertical, before the nuclei have time to move (Franck–Condon): the σg → σu* transition of H₂⁺ breaks the molecule apart; the transition of H₂ to a triplet state changes the spin — forbidden for a photon, allowed for electron impact. H₂ absorbs photons only in the vacuum ultraviolet (from ≈ 11 eV).

CO₂ and the greenhouse effect

The linear three-atom CO₂ has three kinds of vibration: the symmetric stretch $ν_{1}$ (1333 cm⁻¹) — the dipole does not change, IR inactive; the bend $ν_{2}$ (667 cm⁻¹, 15 µm) and the asymmetric stretch $ν_{3}$ (2349 cm⁻¹, 4.26 µm) — IR active. By Wien's law the Earth's surface (≈ 288 K) radiates around ≈ 10 µm, and the 15 µm band falls in a strong part of this radiation — CO₂ traps part of the heat leaving the Earth. In a cell the absorption obeys the Beer–Lambert law: $T = e^{- σ N L}$ — if the concentration doubles, a path half as long gives the same absorption. Model: the band strengths are approximate (≈ 9·10⁻¹⁷ and 8·10⁻¹⁸ cm/molecule), the lines are smoothed with a resolution of 2 cm⁻¹, and “hot” bands and other gases are not taken into account.

Raman (inelastic) scattering

The electric field of light induces a dipole $p = α E$ in a molecule. If the polarisability $α$ changes as the molecule vibrates, lines at ${\tilde{ν}}_{0} \pm {\tilde{ν}}_{v i b}$ appear in the scattered light in addition to the laser frequency. In the Stokes line the molecule takes energy from the photon, in the anti-Stokes line it gives energy — for this the molecule must already be excited, so the anti-Stokes line is very weak and grows with temperature. IR activity requires a change of dipole, Raman activity a change of polarisability: in CO₂, which has a centre of symmetry, $ν_{1}$ is seen only in Raman, and $ν_{2}$ and $ν_{3}$ only in IR (the mutual exclusion rule). In gases there are also rotational Raman lines beside the Rayleigh line ( $Δ J = \pm 2$ , shift $4 B (J + \frac{3}{2})$ ), and in nitrogen, because of nuclear spin, the intensities of even and odd J are in the ratio 2 : 1. To record the spectrum, a notch filter suppresses the strong Rayleigh line, a monochromator spreads the light by wavelength, and a photomultiplier or CCD records it. Model: the relative line intensities are approximate.

Did you know?

The first quantum-mechanical theory of the hydrogen molecule was created in 1927 by Walter Heitler and Fritz London — this was the beginning of the explanation of the chemical bond by quantum mechanics.
Raman scattering was observed in 1928, almost simultaneously, in India by C. V. Raman and K. S. Krishnan (in liquids) and in Moscow by G. S. Landsberg and L. I. Mandelstam (in a quartz crystal); Raman received the Nobel Prize in Physics in 1930.

Inside the experiment

The experiment starts from a real-life situation: you predict the result, change the controls, run the experiment and draw a conclusion. Inside are missions, theory, a quiz and a guided tour in which the camera moves and explains the key points.

More experiments in this section

Spin, the Zeeman effect and MRI — Year 1
Many-electron atoms — Year 1
Bands in solids — Year 1
Schrödinger and quantum tunnelling — Year 3
Quantum dots and the potential well — Year 3
Superconductivity — Year 3