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For a diatomic molecule, let $V(r)$ be the electronic potential energy at bond length $r$, with the atomic nuclei treated as fixed while finding the electronic energy. Choose zero at infinitely separated chemical atoms in the specified electronic quantum states. If the minimum occurs at $r_e$, the bond potential-well depth isThe bottom of this curve describes an equilibrium bond length, not an attainable motionless vibrational quantum state. Changing the zero of potential energy shifts both endpoints equally and leaves $D_e$ unchanged. A quantum harmonic oscillator approximates the curve near its minimum but rises without limit, so that local approximation alone cannot supply a dissociation asymptote. A Morse potential supplies both a minimum and a finite separated-fragment limit.^[ref]

A heteronuclear diatomic molecule consisting of carbon and sulfur. Its permanent electric dipole moment permits an electric-dipole rotational spectrum. The linear rigid rotor model relates its rotational constant to its reduced mass and bond length.

A diatomic molecule contains two chemical atoms; hydrogen chloride and carbon monoxide are examples. Their fixed-separation moment of inertia is $I=\mu r^2$ in the linear rigid rotor model.

The ground-state dissociation energy starts at the lowest vibrational quantum state, with the lowest allowed rotational quantum state and the fragment states explicitly specified. For a diatomic molecule whose fragments are ground-state chemical atoms, omitting any additional rotational correction,where $E_{\mathrm{ZP}}$ is the vibrational zero-point energy measured upward from the well minimum. The initial molecule already has this energy, so less additional energy is needed to reach the same final limit.^[ref]

For fragments that are themselves molecules, include their zero-point energies too. Define $D_e$ as the electronic energy difference between separately relaxed products and reactant; then, with consistent rotational conventions,Thus “subtract the parent's zero-point energy” is sufficient for atomic products but generally incomplete for molecular products. An excited final quantum state raises its channel threshold by its excitation energy; that threshold does not become the ground-product $D_0$ just because a detector selects it.

A diatomic molecule whose two chemical atoms belong to the same chemical element.

The gaseous hydrogen halide consisting of one hydrogen atom and one bromine atom. Its molecular vibration can be studied through its vibrational fundamental and weak vibrational overtones. The aqueous acid and the isolated diatomic molecule should be distinguished when specifying a dissociation process.^[ref]

The diatomic molecule containing one hydrogen atom and one iodine atom. Its measured spectroscopic wavenumbers include anharmonic corrections beyond the two-term Morse oscillator model.^[ref]

Near a stable equilibrium bond length $r_e$, a Taylor expansion of a diatomic molecule's potential energy gives $U(r)\simeq U(r_e)+k(r-r_e{)}^2/2$, with $k=U^{\prime \prime }(r_e)>0$. Treat the internal stretch as a harmonic oscillator of reduced mass, initially omitting molecular rotation. If replacing an isotope leaves this local potential energy curvature unchanged, thenHere $c$ is the speed of light and ${\tilde{\nu }}_{\mathrm{harm}}$ a spectroscopic wavenumber; use consistent length units for $c$ and ${\tilde{\nu }}_{\mathrm{harm}}$. The quantum harmonic oscillator gives adjacent energy eigenvalues separated by $h\nu$, with $h$ the Planck constant. Consequently a vibrational photon absorption can probe the same frequency scale, subject to electric-dipole selection rules.

For an explicitly approximate isotope example, assign the atomic masses of ${}^1\mathrm{H}$, ${}^2\mathrm{H}$, ${}^{35}\mathrm{Cl}$, and ${}^{37}\mathrm{Cl}$ the mass-number values $1u,2u,35u,37u$, respectively, with $u$ the unified atomic mass unit. These are stipulated approximations to individual atomic masses, not measured masses of molecules or natural-abundance averages. They giveThe symbol $\mathrm{D}$ means deuterium, ${}^2\mathrm{H}$. Doubling the light partner's approximate atomic mass therefore produces a large predicted shift, whereas changing the already heavy chlorine partner by two mass units produces a much smaller one. A quantitative comparison with infrared spectroscopy requires actual isotope-specific atomic masses, fitted band parameters, and uncertainties, rather than identifying any arbitrary peak with ${\tilde{\nu }}_{\mathrm{harm}}$.

A linear molecule has collinear atomic nuclei in its idealized molecular geometry. Both a diatomic molecule and carbon dioxide can supply linear rigid rotor examples, although their electric-dipole selection rules need not give an observable permanent-electric dipole spectrum.

For a diatomic molecule, let $m_1,m_2$ be the two masses and $r$ their fixed separation. The distances from the centre of mass are $r_1=m_{2}r/(m_1+m_2)$ and $r_2=m_{1}r/(m_1+m_2)$. A perpendicular axis through the centre of mass therefore has moment of inertiaThe reduced mass $\mu$ here is not the electric dipole moment, for which we use $\mathbf{d}$. For a longer linear molecule, replace $\mu r^2$ by ${\sum }_i{m}_i{\rho }_i^2$, where ${\rho }_i$ is each distance along the axis from the centre of mass. In the point-mass idealization, the axial moment of inertia is zero; turning about that axis changes no nuclear positions, so no independent axial rotational degree of freedom is included.

For the field-free linear rigid rotor, assume a fixed internal quantum state with no electronic angular momentum or electronic spin, and omit couplings to nuclear spin. The rotational Hamiltonian is ${\hat{H}}_{\mathrm{rot}}={\hat{\mathbf{L}}}^2/(2I)$. Its orientation wavefunctions are spherical harmonics $Y_{JM}$, and its angular momentum eigenstates satisfywhere $J=0,1,2,\dots$ and $M=-J,-J+1,\dots ,J$. Each $E_J$ belongs to $2J+1$ angular momentum eigenstates distinguished by the magnetic quantum number $M$, an instance of degenerate energy levels. The Planck constant is $h=2\pi \hslash$; $B$ is the frequency-valued rotational constant. Notice that the energy eigenvalues scale as $J(J+1)$, not $J^2$.

Subtracting consecutive energy eigenvalues givesThus increasing level gaps produce a regularly spaced series of lines at $2B,4B,6B,\dots$, with spacing $2B$. Allowed positions do not guarantee that every line is visible: electric-dipole selection rules, occupation of the initial quantum state, and instrument response also matter.

Recovering molecular geometry from rotational constants is an inverse problem. In the synthetic assignment, $B=5\times 10^9\mathrm{Hz}$ gives $I=h/(8{\pi }^{2}B)\simeq 1.6784\times 10^{-45}\mathrm{kg}{\mathrm{m}}^2$. For a diatomic molecule with known masses, rotational constant explains how to infer an effective bond length. For a longer linear molecule, one $I$ cannot determine every bond length: several different arrangements can have the same moment of inertia.

For a fixed three-atom linear model with positions $0,a,a+b$ and positive masses $m_1,m_2,m_3$, subtracting the centre of mass givesThis is one equation for two unknown bond lengths. Even fixing the order of the chemical atoms leaves a family of solutions. Additional isotope substitutions change known masses and supply independent equations, provided the effective molecular geometry can be treated as shared. For nonlinear molecules, three principal moments of inertia give frequency-valued constants $A=h/(8{\pi }^2{I}_a)$, $B=h/(8{\pi }^2{I}_b)$, and $C=h/(8{\pi }^2{I}_c)$; a single evenly spaced comb is generally insufficient.

Vibrational averaging, unresolved hyperfine structure, internal motion, and centrifugal distortion limit structural inference. A ground-vibrational fitted rotational constant is not automatically the equilibrium constant. Ordinary field-free rotational spectroscopy also gives identical spectra for opposite enantiomers under the usual parity-symmetric molecular Hamiltonian; assigning composition and conformation does not by itself determine handedness. Molecular identification from a rich fingerprint can therefore succeed even when a unique full molecular geometry cannot be reconstructed.

Molecular vibration changes internal separations and bond angles, unlike the fixed internal molecular geometry of a rigid rotor. Its coupling to molecular rotation makes measured rotational constants depend on the vibrational quantum state.

A diatomic molecule has one internal stretching coordinate: its two atomic nuclei approach and separate about an equilibrium bond length. The diatomic bond stretch turns this motion into a particularly useful model. The local potential energy curvature supplies the restoring force, while the reduced mass accounts for both moving partners. Thus the same chemical bond can vibrate more slowly after an isotope substitution even when its electronic structure is nearly unchanged.

This separation makes infrared spectroscopy informative: transition frequencies constrain chemical bond stiffness, and isotope scaling of diatomic vibration tests the mass dependence. The University of Florida HCl and DCl FTIR experiment illustrates the measurement. Detectability also requires vibrational infrared activity; having vibrational energy eigenvalues alone does not guarantee an infrared absorption line.

A diatomic molecule has $3N-5=1$ vibrational normal mode: the bond stretch. For hydrogen chloride, opposite longitudinal displacements change the electric dipole moment, so the mode has vibrational infrared activity. The following original sketches show one phase of this motion and the three water modes; arrow lengths are schematic, not mass-weighted amplitudes.

BeLi has three valence electrons: two occupy a bonding sigma orbital and the third its higher partner. The elementary bond order is therefore $1/2$, with one unpaired electron: expect a relatively weak, paramagnetic, doublet diatomic molecule. Because Be attracts valence charge more strongly than Li, the bond is polar toward Be. This is a qualitative prediction, not a numerical dissociation energy.

This is the elementary picture of the lithium beryllium diatomic molecule.

Only ${}^1{\Delta }_g\to {}^1{\Delta }_u$ is electric-dipole allowed within the listed terms. For a diatomic molecule, the electric-dipole selection rule requires $\Delta S=0$, $g\leftrightarrow u$, and $\Delta \Lambda =0,\pm 1$. The lower singlet of part (a) is ${}^1{\Delta }_g$. Every candidate from the same configuration fails the inversion rule, while the ${}^1{\Sigma }_u^{\pm }$ terms of part (b) require $|\Delta \Lambda |=2$. Triplet destinations fail the spin rule. The surviving $\Delta \to \Delta$ transition uses the axial dipole component, so it is a parallel electronic transition. Weak spin-orbit or other intensity-borrowing transitions are outside this leading electric-dipole classification.

A diatomic molecule has one internal bond-stretching vibrational coordinate. Near its equilibrium separation, approximate the bond by a spring of stiffness $K$ and use the reduced mass $\mu =m_1{m}_2/(m_1+m_2)$. Its frequency is $\omega =\sqrt{K/\mu }$ and its quantum harmonic oscillator levels are ${ϵ}_r=(r+1/2)\hslash \omega$.

For independent molecules in thermal equilibrium, set $x=\hslash \omega /(k_{B}T)={\theta }_v/T$. The single-molecule canonical partition function is a geometric series:Using $U=-\partial \log (z_v^N)/\partial \beta$, where $\beta =1/(k_{B}T)$, givesPer mole replace $N{k}_B$ by $R$. The zero-point energy is present in $U_v$ but is temperature-independent and contributes nothing to the heat capacity.

For $T\ll {\theta }_v$, occupation of the first excited vibrational state is exponentially suppressed: $C_{V,v}\simeq N{k}_B{x}^2{e}^{-x}\to 0$. The vibrational degree of freedom is frozen out. For $T\gg {\theta }_v$, $C_{V,v}\to N{k}_B$. This recovers classical equipartition: the quadratic kinetic term and quadratic elastic term each contribute $k_B/2$ per molecule.

The vibrational heat capacity of a diatomic gas must be distinguished from the total heat capacity. In a dilute gas with rotations already active and vibration still frozen, translation contributes $3R/2$ and rotation $R$, so molar $C_V\simeq 5R/2$. Once vibration is active it adds $R$, giving $C_V\simeq 7R/2$. For an ideal gas of a fixed number of intact molecules, $C_P=C_V+R$, so the corresponding values are $7R/2$ and $9R/2$. Very low temperatures also freeze rotations, while sufficiently high temperatures bring anharmonicity, dissociation or electronic excitations; the harmonic plateau is restricted to the range where those effects are negligible.

The reduced mass is the effective mass for the two-particle relative coordinate of two bodies with constant positive masses $m_1,m_2$:Both bodies can respond to their mutual force, so their relative acceleration is greater than if one were fixed. Consequently $0<\mu <\min (m_1,m_2)$. This explains why using either individual mass, or their sum, generally gives the wrong internal oscillation rate. The same reduced mass describes a two-body problem, a free two-mass spring, and the approximate molecular vibration of a diatomic molecule.^[ref]

For equal masses $m_1=m_2=m$, $\mu =m/2$. Two $1\mathrm{kg}$ bodies therefore have $\mu =0.5\mathrm{kg}$: in their stationary centre of mass frame a relative speed $2m/s$ means individual speeds $1m/s$ and total kinetic energy $1\mathrm{J}$. With $m_1=1\mathrm{kg}$ and $m_2=100\mathrm{kg}$, instead $\mu =100/101\mathrm{kg}\approx 0.990099\mathrm{kg}$. At the same relative speed, the individual speeds are $200/101$ and $2/101m/s$, and the total kinetic energy is $200/101\mathrm{J}$. These are calculations for stipulated masses and velocities.

For fixed $m_1$ and $q=m_2/m_1>0$, $\mu /m_1=q/(1+q)$. The heavy-partner limit $q\to \infty$ gives $\mu \to m_1$ from below; the relative error incurred by replacing $\mu$ with $m_1$ is $(m_1-\mu )/\mu =1/q$. At $q=100$ this is $1\%$. Treating the heavy body as fixed is thus a controlled approximation to internal motion, rather than an exact consequence of its being heavier.

The angular momentum associated with end-over-end molecular rotation. In an open-shell diatomic molecule it is useful to distinguish $\mathbf{N}$ from the total $\mathbf{J}=\mathbf{N}+\mathbf{S}$, which also contains the electronic spin angular momentum.

Spin isomers of hydrogen are two forms of molecular hydrogen, ${}^1{\mathrm{H}}_2$, distinguished by the combined nuclear spin of their two protons. They have the same chemical formula and electronic ground state, but different allowed molecular rotation states. Parahydrogen has total nuclear spin zero and even rotational quantum number $J$; orthohydrogen has total nuclear spin one and odd $J$. The difference makes heat capacity a probe of fermionic exchange symmetry, and makes cooling history matter for a tank of molecular hydrogen.

Assume the electronic ground state $X{}^1{\Sigma }_g^{+}$ and the vibrational ground state of a free molecule. The electronic factor is symmetric under exchanging the two protons, and the vibrational factor depends only on their separation, so it is also symmetric. These assumptions concern the exchange of protons; the electronic wavefunction separately obeys fermionic exchange symmetry under exchange of electrons. Neglect electronic excitation, vibrational excitation, external fields, and small interactions involving nuclear spin when calculating rotational energy eigenvalues.

The two protons are identical particles with spin $1/2$. Their complete wavefunction must change sign under their particle exchange operator. Interchanging the two ends reverses the internuclear direction $\hat{\mathbf{r}}$, and the spherical harmonic rotational factor satisfiesThe antisymmetric spin-one-half singlet state therefore requires symmetric, even-$J$ rotation; the symmetric spin-one-half triplet state requires antisymmetric, odd-$J$ rotation. This is exchange symmetry of two identical spin-one-half fermions applied inside a diatomic molecule, not an extra rotational force. A linear rigid rotor may omit spin-dependent splittings while still requiring this restriction on its allowed quantum states.

Take an invented diatomic molecule with a Morse potential, setting the separated-atom energy to zero:Here $a$ is an inverse length, $r_e$ the equilibrium bond length, and ${\omega }_e=a\sqrt{2D_e/\mu }$, with reduced mass $\mu$. These stipulated parameters describe a simple model, not a measured chemical substance. For the one-dimensional Morse oscillator, energy eigenvalues above the minimum areConsequently,The harmonic estimate $A/2=0.1000\mathrm{eV}$ is slightly larger than this model's exact zero-point energy. The physical bound-level restriction matters: the quadratic expression is not an infinite ladder of vibrational quantum states.

For a separate thermal illustration using this same stipulated $D_0$, assume an ideal gas, fully active molecular rotation of the parent, negligible thermal molecular vibration, and negligible electronic excitation for all species. Each monatomic product contributes $\delta h=5RT/2$; the diatomic molecule contributes $7RT/2$. Thus the net increment is $3RT/2$, with molar gas constant $R$. At $T=300\mathrm{K}$,This last result follows from the stated thermal model, not a universal correction for every chemical bond. Frozen rotations, accessible excited states, different fragment composition, or a condensed phase change the thermal increments.