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An isolated boron trifluoride molecule has trigonal planar molecular geometry and the D3h molecular point group. Its three equal fluorine contributions give $I_{\parallel }=3m_{\mathrm{F}}{R}^2$ and $I_{\perp }=3m_{\mathrm{F}}{R}^2/2$ in the rigid rotor approximation.

Real chemical bonds can stretch as molecular rotation increases. The resulting larger moment of inertia lowers the energy eigenvalues relative to a rigid rotor. Within one vibrational quantum state, a leading effective correction isBoth $B_v$ and $D_v$ are in frequency units here; $v$ labels the vibrational quantum state. For positive $D_v$, the adjacent-line spacing becomesso the departures grow with $J$. Molecular vibration also changes the effective rotational constant: even the lowest vibrational quantum state does not have motionless atomic nuclei.^[ref]

The quartic correction is a limited expansion, not a model to extrapolate to arbitrary $J$. It is small relative to the rigid contribution only while $(D_v/B_v)J(J+1)\ll 1$; eventually higher terms, changed internal motion, or chemical bond breaking matter. In the synthetic plot, the largest upper-state label is $J=5$, giving a stipulated correction ratio of $0.03$. No universal maximum valid $J$ follows without a particular molecule and precision target.

The quantum numbers also have limits of meaning. Here $J$ describes spin-free rotation. In a molecule with electronic spin, rotation alone is commonly labelled $N$, while $J$ includes electronic angular momentum; including nuclear spin introduces additional total-angular momentum labels and resolved splittings. An applied electric field can mix different field-free $J$ states through the Stark effect. For identical atomic nuclei, the Pauli exclusion principle can restrict combinations of rotational and nuclear-spin states. The simple $2J+1$ degeneracy and every-integer-$J$ ladder therefore describe the stated idealization, not all observed sublevels.^[ref]

A permanent electric dipole moment gives evidence of separated positive and negative electric charge in an overall neutral molecule. This is a more direct probe of chemical polarity than an electronegativity table. For hydrogen chloride, NIST lists about $1.109$ debye for the lowest vibrational state and $1.093$ debye for the equilibrium value. They describe different averaging conventions, not two contradictory measurements. The NIST WebBook traces the vibrational value $1.1085$ debye to molecular-beam electric-resonance measurements in 1970.^[ref][ref]

How does molecular beam electric resonance reveal an electric dipole moment? A source and apertures form a beam of molecules in vacuum. A first nonuniform electric field, A, selects quantum states through their different Stark effects. In a central region, C, electrodes supply a nearly uniform electric field and applied radiation drives transitions at selected frequencies. A second nonuniform electric field, B, analyzes the outgoing quantum states; a detector measures the rate of arriving molecules. In a transmission-loss arrangement, resonant radiation transfers some molecules to a quantum state rejected by B, so the measured arrival rate dips. The observable is a change in detected molecules as the radiation frequency is scanned, rather than light absorption itself. This general apparatus description is documented in a later experimental thesis; it is not a reconstruction of the exact 1970 HCl instrument.^[ref]

The inference comes from how transition frequencies change with the known electric field. The interaction $-\mathbf{\mu }\cdot \mathbf{E}$ couples the electric dipole moment to the electric field; fitting the quantum mechanics of the relevant rotational states extracts its magnitude. A free rotating molecule need not show a first-order shift proportional to $|\mathbf{E}|$: the weak-field Stark effect can be quadratic. The diagram uses a simplified rigid rotor, with $B$ an energy constant, to illustrate this behavior. It omits the fine details of actual spectra.

Useful controls include scans with the radiation off to measure the baseline arrival rate, repeated scans at several calibrated electric fields, and agreement between independently assigned transitions. These are methodological checks, not claims about undocumented steps of the 1970 experiment. Field calibration and state assignment limit the inferred electric dipole moment. The published 1970 abstract confirms analysis of HCl and DCl beam-resonance spectra and combination with infrared spectroscopy intensities to obtain a distance-dependent electric dipole moment; its full apparatus and calibration record were unavailable here.^[ref]

Even an accurately measured electric dipole moment is not an absolute partial atomic charge. Replacing a hydrogen chloride molecule by two point electric charges $\pm q$ separated by bond length $r$ gives $|\mu |=|q|r$. Using $1.109$ debye and $r=1.27455\times 10^{-10}\mathrm{m}$ gives $|q|\approx 0.181$ times the elementary electric charge. This is an effective two-point model, combining a vibrationally averaged electric dipole moment with an equilibrium bond length, not a unique partition of the actual electron density or a measured “18.1 percent ionic bond”.^[ref]

For a molecule with several chemical bonds, molecular geometry can cancel opposing contributions to the total electric dipole moment. Carbon dioxide illustrates why an overall nonpolar molecule need not have unpolarized individual chemical bonds. An electronegativity comparison helps anticipate local bond polarization; the measured molecular electric dipole moment tests the entire electric charge distribution.

The apparatus and dipole inference are described under dipole evidence for bond polarization. The rigid rotor supplies a general starting model for rotational quantum states; rotational spectroscopy measures their transition frequencies, while molecular beam electric resonance can select and analyze states through their Stark effects.

An actual implementation gives substance to the apparatus description above. In Johannes Jacobus van Vaals's experiments, heated samples expanded with argon through a nozzle, then molecular beam skimmers formed the molecular beam in vacuum. Nonuniform electric fields selected and analyzed quantum states; parallel plates housed the microwave interaction. A surface ionization detector supplied electric current to an electronic amplifier and a lock-in amplifier. Resonant transfer into a rejected quantum state reduced the detected flux. Phase locking stabilized the microwave frequencies; an absorber suppressed reflections that could distort spectral lines. The thesis reports preliminary electric dipole moment components for gaseous potassium cyanide from measurements in nonzero electric fields. Such inference uses the Stark effect, rather than measuring electronegativity. The source notes restrictions from electric-dipole selection rules and molecular beam transit time; its preliminary dipole report is not a full calibration record. These are documented features of this apparatus, not proposed controls for the earlier hydrogen chloride experiment.^[ref]

Molecular rotation changes a molecule's orientation. Its rotational kinetic energy depends on the arrangement and masses of its chemical atoms. A rigid rotor freezes internal distances to isolate this motion; rotational spectroscopy tests the resulting energy eigenvalues through transitions between quantum states. The molecular beam electric resonance experiment provides a complementary probe of the coupling to an electric field.

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.

Assume a single-temperature optically thin local thermodynamic equilibrium population, negligible line absorption/stimulated corrections, and that the measured line luminosity is from the remnant's CO. Each spontaneous $J=1\to 0$ decay emits $h{\nu }_{10}$, givingFor a rigid rotor, $E_J/k_B=T_{r}J(J+1)$ with $T_r=h{\nu }_{10}/(2k_B)=2.767$ K. The degeneracy is $g_J=2J+1$, and the rotational partition function isAt 20 K the direct sum gives $Z\simeq 7.57$, $f_1=0.3005$; the approximation $Z\simeq T/T_r+1/3$ is already adequate. Thus, with molecular mass $m_{\mathrm{CO}}\simeq 28m_u$,The remnant age does not enter this optically thin population count. It becomes relevant when estimating emitting volume, density, expansion and optical depth to test the assumptions.

The expanded feature resolves the Q branch: each line has $\Delta J=0$ but starts from a different lower rotational level. In the simple rigid rotor band model,Because the upper rotational constant is larger, the lines extend towards higher wavenumber. Their adjacent spacings areso they crowd together at the low-frequency edge and become increasingly separated. A strictly resolved bending-band analysis may distinguish the upper parity component sampled by Q lines from that sampled by P/R lines; it is not necessary to impose identical high-resolution constants on every branch in this sketch.

The intensity envelope reflects rotational state populations and the rotational transition strength. For this perpendicular Q branch, the usual rotational line-strength factor gives the approximate dependenceAt small $J$, the increasing magnetic degeneracy makes the line intensity rise; at large $J$, Boltzmann suppression dominates. The maximum is therefore at an intermediate $J$. This is a population envelope across distinct transitions, rather than one intrinsically broad vibrational line.

For independent internal rotational states, the rotational factor in the many-molecule partition function is $q_{\mathrm{rot}}^N$. The indistinguishability factor has already been included in the translational part and is not applied a second time to the internal factor. ThusA nonlinear rigid rotor has three rotational quadratic degrees of freedom; in the high-temperature limit, $q_{\mathrm{rot}}\propto T^{3/2}$. ThereforeThis is the rotational entropy contribution, with the appropriate rotational symmetry number already inside $q_{\mathrm{rot}}$. It has no ideal-gas pressure dependence.

Chloroform has the C3v molecular point group. Its three identical ${}^{35}\mathrm{Cl}$ atoms are interchanged by three proper rotations, so $\sigma =3$. The three mirror operations do not increase the rotational symmetry number to six. Isotopic identity matters for this count.

Using the supplied wavenumber form of $k_{\mathrm{B}}$,All are far below $298\mathrm{K}$, justifying the high-temperature nonlinear rigid rotor expression. ThenandThe calculation treats the molecule as rigid and separates electronic, vibrational and nuclear-spin contributions from the requested rotational term.

The gross requirement is an anisotropic polarizability tensor, so that changing molecular orientation changes the induced response. A permanent electric dipole moment is not required; a homonuclear diatomic can therefore have a rotational Raman scattering spectrum.

For the inelastic lines of a linear rigid rotor, the detailed rule is $\Delta J=\pm 2$, with $J\ge 0$. The $\Delta J=0$ contribution is unshifted Rayleigh scattering, not a rotational Stokes Raman scattering line. Stokes Raman scattering has $\Delta J=+2$ because the molecule gains rotational energy.

A useful vibration-rotation expression, with energies measured from the potential minimum, isHere $v=0,1,\dots$ is the vibrational quantum number, $J=0,1,\dots$ the rotational angular-momentum quantum number, ${\tilde{\nu }}_e$ the harmonic spectroscopic wavenumber, $x_e$ a dimensionless anharmonic coefficient, and $B_v$ the vibrational-state-dependent rotational constant. $B_e$ is its equilibrium value and ${\alpha }_e$ describes vibration-rotation coupling. All wavenumber constants use the same inverse-length units; $h$ is the Planck constant and $c$ the speed of light.

The elementary harmonic rigid rotor result follows by setting $x_e={\alpha }_e=0$, with ${\tilde{\nu }}_e=(2\pi c{)}^{-1}\sqrt{k/\mu }$ and $B_e=h/(8{\pi }^{2}cI)$, where $I=\mu r_e^2$. The molecule's electronic energy would add a common constant.

For a linear rigid rotor in the high-temperature regime, $q_{\mathrm{rot}}\propto T$. Its molar rotational entropy isThe rotational symmetry number must already be included in $q_{\mathrm{rot}}$.

Write $M=m_A+m_B$ and $\mu =m_A{m}_B/M$. An atom has no rotational or vibrational internal factor under the stated electronic assumption. The transition-state moment of inertia is $I=\mu R^2$, so the high-temperature rigid rotor result isThe translational factors therefore giveSubstituting into the transition state theory expression and simplifying,For distinct atoms, $\sigma =1$. This is a number-density rate coefficient, with units of volume per time. A rate law in molar concentrations uses $N_A{k}_{2nd}$. The derivation assumes a classical high-temperature rotor at the dividing surface and a transmission coefficient of one.

Exchange constraint. Exchange of two identical particles cannot change physical predictions. For the usual three-dimensional bosonic or fermionic state space, the exchange operator obeys $P_{12}^2=1$, so its allowed eigenvalues are $\pm 1$. bosons use the symmetric overall wavefunction, $P_{12}\Psi =+\Psi$; fermions use the antisymmetric overall wavefunction, $P_{12}\Psi =-\Psi$. Exchange includes the spatial coordinates and all internal labels. The identity $P_{12}^2=1$ alone does not prove which particle species has which statistics; that assignment is the Spin-statistics theorem.

Factored fermion states. If $\Psi ={\psi }_{\mathrm{space}}{\chi }_{\mathrm{spin}}$ and both factors have definite exchange symmetry, their exchange signs must multiply to $-1$. Thus a symmetric spatial factor requires an antisymmetric spin factor, and an antisymmetric spatial factor requires a symmetric spin factor. For two spin-one-half particles, the former spin factor is the singlet $(|\uparrow \downarrow ⟩-|\downarrow \uparrow ⟩)/\sqrt{2}$; the latter factors are the three triplet states. A more general entangled spin-space state need not factor, but must still be antisymmetric overall.

hydrogen spin isomers. The two protons are identical spin-one-half fermions. parahydrogen has total nuclear spin $I_{\mathrm{n}}=0$, the antisymmetric singlet; orthohydrogen has $I_{\mathrm{n}}=1$, the symmetric triplet. Here $I_{\mathrm{n}}$ is a spin quantum number, distinct from the moment of inertia used below.

Rotational restriction. In the electronic and vibrational ground state, these two factors are symmetric under nuclear exchange. Exchanging the protons reverses the internuclear axis, $(\theta ,\varphi )\mapsto (\pi -\theta ,\varphi +\pi )$. The spherical harmonic rotational state contributes the exchange sign $(-1{)}^{\ell }$. Overall antisymmetry therefore requiresHence parahydrogen has even $\ell$; orthohydrogen has odd $\ell$. The $\ell$ here labels molecular rotation, rather than the angular momentum obtained by also coupling the nuclear spins.

Conversion heat. The ortho-para conversion heat follows from the linear rigid rotor approximation, $E_{\ell }={\hslash }^2\ell (\ell +1)/(2I)$ with $I=\mu r^2\simeq m_p{r}^2/2$. Thus conversion of one molecule from $\ell =1$ to $\ell =0$ releasesNeglecting the small electron contribution to the molecular mass, a sample of mass $M$ contains $M/(2m_p)$ molecules. Only the initial orthohydrogen fraction releases this energy, givingThe estimate uses $r=74$ pm, $m_p\simeq 1.67262\times 10^{-27}$ kg and $\hslash \simeq 1.05457\times 10^{-34}$ J s. Handbook-level rounding and the rigid rotor approximation justify about three significant figures, not a high-precision thermochemical prediction.^[ref]

Storage consequence. The internal conversion is a delayed heat source even in a well insulated container, causing warming or additional evaporation of the liquid. Insulation reduces external heat transfer but does not remove this source. Converting to predominantly parahydrogen during liquefaction, while removing the released heat, reduces the subsequent storage heat load. No evaporation fraction can be calculated without a latent heat and a model for how the heat is removed.

The rotational kinetic energy is $K=L^2/(2I)$. For a three-dimensional linear rigid rotor, angular momentum operator replaces the squared magnitude by ${\hslash }^{2}l(l+1)$, where $l=0,1,2,\dots$. HenceThe three lowest distinct energy levels are thereforeTheir magnetic quantum degeneracy values are $1,3,5$. A single moment of inertia specifies this rigid rotor model; an arbitrary asymmetric body would require its other principal moments.

For a reversible adiabatic process of an ideal gas, $\delta Q=0$ and $dU=-pdV$. At fixed composition and constant sample heat capacity at constant volume $C_V$,Mayer's relation gives $n_{\mathrm{mol}}R=C_p-C_V$, so $T{V}^{\gamma -1}$ is constant andThe finite-path power law requires approximately constant heat capacities as well as reversibility. When rotational or vibrational populations change appreciably, use $\int C_V(T)dT/T+n_{\mathrm{mol}}R\ln V=\mathrm{constant}$ instead; the local isentropic pressure exponent remains $\gamma (T)$.

For an intact heteronuclear diatomic gas, quantum excitation progressively activates molecular motion. With rotational and vibrational modes frozen, the three translational degrees of freedom give $C_{V,\mathrm{m}}=3R/2$ and $\gamma =5/3$. Two active rotations add $R$, giving $C_{V,\mathrm{m}}=5R/2$ and $\gamma =7/5$. One active harmonic vibration contributes a further $R$, through kinetic and potential terms, giving $C_{V,\mathrm{m}}=7R/2$ and $\gamma =9/7$.

For molecular moment of inertia $I={\mu }_{\mathrm{red}}{r}_e^2$, the rigid rotor levels are $E_J={\hslash }^{2}J(J+1)/(2I)$, so the rotational excitation scale isThe first gap corresponds to $2{\Theta }_{\mathrm{rot}}$, so the crossover occurs at temperatures of that order. For bond force constant $k_s$ and reduced molecular mass ${\mu }_{\mathrm{red}}$, the vibrational scale isTypically ${\Theta }_{\mathrm{vib}}\gg {\Theta }_{\mathrm{rot}}$. Real gases may condense at the low-temperature end or dissociate and excite electronic states at the high-temperature end; the three plateau values describe the ideal intact-molecule model.

For the piston apparatus, mechanical equilibrium requires $p_{0}A=p_{\mathrm{air}}A+mg$, hence $p_0=p_{\mathrm{air}}+mg/A$. Let $x$ be a small upward displacement, so $V=V_0+Ax$. The local isentropic compressibility givesConsequently the adiabatic gas-spring piston period isThis neglects friction, gas inertia, leakage and dissipation, with internal equilibration much faster than the oscillation and external thermal equilibration much slower. The small oscillation probes $\gamma$ locally; it need not keep $\gamma$ constant over a large temperature interval.

With $p_{\mathrm{air}}=101325\mathrm{Pa}$, the supplied dimensions give $p_0=101351.4\mathrm{Pa}$ and $B=2.20566\times 10^{-5}{\mathrm{s}}^2$. ThusHelium is close to the monatomic ideal gas value $5/3$; NO is close to the diatomic gas value $7/5$, with rotations active and vibration largely frozen. The numbers are approximate experimental inferences, rather than exact rational values.

A rigid rotor restricted to one angular coordinate $\varphi$. Single-valued spatial states $e^{im\varphi }$ have $m\in \mathbb{Z}$ and energy $E_m={\hslash }^2{m}^2/(2I)$, with moment of inertia $I$. The zero-angular-momentum level has state degeneracy one; every nonzero $|m|$ has two directions. For labelled molecular ends its high-temperature rotational partition function is $\sqrt{\pi T/{\theta }_{\mathrm{rot}}}$, where ${\theta }_{\mathrm{rot}}={\hslash }^2/(2I{k}_{\mathrm{B}})$. Equivalent orientations and nuclear spin statistics require a further molecular counting convention. One quadratic angular momentum gives rotational heat capacity $k_{\mathrm{B}}/2$ per molecule, rather than $k_{\mathrm{B}}$ for an unrestricted linear rigid rotor.

A rigid rotor models rotational motion with fixed internal separations. Its use in the synthetic Stark effect figure omits vibrational motion and additional interactions in real molecules.

Rigidity fixes the molecular geometry, not the orientation, and does not require a spherical mass distribution. Remove motion of the centre of mass, choose principal axes of the inertia tensor, and write the classical rotational kinetic energy asHere $I_a,I_b,I_c$ are the principal moments of inertia, ${\omega }_a,{\omega }_b,{\omega }_c$ components of angular velocity, and $L_a,L_b,L_c$ components of angular momentum. Quantization replaces these last components by angular momentum operators. Different principal moments of inertia produce different spectra: the single-constant linear rigid rotor is a special case, not a formula for every molecule.^[ref]

The entropy contribution from populated molecular rotational states. A nonlinear rigid rotor has $q_{\mathrm{rot}}\propto T^{3/2}$ at high temperature, giving $S_{\mathrm{rot},\mathrm{m}}=R(\ln q_{\mathrm{rot}}+3/2)$. Its rotational symmetry number removes redundant orientations.