Past exam of the chemistry A course of the University of Cambridge / 2021 / Paper 1 / A3

Place the carbon and sulfur nuclei on a line separated by $r$, with masses $m_{\mathrm{C}}$ and $m_{\mathrm{S}}$. The center of mass lies between them and closer to the heavier sulfur nucleus. In the linear rigid rotor approximation, $r$ remains fixed while the molecule rotates.

Measure $x$ from the carbon nucleus toward the sulfur nucleus. The definition of the center of mass givesTherefore the requested distance isThe other distance is $r_2=r-r_1=m_{\mathrm{C}}r/(m_{\mathrm{C}}+m_{\mathrm{S}})$, consistent with $m_{\mathrm{C}}{r}_1=m_{\mathrm{S}}{r}_2$.

For an axis through the center of mass perpendicular to the bond, the point-mass moment of inertia isThe prefactor is the reduced mass $\mu$; this is the standard reduction of a two-body linear rigid rotor to $I=\mu r^2$.

Using the stipulated integer isotope masses givesThe rotational constant in wavenumber units isHere $c$ must be expressed in $\mathrm{cm}{\mathrm{s}}^{-1}$ when $I$ is in SI units. Equivalently calculate $B=h/(8{\pi }^{2}I)$ in $\mathrm{Hz}$ first, then divide by $c$ in $\mathrm{cm}{\mathrm{s}}^{-1}$.

The rotational state populations start at $N_0/N_0=1$. Their $2J+1$ statistical weight factor initially favors higher $J$, while the Boltzmann factor eventually suppresses them, producing one maximum followed by a tail tending to zero. The physically allowed values occur at integer $J$; a continuous curve only guides the eye.

Set ${\theta }_{\mathrm{rot}}=hc\tilde{B}/k_{\mathrm{B}}$, the rotational temperature. For a continuous nonnegative variable $J$,Equating this derivative to zero gives the most populated rotational levelThe second derivative is $-4/(2J+1{)}^2-2{\theta }_{\mathrm{rot}}/T<0$, so this is a maximum when $J_{\ast }\ge 0$. At lower temperatures the constrained continuous maximum is $J=0$. For the physical integer maximum compare the nearby allowed levels explicitly.

The calculated rotational constant gives ${\theta }_{\mathrm{rot}}=1.18721\mathrm{K}$. At room temperature, taken as $300\mathrm{K}$, the continuous maximum is $J_{\ast }=10.7404$; comparing the adjacent integer populations gives $J_{\max }=11$. Taking $298.15\mathrm{K}$ instead gives $J_{\ast }=10.7057$ and the same integer answer.

At $8.7\mathrm{K}$, $J_{\ast }=1.41417$. In particular,and $N_1>N_0$, so $J_{\max }=1$. The integer comparison avoids treating a noninteger angular momentum quantum number as a physical state.

Low rotational energy spacings can be excited in cold interstellar gas when electronic and most vibrational excitations are frozen out. A polar molecule such as carbon monosulfide has allowed electric dipole transition lines whose frequencies depend on its moment of inertia, providing a distinctive pattern for rotational spectral line assignment. Several matching spectral lines and their isotope shifts give a much stronger identification than one coincidence. The method is especially effective for polar species; a molecule without a permanent electric dipole moment lacks an ordinary pure electric-dipole rotational spectrum.