Effects of Nuclear Spin: Rotational Partition Function and Degeneracies

Abstract

In previous chapters on diatomic and polyatomic spectra, we discussed quantized angular momentum as it relates to a rotating molecule and its effect on Boltzmann statistics. Rotational angular momentum is, in fact, only one kind of angular momentum that must be considered when evaluating real molecules and their spectra. For example, electrons also have quantized angular momentum as they orbit nuclei (called orbital angular momentum, which we will discuss further in Chaps. 9 and 10).

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In previous chapters on diatomic and polyatomic spectra, we discussed quantized angular momentum as it relates to a rotating molecule and its effect on Boltzmann statistics. Rotational angular momentum is, in fact, only one kind of angular momentum that must be considered when evaluating real molecules and their spectra. For example, electrons also have quantized angular momentum as they orbit nuclei (called orbital angular momentum, which we will discuss further in Chaps. 9 and 10). Additionally, individual subatomic particles like nuclei and electrons have their own intrinsic, quantized, angular momentum, analogous to the angular momentum of a spinning top or planet; hence, the common name for this property is spin. Because of the added complexity that nuclear spin can introduce, we have not discussed its effects up to this point; however, nuclear spin can, in symmetric molecules, change the degeneracy of some states and thus affect molecular statistics. This chapter will introduce the effects of nuclear spin, when they need to be considered, and how one can incorporate them into the statistics of linear and nonlinear molecules.

5.1 Introduction

Recall that the population fraction, F _i , of a given molecular energy level E _i is

$$\displaystyle{ F_{i} = \frac{g_{i}\exp \left (- \frac{\epsilon _{i}} {kT}\right )} {Q}, }$$

(5.1)

where the partition function, Q, is given by the summation over all energy levels

$$\displaystyle{ Q =\sum _{i}g_{i}\exp \left (- \frac{\epsilon _{i}} {kT}\right ) }$$

(5.2)

When evaluating partition functions and energy level degeneracies, g _i , for asymmetric molecules, e.g. heteronuclear diatomics like NO and polyatomics like HCN, we usually consider only three internal energy modes: electronic, vibrational, and rotational. Thus, the partition function is

$$\displaystyle{ Q = Q_{\mathrm{rot}}Q_{\mathrm{vib}}Q_{\mathrm{elec}} }$$

(5.3)

We have thus far neglected nuclear spin (even though it exists) for such cases because there is no coupling between nuclear spin and rotation and because the nuclear terms cancel in expressions for population fraction in specific rotational states.

However, in molecules with certain symmetric placement of equivalent nuclei, e.g. O₂ and NH₃, we must consider the total partition function, including the nuclear terms:

$$\displaystyle{ Q = Q_{\mathrm{rot}}Q_{\mathrm{vib}}Q_{\mathrm{elec}}Q_{\mathrm{nuc}} }$$

(5.4)

Here again, we have invoked the Born–Oppenheimer approximation: the energy associated with each mode is independent or separable. As outlined below, we must include the nuclear terms for symmetric molecules because of their coupling to the rotational levels. In these cases, we use an effective rotational partition function defined by

$$\displaystyle{ Q'_{\mathrm{rot}} = Q_{\mathrm{rot}}Q_{\mathrm{nuc}}, }$$

(5.5)

and likewise, an effective rotational degeneracy

$$\displaystyle{ g'_{\mathrm{rot}} = g_{\mathrm{rot}}g_{\mathrm{nuc}}. }$$

(5.6)

5.2 Nuclear Spin and Symmetry

Since nuclear energies are quite large for most if not all practical conditions (E _nuc ≫ kT for excited nuclear states), we are only interested in the lowest (ground) nuclear energy level. It follows that

$$\displaystyle{ Q_{\mathrm{nuc}} = g_{\mathrm{nuc}} }$$

(5.7)

since $\exp (-E/kT) \approx 0$ for all but the ground level. In general, the ground nuclear energy level has a degeneracy associated with the spin quantum number, I (not to be confused with the moment of inertia I _A , I _B , etc.) of the nuclei (Table 5.1). For a single nucleus, the number of degenerate spin states is given by

$$\displaystyle{ g_{\mathrm{nuc}} = 2I + 1 }$$

(5.8)

For example, a¹⁴N nucleus has I = 1 and a corresponding nuclear spin degeneracy g _nuc = 3. For a molecule made up of L atoms, the total nuclear partition function is formed by the product of the terms for the individual nuclei

$$\displaystyle{ Q_{\mathrm{nuc}} =\prod _{ n=1}^{L}\left (2I_{ n} + 1\right ) }$$

(5.9)

where I _n is the spin of the nth nucleus.

Table 5.1 Spin of several nuclei

Nuclear spin states and rotational states couple through the symmetry properties of their respective wave functions. A wave function that describes any state composed of two or more identical particles is either symmetric or antisymmetric with respect to those particles. For rotational states, the symmetry property is determined by the resulting wave function if a pair of identical nuclei were interchanged. For symmetric states, the sign of the wave function remains unchanged [+] upon interchange of the nuclei; for antisymmetric states, the sign changes [−]. For example, if a molecule had two nuclei, x and y, its rotational wave function is symmetric if $\Psi _{\mathrm{rot}}(x,y) = \Psi _{\mathrm{rot}}(y,x)$; it would be antisymmetric if $\Psi _{\mathrm{rot}}(x,y) = -\Psi _{\mathrm{rot}}(y,x)$.

In rotationally symmetric molecules like O₂ and NH₃, nuclear spin states (with their own symmetry properties) can only pair with rotational states having a compatible symmetry, either symmetric or antisymmetric. Which one is compatible depends on the nature of nuclei (i.e., whether the system is a boson or fermion).

The rules governing symmetry compatibility depend on which statistics apply to the nuclear system. For nuclear systems that follow Fermi–Dirac statistics (fermions), the total molecular wave function must be antisymmetric. This symmetry is made up of a combination of a symmetric and antisymmetric states (an overall sign change comes from multiplying a function having no sign change with a function having a sign change, i.e. $[+] \times [-] = [-]$). For example, antisymmetric nuclear spin states combine (only) with symmetric rotational levels to produce an overall antisymmetric state. For Bose–Einstein nuclear systems (bosons), the overall wave function must be symmetric. Thus symmetric states combine only with symmetric states, and antisymmetric with antisymmetric ($[+] \times [+] = [-] \times [-] = [+]$).

For a single nucleus, Fermi–Dirac statistics are associated with nuclei having half-integral spins ($I = 1/2,3/2,\ldots$) and Bose–Einstein statistics hold for nuclei with integral spins (0, 1, 2, …). The important question now becomes, what is the symmetry character of the rotational levels? The most powerful tool for attacking symmetry questions like these is called group theory. Since group theory requires more explanation than can be given here, results for some of the most important (and common) molecular configurations are presented below.

5.3 Case I: Linear Molecules

5.3.1 Asymmetric (e.g., CO and N ₂ O)

For linear molecules without symmetry, there is no coupling between nuclear spin and molecular rotation. Thus, the effective rotational expressions are simply the standard expressions multiplied by a constant term,

$$\displaystyle{ Q'_{\mathrm{rot}} = \frac{T} {\sigma \theta _{r}} \prod _{n=1}^{L}(2I_{ n} + 1) }$$

(5.10)

and

$$\displaystyle{ g'_{\mathrm{rot}} = (2J + 1)\prod _{n=1}^{L}(2I_{ n} + 1) }$$

(5.11)

For asymmetric linear molecules, the symmetry factor σ = 1; this factor represents the number of ways that the molecule may be rotated into an identical configuration. As stated initially, the nuclear spin terms can be omitted from the partition functions and degeneracies for asymmetric molecules since they cancel in expressions for the population fraction.

Consider CO :

$$\displaystyle\begin{array}{rcl} Q'_{\mathrm{rot}}& =& \frac{T} {\theta _{r}} \prod _{n=1}^{2}(2I_{ n} + 1) = \frac{T} {\theta _{r}} (2I_{\mathrm{C}} + 1)(2I_{\mathrm{O}} + 1) = \frac{T} {\theta _{r}} {}\\ g'_{\mathrm{rot}}& =& (2J + 1)(2I_{\mathrm{C}} + 1)(2I_{\mathrm{O}} + 1) = 2J + 1 {}\\ \end{array}$$

Thus the fractional population in J is, as we found for an asymmetric rigid rotor without consideration of nuclear spin,

$$\displaystyle{\frac{N_{J}} {N} = \frac{(2J + 1)\exp (-F(J)hc/kT)} {T/\theta _{r}} }$$

Consider N ₂ O :

$$\displaystyle\begin{array}{rcl} Q'_{\mathrm{rot}}& =& \frac{T} {\theta _{r}} \prod _{n=1}^{3}(2I_{ n} + 1) = \frac{T} {\theta _{r}} (2I_{\mathrm{N}} + 1)^{2}(2I_{\mathrm{ O}} + 1) = 9\frac{T} {\theta _{r}} {}\\ g'_{\mathrm{rot}}& =& (2J + 1)(2I_{\mathrm{N}} + 1)^{2}(2I_{\mathrm{ O}} + 1) = 9(2J + 1) {}\\ \end{array}$$

Both Q _rot and g _rot are larger by a factor of 9 owing to nuclear spin, but these terms cancel in forming the Boltzmann fraction,

$$\displaystyle{\frac{N_{J}} {N} = \frac{(2J + 1)\exp (-F(J)hc/kT)} {T/\theta _{r}}.}$$

Recall that the term 2J + 1 is the degeneracy for energy level F(J) associated with the number of possible directions (orientations) of the angular momentum vector of the molecular rotation. In the presence of an applied electric or magnetic field, these 2J + 1 states may be slightly separated in energy. In the absence of such fields, the states all have the same energy, but the degeneracy remains.

5.3.2 Symmetric (e.g., O₂, CO₂, and C₂H₂)

Now the rotational partition function contains a non-unity value for the symmetry factor, σ. For linear and symmetric molecules, σ = 2 because these molecules are indistinguishable upon rotation about the B axis.

$$\displaystyle{ Q'_{\mathrm{rot}} = \frac{1} {2} \frac{T} {\theta _{r}} \prod _{n=1}^{L}(2I_{ n} + 1) }$$

(5.12)

The degeneracy is determined as follows. The overall symmetry required by the nuclear statistics must first be determined. Then we multiply the degeneracy (2J + 1) of a rotational level, with a given symmetry, by the degeneracy of the appropriate nuclear spin state.

In a linear molecule, the nuclear statistics of the overall nuclear system are controlled by the number of fermions (half-integral spin nuclei) on one side of the center of the molecule (if a central nucleus exists, it is ignored). For an odd number of fermions, the system behaves according to Fermi–Dirac statistics and the overall symmetry is antisymmetric. Conversely, for an even number of fermions (including zero or none), Bose–Einstein statistics and an overall symmetric wave function are required.

The degeneracy of the symmetric nuclear spin states is given by

$$\displaystyle{ g_{\mathrm{nuc,symm}} = \frac{(2I_{C} + 1)} {2} \left [\prod _{m=1}^{M}(2I_{ m} + 1)^{2} +\prod _{ m=1}^{M}(2I_{ m} + 1)\right ] }$$

(5.13)

and the degeneracy of the antisymmetric states is given by

$$\displaystyle{ g_{\mathrm{nuc,asymm}} = \frac{(2I_{C} + 1)} {2} \left [\prod _{m=1}^{M}(2I_{ m} + 1)^{2} -\prod _{ m=1}^{M}(2I_{ m} + 1)\right ] }$$

(5.14)

where I _m is the nuclear spin of the mth nucleus on one side of the center of the molecule, $M = (L - 1)/2$, and I _C is the nuclear spin of the central nucleus if one is present (not to be confused with either the nuclear spin of a carbon atom, I _C, or the moment of inertia about the C-axis of a nonlinear molecule, I _C ). If no central nucleus exists (e.g., in a homonuclear diatomic), the term (2I _C + 1) is replaced by unity and $M = L/2$ [see Eq. (5.16)]. For a symmetric molecule made up of nuclei with spin I = 0, it can be seen that only symmetric spin states exist.

The symmetry of a rotational state depends on its rotational quantum number as well as the structure of the electronic manifold within which it exists (Table 5.3). For rotational levels in electronic manifolds designated $\Sigma _{g}^{+}$ and $\Sigma _{u}^{-}$, levels with even N are symmetric, and those with odd N are antisymmetric. (Note, here the quantum number N describes molecular rotations, while J includes contributions from electron spin. For molecular states with no electron spin, J = N.) For rotational levels in $\Sigma _{u}^{+}$ and $\Sigma _{g}^{-}$ electronic manifolds, the reverse is true; even N are antisymmetric and odd N are symmetric. These four electronic configurations represent the ground electronic structures of most common linear symmetric molecules. For other electronic configurations (e.g., $\Pi$ and $\Delta$), it turns out that each rotational level consists of two nearly degenerate states, one symmetric and the other antisymmetric, and like asymmetric molecules, the nuclear spin effects can usually be ignored.

Table 5.3 Sample species and ground state configurations

Combining the above, we have

$$\displaystyle{ g'_{\mathrm{rot}} = (2J + 1)\frac{(2I_{C} + 1)} {2} \left [\prod _{m=1}^{M}(2I_{ m} + 1)^{2} \pm \prod _{ m=1}^{M}(2I_{ m} + 1)\right ] }$$

(5.15)

where the choice between adding or subtracting $(+/-)$ the two products is determined from Table 5.2

Table 5.2 Key to addition or subtraction of nuclear degeneracies in Eqs. (5.15) and (5.16) for different rotational statistics and electronic manifold configurations

. For homonuclear diatomics, as described above, the rotational degeneracy including nuclear spin effects reduces to

$$\displaystyle{ g'_{\mathrm{rot}} = (2J + 1)\frac{1} {2}[(2I + 1)^{2} \pm (2I + 1)]. }$$

(5.16)

Consider H ₂ (molecular hydrogen comprised of two¹H atoms):

$$\displaystyle{Q'_{\mathrm{rot}} = \frac{1} {2} \frac{T} {\theta _{r}} (2I_{\mathrm{H}} + 1)^{2} = 2\frac{T} {\theta _{r}} }$$

Since there is one fermion on either side of the axis of symmetry, Fermi statistics apply. And, since the ground state of H₂ is $X^{1}\Sigma _{g}^{+}$,

$$\displaystyle\begin{array}{rcl} g'_{\mathrm{rot}}& =& (2J + 1)\frac{1} {2}\left [(2I_{\mathrm{H}} + 1)^{2} \pm (2I_{\mathrm{ H}} + 1)\right ]\left \{\begin{array}{cc} -,&J\mathrm{\ even} \\ +,& J\mathrm{\ odd}\end{array} \right \} {}\\ & =& (2J + 1)\frac{1} {2}\left \{\begin{array}{cc} 2,&J\mathrm{\ even}\\ 6, & J\mathrm{\ odd }\end{array} \right \} {}\\ & =& (2J + 1)\left \{\begin{array}{cc} 1,&J\mathrm{\ even}\\ 3, & J\mathrm{\ odd }\end{array} \right \} {}\\ \end{array}$$

so that the effective rotational degeneracy alternates with even and odd J.

Consider O ₂ :

$$\displaystyle{Q'_{\mathrm{rot}} = \frac{1} {2} \frac{T} {\theta _{r}} (2I_{\mathrm{O}} + 1)^{2} = \frac{T} {2\theta _{r}}\ \ \ (I_{\mathrm{O}} = 0)}$$

Since there are no fermions (I = 0), Bose–Einstein statistics apply. And, since we have a $^{3}\Sigma _{g}^{-}$ state,

$$\displaystyle\begin{array}{rcl} g'_{\mathrm{rot}}& =& (2J + 1)\left (\frac{1} {2}\right )\left \{\begin{array}{cc} 1 - 1,&J\mathrm{\ even}\\ 1 + 1, & J\mathrm{\ odd }\end{array} \right \} {}\\ & =& \left \{\begin{array}{cc} 0, &J\mathrm{\ even}\\ 2J + 1, & J\mathrm{\ odd }\end{array} \right \} {}\\ \end{array}$$

Therefore, only the odd J states are populated! Even J states do not exist for O₂ in its ground electronic state.

Consider CO ₂ :

$$\displaystyle{Q'_{\mathrm{rot}} = \frac{1} {2} \frac{T} {\theta _{r}} (2I_{\mathrm{C}} + 1)(2I_{\mathrm{O}} + 1)^{2} = \frac{T} {2\theta _{r}}}$$

The number of fermions on one side of the axis is zero, therefore the molecule obeys Bose statistics.

$$\displaystyle\begin{array}{rcl} g'_{\mathrm{rot}}& =& (2J + 1)\frac{1} {2}(2I_{\mathrm{C}} + 1)\left [(2I_{\mathrm{O}} + 1)^{2} \pm (2I_{\mathrm{ O}} + 1)\right ]\left \{\begin{array}{cc} +,&J\mathrm{\ even} \\ -,& J\mathrm{\ odd}\end{array} \right. {}\\ & =& (2J + 1)\frac{1} {2}\left \{\begin{array}{cc} 2,&J\mathrm{\ even}\\ 0, & J\mathrm{\ odd }\end{array} \right \} {}\\ & =& (2J + 1)\left \{\begin{array}{cc} 1,&J\mathrm{\ even}\\ 0, & J\mathrm{\ odd }\end{array} \right \} {}\\ \end{array}$$

Therefore, only the even J states are populated! Odd J states do not exist for CO₂ in its ground electronic state.

Consider C ₂ H ₂ :

$$\displaystyle{Q'_{\mathrm{rot}} = \frac{1} {2} \frac{T} {\theta _{r}} (2I_{\mathrm{C}} + 1)^{2}(2I_{\mathrm{ H}} + 1)^{2} = 2\frac{T} {\theta _{r}} }$$

There is one fermion on either side of the axis of symmetry, so Fermi statistics apply.

$$\displaystyle\begin{array}{rcl} g'_{\mathrm{rot}}& =& (2J + 1)\frac{1} {2}\left [(2I_{\mathrm{C}} + 1)^{2}(2I_{\mathrm{ H}} + 1)^{2}\right. {}\\ & & \mbox{ }\left.\pm (2I_{\mathrm{C}} + 1)(2I_{\mathrm{H}} + 1)\right ]\left \{\begin{array}{cc} -,&J\mathrm{\ even}\\ +, & J\mathrm{\ odd }\end{array} \right. {}\\ & =& (2J + 1)\frac{1} {2}\left \{\begin{array}{cc} 2,&J\mathrm{\ even}\\ 6, & J\mathrm{\ odd }\end{array} \right \} {}\\ & =& (2J + 1)\left \{\begin{array}{cc} 1,&J\mathrm{\ even}\\ 3, & J\mathrm{\ odd }\end{array} \right \} {}\\ \end{array}$$

5.4 Case II: Nonlinear Molecules

5.4.1 Asymmetric Rotor (e.g., CHFClBr and N ₂ H ₄)

Recall that asymmetric rotor molecules have three non-zero and unequal moments of inertia, $I_{A}\neq I_{B}\neq I_{C}$. The general expression for the rotational partition function, including contributions from all three axes of rotation and nuclear spin is

$$\displaystyle{ Q'_{\mathrm{rot}} = \frac{1} {\sigma } \sqrt{\pi \left (\frac{T} {\theta _{A}}\right )\left (\frac{T} {\theta _{B}}\right )\left (\frac{T} {\theta _{C}}\right )}\prod _{n=1}^{L}(2I_{ n} + 1) }$$

(5.17)

The degeneracy is

$$\displaystyle{ g'_{\mathrm{rot}} = \frac{1} {\sigma } (2J + 1)\prod _{n=1}^{L}(2I_{ n} + 1) }$$

(5.18)

Again, the nuclear spin terms can be omitted since there is no nuclear spin-rotation coupling, and the nuclear terms cancel in expressions for the population fraction.

Consider ¹ H ² HO (HDO):

$$\displaystyle\begin{array}{rcl} Q'_{\mathrm{rot}}& =& \sqrt{ \frac{\pi T^{3 } } {\theta _{A}\theta _{B}\theta _{C}}}\,(2I_{\mathrm{O}} + 1)(2I_{\mathrm{H}} + 1)(2I_{\mathrm{D}} + 1) \\ & =& 6\,\sqrt{ \frac{\pi T^{3 } } {\theta _{A}\theta _{B}\theta _{C}}} \\ g'_{\mathrm{rot}}& =& 6(2J + 1) {}\end{array}$$

(5.19)

5.4.2 Symmetric Top

Recall that symmetric top molecules have two equal moments of inertia, $I_{A}\neq I_{B} = I_{C}$. The general expression for the rotational partition function, including contributions from all three axes of rotation and nuclear spin is

$$\displaystyle{ Q'_{\mathrm{rot}} = \frac{\sqrt{\pi }} {\sigma } \sqrt{\frac{T} {\theta _{A}}}\left (\frac{T} {\theta _{B}}\right )\prod _{n=1}^{L}(2I_{ n} + 1) }$$

(5.20)

where σ depends on the group symmetry.

• (a) $\mathit{\mathbf{C}}\mathbf{_{3}}\mathit{_{v}}\mathbf{\ Group\ Symmetry\ (e.g.\ NH_{3}\ and\ CH_{3}Cl)}$

  $$\displaystyle{ Q'_{\mathrm{rot}} = \frac{\sqrt{\pi }} {3} \sqrt{\frac{T} {\theta _{A}}}\left (\frac{T} {\theta _{B}}\right )\prod _{n=1}^{L}(2I_{ n} + 1) }$$

  (5.21)

  The symmetry factor for C _3v group symmetry is σ = 3 due to the threefold rotational symmetry about the A-axis. The degeneracy is dependent on whether the quantum number K is divisible by 3.

  For K divisible by 3 (including K = 0)

  $$\displaystyle{ g'_{\mathrm{rot}} = (2J + 1)\left [\prod _{c=1}^{C}(2I_{ C} + 1)\right ]\frac{2I_{NC} + 1} {3} (4I_{NC}^{2} + 4I_{ NC} + 3) }$$

  (5.22)

  For K not divisible by 3

  $$\displaystyle{ g'_{\mathrm{rot}} = (2J + 1)\left [\prod _{c=1}^{C}(2I_{ C} + 1)\right ]\frac{2I_{NC} + 1} {3} (4I_{NC}^{2} + 4I_{ NC}) }$$

  (5.23)

  where I _C  is the nuclear spin of the cth nucleus on the threefold symmetry axis and I _NC is the nuclear spin of one of the nuclei off the threefold symmetry axis of the molecule (e.g., H in NH₃). The degeneracy is independent of the statistics of the nuclei. Thus, for a given J, the K degeneracies vary like x: x: x′: x: x: x′: …, with x < x′.

  Consider CH ₃ Cl as an example of a molecule with C _3v symmetry.

  $$\displaystyle\begin{array}{rcl} Q'_{\mathrm{rot}}& =& \frac{\sqrt{\pi }} {3} \sqrt{\frac{T} {\theta _{A}}}\left (\frac{T} {\theta _{B}}\right )(2I_{\mathrm{C}} + 1)(2I_{\mathrm{Cl}} + 1)(2I_{\mathrm{H}} + 1)^{3} {}\\ & =& \frac{32\sqrt{\pi }} {3} \sqrt{\frac{T} {\theta _{A}}}\left (\frac{T} {\theta _{B}}\right ) {}\\ g'_{\mathrm{rot}}& =& (2J + 1)\frac{1} {3}(2I_{\mathrm{C}} + 1)(2I_{\mathrm{Cl}} + 1) {}\\ & & (2I_{\mathrm{H}} + 1)\left \{\begin{array}{rl} 4I_{\mathrm{H}}^{2} + 4I_{\mathrm{H}} + 3,&K\mathrm{\ div.\ by\ 3} \\ 4I_{\mathrm{H}}^{2} + 4I_{\mathrm{H}},&K\mathrm{\ not\ div.\ by\ 3}\end{array} \right \} {}\\ & =& (2J + 1)\frac{8} {3}\left \{\begin{array}{rl} 6,&K\mathrm{\ div.\ by\ 3}\\ 3, &K\mathrm{\ not\ div.\ by\ 3}\end{array} \right \} {}\\ & =& (2J + 1)8\left \{\begin{array}{rl} 2,&K\mathrm{\ div.\ by\ 3}\\ 1, &K\mathrm{\ not\ div.\ by\ 3}\end{array} \right \} {}\\ \end{array}$$

• (b) $\boldsymbol{D_{3_{h}}}$ Group Symmetry (e.g., BCl ₃)

  The effective partition function is the same as given for the $C_{3_{v}}$ case above,

  $$\displaystyle{ Q'_{\mathrm{rot}} = \frac{\sqrt{\pi }} {3} \sqrt{\frac{T} {\theta _{A}}}\left (\frac{T} {\theta _{B}}\right )\prod _{n=1}^{L}(2I_{ n} + 1) }$$

  (5.24)

  For levels with K = 0, there is now an alternation in nuclear degeneracy as J (or N) increases. Depending on the appropriate nuclear statistics and rotational quantum number, the degeneracy is given by:

  For integral I _NC and even N or half-integral I _NC and odd N:

  $$\displaystyle{ g'_{\mathrm{rot}} = (2J + 1)\left [\prod _{c=1}^{C}(2I_{ C} + 1)\right ]\frac{2I_{NC} + 1} {3} (2I_{NC} + 3)(I_{NC} + 1) }$$

  (5.25)

  For integral I _NC and odd N or half-integral I _NC and even N:

  $$\displaystyle{ g'_{\mathrm{rot}} = (2J + 1)\left [\prod _{c=1}^{C}(2I_{ C} + 1)\right ]\frac{2I_{NC} + 1} {3} (2I_{NC} - 1)I_{NC} }$$

  (5.26)

  For levels with K ≠ 0, Eqs. (5.22) and (5.23) apply for K divisible by 3 and not divisible by 3, respectively.

  Consider BCl ₃ as an example of a molecule with D _3h symmetry.

  $$\displaystyle\begin{array}{rcl} Q'_{\mathrm{rot}}& =& \frac{\sqrt{\pi }} {3} \sqrt{\frac{T} {\theta _{A}}}\left (\frac{T} {\theta _{B}}\right )(2I_{\mathrm{B}} + 1)(2I_{\mathrm{Cl}} + 1)^{3} \\ & =& 128\sqrt{\pi }\sqrt{\frac{T} {\theta _{A}}}\left (\frac{T} {\theta _{B}}\right ) {}\end{array}$$

  (5.27)

  Since $I_{NC} \equiv I_{\mathrm{Cl}} = 3/2$, the system follows Fermi statistics.

  $$\displaystyle\begin{array}{rcl} g'_{\mathrm{rot}}& =& (2J + 1)(2I_{\mathrm{B}} + 1)\frac{1} {3} {}\\ & & (2I_{\mathrm{Cl}} + 1)\left \{\begin{array}{rl} (2I_{\mathrm{Cl}} - 1)I_{\mathrm{Cl}},&K = 0,J\mathrm{\ even} \\ (2I_{\mathrm{Cl}} + 3)(I_{\mathrm{Cl}} + 1),&K = 0,J\mathrm{\ odd} \\ 4I_{\mathrm{Cl}}^{2} + 4I_{\mathrm{Cl}} + 3,&K\neq 0,\ K\mathrm{\ div.\ by\ 3} \\ 4I_{\mathrm{Cl}}^{2} + 4I_{\mathrm{Cl}},&K\neq 0,\ K\mathrm{\ not\ div.\ by\ 3}\end{array} \right \} {}\\ & =& 8(2J + 1)\left \{\begin{array}{rl} 3,&K = 0,J\mathrm{\ even} \\ 15,&K = 0,J\mathrm{\ odd} \\ 18,&K\neq 0,K\mathrm{\ div.\ by\ 3} \\ 15,&K\neq 0,K\mathrm{\ not\ div.\ by\ 3}\end{array} \right \} {}\\ & =& 24(2J + 1)\left \{\begin{array}{rl} 1,&K = 0,J\mathrm{\ even} \\ 5,&K = 0,J\mathrm{\ odd} \\ 6,&K\neq 0,K\mathrm{\ div.\ by\ 3} \\ 5,&K\neq 0,K\mathrm{\ not\ div.\ by\ 3}\end{array} \right \} {}\\ & & {}\\ \end{array}$$

5.4.3 Others (e.g.,$\mathbf{C_{6}H_{6},CH_{4},and\ P_{4}}$)

These molecules must be considered by symmetry group. See, for example, Herzberg [1, vol. 2, Chap. 2] and [2, vol. 3, Chap. 1].

5.5 Exercises

• 1. Ammonia, NH₃, is a symmetric top molecule with C _3v symmetry, which means σ = 3, and the molecule can be rotated into itself three different ways.

  • (a) Evaluate the relative strengths of the microwave absorption lines for J″ = 1 and J″ = 2 at 300 K assuming that these relative strengths follow Boltzmann statistics, i.e., evaluate ($N_{J=1}/N_{J=2}$). Use θ _A  = 9 K and θ _B  = 14 K in your calculations. Hint: don’t forget to sum over the allowed values of K.

  • (b) Calculate the effective (including nuclear spin) rotational partition function for ammonia at 300 K.