6-4. Show that rotational transitions of a diatomic molecule occur in the microwave region or the far infrared region of the spectrum. see section 6.2 in McQuarrie (pg. 274) for typical parameter values6.2 The Rigid Rolator Is a Model for a Rotating Diatomie Molecule The allowed energies of a rigid rotator are given by Equation 6.25. We will shoalr in Section 6.? that the selection rule for the rigid rotator says that transitions are ailowed only from adjacent states or that a; =:|:1 {6.27) In addition to the requirement that of = :i: l, the molecule must also possess a perma- nent dipole moment to absorb electromagnetic radiation. Thus, HCI has a pure rotational 6.2. The Rigid Rotator Is a Model for a Rotating Diatomic Molecule 273 spectrum, but N2 does not. In the case of absorption of electromagnetic radiation, the molecule goes from a state with a quantum number / to one with J + 1. The energy difference, then, is AE = E/1 - E, = [( + 1)(J +2) - J(J + 1] h2 JU+1=- ,(J + 1) J= 0, 1, 2, ... (6.28) The energy levels and the absorption transitions are shown in Figure 6.2. Using the Bohr frequency condition AE = hvos the frequencies at which the absorption transitions occur are h Vobs = -(J + 1) J = 0, 1, 2. ... (6.29) 4721 AE = 8B 3 Rotational energy levels AE = 6B 2 AE = 4B AE = 2B 0 Spectrum 0 2B 4B 6B 8B FIGURE 6.2 The energy levels and absorption transitions of a rigid rotator. The absorption transitions occur between adjacent levels, so the absorption spectrum shown below the energy levels consists of a series of equally spaced lines. The quantity B is h/8x c/ (Equation 6.33).274 Chapter 6 / The Rigid Rotator and Rotational Spectroscopy The reduced mass of a diatomic molecule is typically around 10 25 to 10-26 kg, and a typical bond distance is approximately 10 10 m (100 pm), so the moment of inertia of a diatomic molecule typically ranges from 10-45 to 10 46 kg.m . Substituting 1 =5 x 10 4 kg.m into Equation 6.29 shows that the absorption frequencies are about 2 x 10 0 to 10 Hz (cf. Problem 6 4). By referring to Figure 1.11 in Problem 1-1, we see that these frequencies lie in the microwave region. Consequently, rotational transitions of diatomic molecules occur in the microwave region, and the study of rotational transitions in molecules is called microwave spectroscopy. It is common practice in microwave spectroscopy to write Equation 6.29 as Vobs = 2B (J + 1) J = 0, 1, 2, ... (6.30) where h B = (Hz) (6.31) 877 21 is the rotational constant expressed in units of hertz. Furthermore, the transition fre- quency in Equation 6.29 is commonly expressed in terms of wave numbers (cm-) rather than hertz (Hz). If we use the relation @ = v/c, then Equation 6.30 becomes Wobs = 2B (J + 1) J=0, 1, 2, .. . (6.32) where B is the rotational constant expressed in units of wave numbers: B = = h (cm -) (6.33) 872cl The tilde here emphasizes that B has units of wave numbers. From either Equation 6.30 or 6.32, we see that the rigid-rotator model predicts that the microwave spectrum of a diatomic molecule consists of a series of equally spaced lines with a separation of 2B Hz or 2B cm , as shown in Figure 6.2. From the separation between the absorption frequencies, we can determine the rotational constant and hence the moment of inertia of the molecule. Furthermore, because I = ul , where / is the internuclear distance or bond length, we can determine the bond length. This procedure is illustrated in Example 6-2. EXAMPLE 6-2 To a good approximation, the microwave spectrum of H"Cl consists of a series of equally spaced lines, separated by 6.26 x 10# Hz. Calculate the bond length of H CI. SOLUTION: According to Equation 6.30, the spacing of the lines in the microwave spectrum of HS'CI is given by h 2B = - 4x7 216.3. Rotational Transitions Accompany Vibrational Transitions 275 and so h = 6.26 x 10 Hz 4x21 Solving this equation for / , we have 6.626 x 1034 J.S 1 = 4x 2(6.26 x 101 1 5-1) =2.68 x 10 47 kg.m The reduced mass of H-Cl is (1.00 amu ) (35.0 amu) N = (1.661 x 10-27 kg .amu-) = 1.66 x 10-27kg 36.0 amu Using the fact that / = ud, we obtain 1 = 2.68 x 10 47 kg.m2\\ 1/2 = 1.29 x 10-m = 129 pm 1.661 x 10-27 kg Problems 6-5 through 6-8 give other examples of the determination of bond lengths from microwave data