ROTATIONAL ENERGIES AND SPECTRA: . LINEAR MOLECULE SPECTRA: Employing the last equation twice ΔE=...

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ROTATIONAL ENERGIES AND SPECTRA:

Transcript of ROTATIONAL ENERGIES AND SPECTRA: . LINEAR MOLECULE SPECTRA: Employing the last equation twice ΔE=...

Page 1: ROTATIONAL ENERGIES AND SPECTRA: . LINEAR MOLECULE SPECTRA:  Employing the last equation twice  ΔE= E J+1 – E J = hB(J+1)(J=2) – hBJ(J+1)  Or: ΔE.

ROTATIONAL ENERGIES AND SPECTRA:

Page 2: ROTATIONAL ENERGIES AND SPECTRA: . LINEAR MOLECULE SPECTRA:  Employing the last equation twice  ΔE= E J+1 – E J = hB(J+1)(J=2) – hBJ(J+1)  Or: ΔE.

LINEAR MOLECULE SPECTRA:

Employing the last equation twice ΔE= EJ+1 – EJ = hB(J+1)(J=2) – hBJ(J+1) Or: ΔE = 2hB(J+1) for a transition from the

Jth to the (J+1)th level. Using ΔE = hν gives us ν = 2B(J+1) for rotational transitions.

Page 3: ROTATIONAL ENERGIES AND SPECTRA: . LINEAR MOLECULE SPECTRA:  Employing the last equation twice  ΔE= E J+1 – E J = hB(J+1)(J=2) – hBJ(J+1)  Or: ΔE.

ROTATIONAL ENERGIES – LINEAR MOLECULES:

J Value J(J+1) Value Rotational Energy

EJ+1 - EJ ν =(EJ+1 – EJ)/h

0 0 01 2 2hB 2hB 2B2 6 6hB 4hB 4B3 12 12hB 6hB 6B4 20 20hB 8hB 8B5 30 30hB 10hB 10B6 42 42hB 12hB 12B

Page 4: ROTATIONAL ENERGIES AND SPECTRA: . LINEAR MOLECULE SPECTRA:  Employing the last equation twice  ΔE= E J+1 – E J = hB(J+1)(J=2) – hBJ(J+1)  Or: ΔE.

LINEAR MOLECULE ROTATIONAL TRANSITIONS: J = 4

J = 3

J = 2 J = 1 J = 0

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LINEAR MOLECULE ROTATIONAL SPECTRUM:

Intensity J = 4←3

J = 1←0 2B

Absorption Frequencies →

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LINEAR MOLECULE SPECTRA: Given a molecular structure we can predict

the appearance of a rotational (microwave) spectrum by calculating (a) the moment of inertia and (b) the value of the rotational constant. A linear molecule can be treated as a series of point masses arranged in a straight line.

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ROTATIONAL SPECTRA DIATOMICS:

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MOMENTS OF INERTIA – DIATOMICS:

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MOMENTS OF INERTIA – DIATOMICS:

Page 10: ROTATIONAL ENERGIES AND SPECTRA: . LINEAR MOLECULE SPECTRA:  Employing the last equation twice  ΔE= E J+1 – E J = hB(J+1)(J=2) – hBJ(J+1)  Or: ΔE.

CLASS EXAMPLE CALCULATIONS:

We will calculate moments of inertia for 14N2, 14N15N and 12C16O2 using, where possible, symmetry arguments to simplify the arithmetic.

Data: r(N≡N) = 1.094 Å (109.4pm) and r(C=O) = 1.163 Å

Masses: 14N (14.00307u), 15N(15.00011u), 12C(12.00000u) and 16O(15.99492u).

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SPECTRA OF CARBON MONOSULFIDE:

In rotational spectroscopy, energy level separations become smaller as atomic masses increase and as molecular dimensions increase. On the next slide the effect of mass alone is illustrated for the 12C32S and 12C34S molecules which have, of course, identical bond distances(almost!).

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12C32S AND 12C34S ROTATIONAL SPECTRA:

Transition 12C32S (Frequency (MHz) 12C34S Frequency (MHz)

J=1←0 48990.97 48206.92

J=2←1 97980.95 96412.94

J=3←2 146969.03 144617.11

J=4←3 195954.23 192818.46

J=5←4 244935.74 241016.19

J=6←5 293912.24 289209.23

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EFFECTS OF MASS AND MOLECULAR SIZE:

The slide that follows gives B values for a number of diatomic molecules with different reduced masses and bond distances. What is the physical significance of the very different frequencies seen for H35Cl and D35Cl? All data are taken from the NIST site.

http://www.nist.gov/pml/data/molspec.cfm

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REDUCED MASSES AND BOND DISTANCES:Molecule Bond

Distance (Å)B Value (MHz)

H35Cl 1.275 312989.3H37Cl 1.275 312519.1D35Cl 1.275 161656.2H79Br 1.414 250360.8H81Br 1.414 250282.9D79Br 1.414 127358.124Mg16O 1.748 17149.4107Ag35Cl 2.281 3678.04

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REAL LIFE – WORKING BACKWARDS? In the real world spectroscopic experiments

provide frequency (and intensity) data. It is necessary to assign quantum numbers for the transitions before molecular (chemically useful) information can be determined. Sometimes “all of the data” are not available!

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SPECTRUM TO MOLECULAR STRUCTURE: Class Example: A scan of the microwave

(millimeter wave!) spectrum of 6LiF over the range 350 → 550 GHz shows lines at 358856.2 MHz, 448491.1 MHz and 538072.7 MHz. Assign rotational quantum numbers for these transitions. Determine a B value and the bond distance for 6LiF. Are the “lines” identically spaced?