Physical Chemistry 2 nd Edition Thomas Engel, Philip Reid Chapter 19 The Vibrational and Rotational Spectroscopy of Diatomic Molecules

Physical Chemistry 2Physical Chemistry 2ndnd Edition EditionThomas Engel, Philip Reid

Chapter 19 Chapter 19 The Vibrational and Rotational Spectroscopy

of Diatomic Molecules

© 2010 Pearson Education South Asia Pte Ltd

Physical Chemistry 2nd EditionChapter 19: The Vibrational and Rotational Spectroscopy of Diatomic Molecules

ObjectivesObjectives

• Describe how light interacts with molecules to induce transitions between states

• Discuss the absorption of electromagnetic radiation



OutlineOutline

1. An Introduction to Spectroscopy2. Absorption, Spontaneous Emission, and

Stimulated Emission3. An Introduction to Vibrational

Spectroscopy4. The Origin of Selection Rules5. Infrared Absorption Spectroscopy6. Rotational Spectroscopy



19.1 An Introduction to Spectroscopy19.1 An Introduction to Spectroscopy

• Spectroscopy are tools chemists have to probe the species at an atomic and molecular level.

• The frequency at which energy is absorbed or emitted is related to the energy levels involved in the transitions by

12 EEhv




• 19.1 Energy Levels and Emission Spectra




• During vibration, oscillator will absorb energy in both the stretching and compression.

• The molecule can absorb energy from the field during oscillation.



Band name Abbr ITU bandFrequencyandwavelength in air

Example uses

subHertz subHz 0< 3 Hz

> 100,000 km

Natural and man-made electromagnetic waves (millihertz, microhertz, nanohertz) from earth,

ionosphere, sun, planets, etc[citation needed]

Extremely low frequency

ELF 13–30 Hz

100,000 km – 10,000 km

Communication with submarines

Super low frequency SLF 230–300 Hz

10,000 km – 1000 km

Communication with submarines

Ultra low frequency ULF 3300–3000 Hz

1000 km – 100 kmCommunication within mines

Very low frequency VLF 43–30 kHz

100 km – 10 kmSubmarine communication, avalanche beacons,

wireless heart rate monitors, geophysics

Low frequency LF 530–300 kHz

10 km – 1 kmNavigation, time signals, AM longwave broadcasting,

RFID

Medium frequency MF 6300–3000 kHz1 km – 100 m

AM (medium-wave) broadcasts

High frequency HF 73–30 MHz

100 m – 10 mShortwave broadcasts, amateur radio and over-the-

horizon aviation communications, RFID

Very high frequency VHF 830–300 MHz10 m – 1 m

FM, television broadcasts and line-of-sight ground-to-aircraft and aircraft-to-aircraft communications. Land

Mobile and Maritime Mobile communications

Ultra high frequency UHF 9300–3000 MHz1 m – 100 mm

Television broadcasts, microwave ovens, mobile phones, wireless LAN, Bluetooth, GPS and two-way radios such as Land Mobile, FRS and GMRS radios

Super high frequency SHF 103–30 GHz

100 mm – 10 mmMicrowave devices, wireless LAN, most modern radars

Extremely high frequency

EHF 1130–300 GHz

10 mm – 1 mmRadio astronomy, high-frequency microwave radio

relay

Terahertz THz 12300–3,000 GHz1 mm – 100 μm

Terahertz imaging – a potential replacement for X-rays in some medical applications, ultrafast molecular

dynamics, condensed-matter physics, terahertz time-domain spectroscopy, terahertz

computing/communications





19.2 Absorption, Spontaneous Emission, and Stimulated 19.2 Absorption, Spontaneous Emission, and Stimulated EmissionEmission

• The 3 basic processes by which photon-assisted transitions occur are absorption, spontaneous emission and stimulated emission.




• In absorption, the incident photon induces a transition to a higher level.

• In emission, a photon is emitted as an excited state relaxes to one of lower energy.

• Spontaneous emission is a random event and its rate is related to the lifetime of the excited state.




• At equilibrium,

where = radiation density at frequency ν

= rate coefficient

• Einstein concluded that3

32

21

212112

16 and

c

hv

B

ABB

221221112 NANvBNvB



Example 19.1Example 19.1

Derive the equations

using these two pieces of information: (1) the overall rate of transition between levels 1 and 2 is zero at equilibrium, and (2) the ratio of N2 to N1 is governed by the Boltzmann distribution.

33221212112 /16/ and chvBABB



SolutionSolution

The rate of transitions from level 1 to level 2 is equal and opposite to the transitions from level 2 to level 1. This gives the equation .The Boltzmann distribution function states that

221221112 NANBNB

kThveg

g

N

N /

1

2

1

2



SolutionSolution

In this case . These two equations can be solved for , giving . Planck has showed that

For these two expressions to be equal

3

21/ 3 /

12 21

8 1

1hv kT hv kT

A hvv

B e B c e

21/

1221 / BeBA kThv v12 gg

3323321212112 /16/8/ and chvchvBABB



19.3 An Introduction to Vibrational Spectroscopy19.3 An Introduction to Vibrational Spectroscopy

• The vibrational frequency depends on two identity vibrating atoms on both end of the bond.

• This property generates characteristic frequencies for atoms joined by a bond known as group frequencies.




A strong absorption of infrared radiation is observed for 1H35Cl at 2991 cm-1.a. Calculate the force constant, k, for this molecule.b. By what factor do you expect this frequency to shift if deuterium is substituted for hydrogen in this molecule? The force constant is unaffected by this substitution.



SolutionSolution

a. We first write . Solving for k,

b. The vibrational frequency for DCl is lower by a substantial amount.

mN

ck

khhchvE

/3.51610661.1977.35

969.34008.1100299110998.244

and 2

272822

2

// khchvE

1.0078 36.9830.717

2.0140 35.977HCl

DCl

H cl D C

D cl H C

m m m m

m m m m



19.3 An Introduction to Vibrational 19.3 An Introduction to Vibrational SpectroscopySpectroscopy

• 19.2 The Morse Potential

• The bond energy D0 is defined with respect to the lowest allowed level, rather than to the bottom of the potential.

• The energy level is

2 21 1

2 4 2ne

hvE hv n n

D

( ) 2( ) [1 ]ex xeV x D e

max

max

0 0

2 2

max max

2 2

0

1 1

2 4 2

1 10 0

2 4 2

n

n ee

e

D E E

hvE hv n n D

D

hvE hv

D



19.3 An Introduction to Vibrational 19.3 An Introduction to Vibrational SpectroscopySpectroscopy

• Parameters for selected model are shown.



19.4 The Origin of Selection Rules19.4 The Origin of Selection Rules

• The transition probability from state n to state m is only nonzero if the transition dipole moment satisfies the following condition:

where x = spatial variable μx = dipole moment along the electric field direction

0* dxxx nxmmnx



19.5 Infrared Absorption Spectroscopy19.5 Infrared Absorption Spectroscopy

• Atoms and molecules possess a discrete energy spectrum that can only be absorbed or emitted which correspond to the difference between two energy levels.

• Beer-Lambert law states that

where I(λ) = intensity of light leaving the cell

I0(λ) = intensity of light passing dl distance l = path length

ε(λ) = molar absorption coefficient

MleI

I

0




The molar absorption coefficient for ethane is 40 (cm bar)-1 at a wavelength of 12 μm. Calculate in a 10-cm-long absorption cell if ethane is present at a contamination level of 2.0 ppm in one bar of air. What cell length is required to make ?

0/ II

90.0/ 0 II



SolutionSolution

Using

This result shows that for this cell length, light absorption is difficult to detect. Rearranging the Beer-Lambert equation, we have

cm

I

I

Ml 3

60

103.190.0ln100.240

1ln

1

MleI

I

0

0.19992.00.1100.240exp 6

0

I

I



19.5 Infrared Absorption Spectroscopy19.5 Infrared Absorption Spectroscopy

• Coupled system has two vibrational frequencies: the symmetrical and antisymmetric modes.

• For symmetrical and asymmetrical, the vibrational frequency is

1

2

1 kvsymmetric

21 2

2

1 kkv ricqntisymmet



19.6 Rotational Spectroscopy19.6 Rotational Spectroscopy

• 19.3 Normal Modes for H2O

• 19.4 Normal Modes for CO2

• 19.5 Normal Modes for NH3

• 19.6 Normal Modes for Formaldehyde




Using the following total energy eigenfunctions for the three-dimensional rigid rotor, show that the J=0 → J=1 transition is allowed, and that the J=0 → J=2 transition is forbidden:

The notation is used for the preceding functions.

1cos316

5,

cos4

3,

4

1,

2/10

2

2/10

1

2/10

0

Y

Y

Y

jMjY



SolutionSolution

Assuming the electromagnetic field to lie along the zaxis, , and the transition dipole moment takes the form

For the J=0 → J=1 transition,

cosz

03

3

3

cos

2

3sincos

4

3

0

32

0

22

0

10

ddz

dYYd JJz sincos,cos, 0

0

2

0

02

0

0



SolutionSolution

For the J=0 → J=2 transition,

The preceding calculations show that the J=0 → J=1 transition is allowed and that the J=0 → J=2 transition is forbidden. You can also show that is also zero unless MJ=0 .

04

1

4

1

8

5

2

cos

4

cos3

8

5sincos1cos3

8

5

0

242

0

22

0

20

ddz




• For vibrational spectroscopy, we have to change the symbol for the angular momentum quantum number from l to J.

• Thus the dependence of the rotational energy on the quantum number is given by

where rotational constant is

118

12 2

2

2

2

JhcBJJJr

hJJ

r

hE

oo




• We can calculate the energy corresponding to rotational transitions

hcBJJJr

hJJ

r

hE

JJhcBJJr

hJJ

r

hE

JJEJEE initialfinal

212

12

1for and 1212

212

1for

20

2

20

2

20

2

20

2




Because of the very high precision of frequency measurements, bond lengths can be determined with a correspondingly high precision, as illustrated in this example. From the rotational microwave spectrum of 1H35Cl, we find that B=10.59342cm-1. Given that the masses of 1H and 35Cl are 1.0078250 and 34.9688527 amu, respectively, determine the bond length of the 1H35Cl molecule.



SolutionSolution

We have

2 20

34

0 22 27

10

8

6.6260755 10

8 1.0078250 34.96885278 1.66054 10 10.59342

1.0078250 34.9688527

1.274553 10

hB

cr

hr

cBc

m




• To excite various transitions consistent with the selection rule , we have

1 initialfinal JJJ





19_16fig_PChem.jpg19_16fig_PChem.jpg



Δ 1J Δ 1J Δ 0J

Δn 1

P,Q,R branches of rotational spectrum

, R: , Q:

(vibrational ).




• 19.7 Rotational Spectroscopy of Diatomic Molecules

• 19.8 Rotational-Vibrational Spectroscopy of Diatomic Molecules

• The ratio for value of J relative to the number in the ground state (J=0) can be calculated using the Boltzmann distribution: kTJJhkTJJ eJe

g

g

n

nJ 2/1/

00

20 12



Rotational Raman SpectraRotational Raman Spectra

E

The molecule can be made anisotropically polarized andRaman active.

Selection Rules:

2,0J

2,1,0J 0K

Linear rotors

Symmetrical rotors



Proof of Rotational Raman Proof of Rotational Raman Selection RulesSelection Rules

tcosE)t(E iind t2cos R0

t)2cos(t)2cos(E2

1tcosE

tcost2cosEtcosE

)tcosE()t2cos(

RiRii0

iRi0

iR0ind

cossinxx,ind sinsinyy,ind coszz,ind

cossinEEx sinsinEE y

cosEEz

2

//2

z//yxind cosEsinEcosEsinsinEcossinE

E),(53

4

3

2

3

10,2

21

//ind

i,Jif,Jf M,JM,J// YY)3

2

3

1(

i,Jif,Jf M,JM,J YY Selection rules



A Typical Rotational Raman A Typical Rotational Raman Spectrum (Linear rotors)Spectrum (Linear rotors)

0, 2J (Linear rotors)

)3J2(B2~)J(F)2J(F~)J1J(~ii

Stokes lines

)1J2(B2~)2J(F)J(F~)J1J(~ii

Anti-Stokes lines



19_21fig_PChem.jpg19_21fig_PChem.jpg

Vibrational Raman effect, Δ n=+1,-1







Documents

Physical Chemistry 2 nd Edition Thomas Engel, Philip Reid Chapter 19 The Vibrational and Rotational Spectroscopy of Diatomic Molecules