Ir Absorption 53w67

Lecture V Infrared (IR)

Lecture Outline  Theory

 What is a vibration?  Morse potential energy curves!!  Why molecules (bonds) are IR active?

 IR Techniques  FT-IR  ATR  PAS

 Instrumentation

 FT-IR  ATR  FT-IR microscope

 Sample Analysis  Applications

Infrared Spectroscopy IR spectroscopy is the study of the interaction of infrared light with matter. This interaction can provide structural information, quantification, and identification. The fundamental measurement obtained in IR spectroscopy is an IR spectrum which is a plot of measured IR intensity (absorbance or transmittance) versus wavelength (or wavenumber) of light. The instrument used to make these measurements is called an infrared spectrometer. There are several types. IR spectroscopy is sensitive to the presence of chemical functional groups (chromophores) in a sample. A functional group is a structural fragment within a molecule.

Why molecules absorb IR? In order to answer this question, need to know what is the makeup of a chemical bond. Also, how can you classify this chemical bond, i.e. is it a spring? What is the geometry of the molecule and how does this affect its interaction with IR? Several approaches need to be addressed. Classical or mechanical description of the bond. Quantum mechanical description of the bond.

Infrared sub-regions Region

Transition

Wavelength mm

Wavenumber cm-1

Near IR (NIR)

overtones

0.75-2.5

13,3004,000

Fundamental vibs, rots

2.5-25

4000-400

Far IR

25-1000

400-10

skeletal, rots

1.1

3393.03

1

0.6

597.862

1017.01

1164.87

1734.44

0.7

1409.15

1590.44

0.8

2925.02 2854.31

0.5

0.4

405.003

Transmittance

0.9

9.spc 4000

3800

3600

3400

3200

3000

2800

2600

2400

2200

2000

Wavenumbers

1800

1600

1400

1200

1000

800

600

400

LCAO

H2 Molecule

Making Chemical Bonds

Molecular Vibrations

Chemical bonds  Consider the hydrogen chloride (H-Cl) molecule whose structure is shown to the right. The H-Cl molecule is shown as a “ball and spring” model where the atoms in the molecule are represented by balls and the chemical bond by a spring. Because there is a difference in electronegativity between the two atoms, the electrons in the H-Cl bond have a higher probability of being found on the Cl atom. As a result, the chlorine has a partial negative charge on it as indicated by the dand the hydrogen atom d+.

Dipole moment Subsequently the H-Cl molecule contains two charges separated by a distance, r. This phenomenon is known as a dipole moment. The dipole moment is a measure of the charge asymmetry of a molecule. The magnitude of the dipole moment is calculated from the following equation: m=qr where q = charge, r = distance.

Electric field interaction The alternating electric field interacts with the changing dipole moment. If the electric vector polarity is positive, it will repel the partial positive charge on the hydrogen atom because like charges repel. Consequently, the hydrogen atom will move away from the electric vector and the H-Cl bond will shorten. Once the polarity of the electric vector changes to being negative, it will attract the hydrogen atom because opposite charges attract. Therefore, the H-Cl bond will get shorter and longer and shorter and longer. During this process, the molecule vibrates at the same frequency as the electric vector, and the energy of the light beam is transferred to the molecule. Hence an infrared absorbance takes place.

Dipole Moments

Electromagnetic wave Two fields, electric and magnetic

Electric field component

IR Absorbance Before the IR beam and molecule interact, the H-Cl molecule is at rest, and the IR photon has energy equal to hn. After the interaction, the photon has been absorbed, and its energy deposited into the H-Cl molecule as bond stretching motion. Energy is conserved in the reaction since all the photon’s energy has been transferred to the molecule as vibrational energy. We detect the absorbance of the photon by a decrease in infrared intensity at the wavenumber of the light absorbed, giving an absorption feature in the IR spectrum of the molecule. Can say that the dipole moment is the handle on the molecule that the IR light grasps so it can interact with the molecule.

HCl 100

80

60

40

20

5000

4500

Transmi ssi on / Wavenu mber (cm-1) Fi le # 1 = HCL Hi gh resol uti on IR gas phase spectrum of HCL

4000

3500

3000

2500

2000

1500

1000

Page d X-Zoom CURSOR 3/4/199 2 2:23 PM Res=.001

High resolution HCl, fine structure 100

90

80

70

%Transmittance

60

50

40

30

20

10

0 3250

3200

3150

3100

3050

3000

2950 2900 2850 Wav enumber (cm-1)

2800

2750

2700

2650

2600

2550

2500

Requirements for IR Absorption 2 m 1  m m = m o + (r - re )( ) o + (r - re ) 2 ( 2 ) o + .... r 2  r

  m   R =  m o + r - re    jd  r o   * i

shows that the transition probability is related to a change in dipole moment.

Classical harmonic oscillator  Consider a mass m attached to a stationary object through a spring. As the mass is displaced from equilibrium position, it will return with a restoration force that is proportional to its displacement from equilibrium. This spring is said to obey Hooke’s law.  F = -kx. k is the force constant. Newton’s law: F= m d2x/dt2.  Combine equations: -kx = m d2x/dt2 Can integrate and get the potential energy of this oscillator.  V=1/2 kx2

M

Classical harmonic oscillator This equation is that of a parabola or simple harmonic oscillator. V=1/2 kx2. Frequency of vibration is: n =1/2p(f/m)1/2

V

x

Classical harmonic oscillator, diatomic Now consider a diatomic molecule of two masses, m1 and m2.

m1

m2

Fundamental stretch frequency n

1 K 2 pc m

m1 m

m 1m 2 m1 + m2

or m 

m2

M 1M 2 , m a sses of a t om s in a m u (M 1 + M 2 )(6.02 x 10 23 )

7.76 x 10 11 n 2 pc

K m

n(cm -1 )  4 .12

K wh er e K = for ce con st a n t in dyn es / cm m

Quantum mechanical approach to harmonic oscillator Solving the Schrodinger equation gives:

1 h Ev  ( v + ) 2 2p

k

m

h E  Ev +1 - Ev  2p

Shows that the spacings between levels is constant.

k

m

Relationship

Equation is similar to the classical case. This equation shows that the transition is quantized. However, in a diatomic or larger molecule, don’t have a simple harmonic oscillator. Instead, anharmonic oscillator in which we need to add more onto the simple harmonic oscillator expression. This contributes to overtones and Near IR.

1 1 2 Ev  hce (v + ) - hcxe (v + ) + ...... 2 2

Vibrational Energy Diagram

Conditions for IR absorption Need quantized energy Mode must have a change in dipole moment

Fundamental Frequencies for non-diatomics For linear triatomic, symmetric:

 1  n  1303 f    m1  -

-

n as

 1 2    1303 f  +  m1 m2 

m1

m2

Bent tri-atomic Angle a

-

 1

1 + cosa   m2 

 1

1 - cosa   m2 

n s  1303 f  +  m1 -

n as  1303 f  +  m1

m1

m2

Predictions of vibrational modes.

3n-5 (linear molecule) vibrational modes. 3n-6 for non-linear. Can use group theory to determine which modes will be IR and Raman active. Depends on what point group the molecule belongs to. These point groups or character tables can assist in determine the active modes of IR or Raman.

Water molecule

H

H

H

O

O

O

H

H

H

Water vibrations Asymmetric stretch

H

H

H

O

O

O

H

H

H

Symmetric stretch

Bending or scissors stretch All modes IR active!!!

Modes for CO2 symmetric stretch no IR

asymmetric stretch, IR

Bending, 667 cm-1

Bending, out of plane, 667 cm-1 +

-

+

Point groups

Multi-vibrations

FT-IR (Non dispersive) The ac component of the detector signal S(x) as a function of mirror displacement is related to the source spectrum  4p x  S ( x)  K cos 

c

 4p x   K  cos    

 

FT-IR where K is the constant that includes detector response and geometrical factors, x is the mirror displacement. The signal S(x) measured vs. displacement x is called the interferogram. If the mirror is moved at a constant rate (r= dx/dt), the detector signal oscillates with a frequency f=2rv/c.

FT-IR The ac signal is the integral over all frequencies: -

+

-

S ( x)    - cos(4pxn )dn -

n

This is a Fourier integral whoseFourier transformis given by : +

n   S ( x) cos(4 xp n )dx -

Interferogram: Interferogram  2pd  I(d )  B( ) cos     



 B( ) cos 2pd



where d is the difference between moveable mirror and stationary mirror =2(OM-OS) in diagram below.

FT-IR Fourier Transform: 

 a n 0

n

sin nx  + bn cosnx 

2p  x2 - x1 where the coefficients an and bn are determined to reconstruct the curve.

FT

BRITISH MODERN BN.0 (2) 0.1

0.02

BRITISH MODERN BN.0 (1)

0.09 0

0.08 0.07

-0.02

0.06 0.05 -0.04

0.04 0.03

-0.06

0.02 -0.08

0.01

4000

3750

3500

3250

3000

2750

2500

2250

2000

1750

1500

1250

1000

750

500

250

4000

3750

3500

3250

3000

2750

2500

2250

2000

Wavenumbers

1750

1500

1250

1000

750

500

Advantages of FT-IR 1. Speed of analysis 2. Small samples can be analyzed i.e. use a microscope 3. Signal to noise is 100,000 : 1. 4. Data manipulation

Quantitative Analysis

IR Io ()

I () sample

V1 V0

Beer’s law





0

d



 - k  db 0

 ln  - kb 0 and

 = 0e - kb

Io

I

b

c

Concentration Determination

Concentration

C = (slope)(A) + intercept

0

A

Sample Analysis Procedure Run blank spectrum For solids, grind in a “wiggle bug” with Nujol and then press a pellet under high pressure. Liquids can be analyzed in special quartz cell. For solids, can use a diamond anvil, microscope, or ATR. Gases can be analyzed in special gas cell.

S/N is very important

1 0.955 0.9 0.95 0.8 0.945

0.94

Transmittance

Transmittance

0.7

0.6

0.5

0.935

0.93

0.925 0.4 0.92 0.3 0.915 0.2

0.1

0.91

TUMIX.SPC 4000

3750

0.905 3500

3250

3000

2750

2500

2250

2000

Wavenumbers

1750

1500

1250

1000

750

500

TUMIX.SPC 4000

3950

3900

3850

3800

3750

3700

3650

3600

3550

Wavenumbers

3500

3450

3400

3350

3300

3250

Background spectra

1

0.1

BRITISH MODERN BN.0 (1) 0.95

0.09

0.9

0.08

0.85

Transmittance

0.07 0.06 0.05 0.04

0.8

0.75

0.7

0.03

0.65

0.02

0.6

0.01 0.55

4000

3750

3500

3250

3000

2750

2500

2250

2000

Wavenumbers

1750

1500

1250

1000

750

500

BRITISH MODERN BN.5 (3) 4000

3750

3500

3250

3000

2750

2500

2250

2000

Wavenumbers

1750

1500

1250

1000

750

500

Instrumentation Dispersive Detector

Sample

Source

Dispersing

Computer

Instrumentation FT-IR (non-dispersive)

Sample

Source

Detector

Interferometer

Computer

Instrumentation Sources: based on blackbody radiator and using the Wien displacement law: max =3000 /T (K) W filament lamp for Near-IR Nernst glower about 1500 C is fused mixture of Zr, Y, and Th oxides. Good for Mid-IR. See figure below. Globar is rod of SiC about 1300 C. Also good for Mid-IR. High pressure Hg arc lamp for Far-IR. Tuneable dye lasers can be used for non-dispersive work where an intense source is needed to monitor one wavelength.

Nernst glower

More instrumentation Dispersing element: Prism, filter, or monochromator (dispersive)

Detectors Photoconductive cells Thermopile, thermister, pyroelectric, semiconductor (Ge and InGaAs), Golay, DTGS

More detectors Photoconductive detectors have higher sensitivity and are based on the photoconductive properties of mercury cium telluride (MCT). Operate at liquid nitrogen temperatures. Increases the signal/noise. Less sensitive by a factor of 10 are the pyroelectric detectors such as deuteriated triglycine sulphate (DTGS) or lithium tantalate (LiTaO3). Rapid response detectors that are used with FTIR.

Michelson Interferometer

Modern Day FT-IR

Bruker Equinox 55

Demountable cells for liquids or gases

liquids

gases

Different IR modes

Attenuated Total Reflectance (ATR) ATR is a great method for analyzing solids non-destructively. Also can analyze liquids. It is based on having the sample in close with a highly reflecting surface such as a diamond. The IR beam will partial penetrate the sample and totally reflect giving a good IR spectrum.

ATR Equation

 dp  2 2 1/ 2 2pn p (sin  - n sp ) Where nsp and np are the ratio of the refractive indices of the sample vs the and the respectively. Theta is the angle of incidence and  is the wavelength. For ZnSe and a polymer sample, depth is 2.0 mm at 1,000 cm-1.

SensIR ATR

Jasco IR Microscope  The optical system is shown below. The interference light from the FT/IR main unit is introduced to the main unit by the plane mirror M1. The changeover mirror M2 reflects the light upward in reflection measurement and downwards in transmission measurement. In reflection measurement, the light reflected upward is reflected towards the sample by the beam splitter BS and is converged onto the sample surface by the upper Cassegrain mirror CM1. The light reflected by the sample is collected by the same Cassegrain mirror CM1 and is focused through the aperture AP to the detector D which is a high sensitivity infrared detector (MCT detector). In transmission measurement, the light reflected downwards is focussed onto the sample by the Cassegrain mirror CM2 below. The light transmitted by the sample is collected by the Cassegrain mirror CM1 above (as in reflection measurement) and is again detected by the high sensitivity infrared detector D.

Advantages of IR A universal technique i.e. solids, liquids, gases, semi-solids, powders, and polymers can be routinely analyzed. IR spectra are information rich; peak positions, intensities, widths, and shapes in a spectrum all give useful information about the analyte. IR is relatively fast and easy technique. Most samples can be prepared and scanned in less than five minutes. IR is very sensitive. Micro to nano gram quantities can routinely be detected.

Disadvantages of IR Homonuclear compounds don’t absorb. Aqueous solutions difficult to analyze because the strong absorbance of water. Some compounds give broad bands that interfere with other compounds. Complex mixtures difficult. Dark (black) compounds often absorb the IR beam completely, i.e. 0% transmittance.

Software Robust software is needed to manipulate the FT-IR spectrum. This includes smoothing, peak area determination, subtraction, switching back and forth between absorbance and transmittance. Can be used to create spectral libraries for searching. Program to identify peaks and use correlation diagrams to determine structure. (Go to KnowItAll)

How to purchase? Depends on needs? Solid samples only, ATR-FT-IR. Ease of use, resolution, S/N, vacuum or N2 purged system. Range of instrument? NIR for overtones Mid IR for fundamentals

Software and data manipulation ease Ruggedness of instrument for field sampling Cost: $15,000 - $90,000

PE Spectrum Spotlight 300 FT-IR Imaging System

IR Imaging

Visible image of an inclusion in a polypropylene film. Sample size is approximately 450 x 450 m

 IR image showing the

distribution of a carbonyl contaminant through the inclusion. Image contains more than 5000 spectra at 6.25 m pixel size. Total data collection time was 90 seconds.

Imaging another view

Fingerprint analysis

Closer examination in right figure reveals that the spectrum at one of the spots has CH3, CH2, NH and OH absorptions typical of proteins, while neighbouring spectra just show finger grease.

Instrument Vendors SensIR (microscope, ATR) Bruker Schmazhu Agilent Perkin Elmer Digilab Nicolet Buck Scientific (inexpensive)

FT-IR Applications Forensics Drugs, fibers, paint

Cell biology Proteins Polymers Structures, film layers

Lubricants Food analysis (NIR)

Summary For IR to occur, the vibration must undergo a change in dipole moment. Must have the correct energy to couple to this dipole moment so that the bond can be promoted to a higher vibrational energy level. As a result, will get an absorption. Structure of molecule is important to determine the absorption characteristics. FT-IR is very rapid and sensitive. In addition, spectra can be manipulated easily.

Raman Spectroscopy Review molecule polarization Raman instrumentation What is the difference between Raman and IR? What does a Raman spectrum look like? Some applications