IR

Question 1

How do atoms move

Answer 1

Atoms in a molecule do not maintain fixed positions - they vibrate back and forth.

Question 2

The bond length is

Answer 2

is an average value.

Question 3

A diatomic molecule motion

Answer 3

A diatomic molecule, e.g. H-Cl can only

undergo a STRETCHING vibration.

Question 4

More complex molecules motion

Answer 4

More complex molecules exhibit both STRETCHING and BENDING vibrations.

Question 5

Vibrational motion and infrared light

Answer 5

Vibrational motion is excited within a single electronic state by infrared light.

Question 6

Vibrations of bonds involving HYDROGEN are

Answer 6

Vibrations of bonds involving HYDROGEN are very significant as atoms of low mass move a lot more in comparison to atoms of higher mass.

Question 7

diatomic molecule

Answer 7

Treat a diatomic molecule as two masses joined by a spring

Question 8

Hooke’s Law

Answer 8

The force to compress or extend a spring is:
F(r) = – k (r - re)
k spring constant 
r actual length
re equilibrium length

Question 9

Units of force constant are:

Answer 9

Units of force constant are: N m-1

Question 10

Strong spring - _______ force constant

Answer 10

Strong spring - large force constant

Question 11

Weak spring - _________ force constant

Answer 11

Weak spring - small force constant

Question 12

Strong bonds have ________ force constants.

Answer 12

Strong bonds have large force constants.

Question 13

Potential energy in the spring, V(r)

Answer 13

V (r) = ½ k(r–re)^2

Question 14

The vibrational frequency is given by:

Answer 14

v = 1/2(pi) radical (k/μ)
Hz

SI Units
• Hz (or s-1) for v
• N m-1 for k
• kg (per molecule) for μ

Question 15

μ is the reduced mass

Answer 15

μ= (m1m2) / m1 + m2.

kg

Question 16

We frequently use masses

Answer 16

We frequently use masses in g mol-1 (e.g. H=1, C=12, O=16 …)
so
μ= ((m1m2) / m1 + m2)* u
Where u is the unified mass constant u = 1.66 x 10^-27 kg

Question 17

The SI unit of wavenumber is

Answer 17

The SI unit of wavenumber is m^-1

But cm^-1 is more common.

Question 18

wavenumber

Answer 18

v = 1/lambda

Question 19

wavenumber

Answer 19

cm^-1

v = 1 k
——– x radical ( ____)
200(pi) c μ

Question 20

The frequency is:

Answer 20

The frequency is:

1. directly proportional to the force constant
2. inversely proportional to the reduced mass

Question 21

Energy levels of a harmonic oscillator
classically:
quantum mechanically:

Answer 21

Classically the harmonic oscillator can have any energy.
Quantum mechanically it only has discrete equally spaced energy levels given by:
Ev = (v + ½) hV
V ( vib. freq)
v = 0, 1, 2,……. energy level

Question 22

Selection rule

Answer 22

Absorption of light only occurs for ∆v = ± 1

This means that each transition has the same energy change.

Question 23

what determines whether a molecule will absorb the infrared radiation?

Answer 23

Interaction with light

Oscillating dipole

Question 24

Interaction with light

Answer 24

1. Light has an oscillating electric field.
2. The molecule must also possess an oscillating electric field (at
   the same frequency).
3. Photon energy must equal the difference between energy levels.

Question 25

Oscillating dipole

Answer 25

1. Heteronuclear diatomics have a dipole due to the uneven electron distribution around each atom.
2. The dipole is the product of the charge  distance apart.
3. Bond vibrations change the distance and hence the dipole
   oscillates. This creates an oscillating electric field.

Question 26

Infrared inactive

Answer 26

1. Homonuclear diatomics (H2, N2, Cl2 etc.) do not have a dipole and cannot absorb light in the IR to excite bond vibrations. (The vibrational motion of homonuclear diatomics can be probed using RAMAN spectroscopy).
2. In many molecules, some vibrations are IR inactive, some are IR active. The requirement is a net oscillating dipole, e.g. CO2

Question 27

Characteristic Vibrations of H2O - all IR active

Answer 27

symmetric stretch n = 3657 cm-1
bend n = 1595 cm-1
asymmetric stretch n = 3756 cm-1

Question 28

Number of vibrational modes

Answer 28

Linear molecule with n atoms - number modes
= 3n–5
e.g. CO2 n = 3 so number vib modes = 3x3–5 = 4

Non-linear molecule with n atoms - number modes
= 3n–6
e.g. CH4 n = 5 so number vib modes = 3x5–6 = 9

Question 29

IR spectra are complex

Answer 29

The number of vibrational modes increases very rapidly with n, PLUS many of these vibrations may occur at the same frequency and so not all of the possible bands are seen as independent absorptions.

Question 30

Stretching vibrations can be:

Answer 30

Stretching vibrations can be: (i) symmetric

(ii) asymetric

Question 31

Bending vibrations can be:

Answer 31

Bending vibrations can be:

i) scissoring (in-plane
(ii) rocking (in-plane)
(iii) wagging (out-of-plane) (iv) twisting out of plane

Question 32

Infrared INACTIVE vibrations

Answer 32

Infrared INACTIVE vibrations
• Not all vibrations result in absorption bands.
• For a vibration to cause IR absorption the dipole moment of the molecule must change when the vibration occurs.

Question 33

IR spectra of Organic/Biological molecules

Answer 33

• Each stretching and bending vibration of a bond is associated with radiation whose frequency exactly matches the frequency of the vibration of the bond.
• Different functional groups, e.g. C–C, C=C, C=O, N–H … will have different characteristic frequencies
• Therefore, by determining the frequencies of light (IR) that are absorbed by a particular compound we can determine what kind of bonds it has.

Question 34

IR spectra

Answer 34

IR spectra are often shown as % Transmittance vs Wavenumber.

Question 35

fingerprint region

Answer 35

The fingerprint region is an area of complex overlaps where it is impossible to assign all (any?) of the bands. It can act as a unique ‘fingerprint’ for a given molecule.

Question 36

Middle IR spectrum can also be roughly divided into:

Answer 36

1. FUNCTIONAL GROUP REGION, v = 4000 - 1200 cm-1

2. FINGERPRINT REGION, v = 1200 - 400 cm-1

Question 37

stretching vs bending

Answer 37

It takes more energy to stretch a bond than bend it

Therefore stretching bands are found in the
4000-1200 cm-1 (functional group region)

Bends are typically observed in the fingerprint region

Question 38

functional group frequency

Answer 38

to be done later