Download Spectroscopy questions for midterm or semi-final exam, 2016

Electric properties of material I. and II.
Preparation for test writing
1.Dielectric
1a. Insulator material of low electric conductivity. It contains no mobile charged
particle.
2.Give two definitions for electric field strength and effective electric field strength
of a parallel plate capacitor.


Eeff  E  Epolar 
d
k 0
Potential difference between two metal sheets of capacitor divided by the distance
of the sheets. Effective field strength is proportional to the charge density on one of
the capacitor plates.
2a. E 
3. How to characterize homogeneous and inhomogeneous electric field?
3a.By the arrangement of force field lines. When lines run parallel the field is
homogeneous. When lines are bent the field is inhomogeneous, a problem occurs at
the edge of capacitor.
4. Dipole moment vector and its SI unit of a multi charge system.
4a. μ  qi  ri
Vectors are in boldface letters: µ, the resultant dipole moment and ri the distance of
one of the component charges. SI unit of µ is C/m.
5. What is the vector orientation of µ and E?
5a. Dipole moment vector points from negative to positive charges, while electric
field strength vector points toward the negative charges.
1
6. Polarization.
6a. Polarization can be defined by equation P 
1 i
N
 μi  μ
V N 1
V
We add up all the dipole vectors by magnitude and orientation in a volume V and
the two quantities are multiplied polarization density or simply polaziation we
get.
Other words: Polarization density, or simply polarization is the vector field that
expresses the density of permanent or induced electric dipole moments in
a dielectric material
7. Polarizability
7a. At low field strength μ induced is linearly dependent on E,
μinduced  α  E
unit α: C2m2J-1
with a proportionality coefficient, α, the polarizability characteristic to the
molecule, referred to as the deformability of electron cloud. Polarizability change
with frequency of electromagnetic radiation is used in Raman spectroscopy.
8. Compare the dipole moments of dichlorobenzene derivatives.
8a. para-dichloro benzene, µ = 0, two vectors with the same magnitude both are
localised in a straight line with the contrary direction.
ortho-dichloro benzene has grater moment than meta-dichloro benzene. (Show it by
making a drawing.)
9. Compare equations Pm 
 r 1 M NA 
 
, and
r  2 
3 0
r 1
NA 
2 
 

Pm 
V 
 r  2 m 3 0 
3kT 
The first eq. refers to nonpolar molecules µ = 0 (Clausius-Mossotti). The second
eq. is named after P. Debye. Can be applied polar molecules in a narrow
2
temperature range. If T is high enough the randomizing influence of thermal
agitation sets µ = 0.
The Clausius-Mossotti relation connects the relative permittivity εr of
a dielectric to the polarizability α of the atoms or molecules constituting the
dielectric. The relative permittivity is a bulk (macroscopic) property and
polarizability is a microscopic property of matter; hence the relation bridges the
gap between a directly-observable macroscopic property with a microscopic
molecular property.
10. Polarization types and their distinction
10a. The different polarizations can be distinguished by using up their frequency
dependence. 1. Orientation polarization: resonancia achieved in radio frequency
and microwave range of EM waves. Under an external electric field, the dipoles
rotate to align with the electric field causing orientation polarization to occur. In the
higher frequency range there is no resonance between molecules and the field
observed.
2.Distortion polarization
Electron and distortion polarization are independent of dipole orientation generated
by EM waves. Resonance frequencies (1012 – 1014 s-1). identical the vibrational
frequencies that fall into infrared.
3.Electron polarization
The electron polarization is a property of electrons. They lightly follow the
frequency of photons in the visible and ultraviolet range. (1014 – 1017 s-1).
Total polarization can be measured in the microwave region, electronic polarization
exclusively can be observed in the vis or uv ranges.
11. Explain by using figure below
3
the relation of phase velocity in the upper and lower medium. Derive the refractive
index of two media by using angles.
11a. The velocity ratio of monochromatic light can be phrased by the ratio of
refractive indices
vi
n
 t
v t ni
when the beam advances from optically less denser to optically denser media.
vincident is greater than vtransmitted because the incident light runs in the less denser
medium.
According to the Schnell’s law
sin θi  ni  sin θt  nt
 nr 
nt sin θi


ni sin θt
nr is referred to the first medium, and has a value greater than one, because
sin θi  sin θt .
12. What does the refractive index depend on?
4
12a.The refractive index depends on the wavelength of light beam and the
temperature of the system probed. The wavelength dependence of  nr  is the
dispersion. In order to compare the results measured on different Abbe
refractometers you need to use identical wavelength, e.g. sodium D-line, λ =
589 nm .
13. How to give Debye equation in terms of refractive index?
13a. Maxwell law states for nonmagnetic materials:
nr2  εr
refractive index squared equals the relative permittivity. Substituting this into
Debye equation
NA 
2 
 

Pm  2
 Vm 
3 0 
3kT 
nr  2
we another function applicable to calculate polarizability and dipole moment.
The experimental determination of refractive index easier than that of relative
permittivity That is the advantage of this function.
nr2  1
14. What type of interaction in solution is characterized? How does it characterize?
14a. The potential energy vs. distance function is a good tool for characterization of
strength of different interactions in solution. The ion – dipole interaction is
described by taking the sum of repulsive and attractive interactions.
Starting arrangement of charges and equation:
5



q1q2
1
1 


V  Vr  Va 

4 0  r  l r  l 



2
2
The pot. energy distance function for ion - dipole interaction:
V 
q21
4r 2 0
Boltzmann distribution, Selection rules
Preparation for test writing
1.Give the distribution function for a multicomponent and a binary (two
component) system.
1a.
nj
N

exp  ε j / kT 
 exp- εi / kT 
i
It means that we have j number of components, sitting on εj energy level. Each of
them has their own level for waiting a jump to another εj energy level.
n2
 ε ε 
 exp   2 1 
n1
kT 

The distribution function for a binary system.
2. How does the ratio of
n2
change when ε2  ε1 difference increases?
n1
2a. The increase of difference in the negative exponent decreases the ratio, which
means that only a smaller amount of particle is present on the upper level. At the
same time, the level distance is also increases.
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3. How many out of every 20 000 molecules is found to be at the first vibration
excited state in a time average?
Data:
RT = 298.15·8.314 = 2478.9 Jmol-1, for CO molecules levels Δε=25 kJ mol-1.
We have a system containing 20 000 molecules.
 25 kJ mol-1  10
nv 1
  e  5 10  5
3a.
 exp 
-1 
nv  0
 2.48 kJ mol 
If we have a system containing 20 000 molecules, than
nv 1  20 000  5  10 5  1
It means that one out of every 20 000 molecules is found to be at the first vibration
excited state in a time average.
4. Calculate the number of microstates for the case eight particles having nine
possible energy levels. Use the table attached here.
Table. The possible arrangements of 8 atoms and 9 allowed energy levels:
Case
x
n0
n1
n2
n3
n4
n5
n6
n7
n8
W
Data 4! = 24, 5! = 120, 6! = 720, 7! = 5040, 8! = 40320.
4a. Let’s fill the boxes by 8 atoms in a way the sum of energies taken up by atoms
is 8 units altogether.
Case
x
n0
5
n1
1
n2
0
n3
1
The number of microstates,
W
n4
1
W
n5
0
n6
0
n7
0
n8
0
N!
, and in our case
n0 ! n1! n2 !...
8!
40320

 336
5!1!0!1!1!0!0!0!0! 120
5. Born-Oppenheimer approximation
7
W
5a. The quantized energies of a molecule do not interact each other, because of
great transition energy differences, they can handle separately. That is why the
additivity can be written as
ΔE = ΔEelectronic+ ΔEvibration + ΔErotation
and
ΔEelectronic > ΔEvibration > ΔErotation The inequality tells us the direction of energy
decrease.
6 Transition integral.
f,i   Ψ final μˆ Ψ initial dτ  0
6a. The value of transition integral determines a certain transition available or not.
If the transition dipole moment, the moment difference between final and initial
state is zero we are unable to measure the energy change of a transition. We say the
transition is forbidden.
7. Calculate the energy of 285 nm photon (c = 3 108 m/s, h = 6,63 10-34 Js). Using
the Planck equation.
7a. Ephoton = h ν = (h c) / λ
8.Reproduce the energy level diagram of different electronic and vibration
transitions.
8a. Superposition principle should be applied.
9.Vibration and rotation selection rule.
9a. Selection rule for vibration is:
dμ
0
dQ
8
where Q is the coordinate of the vibration. For a diatomic: Q = R (bond length).
The vibrational transition requires a dipole moment change,
Selection rule for rotation is:
The physical requirement for observation of a pure rotational spectrum is an
existing permanent dipole.
There is no dipole moment change when rotation axis coincides the direction of the
bond.
File: SP0506
Beer Lambert law, Beer Lambert usage
Preparation for test writing
1. Flux of photons, intensity of photons, transmittance, absorbance
1a.  
np
The equation defines flux of photons which crosses an area
q t
q, during time t .
I    h 
h  np
q t
Flux can be transformed to intensity , when flux of photons is multiplied
by h . This intensity refers to a monochromatic source if ν is taken as
constant.
The ratio of outgoing and incoming intensities is named as transmittance.
This intensity function can be measured directly by spectrophotometer.
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I
I0
T
ln
I0
 A    c  l Absorbance is a logrithmic intensity ratio, which is
I
concentration, c thickness, l and absorption coefficient, ε dependent.
ε is the molar absorbance if concentration is given in mol dm-3.
Its linear concentration dependence is a good tool to determine concentrations.
A       c  l The absorbance is wavelength dependent.
2. Draw a concentration dependence of a spectral band. Note the measuring
conditions of band series.
2a. One can use the wavelength range where the band intensity is meaningful.
Thickness of cuvette should be constant for all the concentration. With increasing
concentration the band intensities should be greater and greater.
0,5
Concentration series
b = const.
 = const.
Absorbance
0,4
0,3
0,2
0,1
0,0
200
300
400
500
600
Wavelength / nm
Draw something like this.
3. What an experimental arrangement can be seen on figure below?
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Find the relationship between I0 and I.
3a. The figure is about an experimental arrangement of a monochromatic light
intensity decreasing from I0 incoming intensity to I outgoing intensity. The light
beam crosses the sample placed in a cuvette. To observe this effect the sample
should contain at least one optically active material which absorbs light at the
wavelength of incoming beam.
4. Deduce the Beer Lambert law!
4a. The derivative 
dI
is proportional to a material coefficient, ε the outgoing
dl
intensity, I and the concentration of light absorbing material c in the solution. It is
given as:

dI
 I   c
dl
Integrating this equation between limits of intensity, and limits of thickness
l
1

d
I



c
I 0 I
0 dl
I
we get
 ln I  ln I 0     c  l
After simplification
ln
I0
 A   c l
I
we name the logarithmic ratio as absorbance. The function form:
A       c  l
Absorbance is directly proportional to the concentration and thickness of solution.
The molar absorbance is wavelength and solution matrix dependent.
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5. Riboflavine in dilute acetate solution shows a maximum in absorption spectrum
at 444 nm.
Calculate the wavenumber in cm-1 and frequency of light at this peak.
Calculate the energy of corresponding electronic transition for one molecule and
one mole of riboflavin.
Data: h = 6.626 10-34 Js, c = 3.00 108 ms-1, NA = 6.022 1023, λ = 444 nm
5a. First transform data to proper units:
λ = 444 10-9 m = 444 10-7 cm
the wavenumber
~

1


1
 22522.5 cm-1
7
444 10
the frequency
3 108
14 -1
 

6
.
756
10
s
 444 109
c
For one molecule E  h   6.626 10-34  6.756 1014  4.476 1019 J
19
For one mole Emolar  E  N A  4.476 10
269.4552 kJ/mol.
 6,02 1023  269455.2 J/mol =
6. Dissociation equilibrium constant, K is given by equilibrium concentrations at dT
H   A
= 0 and dp = 0. K 
. Find K in terms of α, the dissociation degree.
HA 
  
6a. The relationship between the measured concentration of analyte (c) and
equilibrium (e.g. [A-]) concentrations
[A-] = [H+] = c∙α and [HA] = (1-α)∙c . Doing the substitutions
c 2 2
c 2
K

1   c 1  
7. The optical arrangement and main elements working principle of a double beam
spectrophotometer.
7a.
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Sources: deuterium lamp UV range, tungsten lamp VIS range
Monochromator:
produces monochromatic radiation
Sector mirror: changes beam from reference to sample at a given wavelength
Sample and reference cuvettes:
solvent absorbance is eliminated
Detector:
(phototube) converts intensity to electric current or voltage
Ratio:
is proportional to I/Io
8. Why we use reference system?
8a. Interaction between light and medium. The disturbing interactions should be
subtracted by preparing proper reference solution.
Proper reference solution contains all the material in the same concentration as in
the sample solution except the optically active material/materials to be studied.
9. Explain the cause the deviation from BL law.
9a. using high analyte concentration:
causing strong electrostatic interactions between molecules,
increases refractive index, consequently photons by-pass the detector, greater
absorbance is detected.
if we have a system in chemical equilibrium, equilibrium may shift at high
concentrations
10. Give an explanation for the effect of non-monochromatic radiation.
10a. Monochromators do not produce a single wavelength radiation. The
“monochromatic” beam contains several wavelegths.
Calibration curve is constructed at band A shows a straight line, while that is
constructed at band B is bent.
Explanation: There is a flat part of absorption band in the vicinity of its maximum
(band A), therefore the average wavelength has a small standard deviation. Molar
absorbance is hardly shifted at the peak, while there is a great change in molar
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absorbance at the sides. A small shift in wavelength causes a great change in molar
absorbance.
11. Chromophore, red shift
11a. A chromophore is the part of a molecule responsible for its absorbance. For
example in acetone CH3- CO -CH3 the carbonyl group. The chromophore is a
region in the molecule where the energy difference between two
different molecular orbitals falls within the range of the visible spectrum.
Conjugation of π bonds causes red shift of λmax
Original molecule without conjugation: CH2=CH-CH2-CH2-CH=CH2 λmax =185
nm. Molecule with conjugation: CH2=CH-CH2=CH-CH=CH2 λmax =217 nm
File: SP0708
Luminescence, PhotoelectronSp
Preparation for test writing
1.Singlet and triplet electronic spin states and transitions. Make an arrow diagram
and explain it.
1a. Electrons in molecular orbitals are paired, according to Pauli exclusion
principle. In ground state the two electrons are paired. In a singlet transition
electron preserve original orientation. In a triplet transition one of the electrons flip
in spin. Electrons are unpaired, extremely low probability geometry. From singlet
transition it can be gained by intersystem crossing.
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2. Time domain of fluorescence and phosphorescence.
2a. After irradiation stops the fluorescent intensity abruptly decreases to zero.
Lifetime of fluorescence: 10-5 – 10-9 s.
After irradiation stops
Lifetime of phosphorescence: 10-5 – 10 s.
After irradiation phosphorescent light is visible for a long time. Triplet states,
resulting from a spin interconversion process in the excited state, have quite long
lifetimes, sometimes to beyond 1 millisecond. That is why phosphorescent light is
visible for a long time.
3. Compare in terms of wavelength and energy of absorbed and emitted fotons in
the fluorescence process!
3a. The ground singlet state molecule, S0, goes into first excited singlet state, S1,
Fluorescence may appear as an approximate mirror image of the absorption at a
lower energy, and hence longer wavelength than the absorption, the difference (in
wavenumbers) between the two corresponding bands being known as the Stokes
shift.
4. Detail the energy releasing process in phosphorescence!
4a. Phosphorescence may be defined as the emission occurring due to the radiative
transition between two states of different spin multiplicity, for example between T 1
and S0.
15
The intersystem crossing process takes place where two electronic states T1 and S1
have the same energy. Relax to ground state with a second flip in spin to satisfy the
singlet ground state.
5. The photoionization process.
5a. M + hν = M+ + eFrom monochromatic source a photon ionizes an atom or molecule, depending its
energy. In XPS the photon is absorbed by an atom in a molecule, leading to ionization
and the emission of a core (inner shell) electron.
In UPS the photon interacts with valence levels of the molecule, leading to ionisation
by removal of one of these valence electrons.
6. Apply the energy conservation law to the ionization process according to
Koopman’s theorem! Which term of equation is measured?
6a. Photoelectron spectroscopy is based upon a single photon in/electron out
process. The law of energy conservation says
16



12



h

m
v

E
M

E
M
e
e
2
The total energy of photon is the sum of the binding energy, BE




and the kinetic energy of electron. By measuring the


BE

E
M

E
M
velocity of photon the kinetic energy can be determined. The photon’s energy is
well known, therefore binding or ionization energy can be determined.
7. How does electron energy analyser work?
7a. . PES uses monochromatic sources of radiation. E.g. Mg Kα radiation:
hν = 1253.6 eV Sample is placed in a high vacuum chamber (P ~ 10−8 millibar) for
excluding other ionization (oxygen, nitrogen or other components from air)
processes than the studied one.
The preferred option for photoemission experiments is a concentric hemispherical
analyser (CHA) which uses an electric field between two hemispherical surfaces to
disperse the electrons according to their kinetic energy. The electric field is
increased gradually and the greater the field the higher velocity electron get into the
window before the detector.
The higher the velocity measured the lower the ionization energy of photoelectron.
8. Characterize XPS method!
8a. The intensity of the peaks is related to the concentration of the element within
the sampled region. The most commonly employed x-ray sources are those giving
rise to:
Mg Kα radiation: hν = 1253.6 eV
Al Kα radiation: hν = 1486.6 eV
Each element will give rise to a characteristic set of peaks in the photoelectron
spectrum at kinetic energies determined by the photon energy and the respective
ionization energies. Core ionization energies are characteristic of the individual
atom rather than the overall molecule.
The presence of peaks at particular energies therefore indicates the presence of a
specific element in the sample under study.
9. Characterize UPS method!
9a. In UPS the photon interacts with valence levels of the molecule, leading to
ionisation by removal of one of these valence electrons. After ionization process
occurred the electrons remaining rearrange their distribution.
17

When ejected electron arises from bonding orbital, a vibration term, E v should be
added to energy balance equation.
1
h  me ve2  I  Ev
2
File: SP0911
Vibrational, Rotational, NMR
Preparation for test writing, (do not forget to explain your notation).
1.Describe the vibration frequency vs. force constant function given for case of a
mechanical system, and that of a molecular bond!
1a.
The vibration frequency of both mechanical and molecular system is proportional
to force constant, k. The vibration frequency of mechanical system is inversely
proportional to the mass of the ball.

1
k

2 m
18
For molecular system m is replaced by μ, which is given as
1


1
1
.Where

m1 m2
m1 and m2 are the vibrating atoms.
2.The quantum mechanical energy term , its description: quantum jump, vibrational
energy equation, energy level distances.
2a. The force law F = -kx is valid. The energy levels are equidistant. The change in
vibrational quantum number, Δv is restricted to one. The system’s vibrational
energy can be increased or decreased by unit amount of energy independently the
positions of levels. The vibrational energy difference between two successive levels
is given: Ev  v  1 / 2   h
3. The anharmonic oscillator model, its descrition: quantum jump, vibrational
energy equation, level distances in energy.
E
2
3
~
  v  0 v  1 / 2  xe v  1 / 2  ye v  1 / 2
hc


3a. To draw a figure is essential.
Vv
Harmonic
oscillator
Anharmonic
oscillator
2
1
0
r0
Bond distance
The energy distance function is an imperfect parabola. The function can be
approximated a power series. The change in vibrational quantum number, Δv can
take the values ±2, ±3, … The levels are not equidistant.
4. How do you calculate the number of normal vibration?
19
4a. The total number of mechanical freedom for an N-atomic molecule is 3N.
Subtracting the 3 translational freedom 3N -3 means the remaining freedom is
vibrational and rotational. To give the vibrational one we subtract 3 rotational for
non-linear molecules and 2 for linear molecules. The results:
3N – 6 non-linear
3N – 5 linear
5.Calculate the number of normal vibrations for ethylamine, propan-1-ol, CS2
5a. For CH3CH2-NH2 the number of atoms are 10, applying formula 3N-6 for this
angle molecule the number of normal vibration modes: 24.
For CH3CH2-CH2-OH the number of atoms are 12, applying formula 3N-6 for this
angle molecule the number of normal vibration modes: 30.
For CS2 the number of atoms are 3, applying formula 3N-5 for this linear molecule
the number of normal vibration modes: 4.
6. Alkanes. What kind of stretching vibrations are characteristic in the IR spectra of
alkanes?
6a.The C-H stretching vibrations originating from CH3 and -CH2- groups are
characteristic. The asymmetric and symmetric stretchings can be found in the
higher energy part, around 2800 - 3000 cm-1 in the IR spectrum.
7. Compare the band positions of stretchings and bendings of alkanes!
7a. For a given functional group e.g. C-H group, the stretching vibrations always
found at a far more higher wavenumber than that of the bending vibrations. The
dipole moment change is greater for stretching vibrations.
Example: one of CH stretching vibrations in n-hexane around 2900 cm-1, one of CH
bending vibrations in n-hexane around 1470 cm-1.
8. Definitions: torque, moment of inertia, angular momentum, rotational energy.
8a. Torque: m·r, where m is the rotating mass spinning at a perpendicular distance
from rotational axis.
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Moment of inertia, I: I   mi ri2
i
Angular momentum: J = I·ω, where ω is the angular velocity.
Rotational energy: E 
I 2
2
9. Explain the meaning of parameters and variables, Erot  J   J  J  1 
h2
8 2  I
,
9a.
Rotational energy is quantized as J  J  1  by quantum number J. The only
variable here is I the moment of inertia which changes from molecule to molecule.
The greater the extent of molecule the higher the value of moment of inertia.
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