Quiz 2 18/02/14

1. If the wave length of a light is 2.5 micron, what is its corresponding wave number (cm -1)?

a. 4000 b. 4050 c. 3000 d. 3050

2. What is the range of uv-visible radiation?

a. 200-700 mµ b. 200-800 nm c.200-800µ d.150-750nm

3. In which spectroscopic techniques the change in vibrational and rotational movements can be
detected?

a. UV-Vis b. NMR c. IR d. MS

4. Which of the following is a forbidden transition?

a. α-α* b. π-π* c. n- π* d. Both a & b

5. In the past, IR spectra had to be aquired one wavelength at a time, which took a long time. Today
quick spectra is due to the

a. the Fourier Transfer Algorithm allows us to scan all frequencies at once

b. light is faster today that it used to be
c. absence of broad spectrum of wavelength
d. none of the above

6. What is spectroscopy? What is its importance?

Spectroscopy is the use of the absorption, emission, or scattering of electromagnetic radiation by

matter to qualitatively or quantitatively study the matter or to study physical processes.

Spectroscopy is used in analytical chemistry for the identification of substances through the
spectrum emitted from or absorbed by them.

7. Account for the fact that the n-π* absorption of acetone occurs at 279 nm in n-hexane, 270 nm in
ethanol and at 265 nm in water.

 Peaks resulting from n-π* transitions are shifted to shorter wavelength(blue shift) with
increasing solvent polarity
 Because of the increased solvation of the lone pair in the ground state
 Which lowers the energy of the n orbital

The ultraviolet spectrum of benzonitrile shows a primary absorption band at 224 nm and a secondary
band at 271.

a) If a solution of benzonitrile in water , with a concentration of 0.0001 molar, is examined at a

wavelength of 224 nm, the absorbance is determined to be 1.30. The cell length is 1 cm.
What is the molar absorptivity of this absorption band?
Beer–Lambert law, A=Єbc
Є=A/bc
 =1.30/(0.0001M*1cm)
=13000M-1cm-1

b) If the same solution is examined at 271 nm, what will be the absorbance reading
(Єmax = 1000)?

Beer–Lambert law, A=Єbc

=1000M-1cm-1*1cm*0.0001M
=0.1

What will be the intensity ratio?

Transmittance (T) is the intensity ratio

A = -log T

T = 10-A

T = 10-0.1

T = 0.794

8. The UV spectrum of acetone shows absorption maximum at 166, 189 and 279 nm.

a. What type of transitions is responsible for each of these bands?

166 n→

189  →

279 n→

b. Explain three applications each of UV-VIS and IR spectroscopy.

Lecture notes

9. What is the range of IR radiation?

10. Give four(4) comparison of the Dispersive IR and Fourier Transform IR.

Lecture notes

Which of the following wavelength ranges is associated with UV spectroscopy?

Correct answer:

b) 400 - 100nm

Feedback:

0.8 - 500µm is in the IR spectroscopy range, 380 - 750nm is the visible spectroscopy range and
0.01 10nm is the X-ray range.
Page reference: 299

Which of the following compounds does not absorb light in the UV/visible spectrum?

Correct answer:
c) Chloral hydrate

Feedback:

In order to absorb light in the UV/visible spectrum, a molecule must possess a chromophore e.g.
a benzene ring or conjugated double bonds. Chloral hydrate does not possess a chromophore.

20 Carbamazepine tablets were found to weigh 10.000g in total. The tablets were ground to a
fine powder using a pestle and mortar. A 0.3000g sample of the powder was boiled with 25ml
ethanol for a few minutes. The hot mixture was stirred in a closed flask for 10 minutes and
filtered through sintered glass. The flask and filter were washed with ethanol and the washings
combined with the filtrate. The cooled mixture was made up to 100ml with ethanol. 5ml of the
ethanol extract was diluted to 250ml with ethanol and the absorbance of the resulting solution
was found to be 0.588 absorbance units at a wavelength of 285nm. What was the content of a
single tablet if the A1%,1cm was 490?

Your answer:

d) 120mg

Correct answer:

c) 100mg

Feedback:

Using the Beer-Lambert law expressed in grammes and following the principles involved in the
example calculation on page 302 should enable you to arrive at the correct answer of 100mg
(don't forget about the dilutions).

A series of 3 coloured glass plates of equal thickness are placed in a light beam. Each sheet
absorbs one quarter of the light incident upon it. What is the intensity of the light transmitted by
the third glass plate?

Your answer:

b) 42.19%

Feedback:

As each plate absorbs one quarter of the incident light, it is simply a matter of multiplying by
0.75 on three consecutive occasions.

n infrared spectroscopy which frequency range is known as the fingerprint region?

Your answer:

a) 400 - 1400cm-1

Correct answer:

b) 1400 - 900cm-1

Feedback:

The fingerprint region of an infrared spectrum, which is characteristic for each individual
compound, is between 1400 and 900cm-1.
Page reference: 305

n which region of the infrared spectrum would you expect to find a peak characteristic of a triple
bond stretch?
Your answer:

b) 2500 - 2000cm-1

Feedback:

Carbon - carbon and carbon - nitrogen triple bond stretching absorbs in the 2500 - 2000cm-1
range.

In which region of the electromagnetic

spectrum does an absorption at 600 nm
come?

Your Answer: Visible.

2. A solution of 0.001 mol dm–3

NiSO4 is placed in an optical
cell of pathlength 1 cm, and the
absorption spectrum is recorded.
The absorptions have
characteristic λmax and εmax
values. What are the correct
units of εmax?

Your Answer: dm3 mol–1 cm–1

3. In which region of the

electromagnetic spectrum does
an absorption at 177 nm come?

Your Answer: Visible.

Correct Answer: Vacuum-UV.

4. What does the notation σ*←n

mean?

Your Answer: Absorption;

transition
from a
quantum level
n to σ* MO.
Correct Answer: Absorption;
transition
from a non-
bonding MO
to σ* MO.

5. How do values of λmax for the

π*←π transitions vary among a
series of conjugated polyenes?

Your Answer: Values shift to

longer
wavelength as
 the number of
C=C double
bonds increases.

6. What is a red shift?

Your Answer: The shifting

of an
absorption to
shorter
wavelength.
Correct Answer: The shifting
of an
absorption to
lower energy.

7. What is a chromophore?

Your Answer: A group of

atoms in a
compound
responsible for
the absorption of
electromagnetic
radiation.

8. Which of the following is the

principal chromophore in an
azo-dye?

Your Answer: C≡N

Correct Answer: N=N

9. Lycopene (λmax = 469 nm) is

present in tomatoes. What
colour of light does lycopene
absorb?

Your Answer: Blue.

10. In which of the following are

the π-electrons not delocalized?

Your Answer: Allyl anion.

Correct Answer: Hepta-1,6-
diene.

11. All but one of the following

groups can donate a lone pair of
electrons into the carbon π-
 system of a conjugated polyene.
Which is the odd one out?

Your Answer: SMe

Correct Answer: Me

12. Which statement is incorrect

about the allyl anion and buta-
1,3-diene?

Your Answer: In both, the C

atoms are all
sp2
hybridized.
Correct Answer: Both have
four π MOs.

13. Which of the following

arrangements of carbon-carbon
double bonds along the
backbone of a polyene does not
correspond to a conjugated
system?

Your Answer: –C=C–C=C–

Correct Answer: –C=C–C–
C=C–

14. An aqueous solution of a dye

has a strong absorption with
λmax= 464 nm. The colour of the
solution is:

Your Answer: green.

Correct Answer: orange.

15. Which of the following

absorption maxima is not in the
visible range of the electronic
spectrum?

Your Answer: λmax = 480 nm

Correct Answer: λmax = 250 nm

16. Which of the following

statements is consistent with an
electronic absorption being
broad?

Your Answer: Hydrogen

bonding
 causes an
electronic
absorption to
be broad.
Correct Answer: An electronic
absorption
includes
vibrational
and rotational
structure.

17. Two solutions of the same

compound were made up.
Solution A was of concentration
0.98 × 10–4 mol dm–3, and
solution B was 1.66 × 10–2 mol
dm–3. The electronic spectrum
of solution A contained one
absorption at λmax = 230 nm,
while the spectrum of solution B
contained absorptions at λmax =
230 and 365 nm. The difference
in the spectra was because:

Your Answer: the value of

εmax depends
on
concentration.
Correct Answer: εmax for the
absorption at
365 nm is
much smaller
than εmax of
the band at
230 nm.

18. A compound X is characterized

in its electronic spectrum by an
absorption with λmax = 217 nm
(εmax = 21 000 dm3 mol–1 cm–1).
Of the compounds given below,
X is most likely to be:

Your Answer: β-carotene.

Correct Answer: buta-1,3-
diene.

Question 1
How is the wavelength controlled in an FTIR spectrometer?
a. By a Michelson Interferometer
b. By a computer
c. By a laser
d. By calibration with a standard sample

The wavength is very accurately controlled by a laser.

--------------------------------------------------------------------------------
Question 2
What type of technique is FTIR spectroscopy?
a. A dispersive technique?
b. An emission technique
c. An absorbance technique
d. A UV-Vis technique
e. A reflectance technique

FTIR is an absorbance technique.

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Question 3
How is the detector on the Mattson RS/1 FTIR spectrometer cooled?
a. With water
b. With liquid nitrogen
c. With a fan
d. By Peltier cooling
e. It's not.

It has a room temperature detector.

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Question 4
What does the spectrum of Nitrogen(N2) look like?
a. The same as that of air
b. It has only p- and r-branches
c. The same as that of carbon monoxide
d. It doesn't have one!

There is no dipole change for vibration of N2, therefore it has no spectrum.

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Question 5
What are the selection rules for FTIR absorption?

A: Absorption only occurs for symmetrical molecules

B: Absorption only occurs with a dipole change

C: Absorption requires an odd vibrational quantum number

D: Absorption requires promotion to a new quantum level

1. C and D
2. All A,B,C AND D
3. A and D
4. A, B and C
5. B and D

Infrared absorption requires a dipole change for the vibration, AND sufficient energy to
promote the molecule to a new vibrational quantum state.

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Question 6
What occurs when the moving mirror in an FTIR spectrometer is the same distance from the
beamsplitter as the static mirror?
a. Constructive interference
b. Destructive interference
c. Radio interference
d. The spectrum is measured
When the 2 mirrors are at the same distance, the infrared travel the same distance, and
therefore constructive interference occurs.

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Question 7
Which infrared technique can measure two spectra at once?
a. FTIR spectroscopy
b. FTIR-ATR spectroscopy
c. Dispersive infrared spectroscopy
d. FTIR microscopy
e. none

FTIR is single-beam wheras dispersive is double beam.

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Question 8
Which spectrometer is the more complicated: dispersive or FTIR?
a. Dispersive
b. FTIR
c. neither

The FTIR is much more complicated than most dispersive spectrometers

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Question 9
In what region of the spectrum does infrared radiation occur?
a. At the low-energy end
b. Between the visible and ultraviolet regions
c. Between the visible and microwave regions
d. Between the visible and x-ray regions
e. At the high-energy end

Infrared is just lower in energy than visible light. It occurs between the visible and
microwave regions.

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Question 10
What occurs when a molecule absorbs infrared radiation?
a. It warms up
b. It flies around
c. It spins faster
d. It vibrates faster
e. It emits light

Either c or d. In this course we discuss vibrational spectroscopy, but rotational

spectroscopy is also possible.

What are the units most commonly used to identify bands in FTIR spectroscopy?
a. Wavenumbers(cm-1)
b. Wavelength(m)
c. Frequency(Hz)
d. Wavelength(micrometres)

The most common unit is the wavenumber, cm-1.

What are interference fringes caused by?

a. Not putting the sample in properly
b. Gasses in the atmosphere that absorb infrared.
c. Reflection of infrared by the sample.
d. Impurities in the sample.
They're caused by internal reflection in your sample.

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Question 6
What is the correct way to display spectra?
a. In absorbance.
b. In transmission.
c. Either absorbance or transmission
d. As raw data.

You may display in absorbance or transmission.

If you measured a spectrum and it was saturated, what could you do to improve the spectrum?
a. Reduce the iris setting
b. Use a thinner sample.
c. Scan for a shorter period of time.
d. Increase iris setting
e. Reduce iris setting OR use a thinner sample

Either reducing the iris setting OR using a thinner sample would help.

This set of Analytical Instrumentation Multiple Choice Questions & Answers (MCQs) focuses
on “Fourier Transform Infra-red Spectrometers”.

1. Which of the following is not true about Fourier Transform Infrared (FTIR) spectrometer?
a) It is of non-dispersive type
b) It is useful where repetitive analysis is required
c) Size has been reduced over the years
d) Size has increased over the years
View Answer

Answer: d
Explanation: FTIR is of non-dispersive type of instruments and is used for repetitive analysis.
Initially, it was bulky and the cost was high and hence, it was limited to a special application.
Now, the size has been reduced.

2. In the most widely used beam splitter, a thin film of ________ is sandwiched between two
plates of low refractive index solid. Fill the blank with a suitable option.
a) Mylar
b) Silicon carbide
c) Ferrous oxide
d) Silver chloride
View Answer

Answer: a
Explanation: Generally, mylar is the thin film used in beam splitters. Beam splitter has to pass
50% of the radiation and reflect 50% of the radiation.

3. Which of the following is not the function of the drive mechanism in Fourier Transform
Infrared Spectrophotometer?
a) Movement of mirror to obtain a satisfactory interferogram
b) Acquire a good interferogram pattern
c) Allow 50% of the beam to pass
d) Keep the speed of the moving mirror constant
View Answer

Answer: c
Explanation: Drive mechanism does the functions specified in the other options. It the function
of the beam splitter to allow 50% of the beam to pass through.

4. Only pyroelectric transducer or pyroelectric crystals are used as detectors in Fourier

Transform Infrared Spectrophotometer (FTIR). What is the main reason for other types of
thermal detectors are not being used in FTIR spectrophotometer?
a) Less accuracy
b) Slower response
c) Less precision
d) Less sensitivity
View Answer

Answer: b
Explanation: Other thermal detectors are not employed due to slower response of the detectors.
Pyroelectric transducers or pyroelectric crystals are chosen for their high speed, accuracy,
precision, sensitivity and resolution.

5. Which of the following is not the advantage of Fourier Transform Spectrometers?

a) Signal to noise ratio is high
b) Information could be obtained on all frequencies
c) Retrieval of data is possible
d) Easy to maintain
View Answer

Answer: d
Explanation: The instrument is not easy to maintain and it is a disadvantage. The other
disadvantage is that the cost is high.

6. Which of the following has to be computed to determine transmittance and absorbance at

various frequencies?
a) Ratio of signal and noise
b) Ratio of sample and reference spectra
c) Sample spectra
d) Reference spectra
View Answer

Answer: b
Explanation: The ratio of sample and reference spectra needs to be computed to determine
transmittance and absorbance. First, reference interferogram is obtained and then the sample is
obtained.

7. Computer accepts analog signals directly.

a) True
b) False
View Answer

Answer: b
Explanation: Computer does not accept analog signals directly. An A/D converter is required to
feed the signals to the computer.

8. Which of the following is the reference that is generally used in FTIR interferometer?
a) Air
b) NaCl solution
c) Alcohol
d) Base solution
View Answer

Answer: a
Explanation: Air is generally used as the reference. It is scanned for about 20 to 30 times and the
results are stored in a computer.

9. In Michelson Interferometer, if the reflected and transmitted beams are in phase at the beam
splitter, then maximum intensity will reach the detector.
a) True
b) False
View Answer
Answer: a
Explanation: If the reflected and transmitted beams are in phase at the beam splitter then
maximum intensity will reach the detector. If they are out of phase, then minimum intensity will
reach the detector.

10. The diagram given below is the representation of Fourier transform interferometer. Identify
the missing block in the block diagram.

a) Pyroelectric crystal
b) Display
c) High speed plotter
d) A/D converter
View Answer

Answer: d
Explanation: Computer does not accept analog signals directly. An A/D converter is required to
feed the signals to the computer.
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11. In Michelson’s interferometer, the frequency of the detector output can be determined by
translating the _________ of movable mirror and the ___________ of monochromatic radiation.
a) Velocity, wavelength
b) Thickness, intensity
c) Length, velocity
d) Angle, intensity
View Answer

Answer: a
Explanation: The frequency can be determined by translating the velocity of the movable mirror
and the wavelength of monochromatic radiation. The reflected beam passes towards the movable
mirror.

12. In Michelson’s interferometer, the __________ of the detector output will depend upon the
intensity of incoming radiation.
a) Velocity
b) Frequency
c) Amplitude
d) Phase
View Answer

Answer: c
Explanation: In Michelson’s interferometer, the amplitude of the detector output will depend
upon the intensity of incoming radiation. If a movable mirror is moved uniformly, the output will
be a sine wave.

13. Why is the computer necessary in Fourier Transform Spectrometer?

a) To display the detector output
b) To process the detector output
c) To determine the amplitude
d) To determine the frequency
View Answer

Answer: b
Explanation: The computer is necessary to process the output of the detector. The Fourier
Transform of output is determined using software using computers.
Question 1.
What is the relationship between wavelength and wavenumber ?
(Hint: Wavenumber is measured in cm-1).
a) Wavenumber = 1 / wavelength in centimeters
b) Wavenumber = 1 / wavelength in nanometers
c) Wavelength in nanometers = 1 / wavenumber
Answer: a

Question 2.
For a molecule to absorb IR, why must the molecule's vibrations cause fluctuations in the dipole
moment of the molecule?

a) Because a change in dipole moment lowers the energy required for electronic transitions
b) Because for absorption to occur, the radiation must interact with the electric field caused by
changing dipole moment
c) Because fluctuations in the dipole moment allow the molecule to deform by bending and
stretching
Answer: b

Question 3.
Why are rotational transitions of little use to a spectroscopist?

a) Because the energy required to induce a rotational transition is so small that it cannot be
measured
b) Because rotational transitions are extremely rare
c) Because, in liquids and solids, spectral lines corresponding to rotational transitions are
broadened into a continuum as the result of molecular collisions and other interactions
Answer: c

Question 4.
The energy of a vibration (corresponding to the absorption wavelength) can be influenced by
other vibrations in the molecule.
false