Instrumental Techniques Test On UV-vis And NMR
Test to practice.
- 1.
In quantitative analysis by UV-VIS spectroscopy, the error in the determination of the concentration will be lower if we work with:
- A.
Big cuvettes
- B.
Highly diluted solution
- C.
In the range of intermediate transmittances.
- D.
High absorbance
Correct Answer
C. In the range of intermediate transmittances.Explanation
In UV-VIS spectroscopy, the concentration of a sample can be determined by measuring the transmittance of light through the sample. When working with intermediate transmittances, the absorbance of the sample is neither too low nor too high. This allows for a more accurate determination of the concentration because it falls within the linear range of the instrument's response. If the transmittance is too low (high absorbance), the instrument may not be able to accurately measure it due to limitations in its detection range. Conversely, if the transmittance is too high, the instrument may not be sensitive enough to accurately measure the small changes in transmittance. Therefore, working in the range of intermediate transmittances reduces errors in concentration determination.Rate this question:
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- 2.
Select the correct statement related to UV-VIS spectroscopy
- A.
Only coloured compounds can be analyzed
- B.
Alkanes and alcohols can be used as solvents
- C.
Qualitative analysis is the main application
- D.
After absorption the molecules are decomposed
Correct Answer
B. Alkanes and alcohols can be used as solventsExplanation
Alkanes and alcohols can be used as solvents in UV-VIS spectroscopy. UV-VIS spectroscopy is a technique used to analyze the absorption of ultraviolet and visible light by molecules. Solvents are used to dissolve the sample and create a solution for analysis. Alkanes and alcohols are commonly used solvents in UV-VIS spectroscopy due to their ability to dissolve a wide range of compounds.Rate this question:
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- 3.
A chromophore in UV-VIS spectroscopy is
- A.
A molecule of a coloured compound
- B.
A molecule of a transparent compound
- C.
A structural element of a molecule that causes changes in its capacity of UV-VIS absorption
- D.
The part of a molecule that absorbs UV-VIS radiation
Correct Answer
D. The part of a molecule that absorbs UV-VIS radiationExplanation
In UV-VIS spectroscopy, a chromophore refers to the part of a molecule that absorbs UV-VIS radiation. This absorption occurs due to the presence of certain structural elements within the molecule that have the ability to absorb specific wavelengths of UV or visible light. These chromophores are responsible for the color or transparency of a compound and play a crucial role in determining its UV-VIS absorption capacity.Rate this question:
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- 4.
Sort the following spectral zones according to their energies:
- A.
Gamma ray > Visible > Ultraviolet > Infrared> Radio frequency
- B.
Radio frequency > Infrared > Ultraviolet > Visible > Gamma ray
- C.
Gamma ray > Ultraviolet > Visible > Infrared > Radio frequency
- D.
Ultraviolet > Visible > Infrared > Gamma ray > Radio frequency
Correct Answer
C. Gamma ray > Ultraviolet > Visible > Infrared > Radio frequencyExplanation
The given answer is correct because it correctly arranges the spectral zones in order of their energies, from highest to lowest. Gamma rays have the highest energy, followed by ultraviolet, visible, infrared, and radio frequency waves.Rate this question:
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- 5.
Select the correct statement related to UV-VIS spectroscopy
- A.
All transitions in the molecules are type n---->π*.
- B.
The previous transitions are those with the lowest energy and the highest λ.
- C.
Molecules that absorbs UV-VIS radiation change their structure as a result of the absorption
- D.
Transitions type n--->π*are the most probable
Correct Answer
D. Transitions type n--->π*are the most probableExplanation
Transitions type n-->π* are the most probable in UV-VIS spectroscopy. This means that the most common type of transition that occurs in molecules when they absorb UV-VIS radiation is from a non-bonding orbital (n) to an anti-bonding π* orbital. This transition is more likely to happen compared to other types of transitions.Rate this question:
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- 6.
The comparison of the UV-VIS spectra of (CH2 = CH2) and 1,3-butadiene (CH2 = CH – CH = CH2) shows:
- A.
Two equal spectra
- B.
The spectrum of the first compounds has only one band and the spectrum of the second one has two bands because it has two chromophores.
- C.
The maximum absorption in the butadiene spectrum appears at higher wavelength
- D.
None of them absorbs UV-VIS radiation
Correct Answer
C. The maximum absorption in the butadiene spectrum appears at higher wavelengthExplanation
The correct answer is that the maximum absorption in the butadiene spectrum appears at a higher wavelength. This suggests that butadiene absorbs light at a longer wavelength compared to (CH2 = CH2). This could be due to the presence of conjugated double bonds in butadiene, which allows for a delocalization of electrons and results in a lower energy transition. As a result, the absorption of light by butadiene occurs at a longer wavelength compared to (CH2 = CH2).Rate this question:
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- 7.
A photomultiplier is a transducer that converts:
- A.
Intensity of radiation into electricity
- B.
Electricity into intensity of radiation
- C.
Wavelength into electricity
- D.
Frequencies into energies
Correct Answer
A. Intensity of radiation into electricityExplanation
A photomultiplier is a device that converts the intensity of radiation into electricity. It consists of a photocathode that emits electrons when exposed to light, and a series of dynodes that amplify the electron signal through a process called electron multiplication. This amplified signal is then converted into an electrical current that can be measured. Therefore, the correct answer is that a photomultiplier converts the intensity of radiation into electricity.Rate this question:
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- 8.
A quartz prism is a monochromator of UV radiation because:
- A.
It absorbs all wavelengths except the transmitted one
- B.
It reflects all wavelengths except the transmitted one
- C.
It is transparent to that radiation and the refractive index highly changes with different wavelengths in that zone
- D.
It emits a monochromatic radiation
Correct Answer
C. It is transparent to that radiation and the refractive index highly changes with different wavelengths in that zoneExplanation
A quartz prism is a monochromator of UV radiation because it is transparent to that radiation and the refractive index highly changes with different wavelengths in that zone. This means that when UV radiation passes through the prism, the different wavelengths of the radiation will experience different levels of refraction, causing them to separate. This allows the prism to be used as a monochromator, as it can isolate specific wavelengths of UV radiation.Rate this question:
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- 9.
Which of the following combinations are adequate for UV molecular absorption spectrophotometers?
- A.
Glass cuvette and Xe arc lamp
- B.
Quartz cuvette and hollow cathode lamp.
- C.
Quartz cuvette and deuterium lamp.
- D.
Fuse silica cuvette and sodium vapor lamp
Correct Answer
C. Quartz cuvette and deuterium lamp.Explanation
UV molecular absorption spectrophotometers require a combination of a quartz cuvette and a deuterium lamp. Quartz cuvettes are used because they have high transmission in the UV range. Deuterium lamps are used as the light source because they emit a broad spectrum of UV light, which is necessary for UV absorption measurements. The other combinations mentioned are not suitable because glass cuvettes have limited UV transmission, Xe arc lamps do not emit a broad UV spectrum, and fuse silica cuvettes and sodium vapor lamps are not commonly used in UV molecular absorption spectrophotometry.Rate this question:
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- 10.
Select the effects of increasing solvent polarity in an analyte solution with cromophore:
- A.
Hypsochromic shift for π → π* transitions
- B.
Bathochromic shift for π → π* transitions
- C.
Does not fit Beer ́s law
- D.
Changes in the color of the solution
Correct Answer
B. Bathochromic shift for π → π* transitionsExplanation
Increasing solvent polarity in an analyte solution with a chromophore causes a bathochromic shift for π → π* transitions. This means that the absorption wavelength of the analyte increases, resulting in a shift towards longer wavelengths. This shift is due to the increased stabilization of the excited state of the chromophore by the polar solvent, leading to a lower energy transition.Rate this question:
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- 11.
For the spectrophotometric analysis of a yellow solution containing aniline, which colored filter should I use?
- A.
Yellow
- B.
Blue (complementary to yellow)
- C.
Transparent to visible radiation
- D.
Red
Correct Answer
B. Blue (complementary to yellow)Explanation
The correct answer is blue (complementary to yellow) because when a yellow solution is analyzed using a spectrophotometer, the best filter to use is the one that is complementary to the color of the solution. In this case, yellow is complementary to blue, so using a blue filter will allow for accurate analysis of the solution.Rate this question:
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- 12.
Which one do you like? Select the true statement regarding UV spectrophotometer components
- A.
Flame is a source of radiation
- B.
Monochromators or filters are adequate for wavelength selection
- C.
Detectors give information about absorbed radiation frequency
- D.
Intensity of the radiation can be measured by prims
Correct Answer
B. Monochromators or filters are adequate for wavelength selectionExplanation
Monochromators or filters are used in UV spectrophotometers to select specific wavelengths of light. These components allow only a narrow range of wavelengths to pass through, ensuring accurate and precise measurements. Detectors, on the other hand, measure the intensity of the radiation that is absorbed by the sample. The statement about the flame being a source of radiation is not relevant to UV spectrophotometers. There is no information provided about "prims," so it cannot be determined if they can measure the intensity of radiation.Rate this question:
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- 13.
Select the false statement:
- A.
Every UV-VIS absorbent molecule is fluorescent
- B.
Fluorescence is the emission of radiation by deactivation from a singlet state
- C.
Phosphorescence is less frequent than fluorescence
- D.
Any molecule must have a cromophore group and be rigid to be fluorescent
Correct Answer
A. Every UV-VIS absorbent molecule is fluorescentExplanation
The statement "Every UV-VIS absorbent molecule is fluorescent" is false. While many UV-VIS absorbent molecules may be fluorescent, not all of them are. Fluorescence is a specific property of certain molecules, where they absorb light of a particular wavelength and then emit light of a longer wavelength. UV-VIS absorbent molecules may also exhibit other types of behavior such as phosphorescence or non-radiative relaxation. Therefore, it is incorrect to say that every UV-VIS absorbent molecule is fluorescent.Rate this question:
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- 14.
Select the false statement whether comparing fluorescence and phosphorescence
- A.
Both processes involve molecular deactivation
- B.
Both processes involve changes in the electron distribution of the molecule
- C.
Both processes occur at different wavelengths
- D.
Radiation emitted by fluorescence shows higher wavelength than phosphorescence.
Correct Answer
D. Radiation emitted by fluorescence shows higher wavelength than phosphorescence.Explanation
Fluorescence and phosphorescence are both processes that involve molecular deactivation and changes in the electron distribution of the molecule. However, the radiation emitted by fluorescence has a shorter wavelength than phosphorescence.Rate this question:
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- 15.
Electronic multiplicity state for a molecule could be:
- A.
Triplet, with coupled electron spin
- B.
Singlet, if total spin is 0
- C.
Triplet, if total spin is 3
- D.
Singlet, if it contains any decoupled electron
Correct Answer
C. Triplet, if total spin is 3Explanation
The electronic multiplicity state for a molecule can be determined by the total spin of the electrons. In this case, if the total spin is 3, the molecule will be in a triplet state. The triplet state indicates that the electrons in the molecule have coupled electron spin, meaning their spins are aligned. Therefore, the correct answer is that the electronic multiplicity state for a molecule is triplet if the total spin is 3.Rate this question:
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- 16.
Indicate the errors:
- A.
Fluorescent is a non-radiative relaxation
- B.
Phosphorescence is an absorption process, whether fluorescence is a type of emission
- C.
Fluorescence is a process on which electron states show the same multiplicity
- D.
Fluorescent molecules must have an UV-VIS chromophore
Correct Answer(s)
A. Fluorescent is a non-radiative relaxation
B. Phosphorescence is an absorption process, whether fluorescence is a type of emissionExplanation
The given answer contains multiple errors. Firstly, "Fluorescent" should be corrected to "Fluorescence" in the first statement. Secondly, "whether" should be replaced with "while" in the second statement. Lastly, the third statement is incorrect as fluorescence is a process in which electron states show different multiplicity, not the same.Rate this question:
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- 17.
Among several factors enhancing fluorescence, there is a wrong statement:
- A.
Low temperatures
- B.
Viscous solvents
- C.
Molecular flexibility
- D.
UV-VIS chromophore
Correct Answer
C. Molecular flexibilityExplanation
Molecular flexibility does not enhance fluorescence. In fact, rigidity of the molecule is often associated with increased fluorescence. This is because rigid molecules have restricted vibrational and rotational motions, which prevent non-radiative decay pathways and promote fluorescence emission. On the other hand, molecular flexibility allows for more conformational changes and internal motions, which can result in non-radiative decay and decrease fluorescence efficiency. Therefore, molecular flexibility is not a factor that enhances fluorescence.Rate this question:
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- 18.
Select the most appropriated technique for the analysis of Ca, Fe, Mg and Zn in urine samples:
- A.
AES-ICP
- B.
FAAS
- C.
FAES
- D.
GFAAS
Correct Answer
A. AES-ICPExplanation
AES-ICP (Atomic Emission Spectroscopy-Inductively Coupled Plasma) is the most appropriate technique for the analysis of Ca, Fe, Mg, and Zn in urine samples. AES-ICP is a highly sensitive and accurate method that allows for the simultaneous analysis of multiple elements. It uses an inductively coupled plasma as the ionization source and measures the emission of light from excited atoms to determine the concentration of elements in the sample. This technique is commonly used in analytical chemistry for the determination of trace elements in various matrices, including biological samples like urine.Rate this question:
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- 19.
Select the false sentence:
- A.
Higher number of elements could be analyzed with atomic absorption spectroscopy than with flame photometry
- B.
Atomic emission spectroscopy with plasma needs a hollow cathode lamps for the excitation of elements
- C.
Hollow cathode and EDL are monochromatic lamps
- D.
Simultaneous analysis can be done with plasma atomic emission spectroscopy
Correct Answer
B. Atomic emission spectroscopy with plasma needs a hollow cathode lamps for the excitation of elements -
- 20.
For determining potassium in a sample is required:
- A.
Flame atomic absorption spectroscopy
- B.
Flame photometry
- C.
Addition of Lanthanum for minimizing ionization
- D.
The use of microwave electromagnetic radiation
Correct Answer
B. Flame photometryExplanation
Flame photometry is the technique used to determine the concentration of potassium in a sample. This method involves the measurement of the intensity of light emitted by potassium atoms when they are excited by a flame. The intensity of the emitted light is directly proportional to the concentration of potassium in the sample. This technique is commonly used in analytical chemistry laboratories for the quantitative analysis of various elements, including potassium.Rate this question:
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- 21.
Select the false sentence:
- A.
The use of a wavelength selector is necessary in atomic spectroscopic techniques
- B.
In atomic emission spectroscopy there is a relationship between intensity of radiation and concentration
- C.
Atomic spectroscopic techniques can only be used for quantitative analysis
- D.
Plasma atomic emission spectroscopy is more sensitive than flame atomic absorption spectroscopy
Correct Answer
A. The use of a wavelength selector is necessary in atomic spectroscopic techniquesExplanation
The statement "The use of a wavelength selector is necessary in atomic spectroscopic techniques" is true. In atomic spectroscopic techniques, a wavelength selector is used to select a specific wavelength of light for analysis. This is important because different elements emit or absorb light at different wavelengths, so the wavelength selector allows for the identification and quantification of specific elements in a sample.Rate this question:
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- 22.
From the 1H-NMR spectrum of propane aldehyde we can affirm that:
- A.
The relative signal intensities are 2:2:1
- B.
The signal multiplicities are quadruplet, doblet, singlet
- C.
Aldehyde proton absorbs at the highest field
- D.
Methyl protons are the ones with the highest shield constant
Correct Answer
D. Methyl protons are the ones with the highest shield constantExplanation
The answer is that methyl protons are the ones with the highest shield constant. This is because the statement implies that the methyl protons are the most shielded from the external magnetic field, which is reflected in their higher shield constant.Rate this question:
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- 23.
Proton NMR signals appear at different chemical shifts depending on:
- A.
The physical state of the molecule
- B.
The solvent used
- C.
Their shielding degree
- D.
There are no differences in the position of their signals
Correct Answer
C. Their shielding degreeExplanation
Proton NMR signals appear at different chemical shifts depending on their shielding degree. Shielding refers to the ability of surrounding electron clouds to shield the protons from the external magnetic field. Protons that are more shielded will experience a lower effective magnetic field and will appear at a higher chemical shift, while less shielded protons will experience a higher effective magnetic field and will appear at a lower chemical shift. Therefore, the shielding degree of the protons determines the position of their signals in the proton NMR spectrum.Rate this question:
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- 24.
The 13C-NMR spectrum of phenol ...
- A.
Has just one signal
- B.
Has six signals
- C.
It does not give any signal because it is made up of atoms of normal C (12C), a nucleus that is not active in NMR
- D.
Has four signals
Correct Answer
D. Has four signalsExplanation
The correct answer is that the 13C-NMR spectrum of phenol has four signals. This is because phenol contains four different carbon environments: the carbon in the hydroxyl group, the carbon in the aromatic ring attached to the hydroxyl group, the carbon in the other aromatic ring, and the carbon in the methyl group. Each of these environments will give rise to a separate signal in the 13C-NMR spectrum.Rate this question:
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- 25.
Under a magnetic field, all proton nuclei randomly spinning:
- A.
They will stop their movement
- B.
They will change their direction
- C.
They will be aligned with the magnetic field
- D.
They will rotate 90° far from the induced field
Correct Answer
C. They will be aligned with the magnetic fieldExplanation
Under a magnetic field, all proton nuclei will be aligned with the magnetic field. This is because protons have a positive charge and therefore experience a force when placed in a magnetic field. This force causes the protons to align themselves parallel to the magnetic field lines.Rate this question:
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- 26.
In NMR, we can say for protons with signals at high chemical shifts:
- A.
They are high shielded
- B.
They give resonance signals at high field
- C.
They have a strong bond
- D.
They give resonance signals at low field
Correct Answer
D. They give resonance signals at low fieldExplanation
Protons with signals at high chemical shifts give resonance signals at low field. This is because high chemical shifts indicate that the protons are experiencing a stronger magnetic field from the surrounding electron density. In NMR spectroscopy, protons resonate at different frequencies depending on their chemical environment. Protons with high chemical shifts are shielded from the external magnetic field by the surrounding electron density, causing them to resonate at lower frequencies or lower fields.Rate this question:
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- 27.
Magnetic coupling causes the peaks in 1H NMR spectra to be split into:
- A.
Two peaks
- B.
Multiple peaks equal to the number of hydrogens on surrounding atoms
- C.
Multiple peaks equal to the number of surrounding carbon atoms
- D.
Multiple peaks equal to the number of hydrogens on surrounding atoms, plus one
Correct Answer
D. Multiple peaks equal to the number of hydrogens on surrounding atoms, plus oneExplanation
In 1H NMR spectra, the peaks are split into multiple peaks equal to the number of hydrogens on surrounding atoms, plus one. This is known as the n+1 rule. The splitting occurs due to the magnetic coupling between the hydrogen atoms and their neighboring hydrogen atoms. Each hydrogen atom experiences a slightly different magnetic environment due to the presence of its neighboring hydrogens, which leads to the splitting of the peaks. The number of peaks is determined by the number of hydrogens on the surrounding atoms, and the "+1" accounts for the hydrogen atom itself.Rate this question:
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- 28.
How many signals does the aldehyde (CH3)3CCH2CHO have in 1H-NMR and 13C- NMR spectra?
- A.
Five 1H signals and six 13C signals
- B.
Three 1H signals and four 13C signals
- C.
Five 1H signals and four 13C signals
- D.
Three 1H signals and four 13C signals
Correct Answer
B. Three 1H signals and four 13C signalsExplanation
The aldehyde (CH3)3CCH2CHO has three different hydrogen environments, resulting in three distinct 1H signals in the 1H-NMR spectrum. Additionally, there are four different carbon environments, leading to four distinct 13C signals in the 13C-NMR spectrum.Rate this question:
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- 29.
From the previous aldehyde, select the true statement:
- A.
The two protons in the methylene group are singlets
- B.
Protons in the methyl groups are singlets
- C.
The integration ratio is 6:1:2:3
- D.
The signals in the spectrum will appear at 1.15, 4.05, 1.21, 3.15 and 8.48 ppm
Correct Answer
B. Protons in the methyl groups are singletsExplanation
This statement is true because singlets refer to a single peak in the NMR spectrum, indicating that the protons in the methyl groups are not split by neighboring protons.Rate this question:
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- 30.
Select the correct statement about the 1H-NMR spectrum of paracetamol:
- A.
There will be four signals
- B.
All signals will be singlets
- C.
All aromatic protons will absorb the same frequency
- D.
There will be five signals
Correct Answer
D. There will be five signalsExplanation
The correct statement about the 1H-NMR spectrum of paracetamol is that there will be five signals. This means that there are five different types of hydrogen atoms in the molecule, each producing a unique signal due to its chemical environment. This information can be used to identify and analyze the structure of the compound.Rate this question:
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Mar 21, 2023Quiz Edited by
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May 28, 2020Quiz Created by
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