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Test: Molecular Spectroscopy - Chemistry MCQ


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30 Questions MCQ Test Physical Chemistry - Test: Molecular Spectroscopy

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Test: Molecular Spectroscopy - Question 1

The molecule which is IR-inactive but Raman–active is:

Detailed Solution for Test: Molecular Spectroscopy - Question 1

Raman activity refers to a molecular vibration that can be observed using Raman spectroscopy, while infrared (IR) activity pertains to vibrations detectable by infrared spectroscopy. Some molecules may show characteristics in one type of spectroscopy but not the other.

To determine which molecule is IR-inactive but Raman-active, consider the following:

  • N2: A homonuclear diatomic molecule, which does not have a dipole moment. Therefore, it is IR-inactive and can be Raman-active due to its vibrational modes.
  • HCl: This molecule is IR-active because it has a dipole moment when it vibrates, making it detectable in infrared spectroscopy.
  • SO2: This molecule is also IR-active as its vibration changes the dipole moment, allowing it to be detected in IR spectroscopy.
  • Protein: Proteins typically exhibit both IR and Raman activity due to their complex structures and varying functional groups.

In conclusion, among the options provided, N2 is the molecule that is IR-inactive but can be Raman-active.

Test: Molecular Spectroscopy - Question 2

The rotational (microwave) spectrum of a rigid diatomic rotor consists of equally spaced lines with spacing equal to:

Detailed Solution for Test: Molecular Spectroscopy - Question 2

The rotational (microwave) spectrum of a rigid diatomic rotor features lines that are evenly spaced. The spacing between these lines is determined by the rotational constant, denoted as B.

  • The energy levels of a rigid diatomic rotor are quantised.
  • Each transition between energy levels results in a spectral line.
  • These lines are spaced equally, meaning the difference in energy corresponds to the constant B.
  • The spacing is specifically 2B for the rotational transitions.

Thus, the spacing between the lines in the spectrum reflects the fundamental properties of the diatomic rotor's motion.

Test: Molecular Spectroscopy - Question 3

Upon application of a weak magnetic field, a line in the microwave absorption spectrum of rigid rotor splits into 3 lines. The quantum number (J) of the rotational energy level from which the transition originates is:

Detailed Solution for Test: Molecular Spectroscopy - Question 3

   J=1

Test: Molecular Spectroscopy - Question 4

The vibrational frequency of a homonuclear diatomic molecule is v. The temperature at which the population of the first excited state will be half that of the ground state is given by.

Detailed Solution for Test: Molecular Spectroscopy - Question 4


Test: Molecular Spectroscopy - Question 5

The vibrational rotational spectrum is observed in …………….. region.

Detailed Solution for Test: Molecular Spectroscopy - Question 5

The vibrational rotational spectrum is observed in the near infrared region.

The vibrational rotational spectrum is a key aspect of molecular spectroscopy, particularly in the near infrared part of the electromagnetic spectrum. This region is crucial for understanding molecular vibrations and rotations.

  • The near infrared region typically spans wavelengths from approximately 0.75 to 2.5 micrometres.
  • In this range, molecules absorb light, causing transitions in their vibrational and rotational energy levels.
  • This absorption leads to unique spectral signatures that can help identify different molecules.
  • Applications include chemical analysis, environmental monitoring, and studying biological processes.

By examining these spectra, scientists can gather important information about the structure and dynamics of various substances.

Test: Molecular Spectroscopy - Question 6

In a rotational spectrum, transitions are only observed between rotational levels of ΔJ =

Detailed Solution for Test: Molecular Spectroscopy - Question 6

In a rotational spectrum, transitions are only observed between rotational levels of

  • ΔJ = ±1
  • ΔJ = ±2
  • ΔJ = +1
  • ΔJ = +3

This means that during transitions, molecules can change energy levels by either increasing or decreasing their rotational quantum number by 1 or 2. The most common transitions observed are those with a change of +1 or -1, as these are energetically favourable processes.

Test: Molecular Spectroscopy - Question 7

Vibrational transitions are always accompanied by ………… transitions.

Detailed Solution for Test: Molecular Spectroscopy - Question 7

Vibrational transitions in molecules are always accompanied by rotational transitions. This is because:


  • Molecular vibrations change the distribution of mass, affecting rotational energy levels.

  • Vibrational energy levels are closely spaced, and transitions often overlap with rotational levels.

  • Combined vibrational and rotational transitions are observed in the infrared region of the electromagnetic spectrum.

  • This coupling is due to the molecular structure, enabling simultaneous changes in both vibrational and rotational states.


  •  
Test: Molecular Spectroscopy - Question 8

The increase in rotational energy shows absorption spectrum in.

Detailed Solution for Test: Molecular Spectroscopy - Question 8

The increase in rotational energy is associated with the absorption spectrum in the microwave region. This occurs because the energy transitions involved in rotation are relatively low compared to other forms of energy transitions.

  • Microwave radiation has longer wavelengths and lower frequencies.
  • This type of radiation specifically interacts with the rotational states of molecules.
  • As molecules absorb microwave energy, they undergo rotational transitions.
  • These transitions lead to an increase in rotational energy.

In summary, the absorption spectrum linked to increased rotational energy is primarily found in the microwave region.

Test: Molecular Spectroscopy - Question 9

The increase in vibrational energy leads to absorption spectrum in:

Detailed Solution for Test: Molecular Spectroscopy - Question 9

The increase in vibrational energy results in an absorption spectrum, which can be detected in specific regions of the electromagnetic spectrum.

  • Infrared (IR) region: This region is directly associated with vibrational transitions. Molecules absorb IR radiation as their bonds stretch and bend.
  • Visible region: Although not primarily linked to vibrational energy, some molecules can absorb light in this range, resulting in colour.
  • Microwave region: Absorption in this region typically involves rotational transitions rather than vibrational energy changes.
  • Ultraviolet (UV) region: This region is more relevant for electronic transitions than for vibrational energy absorption.

In summary, the most significant absorption due to increased vibrational energy occurs in the IR region.

Test: Molecular Spectroscopy - Question 10

The frequency of UV radiation is greater than:

Detailed Solution for Test: Molecular Spectroscopy - Question 10

UV radiation has a higher frequency than:

  • Microwaves

  • Infrared (IR) radiation

  • Thus, it surpasses both categories.

In summary:

  • Frequency comparison: UV radiation exceeds both microwaves and IR radiation.
  • Implication: This indicates that UV radiation carries more energy.

Understanding these differences is crucial for various applications, such as:

  • Safety measures in sun exposure.
  • Technological advancements in communication.
Test: Molecular Spectroscopy - Question 11

State which of the following molecules can show a pure rotational microwave spectrum:

Detailed Solution for Test: Molecular Spectroscopy - Question 11

Only certain molecules can exhibit a pure rotational microwave spectrum. This occurs when the molecule has a permanent dipole moment and is not symmetrical in all dimensions.

  • N2: This molecule is diatomic and homonuclear, meaning it has no dipole moment. Therefore, it cannot produce a rotational microwave spectrum.
  • CO2: Although this molecule is linear and has a dipole moment, its symmetrical structure leads to no pure rotational transitions in microwave spectroscopy.
  • OCS: This molecule is polar and has a permanent dipole moment, making it capable of showing a pure rotational microwave spectrum.
  • HCl: This molecule is also polar and exhibits a dipole moment, allowing it to show a pure rotational microwave spectrum.

In summary, among the given molecules, HCl can show a pure rotational microwave spectrum due to its polar nature, while N2 and CO2 cannot.

Test: Molecular Spectroscopy - Question 12

Which of the following diatomic molecule will not give a rotation in spectrum:

Detailed Solution for Test: Molecular Spectroscopy - Question 12

Diatomic molecules can exhibit different behaviours in spectroscopy. Some molecules will show rotational spectra, while others will not.

  • N2 (Nitrogen):
  • This molecule is homonuclear and symmetrical, which means it does not have a permanent dipole moment. As a result, it does not produce a rotational spectrum.

  • CO (Carbon Monoxide):
  • CO is a heteronuclear molecule and has a significant dipole moment. This characteristic allows it to exhibit a rotational spectrum.

  • NO (Nitric Oxide):
  • Similar to CO, NO is also heteronuclear and has a dipole moment, enabling it to produce a rotational spectrum.

  • HF (Hydrogen Fluoride):
  • HF is another heteronuclear molecule with a permanent dipole moment, which allows it to display a rotational spectrum.

In summary, the only diatomic molecule among the options provided that does not give a rotational spectrum is N2.

Test: Molecular Spectroscopy - Question 13

The selection rule of the translation energy levels in the Raman spectrum is:

Detailed Solution for Test: Molecular Spectroscopy - Question 13

The selection rule for translation energy levels in the Raman spectrum is an important principle in understanding molecular vibrations and their interactions with light. This rule defines how energy levels change during the Raman scattering process.

  • The primary rule states that a change in rotational quantum number, ΔJ, must be:
    • ±1 - indicating that the rotational state can increase or decrease by one unit.
  • Other possible changes include:
    • ±2 - which does not typically apply to Raman transitions.
    • +1 - representing an increase in rotational energy.
    • +2 - which is also not relevant for Raman processes.

In summary, the relevant selection rule for Raman spectroscopy is a change in the rotational quantum number of ΔJ = ±1, allowing transitions that either increase or decrease the energy level by one. This is crucial for interpreting Raman spectra effectively.

Test: Molecular Spectroscopy - Question 14

The population (N) distribution over states (n) of a diatomic molecule corresponds to:

Test: Molecular Spectroscopy - Question 15

The rotational spectrum of HI is found to contain a series of lines with a separation of 12.8 cm–1 . Moment of for the molecule is:

Detailed Solution for Test: Molecular Spectroscopy - Question 15

The rotational spectrum of HI reveals a series of spectral lines with a separation of 12.8 cm–1. This separation indicates the presence of quantised rotational energy levels in the molecule. To determine the moment of inertia for the HI molecule, we can use the formula relating the rotational constant to the moment of inertia:

  • Rotational Constant (B): The value of the rotational constant is derived from the line separation.
  • Moment of Inertia (I): The moment of inertia can be calculated using the formula: I = h/(8π²cB), where h is Planck's constant and c is the speed of light.
  • Here, we can deduce that the moment of inertia for the HI molecule is approximately 4.36 x 10-40 g cm2.

Thus, based on the analysis of the rotational spectrum, the calculated moment of inertia confirms that the value is 4.36 x 10-40 g cm2.

Test: Molecular Spectroscopy - Question 16

The bond length in CN+ is 0.129 nm. The position of second line in the microwave spectrum is:

Detailed Solution for Test: Molecular Spectroscopy - Question 16

First calculate the reduced mass μ of CN+:
μ = (12×14)/(12+14) ≈ 6.462 u, then μ = 6.462×1.6605×10−27 kg.
Rotational constant B in plain text:
B = h/(8*pi*pi*c*mu*r2).
Using h = 6.626×10−34 J·s, c = 2.998×108 m/s, r = 0.129×10−9 m and the μ above gives B ≈ 1.566 cm−1.
Microwave lines occur at ν(J→J+1) = 2B(J+1).
For the second line (J = 1 → 2): ν = 2B×2 = 4B = 4×1.566 = 6.264 cm−1.
Therefore, the second line appears at 6.264 cm−1.

Test: Molecular Spectroscopy - Question 17

Among the singlet (S), doublet (D) and triplet (T). electronic states, phosphorescence involves transition between:

Detailed Solution for Test: Molecular Spectroscopy - Question 17

Phosphorescence involves transitions between electronic states, primarily characterised as singlet (S), doublet (D), and triplet (T) states. In this process:

  • The transition occurs from a triplet state (T) to a singlet state (S).
  • This transition is significant because it explains the delayed emission of light often observed in phosphorescent materials.
  • During this transition, the system undergoes a change in spin multiplicity, which is a key feature of phosphorescence.

Thus, the correct understanding of phosphorescence is critical for applications in lighting and displays, where materials can emit light over extended periods after being excited.

Test: Molecular Spectroscopy - Question 18

For pure vibrational spectra, the selection rule is:

Detailed Solution for Test: Molecular Spectroscopy - Question 18

For pure vibrational spectra, the selection rule is:

  • A: 0
  • B: ±1
  • C: 0, ±1
  • D: ±1, 2

The correct answer is B.

Test: Molecular Spectroscopy - Question 19

The Jmax for a rigid diatomic molecule for which at 300K, the rotational constant is 1.566 cm–1, is:

Detailed Solution for Test: Molecular Spectroscopy - Question 19

The maximum rotational quantum number (Jmax) for a rigid diatomic molecule at 300K can be determined using the given rotational constant.

For a rigid diatomic molecule:

  • The rotational constant is 1.566 cm–1.
  • The relationship between the rotational constant and the maximum rotational quantum number is critical.
  • At a temperature of 300K, the energy levels are populated according to the Boltzmann distribution.
  • The maximum quantum number can typically be calculated using the formula:
  • Jmax is influenced by both the temperature and the rotational constant.

Based on these factors, we find that the maximum rotational quantum number for the molecule is 8.

Test: Molecular Spectroscopy - Question 20

The inter nuclear distance for H35 Cl is 0.13 nm and the force constant is 500 Nm–1. The first excitat ion energy in cm–1 of H35 Cl molecule in the harmonic approximation is:

Detailed Solution for Test: Molecular Spectroscopy - Question 20

Test: Molecular Spectroscopy - Question 21

The normal modes of vibrations of N2O is:

Detailed Solution for Test: Molecular Spectroscopy - Question 21

The normal modes of vibration of N2O refer to the distinct patterns in which the molecule can oscillate.

The analysis of these modes reveals:

  • Each mode corresponds to a specific vibrational energy level.
  • The total number of normal modes for a nonlinear molecule like N2O is determined by the formula: 3N - 6, where N is the number of atoms.
  • For N2O, which consists of three atoms, the calculation is:
    • 3 (for each atom) x 3 = 9
    • 9 - 6 = 3 normal modes of vibration.

Thus, the normal modes of vibrations for N2O total three.

Test: Molecular Spectroscopy - Question 22

The fundamental vibration frequency and rotational constant of carbon monoxide molecule are 6.5 x 1013 s-1 and 1.743 x 1011 s-1 respectively. The rotational will have the same energy as ir would have in its first vibrational states with no rotational energy is:

Detailed Solution for Test: Molecular Spectroscopy - Question 22

The fundamental vibration frequency of the carbon monoxide molecule is 6.5 x 1013 s-1. Its rotational constant is 1.743 x 1011 s-1. In quantum mechanics, the energy of rotational states is closely related to vibrational states.

When considering the energy of the molecule, it is important to note that the rotational energy in its first vibrational state can be calculated based on these constants. The energy levels of a molecule can be thought of as a combination of its vibrational and rotational energies.

  • The first vibrational state corresponds to the lowest energy level of vibration.
  • Rotational energy levels depend on the rotational constant, which influences the spacing of these energy levels.
  • In this scenario, the rotational energy aligns with the vibrational states.

Hence, based on the given data, the rotational energy will match that of the first vibrational state. This relationship aids in determining the overall energy levels of the carbon monoxide molecule.

Test: Molecular Spectroscopy - Question 23

Calculate the frequency at which the transition between the two spin states of a free electron may be observed at a field strength of 0.3000 T:

Detailed Solution for Test: Molecular Spectroscopy - Question 23

Correct Answer :- a

Explanation : g = 2.0025

uN = 9.2741 * 10-24J

B = 0.300 T

v = (uN * (g) * B)/h

v = [(9.2741 * 10-24) (2.0025) * (0.300)]/(6.626 * 10-34)

= 8.408 * 109 s-1

Test: Molecular Spectroscopy - Question 24

The different types of energies associated with a molecule are __________.

Detailed Solution for Test: Molecular Spectroscopy - Question 24

The different types of energies associated with a molecule are electronic energy, vibrational energy, rotational energy and translational energy.

Test: Molecular Spectroscopy - Question 25

The pure rotational (microwave) spectrum of the gaseous molecule CN consists of a series of equally spaced line separated by 3.7978 cm–1. The inter nuclear distance of the molecule is [Molar masses are 12C=12.011 and 14N=14.007 g mol–1]:

Detailed Solution for Test: Molecular Spectroscopy - Question 25

The pure rotational (microwave) spectrum of the gaseous molecule CN shows a series of evenly spaced lines. These lines are separated by 3.7978 cm–1. To find the inter-nuclear distance of the CN molecule, we can use the provided molar masses:

  • Carbon (C): 12.011 g mol–1
  • Nitrogen (N): 14.007 g mol–1

The inter-nuclear distance can be assessed from the rotational spectrum and is commonly measured in picometres (pm). Here are the possible distances:

  • 130 pm
  • 117 pm
  • 150 pm
  • 93 pm

The correct inter-nuclear distance for the CN molecule is 117 pm.

Test: Molecular Spectroscopy - Question 26

Rotational quantum number for the rotational spectrum of diatomic molecule for maximum populated level is given by:

Detailed Solution for Test: Molecular Spectroscopy - Question 26

Rotational quantum number for the rotational spectrum of a diatomic molecule at the maximum populated level can be determined using specific equations and principles of quantum mechanics.

The rotational spectrum is influenced by several factors:

  • Quantum Number (J): This number indicates the rotational energy levels of the molecule.
  • Population Distribution: At thermal equilibrium, the distribution of these levels follows the Boltzmann distribution, where higher energy levels are less populated.
  • Temperature Effect: As temperature increases, higher rotational levels become more populated.

The rotational energy levels are given by:

  • The energy of a level is proportional to J(J+1), where J is the rotational quantum number.
  • For diatomic molecules, the maximum populated level can be identified by the condition that the population in that level is greatest.

In practical terms, to find the maximum populated level, one would typically look for the rotational quantum number J that corresponds to the highest population at a given temperature. This often leads to determining the value of J using statistical mechanics principles.

Test: Molecular Spectroscopy - Question 27

Which one of the following exhibits rotation spectra:

Detailed Solution for Test: Molecular Spectroscopy - Question 27

- Rotational spectra occur in molecules with a permanent dipole moment.
- H2 and N2 are homonuclear diatomic molecules with no dipole moment.
- CO is a heteronuclear diatomic molecule with a permanent dipole moment, allowing it to exhibit rotational spectra.
- CO2 is a linear molecule but symmetric, so it lacks a permanent dipole moment and does not exhibit rotational spectra.
- Therefore, CO (option C) is the correct answer as it can exhibit rotational spectra.

Test: Molecular Spectroscopy - Question 28

The microwave spectrum of a molecule yields three rotational constants. The molecule is:

Detailed Solution for Test: Molecular Spectroscopy - Question 28

Test: Molecular Spectroscopy - Question 29

The Q band in the vibrational spectrum of acetylene is observed in the:

Detailed Solution for Test: Molecular Spectroscopy - Question 29

The Q band in the vibrational spectrum of acetylene is associated primarily with molecular vibrations.

  • C–C stretching mode: This mode involves the vibration along the bond between two carbon atoms in acetylene.
  • C–H symmetric mode: In this mode, the hydrogen atoms move in harmony, which contributes to the overall vibrational spectrum.
  • Bending mode: This refers to the angle changes between bonds, affecting the molecule's shape.
  • C–H antisymmetric stretching mode: In contrast to the symmetric mode, this involves hydrogen atoms moving in opposite directions.

Among these, the most significant contribution to the Q band comes from the C–C stretching mode, as it generates distinct spectral features that are crucial for identifying acetylene.

Test: Molecular Spectroscopy - Question 30

Which of the following spectroscopic techniques will be useful to dist inguish between M–SCN and M–NCS binding modes:

Detailed Solution for Test: Molecular Spectroscopy - Question 30

To distinguish between M–SCN and M–NCS binding modes, the following spectroscopic technique is particularly useful:

  • IR (Infrared Spectroscopy):

Infrared spectroscopy is effective for analysing molecular vibrations and can provide distinct absorption bands for different functional groups.

  • M–SCN typically shows specific IR absorption features associated with the thiocyanate group.
  • M–NCS will have different IR characteristics due to the orientation of the thiocyanate ligand.

This ability to observe unique absorption patterns allows for the identification of the binding modes effectively.

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