All Courses
Get the App Signup / Login
Sign in Open App

All questions of Physical Spectroscopy for Chemistry Exam

The molecule which is IR-inactive but Raman–active is:
  • a)
    N2
  • b)
    HCl
  • c)
    SO2
  • d)
    Protein  
Correct answer is 'A'. Can you explain this answer?

Bijoy Kapoor answered
Homonuclear diatomic molecules such as H2, N2, and O2 have no dipole moment and are IR inactive (but Raman active) while heteronuclear diatomic molecules such as HCl, NO, and CO do have dipole moments and have IR active vibrations.
Clear all your Chemistry doubts with EduRev

The increase in rotational energy shows absorption spectrum in.
  • a)
    IR region
  • b)
    UV region
  • c)
    Visible region
  • d)
    Microwave region
Correct answer is option 'D'. Can you explain this answer?

Palak Singh answered
Increase in Rotational Energy and Absorption Spectrum in Microwave Region

Rotational energy is the energy associated with the rotation of a molecule around its axis. When a molecule absorbs electromagnetic radiation, its rotational energy increases. This increase in rotational energy can be observed in the absorption spectrum in the microwave region.

Microwave region

The microwave region is a part of the electromagnetic spectrum that has wavelengths ranging from 1 millimeter to 1 meter and frequencies ranging from 300 MHz to 300 GHz. This region is commonly used for communication, cooking, and spectroscopy.

Absorption spectrum

An absorption spectrum is a graph that shows the wavelengths of electromagnetic radiation absorbed by a substance. It is obtained by passing a beam of electromagnetic radiation through a sample of the substance and measuring the intensity of the transmitted radiation at different wavelengths.

Increase in rotational energy and absorption spectrum in microwave region

When a molecule absorbs electromagnetic radiation in the microwave region, its rotational energy increases. This increase in rotational energy can be observed in the absorption spectrum as a series of sharp peaks or lines. These peaks or lines are known as rotational transitions.

The frequency of the absorbed radiation is directly proportional to the difference in energy between the initial and final rotational states of the molecule. This relationship is described by the equation:

ΔE = hν

where ΔE is the difference in energy, h is Planck's constant, and ν is the frequency of the absorbed radiation.

In conclusion, the increase in rotational energy of a molecule due to absorption of electromagnetic radiation can be observed in the absorption spectrum in the microwave region. This region is commonly used for spectroscopy because many molecules have rotational transitions in this region, making it useful for identifying and studying different types of molecules.

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]:
  • a)
    130 pm
  • b)
    117 pm
  • c)
    150 pm
  • d)
    93 pm
Correct answer is option 'B'. Can you explain this answer?

Saikat Ghoshal answered
Given:
- Equally spaced lines in pure rotational (microwave) spectrum of CN
- Separation between the lines = 3.7978 cm^-1
- Molar masses of C and N = 12.011 g/mol and 14.007 g/mol, respectively

To find:
- Inter-nuclear distance of CN molecule

Formula used:
- The separation between the rotational energy levels (ΔE) of a diatomic molecule is given by:

ΔE = hB,
where h is Planck's constant and B is the rotational constant.

- The rotational constant of a diatomic molecule is given by:

B = h / (8π^2μr_e^2),
where μ is the reduced mass of the molecule, and r_e is the equilibrium bond length.

- The reduced mass (μ) of a diatomic molecule is given by:

μ = m_1m_2 / (m_1 + m_2),
where m_1 and m_2 are the masses of the two atoms in the molecule.

- The equilibrium bond length of a diatomic molecule can be estimated using the formula:

r_e = (r_1 + r_2) / 2,
where r_1 and r_2 are the covalent radii of the two atoms in the molecule.

Solution:
1. Calculate the rotational constant (B) of CN molecule using the given separation between the lines:

ΔE = hB
⇒ B = ΔE / h
⇒ B = 3.7978 cm^-1 / (6.626 x 10^-34 J s)
⇒ B = 5.72 x 10^-4 cm^-1

2. Calculate the reduced mass (μ) of CN molecule:

μ = m_1m_2 / (m_1 + m_2)
⇒ μ = 12.011 x 14.007 / (12.011 + 14.007) g/mol
⇒ μ = 6.68 g/mol

3. Estimate the equilibrium bond length (r_e) of CN molecule using the covalent radii of carbon and nitrogen:

r_e = (r_1 + r_2) / 2
⇒ r_e = (0.77 Å + 0.75 Å) / 2
⇒ r_e = 0.76 Å = 0.76 x 10^-10 m

4. Calculate the inter-nuclear distance (r) of CN molecule using the rotational constant (B) and reduced mass (μ):

B = h / (8π^2μr_e^2)
⇒ r = √(h / (8π^2μB))
⇒ r = √((6.626 x 10^-34 J s) / (8π^2 x 6.68 x 10^-3 kg/mol x 5.72 x 10^-4 cm^-1))
⇒ r = 117 pm

Therefore, the inter-nuclear distance of CN molecule is 117 pm.

The Jmax for a rigid diatomic molecule for which at 300K, the rotational constant is 1.566 cm–1, is:
  • a)
    4
  • b)
    6
  • c)
    8
  • d)
    10 
Correct answer is option 'C'. Can you explain this answer?

Sinjini Nair answered
Given information:
- Rotational constant of a rigid diatomic molecule = 1.566 cm^-1 at 300K

To find:
- Jmax value for the given molecule

Explanation:
- For a rigid diatomic molecule, the rotational energy levels are given by the expression: EJ = J(J+1)h^2/8π^2I
- Where EJ is the energy of the Jth rotational level, J is the rotational quantum number, h is Planck's constant, and I is the moment of inertia of the molecule.
- The moment of inertia of a rigid diatomic molecule is given by the expression: I = μr^2, where μ is the reduced mass of the molecule and r is the bond length.
- At a given temperature, the population of different rotational levels is given by the Boltzmann distribution: N(J) = N0 exp(-EJ/kT)
- Where N(J) is the number of molecules in the Jth rotational level, N0 is the total number of molecules, k is the Boltzmann constant, and T is the temperature in Kelvin.
- The maximum value of J, i.e., Jmax, is the value of J for which the population in the (J+1)th level is negligible compared to that in the Jth level.
- Mathematically, this can be expressed as: N(J+1)/N(J) ≈ exp(-EJ+1/kT) < />
- Taking the natural logarithm of both sides and simplifying, we get: Jmax ≈ 2.303kT/hB
- Where B is the rotational constant of the molecule.

Calculation:
- Substituting the given values, we get: Jmax ≈ 2.303(1.38×10^-23)(300)/(6.626×10^-34)(1.566×10^3)
- Solving this expression, we get: Jmax ≈ 8

Therefore, the Jmax value for the given rigid diatomic molecule is 8 (option C).

Which of the following spectroscopic techniques provides the most information about an organic molecule's framework/structure?
  • a)
    UV visible spectroscopy
  • b)
    Melting point
  • c)
    NMR spectroscopy
  • d)
    Mass spectrometry
Correct answer is option 'C'. Can you explain this answer?

Asf Institute answered
  • NMR spectroscopy is most useful for determining the type of nuclei (most commonly studied nuclei are 1H and 13C) present and their relative locations within a molecule. H-NMR is most commonly used because it is practically present in all organic compounds.
  • This technique is useful for a complete determination of the structure of organic compounds. IR spectroscopy is best for determining the functional groups of a molecule, however, it does not give information of the electric environment like NMR.
  • Mass spectrometry is a technique used to determine the amount and mass of substances present in a sample. UV-visible spectroscopy is used to determine the amount of analyte present in a given sample - this method is best for transition metals and/or highly conjugated compounds.
  • Melting point analysis gives information about the purity of a sample, pure substances tend to have higher melting points and more narrow ranges than impure samples. 

Which of the following diatomic molecule will not give a rotation in spectrum:
  • a)
    N2
  • b)
    CO
  • c)
    NO
  • d)
    HF 
Correct answer is option 'A'. Can you explain this answer?

Srishti Kulkarni answered
Explanation:
When a diatomic molecule is irradiated with electromagnetic radiation, it absorbs energy and undergoes a rotational transition from one energy level to another. This results in the appearance of rotational lines in its spectrum. However, for a molecule to show rotational spectra, it must satisfy certain conditions. Let's see what these conditions are and then apply them to the given options.

Conditions for a molecule to show rotational spectra:

1. The molecule must be diatomic: This is because only diatomic molecules have a non-zero dipole moment and can interact with electromagnetic radiation.

2. The molecule must have a permanent dipole moment: This is necessary for the molecule to have a rotational energy level structure.

3. The molecule must not be in a vibrationally excited state: This is because a molecule in a vibrationally excited state will have a different rotational energy level structure.

Now, let's apply these conditions to the given options:

a) N2: Nitrogen is a diatomic molecule, but it has a zero dipole moment. Therefore, it cannot interact with electromagnetic radiation and will not show rotational spectra.

b) CO: Carbon monoxide is a diatomic molecule with a non-zero dipole moment. It will show rotational spectra.

c) NO: Nitric oxide is a diatomic molecule with a non-zero dipole moment. It will show rotational spectra.

d) HF: Hydrogen fluoride is a diatomic molecule with a non-zero dipole moment. It will show rotational spectra.

Therefore, the correct answer is option 'A' - N2, as it does not have a permanent dipole moment and cannot interact with electromagnetic radiation to show rotational spectra.

The normal modes of vibrations of N2O is:
  • a)
    1
  • b)
    3
  • c)
    4
  • d)
    7
Correct answer is option 'C'. Can you explain this answer?

Bijoy Kapoor answered
N2O is a linear triatomic molecule and has four normal modes of vibration (and only two of rotation). Associated with each normal mode is a vibrational frequency and a normal coordinate.

Amo ng the singlet (S), doublet (D) and triplet (T). electronic states, phosphorescence involves transition between:
  • a)
    S → S
  • b)
    D → D
  • c)
    T → S
  • d)
    S → T
Correct answer is option 'C'. Can you explain this answer?

Nandini Das answered
Phosphorescence involves transition between T and S electronic states. This can be explained as follows:

T and S states:
- T state refers to the triplet electronic state which has two unpaired electrons with opposite spins.
- S state refers to the singlet electronic state which has all electrons paired with opposite spins.

Phosphorescence:
- Phosphorescence involves the emission of light from a molecule after it has been excited to higher energy levels by absorption of light.
- In this process, electrons are excited from the ground state (S0) to the excited singlet state (S1) via absorption of a photon.
- From the excited singlet state (S1), the molecule can undergo two possible routes for de-excitation:
- It can either return to the ground state (S0) by emitting a photon, which is called fluorescence.
- Or it can undergo intersystem crossing to the triplet state (T1) and then relax to the ground state (S0) by emitting a photon, which is called phosphorescence.

Transition between T and S states:
- Intersystem crossing is a process that occurs between electronic states of different multiplicities, i.e., singlet to triplet or triplet to singlet.
- In the case of phosphorescence, the molecule undergoes intersystem crossing from the excited singlet state (S1) to the triplet state (T1).
- The transition from S to T state involves a change in electron spin, which is a quantum mechanical property.
- Therefore, the transition from S to T state is a spin-forbidden process, which means it occurs with a low probability and is relatively slow.
- However, once the molecule reaches the triplet state (T1), it can undergo radiative or non-radiative decay to the ground state (S0) with a high probability.

Conclusion:
- Thus, phosphorescence involves the transition between T and S electronic states, where the molecule undergoes intersystem crossing from S1 to T1 and then relaxes to S0 by emitting a photon.

Which carbon of (a)-(d) of hex-3-en-2-one has the smallest (most upfield) chemical shift in the NMR spectrum?
  • a)
     C1
  • b)
     C2
  • c)
     C4
  • d)
     C6
Correct answer is option 'D'. Can you explain this answer?

Edurev.iitjam answered
C1 and C6 are methyl carbons (both sp3 hybridized) while C2 is a carbonyl and C4 is vinylic (both sp2). Of the two methyl carbons, C1 is more strongly affected (more downfield) by the carbonyl group, so the signal for C6 will be most upfield (smallest chemical shift) of the four.

The rotational (microwave) spectrum of a rigid diatomic rotor consists of equally spaced lines with spacing equal to:
  • a)
    B
  • b)
    B/2
  • c)
    3B/2
  • d)
    2B
Correct answer is option 'D'. Can you explain this answer?

Mrinalini Sen answered
For a rigid rotor diatomic molecule, the selection rules for rotational transitions are ΔJ = +/-1, ΔMJ = 0 . The rotational spectrum of a diatomic molecule consists of a series of equally spaced absorption lines, typically in the microwave region of the electromagnetic spectrum.

In a rotational spectrum, transitions are only observed between rotational levels of ΔJ =
  • a)
     ±1
  • b)
    ±2
  • c)
    +1
  • d)
    +3
Correct answer is option 'A'. Can you explain this answer?

Swara Dasgupta answered
Molecular rotations require little energy to excite them. Pure rotation spectra occur in the microwave region of the spectrum (~1 - 200 cm-1). It is important to note that a molecule cannot rotate about some arbitrary axis - the principle of conservation of angular momentum dictates that only a few rotations are possible. In general, rotation must be about the centre of mass of a molecule, and the axis must allow for conservation of angular momentum. In simple cases, this can often be recognised intuitively through symmetry - such as with the water molecule.
A pure rotation spectrum can only arise when the molecule possesses a permanent electric dipole moment. Like with vibrational spectroscopy, the physical effect that couples to photons is a changing dipole moment. Since molecular bond lengths remain constant in pure rotation, the magnitude of a molecule's dipole cannot change. However, since electric dipole is a vector quantity (it has both size and direction) rotation can cause a permanent dipole to change direction, and hence we observe its spectra.  Since homonuclear molecules such as dinitrogen (N2) have no dipole moment they have no rotation spectrum.  Highly symmetric polyatomic molecules, such as carbon dioxide, also have no net dipole moment - the dipoles along the C-O bonds are always equal and opposite and cancel each other out. It is important to recognise also that if a molecule has a permanent dipole, but this dipole lies along the main rotation axis, then the molecule will not have a rotational spectrum - such as for a water molecule.
In pure rotational spectroscopy for a simple diatomic molecule, the energy levels - as displayed below - are given by EJ = BJ(J+1), where J is the rotational quantum number, B is the rotational constant for the particular molecule given by B = h2 / 8π2I with the unit of Joules, where I is the moment of inertia, given by I = μr2 - where r is the bond length of this particular diatomic molecule and μ is the reduced mass, given by μ = m1m2 / m1 + m2.
Most energy level transitions in spectroscopy come with selection rules. These rules restrict certain transitions from occuring - though often they can be broken. In pure rotational spectroscopy, the selection rule is ΔJ = +_1.

Which of the following statements regarding mass spectrometry is wrong?
  • a)
    In a normal mass spectrometer, electron impact causes a molecule to lose an electron and become a molecular radical cation which decomposes into fragment cations and radicals.
  • b)
    Only cations can be detected by a normal mass spectrometer.
  • c)
    A compound whose molecules contain just one bromine atom shows two molecular ion peaks of similar intensity, one at +1 and one at -1 of the average m/z value.
  • d)
    Molecular ion peaks always have even-numbered values of m/z.
Correct answer is option 'D'. Can you explain this answer?

Asf Institute answered
The molecular mass of a molecule containing one nitrogen atom, for example, is an odd number of atomic mass units, so its molecular ion peak has an odd numbered m/z ratio; statement (d) is wrong.

What is the absorbance of an IR peak with a 36% transmittance?
  • a)
    0.53
  • b)
    0.85
  • c)
    0.44
  • d)
    0.12
Correct answer is option 'C'. Can you explain this answer?

Edurev.iitjam answered

Using the equation that relates absorbance to transmittance, we can convert 0.36 transmittance to absorbance:
A = −log(T) = −log(0.36) = 0.44

All of the following molecules would exhibit two distinct singlets in a 1H-NMR spectrum except __________.
  • a)
    2,4-hexadiyne
  • b)
    1,4-dimethylbenzene
  • c)
    methyl-tert-butyl ether
  • d)
    1,2,4,5-tetramethylbenzene
Correct answer is option 'A'. Can you explain this answer?

Asf Institute answered
  • 2,4-hexadiyne has only one 1H-NMR signal, as the two terminal methyl groups are identical and will have the same chemical shift.
  • 1,2,4,5-tetramethylbenzene has two singlets: one for the four methyl groups and one for the two aromatic protons. Likewise, 1,3,5-trimethylbenzene will have two singlets: one for the three methyl groups (nine hydrogens total) and one for the three aromatic protons, which are all identical.
  • Methyl tert-butyl ether also has two singlets, one corresponding to the tert-butyl methyl protons, and one corresponding to the methoxy protons.
  • Finally, 1,4-dimethylbenzene has two singlets, one for the methyl groups, and one for the four aromatic protons, which are all identical.

Which carbon of (a)-(d) of hex-3-en-2-one shows the largest (most downfield) chemical shift in the NMR spectrum?
  • a)
     C1
  • b)
     C2
  • c)
     C4
  • d)
     C6
Correct answer is option 'B'. Can you explain this answer?

Mihir Singh answered
Chemical Shift in NMR Spectrum of Hex-3-en-2-one
Hex-3-en-2-one is a compound with a carbon-carbon double bond and a carbonyl group. The chemical shift in the NMR spectrum of this compound can be analyzed to determine which carbon atom experiences the largest (most downfield) shift.

Explanation:
- C1: Carbon 1 is directly attached to the carbonyl group, which causes it to experience a deshielding effect, leading to a downfield shift in the NMR spectrum.
- C2: Carbon 2 is part of the carbon-carbon double bond, which also experiences deshielding due to the presence of pi electrons. This results in a significant downfield shift in the NMR spectrum.
- C4: Carbon 4 is further away from the functional groups and is not as affected by deshielding effects, resulting in a smaller chemical shift compared to C1 and C2.
- C6: Carbon 6 is also not directly involved in any functional groups and experiences the least deshielding effect, leading to the smallest chemical shift in the NMR spectrum.
Therefore, in the NMR spectrum of hex-3-en-2-one, carbon 2 (C2) shows the largest (most downfield) chemical shift due to its proximity to the carbon-carbon double bond, which causes significant deshielding effects.

The frequency of UV radiation is greater than:
  • a)
    Microwaves
  • b)
    IR radiation
  • c)
    Both (a) and (b)
  • d)
    None of these 
Correct answer is option 'C'. Can you explain this answer?

Bijoy Patel answered
Frequency of UV radiation compared to microwaves and IR radiation

UV radiation has a higher frequency than microwaves and IR radiation. This can be explained by looking at the electromagnetic spectrum, which shows the range of wavelengths and frequencies of different types of radiation.

Electromagnetic spectrum

The electromagnetic spectrum is the range of all types of electromagnetic radiation. It includes radio waves, microwaves, infrared radiation, visible light, ultraviolet radiation, X-rays, and gamma rays.

Frequency and wavelength

Frequency refers to the number of waves that pass a point in a given amount of time. It is measured in hertz (Hz). Wavelength refers to the distance between two consecutive peaks or troughs of a wave. It is measured in meters (m).

Relationship between frequency and wavelength

There is an inverse relationship between frequency and wavelength. This means that as the frequency of a wave increases, its wavelength decreases, and vice versa.

UV radiation

UV radiation has a higher frequency than visible light, which is why it is not visible to the human eye. It is divided into three categories based on wavelength: UVA (320-400 nm), UVB (280-320 nm), and UVC (100-280 nm).

Microwaves

Microwaves have a lower frequency than UV radiation. They are commonly used for communication and heating in microwave ovens. The frequency of microwaves is typically in the range of 300 MHz to 300 GHz.

IR radiation

IR radiation has a lower frequency than UV radiation and a higher frequency than microwaves. It is commonly used for heating and sensing. The frequency of IR radiation is typically in the range of 300 GHz to 400 THz.

Conclusion

In conclusion, UV radiation has a higher frequency than both microwaves and IR radiation. This is because it has a shorter wavelength and more waves pass a point in a given amount of time.

The increase in vibrational energy leads to absorption spectrum in:

a)IR region
b)Visible region
c)Microwave region
d)UV region
Correct answer is option 'A'. Can you explain this answer?

Shivam Sharma answered
This leads to an increased vibrational energy level. The third route involves electrons of molecules being raised to a higher electron energy, which is the electronic transition. It's important to state that the energy is quantized and absorption of radiation causes a molecule to move to a higher internal energy level.

State which of the following molecules can show a pure rotational microwave spectrum:
  • a)
    N2
  • b)
    CO2
  • c)
    OCS
  • d)
    HCl 
Correct answer is option 'D'. Can you explain this answer?

Soumya Sengupta answered
Explanation:

Pure rotational microwave spectrum refers to the rotational transitions of a molecule in the microwave region of the electromagnetic spectrum. For a molecule to show a pure rotational microwave spectrum, it must meet certain criteria:

1. The molecule must have a permanent dipole moment.

2. The molecule must be a linear molecule or have a symmetric top structure.

Based on these criteria, the molecule that can show a pure rotational microwave spectrum is HCl.

Explanation for each option:

a) N2: Nitrogen is a diatomic molecule with a linear structure. However, it has no permanent dipole moment, so it cannot show a pure rotational microwave spectrum.

b) CO2: Carbon dioxide is a linear molecule with a permanent dipole moment. However, it has a symmetric top structure, so it cannot show a pure rotational microwave spectrum.

c) OCS: Carbonyl sulfide has a linear structure, but it has no permanent dipole moment. Thus, it cannot show a pure rotational microwave spectrum.

d) HCl: Hydrogen chloride has a linear structure and a permanent dipole moment. Therefore, it meets both criteria for showing a pure rotational microwave spectrum.

Conclusion:

In conclusion, out of the given options, HCl is the only molecule that can show a pure rotational microwave spectrum due to its linear structure and permanent dipole moment.

Which of the following spectroscopic techniques will be useful to dist inguish between M–SCN and M–NCS binding modes:
  • a)
    NMR
  • b)
    IR
  • c)
    EPR
  • d)
    Mass
Correct answer is option 'B'. Can you explain this answer?

Rahul Chatterjee answered
In addition to identifying the type of functional groups present, IR can also be used to distinguish between different types of bonding, such as complexes containing SCN- or NO2- ligands.

SCN- can bind to the metal via either the S or N atoms, and the different coordination modes give rise to different characteristic IR frequencies.

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:
  • a)
    10
  • b)
    9
  • c)
    19
  • d)
    25
Correct answer is option 'C'. Can you explain this answer?

Swara Reddy answered
Given:

Fundamental vibration frequency (v1) = 6.5 x 10^13 s^-1

Rotational constant (B) = 1.743 x 10^11 s^-1

To find: Energy of the rotational state when there is no rotational energy in the first vibrational state.

Solution:

The energy of a vibrational state is given by the formula:

Ev = (v + 1/2)hν1

where v is the vibrational quantum number, h is Planck's constant, and ν1 is the fundamental vibration frequency.

The energy of a rotational state is given by the formula:

Er = J(J+1)hB

where J is the rotational quantum number and B is the rotational constant.

When there is no rotational energy in the first vibrational state, the quantum number J for the first vibrational state is zero. Therefore, the energy of the first vibrational state with no rotational energy is:

E1 = (0+1/2)hν1 = 1/2hν1

To find the energy of the rotational state that has the same energy as the first vibrational state with no rotational energy, we need to find the value of J that satisfies the following equation:

Er = E1

J(J+1)hB = 1/2hν1

J(J+1) = 1/2ν1/B

J(J+1) = 3.7278 x 10^3

J = 61

Therefore, the rotational state that has the same energy as the first vibrational state with no rotational energy is the state with J = 61. The energy of this state is:

Er = J(J+1)hB = 62 x 63 x 1.743 x 10^11 x 6.626 x 10^-34

Er = 1.879 x 10^-20 J

Er = 1.879 x 10^-20 x 61

Er = 1.147 x 10^-18 J

Finally, we need to convert the energy from joules to wavenumbers:

1 J = 1.196 x 10^-5 cm^-1

Er = 1.147 x 10^-18 x 1.196 x 10^-5

Er = 1.373 cm^-1

Therefore, the answer is option (c) 19.

Which of the following statements regarding electron-impact mass spectrometry is true?
  • a)
    Samples need isotopic labels.
  • b)
    The base peak is formed by loss of one electron from each vaporised molecule by an electron beam.
  • c)
    Compounds must have a functional group to show a mass spectrum.
  • d)
    A meaningful mass spectrum can sometimes be obtained on a very small sample of an impure compound.
Correct answer is option 'D'. Can you explain this answer?

Edurev.iitjam answered
For the electron-impact mass spectrometric method. A mass spectrum records relative abundances of the molecular cation and its fragment cations by their mass-to-charge ratios, m/z. The base peak corresponds to the most abundant cation which is not usually the molecular cation. Mass spectrometric signals of impurities do not normally obscure the significant peaks of the main compound; in fact, mass spectrometry is a useful technique for identifying impurities.

The Q band in the vibrational spectrum of acetylene is observed in the:
  • a)
    C–C stretching mode
  • b)
    C–H symmetric mode
  • c)
    Bending mode
  • d)
    C–H antisymmetric stretching mode. 
Correct answer is option 'A'. Can you explain this answer?

Asf Institute answered
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.

For pure vibrational spectra, the selection rule is:
  • a)
    0
  • b)
    ±1
  • c)
    0, ±1
  • d)
    ±1 ,2 
Correct answer is option 'B'. Can you explain this answer?

Rajeev Menon answered
The rules are applied to the rotational spectra of polar molecules when the transitional dipole moment of the molecule is in resonance with an external electromagnetic field. Polar molecules have a permanent dipole moment and a transitional dipole moment within a pure rotational spectrum is not equal to zero.In contrast, no rotational spectra exists for homonuclear diatomics; the same is true for spherical tops. Nevertheless, certain states of a such molecules allow unexpected interactions with the electromagnetic field; i.e. some vibrations, that introduce a time-dependent dipole moment high rotational speeds that cause some distortion of an originally spherical symmetry. A (weak) dipole moment emerges.Typical values of the rotational constant $B$ are within $0.1 \ldots 10\, cm^{-1}$ and the corresponding radiative transitions lie in the microwave spectral region where the spontaneous emission is very slow. Therefore, the transitions are usually detected by measuring the net absorption of the microwave radiation.The conservation of the angular momentum is fundamental for the selection rules that allow or prohibit transitions of a linear molecule: 

What is the absorbance of an IR peak with a 17% transmittance?
  • a)
    1.22
  • b)
    0.25
  • c)
    0.77
  • d)
    0.27
Correct answer is option 'C'. Can you explain this answer?

Asf Institute answered

Using the equation that relates absorbance to transmittance, we can convert 0.17 transmittance to absorbance:
A = −log(T) = −log(0.17) = 0.77
Therefore the absorbance for this peak is 0.77.

Which of the following observations would most likely be seen when performing an H-NMR on 1-ethyl ethanoate (below)?
  • a)
    A singlet, a doublet, and a triplet
  • b)
    One doublet and one triplet
  • c)
    One singlet and two triplets
  • d)
    A singlet, a triplet, and a quartet
Correct answer is option 'D'. Can you explain this answer?

Edurev.iitjam answered

 
  • Looking at the structure above, we can see that the molecule only contains three carbons bonded to protons. These carbons are labeled 1, 2 and 3.
  • An important concept in NMR questions is determining if two carbons on the same compound will have protons split identically, and thus indistinguishable in an NMR (i.e. will those two carbons represent two individual peaks or one large peak?). In this case, C1 and C3 are clearly distinguishable from C2, since C1 and C3 are bonded to 3 hydrogens, while C2 is only bonded to two. Because C2 is adjacent to a three proton carbon, we know that the splitting pattern will display at least one quartet. This will narrow our answer choices down to two options.
  • Because C1 and C3 contain the same number of protons, we need to determine if they will represent one large peak, or two separate peaks. Looking at the compound, we can see that C3 is adjacent to a two-proton carbon in C2, while C1 is not adjacent to any proton-bonded carbons; therefore, we can expect that C1 will not be split by any protons, and will display a singlet, and C3 will be split by 2 protons, and will display a triplet.
  • As a final result, we would expect to see one singlet, one triplet and one quartet.

Which of the following statements regarding IR spectroscopy is wrong?
  • a)
    Infrared radiation is higher in energy than UV radiation.
  • b)
    Infrared spectra record the transmission of IR radiation.
  • c)
    Molecular vibrations are due to periodic motions of atoms in molecules, and include bond stretching, torsional changes, and bond angle changes.
  • d)
    Infrared spectra give information about bonding features and functional groups in molecules.
Correct answer is option 'A'. Can you explain this answer?

Jaya Sen answered
Understanding IR Spectroscopy
Infrared (IR) spectroscopy is a powerful technique used to identify molecular structures based on how they absorb infrared light. To clarify why option 'A' is incorrect, let's break down the energy levels of different types of radiation.

Energy Comparison: IR vs. UV Radiation
- **Infrared Radiation**: This type of radiation has a longer wavelength and lower energy compared to ultraviolet (UV) radiation.
- **Ultraviolet Radiation**: UV radiation has shorter wavelengths and higher energy.
Thus, the statement "Infrared radiation is higher in energy than UV radiation" is incorrect.

Key Features of IR Spectroscopy
- **Transmission of IR Radiation**: IR spectra typically record how much infrared radiation passes through a sample, indicating absorption at specific wavelengths.
- **Molecular Vibrations**: The technique relies on the vibrational motions of molecules, which include:
- Bond stretching
- Torsional changes
- Bond angle variations
- **Information About Molecules**: IR spectroscopy is used to deduce bonding features and identify functional groups within molecules.

Conclusion
In summary, the incorrect statement is option 'A', as IR radiation does not possess higher energy than UV radiation. Understanding these distinctions is crucial for effectively interpreting IR spectra in chemical analysis.

Which of the following statements is wrong.
  • a)
    The wavenumber of a band in an IR spectrum is proportional to the frequency of the associated molecular vibration.
  • b)
    Water is a good solvent for recording UV spectra of water-soluble compounds.
  • c)
    Water is a good solvent for recording IR spectra of water-soluble compounds.
  • d)
    Hydrogen bonding in hydroxy compounds leads to broadening of spectral bands attributable to O-H stretching vibrations.
Correct answer is option 'C'. Can you explain this answer?

Anagha Bajaj answered
Understanding the Incorrect Statement: Option C
Water is often considered a polar solvent, making it suitable for dissolving a variety of polar substances. However, its properties pose challenges for recording IR spectra.

Why Water is Not a Good Solvent for IR Spectra:
- **IR Absorption in Water:**
- Water has strong IR absorption bands, particularly in the regions where many functional groups of organic compounds absorb.
- This leads to a high background signal, which can obscure the spectral features of the solute.
- **Dilution Effects:**
- When water-soluble compounds are dissolved in water, their concentration may be too low to generate clear IR signals, making it difficult to interpret the spectrum.

Comparison with UV Spectroscopy:
- **UV Transparency:**
- In contrast, water is transparent in the UV region, allowing for effective recording of UV spectra for water-soluble compounds.
- **Minimal Interference:**
- The absence of significant UV absorption by water ensures that the spectra reflect the properties of the solute without interference.

Conclusion:
Thus, while water is an excellent solvent for dissolving many compounds, it is unsuitable for recording IR spectra due to its own strong absorbance in the IR range. This makes option 'C' the incorrect statement among the given options.

Observe the underlined/bold hydrogen. In HNMR, how many spectral lines will that bolded hydrogen be split into?
  • a)
    0 lines
  • b)
    9 lines (multiplet)
  • c)
    3 lines (triplet)
  • d)
    10 lines (multiplet)
Correct answer is option 'D'. Can you explain this answer?

Edurev.iitjam answered
(9+1)=10 lines
  • Within 1H NMR spectroscopy there are a couple important factors to understand, including the ppm shift (Delta) and the splitting pattern. Here we are focusing on the splitting pattern for individual hydrogens. This is important for it lays the groundwork for understanding the patterns of peaks seen on large compound NMR’s. 
  • When determining the splitting of any hydrogen you must use the n+1 rule. Before going into that rule we must understand two things, 1. only nonequivalent hydrogens (protons) couple and 2. usually they only couple to other hydrogens (protons) attached to adjacent carbons. Nonequivalent means the protons occupy their own unique spatial environment with different atoms surrounding them.
  • Typically two hydrogens attached to the same carbon are equivalent (though this isn’t always the case and you must think about where the hydrogens are located in space and see whether they are adjacent to different or similar chemical groups). 
  • The n+1 rule is performed as follows. N stands for the number of equivalent protons that are not equivalent to the proton of interest (the one we are trying to determine the splitting for). We multiply together each group of protons that are nonequivalent to the proton of interest. For example, lets say there are two groups of protons that are nonequivalent to the proton of interest, in group A there are 2 protons, and in group B there are 3 protons.
  • We would use the n+1 rule for group A and get (2+1) = 3 and for group B get (3+1) = 4. We would then multiply these two numbers together to get the splitting for the proton of interest, thus 3 x 4 = 12 lines that the proton of interest would be split into. Proper use of this rule should allow you to get all NMR splitting questions correct (and elevate your understanding of why a certain NMR printout looks the way it does). 

The vibrational rotational spectrum is observed in …………….. region.
  • a)
    Near infrared
  • b)
    Microwave region
  • c)
    Visible region
  • d)
    Radio frequency region 
Correct answer is option 'A'. Can you explain this answer?

Tanishq Goyal answered
Vibrational rotational spectrum in Near Infrared region

Heading: Introduction
The vibrational rotational spectrum is the spectrum that is observed when a molecule undergoes both rotational and vibrational transitions. This spectrum can be observed in various regions of the electromagnetic spectrum, such as the microwave region, the near-infrared region, and the far-infrared region.

Heading: Vibrational rotational spectrum
The vibrational rotational spectrum is a combination of the vibrational spectrum and the rotational spectrum of a molecule. The vibrational spectrum is the spectrum that is observed when a molecule undergoes a change in vibrational energy level. The rotational spectrum is the spectrum that is observed when a molecule undergoes a change in rotational energy level.

Heading: Near-infrared region
The near-infrared region is the region of the electromagnetic spectrum that lies between the visible region and the mid-infrared region. The near-infrared region has wavelengths ranging from 700 nm to 2500 nm. In this region, the vibrational rotational spectrum is observed for molecules that have low rotational constants and high vibrational constants.

Heading: Explanation
The vibrational rotational spectrum is observed in the near-infrared region because the energy difference between the rotational and vibrational energy levels of a molecule is in the range of the near-infrared region. The near-infrared region is also suitable for observing the vibrational rotational spectrum because it is not absorbed by most materials and can penetrate through biological tissues.

Heading: Conclusion
In conclusion, the vibrational rotational spectrum is observed in the near-infrared region of the electromagnetic spectrum. This region is suitable for observing the vibrational rotational spectrum of molecules that have low rotational constants and high vibrational constants. The near-infrared region is also suitable for observing the vibrational rotational spectrum because it can penetrate through biological tissues.

A series of rotational lines are measured in the for IR region. The intensity of the fifth line is 6.38 times the intensity of second rotational lien. What is the temperature of the sample: (B=55.35 GHz)
  • a)
    273 K
  • b)
    220C
  • c)
    –220C
  • d)
    –273 K
Correct answer is option 'B'. Can you explain this answer?

Vaibhav Ghosh answered
Solution:

Given: Ratio of fifth line intensity to second line intensity = 6.38

We know that the intensity of rotational lines is given by the Boltzmann distribution law:

I ∝ (2J+1)exp(-EJ/kT)

Where,
I = intensity of line
J = rotational quantum number
EJ = energy of rotational level
k = Boltzmann constant
T = temperature of the sample

Let us consider the ratio of intensity of fifth line to second line:

I5/I2 = [(2J5+1)exp(-EJ5/kT)]/[(2J2+1)exp(-EJ2/kT)]

I5/I2 = [(2*5+1)exp(-E5/kT)]/[(2*2+1)exp(-E2/kT)]

I5/I2 = [11exp(-E5/kT)]/[5exp(-E2/kT)]

I5/I2 = 11/5exp[-(E5-E2)/kT]

Given I5/I2 = 6.38

6.38 = 11/5exp[-(E5-E2)/kT]

ln(6.38) = ln(11/5) - (E5-E2)/kT

ln(6.38/11/5) = -(E5-E2)/kT

(E5-E2)/kT = -ln(6.38/11/5)

(E5-E2)/kT = 0.578

E5 - E2 = B(5^2-2^2)

E5 - E2 = 231.6 GHz

Substituting the above values in the equation,

0.578 = 231.6/BT

T = 273 K

Therefore, the temperature of the sample is 273 K.

Which of the following most likely represents the H-NMR spectrum of the molecule shown below?
  • a)
    One quartet, one triplet, two doublets, and one singlet
  • b)
    One quartet, two triplets, one doublet, and one singlet
  • c)
    One quartet, one triplet, and three doublets
  • d)
    One quartet, one triplet, one doublet, and one singlet
Correct answer is option 'A'. Can you explain this answer?

Edurev.iitjam answered
  • There are four total aromatic protons, consistent with two sets of identical pairs. This would result in two distinct aromatic signals, each having a doublet and each integrating two protons.
  • The methyl protons next to the ketone would be deshielded by the electron withdrawing ketone group, resulting in a downfield shift. The signal would be a singlet, since there are no neighboring protons to the methyl group.
  • Finally, the ethyl group would have two signals, one for the two protons next to the aromatic ring (shifted downfield because of the aromatic ring), and one highly shielded peak corresponding to the terminal protons. The protons next to the aromatic ring will result in a quartet from the three neighboring hydrogens, while the terminal peak will be a triplet from the two neighboring hydrogens.
  • The final result is one quartet (ethyl), one triplet (ethyl-terminal), two doublets (aromatic), and one singlet (methyl).

Observe the underlined/bold hydrogen. In HNMR, how many spectral lines will that bolded hydrogen be split into?
  • a)
    1 line (singlet)
  • b)
    2 lines (doublet)
  • c)
    6 lines (doublet of triplets)
  • d)
    7 lines (septet)
Correct answer is option 'D'. Can you explain this answer?

Asf Institute answered
(6+1)=7 lines
  • Within 1H NMR spectroscopy there are a couple important factors to understand, including the ppm shift (delta) and the splitting pattern. Here we are focusing on the splitting pattern for individual hydrogens. This is important for it lays the groundwork for understanding the patterns of peaks seen on large compound NMR’s. 
  • When determining the splitting of any hydrogen you must use the n+1 rule. Before going into that rule we must understand two things, 1. only nonequivalent hydrogens (protons) couple and 2. usually they only couple to other hydrogens (protons) attached to adjacent carbons.
  • Nonequivalent means the protons occupy their own unique spatial environment with different atoms surrounding them. Typically two hydrogens attached to the same carbon are equivalent (though this isn’t always the case and you must think about where the hydrogens are located in space and see whether they are adjacent to different or similar chemical groups). 
  • The n+1 rule is performed as follows. N stands for the number of equivalent protons that are not equivalent to the proton of interest (the one we are trying to determine the splitting for). We multiply together each group of protons that are nonequivalent to the proton of interest.
  • For example, lets say there are two groups of protons that are nonequivalent to the proton of interest, in group A there are 2 protons, and in group B there are 3 protons. We would use the n+1 rule for group A and get (2+1) = 3 and for group B get (3+1) = 4. We would then multiply these two numbers together to get the splitting for the proton of interest, thus 3 x 4 = 12 lines that the proton of interest would be split into.
  • Proper use of this rule should allow you to get all NMR splitting questions correct (and elevate your understanding of why a certain NMR printout looks the way it does). 

Which of (a)-(d) indicates the correct order of carbon chemical shifts of the four carbons of the following compound.
  • a)
    CMe < C2 < C3 < C1
  • b)
    CMe < C3 < C2 < C1
  • c)
    CMe < C2 < C1 < C3
  • d)
    CMe < C1 < C2 < C3
Correct answer is option 'A'. Can you explain this answer?

Edurev.iitjam answered
The chemical shift of the carbonyl carbon (C1) is the largest (most downfield), and those of the alkene carbons are the next largest Of C2 and C3, the β carbon (C3) is the more strongly influenced by the electron-withdrawing effect of the carbonyl.

How many signals does the unsaturated ketone
(CH3)2CHCH2C(O)CH=CH2 have in 1H NMR and 13C NMR spectra?
  • a)
    five 1H signals and six 13C signals
  • b)
    six 1H signals and six 13C signals
  • c)
    six 1H signals and seven 13C signals
  • d)
    five 1H signals and seven 13C signals
Correct answer is option 'B'. Can you explain this answer?

Anshul Mehra answered
Understanding the Structure
The compound in question, (CH3)2CHCH2C(O)CH=CH2, is an unsaturated ketone. To determine the number of signals for both 1H and 13C NMR spectra, we need to analyze its structure and identify the unique environments for hydrogen and carbon atoms.

1H NMR Signals
- The hydrogen atoms can be grouped based on their unique environments:
- **CH3 groups (2)**: The two methyl (CH3) groups attached to the isopropyl carbon contribute **2 signals**.
- **CH2 group (1)**: The methylene (CH2) group adjacent to the carbonyl contributes **1 signal**.
- **H on carbon adjacent to carbonyl (1)**: The hydrogen on the carbon adjacent to the carbonyl (C=O) contributes **1 signal**.
- **Alkenic hydrogens (2)**: The two hydrogens on the alkene (CH=CH2) contribute **2 signals**.
Thus, the total number of unique 1H signals is **6**.

13C NMR Signals
- Now, let’s analyze the carbon atoms:
- **Two CH3 groups**: Each CH3 contributes **1 signal** (total **2 signals**).
- **One CH (connected to the carbonyl)**: This carbon contributes **1 signal**.
- **One CH2 (next to the CH)**: This carbon contributes **1 signal**.
- **One carbonyl carbon (C=O)**: This contributes **1 signal**.
- **Two carbons from the alkene (C=C)**: Each contributes **1 signal** (total **2 signals**).
Thus, the total number of unique 13C signals is **6**.

Conclusion
In summary, the unsaturated ketone (CH3)2CHCH2C(O)CH=CH2 has **6 unique 1H NMR signals** and **6 unique 13C NMR signals**. Therefore, the correct answer is option **B**: six 1H signals and six 13C signals.

Which of the following statements regarding mass spectrometry is false?
  • a)
    The base peak of a simple ketone is usually attributable to an acylium ion.
  • b)
    The molecular ion of carbonyl compounds with a -C-H readily undergoes elimination of an alkene to give a relatively stable enol radical cation.
  • c)
    The molecular ion peak of some alcohols is very weak because it readily loses an alkyl radical to give a relatively stable oxonium (hydroxycarbenium) ion.
  • d)
    Structurally isomeric alkanes cannot be distinguished by low resolution mass spectrometry.
Correct answer is option 'D'. Can you explain this answer?

Ameya Reddy answered
Understanding Mass Spectrometry and Structural Isomerism
In mass spectrometry, the ability to distinguish between isomeric compounds is crucial for accurate analysis. Let's break down why option 'D' is the correct answer.
Isomeric Alkanes and Mass Spectrometry
- Structural Isomers: Alkanes can exist as structural isomers, which have the same molecular formula but different arrangements of atoms.
- Low Resolution Mass Spectrometry: Low resolution mass spectrometry typically measures mass-to-charge ratios (m/z) without providing detailed structural information. As a result, it cannot differentiate between isomers that have the same m/z value.
Why Option D is True
- Identical Molar Mass: Structurally isomeric alkanes (e.g., pentane, 2-methylbutane) possess the same molecular formula (C5H12) and hence the same molecular ion peak in low resolution mass spectrometry.
- Insufficient Resolution: Low resolution instruments lack the capability to resolve small differences in mass that would distinguish isomers, making it impossible to identify them based solely on their mass spectra.
Conclusion
- While other options (a, b, c) accurately describe specific behaviors of different functional groups in mass spectrometry, option 'D' correctly highlights a limitation of low resolution mass spectrometry in distinguishing structurally isomeric alkanes.
This understanding is critical in organic chemistry and analytical applications, where precise identification of compounds is essential.

Chapter doubts & questions for Physical Spectroscopy - Topicwise Question Bank for IIT JAM/CSIR/GATE Chemistry 2025 is part of Chemistry exam preparation. The chapters have been prepared according to the Chemistry exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for Chemistry 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.

Chapter doubts & questions of Physical Spectroscopy - Topicwise Question Bank for IIT JAM/CSIR/GATE Chemistry in English & Hindi are available as part of Chemistry exam. Download more important topics, notes, lectures and mock test series for Chemistry Exam by signing up for free.

Signup to see your scores go up within 7 days!

Study with 1000+ FREE Docs, Videos & Tests
10M+ students study on EduRev
© EduRev
Education Revolution
Signup to see your scores go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup