Molecular Spectroscopy Questions and Answers

Q: In rotational fine structure of vibration-electronic transitions of a diatomic molecule, the band head will appear in R-branch if :

A band head forms in the R branch if B′ \(\lt\) B″.In the rotational fine structure of an electronic band, the line positions follow\[ \begin{aligned} \bar{\nu} &= \bar{\nu}_{0} + (B'+B'')m \\ &+ (B'-B'')m^{2} \end{aligned} \]with m = J + 1 for the R branch (m positive) and m = −J for the P branch. If B′ \(\lt\) B″, which is common because a bond usually lengthens on excitation, the m2 term is negative. In the R branch the lines first move to higher wavenumber, then turn back once the negative quadratic term wins: they pile up at a band head, and the band is degraded (shaded) to the red.If B′ \(\gt\) B″, the head forms in the P branch instead. If B′ = B″, there is no quadratic term and no head, and B′ = −B″ is physically impossible, since rotational constants are positive.

Q: In electronic spectra of molecules, the vibrational quantum number leading to convergence of vibrational spectral lines is given by :

The convergence occurs at v = 1/(2xe) − 1.For an anharmonic oscillator the vibrational levels are\[ \begin{aligned} \varepsilon_{v} &= \bar{\omega}_{e}(v+\tfrac{1}{2}) \\ &- \bar{\omega}_{e}x_{e}(v+\tfrac{1}{2})^{2} \end{aligned} \]The spacing between neighbouring levels shrinks as v rises. Setting the spacing to zero gives the level where the progression converges:\[ \begin{aligned} \Delta\varepsilon &= \bar{\omega}_{e}[1 - 2x_{e}(v+1)] = 0 \\ v_{max} &= \frac{1}{2x_{e}} - 1 \end{aligned} \]This is the form given in Banwell. The other options drop the factor 2, or put the wavenumber \(\bar{\omega}_{e}\) in the denominator, which makes v carry units of cm instead of being a pure number.

Q: The excess population of nuclear spins in lower state is given by :

The excess population is (nlower − nupper)/(nlower + nupper).In NMR, nuclear spins split into two levels in the magnetic field. The net absorption signal depends on how many more spins are in the lower level than in the upper one, expressed as a fraction of all the spins: the fractional population difference, or polarisation. From the Boltzmann distribution it is tiny, about γℏB0/2kT, roughly 10−5 for protons in common magnets, which is why NMR is an insensitive technique.The first option is simply the fraction of spins in the lower level, not an excess. The second is the reciprocal of the excess. The third has the sign reversed, giving a negative number for the normal case where the lower level is more populated.

Q: Degeneracy in rotational levels for H-atom along z-axis for l = 0, 1, 2, 3 are :(Given : \(\dfrac{\hbar^{2}}{2I}\) = 1.2 × 10−21 J)

The degeneracies are 1, 3, 5, 7.For angular momentum quantum number l, the projection quantum number ml takes the values −l, ..., 0, ..., +l, so each level is (2l + 1)-fold degenerate:\[ \begin{aligned} l=0 &: 1,\quad l=1: 3 \\ l=2 &: 5,\quad l=3: 7 \end{aligned} \]The value of ℏ2/2I fixes the energies, E = l(l + 1)ℏ2/2I, but not the degeneracy. 1, 2, 3, 4 counts l + 1; 2, 4, 6, 8 doubles that; and 1, 4, 9, 16 is (l + 1)2, the degeneracy of the hydrogen shells n = 1 to 4, not of a single l level.

Q: Every alternate line in Raman spectra of a symmetric top molecule has increased intensity because of :

The alternation arises from S-branch lines overlapping every alternate R-branch line.For a symmetric top the rotational Raman selection rules are ΔJ = 0, ±1, ±2 (with ΔK = 0). The ΔJ = ±1 transitions (R branch) give lines spaced 2B apart, and the ΔJ = ±2 transitions (S branch) give lines spaced 4B apart. Every S-branch line falls exactly on top of every second R-branch line, so alternate lines have extra intensity.Nuclear spin statistics cause intensity alternation in molecules with equivalent nuclei such as H2 or CO2, but that is not the cause here. ΔK = 0 is a selection rule, not a cause of alternation, and 'combined rotational Raman spectra' names nothing specific.

Q: Assign the spectral lines given in the table to different vibrational modes of A2B type molecule :\(\bar{v}\) (cm−1)Description(1)590IR-active (PQR branches)(2)1290Raman active, polarizedIR-active (PR branches)(3)2220Raman active, depolarizedIR-active (PR branches)

The assignment is 1 bending, 2 symmetric stretching, 3 asymmetric stretching.The PR band contours of modes 2 and 3 and the PQR contour of mode 1 show the molecule is linear: stretching modes are parallel bands (P and R only) and the bending mode is a perpendicular band with a Q branch. Bending requires the least energy, so the lowest frequency, 590 cm−1, is the bend.Of the two stretches, the totally symmetric one gives a polarized Raman line, so 1290 cm−1 (polarized) is the symmetric stretch, and the depolarized 2220 cm−1 band is the asymmetric stretch, which also lies higher in energy. These values match nitrous oxide, N2O, a linear A2B molecule without a centre of symmetry, which is why both stretches appear in both spectra. The other options misplace the bend or swap the stretches.

Q: The molecular population is :

The population of rotational level J is proportional to (2J + 1)e−E/kT.Two factors combine. Level J has 2J + 1 degenerate states (MJ = −J ... +J), so it can hold proportionally more molecules; and each state is populated according to the Boltzmann factor e−E/kT, which falls with energy. Their product rises at first and then falls, which is why rotational spectra show a maximum in intensity at an intermediate J.Option A omits the Boltzmann factor and inverts the degeneracy. Option B divides by the Boltzmann factor, so higher levels would be more populated, and option D both inverts the degeneracy and reverses the sign of the exponent.

Q: The spectra shown below is due to :

The band is due to perpendicular vibrations in a linear polyatomic molecule.A vibration-rotation band normally has a P branch (ΔJ = -1) and an R branch (ΔJ = +1), the two broad humps. A sharp, intense central Q branch (ΔJ = 0) appears only when the vibration carries angular momentum about the molecular axis, which is the case for a perpendicular vibration of a linear polyatomic molecule, such as the bending mode of CO2 or HCN.A diatomic molecule's vibration-rotation band (and a parallel band of a linear molecule) has only P and R branches with a gap at the centre, not a spike. Nuclear spin statistics alter the intensities of alternate lines, and breakdown of the Born-Oppenheimer approximation does not produce a Q branch.

Q: For a molecule to be Raman active :

A vibration is Raman active when the molecular motion causes some change in the polarizability.Raman scattering arises from the dipole induced by the light's electric field, which depends on the polarizability. Only if the polarizability changes during the vibration or rotation does the scattered light carry the shifted frequencies.Merely having polarizability is not enough, since every molecule has some. A change in dipole moment is the selection rule for infrared absorption, and a permanent dipole moment is the condition for a pure rotational (microwave) spectrum. Homonuclear diatomics such as N2 have no dipole yet are Raman active.

21.

In rotational fine structure of vibration-electronic transitions of a diatomic molecule, the band head will appear in R-branch if :

A.\(B' \lt B''\)
B.\(B' = B''\)
C.\(B' \gt B''\)
D.\(B' = -B''\)

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: A

A band head forms in the R branch if B′ \(\lt\) B″.

In the rotational fine structure of an electronic band, the line positions follow

\[ \begin{aligned} \bar{\nu} &= \bar{\nu}_{0} + (B'+B'')m \\ &+ (B'-B'')m^{2} \end{aligned} \]

with m = J + 1 for the R branch (m positive) and m = −J for the P branch. If B′ \(\lt\) B″, which is common because a bond usually lengthens on excitation, the m² term is negative. In the R branch the lines first move to higher wavenumber, then turn back once the negative quadratic term wins: they pile up at a band head, and the band is degraded (shaded) to the red.

If B′ \(\gt\) B″, the head forms in the P branch instead. If B′ = B″, there is no quadratic term and no head, and B′ = −B″ is physically impossible, since rotational constants are positive.

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22.

In electronic spectra of molecules, the vibrational quantum number leading to convergence of vibrational spectral lines is given by :

A.\(\frac{1}{x_{e}} - 1\)
B.\(\frac{1}{\bar{\omega}_{e}} - 1\)
C.\(\frac{1}{2x_{e}} - 1\)
D.\(\frac{1}{\bar{\omega}_{e} x_{e}} - 1\)

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: C

The convergence occurs at v = 1/(2x_e) − 1.

For an anharmonic oscillator the vibrational levels are

\[ \begin{aligned} \varepsilon_{v} &= \bar{\omega}_{e}(v+\tfrac{1}{2}) \\ &- \bar{\omega}_{e}x_{e}(v+\tfrac{1}{2})^{2} \end{aligned} \]

The spacing between neighbouring levels shrinks as v rises. Setting the spacing to zero gives the level where the progression converges:

\[ \begin{aligned} \Delta\varepsilon &= \bar{\omega}_{e}[1 - 2x_{e}(v+1)] = 0 \\ v_{max} &= \frac{1}{2x_{e}} - 1 \end{aligned} \]

This is the form given in Banwell. The other options drop the factor 2, or put the wavenumber \(\bar{\omega}_{e}\) in the denominator, which makes v carry units of cm instead of being a pure number.

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23.

The excess population of nuclear spins in lower state is given by :

A.\(\frac{n_{lower}}{n_{lower} + n_{upper}}\)
B.\(\frac{n_{lower} + n_{upper}}{n_{lower} - n_{upper}}\)
C.\(\frac{n_{upper} - n_{lower}}{n_{upper} + n_{lower}}\)
D.\(\frac{n_{lower} - n_{upper}}{n_{lower} + n_{upper}}\)

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: D

The excess population is (n_lower − n_upper)/(n_lower + n_upper).

In NMR, nuclear spins split into two levels in the magnetic field. The net absorption signal depends on how many more spins are in the lower level than in the upper one, expressed as a fraction of all the spins: the fractional population difference, or polarisation. From the Boltzmann distribution it is tiny, about γℏB₀/2kT, roughly 10⁻⁵ for protons in common magnets, which is why NMR is an insensitive technique.

The first option is simply the fraction of spins in the lower level, not an excess. The second is the reciprocal of the excess. The third has the sign reversed, giving a negative number for the normal case where the lower level is more populated.

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24.

Degeneracy in rotational levels for H-atom along z-axis for l = 0, 1, 2, 3 are :

(Given : \(\dfrac{\hbar^{2}}{2I}\) = 1.2 × 10⁻²¹ J)

A.1, 2, 3, 4
B.1, 3, 5, 7
C.2, 4, 6, 8
D.1, 4, 9, 16

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: B

The degeneracies are 1, 3, 5, 7.

For angular momentum quantum number l, the projection quantum number m_l takes the values −l, ..., 0, ..., +l, so each level is (2l + 1)-fold degenerate:

\[ \begin{aligned} l=0 &: 1,\quad l=1: 3 \\ l=2 &: 5,\quad l=3: 7 \end{aligned} \]

The value of ℏ²/2I fixes the energies, E = l(l + 1)ℏ²/2I, but not the degeneracy. 1, 2, 3, 4 counts l + 1; 2, 4, 6, 8 doubles that; and 1, 4, 9, 16 is (l + 1)², the degeneracy of the hydrogen shells n = 1 to 4, not of a single l level.

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25.

Every alternate line in Raman spectra of a symmetric top molecule has increased intensity because of :

A.Existence of nuclear spin
B.Overlapping of S-branch lines with every alternate R-branch line
C.Selection rule \(\Delta\)K = 0
D.Combined rotational Raman spectra

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: B

The alternation arises from S-branch lines overlapping every alternate R-branch line.

For a symmetric top the rotational Raman selection rules are ΔJ = 0, ±1, ±2 (with ΔK = 0). The ΔJ = ±1 transitions (R branch) give lines spaced 2B apart, and the ΔJ = ±2 transitions (S branch) give lines spaced 4B apart. Every S-branch line falls exactly on top of every second R-branch line, so alternate lines have extra intensity.

Nuclear spin statistics cause intensity alternation in molecules with equivalent nuclei such as H₂ or CO₂, but that is not the cause here. ΔK = 0 is a selection rule, not a cause of alternation, and 'combined rotational Raman spectra' names nothing specific.

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26.

Assign the spectral lines given in the table to different vibrational modes of A₂B type molecule :

	\(\bar{v}\) (cm⁻¹)	Description
(1)	590	IR-active (PQR branches)
(2)	1290	Raman active, polarized IR-active (PR branches)
(3)	2220	Raman active, depolarized IR-active (PR branches)

A.1 \(\rightarrow\) bending, 2 \(\rightarrow\) symmetric stretching, 3 \(\rightarrow\) asymmetric stretching
B.1 \(\rightarrow\) bending, 2 \(\rightarrow\) asymmetric stretching, 3 \(\rightarrow\) symmetric stretching
C.1 \(\rightarrow\) symmetric stretching, 2 \(\rightarrow\) asymmetric stretching, 3 \(\rightarrow\) bending
D.1 \(\rightarrow\) asymmetric stretching, 2 \(\rightarrow\) bending, 3 \(\rightarrow\) symmetric stretching

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: A

The assignment is 1 bending, 2 symmetric stretching, 3 asymmetric stretching.

The PR band contours of modes 2 and 3 and the PQR contour of mode 1 show the molecule is linear: stretching modes are parallel bands (P and R only) and the bending mode is a perpendicular band with a Q branch. Bending requires the least energy, so the lowest frequency, 590 cm⁻¹, is the bend.

Of the two stretches, the totally symmetric one gives a polarized Raman line, so 1290 cm⁻¹ (polarized) is the symmetric stretch, and the depolarized 2220 cm⁻¹ band is the asymmetric stretch, which also lies higher in energy. These values match nitrous oxide, N₂O, a linear A₂B molecule without a centre of symmetry, which is why both stretches appear in both spectra. The other options misplace the bend or swap the stretches.

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27.

The molecular population is :

A.Proportional to \((2J + 1)^{-1}\)
B.Proportional to \(\dfrac{(2J + 1)}{e^{-E/kT}}\)
C.Proportional to \((2J + 1)e^{-E/kT}\)
D.Proportional to \((2J + 1)^{-1}e^{E/kT}\)

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: C

The population of rotational level J is proportional to (2J + 1)e^−E/kT.

Two factors combine. Level J has 2J + 1 degenerate states (M_J = −J ... +J), so it can hold proportionally more molecules; and each state is populated according to the Boltzmann factor e^−E/kT, which falls with energy. Their product rises at first and then falls, which is why rotational spectra show a maximum in intensity at an intermediate J.

Option A omits the Boltzmann factor and inverts the degeneracy. Option B divides by the Boltzmann factor, so higher levels would be more populated, and option D both inverts the degeneracy and reverses the sign of the exponent.

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28.

The spectra shown below is due to :

An absorption band against wavenumber in cm-1 made of two broad humps of similar height on either side of a sharp, intense central spike

A.Perpendicular vibrations in a linear polyatomic molecule
B.Vibrational-rotational transitions in a diatomic molecule
C.Effect of nuclear spin homonuclear diatomic molecule
D.Breakdown of Born-Oppenheimer approximation

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: A

The band is due to perpendicular vibrations in a linear polyatomic molecule.

A vibration-rotation band normally has a P branch (ΔJ = -1) and an R branch (ΔJ = +1), the two broad humps. A sharp, intense central Q branch (ΔJ = 0) appears only when the vibration carries angular momentum about the molecular axis, which is the case for a perpendicular vibration of a linear polyatomic molecule, such as the bending mode of CO₂ or HCN.

A diatomic molecule's vibration-rotation band (and a parallel band of a linear molecule) has only P and R branches with a gap at the centre, not a spike. Nuclear spin statistics alter the intensities of alternate lines, and breakdown of the Born-Oppenheimer approximation does not produce a Q branch.

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29.

For a molecule to be Raman active :

A.Dipole moment of molecule must change during molecular transitions
B.It must have polarizability
C.Molecular motion must cause some change in molecular polarizability
D.It must have non-zero dipole moment

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: C

A vibration is Raman active when the molecular motion causes some change in the polarizability.

Raman scattering arises from the dipole induced by the light's electric field, which depends on the polarizability. Only if the polarizability changes during the vibration or rotation does the scattered light carry the shifted frequencies.

Merely having polarizability is not enough, since every molecule has some. A change in dipole moment is the selection rule for infrared absorption, and a permanent dipole moment is the condition for a pure rotational (microwave) spectrum. Homonuclear diatomics such as N₂ have no dipole yet are Raman active.

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