In the following table, the left column represents the rigid-rotor type and the right column shows a set of molecules.P. Symmetric rotor (oblate)1. SiH4Q. Symmetric rotor (prolate)2. CH3ClR. Spherical rotor3. C6H6S. Asymmetric rotor4. CH3OH5. CO2The CORRECT match is

The correct match is P-3, Q-2, R-1, S-4.Benzene, C6H6, is a flat symmetric top: the two in-plane moments are equal and the moment about the six-fold axis is the largest, so it is an oblate symmetric rotor.CH3Cl has a three-fold axis along C-Cl; the moment about that axis is the smallest and the other two are equal and larger, a prolate symmetric rotor.SiH4 is tetrahedral, with all three moments equal: a spherical rotor.CH3OH has no axis of three-fold or higher symmetry, so all three moments differ: an asymmetric rotor.CO2 is linear and is the extra item. Options that call SiH4 oblate, benzene spherical, or CO2 or CH3OH prolate misclassify at least one molecule.

Consider the following six vibrational modes:symmetric stretching of CO2, O-H symmetric stretching of H2O, stretching of HCl, stretching of H2, N-H symmetric stretching of NH3, and bending of CO2.Among these modes, if k number of modes are IR active but Raman inactive, l number of modes are IR inactive but Raman active, and m number of modes are both IR and Raman active.k, l, and m, respectively, are

k, l and m are 1, 2 and 3.Symmetric stretch of CO2: no dipole change, polarisability changes: Raman only.Stretch of H2: homonuclear, no dipole: Raman only.Bend of CO2: dipole changes but, by the mutual exclusion rule for a centrosymmetric molecule, Raman inactive: IR only.O-H symmetric stretch of H2O, the HCl stretch and the N-H symmetric stretch of NH3: molecules without a centre of symmetry, both IR and Raman active.That gives k = 1 (IR only), l = 2 (Raman only) and m = 3 (both). The other options swap these counts.

Molecular Spectroscopy Questions and Answers

Q: The bond that gives the most intense band in the infrared spectrum for its stretching vibration is

The most intense stretching band comes from O-H.An infrared band's intensity is proportional to the square of the change in dipole moment during the vibration, (dμ/dr)2. The larger the electronegativity difference across the bond, the larger that change. Oxygen is the most electronegative of the four atoms bonded to hydrogen here, so the O-H stretch, near 3200-3600 cm-1, is strong and broad.N-H is less polar and gives a weaker band. C-H is only slightly polar, and S-H, with sulfur's electronegativity close to that of carbon, gives one of the weakest bands, near 2550 cm-1.

Q: The pure vibrational spectrum of a hypothetical diatomic molecule shows three peaks with the following intensity at three different temperatures.PeakIntensity (arbitrary unit)300 K600 K900 KI1.01.01.0II0.10.10.1III0.020.040.06The CORRECT statement is

Peak III appears at the lowest energy.Peak I is the strongest and does not change with temperature: it is the fundamental, v = 0 to 1, from the ground level that holds almost all molecules. Peak II is ten times weaker and also temperature-independent: it is the first overtone, v = 0 to 2, which lies at almost twice the energy of the fundamental.Peak III grows as the temperature rises, so it starts from a level whose population increases with temperature: it is a hot band, v = 1 to 2. Because of anharmonicity the vibrational levels get closer together as v increases, so the 1 to 2 gap is smaller than the 0 to 1 gap and the hot band lies just below the fundamental.So in energy, peak III lies just below peak I and peak II lies highest. That makes 'peak I at the highest energy' wrong, and neither I nor II is the lowest.

Q: The CORRECT statement regarding the following three normal modes of vibration of SO3 is

The correct statement is (I) is infrared inactive while (II) and (III) are infrared active.A mode is IR active only if it changes the dipole moment. SO3 is trigonal planar (D3h).(I): all three S-O bonds stretch in phase (the symmetric stretch, A1'). The molecule keeps its full symmetry and its dipole stays zero, so this mode is IR inactive (it is Raman active).(II): S moves out of the plane one way and the three O atoms the other (A2''). This creates a dipole along the three-fold axis, so it is IR active.(III): one S-O bond shortens while the other two lengthen, an asymmetric stretch (E'). The in-plane dipole changes, so it is IR active.A zero permanent dipole does not make every mode inactive: what matters is whether the vibration makes the dipole oscillate, so the 'none of the modes' option is wrong, and the options calling (III) inactive or (I) active are wrong too.

Q: The molecule XY2 is microwave active and its vibration-rotation spectrum shows only P and R transitions. In the correct structure,

The structure is Y as the central atom in linear XY2, that is X-Y-Y.A vibration-rotation spectrum with only P and R branches and no Q branch is characteristic of a linear molecule (parallel bands). Being microwave active, the molecule must also have a permanent dipole moment.A linear molecule with X in the middle, Y-X-Y, would be centrosymmetric and have no dipole, so it would be microwave inactive (like CO2). The unsymmetrical linear arrangement X-Y-Y has a dipole, as N2O (NNO) does. Bent molecules show a Q branch in their vibration-rotation bands, which rules out both bent options.

Q: Among the following, the one for which the infrared active vibrational modes are Raman inactive and vice versa is

The molecule is the trans-planar conformer of H2O2.The question describes the rule of mutual exclusion: in a molecule with a centre of inversion, vibrations that are IR active (odd, u) are Raman inactive, and Raman-active (even, g) ones are IR inactive. Planar trans H-O-O-H belongs to \(C_{2h}\), which contains i, so the rule applies.None of the others has a centre of inversion. H2O is \(C_{2v}\), and all three of its modes are both IR and Raman active. Eclipsed ethane is \(D_{3h}\) (it is the staggered form, \(D_{3d}\), that is centrosymmetric). CH4 is \(T_{d}\), and its t2 modes are active in both spectra.

Q: The molecule that does NOT absorb the microwave radiation is

The molecule that does not absorb microwave radiation is CO\(_{2}\).A pure rotational (microwave) spectrum needs a permanent electric dipole moment, so that the rotating molecule presents an oscillating dipole to the radiation. CO2 is linear and symmetric, O=C=O, and its two bond dipoles cancel, so it has no permanent dipole and shows no microwave spectrum.H2O is bent and strongly polar, and the heteronuclear diatomics CO and NO each have a permanent dipole, so all three absorb in the microwave region.

Q: The rotational constant (usually denoted as B) for the oxygen molecule is

The rotational constant is B = 1.2 cm-1.The 16O nucleus has spin zero, and the ground state of O2 is 3Σg-. Nuclear spin statistics then allow only odd rotational levels (J = 1, 3, 5 ...). The first Stokes line is therefore J = 1 to J = 3, not J = 0 to J = 2. With rotational energies BJ(J + 1):\[ \begin{aligned} \Delta\tilde{\nu} &= B(3 \cdot 4 - 1 \cdot 2) = 10B \\ 10B &= 20491 - 20479 = 12\ \text{cm}^{-1} \\ B &= 1.2\ \text{cm}^{-1} \end{aligned} \]2.0 cm-1 is what the usual first-line shift of 6B (J = 0 to 2) would give, which is the trap here. 3.0 and 6.0 come from taking the shift as 4B or 2B.

Choose the allowed transition

A.\(^{1}\Sigma_{g}^{+} \rightarrow {}^{3}\Sigma_{u}^{+}\)
B.\(^{1}\Sigma_{g}^{+} \rightarrow {}^{3}\Sigma_{u}^{-}\)
C.\(^{1}\Sigma_{g}^{+} \rightarrow {}^{1}\Sigma_{u}^{+}\)
D.\(^{1}\Sigma_{g}^{+} \rightarrow {}^{1}\Sigma_{u}^{-}\)

Exam Year(s)

GATE Chemistry - 2012,

Answer: C

The allowed transition is \(^{1}\Sigma_{g}^{+} \rightarrow {}^{1}\Sigma_{u}^{+}\).

Electric dipole transitions in a diatomic molecule obey three rules that matter here:

ΔS = 0: the spin multiplicity must not change;
g ↔ u: parity must change (Laporte rule);
for Σ to Σ transitions, + ↔ + and - ↔ -, never + ↔ -.

\(^{1}\Sigma_{g}^{+} \rightarrow {}^{1}\Sigma_{u}^{+}\) satisfies all three. The two transitions to \(^{3}\Sigma_{u}\) states change the multiplicity from singlet to triplet, so they are spin-forbidden. \(^{1}\Sigma_{g}^{+} \rightarrow {}^{1}\Sigma_{u}^{-}\) keeps the spin and changes parity, but goes from + to -, which is forbidden.

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The bond that gives the most intense band in the infrared spectrum for its stretching vibration is

A.C-H
B.N-H
C.O-H
D.S-H

Exam Year(s)

GATE Chemistry - 2012,

Answer: C

The most intense stretching band comes from O-H.

An infrared band's intensity is proportional to the square of the change in dipole moment during the vibration, (dμ/dr)². The larger the electronegativity difference across the bond, the larger that change. Oxygen is the most electronegative of the four atoms bonded to hydrogen here, so the O-H stretch, near 3200-3600 cm^-1, is strong and broad.

N-H is less polar and gives a weaker band. C-H is only slightly polar, and S-H, with sulfur's electronegativity close to that of carbon, gives one of the weakest bands, near 2550 cm^-1.

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The pure vibrational spectrum of a hypothetical diatomic molecule shows three peaks with the following intensity at three different temperatures.

Peak	Intensity (arbitrary unit)
Peak	300 K	600 K	900 K
I	1.0	1.0	1.0
II	0.1	0.1	0.1
III	0.02	0.04	0.06

The CORRECT statement is

A.Peak I appears at the lowest energy
B.Peak II appears at the lowest energy
C.Peak III appears at the lowest energy
D.Peak I appears at the highest energy

Exam Year(s)

GATE Chemistry - 2022,

Answer: C

Peak III appears at the lowest energy.

Peak I is the strongest and does not change with temperature: it is the fundamental, v = 0 to 1, from the ground level that holds almost all molecules. Peak II is ten times weaker and also temperature-independent: it is the first overtone, v = 0 to 2, which lies at almost twice the energy of the fundamental.

Peak III grows as the temperature rises, so it starts from a level whose population increases with temperature: it is a hot band, v = 1 to 2. Because of anharmonicity the vibrational levels get closer together as v increases, so the 1 to 2 gap is smaller than the 0 to 1 gap and the hot band lies just below the fundamental.

So in energy, peak III lies just below peak I and peak II lies highest. That makes 'peak I at the highest energy' wrong, and neither I nor II is the lowest.

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In the following table, the left column represents the rigid-rotor type and the right column shows a set of molecules.

P. Symmetric rotor (oblate)	1. SiH₄
Q. Symmetric rotor (prolate)	2. CH₃Cl
R. Spherical rotor	3. C₆H₆
S. Asymmetric rotor	4. CH₃OH
	5. CO₂

The CORRECT match is

A.P-1, Q-2, R-3, S-4
B.P-3, Q-2, R-1, S-4
C.P-3, Q-5, R-1, S-2
D.P-5, Q-4, R-3, S-2

Exam Year(s)

GATE Chemistry - 2022,

Answer: B

The correct match is P-3, Q-2, R-1, S-4.

Benzene, C₆H₆, is a flat symmetric top: the two in-plane moments are equal and the moment about the six-fold axis is the largest, so it is an oblate symmetric rotor.
CH₃Cl has a three-fold axis along C-Cl; the moment about that axis is the smallest and the other two are equal and larger, a prolate symmetric rotor.
SiH₄ is tetrahedral, with all three moments equal: a spherical rotor.
CH₃OH has no axis of three-fold or higher symmetry, so all three moments differ: an asymmetric rotor.

CO₂ is linear and is the extra item. Options that call SiH₄ oblate, benzene spherical, or CO₂ or CH₃OH prolate misclassify at least one molecule.

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The CORRECT statement regarding the following three normal modes of vibration of SO₃ is

Three normal modes of trigonal planar SO3: (I) all three S-O bonds stretching in phase; (II) the S atom moving out of plane opposite to the three O atoms; (III) the upper S-O bond shortening while the two lower O atoms move outward

A.(I) and (II) are infrared active while (III) is infrared inactive
B.(I) is infrared inactive while (II) and (III) are infrared active
C.(I) and (III) are infrared inactive while (II) is infrared active
D.None of the modes are infrared active since \(\mathrm{SO_{3}}\) has zero dipole moment

Exam Year(s)

GATE Chemistry - 2022,

Answer: B

The correct statement is (I) is infrared inactive while (II) and (III) are infrared active.

A mode is IR active only if it changes the dipole moment. SO₃ is trigonal planar (D_3h).

(I): all three S-O bonds stretch in phase (the symmetric stretch, A₁'). The molecule keeps its full symmetry and its dipole stays zero, so this mode is IR inactive (it is Raman active).
(II): S moves out of the plane one way and the three O atoms the other (A₂''). This creates a dipole along the three-fold axis, so it is IR active.
(III): one S-O bond shortens while the other two lengthen, an asymmetric stretch (E'). The in-plane dipole changes, so it is IR active.

A zero permanent dipole does not make every mode inactive: what matters is whether the vibration makes the dipole oscillate, so the 'none of the modes' option is wrong, and the options calling (III) inactive or (I) active are wrong too.

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The molecule XY₂ is microwave active and its vibration-rotation spectrum shows only P and R transitions. In the correct structure,

A.X is the central atom in linear XY\(_{2}\).
B.X is the central atom in bent XY\(_{2}\).
C.Y is the central atom in linear XY\(_{2}\).
D.Y is the central atom in bent XY\(_{2}\).

Exam Year(s)

GATE Chemistry - 2024,

Answer: C

The structure is Y as the central atom in linear XY₂, that is X-Y-Y.

A vibration-rotation spectrum with only P and R branches and no Q branch is characteristic of a linear molecule (parallel bands). Being microwave active, the molecule must also have a permanent dipole moment.

A linear molecule with X in the middle, Y-X-Y, would be centrosymmetric and have no dipole, so it would be microwave inactive (like CO₂). The unsymmetrical linear arrangement X-Y-Y has a dipole, as N₂O (NNO) does. Bent molecules show a Q branch in their vibration-rotation bands, which rules out both bent options.

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Consider the following six vibrational modes:

symmetric stretching of CO₂, O-H symmetric stretching of H₂O, stretching of HCl, stretching of H₂, N-H symmetric stretching of NH₃, and bending of CO₂.

Among these modes, if k number of modes are IR active but Raman inactive, l number of modes are IR inactive but Raman active, and m number of modes are both IR and Raman active.

k, l, and m, respectively, are

A.1, 3, and 2
B.3, 1, and 2
C.1, 2, and 3
D.2, 1, and 3

Exam Year(s)

GATE Chemistry - 2024,

Answer: C

k, l and m are 1, 2 and 3.

Symmetric stretch of CO₂: no dipole change, polarisability changes: Raman only.
Stretch of H₂: homonuclear, no dipole: Raman only.
Bend of CO₂: dipole changes but, by the mutual exclusion rule for a centrosymmetric molecule, Raman inactive: IR only.
O-H symmetric stretch of H₂O, the HCl stretch and the N-H symmetric stretch of NH₃: molecules without a centre of symmetry, both IR and Raman active.

That gives k = 1 (IR only), l = 2 (Raman only) and m = 3 (both). The other options swap these counts.

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Among the following, the one for which the infrared active vibrational modes are Raman inactive and vice versa is

A.\(\mathrm{H_{2}O}\)
B.trans-planar conformer of \(\mathrm{H_{2}O_{2}}\)
C.eclipsed conformer of ethane
D.\(\mathrm{CH_{4}}\)

Exam Year(s)

GATE Chemistry - 2026,

Answer: B

The molecule is the trans-planar conformer of H₂O₂.

The question describes the rule of mutual exclusion: in a molecule with a centre of inversion, vibrations that are IR active (odd, u) are Raman inactive, and Raman-active (even, g) ones are IR inactive. Planar trans H-O-O-H belongs to \(C_{2h}\), which contains i, so the rule applies.

None of the others has a centre of inversion. H₂O is \(C_{2v}\), and all three of its modes are both IR and Raman active. Eclipsed ethane is \(D_{3h}\) (it is the staggered form, \(D_{3d}\), that is centrosymmetric). CH₄ is \(T_{d}\), and its t₂ modes are active in both spectra.

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The molecule that does NOT absorb the microwave radiation is

A.CO\(_{2}\)
B.H\(_{2}\)O
C.CO
D.NO

Exam Year(s)

GATE Life Sciences - 2012,

Answer: A

The molecule that does not absorb microwave radiation is CO\(_{2}\).

A pure rotational (microwave) spectrum needs a permanent electric dipole moment, so that the rotating molecule presents an oscillating dipole to the radiation. CO₂ is linear and symmetric, O=C=O, and its two bond dipoles cancel, so it has no permanent dipole and shows no microwave spectrum.

H₂O is bent and strongly polar, and the heteronuclear diatomics CO and NO each have a permanent dipole, so all three absorb in the microwave region.

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Direction for [ Question No: 10 To 10 ] :

A 20491 cm^-1 laser line was used to excite oxygen molecules (made of ¹⁶O only) to obtain the rotational Raman spectrum. The resulting rotational Raman spectrum of oxygen molecule has the first Stokes line at 20479 cm^-1.

10.

The rotational constant (usually denoted as B) for the oxygen molecule is

A.1.2 cm\(^{-1}\)
B.2.0 cm\(^{-1}\)
C.3.0 cm\(^{-1}\)
D.6.0 cm\(^{-1}\)

Exam Year(s)

GATE Chemistry - 2012,

Answer: A

The rotational constant is B = 1.2 cm^-1.

The ¹⁶O nucleus has spin zero, and the ground state of O₂ is ³Σ_g^-. Nuclear spin statistics then allow only odd rotational levels (J = 1, 3, 5 ...). The first Stokes line is therefore J = 1 to J = 3, not J = 0 to J = 2. With rotational energies BJ(J + 1):

\[ \begin{aligned} \Delta\tilde{\nu} &= B(3 \cdot 4 - 1 \cdot 2) = 10B \\ 10B &= 20491 - 20479 = 12\ \text{cm}^{-1} \\ B &= 1.2\ \text{cm}^{-1} \end{aligned} \]

2.0 cm^-1 is what the usual first-line shift of 6B (J = 0 to 2) would give, which is the trap here. 3.0 and 6.0 come from taking the shift as 4B or 2B.

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