Rotational Spectra of Diatomic Molecules MCQ Quiz - Objective Question with Answer for Rotational Spectra of Diatomic Molecules - Download Free PDF

CONCEPT:

Rotational Transitions in Microwave Spectroscopy

• Microwave spectroscopy deals with transitions between rotational levels of molecules.
• For a rigid diatomic rotor, rotational energy levels are given by:

  E_J = B·J(J+1)

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

EXPLANATION:

• The intensity of rotational lines depends on the population of the energy levels, which follows the Boltzmann distribution:

  P_J ∝ (2J+1)·e^{-E_J / kT}

• The most intense transition occurs from the rotational level J_max, which has the maximum population.

• This yields an approximate result: J_max ∝ √(T / B)
• Since B is constant for a given molecule, the rotational quantum number of the most populated state (and thus most intense line) varies as:

  

Therefore, the correct answer is  √T.

Concepts:

• Moment of Inertia (( I )): Moment of inertia is defined as:
  • \(I = \sum m r^2\), where ( m ) is the mass of a molecule and ( r ) is the perpendicular distance from the axis of rotation. It helps classify molecules based on their symmetry and rotation properties.
• Types of Molecules by Moment of Inertia:
  • Spherical Top: Molecules with equal moments of inertia along all three axes (I_x = I_y = I_z), such as SF₆.
    • ​Molecules having point group of O_h, T_d, and I_h are the examples of Spherical top molecules.
  • Prolate Top: Molecules with two equal moments of inertia and one distinct (I_x = I_y ≠ I_z), resembling an elongated shape, e.g., CH₃I.
    • ​Molecules having point group of C_nv are the examples of Prolate top molecule.
  • Oblate Top: Molecules with two equal moments of inertia and one distinct (I_x = I_y > I_z), but flattened in structure, e.g., C₆H₆.
    • ​Molecules having point group of D_nh are the examples of Oblate top molecules.
  • Asymmetric Top: Molecules with all three moments of inertia being different (I_x≠  I_y ≠ I_z), e.g., HCHO or H₂O₂.
    • ​Molecules with C₂ as the highest axis are the examples of Asymmetric Top molecules.

Explanation:

1. H₂O₂ is an asymmetric top molecule with a non-linear and non-planar structure. Its moments of inertia along all three principal axes I_x, I_y and I_z are different, leading to three distinct rotational constants.
2. Other molecules, such as CH₄ (a spherical top) and CHCl₃ (a symmetric top), do not have three unique rotational constants because of their higher symmetry, which results in degenerate or equal moments of inertia along certain axes.

Conclusion:

The correct answer is Option 3 (H₂O₂), as it is an asymmetric top molecule with three distinct rotational constants, making it identifiable through its microwave spectrum.

CONCEPT:

Rotational Constant and Reduced Mass

• The rotational constant B for a diatomic molecule is inversely proportional to the molecule's reduced mass  \(\mu\) :
•  \(B = \frac{h}{8 \pi^2 c \mu r^2}\) 
• Here, h is Planck’s constant, c is the speed of light, and r is the bond length. The reduced mass \( \mu\)  depends on the masses of the atoms in the molecule and can be calculated as:
  •  \(\mu = \frac{m_1 \cdot m_2}{m_1 + m_2}\) 
• The reduced mass (\(\mu \)) increases when a hydrogen isotope (H or D) is replaced with a heavier isotope, and a heavier chlorine isotope e.g., \(^{35}\text{Cl}\) vs. \(^{37}\text{Cl}\) also slightly increases \(\mu\) .
• Thus, the rotational constant B decreases as the reduced mass \( \mu \) increases.

CALCULATION:

• Given molecules: \(H^ {35}Cl\), \(D^ {35}Cl\), and \(D^ {37}Cl\).
• Step 1: Analyze the reduced masses of each molecule:
  • \(H^ {35}Cl\) has the smallest reduced mass because it involves the lightest atoms (H is lighter than D).
  • \(D^ {35}Cl\) has a larger reduced mass than \(H^ {35}Cl\) because deuterium (D) is heavier than hydrogen (H).
  • \(D^ {37}Cl\) has the largest reduced mass because it involves the heaviest isotopes (D and \(^{37}Cl\)).
• Step 2: Determine the order of rotational constants based on reduced mass:
  • Since B is inversely proportional to ( \(\mu\) ), the molecule with the smallest reduced mass will have the largest ( B ).
  • Order of reduced masses: \(H^ {35}Cl < D^ {35}Cl < D^ {37}Cl\)
  • Order of rotational constants B : \(H^ {35}Cl > D^ {35}Cl > D^ {37}Cl\)

CONCLUSION:

The correct option is: Option 1) H³⁵Cl > D³⁵Cl > D³⁷Cl

The correct answer is 2.95.

Concept:

The spacing between adjacent lines in the microwave spectrum is determined by the rotational constant ( B ). For a diatomic molecule, the rotational constant ( B ) in Hz is given by:

\( B = \frac{h}{8 \pi^2 I} \)Where:

• ( h ) is Planck's constant.
• ( I = μ r2 ) is the moment of inertia.
• ( μ ) is the reduced mass of the molecule.
• ( r ) is the bond length.

Explanation:

Given:

• Spacing for H₃₅Cl: (ΔB_HCl = 6.35×10¹¹ , Hz )
• Bond length of D₃₅Cl is 5% greater than that of H₃₅Cl

The rotational constant ( B ) is inversely proportional to the moment of inertia ( I = μr² ). Therefore, spacing ( Δ B ) is given by ( 2B ):

For HCl:\( Δ B_{\text{HCl}} = \frac{h}{4 \pi^2 I_{\text{HCl}}} \)

For DCl: \( Δ B_{\text{DCl}} = \frac{h}{4 \pi^2 I_{\text{DCl}}} \)

Because (I = μr²): \( Δ B_{\text{HCl}} = \frac{h}{4 \pi^2 μ_{\text{HCl}} r_{\text{HCl}}^2} \)  and \( Δ B_{\text{DCl}} = \frac{h}{4 \pi^2 μ_{\text{DCl}} r_{\text{DCl}}^2} \)

Since ( r_DCl = 1.05 r_HCl ): \( Δ B_{\text{DCl}} = Δ B_{\text{HCl}} \left( \frac{μ_{\text{HCl}} r_{\text{HCl}}^2}{μ_{\text{DCl}} (1.05)^2 r_{\text{HCl}}^2} \right) \)

\( Δ B_{\text{DCl}} = Δ B_{\text{HCl}} \left( \frac{μ_{\text{HCl}}}{μ_{\text{DCl}} × 1.1025} \right) ]\)

The reduced masses (μ) of HCl and DCl respectively:

\( μ_{\text{HCl}} = \frac{m_{\text{H}} m_{\text{Cl}}}{m_{\text{H}} + m_{\text{Cl}}} \)

\( μ_{\text{DCl}} = \frac{m_{\text{D}} m_{\text{Cl}}}{m_{\text{D}} + m_{\text{Cl}}} \)

Since ( m_D = 2 m_H ):

\( μ_{\text{HCl}} = \frac{1 × 35}{1 + 35} = \frac{35}{36} m_u \)

\( μ_{\text{DCl}} = \frac{2 × 35}{2 + 35} = \frac{70}{37} m_u \)

Therefore, the ratio \( \frac{μ_{\text{HCl}}}{μ_{\text{DCl}}} \):

\( \frac{μ_{\text{HCl}}}{μ_{\text{DCl}}} = \frac{35/36}{70/37} = \frac{35 × 37}{36 × 70} = \frac{35 × 37}{2520} = \frac{1295}{2520} = 0.5147 \)

Then: \( Δ B_{\text{DCl}} = 6.35 × 10^{11} × \frac{0.5147}{1.1025} = 6.35 × 10^{11} × 0.4667 = 2.965 × 10^{11} \text{Hz}\)

Conclusion:

The corresponding spacing for D³⁵Cl is 2.95 × 10¹¹ Hz.

EXPLANATION:

• CH₄ (Methane):
  • Methane is a tetrahedral molecule with four equivalent hydrogen atoms. All three moments of inertia are equal (I_A = I_B = I_C), making it a spherical rotor, not a symmetric rotor.
• CH₃Cl (Chloromethane):
  • Chloromethane has a C_3v symmetry, with the chlorine atom on one axis and three hydrogen atoms in a plane perpendicular to this axis. The moment of inertia about the C-Cl bond axis (I_A) is different from the moments of inertia about the other two perpendicular axes (I_B = I_C), making it a symmetric rotor.
• CH₂Cl₂ (Dichloromethane):
  • Dichloromethane has a greater asymmetry due to the two chlorine atoms and two hydrogen atoms. The moments of inertia are different along all three axes (I_A ≠ I_B ≠ I_C), making it an asymmetric rotor.
• CCl₄ (Carbon Tetrachloride):
  • Carbon tetrachloride is a tetrahedral molecule like methane. All three moments of inertia are equal (I_A = I_B = I_C), making it a spherical rotor, not a symmetric rotor.

Therefore, the correct option is (CH₃Cl).

Concept:

The energy of rotational transitions for molecules in the gas phase is measured using microwave rotational spectroscopy with the help of interaction between the electric dipole movement of molecules and the magnetic field of microwave photons.

The line spacing is shown below.

Pure microwave Rotational spectra do not have stokes/anti-stokes lines, they come in the Raman spectrum.

In the Raman spectrum, two types of line spectra are present which are called stokes and anti-stokes lines.

• Stokes line is observed in the ground state excitation of molecules whereas the anti-stokes line is observed when the molecule returns to the ground state from its higher excited state.
  

Explanation:-

The adjacent lines are separated by 4 cm⁻¹

Thus, 

 2B=4cm-1

or, B= 2cm-1 

The Rotational Raman spectrum is given as,

The spacing between the Rayleigh line and the first Stokes line is 6B.

So, 6B = 6 × (2cm-1 )

or, 6B =12cm-1

Now, the frequency of the 1st stoke line can be calculated using the formula,

= \(\nu \:of Rayleigh \:line-6B\)

So, the frequency of the 1st stoke line will be

= (30,000-12) cm-1 (As frequency of rayleigh line = 30000 cm-1 and B = 4 cm-1)

= 29988 cm^-1 

Conclusion:-

Here the first Stokes line (in cm−1) appears at 29988 cm^-1.

Concept:

• The basic quantum mechanical model of rotational spectroscopy is "particle on a sphere" or "the rigid rotor model".
• Rotational spectroscopy deals with a system consisting of two atoms connected by a bond of fixed length (simply a diatomic molecule). It can rotate to describe the entire 3D space resulting in an infinite number of circular trajectories which eventually describe a sphere.
• The rotational energy level for a particular state (Jth state) is given by

\({{\rm{E}}_{\rm{J}}}{\rm{ = }}{{{\hbar ^{\rm{2}}}} \over {{\rm{2I}}}}{\rm{J}}\left( {{\rm{J + 1}}} \right)\), where J is a quantum number (J=0,1,2...) and I is the moment of inertia (\(I = \mu {R^2}\), where \( \mu \) is the reduced mass and R is the bond length).

• As \({{\rm{E}}_{\rm{J}}} = hc\overline {{\nu _J}} \)

\(\overline {{\nu _J}} = {{{{\rm{E}}_{\rm{J}}}} \over {hc}}\)

\( = {{{\hbar ^{\rm{2}}}} \over {{\rm{2Ihc}}}}{\rm{J}}\left( {{\rm{J + 1}}} \right)\)

\( = {{{h^2}} \over {8{\pi ^2}Ihc}}{\rm{J}}\left( {{\rm{J + 1}}} \right)\)

\( = {h \over {8{\pi ^2}Ic}}{\rm{J}}\left( {{\rm{J + 1}}} \right)\) let, \(B = {h \over {8{\pi ^2}Ic}}\) 

\( = B{\rm{J}}\left( {{\rm{J + 1}}} \right)\) where B is the rotational constant.

• The energy difference between any two consecutive levels J and J+1 is given by:

\({\overline \nu _{J \to J + 1}} = B\left( {{\rm{J + 2}}} \right)\left( {{\rm{J + 1}}} \right) - B{\rm{J}}\left( {{\rm{J + 1}}} \right)\)

\( = 2B\;{\rm{J}}\left( {{\rm{J + 1}}} \right)\)

• The spectrum depicted below is an absorption spectrum:

• Thus, these spectral lines are equispaced.

Explanation:

• Now, the energy separation of 12C16O rotational energy levels between J"=3 and J" = 9 is 24 cm-1.

E₃ = B x 3(4)= 12 B
E₉ = B x 9(10) = 90B

\(\Delta E \) = 90B - 12B= 78B 

Also \(\Delta E \)=24 cm-1.

24 cm-1= 78B 

B = 24/78 = 0.307

• Thus for 12C16O, 

​\({B_{^{12}{C^{16}}O}} = 2\;{\rm{c}}{{\rm{m}}^{ - {\rm{1}}}}\)

• As the reduced mass changes (\( \mu \)) the value of rotational constant (B) also changes as

\({B_{^{13}{C^{16}}O}} = {{{B_{^{12}{C^{16}}O}} \times {\mu _{^{12}{C^{16}}O}}} \over {{\mu _{^{13}{C^{16}}O}}}}\)​

\( = {{0.307 \times {{12 \times 16} \over {28}}} \over {{{13 \times 16} \over {29}}}}\)

\( = 0.307\times {{12 \times 29} \over {13 \times 29}}\)

\( 0.293\)

• ​So, the value of rotational constant B for 13C16O is closest to 0.298 cm^-1. 

Conclusion:

Thus, the rotational constant of 13C16O in cm-1 is closest to 0.298 cm-1. 

Concept:

• Raman is the scattering phenomenon resulting in decrease or increase of the wavelength due to inelastic interaction of photons with the molecules.
• the set of photons received after scattering are categorized into  stokes and anti-stokes lines. Stokes lines represents the photons of decreased energy/higher wavelength while anti-stokes are photons of higher energy/lower wavelength.
• the general selection rule for  the transition in Raman rotational spectroscopy is \(\Delta J=\pm2\)
• rotational Raman shift value is given by \(B(4J+6)\)
• vibrational line in Raman spectra appears at 

\(v = v_{ex}\pm \overline{v_e}(1-2x_e)\)

Explanation:       

First we have to find the stokes vibrational line for the given irradiation, which is given by:

 \(v_{stokes}= v_{ex}- \overline{v_e}(1-2x_e)\)

putting,

\(v_{ex}=20000\;cm^{-1},\)

\(\overline{v_e}=1600\;cm^{-1} \)

\(x_e=0.01\;\)

we get,

\(v_{stokes}= 20000cm^{-1}-1600cm^{-1}(1- 0.02)\)

we get,   \(v_{stokes}=18432\;cm^{-1}\)

In the spectra, rotational Raman shifts appear at the both side of vibrational peak. The first rotational Raman line appears at a shift of 6B and rest of the lines appear at shift of 10B from the vibrational peak (as shown in figure below)

Considering the information, stokes line will appear at:

\((1)\;v-10B =18432\;cm^{-1} -10\times2\; cm^{-1}=18412\; cm^{-1}\)

\((2)\;v-6B =18432\;cm^{-1}-6\times2\; cm^{-1}=18420\; cm^{-1}\)

(3)\(\; v= 18400 \; cm^{-1}\)

(4)\(\;v+6B= 18432\;cm^{-1}+ 6\times2\;cm^{-1}=18444\;cm^{-1}\)

(5)\(\;v+10B= 18432\;cm^{-1}+10 \times2\;cm^{-1}=18452\;cm^{-1}\)

Conclusion:

Therefore, the stokes lines for given molecules on irradiation at 20,000cm^-1 appears at 18412, 18420, 18432, 18444, 18452 cm^-1

Concept:

• The basic quantum mechanical model of rotational spectroscopy is "particle on a sphere" or "the rigid rotor model".
• Rotational spectroscopy deals with a system consisting of two atoms connected by a bond of fixed length (simply a diatomic molecule). It can rotate to describe the entire 3D space resulting in an infinite number of circular trajectories that eventually describe a sphere.
• The rotational energy level for a particular state (Jth state) is given by

\({{\rm{E}}_{\rm{J}}}{\rm{ = }}{{{\hbar ^{\rm{2}}}} \over {{\rm{2I}}}}{\rm{J}}\left( {{\rm{J + 1}}} \right)\), where J is a quantum number (J=0,1,2...) and I is the moment of inertia (\(I = \mu {R^2}\), where \( \mu \) is the reduced mass and R is the bond length).

• As \({{\rm{E}}_{\rm{J}}} = hc\overline {{\nu _J}} \)

\(\overline {{\nu _J}} = {{{{\rm{E}}_{\rm{J}}}} \over {hc}}\)

\( = {{{\hbar ^{\rm{2}}}} \over {{\rm{2Ihc}}}}{\rm{J}}\left( {{\rm{J + 1}}} \right)\)

\( = {{{h^2}} \over {8{\pi ^2}Ihc}}{\rm{J}}\left( {{\rm{J + 1}}} \right)\)

\( = {h \over {8{\pi ^2}Ic}}{\rm{J}}\left( {{\rm{J + 1}}} \right)\) let, \(B = {h \over {8{\pi ^2}Ic}}\) 

\( = B{\rm{J}}\left( {{\rm{J + 1}}} \right)\) where B is the rotational constant.

• The energy difference between any two consecutive levels J and J+1 is given by:

\({\overline \nu _{J \to J + 1}} = B\left( {{\rm{J + 2}}} \right)\left( {{\rm{J + 1}}} \right) - B{\rm{J}}\left( {{\rm{J + 1}}} \right)\)

\( = 2B\;{\rm{J}}\left( {{\rm{J + 1}}} \right)\)

• The spectrum depicted below is an absorption spectrum:

• Thus, these spectral lines are equispaced.

Explanation:

• The rotational absorption spectrum of H35CI shows the following lines

• Now, the energy separation of H35Cl rotational energy levels between J"=3 and J" = 4 is 83.03 cm1 and the energy separation of H35Cl rotational energy levels between J"=3 and J" = 4 is 104.10 cm-1
• Thus, the energy separation between rotation spectrum lines 3→4 and 4→5 J" = 9 is 

= 104.10 cm-1 - 83.03 cm1 

= 21.07 cm1 

• The energy separation between the two successive spectral lines is 2B.
• Hence, \(2B = {\rm{21.07 \;c}}{{\rm{m}}^{ - {\rm{1}}}}\)

\(B = {{21.07} \over {2}}\;{\rm{c}}{{\rm{m}}^{ - {\rm{1}}}}\)

= 10.53 cm^-1​

\( \approx 10\)

Conclusion:

Thus, the value of the rotational constant in units of cm-1 is estimated as 10

Concept:

• The basic quantum mechanical model of rotational spectroscopy is "particle on a sphere" or "the rigid rotor model".
• Rotational spectroscopy deals with a system consisting of two atoms connected by a bond of fixed length (simply a diatomic molecule). It can rotate to describe the entire 3D space resulting in an infinite number of circular trajectories which eventually describe a sphere.
• The rotational energy level for a particular state (Jth state) is given by

\({{\rm{E}}_{\rm{J}}}{\rm{ = }}{{{\hbar ^{\rm{2}}}} \over {{\rm{2I}}}}{\rm{J}}\left( {{\rm{J + 1}}} \right)\), where J is a quantum number (J=0,1,2...) and I is the moment of inertia (\(I = \mu {R^2}\), where \( \mu \) is the reduced mass and R is the bond length).

• As \({{\rm{E}}_{\rm{J}}} = hc\overline {{\nu _J}} \)

\(\overline {{\nu _J}} = {{{{\rm{E}}_{\rm{J}}}} \over {hc}}\)

\( = {{{\hbar ^{\rm{2}}}} \over {{\rm{2Ihc}}}}{\rm{J}}\left( {{\rm{J + 1}}} \right)\)

\( = {{{h^2}} \over {8{\pi ^2}Ihc}}{\rm{J}}\left( {{\rm{J + 1}}} \right)\)

\( = {h \over {8{\pi ^2}Ic}}{\rm{J}}\left( {{\rm{J + 1}}} \right)\) let, \(B = {h \over {8{\pi ^2}Ic}}\) 

\( = B{\rm{J}}\left( {{\rm{J + 1}}} \right)\) where B is the rotational constant.

• The energy difference between any two consecutive levels J and J+1 is given by:

\({\overline \nu _{J \to J + 1}} = B\left( {{\rm{J + 2}}} \right)\left( {{\rm{J + 1}}} \right) - B{\rm{J}}\left( {{\rm{J + 1}}} \right)\)

\( = 2B\;{\rm{J}}\left( {{\rm{J + 1}}} \right)\)

• The spectrum depicted below is an absorption spectrum:

• Thus, these spectral lines are equispaced.

Explanation:

• Now, the energy separation of 12C16O rotational energy levels between J"=3 and J" = 9 is 24 cm-1.
• From the rotation energy level J"=3 to J" = 9, there are 7 energy levels and 6 spectral lines. These are for

3→4, 4→5, 5→6, 6→7, 7→8, 8→9 transitions.

• The energy separation between the two successive spectral lines is 2B. Thus, the energy separation between rotation energy level J"=3 to J" = 9 is 12B (6 spectral lines = \(2B \times 6 = 12B\)).
• Hence, \(12B = {\rm{24 \;c}}{{\rm{m}}^{ - {\rm{1}}}}\)

\(B = {{24} \over {12}}\;{\rm{c}}{{\rm{m}}^{ - {\rm{1}}}}\)

\( = 2\;{\rm{c}}{{\rm{m}}^{ - {\rm{1}}}}\)

• Thus for 12C16O, 

​\({B_{^{12}{C^{16}}O}} = 2\;{\rm{c}}{{\rm{m}}^{ - {\rm{1}}}}\)

• As the reduced mass changes (\( \mu \)) the value of rotational constant (B) also changes as

\({B_{^{13}{C^{16}}O}} = {{{B_{^{12}{C^{16}}O}} \times {\mu _{^{12}{C^{16}}O}}} \over {{\mu _{^{13}{C^{16}}O}}}}\)​

\( = {{2 \times {{12 \times 16} \over {28}}} \over {{{13 \times 16} \over {29}}}}\)

\( = 2 \times {{12 \times 29} \over {13 \times 28}}\)

\( \approx 1.9\)

• ​So, the value of rotational constant B for 13C16O is closest to 1.9 cm-1. 

Conclusion:

• Thus, the rotational constant of 13C16O in cm-1 is closest to 1.9 cm-1. 

Concept:

• The basic quantum mechanical model of rotational spectroscopy is "particle on a sphere" or "the rigid rotor model".
• Rotational spectroscopy deals with a system consisting of two atoms connected by a bond of fixed length (simply a diatomic molecule). It can rotate to describe the entire 3D space resulting in an infinite number of circular trajectories that eventually describe a sphere.
• The rotational energy level for a particular state (Jth state) is given by

\({{\rm{E}}_{\rm{J}}}{\rm{ = }}{{{\hbar ^{\rm{2}}}} \over {{\rm{2I}}}}{\rm{J}}\left( {{\rm{J + 1}}} \right)\), where J is a quantum number (J=0,1,2...) and I is the moment of inertia (\(I = \mu {R^2}\), where \( \mu \) is the reduced mass and R is the bond length).

• As \({{\rm{E}}_{\rm{J}}} = hc\overline {{\nu _J}} \)

\(\overline {{\nu _J}} = {{{{\rm{E}}_{\rm{J}}}} \over {hc}}\)

\( = {{{\hbar ^{\rm{2}}}} \over {{\rm{2Ihc}}}}{\rm{J}}\left( {{\rm{J + 1}}} \right)\)

\( = {{{h^2}} \over {8{\pi ^2}Ihc}}{\rm{J}}\left( {{\rm{J + 1}}} \right)\)

\( = {h \over {8{\pi ^2}Ic}}{\rm{J}}\left( {{\rm{J + 1}}} \right)\) let, \(B = {h \over {8{\pi ^2}Ic}}\) 

\( = B{\rm{J}}\left( {{\rm{J + 1}}} \right)\) where B is the rotational constant.

• The energy difference between any two consecutive levels J and J+1 is given by:

\({\overline \nu _{J \to J + 1}} = B\left( {{\rm{J + 2}}} \right)\left( {{\rm{J + 1}}} \right) - B{\rm{J}}\left( {{\rm{J + 1}}} \right)\)

\( = 2B\;{\rm{J}}\left( {{\rm{J + 1}}} \right)\)

• The spectrum depicted below is an absorption spectrum:

• Thus, these spectral lines are equispaced.

Explanation:

• Now, the energy separation of 1H35Cl rotational energy levels between J"=2→3 and J" = 5→6 is 18 cm-1.
• From the rotational spectral line J"=2→3 and J" = 5→6, there are 4 spectral lines. These are for

2→3, 3→4, 4→5, and 5→6 transitions.

• The energy separation between the two successive spectral lines is 2B. Thus, the energy separation between rotation energy level J"=2→3 and J" = 5→6 is 6B (4 spectral lines =three energy gap = \(2B \times 3 = 6B\)).
• Hence, \(6B = {\rm{18 \;c}}{{\rm{m}}^{ - {\rm{1}}}}\)

\(B = {{18} \over {6}}\;{\rm{c}}{{\rm{m}}^{ - {\rm{1}}}}\)

\( =3\;{\rm{c}}{{\rm{m}}^{ - {\rm{1}}}}\)

• Thus for 1H35Cl, 

​\({B_{^{1}{H^{35}}Cl}} = 3\;{\rm{c}}{{\rm{m}}^{ - {\rm{1}}}}\)

• As the reduced mass changes (\( \mu \)) the value of rotational constant (B) also changes as

\({B_{^{1}{H^{37}}Cl}} = {{{B_{^{1}{H^{35}}Cl}} \times {\mu _{^{1}{H^{35}}Cl}}} \over {{\mu _{^{1}{H^{37}}Cl}}}}\)​

\( = {{3 \times {{1 \times 35} \over {36}}} \over {{{1 \times 37} \over {38}}}}\)

\( = 3\times {{35 \times 38} \over {36 \times 37}}\)

\( \approx 2.99\)

• ​So, the value of rotational constant B for 1H37Cl in cm-1 is closest to 2.99 cm-1. 

Conclusion:

• Thus, the rotational constant of 1H37Cl in cm-1 is closest to 2.99 cm-1. 

Concept:-

Rotational spectroscopy:

• Rotational spectroscopy is caused by the change of the rotational state of molecules.
• The working principle of spectroscopy basically depends on the absorption of an incident electromagnetic wave when it interacts with the molecule. 
• IR and microwave spectroscopy is based on the change in dipole moments or polarizabilities of the molecule. 
• microwave spectroscopy measures the rotational transitions to get an idea of bond length, bond dipole, and the geometrical structure of molecules. 
• IR spectroscopy helps in the determination of functional groups, the properties of bonds, and the possible structure of the molecule.
• IR spectrum can be recorded for any molecule giving permanent or temporary change in the dipole. But only the molecules having permanent dipole moment can give microwave spectra. 
• Molecules having a permanent dipole moment (polar molecules) only show rotational spectra and are called “microwave active”.
• Examples of microwave-active molecules are HCl, CO, etc.
• Homo nuclear molecules such as H₂, Cl₂, N₂, etc. do not show rotational spectra and are called “microwave inactive”.

Explanation:-

• Among C6H6, N2O, CH4, C2H₂, and NO, only N₂O and NO have permanent dipole moment and thus can give microwave spectra as well as IR spectra and therefore, are considered IR active.
• C6H6 does not have a permanent dipole. As a result, it is a microwave-inactive molecule.
• C2H2 doesn't possess any permanent dipole moment but it shows dipole change for asymmetric stretch and bending vibrations which makes it IR active but microwave inactive.

Conclusion:-

• Hence, N₂O (B) and NO (E) will be microwave-active molecules.

Concept:

The energy of rotational transitions for molecules in the gas phase is measured using microwave rotational spectroscopy with the help of interaction between the electric dipole movement of molecules and the magnetic field of microwave photons.

The line spacing is shown below.

Pure microwave Rotational spectra do not have stokes/anti-stokes lines, they come in the Raman spectrum.

In the Raman spectrum, two types of line spectra are present which are called stokes and anti-stokes lines.

• Stokes line is observed in the ground state excitation of molecules whereas the anti-stokes line is observed when the molecule returns to the ground state from its higher excited state.
  

Explanation:-

The adjacent lines are separated by 4 cm⁻¹

Thus, 

 2B=4cm-1

or, B= 2cm-1 

The Rotational Raman spectrum is given as,

The spacing between the Rayleigh line and the first Stokes line is 6B.

So, 6B = 6 × (2cm-1 )

or, 6B =12cm-1

Now, the frequency of the 1st stoke line can be calculated using the formula,

= \(\nu \:of Rayleigh \:line-6B\)

So, the frequency of the 1st stoke line will be

= (30,000-12) cm-1 (As frequency of rayleigh line = 30000 cm-1 and B = 4 cm-1)

= 29988 cm^-1 

Conclusion:-

Here the first Stokes line (in cm−1) appears at 29988 cm^-1.

Concept:

• The basic quantum mechanical model of rotational spectroscopy is "particle on a sphere" or "the rigid rotor model".
• Rotational spectroscopy deals with a system consisting of two atoms connected by a bond of fixed length (simply a diatomic molecule). It can rotate to describe the entire 3D space resulting in an infinite number of circular trajectories which eventually describe a sphere.
• The rotational energy level for a particular state (Jth state) is given by

\({{\rm{E}}_{\rm{J}}}{\rm{ = }}{{{\hbar ^{\rm{2}}}} \over {{\rm{2I}}}}{\rm{J}}\left( {{\rm{J + 1}}} \right)\), where J is a quantum number (J=0,1,2...) and I is the moment of inertia (\(I = \mu {R^2}\), where \( \mu \) is the reduced mass and R is the bond length).

• As \({{\rm{E}}_{\rm{J}}} = hc\overline {{\nu _J}} \)

\(\overline {{\nu _J}} = {{{{\rm{E}}_{\rm{J}}}} \over {hc}}\)

\( = {{{\hbar ^{\rm{2}}}} \over {{\rm{2Ihc}}}}{\rm{J}}\left( {{\rm{J + 1}}} \right)\)

\( = {{{h^2}} \over {8{\pi ^2}Ihc}}{\rm{J}}\left( {{\rm{J + 1}}} \right)\)

\( = {h \over {8{\pi ^2}Ic}}{\rm{J}}\left( {{\rm{J + 1}}} \right)\) let, \(B = {h \over {8{\pi ^2}Ic}}\) 

\( = B{\rm{J}}\left( {{\rm{J + 1}}} \right)\) where B is the rotational constant.

• The energy difference between any two consecutive levels J and J+1 is given by:

\({\overline \nu _{J \to J + 1}} = B\left( {{\rm{J + 2}}} \right)\left( {{\rm{J + 1}}} \right) - B{\rm{J}}\left( {{\rm{J + 1}}} \right)\)

\( = 2B\;{\rm{J}}\left( {{\rm{J + 1}}} \right)\)

• The spectrum depicted below is an absorption spectrum:

• Thus, these spectral lines are equispaced.

Explanation:

• Now, the energy separation of 12C16O rotational energy levels between J"=3 and J" = 9 is 24 cm-1.

E₃ = B x 3(4)= 12 B
E₉ = B x 9(10) = 90B

\(\Delta E \) = 90B - 12B= 78B 

Also \(\Delta E \)=24 cm-1.

24 cm-1= 78B 

B = 24/78 = 0.307

• Thus for 12C16O, 

​\({B_{^{12}{C^{16}}O}} = 2\;{\rm{c}}{{\rm{m}}^{ - {\rm{1}}}}\)

• As the reduced mass changes (\( \mu \)) the value of rotational constant (B) also changes as

\({B_{^{13}{C^{16}}O}} = {{{B_{^{12}{C^{16}}O}} \times {\mu _{^{12}{C^{16}}O}}} \over {{\mu _{^{13}{C^{16}}O}}}}\)​

\( = {{0.307 \times {{12 \times 16} \over {28}}} \over {{{13 \times 16} \over {29}}}}\)

\( = 0.307\times {{12 \times 29} \over {13 \times 29}}\)

\( 0.293\)

• ​So, the value of rotational constant B for 13C16O is closest to 0.298 cm^-1. 

Conclusion:

Thus, the rotational constant of 13C16O in cm-1 is closest to 0.298 cm-1. 

Concept:

• Raman is the scattering phenomenon resulting in decrease or increase of the wavelength due to inelastic interaction of photons with the molecules.
• the set of photons received after scattering are categorized into  stokes and anti-stokes lines. Stokes lines represents the photons of decreased energy/higher wavelength while anti-stokes are photons of higher energy/lower wavelength.
• the general selection rule for  the transition in Raman rotational spectroscopy is \(\Delta J=\pm2\)
• rotational Raman shift value is given by \(B(4J+6)\)
• vibrational line in Raman spectra appears at 

\(v = v_{ex}\pm \overline{v_e}(1-2x_e)\)

Explanation:       

First we have to find the stokes vibrational line for the given irradiation, which is given by:

 \(v_{stokes}= v_{ex}- \overline{v_e}(1-2x_e)\)

putting,

\(v_{ex}=20000\;cm^{-1},\)

\(\overline{v_e}=1600\;cm^{-1} \)

\(x_e=0.01\;\)

we get,

\(v_{stokes}= 20000cm^{-1}-1600cm^{-1}(1- 0.02)\)

we get,   \(v_{stokes}=18432\;cm^{-1}\)

In the spectra, rotational Raman shifts appear at the both side of vibrational peak. The first rotational Raman line appears at a shift of 6B and rest of the lines appear at shift of 10B from the vibrational peak (as shown in figure below)

Considering the information, stokes line will appear at:

\((1)\;v-10B =18432\;cm^{-1} -10\times2\; cm^{-1}=18412\; cm^{-1}\)

\((2)\;v-6B =18432\;cm^{-1}-6\times2\; cm^{-1}=18420\; cm^{-1}\)

(3)\(\; v= 18400 \; cm^{-1}\)

(4)\(\;v+6B= 18432\;cm^{-1}+ 6\times2\;cm^{-1}=18444\;cm^{-1}\)

(5)\(\;v+10B= 18432\;cm^{-1}+10 \times2\;cm^{-1}=18452\;cm^{-1}\)

Conclusion:

Therefore, the stokes lines for given molecules on irradiation at 20,000cm^-1 appears at 18412, 18420, 18432, 18444, 18452 cm^-1