Molecular Spectroscopy MCQ Quiz in বাংলা - Objective Question with Answer for Molecular Spectroscopy - বিনামূল্যে ডাউনলোড করুন [PDF]

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:-

• EPR spectroscopy is used to study a paramagnetic species with one or more unpaired electrons. While diamagnetic species are always EPR silent.
• Paramagnetic species such as free radicals, diradicals, and metal complexes containing paramagnetic metal centers are always EPR active and exhibit an EPR spectrum.
• For a paramagnetic metal ion with one unpaired electron, the total spin quantum number is

• S=. There are two possible spin states with Ms= and Ms=

• By applying a magnetic field Bo, the interaction between the unpaired electrons and the magnetic field leads to the splitting of the energy levels.
• Therefore, by supplying appropriate microwave radiation to the sample, electron spin transitions between the two energy states occur. The system is then in resonance, and the recording of these transitions represents the EPR spectrum.

Explanation:-

• The number of lines observed in the EPR spectrum is given by

2nI+1,

where n is the number of equivalent nuclei with the spin quantum number I.

• For [Co3(CO)9 Se]-, the number of lines observed in the EPR spectrum for the Co nucleus (l = 7/2 for Co nucleus) for n =3 is,

​= 2 × 3 × 7/2  + 1

= 22 lines.

Conclusion:

Therefore, there are 22 hyperfine lines in the EPR spectrum of [Co₃(CO)₉Se]^-.

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:

Inductively Coupled Plasma Atomic Emission Spectroscopy (ICP-AES):

• Inductively Coupled Plasma Atomic Emission Spectroscopy (ICP-AES) is a type of spectroscopy used for the detection of chemical elements. It is a type of emission spectroscopy that uses inductively coupled plasma to produce excited atoms and ions that emit electromagnetic radiation at a particular wavelength.

Explanation:

• ICP is used for the detection of metals in liquid samples. It is not suitable for the detection of noble gas, halogens, and lighter elements such as H, C, N, and O. Elemental S can be determined using a vacuum monochromator.
• The plasma is sustained and stabilized by inductive coupling from the cooled induction coils at megahertz frequencies, where the temperature ranges from 6000K to 10000K.

Conclusion:

The correct statement for 'Inductively Coupled Plasma Atomic Emission Spectroscopy' is Induction coil stabilizes plasma.

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

• We live in a three-dimensional world and the degree of freedom for a molecule having 'N' number of atoms is 3N.
• For a nonlinear molecule, the three degrees of motion can be assigned to translation: Tx, Ty, Tz.
• Three to Rotational degrees of freedom, Rx, Ry, Rz.
• The rest are called the normal modes which is the displacement of the atoms from their mean position, without the change of center of gravity.
• Hence, the normal modes are given by: 3N - 6.

Explanation:

• Water has three atoms, and thus its normal modes of vibration are:

3N - 6 = 3× 3 - 6 = 3.

• The three modes of vibration are given as:

Hence, the normal mode of vibration of water H2O is options 1,2 and 3.​

Additional Information

• A linear molecule has 3N - 5 modes of vibration, where N is the number of atoms in the molecule. 

The correct answer is 25.96 kJ/mol

Explanation:-

Approximate Zero-Point Energy:

The approximate zero-point energy is given by:

Eapprox = (1/2) × h × c × ν~ × NA

• h = 6.626 × 10-34 J·s
• c = 3 × 108 m/s
• ν~ = 4395 cm-1 = 4395 × 102 m-1
• NA = 6.022 × 1023 mol-1

Substituting the values:

Eapprox = (1/2) × (6.626 × 10-34) × (3 × 108) × (4395 × 102) × (6.022 × 1023)

Eapprox = 26.31 kJ/mol

Exact Zero-Point Energy:

The exact zero-point energy includes a correction for anharmonicity:

Eexact = (1/2) × h × c × ν~ × (1 − (ν~x / ν~)) × NA

• ν~x = 118 cm-1

Substituting:

Eexact = 26.31 × (1 − (118 / 4395))

Eexact = 26.31 × (1 − 0.0269)

Eexact = 26.31 × 0.9731

Eexact = 25.96 kJ/mol

Conclusion:-

 25.96 kJ/mol

Concepts:

• Vibrational Frequency: The vibrational frequency (ν) of a diatomic molecule is given by the formula:
  • \(ν = \frac{1}{2\pi} √{\frac{k}{μ}} \) where ( k ) is the force constant and μ is the reduced mass of the system.
• Reduced Mass (( μ )): The reduced mass is defined as:
  • \(μ = \frac{m_1 m_2}{m_1 + m_2}\)where ( m₁ ) and ( m₂ ) are the masses of the two atoms in the molecule. Changes in ( μ ) directly affect the vibrational frequency.

Explanation:

1. The initial vibrational frequency is expressed as:
   • \( ν_1 = \frac{1}{2\pi} √{\frac{k}{μ_1}} \) where ( μ1 ) is the initial reduced mass.
2. When the reduced mass is halved \(μ_2 = \frac{μ_1}{2} \), the new vibrational frequency becomes:
   • \( ν_2 = \frac{1}{2\pi} √{\frac{k}{μ_2}}\)
3. Substitute ( μ2 = fracμ12 ) into the formula:
   • ​​\(ν_2 = \frac{1}{2\pi} √{\frac{k}{μ_1/2}} \)
   • Simplify: \(ν_2 = \frac{1}{2\pi} √{\frac{2k}{μ_1}}\)
4. Compare ( ν₂ ) to the original frequency ( ν₁ ):
   • \( ν_2 = √{2} ν_1\)This shows that the new frequency is √2 times the original frequency.

Conclusion:

When the reduced mass of a diatomic molecule is halved without changing the force constant, the vibrational frequency increases by a factor of ( √2 ). The correct answer is Option 1.

Concepts:

• Rotational Raman Spectroscopy: Raman spectroscopy observes transitions that occur due to inelastic scattering of light by a molecule. For rotational Raman spectra, the rotational energy levels of a molecule are involved in these transitions.
• Selection Rules: The selection rules for the rotational Raman spectra of diatomic molecules are governed by changes in the rotational quantum number (J) due to interaction with the incident light. The allowed transitions must satisfy the condition:
  • Δ J = ±2
• Reason for Δ J = ± 2: This arises because Raman transitions involve changes in the polarizability of the molecule, which allows for the excitation or relaxation of rotational levels by two units (± 2).

Explanation:

1. In rotational Raman spectroscopy, the polarizability of the molecule must change during the transition. This is the primary criterion for Raman activity.
2. The rotational energy levels of a diatomic molecule are quantized and given by the expression:
   • \(E_J = \frac{ℏ^2}{2I} J(J+1) \) where ( J ) is the rotational quantum number, ( ℏ ) is the reduced Planck's constant, and ( I ) is the moment of inertia of the molecule.
3. The selection rule ( Δ J = ± 2 ) means that the molecule can only transition between rotational levels that differ by two units. This is because the Raman process involves a two-photon interaction (absorption and scattering).
4. As a result, the observed Raman spectrum consists of lines corresponding to these allowed transitions:
   • \(J \rightarrow J+2 \) or \(J \rightarrow J-2 \), leading to Stokes and anti-Stokes lines.

Conclusion:

The correct answer is Option 2, where the selection rule for the rotational Raman spectrum of a diatomic molecule is ( Δ J = ± 2 ).