All questions of Physical Spectroscopy for Chemistry Exam

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.

Spacing between rotation spectrum is 2B. ie 2B=12.8 and B=h/8πsqIc where I is moment of inertia and c is speed of light

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.

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).

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.

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.

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.

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.

C

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.

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.

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.

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.

Understanding the Q Band in Acetylene
The vibrational spectrum of acetylene (C2H2) showcases distinct bands corresponding to different molecular vibrations. The Q band specifically relates to the vibrational transitions occurring due to the C–C bond.
Why the Q Band is Observed in C–C Stretching Mode
- C–C Stretching Mode:
- Acetylene features a triple bond between the two carbon atoms, which allows for a significant amount of energy to be absorbed during vibrations.
- The Q band corresponds to transitions where the molecule transitions between vibrational states associated with this bond.
- Vibrational Modes of Acetylene:
- Acetylene has several vibrational modes, including:
- C–H symmetric stretching
- C–H antisymmetric stretching
- Bending modes
- However, the Q band is specifically linked to the energy changes in the C–C stretching mode.
Importance of C–C Stretching in Spectroscopy
- Energy Levels:
- The stretching of the C–C bond involves significant changes in the dipole moment, making it more detectable in infrared spectroscopy.
- Spectral Significance:
- The Q band provides valuable insights into the molecular structure and bonding, specifically demonstrating the strength and character of the C–C bond in acetylene.
Conclusion
In summary, the Q band in the vibrational spectrum of acetylene is observed in the C–C stretching mode due to its significant energy absorption characteristics and its role in defining the molecular structure. Understanding this helps in analyzing molecular vibrations and their implications in spectroscopy effectively.

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: 

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.

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.

Because we plot graph for n/No vs j value in rotational

Selection Rule for Translation Energy Levels in Raman Spectrum

The Raman effect is a phenomenon in which a molecule scatters incident light, leading to a change in the energy of the scattered photons. This effect can be used to obtain information about the vibrational and rotational energy levels of a molecule. However, it can also provide information on the translation energy levels of a molecule.

The selection rule for translation energy levels in the Raman spectrum is as follows:

- ΔJ = ±2

Here, J represents the rotational quantum number. This selection rule means that the Raman spectrum will only show transitions between translation energy levels that differ by a change in rotational quantum number of ±2.

Explanation of the Selection Rule

The selection rule for translation energy levels in the Raman spectrum is based on the fact that the Raman effect is a second-order scattering process. This means that the change in energy of the scattered photons is proportional to the square of the interaction between the incident light and the molecule.

In the case of translation energy levels, the interaction between the incident light and the molecule is due to the change in the polarizability of the molecule as it moves through the electric field of the incident light. This change in polarizability is related to the molecular geometry and the electronic structure of the molecule.

When a molecule undergoes a transition between translation energy levels, its geometry and electronic structure change, leading to a change in its polarizability. This change in polarizability leads to a change in the scattered light, which can be detected in the Raman spectrum.

However, for the Raman effect to be observable, the change in polarizability must be accompanied by a change in the rotational quantum number of the molecule. This is because the rotational quantum number is related to the orientation of the molecule in space, and a change in orientation leads to a change in the polarizability.

Therefore, the selection rule for translation energy levels in the Raman spectrum is ΔJ = ±2. This means that the Raman spectrum will only show transitions between translation energy levels that differ by a change in rotational quantum number of ±2.

Conclusion

In conclusion, the selection rule for translation energy levels in the Raman spectrum is ΔJ = ±2. This selection rule is based on the second-order scattering process of the Raman effect, which requires a change in the polarizability of the molecule accompanied by a change in the rotational quantum number. Understanding this selection rule is important for interpreting Raman spectra and obtaining information on the translation energy levels of a molecule.