Molecular Spectroscopy MCQ Quiz - Objective Question with Answer for Molecular Spectroscopy - Download Free PDF

GIVEN DATA:

• Vibrational frequency of H₂ = 4395 cm^-1
• Vibrational frequency of H³⁷Cl = 2988 cm^-1
• Assume: All molecules in ground vibrational state (only zero-point energy considered)

Calculate vibrational frequency of HD

$$\begin{gathered} \frac{{\nu }_{HD}}{{\nu }_{H_2}}=\sqrt{\frac{{\mu }_{H_2}}{{\mu }_{HD}}} \\ =\sqrt{\frac{1/2}{2/3}} \\ =\sqrt{\frac{3}{4}}{\nu }_{HD} \\ ={\nu }_{H_2}\times \sqrt{\frac{3}{4}} \\ =4395\times \sqrt{\frac{3}{4}}\approx 3806{\text{ cm}}^{-1} \end{gathered}$$

Calculate vibrational frequency of D³⁷Cl

$$\begin{gathered} \frac{{\nu }_{DCl}}{{\nu }_{HCl}}=\sqrt{\frac{{\mu }_{HCl}}{{\mu }_{DCl}}} \\ \Rightarrow {\nu }_{DCl}={\nu }_{HCl}\cdot \sqrt{\frac{39}{2\times 37}} \\ =2988\cdot \sqrt{\frac{39}{74}}\approx 2140.5{\text{ cm}}^{-1} \end{gathered}$$

Energy change ΔE

$$\begin{gathered} \Delta E=[\sum (ZPE{)}_{\text{products}}-\sum (ZPE{)}_{\text{reactants}}] \\ =\frac{1}{2}({\nu }_{H_2}+{\nu }_{DCl}-{\nu }_{HD}-{\nu }_{HCl}) \\ \Delta E=\frac{1}{2}(4395+2140.5-3806-2988) \\ =\frac{1}{2}(258.5)=129.25 \end{gathered}$$

On rounded) off -130 cm^-1 (negative because products have lower ZPE)

CONCEPT:

Microwave Spectroscopy and Rotational Constant

• For a rigid diatomic molecule, the spacing between successive rotational lines in microwave spectroscopy is given by:

  ΔE = 2B, where B is the rotational constant in cm^-1

• Rotational constant B is related to moment of inertia I by:

  B = ℏ² / (2Ihc)

• Moment of inertia I for a diatomic molecule = μr², where:
  • μ = reduced mass = m₁m₂ / (m₁ + m₂)
  • r = bond length (in meters)

EXPLANATION:

• Given: Spacing ΔE = 8 cm^-1 → So, B = 4 cm^-1
• Use the formula:

  B = ℏ² / (2μr²hc)

  Rewriting: r² = ℏ² / (2μhcB)

• Given:
  • ℏ² / (8π²c) = 2.8 × 10^-44 Js²·m^-1
  • amu = 1.667 × 10^-27 kg
  • masses: X = 5 amu, Y = 40 amu
• Reduced mass:
  • μ = (5 × 40) / (5 + 40) = 200 / 45 = 4.444 amu
  • μ = 4.444 × 1.667 × 10^-27 kg ≈ 7.41 × 10^-27 kg

• Using B = h / (8π²cI) and I = μr²

r = √[ h / (8π²cμB) ]

= √[ 2.8 × 10^-44 / (μ × B) ]

• μ = 7.41 × 10^-27, B = 4 cm⁻¹
• r² = 2.8 × 10^-44 / (7.41 × 10^-27 × 4) ≈ 9.45 × 10^-19 m²
• r ≈ √(9.45 × 10^-19) ≈ 9.72 × 10^-10 m = 0.972 Å

Therefore, the bond length is closest to 0.974 Å.

CONCEPT:

3D Isotropic Harmonic Oscillator

• The energy levels of a 3D isotropic harmonic oscillator are given by:

  E = (n_x + n_y + n_z + 3/2) ℏω

• Here, n_x, n_y, n_z are non-negative integers (0, 1, 2,...).
• The total quantum number N = n_x + n_y + n_z.
• The degeneracy of each energy level is equal to the number of non-negative integer solutions to:

  n_x + n_y + n_z = N

EXPLANATION:

• Given energy = (9/2) ℏω
• Compare with the energy expression:

  (N + 3/2) ℏω = 9/2 ℏω → N = 3

• We need to count the number of integer solutions to:

  n_x + n_y + n_z = 3, where each n ≥ 0

• The number of non-negative integer solutions to this equation in 3 variables is:

  C(3 + 3 - 1, 2) = C(5, 2) = 10

• So the degeneracy of N = 3 level is 10.

Therefore, the correct degeneracy is 10.

Note: The offical answer accoring to CSIR is option 3

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.

Concept:

Rotational Spectroscopy

• Rotational spectroscopy deals with the absorption of electromagnetic radiation by molecules as they undergo rotational transitions.
• These transitions primarily occur in the microwave region of the spectrum.
• The transition can only occur if the molecule has a permanent dipole moment, as the absorption of electromagnetic radiation requires a change in dipole moment.
• The selection rule for rotational transitions is ΔJ = ±1, where J is the rotational quantum number.
• At room temperature, typically only one transition is observed because the population of molecules in the higher rotational energy levels is very low, according to the Boltzmann distribution.

Explanation:

• Statement A: "It lies in the microwave region."
  • Correct, as rotational transitions primarily absorb energy in the microwave region (1 GHz to 1000 GHz).
• Statement B: "The molecules must have permanent dipole moment."
  • Correct, because rotational transitions require a change in dipole moment, which is why molecules with no permanent dipole moment do not exhibit rotational spectra.
• Statement C: "The selection rule is ΔJ = ±2."
  • Incorrect. The correct selection rule for rotational spectroscopy is ΔJ = ±1, not ΔJ = ±2.
• Statement D: "Only one transition is observed at room temperature."
  • Incorrect. At room temperature, many rotational levels of a molecule are populated according to the Boltzmann distribution. Therefore, a series of transitions between these levels are possible, leading to a spectrum of lines, not just one.

Analysis of Answer Choices:

• Correct answer: The correct answer is "A and B only".

Concept: 

In a pure rotational spectrum, For a linear molecule

$\begin{array}{c}\mathrm{\Delta }J=J_2-J_1\\ =\pm 1\end{array}$

The wavenumber can be calculated as, 

$\begin{array}{c}{\nu }_{J_2-J_1}=E\left(J_2\right)-E\left(J_1\right)\\ =2B\left(J_1+1\right)\end{array}$

B= rotational constant, and it can be calculated using the equation,

$B=\frac{h}{8{\mathrm{\Pi }}^{2}Ic}$

h=Plancks constant. 

c=Speed of light.

I=Moment of inertia.

Explanation:

Given the pure rotational line in the spectrum of NO molecule, where thetransition is from J = 1 to J = 2 .

Moment of inertia of NO = 1.6427 × 10-46 kg m²

h = 6.626 × 10-34 J s

c = 3 × 108 m/s

Here, to convert to true energy units,cm^-1 needs to be multiplied by the speed of light and Planck's constant. so it becomes,

h = 6.626 × 10-27 erg sec

c = 3 × 10¹⁰ m/s

The frequency in cm-1 for pure rotational line in the spectrum of NO is given by, 

 $\begin{array}{rl}\nu & =2\left(\frac{h}{8{\mathrm{\Pi }}^{2}Ic}\right)\left(J_1+1\right)\\ & =2\left(\frac{6.626\times {10}^{-27}}{8\times {\left(3.14\right)}^2\times \left(1.6247\times {10}^{-46}\right)\times \left(3\times {10}^{10}\right)}\right)\left(1+1\right)\\ & =6.9\times {10}^{7}c{m}^{-1}\end{array}$

The frequency (in cm^-1) for pure rotational line in the spectrum of NO molecule due to change in the quantum number from J = 1 to J = 2 is $6.9\times {10}^{7}c{m}^{-1}$

Concept:

• The Raman spectroscopy is observed only when there occurs a change in the polarizability of the molecule due to its vibrational or rotational oscillation.
• In the presence of a static electric field, a molecule is polarized and acquired an induced dipole moment.
• In the presence of radiation, the molecule also acquires an induced dipole moment due to the electrical component of the incident light.
• As the electrical component of the radiation varies, the induced dipole also varies and emits radiation of its own frequency ν. This frequency gives rise to the Rayleigh line in the light scattered by the molecule.
• Also, there is a periodic change in the polarizability of the molecule. As the polarizability changes, the induced dipole also changes periodically.
• When these periodic changes are superimposed on the periodic oscillation due to the electrical component of light, the induced dipole moment oscillates with frequency ν , ν + v^' and v - v^'.
• Thus, the molecule emits a frequency of ν , ν + v' and v - v'. These radiations appear as Raman lines in the light scattered by the substance.
• The Raman lines are in general of weak intensities. When a molecule after collision with the incident light occupies a higher vibrational energy level, the molecule gains energy equivalent to the difference in energy levels, this energy is provided by the photon and thus its energy is decreased, and thus a Raman line on the low-frequency side is noticed.
• When the molecule occupies a lower energy level than before the collision, a Raman line of higher frequency is observed.
• The selection rule for Raman spectra is that the difference in rotational energy levels is Δ J =0 and +/- 2.
• The Raman lines appear at wavenumbers ν' = ν^'₀ + B(4J+6) and ν' = ν'0 + B(4J+6), where ν'0 is the fundamental Rayleigh lines, the wavenumber lower than ν'0 are called Stokes lines and higher than it are called anti-Stokes lines.

Calculation:

Given:

• The wavelength ' ν 'of the Rayleigh line = 532 nm = 532 × 10^-9m = 532 × 10-7cm.
• The fundamental wavenumber of vibration is given by: 560 cm−1
• When we go towards the left, we get the Stokes line and when we go right, we get the Anti Stokes line:

• The wavenumber of Rayleigh line is =

${\nu }^{\prime }=\frac{1}{\nu }$

${\nu }^{\prime }=\frac{{10}^7}{532}=18796.99c{m}^{-1}$

• So, Stokes line will be present at wave number = wavenumber of Rayleigh - Wavenumber of Fundamental vibration 

= 18797 - 560 = 18237 cm^-1

• Hence, the Stokes line corresponding to this vibration will be observed at 18237 cm-1.

Concept:

• The working principle of spectroscopy basically depends on the absorption of incident electromagnetic wave when it interacts with the molecule. 
• The IR and microwave spectroscopy is based on the change in dipole moments or polarizabilities of molecule. 
• microwave spectroscopy measures the rotational transitions to get the idea of bond length, bond dipole and the geometrical structure of molecules. 
• IR spectroscopy helps in determination of functional groups, the properties of bond and 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. 

Explanation:

Among N₂, C₂H₂, HCl and H₂O, only HCl and H₂O have permanent dipole moment and thus can give microwave spectra as well as IR spectra and therefore, are considered IR active.

C₂H₂ doesn't posses any permanent dipole moment but it shows dipole change for asymmetric stretch and bending vibrations which makes it IR active but microwave inactive.

N₂ has neither a permanent dipole nor a dipole change for bond vibration. As a result, it is both IR and microwave inactive.

Hence, only C₂H₂ among all the molecules is IR active but microwave inactive.

Conclusion:

C₂H₂ doesn't posses any permanent dipole and shows dipole change due to asymmetric and bending vibrations and thus it is microwave inactive but IR active.

Concept:

• harmonic oscillator shows periodic motion and when displaced from its equilibrium position it experiences some restoring force, F in opposite direction to the motion which is given as:

 $F=-Kx$ (x is the displacement)

• Classically, frequency of oscillations is given by :

 $v=\frac{1}{2\mathrm{\Pi }}\sqrt{\frac{k}{m}}$

           where, K is force constant and u is mass of oscillator

• the quantum harmonic oscillators, however, have discrete allowed energy levels corresponding to different wave functions. The general formula for energy of these allowed levels is:

  $\mathrm{\Delta }E=(v+\frac{1}{2})\bar{w}$        where, v= 0,1,2,3,4,.......

Explanation:

(i) incorrect

we know restoring force (F) , the force which acts to bring the body back to the equilibrium position, is related to potential energy (V) as,

$F=-\frac{\mathrm{d}V}{\mathrm{d}q}$

 $dV=-Fdq$

also,

 $F=-kq$ ( here, q is the displacement )

this gives,

 $dV=kqdq$

 $\int dV=\int kqdq$

 $V=k\frac{q^2}{2}$

So, clearly, potential energy is proportional to the square of displacement. so the given statement is incorrect.

(ii) correct

Zero point energy is the lowest possible energy, a particle in a quantum system can have. For the harmonic oscillator, zero point energy is given by:

 $\mathrm{\Delta }E=\frac{\bar{w}}{2}$

$for\bar{w}=2990c{m}^{-1},\mathrm{\Delta }E=1495c{m}^{-1}$

The value is as per the value in statement, so this statement is correct.

(iii) correct

Vibration frequency of  an oscillator is related to reduced mass(u) and force constant(k) as follows:

 $v=\frac{1}{2\mathrm{\Pi }}\sqrt{\frac{k}{u}}$

Accordingly, vibrational frequency is inversely related to the reduced mass and further the mass of atom. So, the bond of molecule with heavier isotope will vibrate at lower frequency. so the order of vibrational frequency in given statement is correct.

(iv) incorrect

Fundamental vibrational transition occurs for v=0 to v=1. If the anharmonicity is ignored, the energy required for fundamental transition can be written as: 

 $\mathrm{\Delta }E=(2+\frac{1}{2})\bar{w}-(0+\frac{1}{2})\bar{w}$

Overtone is received when the transition occurs from v=0 to v=2

 $\mathrm{\Delta }E=(2+\frac{1}{2})\bar{w}-(0+\frac{1}{2})\bar{w}$

$\mathrm{\Delta }E=2\bar{w}$

  $\mathrm{\Delta }E=2\times 1880c{m}^{-1}=3760cm-1$

the value for overtone in the statement is incorrect

Conclusions:

Zero point energy of  an oscillator is half of its vibrational frequency, which is further depends inversely on the mass of atom/ molecule. Thus out of given statements , only statement (i) and (ii) is correct.

Concept:

In the rotational Raman spectrum of a diatomic molecule, three main types of lines are observed: Rayleigh, Stokes, and anti-Stokes lines. These lines arise due to rotational transitions of the molecule when it interacts with light. The rotational constant B is crucial in determining the positions of these lines.

• Rayleigh Line: The Rayleigh line corresponds to the scattering of light without any change in energy (elastic scattering). This line represents the incident photon energy and occurs at the same wavelength as the incident light.

• Stokes Lines: In the Stokes process, the molecule absorbs energy from the incident photon and transitions to a higher rotational energy level. The energy gap between the Rayleigh line and the first Stokes line is 6B .

• Anti-Stokes Lines: In the anti-Stokes process, the molecule loses energy by transitioning from a higher rotational state to a lower state. The energy gap between the Rayleigh line and the first anti-Stokes line is 6B .

• Energy Gap between First Stokes and First Anti-Stokes Lines: The total energy gap between the first Stokes and first anti-Stokes lines is the sum of the individual transitions, which results in 12B .

Explanation: 

• The energy gap between the Rayleigh line and the first Stokes or anti-Stokes line is calculated as:

  $\mathrm{\Delta }E=6B$ for both the Stokes and anti-Stokes lines.

•  

  

   

  The energy gap between the first Stokes and first anti-Stokes lines is calculated as:

  $\mathrm{\Delta }E=12B$ , where B is the rotational constant.

• The gap between the Rayleigh line and both the first Stokes and first anti-Stokes lines is 6B , representing the transition between the incident photon energy and the rotational transitions in the molecule.

Conclusion:

The energy gap between the first Stokes and first anti-Stokes lines in the rotational Raman spectrum of a diatomic molecule is 12B.

Concept:

The vibrational quantum number, often denoted as "ν", is a quantum mechanical parameter used to describe the vibrational energy levels of molecules. It is a fundamental concept in the field of molecular spectroscopy and quantum mechanics and is particularly important in the study of molecular vibrations using techniques like infrared (IR) spectroscopy and Raman spectroscopy.

Explanation:

We know that, 

$E=(n+\frac{1}{2}){\omega }_e-(n+\frac{1}{2}{)}^2{\omega }_e{X}_e$

Comparing, G(n) with E

$[3000(n+\frac{1}{2}){\omega }_e-50(n+\frac{1}{2}{)}^2{\omega }_e{X}_e]$ $=[(n+\frac{1}{2}){\omega }_e-(n+\frac{1}{2}{)}^2{\omega }_e{X}_e]$

∴ ω_e =3000

ω_eX_e = 50

X_e = 1/60

Now, 

Putting the value of X_e in n_max i.e.

$n_{max}=\frac{1}{2X_e}-1$

⇒n_max = 29

Conclusion:

Therefore, the value of n_max is 29

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,

= $\nu ofRayleighline-6B$

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

${\mathrm{E}}_{\mathrm{J}}=\frac{{\hslash }^2}{2\mathrm{I}}\mathrm{J}\left(\mathrm{J}+1\right)$, 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 ${\mathrm{E}}_{\mathrm{J}}=hc\overline{{\nu }_J}$

$\overline{{\nu }_J}=\frac{{\mathrm{E}}_{\mathrm{J}}}{hc}$

$=\frac{{\hslash }^2}{2\mathrm{I}\mathrm{h}\mathrm{c}}\mathrm{J}\left(\mathrm{J}+1\right)$

$=\frac{h^2}{8{\pi }^{2}Ihc}\mathrm{J}\left(\mathrm{J}+1\right)$

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

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

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

${\bar{\nu }}_{J\to J+1}=B\left(\mathrm{J}+2\right)\left(\mathrm{J}+1\right)-B\mathrm{J}\left(\mathrm{J}+1\right)$

$=2B\mathrm{J}\left(\mathrm{J}+1\right)$

• 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

$\mathrm{\Delta }E$ = 90B - 12B= 78B 

Also $\mathrm{\Delta }E$=24 cm-1.

24 cm-1= 78B 

B = 24/78 = 0.307

• Thus for 12C16O, 

​$B_{{}^{12}{C}^{16}O}=2\mathrm{c}{\mathrm{m}}^{-1}$

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

$B_{{}^{13}{C}^{16}O}=\frac{B_{{}^{12}{C}^{16}O}\times {\mu }_{{}^{12}{C}^{16}O}}{{\mu }_{{}^{13}{C}^{16}O}}$​

$=\frac{0.307\times \frac{12\times 16}{28}}{\frac{13\times 16}{29}}$

$=0.307\times \frac{12\times 29}{13\times 29}$

$0.293$

• ​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 $\mathrm{\Delta }J=\pm 2$
• 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,

$v_{ex}=20000c{m}^{-1},$

$\overline{v_e}=1600c{m}^{-1}$

$x_e=0.01$

we get,

$v_{stokes}=20000c{m}^{-1}-1600c{m}^{-1}(1-0.02)$

we get,   $v_{stokes}=18432c{m}^{-1}$

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:

$(1)v-10B=18432c{m}^{-1}-10\times 2c{m}^{-1}=18412c{m}^{-1}$

$(2)v-6B=18432c{m}^{-1}-6\times 2c{m}^{-1}=18420c{m}^{-1}$

(3)$v=18400c{m}^{-1}$

(4)$v+6B=18432c{m}^{-1}+6\times 2c{m}^{-1}=18444c{m}^{-1}$

(5)$v+10B=18432c{m}^{-1}+10\times 2c{m}^{-1}=18452c{m}^{-1}$

Conclusion:

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