Lec 3: Molecular spectroscopy

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ادلكتور

براهمي مفتاح عبدهللا ا

جامعة املنيا -لكية العلوم-قسم الكميياء

𝜐 ≡ 𝐹 𝐽 = 𝐵𝐽 𝐽 + 1˜ ˜˜

Rotational energy levels get more widely

space with increasing J

Obs:

1

3

2

4

5

6

0

7

J

2B

6B

12B

20B

30B

42B

56B

0

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Energy levels of a rigid diatomic rotor

Since, The allowed rotational energies are

given by;

➢ The wave numbers of the different

rotational levels will be; 0, 2B, 6B,

12B, 20B, 30B (cm-1),… and so on

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1

3

2

4

5

6

0

7

J

2B

6B

12B

20B

30B

42B

56B

2B

4B

6B

8B

10B

12B

14B

𝜐 ≡ 𝐹 𝐽 = 𝐵𝐽 𝐽 + 1˜ ˜˜

B(J1+1)(J1+2)

B(J1+1) J1

J2 = J1+1

J1

∆ 𝜐 =∆ ෨𝐹 𝐽 = 2𝐵(𝐽1+ 1)

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B (J2+1) J2J2

J2-1

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∆ 𝜐 =∆ ෨𝐹 𝐽 = 2𝐵𝐽2˜

B(J2-1)J2˜

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Absorption

Emission

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Rotational Transitions in Rigid Diatomic Molecules

As ෨𝐵 decreases with increasing I ⇒ large molecules have closely spaced energy levels.

2 ෨𝐵 𝐽1+ 1 𝑶𝑹 2 ෨𝐵𝐽2

1. For the transition J = 0 → J = 1

𝜐𝐽=1 − 𝜐𝐽=0 =△ 𝜐 = 2𝐵 − 0 = 2𝐵˜ ˜ ˜ ˜ ˜

2. For the transition J = 1 → J = 2

𝜐𝐽=2 − 𝜐𝐽=1 = △ 𝜐 = 6𝐵 − 2𝐵 = 4𝐵˜ ˜ ˜ ˜ ˜˜

3. For the transition J = 2 → J = 3

𝜐𝐽=3 − 𝜐𝐽=2 = △ 𝜐 = 12𝐵 − 6𝐵 = 6𝐵˜ ˜ ˜ ˜ ˜˜

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= Position of the first line in the spectrum

= Position of the second line in the spectrum

= Position of the third line in the spectrum

✓ Hence, the wave number of the lines observed in the rotational spectrum will be; 2B, 4B, 6B, 8B

(cm-1), …… and so on.

✓ And the various lines in the rotational spectra will be equally spaced (separation between lines =

2B)..

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In general the energy difference between two rotational levels

Rotational Transitions in Rigid Diatomic Molecules

Separation between adjacent lines = 2෩𝑩 So, ෩𝑩 can be

obtained from the spacing between rotational lines.

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Question: Explain that the rotational spectra predicted by rigid rotor model

consists of a series of equally spaced spectral lines in the microwave region?

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Ans.

Example: If the bond length of HCl is 0.127 nm, calculate the spacing

between the rotational spectral lines in cm-1 ( 1 amu = 1.66 × 10-27 Kg)

Selection rules for the diatomic rotor

1. Gross selection rule

➢ For a molecule to exhibit a pure rotational spectrum it must

posses a permanent dipole moment. (otherwise the photon

has no means of interacting “nothing to grab hold of”)

➢ a polar rotor appears to have an oscillating electric dipole.

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8

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+1 absorption.

-1 emission.

2. Specific selection rule

➢ Application of the Schrödinger wave equation shows

only transitions in which J changes by one unit are

allowed (△ 𝑱 = ±𝟏) , all other transitions being

spectroscopically forbidden (△ 𝑱 = ±𝟐,±𝟑, 𝒆𝒕𝒄)

➢ To explain this: light behaves as a particle: photons

have a spin of 1, i.e., an angular momentum of one

unit. When a photon is absorbed, the angular

momentum of the combined system is conserved. If

the molecule is rotating in the same sense as the spin

of the incoming photon, then J increases by 1.

△ 𝑱 = ±𝟏

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From pure rotational spectra of molecules we can obtain:

1. bond lengths

2. Atomic (isotopic) masses

3. isotopic abundances

4. Temperature

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What information can we obtain from a

rotational spectrum

Bond length

➢ Line separation in the rotational spectrum of HCl is 21.2 cm-1

Therefore ෨𝐵 = 10.6 cm-1

➢ IHCl can be found from ෨𝐵 =ℎ

8𝜋2. 𝐼. 𝑐

➢ Next rHCl can be obtained using I =𝑚𝐻𝑚𝐶𝑙

𝑚𝐻 +𝑚𝐶𝑙𝑟𝐻𝐶𝑙2

From microwave spectroscopy, bond

lengths can be determined with a

correspondingly high precision

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Example:

Ans.

𝜐𝐽=0⟶𝐽=1 = 3.8425 𝑐𝑚−1 = 2 ෨𝐵

෩B = 1.92118 cm−1

𝐼 =ℎ

8𝜋2 ෨𝐵𝑐=

6.626 × 10−34

8𝜋2 × 2.99793 × 1010 × ෨𝐵

𝐼 = 14.5695 × 10−47 𝐾𝑔.𝑚2

𝜇 =19.92168 × 26.56136 × 10−54

46.48303 × 10−27= 11.38365 × 10−27 𝐾𝑔

𝑅2 =𝐼𝐶𝑂𝜇

= 1.2799 × 10−20 𝑚2 ⟹ 𝑅𝐶𝑂 = 0.113 𝑛𝑚 = 1.131 Å

The first rotational line in the rotational spectrum of CO is observed at 3.84235 cm-1.

Calculate the rotational constant ( ෨𝐵) and bond length of CO. The relative atomic

weight C =12.00 and O = 15.9994, the absolute mass of H = 1.67343x10-27 kg.

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෨𝐵 =ℎ

8𝜋2𝑐 𝜇𝑟𝐻𝐶𝑙2

𝑟 =ℎ

8𝜋2𝜇𝑐 ෨𝐵=

6.6260755 × 10−34

8𝜋2𝑐1.0078250 × 34.96885271.0078250 + 34.9688527

× (1.66054 × 10−27)(10.59342)

r = 1.274553 × 10−10 𝑚

Example: From the rotational microwave spectrum of 1H35Cl, we find that ෨𝐵 =

10.59342 cm-1. Given that the masses of 1H and 35Cl are 1.0078250 and

34.9688527 amu, respectively, determine the bond length of the 1H35Cl molecule.

Ans.

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From spectra we can obtain:

✓ Bond length

➢ Atomic weights

➢ Isotopic Abundances

Effect of Isotope Substitution

2 ෨𝐵 4 ෨𝐵 6 ෨𝐵 8 ෨𝐵 10 ෨𝐵 12 ෨𝐵

cm-1 spectrum

J = 6

5

4

3

210

12CO13CO

Energy

levels

1. isotopic masses

➢ There is no appreciable change in internuclear distance on isotopic substitution. There is,

however, a change in the total mass and hence in the moment of inertia and ෨𝐵 value for the

molecule.

➢ From 12C16O→ 13C16O, mass increases, ෨𝐵 decreases (1

𝐼), so energy levels lower.

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Effect of Isotope Substitution

✓ Microwave studies can also give directly an estimate of the abundance of isotopes by

comparison of absorption intensities (absorption relative intensities).

Comparison of rotational energy levels of 12CO and 13CO

What is isotopic mass of 13C ?

෨𝐵(12𝐶𝑂) = 1.921175 𝑐𝑚−1

෨𝐵(13𝐶𝑂) = 1.836685 𝑐𝑚−1

𝐵 ∝1

𝐼∝

1

𝜇

𝜇(13𝐶𝑂)

𝜇 (12𝐶𝑂)=

1.921175

1.836685= 1.04600 ⇒ 1.04600 =

13𝐶 × 15.999413𝐶 + 15.9994

×12 + 15.9994

12 × 15.9994

⇒ 13𝐶 = 13.0006 𝑎𝑚𝑢

Example:

for 12CO J=0→ J=1 at 3.84235 cm-1

for 13CO J=0→ J=1 at 3.67337 cm-1

Given : 12C = 12.0000 amu & O = 15.9994 amu

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2. Isotopic abundances

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Effect of Isotope Substitution

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Example: The rotational constants for the four most common isotops of

hydrogen bromide (ignoring tritium) are listed below. Calculate the bond

distance r in each of these molecules neglecting centrifugal distortion and

comment on your results.

1H79Br 1H81Br 2H79Br 2H81Br

෨𝐵/𝑐𝑚−1 8.351 8.347 4.248 4.245

Ref. Chemistry: Quantum Mechanics and Spectroscopy II (J. E. Parker)

Answer

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To convert from atomic mass in g mol−1 or “u” to kilograms we multiply by

the atomic mass constant (u).

Ref. Chemistry: Quantum Mechanics and Spectroscopy II (J. E. Parker)

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➢ The different isotope’ bond lengths do not vary very much, only in the fourth decimal place

for 1H79Br compared with 1H81Br and in the sixth decimal place for 2H79Br compared with

2H81Br.

➢ The substitution of an atom in a molecule with its isotope does not change the corresponding

bond lengths but changes the momentum of inertia

Ref. Chemistry: Quantum Mechanics and Spectroscopy II (J. E. Parker)

Note: The experimental values (Huber and Herzberg 1979 p. 276) are r(1H81Br) = 1.4144,

r(2H81Br) = 1.4145 and r(3H81Br) = 1.4146 Å. the bond lengths calculated by the rigid rotor

model as well as the experimental values are accurate values of an average bond length