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What is Rotational spectroscopy or microwave spectroscopy?

Rotational spectroscopy which is also called microwave spectroscopy provides information about the absorption or emission of electromagnetic radiation typically in the microwave region of the electromagnetic spectrum.

Heteronuclear diatomic molecules are microwave active.

(i)  A Molecule, to give rotational spectra should possess a permanent dipole moment.

(ii)  Heteronuclear diatomic molecules possess a permanent dipole moment. This dipole moment gives rotational spectra.

(iii)  If the energy supplied is small enough during the rotation of molecules, the bond length of the molecule remains unchanged.

(iv) The amount of energy required to change the rotational energy level is of the order of 0.05 ev  ( 2.5×10-4 m) which corresponds to the energy of radiation in the microwave region.            

(v) During rotation, these molecules generate an electric field that interacts with the microwave radiations giving rise to a spectrum.

(vi) This absorption of radiation take place in the infrared or microwave region of the electromagnetic spectrum

(vii) Due to the above reasons, Heteronuclear diatomic molecules are microwave active, and Rotational spectra are known as Microwave spectra.

(viii) Example: HCl, HBr, and CO are microwave active.

Expression for the moment of inertia in a simple diatomic molecule.

Rotational Spectroscopy

Consider a simple diatomic rigid polar molecule of masses ‘m1 and ‘m2 separated by an internuclear distance ‘r’.

-Let  r1 and r2be the distances of these atoms from the center of gravity (G), about which the molecules rotate.

-Moment of inertia of a particle of mass ’m’ revolving around a fixed point at a  distance ‘r’ is given by ‘mr2’.

-For a diatomic molecule, the total moment of Inertia (I) is given by,

I =  m1r12+ m2r22………………………………….…….i)

     Since the system is balanced, moments of both the atoms are equal i.e.

m1r1 =m2r2………………………………..……….…  .ii)

            On substituting vales from eq (ii) in eq (i), we get

                                    I =  m2r2r1+ m1r1r2

I =  r1r2 (m1 + m2) ……………………………….…..(iii)

                                    But,      r = r1 +r2

r=  r1 + m1r1 / m2  …………………..(from eq ii)

 r =  r1(1 + m1/m2 )  

 r =  r1(m1+ m2 /m2)

Therefore r1=  m2/m1+m2 . r      ……………………………… (iv)

Similarly r2 = m1/m1+m2 .  r          …..………..……………… (v)

On substituting the values of r1 and r2 in equation (iii) we get,

                I =     r ( m2/ m1+m2 ) ×  r (m2/m1+m2)   × (m1 + m2)

                Therefore I = m1m2/ m1+m2 . r2               ……………….. eq (vi)

            The term   m1m2/m1+m2 = µ  i.e. reduced mass of the system

Therefore I =µ r2

The frequency separation (∆v) between two successive lines in the rotational spectrum is 2B  i.e ∆ v = 2B.

The rotational spectrum of a diatomic molecule consists of equally spaced lines.

The equation for the frequency separation of the lines in the rotational spectrum of a  diatomic molecule.

The rotational energy EJ of a diatomic molecule is given by,

Where, I =moment of inertia of the molecule

h =Planck s constant

J =rotational quantum number which can have value 0,1,2,3……

Consider a rotational transition from a rotational energy level J to another level (J1)

   (c) Now change its rotation energy (∆E) is given as,

                             ∆E = EJ – EJ’

But accordingly, to the selection rule, ∆J=+1

                             i.e. ∆J= J-J= +1

                             Therefore J= J-1

            By substituting equation of J in eq (iii) we get,

(d) According to quantum theory, energy changes are quantized by the equation,

            ΔE = hυ = hῡc            ……………  ………………….…. ..(v)

  Therefore ῡ = 2BJ

Where ῡ = frequency in wavenumbers of lines in the rotational spectrum.

(e) Thus, from eq(vi), the frequency of lines in the rotational spectrum can be found.

(i) When J=1, i.e for the transition from J=0 to J=1 ( Transition from ground state J=0 to the  first energy level  J=1 takes place)

     Therefore 1 = 2Bm-1

(ii) When J=2, i.e for transition from J=1 to J=2   ( Transition from the  first energy level  J=1 to     second energy level  J=2  takes place )

     2 = 4Bm-1

(ii) When J=3, i.e for transition from J=2 to J=3

      3 = 6Bm-1

(f) The frequency separation between consecutive lines will be,

          ∆  = 21  =4B – 2B = 2B

          ∆  = 32  =6B – 4B = 2B &  so on . Thus the rotational spectrum of a diatomic molecule consist of series of equidistance lines with the separation of 2B.

Thus, the frequency difference between two successive lines in the rotational spectrum is 2B.

(g) From the Moment of Inertia (I) the equilibrium distance ‘r’ or bond length can be  calculated

Where m1 and m2 are masses of atoms in a diatomic molecule.

Diagram showing different rotational energy levels with permitted rotational transition in a domestic molecule

The nature of the rotational spectra:-

  1. The spectral lines are present in the microwave region.
  2. The spectrum consists of equally spaced lines.
  3. The spacing between any two successive lines is constant and is equal to 2B0m-1.
  4. At room temperature, the rotational energy levels up to J = 5 are occupied by the molecules. Hence, all the transitions from J1 = 0 to J1= 5 will be observed.

The effect of isotopic substitution on the rotational energy levels and rotational spectrum level of a diatomic molecule.

(a) The Presence of isotopes in a molecule will bring about a considerable change in reduced mass (µ) of the molecules. However, the intermolecular distance (r) and geometry of the molecule remain unchanged. As reduced mass changes, it will bring about a change in the moment of inertia (I) and Rotational Constant (B).

(b) In the Relation B = h/8π2Ic, the rotational constant B is inversely proportional to the moment of inertia.

(c) Therefore, the heavier isotopic molecule having a larger moment of Inertia will have a small value of B. As a result, the following frequency separation of successive lines which depends on B will be smaller for the heavier isotopic molecule than for the lighter isotopic molecule.

(d) These small differences in the frequency separation have been utilized to determine the isotopic masses.

(e) Consider two isotopic molecules A and B. Let IA and IB be their moment of inertia and πA and πB be their reduced masses. The rotational constants BA and BB will be given by.

This relationship helps to calculate the abundance of isotopes by knowing the shifts in the spectral lines with the isotopic effect.

Limitations of Rotational Spectra

The limitations of Rotational Spectra are as follows,

  1. Heteronuclear molecules and groups which have a permanent dipole moment alone give rotational spectra. eg. HCl, HCN, CH3Cl, etc.
  2. Homonuclear molecules and groups which do not give a permanent dipole moment do not give rotational spectra and thus cannot be studied by using rotational spectroscopy. eg. H2, N2, O2, etc.
  3. Similarly, linear and non-linear polyatomic molecules cannot be studied by using rotational spectroscopy as they do not show rotational spectra. In the case of polyatomic molecules, a single moment of Inertia will not give complete molecular structure
  4. The substances under study should be in the gaseous condition for free rotation of molecules to take place. In solids and liquids, strong intermolecular forces of attraction do not allow the free rotation of molecules. Hence, the substances in solid and liquid conditions cannot be studied.

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