Rotational and Vibrational Spectroscopy

Energy of a rotational level

`E_J=hcBJ(J+1)=hbar^2/(2I) J(J+1)`

The rotational constant, B,

`B=h/(8pi^2cI)`

Moment of Inertia, I,

`I=muR^2`

Reduced mass, `mu`,

`mu=(m_1m_2)/(m_1+m_2)`

A transition is observed in a pure rotation spectrum is the difference in energy between two E_J states, ie.,

`DeltaE_J`

`=E_(J-2)-E_(J-1)` `=hcB(J+1)(J+2)-hcB(J+1)J` `=hcB(J+1)2`

Thus,

`DeltaE_J=2hcB(J+1)`

where J is the quantum number of the lower level.

`DeltaE_J=hnu=hcstackrel(~)nu`

where `stackrel(~)nu` is the wavenumber of an individual transition.

On a spectrum, the difference in wavenumbers between two peaks adjacent is 2B,

diagram

ie., `stackrel(~)nu_2-stackrel(~)nu_1=2B(J+2)-2B(J+1)=2B`

To get Bond Length from `stackrel(~)nu` spacing

spacing = 2B

Generally, `stackrel(~)nu` given in cm^-1, to convert to m^-1 multiply by 100

Spacing, `Deltastackrel(~)nu=2B`

`B=(Deltastackrel(~)nu)/2=h/(8pi^2cI)`

Solve for I, `I=muR^2`

`mu=(m_1m_2)/(m_1+m_2)`

Convert molar masses into gmol^-1 into kg,

`(M_r(gmol^(-1)))/(1000xxN_A)=m(kg)`

Reduced mass,

`mu=N_A^(-1)xx10^(-3) ((m_(r1)xxm_(r2))/(m_(r1)+m_(r2)))`

Bond length, R, is calculated,

`R=sqrt(h/(4pi^2cmuDeltastackrel(~)nu))`

To find which Rotational Level has the largest population:

The Boltzmann Distribution is differentiated,

The Boltzmann distribution:

`N_J/N_0=(g_J/g_0) exp(-(hcBJ(J+1))/(kT))`

`J_(max)=sqrt(T/(2theta))-1/2`

`theta=(hcB)/k`

thus,

`J_(max)=sqrt((kT)/(2hcB))-1/2`

Centrifugal Distortion

`E=hcBJ(J+1)-hcDJ^2(J+1)^2`

`D=(4B^3)/((stackrel(~)nu)^2)`

where `stackrel(~)nu` is the vibrational wavenumber

Vibrational Motion

The energy of a vibrational level,

`E_v=homega_0(v+1/2)`
where `v` is the vibrational quantum number and `omega_0` is the fundamental frequency of vibration

`omega_0=1/(2pi) sqrt(k/mu)`

Calculation of a Force Constant

Example X:

Vibrational transition, `v=0rarrv=1` for CO at `stackrel(~)nu=2143.3cm^(-1)`

We assume the moelcular behaves like a simple harmonic oscillator,

`DeltaE=E_2-E_1`	`=homega_0(1+1/2)-homega_0(0+1/2)`
	`=homega_0` =`h/(2pi) sqrt(k/mu)`

`hcstackrel(~)nu=h/(2pi) sqrt(k/mu)`

`k=(2picstackrel(~)nu)^2mu`

`mu=(m_1m_2)/(m1+m2)` in kg

to convert `mu(kg)` from `mu(gmol^(-1))`,

`mu(gmol^(-1))xx(N_Axx1000)=mu(kg)`

`M_r(C)=12.01gmol^(-1)`

`M_r(O)=16.00gmol^(-1)`

`k`	`=(2pixx2.998xx10^8ms^(-1)xx2143.3xx100m^(-1))^2mu`
	`=1.63xx10^29s^(-2)xx(M_r(C)xxM_r(O))/(M_r(C)+M_r(O))xx1/(N_Axx1000)` `=1856Nm^(-1)` where Nm-1 is kgs-2

Rovibrational Spectroscopy

The origin of the spectrum is centred around the forbidden transition, DeltaJ=0` & `Deltav=1`. Extrapolation between the two points in the centre give the wavenumber for the pure vibrational transition. ie.,

`hcstackrel(~)nu=homega_0`

The lower `stackrel(~)nu` brnach is the P-Branch, corresponding to `Deltav=1` & `DeltaJ=-1`. The higher energy branch, R-Branch corresponds to `DeltaJ=+1` & `DeltaV=+1`.

diagram

diagram

Rotational Energy Level Diagram

diagram - divergent energy levels

Electronic Energy Level Diagram

diagram - convergent energy levels

`E=(hcR_H)/n^2`

Vibrational Energy Level Diagram

diagram diagram