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Learning outcomes from lecture 9

- Be able to qualitatively explain the origin of the Stokes and anti-Stokes line in the Raman experiment
- Be able to predict the Raman activity of normal modes by working out whether the polarizability changes along the vibration
- Be able to use the rule of mutual exclusion to identify molecules with a centre of inversion (centre of symmetry)

Assumed knowledge

- Excitations in the visible and ultraviolet correspond to excitations of electrons between orbitals. There are an infinite number of different electronic states of atoms and molecules.

Which electronic transitions are allowed?

The allowed transitions are associated with electronic vibration giving rise to an oscillating dipole

Electronic spectroscopy of diatomics

- For the same reason that we started our examination of IR spectroscopy with diatomic molecules (for simplicity), so too will we start electronic spectroscopy with diatomics.
- Some revision:
- there are an infinite number of different electronic states of atoms and molecules
- changing the electron distribution will change the forces on the atoms, and therefore change the potential energy, including k, we, wexe, De, D0, etc

1. Unbound

2. Bound

Ground Electronic State

Notice the different shape potential energy curves including different bond lengths…

Depicting other electronic statesThere is an infinite number of excited states, so we only draw the ones relevant to the problem at hand.

we’

De”

we”

re”

re’

Ladders upon ladders…Each electronic state has its own set of vibrational states.

Note that each electronic state has its own set of vibrational parameters, including:

- bond length, re

- dissociation energy, De

- vibrational frequency, we

Notice: single prime (’) = upper state

double prime (”) = lower state

The Born-Oppenheimer Approximation

The total wavefunction for a molecule is a function of both nuclear and electronic coordinates:

(r1…rn, R1…Rn)

where the electron coordinates are denoted, ri , and the nuclear coordinates, Ri.

The Born-Oppenheimer approximation uses the fact the nuclei, being much heavier than the electrons, move ~1000x more slowly than the electrons. This suggests that we can separate the wavefunction into two components:

(r1…rn, R1…Rn) = elec(r1…rn; Ri)x vib(R1…Rn)

Total wavefunction=

Electronic wavefunctionat ×

each geometry

Nuclear wavefunction

The Born-Oppenheimer Approximation

(r1…rn, R1…Rn) = elec(r1…rn; Ri)x vib(R1…Rn)

Electronic wavefunction at ×

each geometry

Total wavefunction =

Nuclear wavefunction

The B-O Approximation allows us to think about (and calculate) the motion of the electrons and nuclei separately. The total wavefunction is constructed by holding the nuclei at a fixed distance, then calculating the electronic wavefunction at that distance. Then we choose a new distance, recalculate the electronic part, and so on, until the whole potential energy surface is calculated.

While the B-O approximation does break down, particularly for some excited electronic states, the implications for the way that we interpret electronic spectroscopy are enormous!

Etot

Spectroscopic implications of the B-O approx.1. The total energy of the molecule is the sum of electronic and vibrational energies:

Etot = Eelec + Evib

Evib

|1

Spectroscopic implications of the B-O approx.- In the IR spectroscopy lectures we introduced the concept of a transition dipole moment:

upper state wavefunction

lower state wavefunction

transition dipole moment

dipole moment operator

integrate over all coords.

using the B-O approximation:

2. The transition moment is a smooth function of the nuclear coordinates.

|1

Spectroscopic implications of the B-O approx.2. The transition moment is a smooth function of the nuclear coordinates. If it is constant then we may take it outside the integral and we are left with a vibrational overlap integral. This is known as the Franck-Condon approximation.

3. The transition moment is derived only from the electronic term. A consequence of this is that the vibrational quantum numbers, v, do not constrain the transition (no Dv selection rule).

Electronic Absorption

There are no vibrational selection rules, so any Dv is possible.

But, there is a distinct favouritism for certain Dv. Why is this?

Franck-Condon Principle (classical idea)

Classical interpretation:

“Most probable bond length for a molecule in the ground electronic state is at the equilibrium bond length, re.”

In the ground state, the molecule is most likely in v=0.

Franck-Condon Principle (classical idea)The Franck-Condon Principle states that as electrons move very much faster than nuclei, the nuclei as effectively stationary during an electronic transition.

Franck-Condon Principle (classical idea)

- The Franck-Condon Principle states that as electrons move very much faster than nuclei, the nuclei as effectively stationary during an electronic transition.

The electron excitation is effectively instantaneous; the nuclei do not have a chance to move. The transition is represented by a VERTICAL ARROW on the diagram (R does not change).

Franck-Condon Principle (classical idea)

- The Franck-Condon Principle states that as electrons move very much faster than nuclei, the nuclei as effectively stationary during an electronic transition.

The most likely place to find an oscillating object is at its turning point (where it slows down and reverses). So the most likely transition is to a turning point on the excited state.

Quantum (mathematical) description of FC principle

approximately constant with geometry

Franck-Condon (FC) factor

μ21 = constant × FC factor

FC factors are not as restrictive as IR selection rules (v=1). As a result there are many more vibrational transitions in electronic spectroscopy.

FC factors, however, do determine the intensity.

Franck-Condon Principle (quantum idea)

In the ground state, what is the most likely position to find the nuclei?

Max. probability at Re

If electronic excitation is much faster than nuclei move, then wavefunction cannot change. The most likely transition is the one that has most overlap with the excited state wavefunction.

2

1

v’ = 0

v” = 0

Positive overlap at edges

overall very small overlap

Negative overlap to left, postive overlap to right

overall zero overlap

Look at this more closely…- Excellent overlap everywhere

Electronic Absorption

There are no vibrational selection rules, so any Dv is possible.

Relative vibrational intensities

come from the FC factor

μ21 = constant × FC factor

Absorption spectrum of binaphthyl

- Example of real spectra showing FC profile

Unbound states (1)

If the excited state is dissociative, e.g. a p* state, then there are no vibrational states and the absorption spectrum is broad and diffuse.

Unbound states (2)

Even if the excited state is bound, it is possible to access a range of vibrations, right into the dissociative continuum. Then the spectrum is structured for low energy and diffuse at higher energy.

Analyzing the spectrum…

All transitions are (in principle) possible. There is no Dv selection rule

Vibrational structure

Analysing the spectrum…

How would you solve this?

(you have too much data!)

1. Take various combinations of v’ and solve for we and wexe simultaneously. Average the answers.

2. Fit the equation to your data (using XL or some other program).

Learning outcomes

- Be able to draw the potential energy curves for excited electronic states in diatomics that are bound and unbound
- Be able to explain the vibrational fine structure on the bands in electronic spectroscopy for bound excited states in terms of the classical Franck-Condon model
- Be able to explain the appearance of the band in electronic spectroscopy for unbound excited states

The take home message from this lecture is to understand the (classical) Franck-Condon Principle

Next lecture

The vibrational spectroscopy of polyatomic molecules.

Week 12 homework

- Vibrational spectroscopy worksheet in tutorials
- Practice problems at the end of lecture notes
- Play with the “IR Tutor” in the 3rd floor computer lab and with the online simulations:
- http://assign3.chem.usyd.edu.au/spectroscopy/index.php

Practice Questions

1. Which of the following molecular parameters are likely to change when a molecule is electronically excited?

(a) ωe (b) ωexe (c) μ (d) De (e) k

2. Consider the four sketches below, each depicting an electronic transition in a diatomic molecule. Note that more than one answer may be possible

(a) Which depicts a transition to a dissociative state?

(b) Which depicts a transition in a molecule that has a larger bond length in the excited state?

(c) Which would show the largest intensity in the 0-0 transition?

(d) Which represents molecules that can dissociate after electronic excitation?

(e) Which represents the states of a molecule for which the v”=0 → v’=3 transition is strongest?

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