Vibrational frequency

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Introduction to Molecular Spectroscopy

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University of Manchester

Introduction to Molecular Spectroscopy

173 оценки

The course introduces the three key spectroscopic methods used by chemists and biochemists to analyse the molecular and electronic structure of atoms and molecules. These are UV/Visible , Infra-red (IR) and Nuclear Magnetic Resonance (NMR) spectroscopies. The content is presented using short focussed and interactive screencast presentations accompanied by formative quizzes to probe understanding of the key concepts presented. Numerous exercises are provided to facilitate mastery of each topic. A unique virtual spectroscopic laboratory is made available to enable students to measure and analyse spectra online. Assessment is via summative quizzes completed during the course period.

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Infrared Spectroscopy

In this module we introduce the theory underpinning infrared (IR) spectroscopy and show examples of analysis using the technique. Transitions between the vibrational energy levels of molecules occurs in the infrared region of the electromagnetic spectrum. We start with the theory underlying vibration using the simple harmonic oscillator model. Analysis of more complex molecules is introduced using group frequencies and number of vibrational modes. You will also be shown how to obtain an infrared spectrum and will have an opportunity to run your own spectrum. At the end of this module you are given a link to view how to obtain an infra red spectrum in the laboratory. Don't forget to complete the end of week laboratory quiz which contributes to your final mark for this course.

Infrared spectroscopy13:15

Vibrational frequency12:01

Example vibrational frequency calculation10:08

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Patrick J O'Malley, D.Sc
Reader
Chemistry

Right, so we're interested also, back to energy, so we've got the energy.

We talk about infrared spectroscopy we're gonna talk about

the frequency, and we're also going to talk about the wavenumber.

And again, I remind you of these important equations,

E = h nu, that's the famous Planck Equation.

Or E = hc over lambda.

C is the speed of light, and lambda, if I write it properly,

is this Greek letter, lambda, which we use for wavelength.

Or E = Page C, nu bar 1 over lambda

lastly is also called new bar.

Nu bar is called wavenumber.

So the vibrations frequency for

that diatomic oscillator had these

= half k and r minus re squared.

Again using mathematics that we're not gonna go into,

you can work out [COUGH] what is the frequency for that vibration.

And the frequency is given by this equation here,

which you'll need to remember.

It's 1 over 2 pi, [COUGH] the square root of k, which we've met.

K is the force constant we talked about.

Divided by mu, and mu is what's know as the reduced mass.

And the reduced mass, worth keeping in SI units, and

we want the reduced mass In kilogram.

Now, as we see here, m1, if we go back to our instant, m1 is just a mass of atom 1.

We're talking about a diatomic here at the beginning, and m2 is the mass of atom 2.

So to reduce mass it's the product of the masses divided by the sum of the masses.

And for quantitive purposes, you need to quote that in kilograms.

And we'll be doing an exercise on that in a few minutes,

which will hopefully make it clear for you.

So the vibrational frequency for the diatomic oscillator is given by this.

And you know all the terms here.

K's the force constant, and mu is the reduced mass, and that's given in hertz.

And again hertz, is the same as seconds to -1.

Right, so hertz is seconds -1 for nu.

The force constant and the SI units we want to keep

it in is newtons per meter and then we want to quote

the reduced mass in kilograms per molecule for mu.

So we'll again do a little exercise on this in a minute,

how do we work out the reduced mass.

The reduced mass you see the periodic table,

it's usually quoted in grams per mole.

So you have say hydrogen is 1, carbon is 12,

oxygen is 16, the atomic masses.

And to work out the reduced mass you simply

put the atomic masses in gram for mol, like this here.

And then you multiplied them by a conversion factor,

which is called a unified mass constant, u.

And this is always given usually in the back of text books or

you guys remember it.

Or if you're asked in the exam,

you've been given this in the data sheet that would accompany the exam.

But it's not that, sometimes even if you forget it, you're quoting here,

you want to convert.

So you had these units here.

And you wanted to quote them in kilograms as I say here they're

expressed in grams per mole.

So if you wanted to in grams per molecule what would you want to divide by?

Going from mols to molecules.

Mol to molecule.

Avogadro's constant, so you divide by that.

So that would give you grams, wouldn't it?

Grams per molecule.

You divided these atomic mass weights here by,

that will give you grams per atom or grams per molecule.

And then you want to convert to kilograms, so

you'd just divide by 1,000 wouldn't you.

So you can look up to see [INAUDIBLE] but this number here.

Is just one over Avogadro's number, multiplied by 10 to the -3.

Right, okay.

So another thing, what you we go on to practical applications of spectroscopy.

The frequency or the energy is almost always given in wavenumber,

which is one over the wavelength but

it's almost always quoted in si units it should be meters -1.

But traditionally it is given to centimeters to the -1.

Because somebody asked me that in the workshop and

it's generally, usually if the number is easy to deal with.

Sometimes it gives nice whole numbers in this region and

therefore people like to quote it in that centimeters to -1.

So that's how it has stayed, if you like.

But you then need to be able to convert, you see, from that.

So again, reminding you here, that nu bar, the wavenumber,

is 1 over the wave length C = lambda times nu.

So, you do the manipulation you can also get nu bar = nu over c.

So, as I said the SI unit should be meters -1, but centimeters -1 is more common.

So, you need to remember this.

That nu bar, given in meters -1 = the nu bar and

centimeters -1 by a 100.

So people will always tend to get that wrong.

And one way I think of it is nu bar is wavenumber,

so it's number of waves in a meter.

So say you had a number of waves as we have here, say in a centimeter.

Then the number of waves in a meter is gonna be greater.

Usually, people divide here.

If you think about converting to meters,

the number of waves in a meter is gonna be 100 times that.

Sorry about that.

So that's going to be, so nu bar in meters -1 = nu bar in centimeters -1 by 100.

And then we have given the.

We get the vibration frequency before, and frequency,

and I think it was nu = 1 over 2 pi the square root of k over mu.

That's in, if you remember, that was in seconds -1.

So if we quoted nu bar, nu bar, which is the wavenumber,

you have to divide by c here, to convert from nu bar to nu.

So nu to nu bar.

So you had 1 over 2 pi c, the square root of k over nu.

And that's in the meters -1.

You divide it by the speed of light c.

And now we want to go to centimeters -1.

So we need to divide by 100 to convert it to centimeters -1.

So these are all kinds of manipulations that you should be able to do.

All right.

Sorry, I've written over here a bit.

So the frequency, After

vibration is directly proportional to the force constant.

We can see that from this equation here.

Or the wavenumber, it's on the top line so it's going to be as the force constant

increases the frequency of the vibration increases and so also does the wavenumber.

And what you also see is that it's inversely proportional to reduce mass.

So the reduced mass is below the line.

So as you increase that you will reduce the frequency,

or the wavenumber of that vibration.

So I think what we have here is, Some examples of that.

So here you have hydrogen and here you have the force constant.

So this is 500, get rid of that.

Okay.

So H2, the force constant in Newtons meters to the -1 is 509.

And then you have the reduced mass here is this 8.31 times 10 to the -28.

So it's got a wavenumber given in centimeters -1 of 4159.

And then you go to HCl, it's got the slightly lower force constant but

you see that the reduced mass is quite high so

it's higher -27 as opposed to 28.

So both of these factors the lower force constants will reduce the wavenumber.

And also the higher reduced mass will reduce the wavenumber as well.

So you can see the value of the wavenumber of that bond is 2890.

And then you go on to Cl2, okay the force constants again,

down a bit so you'd expect a lower value for the wavenumber.

And the reduced mass is quite a bit bigger, it's 10 to the -26.

So now the wavenumber has gone down as well.

And then lastly, you have nitrogen, we talked about nitrogen already.

Nitrogen of course has a triple, has a triple bond, so

the force constant is quite high.

And therefore that will govern the frequency, at least compared with Cl2.

Cl2 and nitrogen have fairly similar reduced masses but

you can see the force constant of the nitrogen is quite high.

So the frequency of the vibration you observe depends on the force constant and

inversely proportional to the reduced mass and

directly proportional to the force constant.

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