 Hello everyone, I once again welcome you all to MSP lecture series on Interpretative Spectroscopy. In my previous lectures, about 8 lectures I gave focusing on UV visible spectroscopy and of course, I am giving more emphasis for inorganic molecules in particular DD transitions. That means 3D, 4D, 5D metals having D1 to D10 electronic configuration show different type of absorption depending upon the nature of the ligands and also the nature of the metals and their oxygen state and also under which ligand field they are situated. So, based on all those things lambda maximum varies and also I showed you how we can simplify all electronic configurations, all metal complex having different electronic configuration into simply 3 or 4 categories. One is D0 and D10, D0 do not have any electrons and D10 is completely filled, unique cases for example, you can consider potassium permanganate, potassium dichromate, in case of D10 you can consider mercury ciodate and then the remaining ones D5 is a unique one, then the remaining 8 electronic configurations are D1, D4, D6 and D9 and then D2, D3, D7 and D8. And there is some similarities I showed you, for example, if you take D1, D4, D6 and D9, one electron is there and one less than half field electronic configuration and one more than half field electronic configuration and then D9 is one less than completely filled electronic configuration. So, these 4 electronic configurations those metal complexes which have invariably show one DD transition and then the second type is D2, D3, D7, D8, the similarities you can conclude in this manner, 2 electrons 2 less than half field 2 more than half field and 2 less than completely filled. So, these show 3 bands in their DD spectra and then of course D5 what happens when you look into selection rule, D5 high spin complexes have 5 electrons in T2G and EG in case of octahedral complexes and in case of tetrahedral complexes we have here 2 and then 3 in T2 system. So, in these cases what happens since the selection rule says delta S equals 0 and delta L equals plus or minus 1 and strictly speaking none of these DD transition should occur they are all prohibited from LePold rule, but what happens when we look into metal complexes, metal ions under the influence of ligand field they lose the degeneracy of D orbitals and some of the orbitals will be lower in energy, some of the orbitals in higher energy because of extensive mixing of D orbitals with S and P characters. As a result what happens they are no longer pure D in nature as a result LePold A allowed transition could be seen and hence we will see DD transitions and selection rule as I mentioned delta S equals 0 and this is quite opposite to what we see in case of nuclear transitions. Nuclear transition if you recall it is a delta S equals plus or minus 1. So, that means a nucleus having spin half will go to spin minus half downward spin whereas, here electron with upward spin should go like this only. So, delta S equals 0 here when you look into D5 they are spin fernan and especially when they have center of symmetry the colors are very very weak, but when we look into tetrahedral complexes they do not have center of symmetry as a result some mixing is there and hence we see some of these compounds having D5 electronic configuration with weak fielding and would be little colored, but why D5 octahedral complexes of having high spin still in the color what happens if we look into the ligands ligand to metal bonds are not rigid, but they will be vibrating with respect to the mean position. So, because of this vibrations at what happens often some of these ligands come out of the equilibrium position as a result these complexes lose center of symmetry and hence mixing would take place and hence they show color. So, this is all about UV visible spectroscopy and again for D1, D4, D6, D9 we can have conveniently one diagram that is called Argyl diagram to explain electronic spectra for both octahedral and tetrahedral complexes similarly we can have another one Argyl diagram to explain spectra of octahedral and tetrahedral complexes having D2, D3, D7, D8 electronic configuration, but when you go to Tanube-Sugano diagram every electronic configuration we have a we should have a separate Tanube-Sugano diagram, but that includes the entire band of ligands that means whether it is a weak field ligand, there is a high field ligand everything is included and, but we should have one Tanube-Sugano diagram per one electronic configuration and also we saw in case of D2, D3, D7, D8 when we have there is a drop in the observed frequency of one of the bands are increasing there because this is due to nephelacetic effect and that can be corrected using Rackha parameters and another important thing one should remember about UV visible spectroscopy is how to write term symbols ok term symbols you should be able to write by simply knowing L azimuthal quantum number sigma L that gives capital L and L can have anywhere between 0 to 1, 2, 3, 4, 5 and then corresponding terms we have SpDFGHI and then 2S plus 1 is called spin multiplicity and then we have J you can take values anywhere between L plus R minus S and here what we should remember is when we identify the ground term that should have highest multiplicity when we look into highest multiplicity maximum number of unpaid electrons will be there. So, the one state which is having highest 2S plus 1 value will be the least energetic and ground state and then due to some reason if we have two terms both having the same multiplicity value 2S plus 1 then we have to consider the L value highest L value will be the least energetic and then let us say we have multiplicity is same and then L value is same in that case we have to consider the J value when you consider J value we have to consider L minus S for substance having less than half electronic configuration and then L plus S for those which is which are having more than half electronic configuration. Then the other one is microstate microstate will give you ground state along with several possible excited states and formula is very simple n factorial over R factorial into n minus R factorial. So, n factorial will tell you the total capacity of the sub shell which we are considering for example, if we consider d orbital total capacity is 10 electrons. So, hence n factorial means 10 factorial if you consider f orbitals we have 14 electron capacity. So, it is 14 factorial and then R is the total number of electrons we are considering. So, d 2 system n factorial is 10 factorial and R is 2. So, this is how you should be able to calculate the microstates and all possible microstates are actually do not show transitions and it is much simplified because we are bringing Lapporte selection rule again and hence it makes understanding and observed spectrum much simpler compared to what we could think of considering all possible microstates. So, this is all about UVS spectroscopy and I would come back again at the end with more interesting problems to make you familiar with solving problems as far as UVS spectroscopy is concerned. Now, let us move on to IR spectroscopy I am sure you are all familiar with infrared spectroscopy. So, here infrared spectroscopy is a very important tool to commits to identify what functional groups exist in unknown samples or samples you have made in the laboratory. The light our eyes see is, but a small part of a broad spectrum of electromagnetic radiation in the visible region I would say and the immediate high energy side of the visible spectrum lies the ultraviolet and on the lower energy side is the infrared. So, you can see here we have visible spectrum here and infrared is here and microwave radiation comes here. So, this is the region we are going to focus as far as infrared spectroscopy is concerned. So, on the having a wavelength range of about 2500 to 16000 nanometer the infrared region is capable of revealing information not easily uncovered through basic means. So, that gives vital information about different type of functional groups present and also the characteristic bands due to the different bonds present in the molecule. The preferred method of infrared spectroscopy is FTIR that means Fourier Transform Inferred Spectroscopy commonly used. Nowadays IR can aid in the identification of unknown compounds and also it can determine the quality or consistency of a sample if it is in the large production industrial scale production. Then how it happens when infrared radiation is passed through the sample some of the radiation is absorbed by the sample and some of the radiation is passed through are transmitted. The resultant spectrum is nothing but the plot of percentage transmission versus frequency in centimeter inverse of the radiation that is passed through in the infrared region. So, the energies of photons associated with the infrared region are not large enough to excite electrons, but they are still strong enough to induce vibrational excitation of covenantly bonded atoms and groups. That means energy is the of photons associated with infrared radiation is so small it cannot really excite an electron from E 1 level to E 2 level higher end level, but on the other hand they can induce vibration of the bonds. That means when we consider a molecule we have covalent bonds these covalent bonds are not rigid sticks or rods, but are more like stiff springs that can be rotated provided it is there is a single bond and they can be stretched or they can be bent. So, these different types of vibration motions are characteristic to a molecules component atoms. So, this information comes directly from infrared spectroscopy. So, all organic inorganic compounds will absorb infrared radiation that corresponds in energy to these vibrations and infrared spectrometers allow chemists to obtain absorption spectra of compounds that are a unique reflection of their molecular structure. So, that means an IR spectrum can be considered as a molecular fingerprint of the sample as no two unique molecular structures produce the same IR spectrum. So, this is very very important it may not be possible to understand all the peaks present and one need not have to worry about interpreting all the absorption peaks that are seen in that region, but what one should remember is this fingerprint region is characteristic of molecules. So, what is the theory involved behind this infrared spectroscopy? So, IR can provide information about the presence of particular functional groups say whether OH group is there, whether CO group is there, whether amine group is there or whether there is a C double bond whether P double bond O or C triple bond N C double bond N all this vital information one can extract from simply by looking into IR spectrum of a given molecule. IR data along with NMR and mass spectrometric data can provide the structural features of a compound without any ambiguity. So, minimum information about the theoretical aspects of IR is sufficient to use in structural analysis. So, I am not going into the details of theoretical aspects as I had mentioned since my attention is to focus on interpretation with minimum theoretical background I try to make you familiar with interpretation of data from various spectroscopic methods that we use routinely while characterizing organic or inorganic molecules. So, IR absorption band of a compound indicate the presence of functional groups this is the most important point one should remember. IR absorption bands of a compound indicate the different type of functional groups present in that molecule. So, narrow and sharp bonds seen in the spectrum are called peaks. So, peaks we are band positions are given in wave numbers as unit of reciprocal centimeter that you know in the IR spectrum IR region between 400 to 4000 centimeter inverse is very important most of the functional groups have absorption in this region of 400 to 4000 centimeter inverse. The frequency divided by C the speed of light provides the wave number this one should remember this indicates for example vibration of frequency 6 into 10 rise to 13 has a corresponding wave number 2000 centimeter minus 1 that is equal to you know simply takes 6 into 10 rise to 13 over velocity of light 3 into 10 rise to 10 centimeter per second. So, that would give you about 2000 wave numbers. So, this is how you can simply calculate the frequency to wave numbers. So, band intensities are reported as follows. So, by looking into the spectrum we have to identify and we have to give a term depending upon the type of intensity we come across for example, if it is strong we have to denote it by S and if it is medium we have to say M and if it is weak we should say W and if it is a sharp we have to say sharp SH. A fundamental vibration is one in which an atom sets a simple harmonic oscillation about its equilibrium position. So, each atom if you consider has 3 degrees of emotional freedom along 3 axis such as x y and z directions. So, that means n atoms have 3 n independent motions, but in a molecule the motions are not independent and motions within the molecule are not independent of each other due to having the bonds with the other atoms present in the molecule. So, as a result 3 motions becomes translations of the molecule. So, where all atoms move simultaneously in the x y and the z direction. So, another 3 are rotations where all atoms rotate in the same phase about the x y and z axis. So, this is very important. So, all atoms move simultaneously in the x y and z direction and all atoms rotate in the same phase about x y and z axis. So, these points are very very important one should remember. So, this leaves now 3 n minus 6 motions in which bond angles and bond distances without altering the center of gravity. That means, it is 3 n minus 5 motions for linear molecules. These are referred to as fundamental vibrations of the molecule and are of 2 types. One is stretching and one is bending. So, that imagine 2 diatomic molecule 2 or 3 balls connected by a spring. If you consider a diatomic molecule imagine 2 balls connected with a spring or diatomic molecule you know 3 atoms are connected with 2 springs. So, something like this. So, let us consider a triatomic molecule such as water which contain 3 into 3 minus 6 3 fundamental vibrations. What are those 3 fundamental vibrations? One is this one symmetrical stretching and then of course, here the stretching frequency nu comes around 36 52 centimeter minus 1 and then the second one is asymmetrical stretching and here we refer this one as nu s and here asymmetrical stretching is referred as nu a s o h or one can also say something like this. So, this indicates we are considering the asymmetrical stretching of o h group and here we are considering the symmetric stretching of o h group and asymmetric stretching value comes around 37 56 centimeter minus 1. The other one is scissoring. So, focus towards scissoring. So, here this is again represented delta s here. So, delta s o h means it indicates the scissoring frequency corresponds to the o h molecule o h group in water this comes around 1596 centimeter minus 1. So, this is a typical example where you can see how 3 n minus 6 equals 3 fundamental vibrations are seen. So, now let us consider a triatomic molecule which is linear we consider triatomic molecule such as water bent of course, the shape is bent still oxygen geometry is tetrahedral, but let us consider a triatomic molecule having linear geometry such as CO 2. So, here if you use the same analogy 3 into 3 minus 5 equals 4 fundamental vibrations are possible. So, what are those symmetrical stretching comes around 1340 centimeter minus 1 that is designated as nu s CO and then asymmetrical stretching again very similar to what we saw in case of water. So, you can see symmetric stretching the directions given here and in asymmetric stretching. So, one is here one is in this direction other one is this one. So, this asymmetrical stretching is there not uniform this is also designated as nu a s CO 2 this comes around 2350 centimeter minus 1 and then we have scissoring, scissoring bending we have here it comes around 666 centimeter inverse and designators as delta s CO 2 and then we have scissoring bending is there. So, you should know the difference scissoring bending this one scissoring bending and here this comes around. So, both are identical. So, they come around 666 centimeter minus 1. So, this is how you can identify 4 fundamental vibration modes in case of triatomic linear molecules such as CO 2. So, now let us consider a CH 2 which is part of a molecule for example, if you consider ethanol, propanol or any aliphatic molecules where we have CH 2. So, in this case what we have is a 3 n plus 6 root does not apply here then we have to see what are the possible vibrational modes for CH 2. So, one is symmetrical stretching that you can see in most of the molecules around 28 53 centimeter minus and designated as nu s CH 2 and then similarly we have a symmetrical stretching we should focus your attention also when I say symmetrical stretching how these atoms are stretched in what direction and then a symmetrical stretching. So, one is in this direction one is in this direction one is in this direction. So, this is called asymmetrical stretching and here it comes around 29 26 centimeter minus 1 and then scissoring we have delta s equals 1465 centimeter minus 1 apart from these three fundamental vibrational modes we also come across wagging. So, here wagging is referred to as gamma s this comes in the range of 1350 to 1150 centimeter minus 1 and then another one is twisting in the twisting it is a sto A s it is called as then this comes around 1350 to 1150 centimeter minus 1 and then the other one is rocking both in the same direction this is rho CH 2 then that comes around 720 centimeter minus 1. So, often the theoretical number of vibrations are not observed and this may be due to symmetric reasons or due to degenerate vibrations what is degenerate vibrations? Vibrations of the same frequency. So, that means although we have clearly seen six possible vibrational modes for CH 2 groups present in a molecule sometimes we may see only three of them and this is the theoretically predicted ones, but on the other hand due to some reason and that is clearly shown here often the theoretical number of vibrations are not observed and this may be due to symmetry reasons or due to degenerate vibrations that means vibrations having the same frequency in that case we might miss one or two and then that might result in the increase in the intensity of few bands. So, let me stop here and come back to you again in my next lecture more details about infrared spectroscopy. So, thank you so much.