 welcoming all to MSP lecture series on interpretive spectroscopy. So we are discussing about UV visible spectroscopy. In my last lecture, I was discussing about the DD transitions and not the conditions. Let's continue knowing more about that. Electronic spectra of transmittal complexes. Here, what would happen is, essentially electrons are exited from the ground state to one of the exited states. Such electronic transitions require high energy compared to other transitions. So during electronic transition, we also come across low energy vibrational rotational transitions because each electronic level is associated with several vibrations levels, which in turn are associated with our having several rotational levels. So in electronic spectra, such transitions are too close in energy to be resolved into separate absorption bands. The energy required for vibrational and exaltation as very minimal compared to the energy required for electronic transition from one level to another one. As a result of that one, what happens? They cause broadening of the absorption band in DD spectra. As a result, the bandwidth observed in DD spectra are in the range of 1000 to 3000 centimeter minus 1. So in a free gaseous metal ion, if you consider, DR builds are degenerate, and hence, no DD transitions are observed. But in a complex, degeneracy is lost because of ligand field splitting the orbitals into two or more levels. For example, if you consider octahedral level, we'll be having octahedral field, and we'll be having T2g, composite of dxy, dxz, and dyz orbitals which is lower in energy, and then higher energy, Eg level, composite of dx, minus y square and dz square levels. But in tetrahedral, opposite happens. So E will be the ground state in the homo level, and T2 will be the lumo level. So the magnitude of this delta O or delta T depends on the nature of the ligands, and affect the energy of electronic transitions, and hence, the frequency of absorption maxima. For example, if you are considering weak field ligands, of course, you can look into spectrochemical series. How we arrived at a position, a given position for a ligand in the spectrochemical series is considering the donor and acceptor properties of the ligands, and if you recall what I taught in advanced transmittal chemistry that all ligands, whatever you come across, can be simply classified into three categories. One is pure sigma donor ligands, such as water and ammonia, all neutral oxygen donor ligands, and also alkylamines and arylamines, on the other hand, the next one is sigma donor and pi donor ligands, such as halogens. Here, what we have is low energy field sigma orbitals, and low energy field pi orbitals are there. When such ligands interact with the metal, what would happen? Both of them will be donating to the metal, and hence, the CFSC, the homo-lumo gap, drops relatively compared to what we saw in case of pure sigma donor ligands, such as ammonia and water. The third category is sigma donor and pi acceptor ligands, example, carbon monoxide, phosphines, and nitrogen heterocycles, and other things. So in this case, what happens? They have low energy field sigma orbitals, and high energy, empty pi orbitals are there. And in this case, because of the sigma donation, and then pi donation back from the metal to the ligand, what happens? The energy gap between homo-lumo increases. As a result, this complexes are more stabilized, and also, in case if you want to have a DD transition, it requires very high energy. So that means the magnitude of delta O depends on the nature of the ligands, and affect the energy of electronic transition, and hence, the frequency of absorption maxima. Extent of splitting is related to the ligand position in the spectrochemical series. So the ligand position spectrochemical series very, very important in electronic spectroscopy. So here, just look into all the orbitals are classified into four categories here. There is some reason behind classifying like this. So first, D0 and D10. D0, D10 complexes are usually colorless, not always. Why they're colored? That I will come back soon. And D0 and D10, you cannot unspeak DD transition. But on the other hand, if you look into D1, D4, D6, D9, they have some similarities. In the same way, if you consider D2, D3, D7, D8, they also have some similarities. And D5 is unique. And what are the similarities between D1, D4, D6, and D9? So D1, we have one electron. D4, one less than half field electronic configuration. And D6, one more than half field electronic configuration. And D9, one less than completely field electronic configuration. So that means here, they have some similarities. And they show very similar spectra when you consider electronic or UV visible spectrometer, when we subject those into absorption. And the other hand, if you look into D2, D3, D7, D8, they have similarities. Because in D2, we have two electrodes. And D3, we have less than two electrodes for half-shell, half-field electronic configuration. And D7, we have two electrodes, more than half-field electronic configuration. And D8, we have two less than completely field electronic configuration. So that means again, they have similarities. And the spectra look almost identical, except for some anomalies here and there. And these compounds are colored. And D5 compounds are little bit pale colored ones, especially when they are tetrahedral compounds. And they're almost colorless when they are octahedral homo-liptic complexes having centrosymmetry. We will learn all those things in detail. But now let's look into HGI2. It is brick-red color. And here, electronic configuration, if you consider mercury, D10 electronic configuration is there. So no DD transition is there, but still it is intensely colored. When you consider potassium permanganate or potassium dichromate, metal exists in their highest possible oxygen states. That means Mn is in plus seven state. All electrons have been removed. And it has a D0 electronic configuration. But still, Km4 is intense purple in color. And similarly, if you look into potassium dichromate, chromium is in plus six state, and D0 system, and also intense orange-red color. And similarly, if you look into bismuth triiodide, we have D10 S2 electronic configuration. It's orange-red in color. And if you look into prussian blue, here we have iron 3 and iron 2 all there. This is also very intense in color. We will see one by one why. What is the origin of color in these compounds as you cannot anticipate any DD transition? So iodide has very high polarizability, which results in anionic charge getting easily transferred to the mercury 2 plus cation. So this process releases some energy which falls in the visible spectrum. Hence, we can say that compounds like MnO4, Hg2, et cetera are colored. That means what we come across is a term called charge transfer transition. When we talk about charge transfer transition, the moment we look into the electronic configuration, we should be able to tell in which direction charge is transferred. For example, if you look into D10, it is metal to ligand charge transfer transition. On the other hand, if you look into KMnO4, it is D0. It is ligand to metal charge transfer transition. And if you look into prussian blue, here we have iron in 3 state and iron in 2 state. So here it is metal to metal charge transfer transition. So these sum of these charge transfer transitions are responsible for the appearance of intense color, the respective colors in these compounds. So now we should come back to the selection rules of electronic transition. So when we were writing microstates, we saw several possible exercise states. And then if electrons are promoted to each state from the ground state, then electronic spectrum would have been very complicated. And elucidation or interpretation would have been very difficult. But however, all these transitions are governed by certain rules. That we call it as selection rules for electronic transition. What are those? So electronic transition may be classified as intense or weak according to the magnitude of epsilon maximum that corresponds to allowed or forbidden transitions. That means what are allowed transitions? These transitions have a maximum value tends to 4 or more. And property of their occurrence is very high. These are generally due to pi, pi star transitions, allowed transitions. And then example, pi, pi star transition in 1, 3 butadiene shows absorption at 217 nanometer. And epsilon has 2,900 represents an allowed transition. Then if allowed transition term comes, there should be forbidden transitions. And here if you look into it, n to pi star transition is forbidden. For this transition, e maximum is generally less than 10 or rise to 4. For example, if you consider n to pi star transition of saturated aldehydes and ketones, which exhibit a weak absorption of low intensity near 285 nanometer and having e maximum value less than 100 because they are forbidden transitions. So that means which transitions are forbidden and which transitions are allowed, we can look into it now. Not all theoretically possible electronic transitions are actually observed. That's what I told you because they are bound by a selection rules. Selection rules are there to distinguish between allowed and forbidden transitions. And allowed transitions occur. Forbidden do occur, but much less common and are of low intensity in nature. So now the first one is leopard selection rule. What is leopard selection rule? The transition should involve delta l equals plus r minus 1. If that is involved, they are called leopard allowed and have high absorbance. That means during electronic transition, azimuthal quantum number should change. It should not be, delta l should not be 0. This is about leopard selection rule. And then what is spin selection rule? During the electronic transition, electron does not change its spin. This electron is getting excited to the higher level. It should stay something like this. This is allowed. And then if this electron, if it goes like this, this is not allowed. So that means what it says is delta s equals 0. That means during electronic transition, electron does not change its spin. If the upward spin is there, it should go and sit in excited level, something like that. It's not allowed to go and sit like this. This is exactly opposite to NMR selection rule where in order to see nuclear spin transition, flipping of nucleus when you are applying radio frequency in a direction perpendicular to the applied magnetic field, the frequency of that one matches the present frequency of the nucleus, then what happens? It would start wabbling and coming out from more and more coming out of the axis of magnetic field and its flips. So there, if you consider in NMR, delta s equals plus or minus 1. One should remember, whereas in case of electronic transition, it is delta s equals 0. In nuclear transition, it's opposite. Upward spin will become downward spin. So here, the spin state doesn't change. That means d d transits are strictly speaking leopard fermedon. That means we should not anticipate any d d transition at all because it says delta l equals 0. And all d rs have the same delta l value. And of course, since they are leopard fermedon, that can be seen from the epsilon value. Molar absorptive coefficient will be in the range of 5 to 10 liters per mole per centimeter minus. When transition metal forms a complex, then why do we observe d d transition? When transition metal forms a complex, M is surrounded by ligands. Mixing of d and p-arbals may occur. And as a result, transition is no longer d d in nature. Of course, you can also recollect from Welles-Born theory about hybridization in case of octahedral complexes. When octahedral compound is formed, the hybridization we are giving is d2 sp3 or sp3 d2. So that means s and 3p and 2d-arbals, dz square and dx-my-square will combine to generate a set of six hybrid-arbals, which are having the properties of all the ligands which are involved in this hybridization. So as a result, what would happen? The d-arbals present in sp3 d2 are no longer d-arbals we observed in free gaseous metal ion. So that means mixing of d-p-arbals may occur as a result, transitions are no longer. When that is split into t2g and eg, they are no longer d-arbals of same property that we saw when it is not kept in the ligand field. So now, because of the mixing, they lose original, so d character. As a result, what happens? It becomes leopard-alloyed, and hence we see d-d transition. So because of slight relaxation in leopard rule, d-d transitions are observed. Mixing is common when complexes lack centrosymmetry. Mixing is even more pronounced when they lack centrosymmetry. So now, let's look into tetrabromomanganate 2 minus. This is tetrahedral. So manganese in plus 2 z is a d5 system. And then, if you look into pentamin chlorocobal 2 plus is octahedral with unsymmetrical substitution. And both the compounds are colored. This compound is also colored. This compound is also colored. So in case of MnBr4 2 minus, if you consider, and it's a tetrahedral compound, we have something like this. So here, if you see d5 system, all the electrons have a spin like this. And here, if you promote one of this electron, it should go here, or here, or here. If it goes here, both the electrons will be having the upward spin. And that is not allowed. That's the reason this is spin-permanent transition. But nevertheless, in tetrahedral compounds, that is allowed. And you can see, they show color because they lack centrosymmetry. And same thing is true in case of this one. We have d7 here. So here, any of these electrons with downward spin can go. And transition can be seen. And here also, it doesn't have centrosymmetry. It's unsymmetrical substitution is there. These compounds are colored. But on the other hand, if you look into homolypti compounds such as hexamin-cobalt 3 plus or hexamin-copal 2 plus, they have centrosymmetry. No mixing of P and D orbits are observed. And as a result, they are not colored or very pale colored in nature. So then, why this color comes, despite having centrosymmetry? That means when the metal to ligand bond vibrates, so that the ligand spends an appreciable amount of time out of their centrosymmetric equilibrium position, as a result, small amount of mixing to occur and low intensity transitions are observed. So that means when they are vibrating with respect to the mean position, it so happens that all the ligands will be at a different position from the central metal atom. As a result, they come out of centrosymmetric equilibrium. And hence, this is essentially due to the mixing of P and D orbitals. And that results in low intensity transitions. So now, you can see the type of transitions. Spin-fibrant transitions will be having very less epsilon value, typically less than 1. And the example is hexa aqua manganese 2 plus. It's a homolypti compound having centrosymmetry. And again, having D5 electronic configuration, the spin-burburden, and also Leporte-Furburden, that is. And if you consider Leporte-Furburden and spin-allowed, they will be having epsilon value anywhere between 1 to 10 and then R 10 to 1,000. For example, if you look into hexa aqua titanium 3 plus, a D1 system here. So here, centrosymmetry is there. Octahedral compounds homolyptic, that's the reason this value is low. And then if you look into tetrachloronucleate 2 minus, this tetrahedral compound, centrosymmetry is not there. And then it is a spin-allowed. And here, the value is about 10 to 1,000. So it's slightly better. And in case of charge transfer, we can see 1,000 to 50,000, for example, manganate ion, potassium per manganate. This is essentially ligand to metal charge transfer transition. So now, let's look into the system and type of spectra we will see. And also, when they are allowed, leopard and spin-allowed. So when a transition is leopard-allowed and spin-allowed, that has to be charge transfer transition. And they will be having very high epsilon values. Example, TiCl6 2 minus. And then, a partly-allowed and some PD mixing is there. We'll see DD transition. Here it is 500. And for example, if you take tetrahedral compounds of bromo and chloro. So tetra-bromo cobaltate or tetra-chloro cobaltate. And then, a leopard, a firmadon, and spin-allowed. We have quite a few. Here it is, again, termed as DD transition. Epsilon value is in the 10 to the 10, where we have centrosymmetry is there. For example, hexa-acquatatin M3-pes or hexa-covalidium 3-plus here. Then, partly leopard-allowed and spin-firmadon are those where we have typical D5 system. Epsilon value is very low, 4 here. And if it is spin-firmadon and leopard-firmadon, the value is very, very minimum, 0.02 epsilon value. In those DD transitions, very weak, and very weak, intense in nature. Intensity is very, very weak. And these are especially having D5 system having centrosymmetry. An example is hexa-acqua manganese 2-plus. Then, let us look into what are the factors that affect crystal-field stabilization energy and absorption wavelength. So, the absorption wavelength is directly proportional to the crystal-field stabilization energy we see. And that is governed by spectrochemical series. I have given here a very extended spectrochemical series here. Most of the ligands have included here. The magnitude of delta depends on the charge on metal. Metal measures in plus one state, plus two state, or plus three state. And also it depends on position in periodic table. That means whether it's a 3D series, 4D series, or 5D series. And also the ligand field strength. All these three factors have an influence on the magnitude of crystal-field stabilization energy. And here, increasing with increasing field strength so wavelength decreases. So that means as crystal-field stabilization increases and homo-lumo gap increases, as a result, what happens? High energy is required, wavelength decreases for those things. And you can see here, in case of octahedral, the energy gap between homo-lumo is very high. And that is slightly less in case of tetrahedral. And here, in case of sway planar, we are considering this transition here, dz squared to dxy level. So now let us consider metal to ligand charge transfer transition. If the metal is in a low oxygen state, that means it is electron rich. And the ligand possess low-laying, empty orbitals. They should have pi star in case of aromatic ligands, or carbon monoxide, or n-heterocyclic carbene, or sigma star in case of phosphine. Then a metal to ligand charge transfer transition occurs. We should remember here, metal is in low oxygen state. When the metal is in low oxygen state, it has more electrons. It is almost vulnerable for oxidation. That means it can readily give electrons. So those metals, when they are combined with ligands having low-laying, empty orbitals, like pi star or sigma star, then a metal to ligand charge transfer is smooth. And you can see MLCT here. MLCT transistors are common for coordination compounds having pi acceptor ligands. Whatever the back donation we call it, metal to ligand back donation, it's nothing but metal to ligand charge transfer. Upon the absorption of light, electrons in the metal orbitals are excited to the ligand pi star orbitals. So metal pi orbitals like DXY, DYZ, and DXZ combine with pi star to generate a set of molecular orbitals, bonding and anti-bonding. And this bonding will be occupying the electrons taken from the metal. This we call it as metal to ligand charge transfer, or metal to ligand back bonding. So metal to ligand charge transfer transitions result in intense bands. Example, if you consider TRIS 2,2-bipridel ruthenium 2-plus, this orange-colored complex is being studied as the exact state resulting from this charge transfer, which has a lifetime of microseconds. And the complex is a versatile photochemical redoxyre agent. So other examples are tetracharmonyl phenanthraline tungsten, and also 2,2-bipridel tricharmonyl iron compounds. If you, several examples you can find, which exhibit metal to ligand charge transfer transition. So let's learn more about charge transfer transitions in my next lecture. Until then, have an excellent time. Thank you.