 Hello everyone. I once again welcome you all to my last lecture in the series of interpretive spectroscopy. Today, let me summarize and conclude about whatever we discussed about various the spectroscopic aspects, problems, and conditions, definitions, and all those things in the last 59 lectures. So this course is all about spectral interpretation and elucidation of the structure of known or unknown samples to understand about spectroscopic methods and their potential in chemistry and other science subjects. So general process for structure elucidation of an unknown sample, we have several avenues here. Of course, before that we have to ensure that we have a pure compound even by physical measurements such as melting point, boiling point, and all those things we should be able to tell whether the compound is pure or not. And first of all our objective is to make the compounds in their purest form and then the purity can be assessed and also the identity can be found out from the spectral interpretation by taking various spectra and always to ascertain the purity and the nature of the molecule, we should always go for more than one type of spectroscopic or analytical means. So then when we have pure compound, we have several options as I mentioned, CHN analysis is there, elemental analysis, mass spectrometry, NMR, and they give you some idea about molecular formula. Before that, once we know whatever we get from mass, mass fragment, molecular ion peak from that one using rule 13 and also hydrogen deficiency index, we should be able to identify. Tentative formula can be identified and then we can identify functional groups using these methods and also substructures again using NMR and eventually X-ray if it is a solid compound and we could be able to get the single crystals. And then molecular formula can also give you about saturation and saturation and also other groups, et cetera. And then we arrive at possible structures and then we have to write all possible isomers if there are any and then again go to the mercy of NMR, mass and IR to identify fragments and also different positions of the chemical shifts and also stretching frequencies and we can write most possible structures and we can confirm it. And then known molecular formula, if anything is there, if the compound is already made or something, we can compare the properties. So this is how the entire process takes place involving more than one spectroscopic means. We use different type of radiations of electromagnetic waves in different instruments. All these things are given here and of course corresponding energy in frequency is also given and also temperature of bodies emitting wavelengths is also given here. It's a very useful electromagnetic spectrum here. For example, the radio waves, microwaves, infrared visible, ultraviolet, x-rays and gamma rays. Of course gamma rays also comes in mass spectroscopy. So now what is important is approximate time scale for structured determination with various techniques. This is very, very important because most of the molecules will be under dynamic process and also if the time scale doesn't suit, then we may not be seeing the information what we are looking for. Electron diffraction 10 raised to minus 20 up to whatever the dynamics that happens up to 10 raised to minus 20 can be analyzed at x-ray 10 raised to minus 18, UV 10 raised to minus 15, visible 10 raised to minus 14 and IR and Raman 10 raised to minus 13 and ESR is coming down 10 raised to minus four to 10 raised to minus eight and NMR 10 raised to minus one to 10 raised to minus nine and first kinetics 10 raised to minus three to 10 to 100 seconds. And then physical separation, they should be stable for more than 100 seconds. Then by looking at the morphology, we should be able to pick and separate the isomers if they have different morphology. This is very, very important. And saying, especially in NMR time scale, if the dynamics is beyond this scale, what happens in order to see NMR for those, we have to either speed up or slow down the dynamic process. For example, that's the reason we go for lower temperature or high temperature NMR studies. And this periodic table shows NMR active nuclei having non-zero nuclear spin. And of course, this color index is very useful in identifying. So given in orange all have I equals spin one and also red I equals spin half. Many of them are say, you can see red one H known ones are 13 C, 31 P, 19 F and all those things, even platinum, tungsten and all those things. And then three by two, we have quite a few. Three by two are there. Even boron, there is one three by two and also one three is also there. And then five by two like this. This is a very useful periodic table with color code for identifying nuclear spin having different values. And then nucleus with negative magnetic ratio will have highest energy for the most positive M values. This you should remember. Let us look into the effect of magnetic field on nucleus in a more classical way. So for this one to understand what would happen. For example, nucleus, they behave like tiny bar magnets. When they are subjected to magnetic field, what happens? Some of them will be aligned with the field and some of them will be opposing the magnetic field. Nevertheless, because this induces a motion because of this induced motion under magnetic field, what happened? They will start possessing with respect to the applied magnetic field, but they would never align with the magnetic field and they'll be at an angle. And then how we move this angle in such a way that we flip it once when you flip it, what happens? We can see nuclear transition has occurred. So in order to do that one, what we do is we apply another magnetic field perpendicular to the applied magnetic field here. So that should have a frequency corresponding to the precision frequency of this nuclei under the influence of the major magnetic field here. And when the resonance occurs, it will be flipping. So this is all about NMR. So this how? If you see this is the axis here, this is the direction of applied magnetic field. They are never aligned, but they will be precision with respect to that one here. You can see that one. And then how we can remove, move this one away from this one, increase the angle so that up to comes here and it will flip. That means we can say that nuclear transition has occurred. We know that delta s equals plus or minus one here. And then the frequency with which they precis about the applied magnetic field is called normal frequency here. So the orientation is not allowed by quantum mechanics. That means aligning is not allowed. They can never align with the magnetic field. That means they can always precis with an angle here. Then precision about B naught here. And then when it is precising, we are applying another magnetic field in a direction perpendicular to the applied magnetic field whose frequency should be the frequency of the precising nucleus. So in that case, what happens? The flipping happens. And then it will, flipping means it's basically nuclear transition happens. That's what we say here. And then we can correlate this one. Since omega equals angular, so 2 pi nu. So then you can say nu equals gamma by 2 pi into B naught. This we should remember. This is a very, very important equation here. Nu equals gamma over 2 pi into B naught. Gamma over 2 pi into B naught. Gamma is garomagnetic ratio. This is how it is related. And gamma is constant here. 2 pi is constant. That means nu is directly proportional to the applied magnetic field. That means increase in the magnetic field increases the frequency, decreases the normal frequency. So this is how in the absence of the magnetic field, they have random orientation. The moment you apply magnetic field, what happens? They will try to align with the magnetic field or opposing the magnetic field. And then you can see very nicely in this cartoon how nicely it is flipping due to the applied frequency in the direction perpendicular to the applied magnetic field in the form of radio waves. So then this equation is very, very important. The energy difference is proportional to the magnetic field strength as mentioned. And this nu equals gamma H over 2 pi into B naught. Of course, H goes nu equals gamma over 2 pi into B naught is the very important one. With this one, most of the problems related to nuclear magnetic resonance can be understood. And this is the constant for each nucleus. And for hydrogen, it is 26.753 radian per tesla per second. That means here in a 14,092 Gauss field, a 60 MHz proton is required to flip a proton. MHz photon is required. That means what we're applying here when the hydrogen is kept in this magnetic field, we are applying a magnetic field perpendicular to this with a frequency of 60 MHz to see electronic transition. That's what it means. So this is its normal frequency. And also the applied magnetic field strength is also 60 MHz. This is the low energy radio frequency we are using. One major difference between EPR and this one is, so here magnetic field is kept constant and we are using radio frequency, we vary. Whereas in case of EPR, microwave radiation is kept constant and magnetic field strength is varied. That's the major difference. Otherwise, more or less they are very similar when it comes to selection rule, everything. And this table is very important. This gives about different nuclei and their spin, natural abundance, and magnetic ratio and also the corresponding NMR frequency. And then the two energy states. So because of this one, when the transition is there, we call alpha state and then it is beta state, we call it. And then the magnetic fields of the spinning nuclei will align either with the external field or against the field. And a photon with the right amount of energy can be observed and cause the spinning photon to flip. As I said, that is in the range of radio frequency. And then as I said, this energy increases with the magnetic field. So you can see here with the increase in the magnetic field strength, the space between these two will also increases. And then to look into the multiplicity of the chemical shifts, we use N plus one rule here. For that one, we can use this Pascal triangle very readily and we have Pascal triangles different for different spin system. This is for i equals half. And whereas this one is for i equals one and also we have for i equals three by two. So you can see here. So these things we can be used appropriately when we are dealing with nuclei of different i values. So this shows about how a typical AB turn it into a A2 system here with variation as the chemical shift difference and the coupling constant vary. A typical AB system will become A2 system here. And then this is how one should be able to find out the chemical shift and coupling constant in a AB system. If you label them as delta one, delta two, delta three, delta four, this midpoint is the chemical shift and the spacing between delta one, delta two or delta three and delta four is called the coupling constant here. Then I have shown a typical compound where non equivalence is there because of heterosubstitution and adjacent carbon atoms. Here the second order comes into picture. They are called second order spin system. And then here how a ABC system, here you can see ABC system with increase in the field strength, they become almost like AB, AMX system or sometime A2-X2 system, depending upon how much difference is there in the chemical and magnetically equivalence here. And these things I've already discussed in detail. You can go back to respective lectures. And then 31p NMR is very useful for studying reactions and reaction mechanism. And especially when we are using phosphins in homogeneous catalysis for different type of organic transformations because they have very different chemical shifts, 100% abundances there. Those quantity of sample is good enough. And also there's no need to use even deuterated solvents that way in whatever the reaction we are doing, we can take the liquid and we can check, we can do continuous reaction, we can do batch reactions and analyze the intermediate through variable temperature or time dependent NMR reactions. As I said here, you can see here, they have very distinct chemical shifts for P3, P5 and various phosphonides, phosphonides, chlorophosphines, et cetera. The extensive list is given here. And also you can also analyze what would happen to donor and acceptor properties simply by looking into the chemical shift positions here. And this is one interesting 15N NMR for this cyclosporine. Here we have 11 different type of nitrogen atoms are there and this beautiful spectrum was obtained in just one hour and because the enriched one, you imagine this is not enriched one. We have very minute percentage of 15N in nitrogen, rest is 14N. And if you want to identify these things in a typical molecule that's not enriched, it would take 10 years to get this kind of spectrum. So you can see how easy to obtain, provided we have right kind of chemicals for labeling and enriching with the required isotopes here. This is one beautiful spectrum of 199 mercury coupled to fluorine atoms. You see, you can see this mercury is coupled. This is a very symmetric and centrosymmetric molecule here. These two fluorines are coupling first to give a triplet and then these two will split each triplet into triplets of triplets. And then we have CF3 groups are there, six are there. They split each line of triplet of triplets into separate and we get this beautiful spectrum here. This is a triplet of triplet of separates. Triplet of triplets of separates, beautiful spectrum here. And then this one is another interesting molecule where we have one, two, three, four, five different type of NMR active nuclei here. I just have shown 31PNMR here how it looks like. It shows 48 lines here. It's a triplet of doublets of doublets of quadrates. And this beautiful spectrum would look like this. Very nice one. So some of these molecules would be very interesting to look into multi-nuclear NMR properties. So then let's move on to UV visible spectroscopy. So this is the very, very important Beer-Lambert's equation we are using here. So a longer path length through the sample will cause more UV light to be absorbed. The greater the concentration of the sample, the more UV light will be absorbed. And UV is the spectrum consists of absorbance A on y-axis and wavelength on the horizontal x-axis. This is a very typical spectrum would look like. And then different type of electronic transitions are shown here. Sigma to sigma star is the highest one. And then pi to pi star. And then N to pi star. How this energy level would change with by changing the functional groups and also having conjugation and all those things we have already discussed here. This we see routinely in all canes. This we see in carbonyls. And this we see in unsaturated compounds pi to pi star. And this one N to sigma star in O, N, S and halogen compounds N to pi star in carbonyls. And then here the total angular momentum quantum number again can be compared very nicely to what we saw in case of nuclear spin in the magnetic field. It's slightly different here. The total angular momentum J is the purple one. Purple one, this is a combination of orbital in S blue and then the spin. So this is apart from going in a orbital surrounding the nucleus. They'll be spinning with their axis with respect to that axis. That is the spin. And then L, the combination J is L plus or minus S. So this you can see very nicely in this color. And then of course microstate and also the term symbols are very important. For microstate we use the equation N factorial or R factorial N minus R factorial. N is the capacity of the sub shell. So D 10 electrons, you have 14 electrons. That's 14 factorial P, it is 6 factorial. And then this would represent the number of electrons. D 7 means it is 7 factorial into 10 minus 7 factorial. And the numerator we have 10 factorial. So this is how you can find out the microstates. And then in case of electronic spectra, electrons may be promoted from one level to another one. That's what exactly happens, electronic transition. So during electronic transition, what happens? The low energy vibration rotation transition will also occur. But the energy difference between vibration rotation is too close to be resolved in an electronic spectrum. As a result, what happens? That results in broadening. And hence the bandwidths will be in the order of 1000 to 3000 centimeter minus 1. So in a free gaseous metal ion DRs degenerate and no DD transients are observed. In a complex degeneracy is lost because of splitting. And also because mixing with the SNP orbitals. As a result, DD transitions are allowed. Strictly speaking, DD transitions are forbidden, they are put a forbidden. The magnitude of delta O depends on the nature of the ligands. And how that going to affect the energy of electrons and hence the frequency of absorption maximum. So this depends not only on the nature of the ligands and also nature of the metals and their position, whether it's 3D, 4D or 5D, and also the charge on them. Extent of splitting is related to the ligand positions in the spectrochemical series because of various donor and acceptor properties. That you know that pure sigma donors, pure sigma donors and pi donors and pure sigma donors and pi acceptors. And then as I mentioned earlier, all DD transitions can be simply classified into four categories, D0, D10. So no transition colorless. If at all color comes, it has to be charge transfer transition, D0 it has to be ligand to metal, D10 it has to be metal to ligand. And D5 is unique. Here it's leopard to forbidden, spin forbidden. As a result, what happened? Those compounds are not colored. If at all color is there in case of tetrahedral complexes, it is pale in color. And then other eight categories, D1, D4, D6, D9, one electron, one less than half field, one more than half field, one less than completely filled. They show invariably one transition. Whereas in case of D2, D3, D7, D8, two electrons, two less than half field, two more than half field, and two less than completely filled show invariably three transitions. In some cases in homolyptic molecules, because of closer space of two transitions, you may see just two transitions. That's what we see in case of vanadium hex aqua vanadium three plus. So for example here, I showed you mercury ciodate. Mercury is an intense system, brick red. Again, it is metal ligand charge transfer, KMNO4 or potassium dichromate. They're 0 valence state, D0s, intense purple or orange. Color is there. That's again ligand to metal charge transfer. In case of bismuth triiodate, orange red, it is metal to ligand charge transfer transition. In case of prussian blue, it is metal to metal charge transfer transition. So since iodide has a very high polarizability, which results in the ionic charge getting easily transferred to the Hg2 plus cation. So this process releases some energy which falls in the visible spectrum. As a result, they appear brick red in color. Hence we say that compounds like KMNO4, mercury, iodide, et cetera are highly colored. So then selection rules are very important. Electronic transition may be classified as intense or weak according to the magnitude of the epsilon, absorptivity coefficient that corresponds to whether they're allowed or forbidden transitions. Allowed transitions, these transitions have an E maximum of 10 raise to four and probability of their occurrence is very high. These are generally due to pi-pi star transitions. And for example, 1-3 butadiene shows absorption at 270 nanometer and then epsilon is about 20,900 represents an all-over transition. What are the forbidden transitions? These are usually related to N2-pi star transition. For these transitions E maximum, epsilon maximum is generally less than 10 raise to four. And then N2-pi star transition of a saturated aldehyde or a ketones exhibit a weak absorption of low intensity, nearly about 295 nanometer and have the value of epsilon maximum less than 100 is a forbidden transition. So then selection rule, it says delta L equals plus or minus one have high absorbance and spin selection rule delta S equals zero. That means during electronic transition they should not change their spin. The upward spin should go to upward spin only. Whereas in case of nucleus, upward spin will become downward spin. So that's the difference. And then DD transits are strictly speaking leopardia forbidden. Because of mixing what happened, they lose their identity and hence we see DD transition. Nevertheless, if you look into leopard delta L value, it is 5 to 10 liters per mole per centimeter, very, very low. That itself indicates these transits are strictly speaking leopardia forbidden. So when transient metal forms a complex, M is surrounded by ligands. Mixing of D and PRs may occur and as a result transitions are no longer true DD in nature because of a slight relaxation in leopard rule DD transits are observed. And also in case of octahedral complex or highly centrosymmetric complexes, what happens when the ligands are vibrating and often they come out of their mean position. As a result what happens, when they come out of mean position, the centrosymmetry will be lagged in those things. As a result what happens? Mixing would occur and they show transitions. For example, tetra bromomagnet is a tetrahedral is colored and then this one heterosubstitute pentamin chlorocobalt is a tetrahedral with unsymmetrical substitution, they are colored. Whereas this one if you look into homolyptic, this one hexa aqua copper or hexa aqua cobalt has centrosymmetry and no mixing of P and D orbits are there and they are not colored. However, 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 do occur and low intensity transits are observed. So that means this is how you can explain still if they show pale color instead of not showing any color at all. And then in case of IR, this is very, very important. How this stretching frequency is related to stretching force constant and the reduced mass of those two atoms involved in a bond. So that means the vibration motions and frequencies of a structure containing several balls of different masses connected by springs. They are not like rigid one. They are almost like connected to a spring and just if you just tap another one they will be moving around. So it shows the vibration something like this. And then here the stretching frequency is related to force constant and reduced mass in this one. This is an ideal equation simplified equation. One can use comfortably to find out if the stretching frequency is given force constant can be found out very easily because others are constant in a diatomic molecule. We have to find out mu is reduced mass. This is M1 M2 by M1 plus M2. For example, if you see here what would happen when they are substituted once. Okay, here observed here is 2150. Here IR inactive. And also they show different type of vibrations wagging and twisting. The twisting mode produces no change in dipole moment and hence IR inactive in symmetrical modes here. And they show different type of vibration motions. This is symmetric stretching and anti-symmetric scissoring, rocking and wagging. So these are all symmetric one. These are all anti-symmetric ones. As I mentioned, this is the equation one can use comfortably. This is the simple derivative from Hooke's law. How they are related. This is another simple equation. Stretching frequency is related to 130.3 into square root of 4 mu. F is force constant, mu is reduced mass. And then here, for example, all the three equations are there. You can always examine because everything is given here. Reduced mass is given. Force constant is given. Frequency is given. Simply apply and then verify. This is for homework or assignment you can keep doing. And of course some of them have shown here. You can just look into it. Couple of examples as I included here for C double bond C also here. And now when it comes to IR carbonyl compounds are very, very important. And carbonyl compounds, you know, that it forms a sigma bond through the lone phase from carbon and giving to appropriate metal orbitals is the D z square or DXY square. And this is the anti-bonding. Pi star would interact with pie symmetry orbitals from the metal, DXY, DYZ or DXZ to generate anti-bonding and bonding. And the bonding will be populated with electrons from the metal that we call it as back bonding. And the same thing we call it as in terms of a spectroscopy charge transfer. These two modes of bonding are mutually reinforcing this metal to carbon bond that is called synergistic effect. And charge removal through pie bonding leads to more extensive sigma bonding and will charge donated through sigma bonding that facilitates further back bonding. So that means sigma donation will make electron deficient and back donation makes electron rich and they'll be acting in tandem. This is called synergistic effect. And then it's very interesting to compare the stretching frequencies in these homolyptic molecules of having different electronic configuration, D 10, D 6, et cetera. And then if you notice here in case of nickel tetra carbonyl, stretching frequency is quite high. That means not much back bonding occurs here. Of course, this entire molecule service and back bonding only as a MO diagram clearly shows that there is no sigma bond. There is no carbonyl electrons donating to nickel to generate nickel to carbon sigma bond. Here it lacks unlike CRCO 6 where we have a well designated chromium to carbon sigma bond whereas here it is not there. So that's the reason these compounds are very volatile and very reactive and very sensitive compounds. They can readily decompose. In fact, N ICO 4 was made with an intention of getting pure nickel for catalytic purpose through decomposition. And this is how this CO bond can vary here. It can become more electron density goes to the pi star. It becomes almost like a ketonic carbonyl group. And then of course here I have shown for different substituted carbonyl groups four, five and three and having different geometric isomers, cis and trans. How many active IR bands are observed for three stretching frequency for CO? For example, here when you have three you can have facial or meridional. In case of meridional you will see three whereas in case of facial we will see two. And of course, these are all for homolyptic molecules. All else should be same L3, but if it is different then they also expected to show three bands here. And in case of this one we have C2V symmetry and we'll see four bands here. We have D4H, we'll see only one band. And here in case of square pyramidal relationship in octahedral C4 symmetry C4V, so we'll see three. So this how you should be able to identify based on the point groups here. So further elaborated for trigonal bipyramidal also and also for tetrahedral compounds here. How it varies when we substitute one or more carbonyl with other ligands. And this is about mass. Mass also, this pyrite table gives about natural abundance of different isotopes that comes very handy in characterizing the compounds through mass spectrometry. So here what we should remember is this is EPR. Now one should remember this equation very well. From that one we should be able to calculate G. G is very similarly, you can compare it to coupling constant we see in case of NMR. And then for a free electron this is 2.0023. And then of course here we consider only the J equals L plus S. Whereas in case of electron spectroscopy we consider both L plus S and L minus S. And we should remember for G H nu over beta H. H is the magnetic field strength. Here also with increase in magnetic field the energy associated with the electron spin also increases and it will go to higher frequency. This is a typical EPR spectrum is shown here. Here we have an electron here, odd electron. So then we have to use 2Ni plus 1 rule and they all of them have I equals 1. So it splits into nine of this intensity. You can see this beautiful spectrum here. And of course here I have not shown these things and hydrogen hyperfine splitting is not shown. This splitting whatever I have shown is due to four of them here with I equals 1. So then this say about hyperspin splitting you can see here first this electron will be split from these two into a quintet because 2Ni plus 1 if you use I equals 1 five will be there. And then we have four hydrogen atoms are there. Four hydrogen atoms with I equals half will split each one into quintet. We'll see quintet of quintets. We can see quintet of quintets here. This is a very, very beautiful, a hyperfine splitting spectrum, EPR spectrum of this anion here. And then for a radical the magnetic field is 38.1 G. The frequency of the microwave is 9600 megahertz. What is the value of G factor? G factor can be simply calculated. And here apart from B naught for G equals H over B naught is a constant that is calculated here. It comes around 71.4484 and then this value is given here. And then divided by magnetic field that you get directly 1.8 here. So this how you can calculate the unknown quantity using the simple equation here in case of EPR. This is again mass spectrometry. This gives various fragments and the corresponding mass. One can see very nicely here. And then of course, if you have hydrogen index deficiency we can find out first we can use 13 rule and then 13 rule will give you approximate molecular formula from molecular formula. Again apply index of hydrogen deficiency and find out the tentative structural formula. And then take the help of data obtained from various spectrometer to elicit the structure. And then isotopic patterns for compounds contained in different elements also shown here. This is very, very important. And also the ratio also one should remember how it shows in case of hydrogen and phosphorus and fluorine you see only one. Whereas in carbon 13 C is there 1.1 and nitrogen 15 N very small O17. And then here is 32 we'll see and 34. So small quantity again 35 and 37 we'll see here. In case of metals we see a patterns very unique for each metal. You can see something like this. If you have these things handy when you have the spectrum you should be able to identify even if the metal is unknown. And rule of 13 is very important as I said. And then take the mass divided by 13 and quotient whatever is there. Quotient will be the number of carbon atoms. Quotient reminder will be the hydrogen and then you can have something like this. And then for heteroatoms what happens if nitrogen is there you take out CH2 12 plus two and oxygen is there takes CH4 and then sulfur is there takes CH2H8 and phosphorus is there CH2H7. So that is 31, 24 plus 7, 31 and then if chlorine is there CH2H11 you take or if hydrogen deficiency is there you can also take CH3H minus. That means every CHL you take take out three carbon atoms add one hydrogen atom. And for bromine I can take out CH6H7 and 127 iodine you take out CH10H7. So this how you can adjust the formula to arrive at tentative formula. For example shown here and possible candidates with heteroatoms and nitrogen rule is also you should remember a molecule with even number of molecular weight must contain either no nitrogen or even number of nitrogen. A molecule with odd number molecular weight must contain an odd number of nitrogen. So you should remember these things when you're using rule 13. So let me conclude now. So far I had discussed several spectroscopy methods and several problems. And of course at some point of time we have to conclude the lecture series and I hope this was very useful and many of you are interested in taking exams, competitive exams to get into higher studies or doctoral work or some of you may be doing doctoral work and all those things. One thing I'm telling you this learning never ends. Learning never ends. And also for students learning is the profession. When the learning is profession your attitude has to be right. How to set your attitude to become a humble student is you have to have for this another five entities such as in this profession of learning we should have sincerity, honesty, dedication, discipline and determination. When you inculcate these five entities in your profession of learning you will be having a right attitude to achieve whatever you want and to make a mark in whatever the subject you just embrace for your higher studies and other things. With this I wish you all the very best. Okay you should remember be greedy for learning and be content with learning as far as you are a student. You should not think about learning. And also you can see here only one letter makes a difference between learning and learning. And in case of course when you earn when you are learning this of enough learning this L will disappear and will assimilate in this one so that you will be earning. So from this point of view I will tell you again be greedy for learning and be content with earning. I wish you all the very best. God bless you. Thank you so much.