 Okay. Hello everyone. Can we start now? Yes, sir. Yeah, I think many of you are still joining in. Yeah, let them come. I'll just set everything by then. Good afternoon, sir. Good afternoon, Ashutosh. Can you see the screen now? Yes, sir. Okay. So guys, today we are going to finish the transition elements, Okay, a little bit. We have discussed already, but we'll start from the beginning. We'll go a bit quick. And then we'll start with aromatic chemistry also. Okay. So aromatic chemistry, there is not much. We have to discuss about the carbines, intermediate, and then the reactions based on that. Okay. I'm a time and reaction and all. And there are various reactions in the carbozoleic acid and easter reaction that we have done already. That is also based on this. That also will discuss once. Okay. So few reactions we have already done. And like I said, carbines plus aromatic chemistry, so aromatic chemistry, we'll discuss about the substitution reactions on the engine ring. So all those things comes under aromatic chemistry and carbines. Okay, so we'll do that in that particular topic. We are starting with transition elements, D and F block elements. After that, we'll start aromatic chemistry. Probably we won't be able to do the entire aromatic chemistry today. So by next class, we'll be able to finish the portions. Okay. So heading right down transition elements or D and F block elements. Transition elements lies from which group? It starts from group three, right? And it goes still group three, two in between S block, P block. So D and F block elements lies between S block and P block. Okay. So this here we have the transition elements. So the first two group is S block, 13th P block starts from group 13. So here we have from three to 12. This is the first thing. All the elements you don't have to memorize, but the elements of third period you must keep in mind. Scandium, titanium, vanadium and all. Okay. Scandium, titanium. Good afternoon, everyone. Scandium, titanium, vanadium, chromium, zinc, copper, all these elements you must remember. Okay, the position, atomic number and all. It's important, right? So the first kind of question they ask from here is related with the general electronic configuration. Okay. General electronic configuration. General electronic configuration. So we have different, different elements, different, different series also we have. The first thing here you see the general electronic configuration for D block elements is N minus 1, D, 1 to 10, NS, 0, 1 or 2. NS, 0, 1 or 2. Okay. This is the general electronic configuration we have. Okay. You have to remember that which elements has no NS electrons. So for that I'll write down one by one here. You copy it down. They have asked question on this one. Okay. So the 3D series, the elements belongs to where the electron goes to 3D orbital. We call it as all those elements we call it as 3D series or we also call it as first transition series. Both are same thing. 3D series or first transition series. Okay. For these elements like Scandium, Titanium and all the electronic configuration for these elements we have argon, then 3D, 1 to 10, 4S, 1 and 2. 4S orbital is not empty here. You see there's no zero electrons possible for this. Similarly, the 4D series or we also call it as second transition series. Second transition series, the electronic configuration for this is scripted, then 4D, 1 to 10, 5S, 0, 1 or 2. Here the question that they have asked here that for which elements the 5S orbital has zero electrons. Okay. So write down for this one. For palladium, the atomic number is 46 and it has zero electron in 5S orbital. The electronic configuration is scripted, 4D, 10 and 5S, 0. This question they have asked once in the exam which element has zero electrons present in 5S orbital. Okay. The next one is the third transition series, transition series, which we also call it as 5D series. And for this, the electronic configuration we have xenon, XE, 4F, 0 to 14, 5D, 1 to 10 and 6S can have 1 or 2 electrons. Okay. Now in this the first element, for that the electronic configuration is important. Okay. The first element is lanthanum, L-A, lanthanum. So electronic configuration for lanthanum which is 57 is the atomic number. The electronic configuration is xenon, 5D, 1, 6S, 2. Right. Here the 4F orbital has zero electron and it is possible only for lanthanum, only for lanthanum. Okay. 4F orbital has zero electron. So these three, two, three electronic configuration you must remember. This question, like I said palladium question they have asked in the exam. So don't forget it. Okay. That's the one thing. Okay. After this we have fourth transition series also, which is 60 series. Okay. In which the electron goes into 60 orbitals. Okay. And that is not much important. Okay. Now, the electronic configuration we have discussed. Okay. There are few exceptions in electronic, in this electronic configuration. For example, for chromium and copper, you know, right. It has completely filled half filled orbital, d orbital and completely filled d orbital chromium and copper electronic configuration, you know already. Okay. The reason also, you know, what is the reason for that? We have discussed many times electronic configuration that half filled and fully filled orbital gain some extra energy. Okay. So that's one important thing here. Just a second. So then we'll finish this portion and then we'll start aromatic chemistry. Okay. Now what are transition elements? Okay. The definition of transition element again on this also they have asked question on this. Okay. So the transition elements are those elements which has incompletely filled d orbital. Okay. So there is a confusion in this that the block elements and transition elements are seen, but it is not safe. D block elements are all the elements which is there in the block, but transition elements are only those elements which has incompletely filled d orbital. So write down the next line. Transition elements are those elements. Write down like this. Just a second. Transition elements, the elements of d block which has which has incompletely filled, which has incomplete d orbital, incomplete d orbitals. Okay. So d 10 configuration, wherever we have those elements are not considered as transition elements. Okay. So for example, you see, for example, you write down zinc cadmium Z and CD and HG are not regarded as regarded as transition elements because when you see the configuration of zinc Z and the configuration is what it is argon 4 s 2 3d 10. Okay. Cadmium you see the same group actually cadmium is the electronic the atomic number is 48 for cadmium and it has crypton 5 s 2 4d 10 and HG the electronic sorry the atomic number is 80 xenon 6 s 2 5d 10. Okay. Now all these examples you see the d orbital has completely filled d orbital you see here 3d 10 4d 10 5d 10. Hence, these are not transition elements. This is one difference we have in transition and d block elements. So we can say all sorry we can say all transition elements are d block elements, but vice versa is not true. Got it. Now we'll see the general characteristics of these elements, general characteristics. The first one, the first one to write down metallic nature. The first one is metallic nature. Okay. All transition elements right down on trans all transition elements are metals all transition elements are metals and they exhibit most of the properties of metals, they exhibit most of the properties of metals. Next line. Next time right on it contains both metallic and covalent bonding contains both metallic and covalent bonding. Next line greater the number of unpaired electrons greater the number of unpaired electrons in d orbital greater the number of covalent bond greater the number of covalent bond example right on chromium molybdenum. And tungsten are very hard metals are very hard metals and why they are hard because maximum number of unpaired d electrons okay. Maximum number of unpaired d electrons and hence they can show maximum covalent bonding maximum covalent bond. Maximum bond they forms hence their strength is high and they are very hard metals correct this example you must remember another point you write down the second one is atomic radius radius okay. Atomic radius there's there's no regular train into this okay because of you know the weak shielding effect of DNF orbital and and the electronic repulsion also so both factor we have to consider right on in 3d series the atomic radius gradually decreases radius here radius decreases radius decreases radius decreases from. Scandium to manganese. Scandium to manganese in 3d series. I'm talking about 3d series here. Scandium to manganese. Then almost same from radius is almost same iron to nickel and then for the last two element the atomic radius increases from copper to. Zinc okay the reason is same effective nuclear charge and the electronic repulsion in the first case from Scandium to manganese the effective nuclear charge is dominating here. Z effective is dominating that's the size decreases here the Z effective and the electronic repulsion is almost same. Electronic repulsion is almost same and hence the size also becomes equal for the last two the electronic repulsion dominates. The Z effective and hence the size increases just a second this order we have also discussed in periodic properties last year if you remember. Right so screening effect. We have discussed that is nothing but the Z effective okay vertical row if you go from top to bottom in vertical row from top to bottom we know as we go down the group. Atomic radius increases okay in this only we have a concept of lanthanide contraction this is important. Lanthanide contraction okay write down in this. Lanthanide contraction right on to this due to presence of 4f orbital due to presence of 4f orbital. Which must be filled due to presence of 4f orbital which must be filled. Before the 5d series of elements before the 5d series elements atomic radius the atomic radius decreases the atomic radius decreases. From the expected value from the expected value as the 4f orbital shows weak shielding effect weak shielding effect okay weak shielding effect. This is observed elements cerium atomic number 58 to lew ttm lew atomic number 71 okay this from cerium to lew ttm the size decreases. This phenomenon is known as lanthanide contraction the decrease in size because of weak shielding effect of f orbital. Okay because of lanthanide contraction what happens the following pair of elements this point you write down they have asked this question in the exam. Because of lanthanide contraction the following pair of elements has almost same size following pair of elements almost same size. They have asked this question in the exam why zirconium and hafnium shows has similar size okay almost same size. Same reason we have for molybdenum and tungsten also okay almost identical in size this one the question they have asked zirconium and hafnium reason is lanthanide contraction okay. Now the next one third property we have is ionization energy we know the definition of ionization energy it is the energy to remove any from the outer most shell of an isolated gaseous atom isolated gaseous atom. Now you see in this one again like atomic size like atomic size it also shows irregular trend okay so it has irregular trend. For example you see the elements has scandium titanium vanadium chromium manganese iron cobalt nickel copper and zinc atomic radius order we have already seen and according to that we know. The size decreases from a scandium to manganese okay so when the size decreases. So tendency to remove the outer most electron is. Also less. The iron increases and since we are talking about the ionization energy of the ionization energy so first electron we are talking about so it is first ionization energy IE one. We know iron to nickel the size is almost same okay the size same it means they have also have same ionization energy IE one is also same. Copper to zinc the size increases hence ionization energy decreases okay size sorry increases ionization energy decreases as size increases we can easily remove the outer most electron hence ionization energy. Decreases it's important to understand the trend of ionization energy in a vertical column okay so in vertical column ionization energy here. Decreases from first to second member member. From first to second member ionization energy decreases the reason we already know size increases okay but the ionization energy. But the ionization energy of third member is more is more. Then the second member the second member and reason behind this is what the reason behind this is lanthanide contraction. Okay in third element third like member we have lanthanide contraction lanthanide contraction and because of that size decreases ionization energy increases for group three elements group third. This question also they have asked in the exam the order of ionization energy is maximum for a scandium then yttrium and then lanthanum this order you must remember it is very important for the elements of group three. But that's the same order the one two three what you said ionization energy of third member is greater than second member right yes but for group three it's the same order yeah is we have a reason for this according to that. The scandium see the first element is scandium right then yttrium and then lanthanum okay so here what happens the atomic radius. For these elements are almost same if you're talking about third and second member okay second and third member the atomic radius is almost same because when you go down the group see I'm talking about this I'll write down here. I'm talking about second to third member now okay so atomic radius is almost same right because the electron is going into the higher shell but at the same time we have lanthanide contraction so elements of second and third period will have almost same size atomic radius but the atomic number differs by what 32. We all we know here that the third elements the third member of this group. Will have 32 more proton correct so since the third element has more proton right that's why the order is this so reason is size is almost same so that is not the factor of ionization energy. Okay ionization energy sentences to lose electron lanthanum has 32 more protons than yttrium right that's why it is a bit difficult to remove electron from lanthanum order will be this for third group for third sorry for third yeah for third group so it is an exception we have here. It is an exception in all other groups what happens we are going fourth group so left to right also they're shifting fifth group left to right we are shifting so in that way the size also decreases if you compare. Right so because of all these things size number of protons the observed ionization energy for group three is scandium yttrium lanthanum. Lanthanum daughter is this I'll give you one general trend here okay like I said there is usually when you consider what the atomic size as you go down the group only you consider atomic size then what should be the ionization energy. As you go down the group atomic size increases so ionization energy should decrease if you are considering only atomic size but that is not the only factor here. Here we have two three more factors like I said. Lanthanite contraction shielding effect okay electronic repulsion many factors are there so all these factors collectively give you a fixed order of ionization energy for various groups so in general I hope you will understand this now if I explain you this point just give me two minutes in engine. If you see this here all those factors that I told you if you consider those factors the order of ionization energy is given for group three elements in general what we say the third 3d series elements will have maximum ionization energy then 4d and then 5d which is nothing but a scandium yttrium and lanthanum. Okay but if the elements belongs to these groups 4 5 6 and 10 if the elements belongs to these group the observed rate observed order of ionization energy it is observed for 5d to be maximum then 4d and then 3d it is just reversed okay but if the element belongs to this group group 7. 8 9 11 and 12 so for these one the elements of 5d series again we have maximum ionization energy then we have 3d and then we have 4d so based on all those factors which is you know theoretically it is very difficult to explain those. Okay but the observed order is this and that is the collective effect of all those effects. Got it so what you have to keep in mind you have to keep these orders in mind for various groups it is very important because you know it is the irregular trend we have so they ask this question in the exam and any other need any other regional exams also they ask this questions. Got it. Sir. Yes. Yes sir. Understood sir. Yes sir. Got it sir. Yeah the first ionization energy the first ionize one more order you write down here first write down this ie3 just a second ie3 the third ionization energy. Write down ie3 for magnies is very high this question also they have asked ie3 for magnies is very high because can you tell me the reason anyone why magnies has very high ionization third ionization energy anyone. Sir because it will have when first two electrons will be removed then half filled. Half exactly half filled 3d configuration correct so that is the reason write down for magnies the third ionization energy is very high because electron removes from the very stable half filled configuration very stable half filled configuration. Okay one last thing you write down for ie1 ie1 the 5d series elements belongs to 5d series will have higher first ionization energy then 3d and it is even higher in case of 4d elements also for ie1 the order is this. This is actually the conclusion of these two. Okay this order we can write except for group three. So group three is has exceptional order. That's why this order they have asked many time in the exam. This order. Scandium, yttrium and latin. Got it. Next one write down melting point and boiling point melting point and boiling point. Write on quickly due to presence of unpaired D electrons due to presence of unpaired D electrons. The elements has tendency to form covalent bond. And hence they have high melting and boiling point. Can you repeat. See basically one point you have to keep in mind more unpaired electron more will be the tendency to form covalent bond more covalent bond more melting and boiling point. Got it. So if they ask you which one of these elements has shows more has more melting or boiling point. You have to count the number of unpaired electrons. Okay, because more number of unpaired electron more will be the covalent bond more covalent bond more melting and boiling point. Okay. Now can you tell me the trend of melting point and boiling point left to write if you go what is the trend of melting point and boiling point. Does it increases up till manganese. And then it decreases. Very good. So as you go left to right first the number of unpaired electron is increasing. And after that the pairing starts. So number of unpaired electron again decreases. So the melting point and boiling point first increases with increasing atomic number right down. There is one exception into this. Okay. So the first one right down the exception manganese has manganese has five un unpaired manganese has five unpaired d electrons, but it exceptionally show lower melting point, but it exceptionally show lower melting point. So number of vanadium and cobalt then that of vanadium and cobalt. This is because of the complex structure of manganese. This one you must remember complex structure of manganese has five electron but lower melting and boiling point then when it even copper tungsten this question they've asked many times in the exam tungsten has highest melting point. 3410 degree Celsius tungsten has highest melting point 3410 degree Celsius. Next, next right down reduction potential standard reduction potential. Sir, did you say tungsten has highest melting point? Yes. Sir, could you repeat that previous exception? The exception I said tungsten, you wrote unpaired electron but it has lower melting and boiling point than vanadium and cobalt. That one, can you repeat that? Yeah, manganese has five unpaired electrons, unpaired d electrons, but it exceptionally show lower melting point than vanadium and cobalt. This is because of its complex structure of manganese. This is because of the complex structure of manganese. Okay. Next right down is standard reduction potential standard reduction potential standard reduction potential actually in solution right down in solution electrode potential depends upon depends upon the following factors. The first one is the enthalpy of sublimation enthalpy of sublimation conversion of solid to gas. The second one is ionization enthalpy ionization enthalpy. And the third one is hydration enthalpy hydration enthalpy. Hydration enthalpy we defined when the solvent is what when the solvent is water. If there is any other solvent, we write here solvation enthalpy enthalpy of solvation. Okay. The general term is solvation hydration we use when water we are using just a second. Now. See what is the total energy change for standard action potential. What is the total energy change first of all, you have a metal solid. Right. It converts into gas. Okay. And for this we required the enthalpy of sublimation. Now this gaseous metal. It loses one electron to forms an ion. And for this we required ionization energy one. And then with this gaseous ion. When put into water H2O it forms M plus aqueous which is the hydration energy that is delta H of hydration. So total energy change delta H total is equals to the sum of all three delta H of sublimation plus delta H of hydration. Okay. So actually in solution the standard action potential depends upon sublimation ionization energy and hydration. These three things. Okay. A smaller the value of delta H total for a particular ion greater will be the stability of the oxidation state. Okay. For any metal ion. Like suppose M plus. Okay. If delta H total is small is small then this oxidation state the metal the oxidation state of this metal is stable. Means suppose one metal can form two three oxidation state the oxidation state which gives minimum delta H total is the stable oxidation state for that particular. Write down this point is smaller the value of delta H total smaller the value of delta H total for a particular oxidation state in aqueous medium in aqueous medium greater will be the stability of oxidation state. So electrode potential is what the electrode potential measures measures all these energy. Okay. It includes the calculation of sublimation energy ionization energy delta H of hydration. Okay. So write down the electrode potential measures all these energy total energy change basically we are considering. Enthalpy ionization energy and hydration energy. Okay. That is the electrode potential. Now the one point into this lower the reduction potential lower the reduction potential means what more negative standard reduction potential. Negative standard reduction potential more negative standard reduction potential. More will be the more will be the stability I on we can write or oxidation state whatever oxidation state. It is actually the same thing electrochemical series we have done in electrode chemistry. Remember that. Yes, sir. Two points are important here you write down because on this you'll get theoretical questions. Okay. The metals which has positive reduction potential value the metals which has positive reduction potential value can displace hydrogen can displace high this point is important can displace hydrogen from dilute acids. As negative reduction next slide copper has negative reduction potential and hence it cannot displace hydrogen. One question you'll get on this incoming CET exam. Can you repeat? Huh? What? Copper has negative reduction potential and hence cannot displace hydrogen from dilute acids. Yeah. Next point. High negative reduction potential chromium has high negative reduction potential and it is unreactive. It is unreactive due to due to the formation of due to the formation of its oxide layer on its surface. This is also this question they have asked many times why chromium is unreactive due to the formation of oxide layer CR203 it forms. Formation of its oxide layer on its surface. Next one. Catalytic properties. Catalytic properties right down next. See we know many deep block elements. Can act as a catalyst in various reaction. For example, platinum. Platinum act as a catalyst for the manufacturing of what H2SO4 what is the name of that method? Formation of S2SO4. Contact process. FE we know is a catalyst in Haber's process. Haber's process. CO2 also behaves as a catalyst when we prepare oxygen from the decomposition of decomposition of KClO3. And I also behaves as a catalyst in hydrogenation of oil. This question they have asked many times platinum also they have asked many times in the exam. Okay. Now let's write down oxidation state. It is also important. Oxidation state. Just a second. I've discussed this one color formation is there. Yeah. Right on oxidation state. Transition element. Transition element. With exceptions of few. Transition elements. With exceptions of few. Large number of oxidation state shows large number of oxidation state. And why various oxidation status shows because it has incomplete d orbital. Okay. Incomplete d orbital. And we know NS orbitals also take part in the reaction. That's why both NS and N-1 d orbital the number of electron present into that. Because of that only it shows variable oxidation state. Okay. So the highest oxidation state I'll give you a small formula here. Highest oxidation state. For 3d series. Highest oxidation state. For 3d series is. N plus 2. N plus 2. Where N is the number of unpaired electron. N is the number of unpaired electron. N plus 2 is the highest oxidation state. For 3d series. And this is not applicable. For chromium and copper. Write down this. Not applicable for. In 3d series. Magnes shows. Maximum oxidation state of. That is important. One note next to write down. In general. The minimum oxidation state is shown by a transition metal. Minimum oxidation state. Is shown by a transition metal. Is equals to. The number of. NS electron. The electron present in s orbital. Is equal to the number of NS electrons. But in this also we have an exception according to this logic. Scandium the minimum oxidation state should be what. Scandium the minimum oxidation state should be 2. Because it has 2 electrons in NS orbital. But in general it shows. Plus 3 oxidation state. Minimum oxidation state for Scandium is plus 3. 4d series. This question they have asked many times. 4d series. The highest oxidation state. Is plus 8. Shown by ruthenium are you. Ruthenium. And for 5d series. The highest oxidation state we have again it is plus 8. Shown by osmium. Right down next as oxidation state increases. As oxidation state increases. The tendency to form covalent bond. Also increases. So could you repeat. As. Oxidation state increases. The tendency to form. Covalent bond. Also increases. So what is it for 5d series. Sorry. What is it for 5d series. 5d series I have told you know. Osmium has maximum oxidation state plus 8. Yes. Osmium. Osmium question they have asked many times in the exam. And overall in the in the prior table. The maximum oxidation is shown by any element is plus 8. That is osmium for 5d series and ruthenium in 4d series. Okay. One more very important comparison we have here. Light on next line. Relative stability of various oxidation state. Relative stability of various oxidation state. Of various oxidation state. Can be explained. On the basis of on the basis of the stability of stability of. D0. Sorry. D0. D5. Or D10 configuration. Okay. For example. Can you tell me which one is more stable here. Titanium 4 plus. Ti 3 plus. Mn plus 2. Mn plus 3. E plus 3. And Fe plus 2. Tell me which one is more stable. Ti 3 plus and. Ti 3 plus Mn 2 plus and. Ti 3 plus Mn 2 plus and. Fe 3 plus. Can you tell me the configuration of Ti 4 plus Ti 3 plus. Why Ti 3 plus it should be Ti 4 plus right. Ti 4 plus. Configuration of Ti 3 plus. If you draw it is 3D0 and 4S0. Yes. This is 3D1 and 4S0. Hence this one is more stable D0 configuration. Here Mn plus 2 the configuration is what? 3D5 4S0. And this is 3D4 4S0. Again D5 configuration is more stable. Fe 3 plus. Again Fe 3 plus has 3D5 configuration. 4S0. Fe 2 plus we have 3D6 configuration. 4S0. D5 is more stable. Okay. Magnetic properties you know. Paramagnetic and diamagnetic. Right on magnetic properties. Paramagnetic diamagnetic. Paramagnetic due to unpaired electron. It is weakly attracted towards the magnetic field. All of you know this correct. Mu effective if they ask you to find out the magnetic moment. Mu effective is root over of N into N plus 2. Bm board magneton when N is the number of unpaired electron. Okay. This formula is also important. Any of you has planning to write NEET exam? Me. Who is this me? Sushant. You are planning to write NEET also? Yeah just like that sir. Good good good. This question in NEET they have asked many times in NEET exam. Root under N into N plus 2. If you see the previous year questions whether it is in chemical bonding or in transition elements. This question they have asked many times. Calculate the board magnetic moment. Okay. Now next write down. Color formation. Color formation. Neat and covalent compounds. Ionic and covalent compounds. Of many transition elements are colored. And color arises because of the excitation of electrons. From low energy to high energy. Right on next line. The color of transition metal. Arises from. Transition metal ion. The color of transition metal ion. Arises from. Arises. Arises due to write down. Arises due to the excitation of electrons from. The excitation of electrons from. Lower energy. Orbitals to higher energy orbitals. See. In coordination compound also we have discussed about it. Color formation of coordination compound. Okay. So if you remember what happens there. There is a. Is splitting of orbitals. Correct. If you can recall. This is splitting of orbitals is there. So electrons when you know. Jump from low energy to higher energy orbitals. So this has to overcome this energy difference delta E. Correct. So the two set of orbitals which splits. There is some difference in energy between the two set of orbitals. So this energy difference. It actually lies in the visible. A region or range you can say. Delta E we can always write at C by lambda. So this and lambda falls in visible range. Hence what happens whenever the electron. Excites into the higher orbital. Okay. It absorbs radiation from visible reason. Whatever the difference in energy we have it absorbs. From the visible reason and jump into the higher energy higher orbital. Okay. And when it jumps it shows the complementary color color. Okay. Opposite color basically basically you understood this. So that energy requires right on in this one line. The energy requires for DT transition. Falls in the visible range falls in the visible range. Hence for excitation. It absorbs radiation from visible region. Absorbs radiation from visible region. And shows complementary color and shows complementary color. What is complementary color complementary color. Here we have V B G Y O. This we call it as Moon shell V. So suppose it absorbs radiation of violet color. Then it shows its the color you know appears will be the complementary of this violet color which is the opposite color of it that is yellow. If it absorbs blue color radiation color will be orange green color will be red. Okay. So opposite color or complementary color means the it shows the opposite color. Okay. So this wheel we call it as Moon shell V. So actually it is very difficult to you know understand logically what colors the compound shows because for that you should know the energy difference first of all. If you know this delta E then you can find out lambda and suppose if you find out lambda also you should know the wavelength of these colors. Then you can say okay take a lambda is green. So color shows will be red. So this is very difficult to calculate. That's why you have to memorize the color of various serve you know metal ions. You understood this. I will give you the you know table for that. What color shows of different metal ions will give you that before that you write down one notes here one note. Write down the note. Mn O4 minus O4 minus is colored is colored in spite of D0 configuration. Mn O4 minus is colored in spite of D0 configuration. And this color is not because of DD transition. This question also they have asked in the exam. The question was why Mn O4 minus shows color since it has D0 configuration. Okay. No electron present in the orbit Mn O4 minus is colored in spite of D0 configuration. It is not because of DD transition, but it is because of charge transfer between manganese and oxygen charge transfer between manganese and oxygen. Electron exchange between manganese and oxygen. So there's no inner transition. There's no DD transition into this. Electron exchange between manganese and oxygen because of that it shows color. The same reason we have for K2 CR 2 or 7 also write down the same reason we have for K2 CR 2 or 7. Similarly, one more point in this you write down the second point. What is the color of AGCL anyone? This is white. What is the color of AGBR AGCL is white correct. AGBR the exact color is pale yellow. GCL yellow. What about AGI? It is black. This four color we have. Okay. For this all these I and you see the configurator. No, sir. AGCL is yellow. AGCL. So what is the difference first and third? Just a second. AG2S I have written it wrong. 2S I'm sorry. This is AG2S. Tell me what? So the third color is the same. AGCL. This is also I have written it by mistake. This is AGI. AGCL AGBR is the AGI yellow. Okay. All these compounds you see we have AG plus ion right. And AG plus ion what is the configuration? Can you tell me? Outermost configuration. So D10. D10. Right. So it is 4D10 5S0 correct. It has completely filled the orbital. Right. So there's no DD transition in this. Again it is DD transition is not possible. So like Mn04 minus and K2CR 207. It shows because of the exchange of electron between manganese and oxygen. Here the color appears because of percentase ionic character. Percentase ionic character. Of these molecules because there is no DD transition. These two examples you must remember the reason of color for these ions. Okay. No DD transition. Still this shows color. The reason is percentase ionic character for this. And the first one it is the charge exchange of electrons between oxygen and manganese. Okay. Now like I said to understand logically what is the color of the given ions or metal ions. Okay. It is very difficult to find out. So you have to memorize the color of various oxidation state of a given element or metal. Okay. Write down here you should copy it down and do memorize it. I'll tell you which one is important in all this. Okay. So write down the color of different hydrated transition metal ion. Getting right down color of different hydrated transition metal ion. First one is green green. We have Ni plus two Fe plus two chromium plus three and vanadium plus two copper plus two cobalt plus three chromium plus two vanadium plus four. There are certain ions which are colorless. This is very important colorless. Okay. And this logically also you can understand except few exceptions that we have ZN plus two Cu plus scandium plus three titanium plus four if you observe for all these ions. There are no unpaired electron no unpaired electrons. That's why it is colorless pink. Pink is cobalt plus two yellow is Fe plus three violet is MN plus three and the last one purple is titanium plus three and all these the important one. We have this colorless. I'll just write ions. I'll tell you colorless all are important. Ni two plus is important. Fe two plus CR three plus CR two plus Cu two plus Fe three plus. Did you miss any of them by any chance because you have written two, three, four is missing. Where numbers are one, two, three, five, six, seven, eight. No, I just written it wrong. Okay. It is four. It is fifth. It is six. It is seven. So cobalt two plus I have given you are spring. Cu two plus is nine. Fe three plus is yellow. That's correct. Okay. These are the general properties of transition elements we have discussed. Okay. There are certain compounds. Okay. Now the leg next thing we are left with certain compounds and their preparation and properties. Okay. There are few tests also in this we have, but that test is, I don't think it's there. Did you see the syllabus transition elements labors? Have you seen that this year? No. Okay. Let it be. I'll just go through it. I'll just check it. Okay. One test is there. Chromite cry test and all whether it is given or not. Anyways, so write down. Preparation and properties of various compounds. Preparation and properties of various compounds. The first one to write down potassium dichromate to CR two or seven potassium dichromate. It is in short you write down there are many theories in this. I'll just give you in short. All those things. Okay. It is an orange crystalline solid. Orange crystalline compound. Paired by chromite or chromite or chromite or is F E CR two or four. Okay. I see this reaction. How it forms K two CR two or seven F E CR two or four. First of all, from F E CR two or four will obtain any two CR two or seven and a two CR two or seven. Okay. And this any two CR two or seven is allowed to react with potassium chloride KCL gives K two CR two or seven plus two molecules of NSEA. Right on next thing two steps are involved here for the preparation of sodium dichromate that is any two CR two or seven. Okay. The entire reaction. I won't write but how do we get this any two CR two or seven to step reaction that the chromite or we have which is F E CR two or four is allowed to react with and a two CO three. I'm not writing down the balance reaction in presence of oxygen. This reaction gives first sodium chromate and a two CR or four and a two CR or four plus F E two or three plus carbon dioxide. This is sodium chromate. The color is yellow. It is yellow in color. Now this sodium chromate we use to convert into sodium dichromate. Okay. The conversion of sodium chromate into sodium dichromate is done with the help of concentrated S two SO four. So any two CR or four when combines with concentrated H two SO four. It gives any two CR two or seven plus two SO four plus H two. This is how we get sodium dichromate from chromite or that is F E CR or four. And then this is allowed to react with KCL gives you K two CR two or seven. Okay. This is the preparation method we have. So important point here is sodium dichromate. So potassium dichromate is obtained from chromite or the formula of chromite or is F E CR. Very important this one. One more thing I'll tell you about this F E CR or four. It actually exists in this form F E O dot CR two or three addition compound of this. It is similar to that. If you remember F E three or four. How F E three or four exist F E three or four exist as F E O dot F E two or three. Yes. Are you there? Yes sir. Yes sir. Okay. Similarly F E because F E three or four they have asked many times in the exam. The oxidation is straight enough. Okay. This they may ask. Okay. Another thing you see this year J means they are they are going to ask five integer type question. Correct. Okay. So five integer type questions you must take care of this kind of questions like suppose they can give you this question and they can ask you if the oxidation is straight off I why what is the value of X plus Y what is the value of X by Y what is the value of X plus Y by two anything they can ask you like this in integer type. Are you getting it? Yes sir. Yeah. So try to figure out where they can ask this integer type questions. Okay. And my analysis is this year the question is going to be a bit tough tougher than the last year because logically if you think you are going to have 25 questions in each subject right 25 into three 75 questions you are going to have. So and the time is three hours time is same number of question is less. So obviously the level of question they'll increase this time. That's what I feel. Okay. So get prepared according to that. Okay. For integer type that will only make the difference why they have done this change because last year if you see 100 percentile there were many students. It's not like 100 200. They were in 1000 100 percentile to reduce this or to reduce the number of students to get the same percentile. They have include this five integer type questions. Okay. This five integer type actually decides your a percentile or rank. Okay. So you have to take care of this integer type questions how you are performing in this test especially this five integer type questions. Is it going good? Are you able to solve how many of you are able to solve all five integer type questions in chemistry. I'm talking about one test. Yes. In that in that test I think most of us got three four maybe right. Yeah. I don't think anyone got full. I don't know but I think three four was the average. Okay. Three four if you are getting it. It's fine. Okay. So keep that in mind. Okay. The integer type is very important. Anyways, coming back to this. Pression method or reaction I have given you. Okay. Reaction they won't ask. Don't bother about it. Just to keep that in mind from it or we'll prepare it to CR two seven from promote. Okay. Properties of this you write down. Properties color I have given you already. It is a orange crystalline compound. Next to write down the yellow color of potassium chromate K2CRO4. Where is K2CRO4? I've given you this. No. Okay. One reaction you write down here. That is K2CRO7. When you heat this K2CRO7. If you heat this it converts into two molecules of K2CRO4. Two molecules of this also we are using and they'll give you oxides of chromium CR2O3 plus three by two or two. Again, the balance reaction is not at all important here. Now this is potassium chromate. This is potassium chromate. Now this potassium chromate when it is okay potassium chromate and this is also or it is yellow in color. Okay. Like sodium chromate potassium chromate also yellow in color. Okay. So what we can say on heating the orange color K2CRO7 converts into yellow color potassium chromate. Even also K2CRO7 in basic medium KOH this again converts into K2CRO4 plus H2O. Okay. So this is yellow color and this is orange. So what you have to memorize again I'm telling you reaction. You don't have to memorize in basic medium the color changes from orange to yellow. Okay. Of K2CRO7 orange to yellow yellow why because potassium chromate forms. Okay. Now for this potassium chromate in acidic medium CRO4 2 minus. Okay. Write down on acidifying the yellow color of chromate potassium chromate or chromate ion on acidifying the yellow color of chromate ion changes into orange color means basically you see basic medium. This reaction goes in this direction when you do this in acidic medium it again converts back into potassium dichromate. Okay. So reaction you write down CRO4 2 minus which is yellow in color in acidic medium H plus this converts into CR2O7 2 minus plus H2O. So in acidic medium it again converts into orange color of dichromate solution. Okay. Now two three reactions they have asked in the exam and for this I'm giving you you have to memorize the reaction here like I said here the reaction is not important color change is important but here you have to memorize the reaction write down with HCl with HCl K2CR2O7 liberates chlorine with HCl K2CR2O7 liberates chlorine. Okay. Reaction you write down I'll write down the reaction here K2 reaction is K2CR2O7 plus 14 HCl. This gives you 2KCl plus CRCl3 plus H2O plus Cl2. So chlorine gas evolves into this. This question they have asked which gas evolves. Answer is chlorine gas evolves into this. Okay. When it reacts with HCl. Similarly when this K2CR2O7 reacts with Ki in acidic medium all these reactions are in acidic medium. The reaction you write down K2CR2O7 plus Ki plus H2SO4 it liberates forms K2SO4 forms CR2SO4 whole price. It also forms water plus iodine gas evolves into this. This is also important iodine gas evolves. Similarly with H2S it oxidizes H2S to sulphur. Okay. Can you draw the structure of K2CR2O7? Can you draw the structure of K2CR2O7? Anyone? One second. Sir, I sent a picture on WhatsApp. Chromium, 2 chromium atom attached with... Let me check. 2 chromium atom attached. No, Gaurav. The structure is this. There will be peroxide linkage. There will be peroxide linkage, right? No, no, no. CR2O7 2 minus is this. It's double bond O. Double bond O. O minus. Double bond O. Double bond O. O minus. K plus attached with this, obviously. CR2O7 2 minus. Okay. Yes, sir. In this only one thing you write down. That is chromide chloride test. Test. After this, okay. Chromide chloride test. Write down. When a mixture of a metal chloride and K2CR2O7 is heated with concentrated S2SO4, when a mixture of metal chloride and K2CR2O7 is heated with concentrated S2SO4, the orange vapors of chromile chloride evolve. The orange vapors of chrom... I think it is chromile chloride, not chromide chloride. Chromile chloride test. M-Y-L write down. Okay. Reaction write down. K2CR2O7 plus H2SO4 plus NaCl. It gives KHSO4 plus NaHSO4 plus 2CRO2Cl2 plus H2O. So this is... This is chromile chloride. We call it as chromile chloride. Now there are few results which is important for this. When this chromile chloride vapors passes through NaOH solution, means the reaction of CRO2Cl2 plus NaOH with NaOH solution it gives yellow solution of sodium chromate NaClO4 yellow solution of this plus NaCl plus H2O. All these are results you must remember, color change, yellow color of this. Na2CRO4 further, when it is allowed to react with lead acetate and hydride it, sorry. It forms PBCRO4 plus CH3COONA. So it forms a precipitate of lead chromate. It is lead chromate, yellow precipitate of it. One last reaction in this write down. Acidified K2CR2O7 reacts with H2O2. Acidified K2CR2O7 reacts with H2O2. Acidified K2CR2O7 reacts with H2O2. H2O2 to give a deep blue solution due to formation of deep blue solution due to formation of CRO5, oxides of chromium. This oxides of chromium, we know the structure of this, it is highly unstable. And hence in this acetic medium itself, it reacts and converts into, converts into sulfate, 2CR2SO4 whole thrice plus 6H2O plus 7. Okay, reaction is not that important. Structure of CRO4, we have already discussed, if you remember last year in Redox reaction, it has butterfly structure, if you remember this. Yes? Yes, sir. What is the oxidation state of chromium here? Can you tell me the oxidation state of chromium? Is it? One second. Plus 6, right? Yes, it is plus 6, very good. It's plus 6. Okay, so we will take a break and then we will start after a break, 2-3-4 compounds more we have to do preparation and properties, and we will finish this. Okay, we will start at 4-10, correct? Okay, sir. Yeah. So when is break till? 4-10. It's break over. Hello. Can we start? Sir. Yeah. Sir. Yes, sir. Okay. So write down the second compound, that is potassium permanganate, KMNO4. KMNO4, right? It is prepared by pyrolysite mineral. Sorry, pyrolysite ore, not mineral. It is prepared by pyrolysite ore. What is the formula of pyrolysite? It is manganese dioxide, MNO2. Reaction is 2MNO2 plus KOH plus O2. This gives you, first it forms 2K2MNO4 plus H2O. K2MNO4 is potassium magnet. We call it as potassium magnet. See, one thing you must know, manganese compound, when react with an oxidizing agent, like here, this oxygen is atmospheric oxygen and oxidizing agent. So manganese compound, when reacts with an oxidizing agent, it usually converts into plus 6 oxidation state. So further what happens, this K2MNO4 is treated with, this is treated with either chlorine or ozone or CO2. And it converts magnet into permanganate. So the reaction is 2K2MNO4. When treated with chlorine, it is also an oxidizing agent. This converts into 2KMNO4 plus KCl. Similarly, when it is heated with plus H2O plus O3, again an oxidizing agent, it converts into 2KMNO4 plus KOHO2. 3K2MNO4 plus CO2, it gives 2KMNO4 plus MNO2 plus 2K2. Can you tell me the oxidation state of manganese in K2MNO4? Plus 6. Plus in K2MNO4? 6. Plus 6. So all these oxidizing agent converts K2MNO4 into KMNO4, that is potassium permanganate. Next thing, a note you write down. In acidic medium, dichromate is stable and in basic medium, chromate is stable. In acidic medium, dichromate is stable, but in basic medium, chromate is stable. If you can remember with this A, B, C, D, it goes like acidic may dichromate and basic may chromate. A, B, C, D, in acidic medium, dichromate is more stable than chromate and in basic medium, chromate is more stable than dichromate. That's what it means. Some properties of this you write down. It is a purple color, purple color crystalline compound, purple color crystalline compound, converts, converts, this reactions are important, converts ferrous salt, ferrous salt, what is the oxidation state? Fe2 plus, ferrous salt into ferric salt. It converts ferric into ferrous into ferric salt, f e3 plus Okay? this, iodine, iodine from iodide KIE, okay. Sulfur also, H2S oxidizes into sulphur and quickly will finish this, can you draw the structure of 4 minus structure of mn, four minus or KMN of four. So one second. What's that sound? Sir, is it three oxygen atoms attached to MN or two double bonded and one is a single bond? No, two double bond oxygen, two single, two single bond oxygen, two single bond oxygen. Is it two single bond oxygen? That's the second. It is KMN of four minus, no? Sir, MN of four two minus, right? So it's three, only one single bond oxygen. I think it's correct. M double bond O double bond O double bond O and O minus. This is a structure. Okay. This is sp three hybridized manganese here. sp three hybridized. Okay. It also eliminates nitrogen and to evolve from hydrazine from NS two NS two. Generally they ask which gas evolves. Okay. Oxalic acid if you have from that it involves CO2 from oxalic acid oxalic acid. These two, three reactions are important. Write down one important point in this. MN of four two minus MN of four two minus. This is important. This is an information actually. Okay. MN of four two minus this one. MN of four two minus dilute alkaline in dilute alkaline, comma water and acidic solution, acidic solution is unstable, is unstable, disproportionates and disproportionates to give disproportionates to give MN of four minus and MN of two. This reaction is important. Okay. This question they may ask you in redox reaction chapter also. MN of four two minus acidic medium. Suppose I am taking four H plus it converts into MN of four minus plus MN of two plus H two O. So this proportionation takes place and it converts into MN of four minus and MN of two. Okay. This is it for potassium per magnet. Okay. Next to write down ferrous sulfate ferrous sulfate ferrous sulfate ferrous sulfate. We also call it as green vitriol green vitriol. It is also known as heracasis H A R A K A S I S. I'll write down again. It's not clear. All are same thing. Green vitriol heracasis ferrous sulfate H A R A K A S I S. Okay. It is formed by the oxidation of iron pyrite oxidation of iron pyrite. What is the formula of iron pyrite? FES2. FES2 reaction is two FES2 plus O2 plus H2O gives two FESO4 plus two H2SO4. This is the reaction. And hydrous form of ferrous sulfate which is FESO4 and hydrous form is colorless FESO4. Okay. There are few reactions. I'll write down this way. You also copy it down same way. We can finish it quickly then. FESO4 we have and when it reacts with KCN. KCN. K2SO4 goes out and it forms K4. FECN6 minus of K2SO4 on exposure of H2 plus O2 H2O plus O2. It forms brownish yellow solution of FeOH hydroxy sulfate. It is the brownish yellow solution on hydrolysis. On hydrolysis it gives FeOH whole twice plus H2SO4. Okay. This is the reactions of FESO4. There are other methods of preparation also. I write down this side the preparation method. One reaction we have already seen. The another reaction you see iron if you take Fe and it reacts with H2SO4. Hydrogen gas evolves in this reaction and it forms FESO4. Another one if you take sorry if you take FeCO3 on reaction with H2SO4 again H2O and CO2 goes out in this reaction forms FESO4. Now iron sulfide FES which is black in color black color of FES reacts with again H2SO4 dilute gives you FESO4. Okay. So reactions of FESO4 is not that important. Okay. From how it forms iron pyrite that you must keep in mind. Common ring you must remember. The most important reaction of FESO4. Can I go to the next page? One minute. So once again. Yes, sir. Okay. The most important reaction of FESO4 is the heating effect of it. Okay. And we'll take hydrated compounds right on the heading. Once again, can you go back to the previous page? Yes, sir. Okay. Now heating effect you write down. See the hydrated form of FESO4. They have asked this question many times. So this reaction heating effect is very, very important. Okay. FESO4.7 H2O when you heat this as 140 degrees Celsius even the temperature here also important. Six molecules of H2O goes out and it forms FESO4.H2O further if you heat it at 300 degrees Celsius and hydrous FESO4 forms and hydrous FESO4 forms and further when you heat this at high temperature higher than 300 degrees Celsius 500 600 like that it dissociates and it forms Fe2O3 plus sulfur dioxide gas forms and sulfur trioxide also forms. Okay. The color of this Fe2O3 is reddish brown. The uses of FESO4 it is used for the making of blue black ink. Okay. It is also used in pentons reagent if you remember pentons reagent that is FESO4 plus or with hydrogen peroxide H2O2 for preparation of more salt also use this more salt. Reaction is FESO4 plus NH42SO4 plus 6H2O gives FESO4 dot NH4SO4 dot 6H2O. This is more salt. Done. So this heating effect these reactions are very, very important. Okay. Must remember that even temperature also you must keep in mind. Next write down copper sulphate. Can we go a bit faster? Yes. So you write a bit fast now. There are many things left. Copper sulphate blue vitriol the formula is CuSO4 dot 5H2. Right. Blue vitriol we also call it as Neela Thotha. Have you heard this name? Neela who? Neela Thotha. Never heard sir. Never ever heard. Sir Ravya has a good story about copper sulphate sir. What? He got suspended I mean. Who? Sir I bought copper sulphate in 7th grade. There was this idiot in our class who ate it thinking it was brandy and because of that I got suspended. When? 8th or 7th or 8th. He had a surgery. He had a minor surgery right. He just puke and he came back some bullshit he was telling. Okay come here. Okay for interesting copper. It is a blue crystalline compound. Blue crystalline compound. Next line you write down. In this 4H2O molecules are ligands. 4H2O molecules are ligands while the 5th water molecule is hydrogen bonded with the sulphate ion. The sulphate ion SO4 2 minus. So you see one thing. 4 water molecules are ligand. So 4H2O molecules are bonded with what? Coordinate bond. Correct. Coordinate bond and the 5th one. The 5th one is bonded with hydrogen bonding which has a 4 2 minus ion. So here also you see if they give you this question like X water molecules are bonded with coordinate bond and Y water molecules bonded with hydrogen bond then what is the value of X plus Y? They can ask you this question in integer type. Got it? So you have to memorize this. Okay reaction you see here. How do we prepare CUSO4? There are 3, 4 reactions and for this again these reactions are not that important. In this also we have heating effect. That is the most important reaction. I'll just write down the preparation method of CUSO4 and it is copper when heated with H2SO4. It gives you CUSO4 in presence of oxygen gas. Water molecule goes out. It gives you CUSO4. Okay? The second reaction we have. CuO. CuO again reacts with H2SO4. Again, H2O goes out and it converts into CuSO4. Okay? Hydroxide of copper CuOH whole twice. All this reaction you see takes place with H2SO4. Okay? So in this also H2O molecules goes out. You'll get CuSO4. Okay? Like this we prepare CuSO4 copper sulphate. The reactions of copper sulphate, we have 2-3 reactions here. When it is heated with NaOH whole twice. Okay? When it is heated with NaOH, not whole twice, NaOH or we can also take Na2CO3. It forms the major product. The most important product here will write CuOH whole twice. It's not Ca. It's CuOH whole twice and the color of this is blue. CuOH whole twice, blue color of it. Okay? When it is heated with ZN or Fe, it gives you copper metal. Cu, solid. ZNSO4 or FeSO4 goes out. When it is heated with KiSO2 with H2O, it converts into iodide, CuI. Okay? Heating effect of CuSO4. Next page, CuI2 here I don't know. This question, CuI2, this question they have, I'm sorry. This question they have asked in JEE. This one a question. CuI2, JEE 2012 most probably. CuI2. Cu2 plus ion. See all these and I'll tell you one thing. I- in this right down, I- ion or you take Cl- or you take Scn- Scn- all these ion converts Cu2 plus to Cu plus. This particular question they have asked in JEE means. Heating effect you write down. The hydrous salt of this CuSO4.5H2O. When it is heated in presence of air, CuSO4 means normal temperature 4H2O. It forms at 100 degrees Celsius. 3H2O molecules goes out. CuSO4.H2O we get. Further if you increase the temperature to 250 degrees Celsius, it forms an hydrous form of copper sulphur and at 750, it dissociates and converts into oxide plus SO2 and all these temperature based reactions important. Okay? So green, blue, we have got 4H2O the one the water attached to hydrogen bounding leaves, right? How do we know which one leaves? You know that we cannot say any one of the water molecule we have and obviously the water, the first water molecule that goes out that you can say it is the one which is attached to the hydrogen bonding that goes out. Yes. Chances are high like that we can say. Understood. So next you write down zinc sulphate. Zinc sulphate, we also call it as white vitriol and for all these guys NCRT you must go through. Okay? They won't ask anything beyond NCRT. Okay? Line by line you have to study. White vitriol. The formula of this is ZNSO4.7H2O. Okay? It is a carol-less crystalline compound, carol-less crystalline compound, soluble in water, isomorphous. You know what is isomorphism? Soluble in water and it is isomorphous with write down. So all this is NCRT, sir. Huh? All this is there in NCRT. No, I am giving you some extra thing also in this, but for exam point of view NCRT at least you must finish line by line. Okay? Further if you have time you can go through any of the book. Okay? But I am not considering, I am not only focusing on NCRT thing. Okay? I am just giving you the other details also. But for preparation purpose at least NCRT you must finish properly. If you have time then you can go through the other book also. Okay? Isomorphous with Ipsum salt. This question also they have asked once. Ipsum salt. Okay? Isomorphism means the formula representation is same like MGS4.7H2O. It looks similar, right? ZNSO4.7H2O, MGS4.7H2O. This kind of compounds are called it as isomorphous compound. Okay? Formula representation is identical. Or so KCL, NACL and all those. So you can say isomorphous. Yes. So you see the reaction of ZNSO4 preparation method and reaction of ZNSO4. The first one what we can do ZN plus H2SO4 minus H2O. It gives you ZNSO4. We can also take all these are similar reaction. You see there we take copper oxide here we have ZNO plus H2SO4. Again H2O goes out forms ZNSO4. Carbonate also we can take for this reaction. ZNCO3 plus H2SO4. Okay? In this H2O and CO2 goes out. H2O and CO2 forms ZNSO4. Okay? The first reaction of this is very important here. NaOH the reaction with NaOH. In this it forms hydroxide ZNOH whole twice which is white precipitated forms which further when it is also in NaOH H2O goes out and it converts into Na2ZNO2. This is a soluble complex complex sodium zincate soluble complex sodium zincate. Okay? Here we have Na2SO4 goes out. Reaction with NaHCO3. H2O molecules and CO2 molecules goes out and it forms ZNCO3 plus Na2SO4. Now the heating effect of this ZNSO4.6 H2O. Okay? 7 H2O 7 H2O. When you heat this at 100 degrees Celsius one water molecules goes out ZNSO4.6 H2O it forms. Further you heat this at 200 degrees Celsius. It forms an hydrous form of ZNSO4 and on 800 degrees Celsius it eliminates sulfur trioxide forms ZNO plus SO3. Yes sir. Okay? Now last thing we'll discuss is photography. Sir, are we going to finish the chapter today? Yeah, yeah, yeah. We'll finish this. There's nothing in F block. In F block we'll have just two, three properties like general electronic configuration and all. There's no compounds into that. Just two, three properties. General electronic configuration we have lanthanide. What is the first element, last element? Actinides. How many elements are there? Some general electronic configuration and that is it. Nothing much. Okay? So write down next photography. Photography pay also they have asked questions. What is photography? It is an art of obtaining the exact impression, impression of an object, of an object through a chemical reaction, through a chemical reaction initiated by what? Initiated by light. So all these are photochemical reactions. Initiated by light. Now the first thing is for photography. The first thing is the preparation of photographic plate. Okay? So in this you write down the copy the same way. Okay? Preparation of photographic plate. Photographic plate is generally obtained by generally made up of silver halide. Okay? So based on it is actually made up of AGX silver halide or we'll write on based on based on the nature of just a second guys based on the nature of AGX. You see in silver halide this question they ask. Okay? Here they ask the question and the question is what we don't use here for photographic plate. We don't use silver fluoride except AGF because AGF is what? AGF is not photosensitive. Okay? So we use only silver halide which is photosensitive. Photosensitive. So here this they have asked this question that which silver halide we do not use for the preparation of photographic plate. Answer is AGF because it is not photosensitive. Okay? So what we do here this reaction we use to carry out for the preparation of silver halide NH4BR plus AGNO3 and it gives AGBR plus NH4NO3. Right? This is we'll get we'll take an emulsion of it. Okay? And that is applied on the celluloid film. You must have seen that that this real no we are not talking about digital camera that real and all we used to a clean it to a movie with the actor where we used to put that into the solution and we'll know when you shake it you'll get the clear picture of that. Yeah, dark room. Yeah, negative. So we are talking about that. It's not about digital camera and all what we are talking about. So this is AGBR emulsion of this emulsion of this we'll take an emulsion of this and that emulsion will applied on applied on the celluloid film that negative we call it as no negative. So this is it is applied on the celluloid film. So this can be activated by the exposure of light. Okay? When light strikes on it, the reaction takes place, it gets activated. Right? That's why we say art of obtaining the exact impression of an object through a chemical reaction initiated by light. Okay? Now, when you get this, when you have this photographic plate, when it is done, then what happens the exposure of photographic plate exposure of photographic plate with the light? Okay? So light what happens when it strikes onto the celluloid film, light, it reduces light right down light reduces AGBR reduces AGBR and effect of light is directly proportional to its intensity. Okay? More intensity, more will be the effect. So this AGBR reduces into AG plus BR2. It reduces. Okay? Now the next step is developing. Only two, three things we have that you have to memorize here in this photographic photography that usually they ask. Okay? First one I told you already developing. Okay? So this developing of the image is carried out in dark carried out in dark. Okay? Now in this in this step, what happens the exposed photographic plate okay? The exposed photographic plate exposed photographic plate is allowed to react with a exposed photographic plate is treated with a reducing agent treated with a reducing agent and hence these reducing agent we also call it as developers. Okay? Developers. The examples of developers sometimes they ask this question also which of these reducing agent developers we use. The first example you write down here potassium ferrous oxalate. Potassium ferrous oxalate. What is the formula of potassium ferrous oxalate? What is oxalic acid? Fiorge twice. So H you remove both hydrogen you remove put A and Fe over there. Correct? Yes sir. Okay. This is one example of developers. Another one one more you write down alkaline solution of pyregalol and q-nol or q-nol not end alkaline solution of pyregalol or q-nol we use for the developing of this photographic plate. Now in this what happens the developed film that you get. Okay? Here you write down the developed film which is nothing but negative. Okay? Developed film is negative. Okay? Why we call it as negative because in this process what happens the object actually reversed. Okay? So the bright part or bright becomes dark or dark becomes bright. Okay? That's why it is negative. So developed film is nothing but negative. Okay? Now the example of then potassium oxalate for developers. What you give any other example for developers as in potassium alkaline solution of pyregalol alkaline solution of pyregalol or q-nol. Okay? Yes sir. Okay. Now did you finish till here? Yes sir. Okay. Now after this after developing the next step is fixing. Fixing. Now in this what we do there are some unused AGBR, silver bromide. So we have to remove that. So fixing means removal of unused AGBR. Okay? And for this we use, this is important here and for this we use sodium thiosulfate solution. Sodium thiosulfate. So they have asked this question also what we use to remove AGBR. Okay? Sodium thiosulfate solution unused AGBR. That is Na2S2O3. Okay? So the reaction of AGBR plus Na2S2O3. Let me tell you this. These reactions are not important. I'm just giving you. You need to memorize this. Okay? It forms a complex Na3AGS2O3 whole twice plus NaBr. Okay? This is fixing. This also carries out in dark. Okay? Now after fixing we have printing. Okay? So photo is almost done. Now we have to print it. Okay? So for printing we use what? So printing paper is what? It is generally bromide paper we use. Silver bromide also we use for this purpose. Bromide paper we use for this purpose. Okay? Now in this, the whole process is repeated again. Okay? The whole process is repeated and the image obtained here is black and white. The image obtained here is black and white. We get black and white image. Okay? And the last step when you get this black and white image then you have to put some color into this for that which is toning. Now for toning the printed image, the printed image dipped in a solution, dipped in a solution containing the salt of gold or platinum or selenium. Okay? Any one of these salts. Okay? So this is the last step of photographing. Okay? So what is the important thing here? For printed solution dipped in a solution of salt of gold, platinum and selenium. Okay? So this is important. Okay? What we use here to remove AGBR? Sodium has sulphate. This is also important. Okay? Image carried out in dark and that image is black and white. Okay? Printing paper we use is bromide paper. This is also important. Reaction, you let it be. You can skip reaction. And for developing, carried out in dark, we use a reducing agent called developers. I have given you two names of, two examples of developers and that is sodium potassium, sorry, potassium ferrous oxalate and alkaline solution of Paragalol and Qnol. Okay? That is important. Emulsion is applied on cellulite film. Okay? Emulsion of AGBR applied on cellulite film. And the very important thing is this one. They have asked this question many times. We don't use AGF because it is insensitive towards light. Understood this? Yes, sir. Okay? So this is it for this chapter. Okay? F block, just of two, three property we have, it is given in the book, like electronic configuration, general electronic configuration you go through. Actonite starts from which element, goes up to which element. Okay? How many members are there? Members of 62 series or not? Lanthanum series and Actonite series we have. Okay? Lanthanum and the starts from 4F orbital films into that. Okay? One in this, Promethium is a artificial radioactive element. P-R-O-M-E-D-H-I-U-M, Promethium is an artificial radioactive element. Yes, sir. Okay? So this is it for this chapter. Okay? NCRT, like I said, said you must finish NCRT properly. Then if you want to study all these things, you can go through any other book. Okay? So next class today, we could not start aromatic chemistry. We'll do it next class. Two more classes I'll take to finish. Maximum two more classes I'll take to finish aromatic chemistry the entire day. Saturday, Saturday you'll be taking a class. Saturday definitely I'll take. I don't think we'll have any class before Saturday because I have other classes and we don't have holiday also these days. Right? Yeah. Do you have any holiday in the week? So Friday is the holiday. Yes, Friday for? Friday. So Friday, Saturday you can take? Yeah, yeah, mostly, most probably I'll take on Friday. Okay? But I'll let you know on Thursday. Yes, sir. Okay? Yes, sir. Okay, thank you. Thank you, sir. Thank you, sir. Thank you, sir. Thank you, sir. I had a question. So I had a question. Yes, tell me. Yeah, in the color of compounds in DD transitions, is the color because of the one second, is the color because of the emissions that emitted light after absorption or is it reflection after absorption? It's reflection. Okay, so it doesn't re-emit light, right? Actually, you know, it actually what happens when the electron jumps to the higher energy state. Okay, so it absorbs some energy. Okay, so that energy will have some, that radiation will have some wavelength. Correct? Yeah. Usually what we say that when electron comes, you know, do the original state, when the electron comes down to the lower energy state, it radiates like emission, right? It radiates different frequency of or wavelength of light, which is there of the complementary nature of the wheel that I've given you. Okay? So the color is there because of the radiation that, you know, radiates from the transition of electrons. The one that absorbs both the absorption layer, that's gone. No, it's not what's reflected after absorption, right? Yes, yes, yes. It's because of reflected. Got it? No, so is it like, let's say the electron is the electron is energized, right? So it jumps to higher state. So the light that isn't absorbed by the electron, is that what gives the color or is it what the electron releases while being de-energized? Yes, yes, yes. The electron, when it comes down to the lower energy level, it radiates some light, some radiation comes out. Okay? So we'll get color corresponding to that radiation. Right. Okay. Thank you, sir. Okay. And usually it is of the complementary nature. That's why I said that if it absorbs green color light wavelength, then the color will be what? The observed color will be red. Okay. Okay, yeah. Yeah, got it, sir. Thank you. Okay, thank you. Bye-bye.