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"d & f-Block Elements: Videos"

00:00:07 [Music] Good morning everybody. Today we'll be starting on another chapter which is D and F block elements. So what are these elements particularly we should know and what are those positions and another name of these block elements are transition elements. So one important definition for this is transition. So these are transition elements and if we consider their position in the periodic table they run from group three to group 11 in the D block. So we all know that when we start from left hand side of the periodic table we


00:01:33 find that group one and group two elements are there and at some point of time we find that group three and up to group 11 will come and these transition elements and if they are present from the occupancy see in the D cell then we consider them as elements of D block. So we'll just go up to group 11. So what about the group 12? So if we ask somebody that what are the group 12 elements? Give some example of these group 12 elements. We know immediately some can say that we have zinc, cadmium


00:02:28 and mercury. So that question immediately comes to us that whether we should include group 12 within these or not that we'll discuss later on. So what about these that the definition of transition? What is the transition? Because of their position. So in the periodic table they present in between S block and the P block elements. So their position in the periodic table is important in the periodic table and it is between here we have the S block elements and on the right hand side we have the P block


00:03:15 elements. So positioning of these elements therefore are important and this position is the transition position from S to P. So if these are S and these are P elements or P block elements. So we basically move or go for a transition from S to P via these elements. So that's why these are known as transition elements. And in terms of their properties, we'll find that the properties are also transitional. These properties are also transitional from S to P. So what is the transitional property? That means these properties


00:04:08 will be in between S block elements and the P block elements. So they are first thing what we consider is their metallic properties. So what are the metallic properties of S and P block elements like sodium, potassium, magnesium and calcium. We know that they are they are corresponding metallic in nature. So when we move from them to these elements, we'll find that they are also highly reactive metallic elements. So, so if we move from these, so which is also very much similar to that of your S block elements.


00:05:08 So they also form like S block elements. As we slightly move from S to these, they will also form typically ionic compounds and we know that the elements on the right hand side including the H hallogens they form the corresponding elements in the P block. But these p block elements are largely coalent. So they will get some property inherited from these p block elements also. And in some cases in the latter part of the periodic table on the right hand side of these DB block elements they will also


00:05:54 form some coalent characters something related to these p block elements. So as we know so s block property is also there. So some amount of P block property like elements of P block as it gives typically our largely coalent compounds. So some of these transition elements also will be responsible for giving typically all of these coalent character to this particular series. So what we'll find now that many of these properties. So these two properties we mostly consider which are the physical


00:06:49 properties. So how we can consider that means physical properties of these elements and the chemical properties since we are considering their properties in relation to S block and pb block elements and what we are doing we are doing in these blocks that we are adding electrons not on the ultimate or the outermost cell but is the penultimate cell is responsible So we have the penultimate cell and the cell is also expanded. When we know that S is filled and P is filled, we get eight electrons. But in this case the D


00:07:35 level is filled, D cell is filled. So we move from 8 to 18 electron occupancy. As a result, many of these physical and chemical properties of these metals when you consider these as the metals, so you they have the metallic property. So metals of these group such as nickel, such as copper. So what are these metals? So they will have some properties which are in common and they give rise to something which is typically the metallic property. That means they are good conductors such as for both these two things. That


00:08:25 means they are good conductors for electricity and heat. Then they can have metallic lasture. They are also hard and strong because when you talk in terms of the metallological behavior of some of these metallions such as iron, iron also comes into this category as the transition element of the DB block element. How we can improve that uh the property which is strong related to the metallic property and in some cases they are also ductile. And another property which is very much related to their physical property is


00:09:24 that they also form alloys. Alloys with other metals. So these group of elements how we can therefore define. So we just go for now the definition because the definition will consider all the species which will be there and we'll have if we have the d electron configuration. So the d electron configuration whether that d electron configuration can have connection with this definition of these transition elements. So that we'll see from here that this not only the D but also we'll be


00:10:14 considering the F block elements in a similar fashion in the latter part of this chapter we we'll first find what are those F block elements. So before that we'll just consider what are these D elements and what are the transition metal ions. So by definition a transition metal is an element whose atom has partially filled D subell. So the occupancy of this D subsell is important and which can give rise to cations with an incomplete D subsell. So if it can give rise to some cations which have incompletely filled D subsell


00:10:52 then that particular metal or that particular element we consider as the D block element. So we have the incompletely filled d orbital. So if we have the D cell or the D orbital then what we'll find that these are all incompletely filled and where these incompletely filled D orbital or D cell in its ground it or in any one of its oxidation states which is therefore four important that the ground state configuration should give us an import dell or any of its oxidation state. So when we'll


00:12:13 consider their possibility of the oxidation states whether it has a one permanent or one very easily accessible oxidation state or it can have the different oxidation state whatever oxidation state you can have whether you have a incompletely filled D cell or not that will typically define whether you are talking about the corresponding element which is a transition element such that the most common practice we know from early days of our schooling that iron is there. We know that iron can have


00:12:47 two plus or iron can have three plus. So one we consider as a common name as the ferrra ion another is known as the feric ion. So any of its ground states basically that means whether it is present in the fer state or feric state we can have incompletely filled d or the d cell or the d orbital which will typically define whether our this f 2+ or f3+ both of them can be considered as the corresponding transition element derive derived ions. So these are all transition ions or transition element ions which can be derived from iron


00:13:33 which is f0. So in a similar way we basically give for the definition for f block and in these two cases we just considered that since the group will start now because here we follow that the transition elements are following from periodic table after calcium. Similarly here these are starting from lanthnanam and actinium. So where the position of this lanthanum and the position of the actinium we should know and based on that we basically consider that following that. So once the lanthanam we reach then the following


00:14:14 electron configuration or the occupancy in the cell which is whether D or F is different and this occupancy will considered as again some type of transition metals but this type the occupancy is not D but it would be occupancy of the F cell. So occupancy for the F cell will basically give us something where we can have a group of elements or group of metal ions which we'll consider as inner transition elements because after D we just get again not a ultimate cell but it is a penultimate cell below this three D


00:14:55 level is the inner transition metals or inner transition metal ions. So initially if we just simply consider a part of the whole periodic table which belongs to the D block elements and those DB block elements are very important to understand that means on the left hand side we have till calcium which has atomic number of 20 and on the right hand side we have the pre-block elements. So in between we have in the period three. So we have period 1, period 2 and period 3. After when we reach period 3 then only the


00:15:34 possibility of coming the corresponding D cell is there. So after calcium we'll be having the element first element will be scandium. So then we have the scandium, titanium, venadium, chromium, manganesees, iron, cobalt, nickel, copper and zinc. So already we have defined this. We are excluding these that means zinc, cadmium, mercury from this list because these are group 12 elements and we cannot consider by that particular definition that means the incomp completely filled dell which cannot be


00:16:10 applied to zinc in its ground state or zinc in its typically available or most commonly available oxidation state which is zinc 2 plus. So we'll just get for period four which is scandium to copper and what are those electronic configuration that will follow that these are occupying 3D levels. So these are also 3D elements or 3D block elements starting from scandium to copper. Similarly, if we go the next period which is period five, we get itrium to zarconium nabbium to ultimately to silver and cadmium.


00:16:53 And similarly period 6 will give you something where we have you see that 57 to 71 these are by definition these are the a block elements and after that a block elements then only we'll get the electron occupancy or electron filling to the D level which is hopnium then tantalum then tungsten to ultimately to gold. So these three we mostly or commonly encountered. So this particular group that means from scandium to gold 79 we see that if we just see in the group level also the group similarity will


00:17:36 also be there. So all these we considered as the triad because these are not naturally occurring elements for the period 7. Only some synthetically prepared elements have been accumulated over there and day by day we are just filling up all these levels. Already we have filled up all these levels till this uh 111 atomic number. But these three periods particularly period 4, period 5 and period 6 are very important to study these things. And we know that one particular that means the clubbing of all these elements we put together


00:18:15 that means the three D elements or 3D block elements or DB block elements for scandium to copper. We'll see that how their properties are changing as we move from scandium to titanium to vanardium to nickel to copper. Similarly, as we move from this particular period 4 to period 5 to period 6, what is changing? We are just changing from 3D to 4D to 5D elements. So down the group that means group four elements, group five elements, group six elements and group seven elements and group eight elements.


00:18:55 So down the group how the properties of all these groups can change because the ultimate electronic configuration will be same such as that of our nickel which is group 10 element and of 3D and of 4D it will be paladium and for 5D it will be platinum. So if we just consider something that means initially we have we don't know much because we are not very much concerned about the corresponding chemistry of the metallic part because it's relation to the metal energy and the metallic part or the


00:19:32 alloy formation but if we take out those two electrons that means two s electrons on the left hand side so s electrons will be losing first so we are remaining with the d electrons for its konic form which is ni2+ plus. So if we have Ni2+ from here, similarly if we can have paladium 2 plus or if we can platinum 2+ in all these cases we'll see that the corresponding configuration in terms of the occupancy in the D level will be 3D some number then 4D some number and then 5D some number. Similarly these


00:20:12 properties that means from iron, ruenium and osmium. But interesting thing is that as we move down from iron to ruinium to osmium and the size of the dell or size of the d orbital are increasing enormously and the corresponding properties and the reactivity patterns are also changing. So the next thing what we'll be seeing that how we can consider that what should be the electronic configuration of say scandium 21 or say platinum 78 we should have some good idea how quickly we can write that it has the ultimate


00:20:48 electronic configuration for scandium scandium zero is 4s2 3d1 that means the first electron is entering in the level which is 3D and that means that we have the unoccupied 3D So scandium by definition falls under that category of transition element. So titanium will be in the similar way that means the 4 s2 3d2. So what we are getting starting from group three to group 11 we are getting d1 d2 d3 d4 d5 d6 d7 d8 and d9 system. So another way of classifying or placing all these in the periodic table is important such


00:21:31 that we quickly consider that in a particular oxidation state we can have electronic configuration which is also known by its positioning in the corresponding group. So what we see that in the long form of the periodic table that means the color which will be telling us that the pink color will be telling us that these are the transition metals. So scandium to corresponding gold. So this group this group and the left hand side we have the corresponding S block elements and the right hand side we have the P block elements in this


00:22:09 side and then the uh group which is inert also we know and as we move from here that means after lanthanam we'll be getting the corresponding occupancy like that 10 electrons in this particular group similarly we have 14 electrons in the F Well, so after lanthanam we get the series from here to here that means serium to lutaceium these are known as lanthnoids. Similarly after actinium whatever element that means the 14 elements will be pushing there due to the occupancy of the corresponding 5F level are known as


00:22:51 correspondingly actinides. So these two groups will be coming over here. Before that we should finish our discussion in relation to the corresponding transition elements and mostly we always concerned about the particular part which is the first transition series because we know very much because these are mostly commonly available on the earth crust because as minerals and ores the corresponding abundances are more even they have present in the biological form in the biological system even in our


00:23:23 body because iron we all know iron is present in our body also and a particular process like the process we call is a mineralization process that mineralization process is responsible for storing iron on the earth crust. Similarly, other process which we can consider in a similar way that is the biommineralization process and that biommineralization process can be considered for storing iron in our body also for the synthesis of hemoglobin and myoglobin like things. So these elements are so important that we should know


00:23:59 very much because they have different interesting properties related to the transition elements. So the definition is that for pink element that we have partly filled D levels and for these two groups for lanthnoids and actides we have the partly filled F cells. So if we just consider that what about the period four transition metals. So we'll be just knowing now that what are those metals because we quickly will see that a particular type of metals what we'll see that we can have these corresponding properties of these


00:24:40 metals particularly how we can store and just now I'm giving some examples that iron we know that iron in a metallic form we know that iron nail we know iron nail or iron seat we know. So use of iron we all know very much. Similarly, if we get something that the corresponding ions Fe2+ and Fe3+ and if I now say that any of them is also present in our blood as hemoglobin and myoglobin. So this particular thing will concern about the corresponding transition metal ions. These are not metals. So the


00:25:32 corresponding property of these thing is that we have the corresponding property of this and how this iron will look like. Some of us have some good information about what the iron nail will look like, iron seat will look like. But what are these particular things will be in solution. So this whether this will be soluble in water medium or any other medium and how they will also look like and similarly some of these elements can also be useful for the alloy formation. So before going into that particular


00:26:05 detail because iron what we know that iron is also present from the ore and minerals because this will all be present on the earth crust. And if they are present as some oxide and in all of our redox classes then the previous classes we have identified that how we can recover iron elemental iron or the metallic iron from all these ores. So this is a typical process which environment does for us the earth is doing for us and we are storing that particular one and when we recover so recovery process is typically the


00:26:59 corresponding metalologgical process. So this is the corresponding metal energy we can have and that giving rise to the iron zero. But how this iron basically iron will look like. Suppose if you are given with some iron powder because it has some important property for this iron as a dust particle type of thing. So how this iron powder will look like. So only some examples for this period 4 transition metal. The first one is a typical example for scandium. This is the metallic scandium. So metallic scandium is there which is


00:27:36 the corresponding group the element and if we put that on a petry d so the metallic form of scandium will look like of this type. Similarly the titanium these are granules. So if we have the corresponding ore from the earth crust so or for the titanium also we know that titanium dioxide ti2 is a typical ore for that. So titanium dioxide is there and from there we just have to go for the corresponding reduction. So the mechanism is there that how to get titanium from ti2. So in the metallic form if we produce


00:28:11 that thing. So in the corresponding granules are forming and the titanium. So we that particular thing will also give us the properties what we are just now discussed that it has a luster it has a strength and all these. So the corresponding metallic properties for all these things will be there. So we get that for the corresponding venadium also venadium we just once we move to venadium venadium will also give us something where the uh color of these things are changing. So if I put something that means how nicely we look


00:28:49 all these things that means from the nature of all these things particularly the color of all these and the metallic lusture of all these things can we can identify it immediately whether it is scandium this is titanium and that is vanardium. So these all have different. So the particular nature of these particular uh units that means the corresponding granules granules the nature of these granules these are not typical powder because some other process we have to go for getting the corresponding powder. Similarly chromium


00:29:23 you see the chromium also look like a chromium powder. So this a chromium powder we can have. So this is a more powdery form. This has less metallic luster type of character. So it is forming a typical powder type of thing. Then manganese manganese you all know the most typical uh corresponding or is the pyrolocite which is manganese dioxide which is plenty in nature. India is also very much rich in having manganesees dioxide or pyrolocite. So uh the mining process basically gives us the mining for manganesees. We take out


00:29:57 that particular ore and the industry the minological industry will give rise to the corresponding manganesees. So if we consider that in some cases we basically get that particular thing that where we can have that particular manganese typically in our hand. So manganesees will be for the particular manganese metallic state. So these metallic state we can use sometime because most of these are as metal they can also react nicely with the acids. So the oxidation process because we all know now that


00:30:30 they can liberate hydrogen from the acid. So direct reaction of all these can liberate hydrogen and the metal will go to the corresponding ions like that iron when it reacts with hydrochloric acid iron powder. So this powder from the petis what we can take. So if we react with hydrochloric acid the corresponding salt what we'll be getting is the feric chloride and hydrogen evolution can take place. So the corresponding iron powder which can also be identified from the ors like hematite and magnetite


00:31:03 cobalt is also very much similar to that of our venardium case. So it has also typical luster. So is a typical globules having a siny appearance. So siny appearance on the surface of this will tell you that this is a cobalt thing. Then nickel. Nickel is also of different nature. That means when we go for the corresponding crystallization from a molten state because all these at high temperature we are getting as in the molten state and when we go down to room temperature they basically crystallize


00:31:34 it out in a typical form. So the metallic nickel will be separated out from in that particular fashion. Similarly this is copper. So copper is also the last pieces which we'll get already we are having 4 + 4 + 8 + 9 elements. We just reached there. Then we can have the zinc. we have the 3D10 arrangement because the jing granules, zinc powders and all these are very important to understand the positioning of this particular as a whether it is a transition element or not. But the zinc will not be a transition element because


00:32:10 in the elemental state or the metallic state it has the corresponding electronic configuration of 4 s2 3d10. So if we just take out those two electrons, so the electrons will go from the 4s level. So the 4s2 electron will go to give you a 4s zero electronic configuration leaving behind with 3d10 electronic configuration. So that 3d10 electronic configuration will give you a field 3D level. So zinc will not consider as a transition element. So already we have discussed how we can jot down these physical properties. So,


00:32:48 so as inferred by the name the transition metals are metals and thus conductors of electricity. So, whatever the species what we are just now have seen as the corresponding metals because in some of our next classes we'll be discussing about the corresponding formation of the transition metals. So if we have the corresponding metal which just now we have seen as the F0. So it has all metallic property inherent to it. But when we move from there to say Fe 2+ or Fe3+ so this is a typical electron transfer


00:33:28 process we all know and this is a oxidation process but whatever things will be producing in solution in water and these will be present in aqua solution. These two are in aqua solution. So these ions in solution. So these are we can consider them as transition metal ions. So whatever we have such as this present in blood if they are present in blood in this particular two forms that is either iron 2 or iron 3 or any other biological system so those are we considered as the transition metal ions. So we should


00:34:19 always be very much particular that you have ions. So these are forming with the corresponding ions not the corresponding metals. So if we just consider that this particular metals and thus they have the good conductors of electricity. So the iron wires and all these things we can have we know aluminum wires like that of we have iron wires. So that particular then we have the good conductor of electricity we are using electrical wires the copper wires and they are highly dense. So they was a high density and high melting points and


00:34:55 boiling points also. So if we consider that corresponding thing that what we get the corresponding properties is due to the progressive filling of the del. But the filling of these levels will give you the corresponding metallic character of these. So the metals when we talk their properties are due to the corresponding zero form that means the iron zero or nickel zero and they have the typical metallic bonding. So we'll not consider all these things in this particular class but we should have some


00:35:28 little bit idea about what is called this metallic bonding. So just now we have seen that in case of 4s element and the 4 P elements we have the ionic bond typical ionic bond and the typical coalent bond and in between we can have for 3D elements in the metallic state they can have metallic bonding and in the metallic bonding case also when we'll consider the typical bonding for the corresponding metal ions there also we'll find some interesting thing for their participation in the corresponding


00:36:00 bonding when they participate as the corresponding transition metal ions but what about in the free form that means in the zero form in the metallic form they also participate for the corresponding deoization of the de electrons and that's why they increase basically their cohesion due to the large number of these electrons because we know that the capacity of d electrons when it is filling in zinc when it is filling completely in cadmium when it is filling completely In mercury they have 10 electrons altogether.


00:36:35 So as we have discussed what is the typical indication that we are not getting as this particular one for the mercury which is a typically a different thing which is the field D level. So all these not only mercury but starting from zinc which is 3D uh field. So is 4 S2 3d10 then cadmium 5 s2 4d10 and then mercury 5 s2 6 d10. So these all will have lower melting point. So the melting point is less boiling point is also less because they have full D subells and they don't participate much in


00:37:16 deoization and sharing of D electrons and they don't have very good DD bonding in relation to increase the corresponding metallic bonding. So the conduction band they form but the corresponding DD bonding will not participate much in forming the corresponding character and as a result the highest one that means the mercury will have a very low melting point of - 38.83° centiggrade or - 37.89° Fahrenheit is a liquid at room temperature. So is basically is failing. So from that particular one that


00:37:53 metallic property is not there but it has other properties though it is in liquid so other metallic property will be there but is not a transition metallic property we just expect from there. So the first 3D series we just now take out partly because we'll be talking about their properties because what about this D block series? So this D block series will be there if we just simply talk in terms of the corresponding appearance of the corresponding oxidation states because just now we have seen from scandium to


00:38:27 zinc how they look like. Now if we take out the corresponding reactivity pattern of these. So the one of the chemical reactivity physical reactivity we all know that how they form alloy what is their metallic luster whether they are conductor all these but what about their corresponding ionization. So ionization is their corresponding reactivity pattern with the acid whether your acid is oxidizing or not. That means the reaction with hydrochloric acid, the reaction with oxidizing acids like nitric acid or sulfuric acid. So that


00:38:59 will give rise immediately whether we are able to get the corresponding salt formation that we have seen that zinc the metallic zinc or the zinc rod. In our previous redox chemistry classes, we have seen that zinc rod can lead to something where the evolution of hydrogen can take place and we can gives rise to the corresponding metallic salt starting from zinc oxide or the zinc itself. So this particular thing that now we can just separate it out quickly from group three to group 11 including the group 12


00:39:34 because we reach ultimately the 3d10. So the atomic number the elements and the configurations. So configurations we always we can have some good idea. So separating out of this candium to zinc and what about this particular possibility is also there. That means we once we know that the zinc then copper then nickel which is a 3D 10 element then we have the 4D then 5D. So zinc copper and then nickel. when you reach down to nickel then downwards we have the paladium and we have the platinum. Similarly when we iron iron is 3d6 4 s2.


00:40:16 So if we go what we are just now seeing that if we know the configuration that means the positioning in the group the corresponding atomic number also. So this we can find out. It's not that you have to memorize all these things but you should know that is atomic number once it is 26 by filling up the electrons from left to right from 21 to 30 where your position of the iron and its electronic configuration. So if it is 3d6 42. So this iron which is 3d6 and 4 s2. So it is in the zero state.


00:41:01 So when it is losing two electron, it is losing three electron. So straight away we will not consider the electron occupancy in this particular level. So we'll straight away we will write that is a 3d6 ion. So ferrus is a 3d6 ion and this particular case what we'll see that we have the most two common oxidation states. So for iron here we'll just simply write as we know that also for our blood in our body that either you can have iron 2 plus or iron 3+ or something which is in between


00:41:44 or something related to it corresponding reduced form of the feric. So this F3PS and so these are most common oxidation states which is very important that how facile they are. That means the formation of these by simply reacting with say dilute hydrochloric acid cold and dilute hydrochloric acid which is aquas. So the reactivity of the iron powder how we have seen that the what is the iron powder. So reactivity of these iron powder will just lead to the evolution of hydrogen. So hydrogen evolution can


00:42:36 take place and the corresponding ions will remain as the chlorides are there. So we having the corresponding thing as the feras chloride and if it is oxidizing because these the redux potential the redox couple between these two are less the e value for these two are also less which is 77 volt so if oxygen is there oxygen is much more oxidizing so it is in air so if we handle everything in aqua solution so already the water present for this aqua solution or the preparation of this hydrochloric acid.


00:43:12 So O2 is there. So this particular aquar solution. So O2 is the oxidizing one. So O2 is oxidizing agent. So that will immediately oxidize this to F3+. So what about this particular electronic configuration? So electron configuration for F3. We have to take out one electron from this 3d6. So it will not be 3d6. It will be 3d 5. So for these two that means we have 3d6 ion and 3d5 ion which are most common oxidation states for 3D level. So since we are talking about 3D. So if we just consider the periodic table.


00:43:56 So in the periodic table we have iron, ruanium and osmium. So which is 3D 4D and 5D and these are having giving rise to some electronic configuration. So if we consider that both of all of them are giving that means the trialent statealent state for iron trivalent state for ruanium and trialent state for osmium. So this is the electronic configuration which is 3d5 for iron. So iron 3+ will be 3d5. So without knowing much or without bothering much about what would be for ruanium. So ruanium will be ruinium 3+.


00:44:50 This is iron 3+. So iron 3+ will be 3d5. So ruinium 3+ will also be 4 D5. Similarly, osmium can go for osmium 3+ which will be 5D5. So this is the advantage of knowing the periodicity of the elements placing them in the periodic table and how quickly we can understand when we talk about the chemistry of these because sometimes we can handle all these in solution. having some test tubes. We can have some test tubes test tube one, test tube two and test tube three. In one case we have fericion in solution. In another case


00:45:37 ruanium in the trivalent state in other case the osmium in the trivalent state. So the generalization of all these things are very important and we know that most of these cases we are removing the electrons from the D level. That means this oxidation the first one electron loss one electron loss the first one electron loss for getting feric ion from ferrra ion is the removal of the electron from the d. So this is much more fasile but if we can have some arrangement and if we can have some stronger oxidizing agent so


00:46:15 then we can find it out whether we will be able to take out the electrons from its corresponding positioning that means whether we can go beyond that means whether we can take out more number of electrons from these that means we can take out one more electron from this level giving 3d4 or 3d3. So those oxidation states we can get and those oxidation states will be termed as unusual oxidation states or uncommon one. So uncommon oxidation states we can have that means beyond this. So two and three


00:46:54 can be there. So we can have 4 plus, we can have 5 plus or we can have 6 plus. But altogether what we can have altogether we can have eight number of electrons, two in S level and six in D level. So if we remove all these electrons from S level or S cell and D level or D cell we'll be getting something which is 8 plus. So whether we'll be getting that particular one is important to discuss whether it is possible for iron and whether it is possible for all other elements. So what we see that getting


00:47:45 all these oxidation states that means + 2 + 3 + 4 + 5 and + 6. So these elements these transition elements therefore occur in variable oxidation states. So they basically occur in variable oxidation states. So one or the other that means as we move that means the corresponding filling off of the D level from scandium to iron we are filling up stepwise one after another one electron 2 electron 3 electron 4 electron 5 electron and 6 electron similarly when we are talking in terms of the removal of electrons from that particular d or d


00:48:36 cell is the corresponding oxidation reaction so solution chemistry for all these metal ions is will be mostly dominated by the presence of the corresponding oxidation states and we should have some good knowledge about the presence of all these oxidation states in our hand. So we get this as the 3D elements or DB block series in the third level. Similarly, we get the next one which is the second D block series which is from Y to CD or itrium to cadmium and again like that of our electronic configuration or the


00:49:16 positioning of all these things. We see that in this particular case also the progressive filling of the D cell is important and we have in some cases we have the mostly from here that means D1 to D9 because this we just move for that if we move this electron to the S level which will be 5 S2 and which will be 3D 4 D9 we get the progressive filling of the D cell so we'll get these and since this particular one we are talking about. So this particular one since the size is increasing instead of writing as


00:49:54 D6 H S2 electronic configuration we can move because these are very close by energy wise these levels are very close by the D level and the S level. So we can move this particular electron to this particular cell. So that is the configuration is now is 47 5S1. So that basically tells us something that whether we can have that removal of single one electron from the S level. So removal of that particular one electron will giving rise to a state where you can have runium in one plus state. So in a particular condition or in a situation


00:50:37 that we'll find afterward that one particular type of compound we call as the organo metallic compounds where we can have some interaction of this metallic state in the zero that means the powders can react with some of the species like that of our metallic state which can interact with that particular thing that means simple carbon monoxide. So 3D container of this paladium is nickel. We all know that nickel can interact with carbon monoxide giving rise to tetracarbon in nickel zero. So nickel will be zero in that particular


00:51:14 case is organometallic compound and that organometallic compound will have electronic configuration. We just consider that electronic configuration in terms of nickel zero. Similarly, if we some think that paladium zero in different organic chemistry reactions the paladium zero the metallic state of paladium is important and the corresponding electronic configuration if we push all these electrons to the D level because this has a extra stability the stabilization we all know that the halffield cell that's why this


00:51:47 particular cell we are writing instead of 5 s2 4d4 we write as 4 d5 5 s1 so one electron we move from S level to D level. So it has some extra stability that means the half field cell and the full field cell. So the paladium in that particular case that this the paladium in the zero state will have a full field state and that fulfilled D level will have a 4d10 electronic configuration and is stable. Similarly the other one that means the 5D block and the 5D block will get the corresponding one from


00:52:27 that uh leatium to len sorry this lanthanum this lanthanum is 71 this not leatium is lanam la it would be la to uh this gold so there we are also having just simply the same type of electronic configuration and same number of electron in the d and the s level but thing is that now we can have the corresponding occupancy of these levels and is changing from one particular level to the other. We are talking same thing related to 3D level, 4D level and 5D level. So the corresponding occupancy of the uh corresponding period that


00:53:05 means the uh period which we are talking about for 3D 4D and 5D. So in this particular case these are the 5D element. So the konic state just now what we have discussed that osmium like iron. So after iron we have ruinium and then we have osmium. So osmium is the conjener of the iron group. So in that particular case we should not forget the group number. Similar fashion we should know the atomic number also nicely and the osmium and the osmium is in plus two oxidation state like that of our iron will be 5d6


00:53:43 electronic configuration. So all these things and the particular type of bonding in all these cases what we see is related to the corresponding occupancy of all these D levels and we'll just be able to consider the different thing that means the 3D 4D and 5D elements. So these 3D 4D and 5D elements. So we have so if we just consider these 3D 4D and 5D elements in its elemental state that means M is in zero state. So physical properties are also changing when we talk about the bond strength. So bond


00:54:27 strength will also be changing and this particular bond strength trend is there. So as we go for bigger and bigger D level or D cell the bond strength trend will be changing and which is a different one which is a reverse one for this. So this trend this trend is reverse to that normally we found for main group elements that means the S block and P block elements. So what we find for main group elements is different for these transition elements. So we find that once we get that that means if we consider that what about


00:55:22 tungsten then which is in the chromium group. So we have chromium, malibdenom and tungsten. So chromium, malibdenom and tungsten we have and in this particular case chromium you know that this has six unpaired electron. Similarly malibdenom will also have six unpaired electron. So we have five d4 and six s2. So all these six electrons if we consider the corresponding property as its zero state that means tungsten in the metallic state. So the tungsten in the metallic state have the six electron and these


00:56:08 six unpaired electron they participate strongly in metallic bonding. So we have large number of electrons which are not possible to get for S level or the P level elements. So large number of electrons are available for these. So as a result they can have also very high electro negativity. So tungsten will have therefore have very high electro negativity and this particular information is also important from our early school days. We know that they can be utilized. So they have a very high melting point and high


00:56:56 boiling point. So tungsten the metallic tungsten will have very high melting and high boiling point. And as a result they can be utilized for making the bulb filament. So the bulb filaments for incandescent lamps we use tungsten as the corresponding material for the making of these bulb filaments. So we can have this that means we just corresponding unpaired electron. So we basically change the corresponding melting point. So we just can see also what about the corresponding melting point trends also.


00:57:39 So as we move from scandium to titanium to ultimately to zinc. So we find that the typical melting point in degrees centigrade will also be changing and which are above 100. So mostly it is above 1 th00and so not sorry above thousand so above 1,000° centigrade and in some cases they can go up to 3,000° centigrade. So one value is 1539° centigrade for scandium. So it is increasing for titanium it is increasing for venadium as well as chromium. But in case of zinc it is less since those number of electrons which are there but


00:58:27 is in the field cell it is not available for that kind of metallic bonding. So minimum we'll be finding over here where the levels are filled. So the melting point minima will be finding over here and melting point maxima will be here for the transition metal lines. So thus we see that number of electrons in the elemental state. We will not forget that all are in the elemental state. That mean the scandium as the metallic scandium, titanium as the metallic scandium. They have the high melting point, high boiling point


00:59:05 and some of these uses related to the metallic state. So the next day we'll be just considering how we get for the corresponding electron transfer reaction for the oxidation that means the availability of the different oxidation states. Okay. Thank you very much.

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Overview

NEET Class 12 Chemistry NCERT Chapter 4 d- and f-Block

In this video, we will cover:

1. d- and f-Block PYQs

2. Previous year Questions d- and f-Block

3. d- and f-Block PYQs

4. PYQs NEET Chemistry

5. Chemistry PYQs

6. Questions NEET Physics

7. NEET Physics Questions

8. Physics NEET PYQs

9. NEET Chemistry Practice Questions

10. Neetbharath.in Tutorials Delhi NEET Chemistry

11. d- and f-Block NEET 2026-2027

12. NEET Chemistry Class 12 Practice Questions

13. Class 12 Chemistry Chapter 4 d- and f-Block NEET Questions

14. Chapter 4 Class 12 Chemistry NEET

15. d- and f-Block Class 12 Chemistry NEET

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17. Chemistry NEET Chapter 4

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20. NEET Mains Concepts Chemistry

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23. NCERT Class 12 Chemistry Chapter 4 NEET Concepts

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25. NEET Chemistry Question

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Chapter 4 d- and f-Block Class 12 NEET Chemistry deals with:

  • In the periodic table the d block consist of the elements of group 3 to 12.
  • The d orbital of the d-block elements in four periods are filled.
  • The three series of the transition metals are 3d series from Sc to Zn, 4d series from Y to Cd and 5dseries from La to Hg.
  • The fourth 6dseries begins from Ac and is incomplete till now.
  • The d-block elements are found in the middle section of s- and p- block elements in the periodic elements.
  • This lead to its name ‘transition’ due to its position between s- and p- block elements.
  • The electronic configuration of d-block elements is(n-1) d1–10 ns1–2. They have two incomplete outershells.
  • Where (n–1) = Inner d orbitals having electrons from 1-10.
  • ns = Outermost orbital may have one or two
  • (n-1) d10 n s2 represents the electronic configurations of Zn, Cd and Hg.
  • They exhibit variable valency that differ by units of one.

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d & f-Block Elements FAQs

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Our 20 NEET Video Lectures span d & f-Block Elements’s syllabus: atomic models (Bohr, Rutherford), quantum numbers, spectral series (e.g., Balmer n₂ = 3 to n₁ = 2), de Broglie wavelength, photoelectric effect—everything NEET Exam tests.

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Absolutely! Each of our 20 NEET Video Lectures includes instant, detailed solutions—e.g., calculate Bohr energy (Eₙ = -13.6 Z²/n²) for He⁺ or spectral wavelength. Learn and fix mistakes fast.

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Definitely! Our 20 NEET Video Lectures are refreshed weekly for NEET Exam, focusing on Bohr’s model, quantum numbers, and more—no syllabus cuts noted, just pure numerical and conceptual prep.

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d & f-Block Elements carries 2-3 questions (8-12 marks) in NEET’s 180-mark Chemistry section. Our 20 mock tests ensure you nail every point with targeted practice.

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Class 12 Chemistry d & f-Block Elements NCERT Chapter 4


Revision of Class 12 Chemistry d & f-Block Elements is a crucial aspect of effective learning. Revision plays a vital role in the learning process and is especially important before exams. Here are some key points you can consider emphasizing in your content:

  1. - Retention and Memory: Regularly reviewing and revisiting the material of Class 12 Chemistry d & f-Block Elements helps reinforce the concepts in student's minds. It strengthens memory pathways, making it easier to recall information during exams and beyond.
  2. - Consolidation of Knowledge: When you revise notes, you are essentially consolidating your knowledge. This means connecting new information with what you already know, making the overall understanding more robust.
  3. - Identifying Knowledge Gaps: Revision allows students to identify any gaps in their understanding or areas where they need further clarification. This gives you a chance to seek help or delve deeper into those topics. For detailed understanding, you can always refer to the videos of d & f-Block Elements Class 12 Chemistry NCERT Chapter 4 on neetbharath.in
  4. - Building Confidence: As you revise d & f-Block Elements Class 12 Chemistry and become more familiar with the content, your confidence in your abilities grows. Confidence is a crucial factor in exam performance as it reduces anxiety and allows you to approach exams with a positive mindset.
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  6. - Spacing Effect: Spacing out revision sessions over time, rather than cramming all at once, has been shown to improve long-term retention. Create a revision schedule leading up to the exams to allow for spaced practice.
  7. - Regular Revision over Cramming: Regular and consistent revision throughout the academic year is very important. Waiting until the last moment to cram everything can be overwhelming and less effective than spaced-out revision.
  8. - Self-Assessment: Assess your understanding periodically through quizzes or self-tests. This helps you to gauge your progress and identify areas that need further attention. Refer Class 12 d & f-Block Elements Online Tests.
  9. - Balanced Approach: Remind students to strike a balance between revision and other activities. Adequate rest, exercise, and relaxation are essential for optimal learning and performance.
  10. - Seeking Help: If you face difficulties during the revision process, Refer the videos of Class 12 Chemistry d & f-Block Elements. Clearing doubts promptly is crucial for a better grasp of the subject matter. You can always ask your doubts on d & f-Block Elements Class 12 Chemistry NCERT Chapter 4. “Ask a Question” section of neetbharath.in

By highlighting the benefits and strategies of effective revision, you can approach your studies more mindfully and achieve better results in your exams. Best of luck bachhon!

Class 12 Chemistry seems to be a quite difficult subject for a lot of students. But, if you get a very good conceptual understanding of the subject, it can be very interesting for you.

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Always remember, it is very important to study with full concentration during Revision. Here are a few tips for you on how to revise with full focus:

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  2. - Set Specific Goals: Before starting your study session, set clear and achievable goals. Break down your study material into smaller tasks, and plan what you want to accomplish during each session. Having specific goals will give you a sense of direction and purpose, making it easier to concentrate.
  3. - Use the Pomodoro Technique: The Pomodoro Technique is a time management method that involves studying in short, focused intervals, typically 25 minutes, followed by a short break of 5 minutes. After completing four sessions, take a longer break of around 15-30 minutes. This technique can improve focus and productivity by providing regular breaks to recharge.
  4. - Stay Organized: Keep your study materials, notes, and resources well-organized. Having everything you need at hand will save time and reduce distractions caused by searching for materials. Use color-coded folders or digital tools to maintain a structured study system.
  5. - Practice Mindfulness and Meditation: Before you begin studying, take a few minutes to practice mindfulness or meditation. Deep breathing exercises and clearing your mind of distractions can help you approach your study session with a calm and focused mindset.

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