COMEDK 2024 Inorganic Chemistry Complete Revision in 1 Video

Summary

In this video by Deeksha Karnataka, students preparing for the COMEDK 2024 examination can find a comprehensive revision of inorganic chemistry. The video covers essential chapters such as coordination compounds, d and f block elements, and chemical bonding. Students are guided on key topics and types of questions frequently asked in the exam, with a focus on high-weightage areas like valence bond theory and electron configurations. Accompanying resources, such as a PDF with notes, are available via a WhatsApp channel to facilitate thorough preparation.

Highlights

• Remember, coordination compounds and d block elements are high-scoring areas. 📈
• Understand valence bond theory; it's a game-changer! 🧠
• Clear your concepts on electron configurations for easy exam questions. 💡
• Practice the summarized notes accessible via the WhatsApp channel for a quick revision. 📲
• Share and discuss the video with peers for an interactive learning experience. 🔄

Key Takeaways

• Ace COMEDK 2024 inorganic chemistry with this full revision video! 🎓
• Cover high-weightage topics like coordination compounds and d block elements. 📚
• Valence bond theory and electron configurations are must-knows for the exam! 🔑
• Revise quickly with summarized notes available in a downloadable PDF. 📄
• Join the WhatsApp channel for more resources and exam tips! 🤝

Overview

The video commences with an engaging introduction, emphasizing the importance of inorganic chemistry in the COMEDK 2024 exam. It highlights the chapters with significant weightage, encouraging students to focus on these areas to score well.

In the core part of the video, key topics such as coordination compounds, chemical bonding, and d block elements are thoroughly discussed. The instructor explains intricate concepts like valence bond theory and electron configurations, making these essential understandings accessible to students.

Concluding the video, viewers are urged to utilize the provided resources, like the downloadable PDF of notes, and join the WhatsApp channel for continuous updates and support. The creator encourages active engagement with the content for effective preparation.

Chapters

• 00:00 - 00:30: Introduction to Inorganic Chemistry Revision The 'Introduction to Inorganic Chemistry Revision' chapter covers a comprehensive review of key concepts in inorganic chemistry targeted at students preparing for the Comet K 2024 examination. The instructor aims to cover all significant chapters, important topics, and commonly asked questions to aid in revising the subject. Students can expect to tackle about 12 to 15 questions from this section in their exam.
• 00:30 - 01:00: Exam Study Tips and PDF Resources In this chapter titled 'Exam Study Tips and PDF Resources', the presenter discusses high-priority topics for an upcoming examination, emphasizing the importance of taking notes on key concepts and exceptions mentioned. Viewers are encouraged to join a WhatsApp channel, where a PDF of the discussed topics will be shared, with a link available in the video description. The chapter is particularly focused on providing a comprehensive revision of inorganic chemistry, which is deemed a crucial area for the exam.
• 01:00 - 01:30: Important Chapters Overview The chapter titled "Coordination Chemistry" is highlighted as a significant topic with high weightage in exams, particularly for the pu2 level. It is expected to contribute three to four questions. Important topics within this chapter include the Wess Theory, notable terms, and the essential 'n clure'. Emphasis is placed on understanding these core areas to excel in this subject.
• 01:30 - 02:30: Coordination Chemistry Topics and Wess Theory The chapter covers key topics in coordination chemistry, including isomerism, bonding theories, and metal carbonyl bonding. It emphasizes the importance of understanding Werner's theory, isomerism, nomenclature, and bonding in metal carbonyls, as potential questions can arise in exams. The chapter also introduces Wess Theory, explaining its application in coordination compounds.
• 02:30 - 04:30: Important Terms in Coordination Chemistry The chapter discusses 'Important Terms in Coordination Chemistry', focusing on two types of valency: primary and secondary. Primary valency is characterized as ionizable and can be satisfied by negative ions, representing the oxidation number of the metal. In contrast, secondary valency is non-ionizable and can be satisfied by negative or neutral molecules, representing the coordination number. The chapter likely aims to clarify these concepts and provide insight into common inquiry formats regarding these terms.
• 04:30 - 06:00: Nomenclature of Coordination Compounds The chapter "Nomenclature of Coordination Compounds" involves understanding questions related to ion production from complexes. An example includes determining the number of ions produced when given a complex containing a metal, ligands, and bromine. Additionally, the chapter touches on the concept of reactions involving compounds like AgCl.
• 06:00 - 08:00: Isomerism in Coordination Compounds The chapter discusses isomerism in coordination compounds, emphasizing how questions related to the theory can determine the arrangement of molecules in a complex. For example, determining the number of chlorines inside or outside a complex by analyzing moles of HCL produced. Important terms like 'lians' (likely meant to be ligands) and 'denticity' (number of ligating groups in a ligand) are highlighted, implying their significance in understanding coordination compounds.
• 08:00 - 12:00: Bonding Theories and Crystal Field Theory The chapter discusses various types of ligands and their ability to donate electrons to metals. It introduces unidentate ligands such as Cl- , H2O, and NH3, bidentate ligands like ethylene-diamine, and polydentate ligands exemplified by EDTA, which is hexadentate. The concept of ambidentate ligands, which have two possible electron-donating sites, is also introduced.
• 12:00 - 15:30: D and F Block Elements This chapter discusses the D and F Block Elements, focusing on their properties, uses, and significance. It covers important concepts like ambidentate ligands, which can donate electrons through two different atoms, though not simultaneously. Examples include SCN, NCS, NO2, and NO. The chapter also explains coordination numbers, defined as the number of ligand atoms directly bonded to a metal atom. Understanding these concepts is crucial for answering relevant questions related to the denticity and coordination behaviors of these elements and their complexes.
• 15:30 - 29:30: Chemical Bonding and Molecular Structure The chapter discusses chemical bonding and molecular structure, focusing on the coordination number and the concepts of homoleptic and heteroleptic complexes. The coordination number is explained with an example involving a central atom bonded to six ligands. Homoleptic complexes are defined as those with only one kind of ligand, while heteroleptic complexes have more than one type. The chapter also emphasizes the importance of nomenclature, providing a brief explanation of the rules and encouraging practice through examples found in the NCERT textbook.
• 29:30 - 35:30: Classification of Elements and Periodic Properties The chapter discusses the classification of elements and their periodic properties, focusing on the importance of practicing nomenclature, particularly in coordination chemistry.
• 35:30 - 36:00: Exam Tips and Conclusion The chapter provides a summary of naming conventions for ligands in coordination chemistry. It explains that anionic ligands typically have names ending in 'o', while neutral ligands have specific names such as 'aqua' for water (H2O), 'amine' for ammonia (NH3), and 'carbonic' for carbon monoxide (CO) groups. The chapter highlights these terminologies from the perspective of preparing for examinations, emphasizing their importance for students to remember.

COMEDK 2024 Inorganic Chemistry Complete Revision in 1 Video Transcription

• 00:00 - 00:30 hello students welcome back to our Channel Dak Karnataka so your Comet K 2024 examination is approaching so in this video we will do the complete revision of the inorganic chemistry okay so I'll be covering all the important or inorganic chemistry chapters we be discussing the important topics and the type of questions that is asked also I'll be discussing so if you watch this video you can revise the complete inorganic chemistry and you can expect like easily 12 to 15 questions from this uh all these four topics because all our
• 00:30 - 01:00 high weightage topics so let us start here and you will be also getting the PDF of this in our WhatsApp channel so join our WhatsApp channel the link will be given in the description section of the video okay so please sit with the notebook and a pen and note down all the important Concepts all the important exceptions and wherever whatever I tell just note it down and revise it thoroughly for your comedk examination okay so one of the very important video complete revision of inorganic chemistry so let us start now with the so in this video we'll be discussing the four
• 01:00 - 01:30 important chapters that is coordination compounds dnf block elements chemical bonding and classification of elements okay so let us start with the first chapter that is coordination chemistry it's a high weightage topic from your pu2 right so you can easily expect three to four questions from this chapter so what are the important topics in this chapter first let us see them so first is your Wess Theory okay so next is your important terms we'll be like questions are asked from these important terms also next is n clure very very important
• 01:30 - 02:00 then you have isomerism then you have bonding theories and the last is bonding in metal carbonal so if by seeing this list I'll say which is the most important thing then you must Mark about burner Theory isomerism nomenclature and bonding in carbonal uh bonding in metal carbonal questions are not asked but we never know the question may be asked so just a very simple concept we can just revise it once okay so now let us start with the Wess Theory so according to wus theory in a coordination compound there
• 02:00 - 02:30 are two types of valency one is your primary valency and secondary valency so what are the properties of primary valencies they are ionizable they can be satisfied by negative ions and what is like if someone ask you what is primary valency it is nothing but the oxidation number of the metal okay now coming to secondary valencies these are non- ionizable satisfied by negative or neutral molecules and the uh this is like what is secondary valency it is the coordination number okay now how questions are asked will they ask you
• 02:30 - 03:00 definition of this no they will ask you like suppose a compound will be given for example we have a complex like this uh a any metal and then you have Co liand 4 and then X leand and then br3 then how many ions we are produced from here right so this complex will be one ion and then three bromine will be another ion so total ions produced will be four so such type of questions are asked and the question like about agcl so prepare this agcl concept okay so when they react with uh when you react a
• 03:00 - 03:30 complex with ag3 how many moles of HCL is produced and based on that we can tell how many chlorines were outside the complex and how many chlorines were inside the complex right so such type of questions application based questions are asked from veres Theory now let us go to the next topic important terms which important terms are there so lians there's like just just for you should know that number of ions or metals that is bonded to the central atom now denticity is important here you should know the number of ligating groups in a
• 03:30 - 04:00 leant how many uh in a legant how many sites are there which can donate the electrons to the metal now if you see un dented Lian your examples Are CL minus H2O NH3 di dented legans your examples are ethan1 to diamine which we write as then poly dented lians you have the example of Eda okay so here you should remember Eda is hexad dented liant okay next coming to ambid dented legans so what do you mean by ambid dented it means like uh there will be two donating
• 04:00 - 04:30 sites in the uh Lian but at a time or at least like at one point of time only one will donate okay so your examples are scn and NCS your example is NO2 and no so you can just know like questions are can be asked which of the following is uh ambid dented Lian what is the denti denticity of these lians okay so on the basis of denticity you should know the examples now coming to coordination number so the number of Lian atoms donated to the metal directly bonded is called as the coordination number so from from here also you can see Cobalt
• 04:30 - 05:00 is bonded to six lians so the coordination number is six now homoleptic and heteroleptic complex these are very simple concept so if only one kind of lians are there then it is homoleptic if more than one kind of so sorry this one nh34 and cl2 okay so if this we have more than one type of legans then we call it heteroleptic okay next coming to nomenclature very very important so I will briefly explain the rules and then we will do two examples you can practice no clature examples from ncrt examp in
• 05:00 - 05:30 the chapter it is there inside the chapter in the text questions also there in the intext questions also there right so practice nomenclature thoroughly okay so if I like in coordination chemistry what you must know is your nomenclature okay so coming to that first rule is the Kat should be named first and then the anine so whether K cat is a complex or anine is a complex or Kat is auxilary Ion doesn't matter first C then anine now the legans are named in the alphabetical order okay so in the complex first the lians will be named
• 05:30 - 06:00 and then the metal will be named name of the anionic lians in ending it with o those of the neutral and cic lians are same expected for water Aman and carbonal for Co and nitril for n okay so what does it say the legans okay the name of the lians will generally end with o for negative anic legans and for neutral lians there are special names like for water it is aqua for NH3 it is amine for Co group it is carbonic and
• 06:00 - 06:30 for n group it is nitril okay okay now coming to this one if more than one lians are there then what prefix we should use mono di TR like that we have to use now if the name of the Lian contains more than one name then we have to use B TR and trra is for giving you the number of lians present next rule is Roman numeral in the parenthesis so after the metal is written its oxidation state is written in Roman numbers in Brackets okay which type of brackets like this bra
• 06:30 - 07:00 okay next if the complex is a cattin so I told you complex can be cattin also can be anine also so if the complex is a cat then the metal is named as same as the element for example if the metal is Ion then it will be ion the metal is Cobalt it will be Cobalt but if the complex is anion okay the anion part is the complex then the metal name should end with a t okay that is very important now let us see one example here you can see this compound okay so this in this case the complex the complex is your
• 07:00 - 07:30 kattin okay so here first you have to name the legans so we have here three aqua and we have three amine group right so based on your no alphabetical order first amine will come so triamine TR Aqua then since complex is the C the metal name will remain as it is so chromium okay then three then we have chloride okay here this should not be written as Tri chloride for auxiliary ions there should not be any suffix or
• 07:30 - 08:00 prefix required now if you come to this one here this is a Lian with more than one name Ethan one to diamine so three are there so we have to use the word t Tris okay tress ethane one to diamine Cobalt then oxidation state and then sulfate here also you need not write Tri sufate it should not be written okay now coming to isomerism next important topic so isomerism is of two types first is your Optical isomerism and then is your structural isomerism in Optical Isom we have geometrical isomerism and stereo
• 08:00 - 08:30 isomerism geometrical isomerism you know sis and trans and in Stereo isomerism in anas are there right now in structural isomerism we have linkage ionization coordination and solvent isomerism very very important topic okay now let us first say geometrical isomerism so in geometrical isomerism if you see if the same lians are on the same side we call it Sis if the lians are on opposite side we call it trans now you can see here this is for coordination number 4 okay now if
• 08:30 - 09:00 coordination number six is there and the legans are of the form like M A4 B2 you can see if the two chlorines are opposite it is trans if two chlorines are lying in the like are not opposite to each other this is the CIS geometry next coming to this one is very important M A2 B2 type okay where this is a by dented liand okay in this case you must remember the CIS form of this one is
• 09:00 - 09:30 optically active Okay CIS form of this one is optically active we'll see that also so CIS form you can see two chlorines are opposite to each other that is transform and when the two chlorines are nearby to each other that is your CIS form okay now coming to the next question uh next uh topic that is Stereo isomerism so if you have the mirror images are um superimpose nons superimposable mirror images then they form in consumers right
• 09:30 - 10:00 so you can see here this is like a form Co en whole3 so these two are nons super imposible mirror images so they will be our in animers and now if you see this one M A2 B A A2 B2 type I told you the CIS form will be your optically active transform will be uh will not be optically active Okay next seeing the next one structural isomerism so here you should know what is linkage isomerism for linkage isomerism we need dented leans okay so if NO2 is there
• 10:00 - 10:30 what is its uh linkage isomer So Co NH3 5 o n o cl2 okay now coming to coordinate isomerism in coordinate what isomerism what happen legans get exchanged so what will be the coordinate isomerism for this one Co I'll write CN whole 6 and we have cr nh36 okay now coming to ionization
• 10:30 - 11:00 isomerism in ionization isomerism when you put the compound in the water if different ions are produced then we call it as ionization isomerism now in this case what we'll get we'll get this complex as one and BR minus as another ion here we'll set this complex as one and sulfate as another ion so formula is same but in water we are getting different ions so we call it as ionization isomerism in solvent isomerism very easy water should be there as a solvent right so you can see here here we have water and here this
• 11:00 - 11:30 water can come out and cl3 can go out so water can be coming like inside the complex and outside the complex you can interchange then we'll get a solvent isomerism okay okay now coming to bonding theories okay so what were the different theories that describe the coordinate Bond or coordination Bond so first theory that came was your Valance Bond Theory according to Valance Bond Theory what happened the the n minus1 d NS NP NS NP and n d okay so these electrons can these orbitals can
• 11:30 - 12:00 participate in the bond formation okay so these orbitals undergo hybridization to yield a set of equivalent mole equivalent hybrid orbitals such as octahedral tetrahedral or Square planer so here are the hybridizations you must remember if the coordination number is four we have two types of hybridization that is sp3 and dsp2 so you can see here there are two types of combination I'll write see you can see if we have n n -1
• 12:00 - 12:30 so I'll write here see for sp3 we need NS and NP orbitals to combine for dsp2 I need n minus1 d NS and NP orbitals to combine for sp3d2 we need NS NP and n d okay and for SP sorry sp3d now same for sp3d2 but for D2 sp3 We need n minus1 d
• 12:30 - 13:00 NS and NP orbitals okay so these are the geometries now what are the geometries produced you can see for coordination number four we have two types of geometry tetrahedral and square planer for sp3 it is tetrahedral for dp2 it is square planer for coordination number five sp3d it is trigonal bipyramidal for coordinate coordination number six sp3d2 octahedral D2 sp3 is also octahedral so whenever you are using n minus 1 D we
• 13:00 - 13:30 call it inner orbital complex okay okay now let us see these are the four examples of the compounds that are given in your ncrt book okay so please remember these examples thoroughly so that in exam you do not have to calculate and find out everything so for Cobalt NH3 it is a strong field liant so that is why we will get D2 sp3 hybridization shape will be octahedral since the coordination number is six and since it is a strong field pairing of electron will take place it will be diamagnetic right now if you see here there next one is cf63 minus it is a
• 13:30 - 14:00 weak field Lian right but coordination number is six so we will get octal Geometry weak field means outer orbital complex sp3d2 so I'll write here this is inner orbital this is outer orbital okay next the magnetic property will be paramagnetic because it has unpaired electrons now if you see nicl4 2 minus this is sp3 hybridization again this is outer orbital complex tetrahedral and paramagnetic and in Ni cn4 2 minus so
• 14:00 - 14:30 this is there's no concept of outer because D orbital is not involved here it is dsp2 square planer and diamagnetic okay next one we'll see so what does crystal field Theory tells us so next after valence Bond Theory the next bonding theory is Crystal field Theory so in Crystal field Theory this one a question was asked you should know metal Lian bond is uh is considered to be ionic ionic arising purely from electrostatic interactions between the metal ion and the Lian okay so this point you should remember according to
• 14:30 - 15:00 uh this Crystal field Theory the metal and lean has a electrostatic force of attraction because of which we consider them as an ionic bond okay next so what happens when lean approaches to the Lian approaches the metal the D orbital split to reduce the repulsions so splitting of D orbitals depends on the strength of the lians okay the splitting can be more splitting can be less depending on the lians okay now coming to this so we have uh Crystal
• 15:00 - 15:30 field Theory Crystal field splitting in octahedral so what happens first initially all the five D orbitals are degenerate now when the liance approaches to reduce the repulsion they go in this way okay so three orbitals come here and two orbitals go down you can see in the diagram here if it is visible these three orbitals are called as your t2g orbital dxy Dy z and d x z okay so what are that I'll write here
• 15:30 - 16:00 dxy d y z and d z x now similarly out of 5 3 went down so two will be up so these two are your D x² - y s and d z s right you can see here now you should also remember the values okay so how much decreas is there so how to remember see total there are five orbitals okay so here three is there three has come down so this will be Min - 2x 5 Delta o and
• 16:00 - 16:30 here there is 2 so it will be 3x 5 Delta o so please remember this is plus this is minus okay next is your Crystal field splitting okay so this is the same one which we have explained next is your Crystal field splitting in tetrahedral complexes so if you see in tetrahedral complex the opposite will happen so here five orbitals were there right after this one two will go down and and three will go up and here we do not use the
• 16:30 - 17:00 word T2 g g is not used so it is only called T2 sorry this one is called E and this one is called T2 right so here the orbitals are your d z s and D x² - Y sare and here it is dxy dy z and d z x okay so Crystal field splitting in tetrahedral is not much asked in the question but in octahedral complex the splitting is very very important okay
• 17:00 - 17:30 now coming to okay here I want to discuss one more thing about the how to write the electronic configuration in strong field and weak field so very important see here if it is a strong field liant okay then splitting will be more okay splitting is high and pairing of electrons will not take place sorry pairing of electrons will take
• 17:30 - 18:00 place okay now for D1 D2 D3 there's no configuration change whether it is a strong field Lian or it is a weak field lean but for D5 D4 to D7 you have to be very careful now we will discuss that suppose D4 is there okay strong field Lian and weak field Lian so here you imagine here we have strong field Lian so splitting will be more okay so we'll draw here in strong strong field liant
• 18:00 - 18:30 the splitting will be more okay and in Weak field Lian the splitting will be less okay so now this is strong field liant and this is weak field liant okay so now we have here three orbitals here two and here also three and here we have two now D4 is there right so how you you should not remember it you have to understand it D4 is there means four electrons 1 2 3 it cannot go there because it requires more energ so it will pair up here so D4 in
• 18:30 - 19:00 strong field will be T2 G4 right but in Weak field what will happen 1 2 3 it is very less energy gap it can easily go there so here it will be T2 G3 and EG G5 now similarly D5 if you see fifth electron will also come here so we will get T2 G5 and for here we will get t2g 3 e G2 now if you see D6 again the sixth electron fifth electron came here six electron will come here so we will get
• 19:00 - 19:30 T2 G6 now for weak field if you see 1 2 3 4 5 was there and now sixth electron will come here so we will get T2 G4 EG G2 now for D7 if you see already T2 G6 has been there now T2 G7 will come here right so 1 2 3 4 5 6 7th electron has no other option it has to go there so we will get here T2 2 G6 EG G1 now if you
• 19:30 - 20:00 see for this one 1 2 3 4 5 6 and 7 I can fill here so I will get T2 G5 and eg2 okay so this D4 to D7 configuration for strong field and weak field you have to remember next color in coordination compound so very basic thing you have to remember there's no like the only thing you have to remember if there is unpaired electron okay if there is unpaired electron then it will be colored okay and if there is no unpaired
• 20:00 - 20:30 electron then it will be colorless how will you know whether it is colored or not for that you must know the uh configuration okay now the color is due to DD transition okay next coming to bonding in metal carbonal I told you this is not a very important topic but we can understand in a very short way so first of all there is a metal carbon Bond so there is a metal carbon Bond okay so carbonal donates electron to
• 20:30 - 21:00 metal that is what is done by every Li every Lian donates electron to the metal what special is here the speciality is the metal also donates back electrons to the carbon right so the metal carbon bond in metal carbon poses both S and P character so this is Sigma Bond and this is pi Bond okay now the MC Sigma Bond the sigma bond which is formed based due to the donation of electrons from the LI to the metal is formed by the donation of lone pair of elect Rons on the carbonal carbon into the vacant orbital
• 21:00 - 21:30 of the metal okay now MC Pi Bond okay this Pi bond is donated by metal to the carbonal is um formed by the donation of a pair of electrons from a filled D orbital of metal into the vacant anti-bonding orbital of carbon monoxide right so you can see here this one okay here um Sigma bond is from carbonal to metal and Pi bond is from metal to carbonal so anti-bonding orbital is used okay so these were the important topics
• 21:30 - 22:00 in coordination chemistry so what you have to focus only one day or like one and a half day is left so focus on your nomenclature isomerism and um D orbital splitting these are the three important topics that you must clear okay so now let us go to the next important topic that is D and F block so here we'll be discussing the important configurations important Trends important exceptions okay and import like the physical properties different dnf block that they
• 22:00 - 22:30 show different physical properties what are the reasons for that the questions can be asked from there right so let us quickly go through this chapter what are the topics first electronic configuration okay this is very very important okay without doing electronic configuration you cannot solve even a single easy question from this one you need to know the magnetic property you need to calculate the color whether it is color or colorless everything you can do only if you know the electronic configuration right so please remember Scandium to Z think all the electronic configuration very very important okay
• 22:30 - 23:00 next important Trends we will discuss next physical properties and why why this physical what is like what are the reasons for this physical properties and then we'll discuss about the F block elements okay okay so the electronic configuration is already there so nothing to discuss here you must remember here so just I'll tell chromium and copper has a special electronic configuration 3D 4 4s2 was the expected value but due to the how field electronic configuration what happens we
• 23:00 - 23:30 get 3d5 and 4s1 similarly for copper we have 3d9 and 4s2 is the expected electronic configuration but the observed is due to 3d10 4s2 so that it becomes a fully filled D orbital okay and we know fully filled electronic configurations are very stable next one more thing here is what is the general electronic configuration n minus1 D1 to 10 n is 1 to 2 okay so n -1 d 1 to 10 n S1 to 2 next so coming to the first Str
• 23:30 - 24:00 that is your atomic radius so first of all atomic radius decreases rapidly from column 3 to 6 so that is from your up to Scandium to chromium the uh atomic size decreases then it remains constant from 7 to 10 so you can say almost from chromium to Nickel it is constant and then it starts rising from 11 so I can write here from Scandium to C chromium then from manganese to nickel and then
• 24:00 - 24:30 copper to zinc it increases okay okay so now what is the reason for that here what happens the here the effective nuclear charge is more powerful okay here what is do do like the effective nuclear charge okay and the repulsion the D electron repulsion both are D electron repulsions are balanced out and in the last one if you see the D electron repulsions are more are more dominating
• 24:30 - 25:00 than the effective nuclear charge so if you can understand effective nuclear charge okay wants to try try to pull the electron towards itself and wants to reduce the size of that at and since we are as we go from the period the D electrons are getting filled so there will be inter electron repulsions between the D electrons so that wants to expand the size so there's a conflict between them so when we are going initially up to you can say up to chromium the effective nucle charge is a winner so what is its effect it wants to
• 25:00 - 25:30 reduce the size right so till chromium it will reduce the size now if you go in this reason you can see there is a like this reason is almost constant right the size is almost constant so at this point what is happening the inter electronic the D repulsions of the D electrons and we have this effective nuclear charge they're almost balanced out on each other now as we go copper and zinc d10 electrons right they're completely filled more repulsion so they are more dominating and the size keep keeps on increasing okay next we'll go ionization
• 25:30 - 26:00 energy so ionization energy increases with the increase in the atomic number up to Fe so as the size increases size decreases the ionization energy will increase okay next in Cobalt and nickel the increasing sharing of D electron compensates for the atomic number increase resulting in the decrease in the ionization energy so again as we say from Cobalt to Nickel as the D electron start filling so the effective nuclear charge decreases so that is why the uh
• 26:00 - 26:30 ionization energy decreases then again copper and zinc shows increasing ionization energy because here it is completely filled d10 orbital right so removing of electron from completely filled orbitals is again very difficult so that is why from nickel to Copper it keeps on increasing okay next metallic character and hardness so the D Block elements have the metallic character like high tensile strength Mal ility docility electrical
• 26:30 - 27:00 and thermal conductivity metallic cluster crystalize in BCC so these are basic things we need not like Focus here here it is important they're very hard and have high enthalpy of atomization and low volatility except copper right so copper is having 3d10 electronic configuration so it doesn't have uh like metallic bonding is very weak so that is why it is enthalpy of atomization is very low now hardness increases with the number of unpaired electrons so more is the
• 27:00 - 27:30 number of unpaired electrons more will be the metallic bonding and more will be the hard hardness of the metal now as you see chromium um volum and tungsten they're having the highest number of unpair electrons so they show the highest like they are the hardest metals and they also have very high melting point and boiling point but if you see zinc copper zinc cadmium and Mercury they are having like no unpaired electrons are there so no um metallic bonding and that is why they are very soft Metals
• 27:30 - 28:00 okay next is your oxidation state so if you see so they show variable oxidation state due to the participation of n and n minus1 d electrons so why both the orbitals participate because of less energy less energy difference okay now the higher oxidation States will be stabilized by chlorine and oxygen the lowest oxidation shown is+ one highest oxidation state in the entire 3D element is shown by Plus 8 that is your osmonian
• 28:00 - 28:30 then this you have to remember very very important which are your basic oxides which are your OTC and which are your acidic oxide so if you remember like a general thing if oxidation state is low then it is basic okay if oxidation state is somewhere in medium then it will be UIC and then oxidation state of if is very high then it will be acidic in nature okay so you have to remember aoic
• 28:30 - 29:00 especially remember all the examples very very important because they will be asking like which of the folling oxides are amphoteric or basic or acidic so you have to remember as many examples as you can okay now coming to the next one electrode potential so depends so here you have to know electrode potential like depends on what three factors it depends on sublimation energy ionization energy and hydration enthalpy next one more thing copper is the only element with a positive electrod potential of 0.34 volt so this one also you remember
• 29:00 - 29:30 okay next coming to physical properties okay we discussed about the chemical properties now some physical properties of transition elements so first is density okay so density increase remains almost the same and then decreases along the period so if you know density formula is what mass by volume volume we can since it is Spar we can like 4X 3 pi r CU okay so density is inversely proportional to radius so initially if
• 29:30 - 30:00 you see the radius was decreasing if radius was decreasing then um if radius will decrease then density will increase so initially decreases then radius remains constant density remains constant then towards the end radius increases so volume will increase so density will decrease right that is what happening now here something you can remember um Scandium in the third 3D series okay Scandium has the lowest density and copper has the high highest density okay and then copper has a
• 30:00 - 30:30 highest density it has different reasons like but you should remember the elements then osmium and iridium these are the highest density among all the D Block elements okay next we come about melting point and boiling point this is also very important to understand the if the metallic bonding is very strong melting point and boiling point will be high and metallic bonding depends on the number of unpaired electrons so chromium mum and tungsten have very high melting point because they have stronger
• 30:30 - 31:00 metallic bonding okay now manganese and technum what happens do although they have unpair electrons but they are half field configuration right so they're very stable they are not involved in metallic bonding so they have low melting point and boiling point now zinc cadmium and Mercury because they have no unpaired electrons they cannot undergo metallic bonding so they also have the lowest uh melting point and boiling point in their respective series okay now coming to colored ion so this is
• 31:00 - 31:30 very easy concept due to DD transition it happens if there is no unpaired electron they will be colorless so questions are asked which of the following are colored which of the following are colorless so you can easily use this concept and find it out next is we know that uh this D Block elements can form complex so what are the reasons for that they can know small ionic size because they have high charge and they have availability of D orbitals so you should know these reasons like question can be asked like 3 four factors will be given which factor
• 31:30 - 32:00 contributes for complex formation right so you should know these basic things next similarly catalic property is Du to the vacant D orbital and they can exhibit variable oxidation States alloy formation this is because of all the um transition elements have almost similar Atomic red ey so it is easy to mix them now interstitial compounds so they um so whenever you have like uh crystal is there there will be some spaces or voids will be there and that voids are filled with non-metallic atoms like here we
• 32:00 - 32:30 have nitrogen hydrogen oxygen carbon okay so when they enter and they form interstitial compounds okay next is very important this one important compounds in um in the compounds in dnf Block so first is your potassium di chromate that is K2 cr207 so if you see here the starting material you should remember many times this question is asked that it is prepared from chrom or okay the formula and the name iron chromite or is there
• 32:30 - 33:00 so this one is converted to sodium chromate then sodium chromate is converted to sodium D chromate then sodium D chromate is converted to potassium D chromate exact relation or equation you need to remember just remember the starting materials and the products of each step okay next is your structure so here question is asked like are all the bonds equivalent or not in D chromate right so you can see that these three bonds under resonance so they are equivalent but these these C
• 33:00 - 33:30 are are not equivalent right and the chromate iion and D chromate iion structure next we'll go to the next important compounds that is km4 so here also you can see the preparation manganese oxide K so the starting material is manganese oxide and it gives you potassium manganate and then from potassium manganate we get potassium diaper manganate okay so these four structures you should remember or the name you have to remember next coming to yes Lites very very
• 33:30 - 34:00 important electronic configurations are there but you may have you I'll recommend or I'll suggest you to remember all the electronic configuration but still if you have no time or you're not able to certain electronic configuration that you must remember is your lanum serium okay and then europium gulium and then you have the itum and lutum and yes these these elements electronic
• 34:00 - 34:30 configuration please remember okay very very important electronic configurations now so what is the general electronic configuration here also it is written 4f 1 to4 5D 0 to 1 and 6s 1 to2 okay now coming to the oxidation state so this is very important slide so common oxidation shown is plus three okay but in certain cases it can show plus4 and plus2 also so what are there if we can get a noble gas configuration so if you see serium okay serium has um 4 F1 5 D1 and 6s2 so
• 34:30 - 35:00 if it loses these all electrons it will have a noble electronic configuration of xenon right so that is why here it can show plus4 oxidation state now it can also show plus two oxidation state if it is getting a half field configuration so europium and terbium 4 plus so if you see europium if you see 4f 4 F7 and 6s2 so plus2 will give you 4 F7 right which is uh which is plus 4 F7
• 35:00 - 35:30 which is your exactly half field so that is why it can show and we have here 4f terbium is 4 F9 and 6s2 so if it loses four electron 2 electron will go here two electron will go here we will get 4 F7 this is also in that case also they can show plus 2 and plus4 state now next one is your yb2 plus if it is a fully filled level is also there then also it can show but General oxid state is plus three in certain cases it can show plus2
• 35:30 - 36:00 and +4 okay now let us see the next one what about atomic radius so we know from lannum to lutum that lanite contraction is there why due to the poor shielding effect of 4f orbitals so the size increases steadily okay now what are about colored compounds they are silvery white in color they show they are all colored compounds due to FF transitions now this is again very important slide lanthanide contraction and its consequences we we have already seen the causes what is the consequences the
• 36:00 - 36:30 atomic size okay of the fourth uh fifth period and sixth period elements are almost same so example you have to remember zirconium and hon harmonium and NB and ta okay next one is your uh due to this what happens due to this they have the similar chemical properties so it is very difficult to separate them and the last one is the basic strength this is very important lutum is the last one okay so effective nuclear charge is very high the electron pairs are very tightly hold it is very difficult to
• 36:30 - 37:00 donate so it will have the least basicity whereas lanum size or the effective nuclear charge is less its loan pair can be easily donated so its basicity is uh less okay it is high okay so now let us see the next chapter from your pu1 this is a very high weightage topic and very easy question the questions are asked from a certain section of the chapter not the entire chapter but we'll discuss the whole chapter but I'll also be telling you which chapter which topics sorry which topics you must Focus from where the questions are must asked okay so let us
• 37:00 - 37:30 start with the next chapter that is chemical bonding and molecular structure okay so first let us see the topics here there are lot of topics in this chapter but again the important topics are very few and very scoring topics first is your Louis structures how to draw the Lou structures second oate rule and its exceptions then we have formal charge calculation ionic bond calent Bond vper Theory Fan's rule then we have dipole moment then we have hydrogen bonding valence Bond Theory and then we have
• 37:30 - 38:00 hybridization and then Bond Bond parameters and then molecular orbital Theory I'll be telling you now which are the most important one so first important one is calent Bond formal charge calculation just we'll practice some formulas and we can do it dipole moment is not very important but sometimes questions are asked like which has a net dipole moment so you should know where it is cancelled and where it is not cancelled then hydrogen bonding also one or two questions have been
• 38:00 - 38:30 asked like intermolecular and intram molecular hydrogen bonding valence Bond theory is s hybridization is important Valance Bond Theory nothing much is there if you know vper Theory then uh just you know have to understand the concept Bond characters is important molecular orbital theory is very important okay now let us start with Lou structure so just what is Louis structure drawing you represent the bonds with the line the lone pairs the non-bonded electrons with the dot so we calculate the total valence electron and
• 38:30 - 39:00 then we assume that every element has octat rule satisfies the octat rule and then we fill the electrons right that is what is Lou structure now octet rule what is it according to octet rule every element wants eight electrons in its valence cell to gain stability right now there are some exceptions to this that you should know so that SPAC is like which have odd number of electrons n o and NO2 they cannot their OCT is not satisfied then we have we have incomplete octed and we have expanded octed so those elements where the number
• 39:00 - 39:30 of electrons of a central atom is less than 8 and there are some elements or some compounds where the number of electrons of the central atom is more than eight right so expanded octed is more than eight in complete OCT is less than it so if you see this one bcl3 okay boron has your three electrons in the valence cell right so it forms bond with three chlorine so if you see there are total six valence electrons for Boron whether it should be eight right so that is
• 39:30 - 40:00 incomplete octed now if you see here sf6 right sulfur has six electrons so we'll draw like this SF so six valence electron form six bonds with FL uh with six flines so the total number of electrons we get is 12 right so you should remember second period elements okay will not show expanded octet will not show expanded octet because
• 40:00 - 40:30 they do not have empty D orbitals the third period elements you can see phosphorus alfha these are all third period elements they can show expanded octed because they have availability of D orbitals okay now let us go next formal charge calculation very very important we'll do one example on this one for example ozone structure if you know resonating structure of ozone is this one right so your task is to calculate the formal charge on all the oxygen o1 O2 and O3 okay so we don't
• 40:30 - 41:00 know this one we'll draw so here you can see 1 2 3 4 so 5 6 7 8 8 electrons are satisfied 1 2 3 4 5 6 so 7 8 and here 1 2 3 4 5 6 7 8 okay now formal charge calculation formula is for every element the total number of valence electron okay total number of valence electron minus the total number of lone pair of electrons how many electrons are there in the lone pair and then plus 2 the total number of bonds means this is negative only this total you have to add
• 41:00 - 41:30 and then subtract from the total valence electron now we will do for this one for o1 oxygen has how many valence electron six right so it will be six minus then you see how many lone pair of electrons are there 1 2 3 4 not pair how many electrons you have to calculate so 6 + 2 that is 4 okay then here we have how many bonds are there not Bond pair of electrons only Bond pairs only Bond pairs so we have two bonds right so this one we will get zero so formal charge of
• 41:30 - 42:00 this oxygen is zero okay now let us calculate for O2 this is not oxygen okay second oxygen so there are six valence electron minus how many lone pairs of electrons are there two okay then how many Bond pairs are there 1 2 and 3 that is 6 - 5 that is 1 so we will get + one here now if we have O3 you can see six valence electron minus how many are there total number of
• 42:00 - 42:30 lone pair of electrons that is six and how many bonds are there that is 1 okay 6 - 7 we have -1 okay + 1 so here it is 0 + 1 minus one okay that is how you can calculate the formal charge now coming to ionic compounds very shortly we can discuss how the ionic compounds are formed by the transfer of electrons ionization energy so the metals has to release one electron the non-metal has to gain one electron and when they combine lattice
• 42:30 - 43:00 energy is released now what are the conditions so ionization energy of the cat should be low very easily we should be able to release the electron or remove the electron from the cation and high electron gain enthalpy of the anion and high lce energy right so means um electron uh the anion should be able or the non-metal should be able to gain the electron it should become stable it should release more energy and when thean and combine there should be loss of energy so that the molecule is stable
• 43:00 - 43:30 so lce energy should be high okay now coming to calent compounds again what is calent compounds it is formed by the sharing of electrons the electrons which participate as called Bond pairs which do not participate as cor are called as lone pairs this is very important because wper theory is completely based on lone pair bond pair right okay now coming to wper Theory so what does wper theory says the shape of the molecule is determined by the repulsions between all the electron pairs present in the valence cell these electron pairs can be
• 43:30 - 44:00 Bond pair can be lone pair okay so the order of repulsions you should remember lone pair lone pair then lone pair bond pair and then we have Bond pair bond pair okay now repulsions among the bond pairs is directly proportional to the bond order and electro negativity difference between the central atom and the other atoms okay yes this is the topic from where definitely every year we get questions a lot of questions are asked this table everyone whether you study chemical bonding or not this table everyone
• 44:00 - 44:30 should remember and go to the exam definitely a question comes from here the structure and the geometry you must remember so if electron pairs are two then it will be definitely two Bond pairs the shape is linear geometry is linear and the example you can remember next coming to three loone three electron pairs it has three Bond Pairs and zero lone pair then we'll get trional planer shape is also trigonal pler examples are bcl3 and BF3 now if we have two lone PA two Bond Pairs and one loone pair then it is a band shape example is s SO2 then we have four lone
• 44:30 - 45:00 Pairs and four Bond Pairs and zero lone pairs then it is tetrahedral CH4 and then three and one you'll have pyramidal 2 and two you'll get bent now if you have five electron pairs what are the possibility 50 4 1 3 2 2 3 right so here we'll get triagonal bam C C S T shape and linear this is important this is important square pamer square planer okay these are all important geometri so please remember them then for coordination number 6 we have 6 0 51
• 45:00 - 45:30 and 42 this table please take a screenshot remember it is there in the ncrt book also very easily available and very very very important table okay so this is just uh just I'll discuss like I have discussed you have to remember the examples properly okay now next one is your fan rule so here the questions are asked like they will give you the compounds and you have to arrange them in the increasing or increasing order of your like calent nature okay so for more
• 45:30 - 46:00 if the covalence is favored by when we can say that the molecule is more calent if the cation size is smaller anion size is larger and the cation or anion anyone should have the large more charge should be more okay like for example if we take um um for example NaCl okay ca cl2 and mgcl2 okay now if you see here first of all all the anions are same you have to check the Cates right so cat
• 46:00 - 46:30 size should be smaller so if you see there in the periodic table like this sodium and magnesium and then calcium so the smallest size is of magnesium right so magnesium will be the least calent nature or the compound with the least calent with the most calent character okay so like this you can find the order and now coming to dipole moment so dipole moment you know the formula is uh Q into d okay dipole moment formula is Q into D
• 46:30 - 47:00 and then uh what else is there here you have to know the units that is d by units and there uh there's more important here formula based questions are not asked like whether the compound is having a net dipole moment or not like for example they'll ask you 14 chlorobenzene and 1 two D chlorobenzene okay this question was asked in the previous year also now if you see here the dipole moment is in this direction this is in this direction equally OPP magnitude and opposite in direction so they will cancel each other now here what happens
• 47:00 - 47:30 it will be having a net dipole moment so this compound is polar or you can say have a net dipole moment this is non-polar compound even though it has a polar bonds okay next is your hydrogen bonding so for hydrogen bonding what is the con like what are the conditions that has to be there so the hydrogen must be attached to the electronegative atom okay Like Oxygen Florine and nitrogen and there can be like two types of hydrogen bonding one is interm molecular and intram molecular here you have to
• 47:30 - 48:00 remember if it is intermolecular hydrogen bonding then the boiling point will increase because the molecules are strongly attracted to each other we need more energy to break the bond so in that case the boiling point will increase and if we have intram molecular hydrogen bonding in that case the boiling point will decrease okay like two compounds are given in one like for example we have orthon nitrophenol okay and paranitrophenol
• 48:00 - 48:30 okay now if you see here here this one can have Oro hydrogen like intr molecular hydrogen bonding and here we can have intermolecular hydrogen bonding right so now if inter molecular hydrogen bonding will be there then it is difficult to separate them and the boiling point will increase so in this case the boiling point will reduce in that case the boiling point will increase okay so next next is your valence Bond Theory so what does the valence Bond Theory says a calent bond
• 48:30 - 49:00 is formed by the overlapping of valence cell atomic orbital of the two atoms having the unpaired electron so when the valence orbitals combine okay then that is called as valence Bond Theory Sigma bond is formed by head and overlapping and Pi bond is formed by sidewise overlapping here you should know Sigma bond is formed from SS overlapping and SP overlapping Pi bond is formed from PP overlap PX PX and py py and here we consider as s pz so there's no hard and
• 49:00 - 49:30 fast rule but as a convention we take that pz orbital is used for Sigma Bond formation but not it's like just a a convention used okay so how it looks SS will look like this sp sp will look like this and we have PX PX means it will look so PX sorry PX PX PX can also form Sigma Bond Okay then if we have one more um we can have here one
• 49:30 - 50:00 more uh Sig like this this will be forming your Pi Bond and we can have like this this can form your two Pi bonds okay now coming to hybridization so according to hybridization what happens the valence Atomic orbitals mix and form new hybrid orbitals so number of atomic orbitals this is very important formed is equal to the number of orbitals taking part in the hybridization so in your syllabus you have three important hybridization sp3 SP2 and sp sp will give you tetrahedral
• 50:00 - 50:30 sp3 trigonal planer and SP is linear here how questions are asked they will give you a structure carbon generally and they will ask you to tell you the hybridization of each carbon right so like for example so how you should know for determining the hybridization you should know that number of bonds number of Sigma bonds plus number of Lone pairs Okay so what number is that coming if it comes out to be four okay then
• 50:30 - 51:00 hybridization is sp3 if it comes out to be three then hybridization is SP2 if this comes out to be two then hybridization is sp okay so for example we have a structure like this we can draw L structure like this okay and you have to tell the hybridization of each carbon now you see if this first carbon has three sigma Bond three sigma Bond three sigma Bond means hybridization is SP2 this one is having also three sigma Bond so this is also SP2 this one is having four Sigma
• 51:00 - 51:30 Bond so it is sp3 okay like this you have to calculate next coming to bond characteristics this is very important here we have to discuss also how to find the bond order with a trick okay so first of all bond strength means what higher the the amount of energy required to break a bond okay now higher the bond order higher will be the bond strength so how to calculate the bond order I'll tell you a trick you have to calculate the total Valance okay total valence electrons if the
• 51:30 - 52:00 total Val valence electrons is 14 then the bond order should be three okay bond order sorry not total valence electron it is total electron okay okay now you have to when you go on the other side this value will decrease by half when you go above 14 or below 14 the value will decrease by half so 14 to 15 it will become 2.5 if 16 it is 2 17
• 52:00 - 52:30 1.5 18 1 13 again 2.5 12 2 11 1.5 10 1 now questions are asked for example you have to compare for N2 n2+ and N2 minus calculate their bond order or Bond length can also be asked now if N2 is there N2 valence total electron is 7 like atomic number is 7 7
• 52:30 - 53:00 + 7 14 so N2 bond order will be 3 now N2 plus means 14 minus 1 we will do 13 so 13 bond order will be 2.5 N2 minus means for so I'll make it into 2 minus for example so then what will there 7 + 7 14 + 2 16 so it bond order will be 2 then what is the bond order order N2 N2 + n and to 2 minus and next you have to see is very important higher the bond order
• 53:00 - 53:30 lesser will be the bond length so if the same thing we want to arrange in the bond length order it will become opposite N2 will have the least one N2 plus and then N2 2 minus okay so this is important this trick you remember questions are asked from bond order especially O2 N2 these are commonly asked so just prepare this formula try these questions it will be very easy to solve and scoring next we come about molecular orbital Theory so according to molecular orbital Theory it is the linear
• 53:30 - 54:00 combination of atomic orbitals okay Atomic orbitals are considered as wave functions here so waves can have constructive interference and destructive interference so if it is constructive interference we get bonding molecular orbital if it is destructive interference we get anti-bonding molecular orbital next point you should remember is anti-bonding molecular orbital have higher energy than bonding molecular orbital now the electronic configuration you should remember if it it is um like 14 or less than 14 the
• 54:00 - 54:30 total number of electrons then Sigma 1 is Sigma star 1 is Sigma 2s Sigma star 2s Pi 2px Pi 2 py Sigma 2 p z Pi star 2px Pi star 2 p y Sigma star 2p Z now if the number of electrons is like greater than 14 for O2 like 16 electrons are there then the electronic configuration will change and we'll get Sigma one so everything is same here what happens this one interchange okay we will get first Sigma 2 PJ then pi2 PX equal to pi2 py other rest part is same okay now
• 54:30 - 55:00 from this one you can also calculate bond order bond order is half of number of bonding electrons minus number of anti-bonding electrons from here questions are also asked to calculate the paramagnetic nature and the diamagnetic nature okay so now we will see one small example for this one suppose you have um N2 okay so N2 has 14 electrons so we have Sigma 1 s Sigma 1 star 1 s Sigma 2s Sigma star 2s right
• 55:00 - 55:30 after that it will come Pi 2 PX Pi 2 pi 2 py right so we can have Pi 2 PX and Pi 2 py so and then we have Sigma 2p Z Now you see 14 electron so I'll fill here 2 4 6 8 here 9 10 10 11 12 13 14 okay so now how to calculate the bond order bond
• 55:30 - 56:00 order will be bonding orbitals how many electrons are there these the ones with star are called as antibonding and the one without star are your bonding orbitals so 1 2 3 4 5 6 7 8 9 10 okay and we have four electrons here by 2 right so that is 6 by 2 that is three bond order is three now how to tell magnetic properties there is no unpaired electron here right so that is why it is a DI magnetic so if there is no unpaired
• 56:00 - 56:30 electron then it is diamagnetic and if there is unpaired electron okay then we call it paramagnetic okay now for example we will write another another example we'll take for oxygen molecule okay so for oxygen the order will change here so I will get here Sigma 1 S2 okay Sigma star 1 S2 Sigma 2 S2
• 56:30 - 57:00 Sigma St 2 S2 then Sigma 2p Z2 will come okay then we have Pi 2p x Pi 2 p y 2 right now we have Pi star 2px okay and Pi star 2p y now if you see till this one we have filled 14 electrons two more electrons are left 15 and 16 so for oxygen total electrons is 16 so if you want to fill the last two
• 57:00 - 57:30 electrons you have to fill like this okay you cannot fill two electrons in one orbital it has to be like if the orbitals are degenerate then all the orbitals should be first filled before starting the pairing up right so we'll get two electrons here so there is unpair electrons and as a result of which this molecule will become paramagnetic and you can calculate the bond order using the same formula okay okay so we have completed chemical bonding again if I want to tell what are the most important topics that you have to revise that is your vper theory very
• 57:30 - 58:00 very important molecular orbital Theory and then calculation of formal charges and then you can practice Louis structures how to do but not many questions are asked from here right so just focus on molecular orbital Theory wper Theory and formal charge calculation okay the key important things okay next coming to the next chapter which is very very easy chapter that is your classification of elements and periodicity in properties let us quickly revise this chap chter some very very very basic things about periodic table we have to understand so what
• 58:00 - 58:30 we'll understand first here what is modern periodic law how to write the iupsc nomenclature for elements with atomic number one to sorry atomic number greater than 100 then classification of elements and then periodicity in properties okay so first one is according to period modern periodic law the functions of an atom depends on the atomic number okay it is not on the atomic mass according to Mandel law it was according to uh like according to the M's law the properties depend on atomic mass but according to Modern law
• 58:30 - 59:00 the properties depend on the atomic number next is um the why there is periodicity because there is a similar configuration repeated over the time and then there is the in modern periodic table it is based that the elements are arranged in the increasing order of their atomic numbers and not in the increasing order of their atomic masses and next we have that there are seven periods and 18 groups n is the principal Quantum for the outermost shell and that tells which group The Element belongs and we have
• 59:00 - 59:30 same number of electrons is present in the outer orbitals to find the electronic configuration okay so these are some basic things you should know seven periods and 18 groups are present now coming to how to write the nomenclature of numbers greater than 100 for example there is an atomic number of 1 03 right so for one what is the name that is UN okay for zero what is it nil okay and then for three it is try so u n
• 59:30 - 60:00 and uh for Tre so the symbol will be u n t okay like this questions uh you can just practice very easy thing this one and now coming to the classification so periodic table has three major groups sorry four major groups s block P block d block and F block so s block is your from U and ns1 122 is the electronic configuration right and they are generally metals okay and so here what should be the one second yeah this
• 60:00 - 60:30 should be generally metals and which elements they belong they are the they contain your group one and group two right next you come to P block elements the general electronic configuration is ns2 to np1 to six they are generally metales and non-metals but some of them are also metals and they consist of Group 13 to group 18 1 to2 and that's at 13 to 18 the middle part is your D Block Elements which has a general configuration of n- 1 d 1 to 10 and N is 1 to 2 here uh here this is um next we have
• 60:30 - 61:00 there generally metals and we have group 3D series 4D series and 5D series and they contain 3 to 12 groups okay so you can see 1 2 2 then 3 to 12 and then 13 to 18 okay next coming to your F block elements n to n minus 2 F1 1 to 14 n is2 n-1 P6 and nus1 d0 this is there for this is for your 6 and seventh period so these elements are generally radioactive and we have two series 4f series and 5f
• 61:00 - 61:30 series okay so this is General classification of the periodic table now we will quickly go to the atomic Properties or periodicity in the properties first one is the atomic radius very simple thing it decreases from left to right along the period it decreases and it increases down the group okay effective nuclear charge increases so size decreases along the group new cells are added so size increases Ines now what happens if you see in 3d Series this is given about 3D
• 61:30 - 62:00 series we have already discussed so there is slight decrease from Scandium to manganes because the effective nuclear charge is more dominant then iron to Nickel it remains constant because both the effects balance each other and then if you see from copper to jinc the um size increases because the shielding effect is more than the effective of the nuclear charge okay now um so in the lanite series what happens there is a steady increase in the size why because of lonite contraction due to the poor shielding of 4f orbitals okay
• 62:00 - 62:30 so you need to remember these two important things now coming to ionization energy so ionization energy increases along the period and decreases down the group okay now what are the factors on which the ionization energy increases atomic size if the size is small ionization energy will be more effective nuclear charge is more ionization energy will be more penetration effect of the orbitals so s orbitals are more in inside the nucleus so if you want to remove the electron from 2s and 2p more electron energy will
• 62:30 - 63:00 be required to remove from 2s because they are more penetrated or in enter towards the nucleus then shielding or screening effect if the shielding effect is poor then ion energy will be more if shielding effect is strong then ion energy will be less then stability of half field and fully field orbitals so if there is any half field or fully field configuration then ionization energy increases okay so like for example if you see nitrogen and carbon nitrogen and oxygen right what is the electronic configuration of
• 63:00 - 63:30 nitrogen 1 S2 2 S2 2 P3 for oxygen it is 1 S2 2 S2 and 2p4 right size is decreasing oxygen is smaller than nitrogen along the period so according to size this oxygen should have more ionization energy but you can see nitrogen has half field electronic configuration as a result of which the um ionization energy of nitrogen is more than that of oxygen okay so this exception you can remember same applies for phosphorus and
• 63:30 - 64:00 sulfur now next one is your electron affinity so electron affinity increases along the period the tendency to attract electron as the size increases effective nuclear charge increases so the tendency to accept electron increases and decreases down the group okay now factors affecting here is again at atomic size so if atomic size will be less more nuclear charge more electron gain enthalpy or electron affinity will be there if effective nuclear charge is more then also more attraction will be there shielding or screening effect if
• 64:00 - 64:30 screening effect is poor nuclear charge will be higher and the electron affinity will increase now coming to electro negativity so in same as electron affinity increases along the period decreases down the group okay now what are the factors affecting here again atomic radius effective nuclear charge oxidation state hybridization so what is hybridization more is the S character okay more s character more electronegative okay now for example if
• 64:30 - 65:00 you see SP 3 SP2 and SP so here 50% is s right here around 33% and here around 25% so more s character so that is why the electro negativity of SP will be carbon with SP hybridization will be higher okay fine so yes so this was the complete inorganic chemistry that we have discussed questions like you can easily EXP 12 to 15 questions from these topics revise them thoroughly note it down and do some questions from this so
• 65:00 - 65:30 I hope you have found this video very helpful just it's like one day just take it and watch it completely and note down all the important points all the exceptions all the trends and DB block what are the properties physical properties what are the reactions preparation method so just note down it everything and revise in two three like one and half hour like 2 hours and the complete inorganic chemistry will be revised so I hope you have found this video very helpful please like share and subscribe our channel for further updates and the PDF of this uh this uh
• 65:30 - 66:00 inorganic revision the all notes will be available in our WhatsApp channel the link of the WhatsApp channel will be there in the description section so download it revise it and then go for the exam all the very best thank you