CSIR DEC 2019: Inorganic Chemistry | Detailed Solution | Section-B (15th Dec)

0:00 hey guys hello everyone and welcome to the channel now in this particular video we are going to solve quotients from inorganic chemistry part which were asked you know CSIR net exam of December 2019 and I'll be taking the exam or that held on 15 top December that means the All India level exam dollar India exam except the sum and makalah region exam and I will be taking two marker portions in this video right because in the last video where I have discussed about the physical chemistry solution now you guys requested me to solve inorganic solid quotients as well so that's why I have made this video right now if you have not watched that physical chemistry video then I will suggest you to watch this video completely at the end you will get as a card over there and you can just click on that to watch the

0:46 solutions of physical chemistry part as well alright so let's start with it so the first question over here was based upon a very simple concept it says that the ion having the highest bond order is okay so the ion which has the highest bond order so the basic trick which we generally apply for calculating the bond order is by calculating the total valence electrons so if you know the total valence electrons you can easily tell the bond order right so if you start it from 14 so if the total valence electron for diatomic molecule or for the molecules basically if the total valence electron is 14 your bond order okay if I write down the bond order over here so the bond order remains three over here okay so for fourteen the bond order remains three for thirteen your

1:31 bond order decreases by by 0.5 value so it will become 2.5 for 12 it becomes 2.0 for 11 it becomes 1 point 5 for 10 it becomes 1.0 and so on okay similarly for 14 it is 3 for 15 it will again decrease by 0.5 so it will become 2 point 5 for 16 a total valence electron it is going to become 2.0 then again for 17 it is going to become 1 point 5 and so on okay for 18 it is going to become 1.0 so it goes like that so by calculating the total number of valence electron you can easily predict the bond order now looking upon the options if I talked about the first one which is and o+ so total number of electrons you are going to get seven electrons from

2:18 nitrogen and you are going to get eight electrons for oxygen and plus one charge is there right so 7 plus 8 and minus 1 minus 1 for that positive charge so overall you will get 14 right so 14 means the bond order for this particular one should be 3 okay so for 14 the bond order becomes 3 all right so obviously these 3 is the highest bond order which you can expect from the molecules but yet let's check the other one as well 402 plus if you see over here for each oxygen 8 electrons so 8 plus 8 and minus 1 because of this positive charge so you are going to get 15 and again the bond order for this is going to become 2.5 ok so this is going to be the bond order here the third part if you see nitrogen so n2 so 7 plus 7 for both the

3:05 nitrogen's and minus 1 because of this positive charge so again it's going to be 13 and for 13 the bond order will again become 2.5 so it is again having a lower bond order and the last one that is C 2 plus so for each carbon you will be having 6 electrons each right so 6 6 and minus 1 because of this plus charge so you will have 1111 means you have a bond order of 1.5 so that means what it means that the highest bond order for this particular question which was asked is 3 and that remains in your end o plus molecule so the correct answer becomes your option number all right so it's very simple very easy question let's go to the next one particular portion it

3:52 says that the expected number of carbonyl bands in the IR spectra of facial complex this is molybdenum pph3 whole try CU whole trace and you have trans molybdenum pph3 whole twice CU whole 4 so now first of all you need to know about like how you can assign these carbonyl bands using the using this structure ok so if you go with the trivial way to find out the type so see you then you will end up doing this quotient wrong ok you have to understand the application of group theory to understand or to find out the spectrum band so to find out the carbonyl bands okay so this is actually a very direct question if you have practiced based upon how how you can assign different IR bands or different

4:38 carbonyl frequencies for different types of mall like metal complexes you could have done this portion very easily so I'll just give you this in very briefs in very like in a very simple way in a very very precise way so I will not be going in much detail so let's first of all draw the complex so the complex will actually looks like something like this all right so this is how both the complexes will look like this is your facial complex and this is your that trans complex which is given to you now ideally if you look upon you will find out that these all carbonyls are active actually they look like equivalent so they should give just one peak okay and many people did it like that only and they thought that the first molecule is going to give you just one peak but this

5:25 question will not be done like that you have to see the point rope in this particular molecule so since we have a c3 group over here you know and you have Sigma V planes you know the point blue for facial complexes right it's C 3 V and for this particular molecule if you look like it will be your D for H ok so that is going to be the point group of this now there is a very basic rule to find out the number of eye airband so to find out the IR bands basically you can just relate the air bands like that it says that lower the symmetry elements or the lower the symmetry higher or more is the number of higher the number of hear bands okay so or your carbonyl

6:12 frequencies if you are talking about carbonyls over here to be very precise so if you have lower symmetry more number of IR bands will be there if you increase the symmetry your number of bands are going to decrease for example it's a very simple molecule if you take a metal octahedral complex m ceo's whole six this is just going to give you one peak because it's a very high symmetry element over here the the symmetry group or the point group over here is octahedral so it's a very high symmetry group which is over here and that's why you will just get one peak if you lower down that symmetry group and if you just make it as MCU six sorry MCU 5l if I make it like that so it is going to give you not two peaks

6:57 now we're now the symmetric group has been changed and it has become c4v okay so if you if you decrease the symmetry groups or if you decrease the symmetry that means the number of higher bands are going to increase so in the same way you can understand from here that in this particular case you're symmetric Paes is quite less okay and that's why the number of bands should be more over here so over here what happens that these equatorial carbonyls they they are going to give you same band whereas this axial carbonyl is going to give you a different band so the total number of bands are two and actually four c3v we calculated along with this like if you have some so for c3v we calculated using

7:43 this particular formula that is n minus one okay so with this n minus 1 you can calculate the number of bands over here so the number of bands for this particular case is going to be two bands okay or two peaks so two peaks you will get for this case whereas for this all these co groups are going to be equivalent to each other because it's it is a higher symmetry element and also it has like all the carbonyls will be equivalent to each other if you rotate it along c for X's so they you are just going to get one peak in this particular case so it was pretty simple to understand this but for this it's you have to remember this that all say shall I so much they give you two peaks and it is because of the lower symmetry element okay because of c3b and if you go for the maratona lie so

8:29 much okay fine so that I'm giving to you guys to find it out okay so try to draw the meridional isomer and try to find out that how many Peaks you are going to get fara for a meridional isomer of the same complex okay so that is for you guys you guys have to do it and tell me in the comment sections below so the correct answer for this portion was your option number three alright so let's go to the next question then what I talking about this particular question now it says that correct order of molar extinction coefficient values of the visible absorption spectra or visual accept option bands for the following species is now you are given with four different types of complexes and it is asked that what is the order of molar extinction coefficient now few things which you need to understand over here is that your molar extinction

9:16 coefficient okay molar extinction coefficient coefficient is actually going to be proportional to your move to your molar absorptivity okay if you know molar absorptivity you can easily tell the molar extinction coefficient from their form right and this molar absorptivity is actually going to be proportional to your intensity of the band okay that how intense the band is intensity of absorption okay intensity of absorption absorption we're now in order to understand that on the basis of your transition metal complexes it is very simple to understand that how

10:02 intense a particular band will be based upon the selection rules right so if the molecule follows the selection rule so in that particular case you can tell that the band will be intense or be so I will just give you quickly as order to identify this so the first one is there are two rules right there is a spin allowed rule or a spin rule and there is a LaPorta rule so both of the rule has to be followed so their spin rule says that the Delta s spin that means the spin during the transition should not change and there should be change in the parity that is called as the Laporte a rule right so if something is spin allowed as well as Laporte allowed both in that case is intensity will be highest then if you have a spin allowed okay it's been allowed but your Laporte

10:48 a does not allow okay so it's not equal to plus minus one in that particular case also you have two types of cases first is a non centrosymmetric molecule okay non centrosymmetric molecules so those which do not have center of symmetry ins symmetry in them they will give more intense band compared to those which are Centro symmetric so here again the formula like the rule is same that is spin allowed but Laporte a not allowed but here it is Centro symmetric okay this is Centro symmetric molecule okay and the least intense is the one which does not follow even the spin rule neither the neither the Laporte rule that means Delta s is not equal to zero and Delta L is not equal to plus minus

11:36 one now let's quickly take some examples for them so for this where spin is all fill our Laporte is also allowed these are generally your PI to PI star transitions or you can say that your charge transfer spectra okay so all types of charge transfer whether it is LM City ml city a ligand to ligand charge transfer intravenous charge transfer all of them are charge transfer spectra and all of them are more intense okay that's why you get very intense color or very intense band in the charge transfer spectra non centrosymmetric means the structures which does not have a center of symmetry example of them is a tetrahedral structure Centro symmetric means those which have a center of symmetry in them for example complete octahedral species for example if metal

12:22 and there are four sorry there are six ligand of the same type attached to this and here if this is the metal and there is four types of ligand attached to that okay and the last one is where your spin allowed and that the transition is neither spin allowed neither neither lip or tail out these are generally your d5 transitions or detent transition where if you cannot do that transition right there's the higher orbital is already filled so these are some of the examples which I have given there are many others which can be discussed but for the current scenario you need to know this okay so this is the order of intensity of pair now based upon that if you understand the intensity you can easily predict the molar extinction coefficient so we are just going to predict the intensity now in the molecule if you look like if you look upon the options given so there is a charge transfer

13:08 which you know it is going to happen in chlorophyll so chlorophyll is going to give you the highest intense band okay so chlorophyll okay chlorophyll chlorophyll is going to give you most intense band then in the other there is no charge transfer spectra happening like neither in NI CL 4 neither in CR h2o 6 neither in ml h 2 6 so what you have to look upon is whether the molecule is Centro symmetric or not so ni CL 4 2 minus is going to be a da transition like sorry D it molecule over here and this chromium over here is going to be a d7 molecule right and this is going to be a d5 molecule no if you look upon them so ya so non

13:54 centrosymmetric so this transition in ni CL 4 is going to be where the spin the spin is not going to change this it's going to be spin allowed transition but there is like it's not lip or tail out right because it's a DD transition which is going to happen over here in that era until species and it's a non centrosymmetric molecule so that's why the next molecule will be ni CL 4 + 2 - alright the next one will be D 7 where there is again spin allowed transitions you can have a spin allowed transition in D 7 as well right if I just show you so one two three four five high spin D seven so five six and seven so you can have a spin allowed transition because the spin is not going to change it is easily going to transit but it's a central

14:40 symmetric molecule so that's why it will be less intense so the other one will be CR h2 whole six + 2 + alright and the least one is going to be the d5 species where the transition is neither spin allowed neither it is Laporte allowed because if you see the d5 structure d5 octahedral a species so it is going to be like this so you cannot have a like same spin transition because over they are already our orbitals are filled and if you want to do the transition you have to change the spin which we don't have to do because it's not following the spin rule so the least intense will be your MN h 2 o 6 2 plus so that in that particular case you can say that your answer is option number 2 so it was pretty simple you just have to look upon

15:27 the intensity order of the adsorption built upon that you can tell the molar absorptivity and molar absorptivity and molar extinction coefficient both are directly proportional to each others with that particular formula a is equals to epsilon C L so you can tell that answer option number 2 is going to be the correct answer for that all right let's go to the next question then this was actually asked from direct reaction of your main group elements and once of this question you don't have to apply any concept and all you it's just based upon whatever you have studied if you remember there you could have answered this question very easily okay it says that consider the four species a b c and one more species is there B and it is said that when a and B reacts in the of potassium iodide and in the acidic medium you are getting C and when you do

16:15 oxidation of a with C that means it's a and C get oxidized in the acidic medium it gives you some compound d so you're asking what is a B C and D so this is actually based upon the redox titration of iodine with with your s 2 O 3 okay so you might have studied this I guess but yeah see actually how does the reaction happens to it based upon Standardization condition so we standardized first of all our methanol fate on the basis or like we standardize the thiosulfate with ki o3 okay so based upon that we are going to do so since ki yo 3 is where in all the options we can just predict that okay it's a reaction of ki u 3 so let us try to analyze the reaction so what actually happens that we you take ki o 3 and you treat it with potassium iodide

17:02 in the presence of your HCL now okay so this is going to give you iodine and then you are also going to get KCl okay so KCl potassium chloride and also you are going to get water as a by-product now this iodine which you get in this particular step this is called as the standardization condition okay as a standardization process your standardization using potassium K I go 3 okay now if you have standardized it you have found iodine now this iodine is actually going to react with n a 2 s 2 O 3 that means basically you can call it as s 2 O 3 2 - and when you react to it

17:48 so you are going to get sodium iodide and with that we'll get an A 2 s 4 okay as 406 or you can say s 406 2 - okay so basically if you look upon the reaction over here so they have already given ki and h+ so these two things ki + h+ is already given to you so that means your ki o 3 is somewhere over here okay and be among a and B something is your ki O 3 then the second thing and the product which are getting is C so your product is see that is iodine so this is going to be your see this is going to be your B and when you are reacting this C with some compound a what is going to be a so this is going to be your C right C reacting with some

18:35 compound a or this is going to be your a so when C reacts with a you are going to get D so this is going to be your D okay now you got all the things so that means a over here will be present but it has nothing to do with the reaction since you are standardizing it so that's what it has to be there in the reaction that's what is kept okay so this s 2 O 3 2 - we'll be there in the reaction so this is how the reaction is actually so you're a comes out to be s 2 O 3 B comes out to be ki O 3 C comes out to be iodine and your d comes out to be s 406 2 - it was a direct reaction asked from your books so option number 3 is going to be the correct answer for this alright now let's go to the next question then alright let's talk about

19:22 this particular question it says that the common hepta City observed for coordination of C 60 or to a metal center is now to answer this particular question it was pretty direct ok if you have studied about C 60 that is buckminsterfullerene you could have easily answered this but if you don't know I am going to explain you a little bit so in the buckminsterfullerene if you look upon this structure so it looks somewhat like this ok if you draw the aromatic structure of this so it looks quite quite equivalent to this now if you want to look upon a certain fraction of it let me just show you a particular fraction and with the help of that we will try to understand this so I guess you guys have already studied about this

20:08 kind of molecule when you are preparing for NMR ok we call it as peripheral aromaticity or or in the organic chemistry also where you have studied about the peripheral or aromaticity and there you might have seen if you have a certain molecule like this ok if you have certain molecule like this in that particular case what happens that the aromaticity off side like ring is going to constitute like that is enough further Marrissa T of the molecule the central double bond or this particular double bond is not taking part in their omote issah t so this kind of effect is called as peripheral aromaticity i guess many of you know about it if you don't know about it I will make a separate video on that it's based upon general organic

20:54 chemistry and it's a very well known and very famous concept so this particular these particular carbon these two carbons which are not involved in the aromatic City they have this particular double bond which can be given to the metal in order to form the complex I so just the way we have a complex like this let's say if you talk about any gentle lamb you have complex like this so here you have a 18 molecule so it's giving electron the PI electrons of this carbon-carbon double bond so in the same way this particular species is going to give this the electron this pi electron of this carbon-carbon double bond so this is going to be called as ETA 2 or the hapticity will be 2 because the electrons between these two carbon is going to be shared in the same way if

21:40 you look upon this buckminsterfullerene there is a part of it which will be available over here in the molecule you can see that this particular part will be available for them for making the app tricity ok so in the complete molecule they will be like you will get many of this kind of part ok you just need to look up on this part ok I just I'll just erase the rest of the part so this much part you just need so in this kind of part only these these particular two carbons are going to constitute this black carbon-carbon double bond and this is going to be shared to the metal in order to make bond or that's why the hapticity becomes 2 now for all of these fragments whatever number of fragments we have in the complete buckminsterfullerene structure everywhere in all of these fragments - 2

22:26 will occur like two of them will be given you don't have to add all of them ok but overall how many carbons or how many like how many PI bonds between the carbons are going to be shared so there are just two carbons between them 1 PI electron is going like 1 PI electron cloud is being shared to the metal so that's why the hapticity becomes 2 over here ok so the PI electrons between the two carbons are only being available for sharing with the metal so the hapticity becomes too so there is direct solution or direct answer of this in many books okay if you have read about books if you have read prepared well you might know the answer like you might know the answer as it is but this is how you can explain or how you can understand this

23:12 that how hapticity becomes - for say 60 metal center right see 62 the metal center now let's go to the next question then alright the next question over here says for an octahedral copper plus two complex depicting axial EPR spectrum where it is given that your G parallel is greater than G perpendicular the geometry of copper plus 2 and orbital containing the unpaired electrons are now see if you have watched my classes if you have watched my videos on YouTube there are many videos on on inorganic chemistry which are already available on the channel I will recommend you to watch them very carefully because many questions if you see they were directly taken from there ok so that this was one

23:58 of that question which was directly taken from my video and if you know that if you know the concept if you have watched my video you could have done this question very very easily so what it says that G parallel is greater than G perpendicular what does that mean it means that electron density electron density along z axis okay is more okay is more than electron density along XY axis that's what it means if G parallel is more than G perpendicular it means what it means that in copper plus 2 which is a DNA species you will have more electrons field along the z axis then along the XY axis so if you if you see this copper plus 2 complex which

24:44 shows jahn-teller distortion so you can say that you you will be filling the electrons like this okay so electrons are going to be filled like this so this is how you are going to fill at all and more electron has to build z axis so that means DZ Square will be lower in energy and DX square Y Square will be higher energies more electron along the z-axis this will be d XZ d why is it then this is going to be D XY so this is how your D orbital is going to split for this okay so this is D orbital and this is how it is going to split out in the case of your copper plus two is showing the jahn-teller distortion now the condition is fulfilled that electron density is more along the z axis what kind of structure it is is it elongated or compressed now you know this right if you have studied

25:30 the concept you know that when if the elongation of the z axis appears when there is a long Gatien in the Z or the axial bonds what happens that the energy along that particular axial direction the energy of that particular orbital along the axial direction decreases so the orbital along the axial position is DZ Square energy of that is lower so this is like this particular splitting of the orbital is for tetragonal a elongated structure right and the unpaired electron is going to be in what so the unpaired electron is there in DX square Y square so the correct answer is option number one tetra kunal elongated then the unpaired electron is in DX is square Y square orbital so this is how this has to be answered let's go to the next question that particular portion

26:16 which was actually asked from your nuclear chemistry and it says identify the following the product of K electron captured by the nucleus what does that mean K electron capture K electron capture means nothing but what happens actually that to insert a nucleus the proton of the nucleus actually it it absorbs or it captures the electron from the K or L level okay you know that there are different orbits like K shell electron shell electron shell so what happens that in electrically neutral atom the inner atomic electrons from the K or L shell are being absorbed by the nucleus and this process is called as your K electron capture so what actually happens that your proton over here is going to absorb that electron from the

27:02 from the K shell okay so it can be from K shell or it can be also from L shell okay so for L shell it will be called as an electron capture both the processes are same it just depends upon which from which shell direct is being absorbed so it can be either case shell or else shell it depends okay so either it is a k electron capture or l electron capture in both the cases proton is going to absorb that electron and it is going to form a neutron and along with that you will get a neutrino okay you will get a neutrino which is going to get formed so Neutron a proton is going to get converted into Neutron and simultaneously emission of an electron neutrino is going to happen okay so the answer will be formation of your neutron and also a neutrino is

27:49 going to get eliminated in this particular reaction so I will just write down so what is happening that a proton is the absorbing electron from the K shell or L shell and it is forming Neutron and what is being eliminated is electron neutrino okay neutrino so that is being eliminated so answer will be option number two which says a and B port alright so this is how this portion has to be there now let's go to the next question all right so let's talk about this particular question over here it says that the most stable vanadium species in the aqueous medium is not once of this particular portion you need to know quite a few things first of all that

28:35 what is the most stable oxidation state of your vanadium okay so the most stable most stable oxidation state of vanadium is actually plus 4 so plus 4 is the most stable oxidation state of vanadium now over here in plus 4 oxidation state it can is like you can easily eliminate few of the examples if you look upon it so in the first example of in the first option your vanadium is in plus 3 state ok the oxidation state is plus 3 and the second option it's in plus 4 in the third option it's in plus 3 right again because oxygen it will give you minus 2 and plus 1 so overall plus 3 and in the last one again it's in plus 4 oxidation state minus 2 from here and plus 2 from here so overall plus 2 so you can easily

29:21 eliminate option number one and option number 3 so you are now left out with option 2 and 4 now the second thing which you should know is that the most stable oxidation state of an idiom is +4 first thing second thing is that the most stable form of vanadium is the observing idioms form okay so the most stable form most stable form of vanadium vanadium is also like it will be also vanadium I am okay also when Adium vanadium 4 I am or it is also called as vanadyl okay so that is also called as

30:07 when a deal so the most stable form will be of the vanadium ion in the plus 4 oxidation state that means vanadium with oxygen so if you look up on the option like you are left out with option 2 & 4 there is just option number 2 which has oxygen and sorry vanadium and oxygen that means you have of the vanadium form in this okay so the correct answer is going to be your option number ii so that's how you have to answer this question all right now let's go to the next question all right let's talk about this particular question over here it says for a complex shown below in non fluxional state the expected 31p and decoupled proton NMR resonance okay so this is phosphorus hydrogen decoupled

30:54 NMR resonance and it is given that the AI value of phosphorus is 1 by 2 so if you look upon the structure over here how many types of phosphorus do we have so first thing is to find out the number of phosphorus okay so you have two types of phosphorus because this is one of the phosphorus now which is on the equatorial position and assist to it you have you have see you pH group whereas this is second type of phosphorus assist which is is your carbonyl group ok so you don't have and and this Co pH group is transferred so both of them are different types of phosphorus so this is phosphorus a and this is going to be phosphorus B now the the the in the EPR respect house or in the NMR spectrum if you see so how many splitting is going to happen so since you have one of this type so 2 and I 2 n I plus 1 I am in sty

31:42 value of the a type of atom and this is going to be 2 into because you just have 1 phosphorus of that type and I value is 1/2 and plus 1 so you are going to get how much you are going to get to that is that means you are going to get doublet because of that similarly for second one also you will get 2 n ib and plus 1 if you put the values so 2 into 1 into half and plus 1 so if you just cancel out so you will get again - that means doublet so what you are getting going to get is doublet and doublet so overall the spectra of the molecule is going to look like to doublets ok so the answer is going to be 2 doublets for this particular model it's it was very simple you just have to see that types of phosphorus and just find out the number of splitting based upon 2 n I plus 1

32:29 formula alright let's go to the next quotient then ok the next question which says the magnitude of bond angle okay so how the magnitude of bond angle is going to change in the gaseous NF 3 s bf 3 and s be CL 3 alright so to answer this particular quotient is a very simple concept which you should know about the bond angle and it says that bond bond angle bond angle is actually directly proportional to electronegativity electronegativity of central atom okay and it's inversely

33:14 proportional to electronegativity so the inversely proportional to so inversely we have to write down like 1 upon so electronegativity electronegativity of surrounding atoms that means if the central atom has more electronegativity the bond angle will be more and if the surrounding atoms have more electronegativity bond angle will be less you have to go for this when the central atom is same okay unless an intelligent central atom is same and unless and until you have different central atom you have to stick to the first two so here you have three types of molecules we have n f3 then you have s bf3 and you have s bcl3

34:04 okay that's what you have now if you look upon the electronegativity order for the central atom central atom is nitrogen and antimony only so if you look upon the nitrogen electronegativity okay electronegativity if I talk about electronegativity so nitrogen has more electronegativity than then the antimony right so that means what the bond angle of your nitrogen containing compound that means nf3 is going to be highest okay so nf3 is going to be highest so by that you can easily cancel out option number two and three now looking upon the other sector where your central atom is same now you have to look upon the electronegativity of subsidiary atoms okay or the surrounding atom so in that particular case the electronegativity order is like this that fluorine has

34:50 more electronegativity than chlorine so if you like to negativity is more the bond angle will be less right so what you can say is that the highest bond angle will be for nf3 then among here flow Dean has more electronegativity right so it's bond angle the the antimony fluorine compound will have lower bond angle alright more electronegativity less bond angle it's like that and chloride one will have higher bond angle so it is going to be like s bcl3 will be more and then you have s bf3 so in that particular case your correct answer becomes your option number four you can understand it in this way also if you want to understand in a more trivial way just like you have been taught in your bachelors or masters it's like when the central atom becomes more

35:37 electronegative electrons will be more concentrated towards its center so if I show see the structures of all of these is going to be same okay it's going to be trigonal or sorry it's going to be pyramidal right the structure is going to be pyramidal according to your VSEPR theory so that means nitrogen and you have these three bonds which you can show them like this okay so the three bonds can be shown like this now you have fluorine over here so if if the center and you have antimony again where you have let us say fluorides like this so yeah so if you have this so if the central atom is more electronegative electrons will be much closer to the central atom in these particular bonds

36:22 right and if they are more closer to is there will be electron-electron repulsion in this and that is going to make the bond angle high okay so these bond angles will become higher over here and that's why they are going to be like the nitrogen compound is having more bond angle so that's the nf3 is higher in there now talking about the one where you have fluoride and chloride for that you can understand in this way that if you have let us say let's say s bcl3 also we have okay so what is going to happen in this particular case so your electron electronegativity over here is more so your electrons are going to be more near to your flow rate of fluorine right so since electrons are away from the central metal atom and here they are not going to be that much closer to chlorine so more electron electron repulsion one thing the second thing is

37:08 that due to bigger size of this chloride ions there will be like inter atomic repulsion so that is going to cause more so SB Steele will have higher bond angle than SF okay so this is other explanation of the same thing or it's much easier to understand it through this particular relation you will never do this quotient wrong whenever it will be asked further than in the exams okay so the correct answer becomes Xerox option number four let's go to the next question then okay the other question which says that the choose the correct statement for oxy myoglobin and cytochrome p450 resting-state okay both of them are in resting state from the following okay there are four statements given to you for that this was actually asked from your bioinorganic chemistry

37:55 okay and i always suggest you to keep on studying by an organic chemistry keep on reading the terms or topics from the by an organic chemistry if you have studied by oxy myoglobin and cytochrome p450 you could have answered this portion very simple okay it was very very simple molecule very very simple potion so if you look upon the structure basically so for oxy myoglobin you have iron and it's attached to pour fire in drink like this okay so you have pour fire in drink or to which it is it is actually being attached okay so I'm not going to draw the molecule but yeah this is how it is going to look like ii also is going to have a central metal atom at as iron all only okay and again the

38:41 power siren ring is attached to this so in both the cases your central metal atom is actually attached to pour fire in ring so both contains dye and iron off four of protoporphyrin okay so option is correct for both of them so i can say is that both of them has six ligand bonded to the metal center from the protein backbone okay now if you look upon and the same fifth ligand okay so if you look upon this structure there is a there is a fifth ligand which is attached to both of them for your oxy myoglobin okay this is oxy myoglobin and this is cytochrome p450 so here if you look upon for your oxy myoglobin the fifth ligand or the fifth - residue

39:26 which is attached to this is your imidazoles ring okay you have imidazole ring which is being attached from the from the protein backbone and for cytochrome p450 it is your cysteine okay so it is cysteine basically so it is here you have Amida's old Amida's old the nitrogen of your Amida's old and over here amid area so this is the emitter the old residue and this is cysteine residue so that means what they do not have same 50 ligands the fifth ligand is not going to be same so option B is wrong okay now whichever option you have where you have option B you can just cancel out them so you can just cancel out one three and four so by that by default you got answer number two as the correct one

40:12 you can check the other as well now the third statement says they contain single active site so yes both of them contain single single active sites you should also know about like you can just differentiate between the molecules or the by organic compounds which you have studied which have single set which have multiple sites like that so both of them are single iron site has the binding site and the last two these statement we say they contain metal in plus 3 oxidation states yes I run over here in oxy myoglobin is in plus 3 state and here again iron in the in the cytochrome p450 is again in plus 3 oxidation state in the resting state of that so option number 2 becomes your correct choice and one more thing there you will also find in some books that oxy myoglobin have a plus 2 oxidation state iron in it but

40:59 that's not like see it's very debatable things in some books you will find it as plus 2 some books you will find it at plus 3 so it's not actually corrected everywhere some there are two theories one says that it's plus 2 one says that it's plus 3 so they will no ever going to ask you in a way that you are going to get stuck in between of that ok just like the way they have given so option D becomes obvious for you so you don't have to think about it much so it's having plus 3 oxidation state you have to consider that for this particular portion right alright let's go to the next question then okay in this question it says that the that the pair of lanthanide ion which shows significant deviation from experimental and calculated magnetic moment considering contribution from the ground

41:44 state only so for this particular thing I guess if you have watched my classes on an academy the free classes which are already there with the video of free videos which are already there I have a have a course over there on lengthen its and actinides there I've explained all the properties okay they have just given all the exceptional cases all the properties of lanthanides and actinides and this question was directly from there so if you know that you could have answered this - this is from the exceptional like exceptional case of length units okay lanthanides so you should know about this that generally your samarium plus 3 should give a should give a very low value of your magnetic moment that should be 0.5 or something and europium plus 3 should

42:31 give you a 0 magnetic moment but the observed one is quite high it gives you around 1.9 or something and gives you it gives you around a 0.9 or something okay so you have a very higher value offer so this is observed sorry this is expected okay and this is observed so these are the two ligands or sorry these are the two lengths tonight's for whom the observed values are very much higher than the expected values so it was a very direct question from lanthanides and actinides directly from the notes directly from the quotient so the correct answer is going to be your option number four okay so fourth option is going to be the correct option alright so that's all for this particular video guys I tried to take

43:17 all the questions from the inorganic portion and try to explain you in the best form now as possible from my side so I hope you guys understood if you guys have any doubt you can ask that in the comment section below one more important thing that if you're new to this channel and if you like my content do subscribe to the channel give it a like and share it with your friends so thank you so much for watching and see you guys in the next video