INORGANIC CHEM TEST-20 VIDEO SOLUTION FOR RE NEET-2026

[00:00:00]Goal Nations [Music] Leading Institute.

[00:00:07]Hello Students, We are going to discuss an organic question of test number 20 of test series for Renee 2026 which was held on 17th June 2026. This is our last test before the final. Now after this we have to fight the final battle on 21st.

[00:00:27]Before that, our last test is the last test of the series, this is test number 20, let us solve its inorganic question, the first question number is of inorganic, that is question number 51 in which of the following pair both species have comparable bond order, okay, the bond order is comparable, in the first option, whose is the bond order of O, O2 ASF6 BaO2, okay, see, O2 ASF6 6, the compound that is formed, if you add one F- in SF5, then the compound that you will get is ASF6-. This is the compound and there is an O2 in it. When you take charge -1 on it, then what is the charge on it? +1 You should know this. Like by the way it is given that O2 PTCl6, so we know that the charge on PTCl6 will be -1, then what will be the charge on this? Will be +1.

[00:01:23]You should know this. If you don't know then you won't be able to make it. A little more concept has been used in this. You must know this.

[00:01:30]Ok? Only then will you be able to go and make it.

[00:01:33]So the charge on top of this will be O2 Plus. The charge on ASF6 will be -1. Arsenic belongs to the nitrogen family. Five bonds were made. An electron is taken to form a F- and a coordinate bond with ASF6-. And with O2+, we had this. This is O2+. So the charge on it is O2+.

[00:01:51]And here it is BaO2 so barium will be +1 anywhere in the first group. Second group will be anywhere so +2 this is always KO2 I have asked once. What is the oxidation state of potassium? What the child does in this is that the charge of X oxygen is -2 and -4 and the value of X is +4. goes wrong.

[00:02:09]Wherever the element of the first group is placed, it will be +1. Wherever the element of the second group is placed, it will be +2. So here potassium will be in +1 and this will be in -1. Meaning it is in peroxide form. So here sorry it is not in peroxide form but in superoxide form.

[00:02:21]I made a mistake O2 - 1 is superoxide.

[00:02:27]Same is the case with barium BaO2. So we don't have to look at O2. We have barium. Barium will always have a charge of +2 because it is a second group element. So what will O2 charge? -2 This is O2 2-. This will be in the form of peroxide.

[00:02:41]In the form of this peroxide, its structure O is like this. Both are charged at -1.

[00:02:46]So let me delete this one. I understand that the charge on O2 is +1 and barium O2 is 2-. I'm erasing a little bit to solve this. Now look, this means one we have O2+ and the other is O2 2-. We have to issue a bond order. Find out the number of electrons. It has 15, it has 16 2 18, the 15 one has a bond order of 2.5 and the 18 one has a bond order of one. Both are not comparable.

[00:03:16]This answer is over, we got it wrong.

[00:03:19]C2 This is C2 given directly.

[00:03:22]CaC2 is the same story again. CaC2 contains calcium carbide. This will have +2 or -2.

[00:03:27]Because calcium is a second group element. Will take +2. So in this we have to take C2 minus. The number of electrons in this is 12, the number of electrons in this is 14. The 12-membered group has a bond order of two, and the 14-membered group has a bond order of three. Again not comparable. Both have different bond orders. So we will not have this option either. Let's go to the third option. Do you know C2H2 C2H2? It has this acidic hydrogen.

[00:03:48]Acidic hydrogen. Both will give H+. H+ will be released. This will become C2- its electron will be taken in the oxidation state. So C2- will become. And CaC2 CC2, calcium +2 and -2 were also mentioned in it.

[00:04:02]So what will happen in CaC2 also? This will be C22-. The number of electrons in both is 14 and its bond order is three. It also has a bond order of three. Same bond order will come for both. This is the right option. The bond order of both is comparable.

[00:04:13]So in CC2, calcium carbon is carbon C22-. And N22-. N22- is. So the number of electrons in C2- is 14 and the number of electrons in N2- is 16. The 14-membered bond has a bond order of three and the 16-membered bond has a bond order of two.

[00:04:28]Again both are different. So this option will also not be ours. I have asked, with whom will this be comparable? So comparative will be third. There is C2- here too. There is also C22 here. In both the number of electrons is 14 and bond order is three.

[00:04:39]So the correct option for 51 will be option number three. Let's move on to the next question.

[00:04:51]Question number 52.

[00:04:54]Which of the following are Bidentate Monoanayan League? be bidentate and have a charge of -1. So in this, Leagued wants to ask whether you know the structure of Leagued etc. or not.

[00:05:04]Ok? I want to know something about ligands. I want to ask you whether you should read the topic Classification of Ligands thoroughly or not. Draw the structure of acetyl acetanoate.

[00:05:11]CH3 C double block O CH double block CO- CH3 This is acetyl acetanoate. If you want to go a little deeper into this, then acetyl acenoate is basically a derivative of acetyl acetone. CH2C + OCH3 It is made from this. It tautomerizes it. So this is the carbonyl carbon. Alpha carbon is also there.

[00:05:34]Alpha carbon is also there. But this Alpha Carbon Active Hydrogen is much better than this. Because the hydrogen of one carbon has two minus same groups. Electrons will be pulled from both sides. This will greatly increase the stability of the carbon ion.

[00:05:46]Therefore this acidic hydrogen is better. In comparison to this alpha carbon. That is why the hydrogen of this carbon will participate in the tautomerism. When its A will not do it. Only A will do it.

[00:05:55]So when we do tautomerism, the compound formed after tautomerism will be CH3 C double bed O CH double bed CO- CH3.

[00:06:06]Now get it protonated.

[00:06:08]What happens if we do this in a base like H+? Its H+ in the base will take and release.

[00:06:11]Carbon ion will be formed. Then the carbon ion creates its resonance. This is what happens step by step through the mechanism.

[00:06:16]And then when H2O is formed, get it protonated. will become OH. Not only this. When you will make its OH then you will involve it, why are you not taking it?

[00:06:24]Why are you taking this in leagued form? Because its enol form is very stable. Its inoalization is so high that sometimes it reaches up to 100%.

[00:06:32]That is why we are taking its more stable form as ligand.

[00:06:36]Ok? That's why I am not taking it.

[00:06:37]Its keto form is less stable. The enol form is more stable. Due to this there are many reasons. The intra molecule will remain H. Due to the presence of intra molecular hydrogen bonding and six member ring extended conjugation, its stability increases. If the minus m group on both sides becomes very strong then it becomes 100% initialized.

[00:06:54]That's why we take it in this form. So acetyl acetone is our league.

[00:06:57]You will have to remember. Ok? would write that ethyl acetone is A. Acetyl acetone is a.

[00:07:01]But in the form of ligand, its enol form works. The anionic form of the enol form works. That too is discussed in the enol form by removing the last H+. So that's why it's a monoanion. The charge is -1 and it acts as a bidentate ligand.

[00:07:14]Forms six membering. That's right sir. This could be an option.

[00:07:17]Not oxylato, oxylato is definitely bidentate but not a monoanion. This is a dye anion. So this is wrong.

[00:07:24]Diethyl glyoxymeto That's right sir. Contains diethyl glyoxymeto. Just look at the structure. CH3 CH3 Double BD N Double BD N OHO- This ligand will behave like diethyl glyoxameto. This is a very good thing. The oxime that is formed is also a concept of organic. When you react any aldehyde or ketone by taking a hydroxyl amine, a derivative of ammonia, then water is released from here and a C double bond NOH is called this, we call this imine bond, C double bond N is called imine bond and a derivative of imine bond is auxin, this is what forms auxin and it behaves like a ligand here but the use of dmg is very good for us in complex compounds, it is useful in the reduction of nic + 2, this is given in its application of complex number, go and read it.

[00:08:16]We use complex compounds in qualitative and quantitative analysis.

[00:08:18]DT and DMG are given in it. Ok? So DMG is used in nickel +2. Take some unknown solution. It is a solution of some unknown salt. I don't know which salt solution is in it.

[00:08:29]You put DMG solution in it.

[00:08:32]And after adding DMG solution, if you get Rosy Rate PPT in it. If a rosy rate ppt is observed then you will confirm that the solution contains Nick +2 ions. So in the qualitative analysis of Nick +2, DMG is used.

[00:08:45]And when DMG reacts with Nick +2 it forms square pin complex.

[00:08:49]In which both the donor atoms are nitrogen.

[00:08:51]Remember this. Because after donating nitrogen atom, it forms intra molecular hydrogen bond. Forms a six member ring.

[00:08:58]Two five-member rings form two chiclet rings and two six-member rings form two separate chiclet rings. Therefore the stability of the compound is exceptionally high due to intra molecular hydrogen bond with the six member ring formation.six member intra molecular hydrogen bond with the six member ring formation makes the stability of the compound exceptionally high.

[00:09:15]Exceptionally it is very high.

[00:09:17]Whatever Tozam told, the major reason is the same in this also. Intramolecular hydrogen bond with a six member ring that eliminates lone pair lone pair repulsion this one. Ok? This is called diethyl glyomato, which has a -1 charge on one oxygen. This is called dmg -1, so this is also -1. This is ox2- and this is dmg -1.

[00:09:39]This is also bydentate, this is also bydentate. This is also bidentate. But this is a dye anion. Mononayan is not. So one and third will be my option. The correct option for 52 will be option number second one and third only. Let's move on to the next question.

[00:10:01]Question number 53 Total number of possible isomers of complex PDNH3 whole twice ONO2 Look, this is a very good question.

[00:10:14]Ok? That's a good question.

[00:10:16]I'm telling you about isomerism again.

[00:10:22]If there is any best topic of complex compound then it is isomerism topic.

[00:10:24]If he wants, he can give you a very good level question in that. Due to which, as there is a little bit about REET, the questions may be asked in a slightly twisted manner.

[00:10:33]What is Twisted? Will ask from the syllabus only.

[00:10:34]How is your preparation level? How have you studied? What would the concept be like in your mind? The concept will be great. There is no need to panic. Studying by rote will always trouble you. There is a very simple formula for studying and reading. If you are studying science and studying it by rote, then the motive of studying science is lost. Therefore, if you have read it with understanding, you will not be too worried and no question will distract you too much. No matter what level of question someone may ask. The concept is in your mind. With all the work done, you can handle any question very well. That's the one from Part A.

[00:11:06]So this is still less. There is less variety in this also. There is not much twisting done to it. But still it is okay in this. Now octahedral has told that when it enters octahedral, then how much total number of geometrical isomers, how much total number of optically inactive and active both can be asked from you separately. What is the total number of stereo isomers? So he can ask you anything. Okay, right? So in that I am telling you that you must keep looking at it once that isomerism of sc isomerism of complex compound. This is definitely an important topic.

[00:11:36]This makes for good questions. Let's look into this.

[00:11:39]So, first we have been asked about isomer possible isomer. We have not been told in structural stereo what is the total number of optically active and inactive. We have to tell the isomer. This means that you have to count both structural and stereo aspects.

[00:11:50]Ok? So, let's look at the structure first.

[00:11:52]So, will ionization occur first in the structure or not? Will not do. There are not two ions. There is only one type of ion. There is no ion outside otherwise ionization will not take place. He will not coordinate. This is not our two different things.

[00:12:01]When the complex cation is a complex anion, then coordination takes place. It will not hydrate.

[00:12:08]Also not possible in hydrides. So one thing left to study in our syllabus is linkage. So look at the linkage o n o these linkages are the adjacent ligands used to exhibit the linkage isomerism. So make its linkage PD whole twice of NH3 whole twice of NO2.

[00:12:24]First, both the ligands, both of them were changed by Emmat. There is another possibility for this. What is the possibility? Hole Twice of PD NH3.

[00:12:34]Change one and leave one. Change one, leave one, and the charge above the PD for PD will all be +2.

[00:12:42]When the charge on everyone in PD is 5s0 4d 8, this will be it. The electronic configuration is that of the Just Like Nick Nick series. Nick Palladium Platinum Let me tell you one more thing.

[00:12:56]What series element is the playdium?

[00:12:58]Tell me. Now rhodium, I have to write the electronic of rhodium +3 of rhodium.

[00:13:03]Rhodium +3 is given in a complex.

[00:13:05]I don't know which series and group element is Rhodium +3? So you will spend time just in writing its atomic number and its electronic number.

[00:13:13]So does he know who owns rhodium in Iran? It is from the Cobalt series, sir. So whatever is of cobalt will be of rhodium. Just the shell will defer to it. So this should be known.

[00:13:22]So, read one more thing.

[00:13:24]All the Legend strings for the Platinum 4D and 5D series are field strings. All leagues are strong field leagues. There is no concept of weak field in this. For 4D and 5D series. When all of them are strong fields then pairing will be done. If pairing is done then hybridization dsp2 all of all and all complex will be a square panel complex. And when there are square foot complexes, both the ligands are different, then they will do cyst transo, right? Look NH3 NH3 ONO ONO Now make another one of this. NH3 ONO ONO NH3 C centrus so it will make two C centrus will show two GI. Square planar complexes are all optically inactive because they already contain a POS. The geometry is planar and there is already a POS in the planar geometry. When POS is on, it will become optically inactive. So the square planar complex does not show optically active properties.

[00:14:09]Shows in a case when the chiral center comes in the legend itself. No.

[00:14:13]But if the complex asks then it is fine. If the Chiral Centre comes to League only then he can do it. Like I read Glyoximeto. Glyoxymato or Glycine Glycineto, what we read is Glycineto, what you have to read is GLY-, so in Glycineto it is CH2 C = OO- NH3 NH2. Now remove one hydrogen from here and get CH3. This is called methyl glycinate. So, the Chiral Center is included in this, right? Then it will become optically active.

[00:14:43]But there is a chance that this question will almost not be asked in your syllabus.

[00:14:45]No matter how hard he tries, he cannot enter you. So what we will read is that the square panel complex does not show optical generally. They are optically inactive because square panel complexes will always have a position along the plane of the molecule. Ok? So optical so show that will do two GI shows. These are also two GI same things, right? So this will also do two GI shows. These two GIs, one C axis and one trans, and this will also show two GIs. Two two legged beans, right? NH3 NH3 Once put NH3 on C and once put NH3 on trans N and N then this will also do two. Apart from this, none are optically active. The square panel complex does not show any optically active.

[00:15:22]So if none of them are optically active then these will be formed. Two, its GI will be made. Two will be made for this, two will be made for this. Total number of isomers. How many exhibits? How many different structures are we going to get? Six. Two from this, two from this, two from this. Therefore the correct option for 53 will be option number six. Option number three. Option number three. Let's move on to the next question.

[00:15:46]Question number 54. If x pentagonal rings and y hexagonal rings are present in fullerene of Z carbon. Absolutely correct. That's ok. Let us discuss a little about fullerene. The general formula of fullerene is CN. Z is given here. I will take Z. Number of pentagonal rings.

[00:16:06]Number of pentagonal rings is equal to 12 and number of hexagonal rings number of hexagonal rings is equal to n/2 - 10 this basically refers to C60.

[00:16:18]Buckminster fluorine has 12.

[00:16:20]This will be 60 / 2 - 10 = 20, this is what it is.

[00:16:25]This is what happens to us. The total number of pentagons is 12. Hexagonalizing n - 2 / 10 What is n here? Number of carbon atoms. Now what is given in this question? Understand that if it is called C60 then it will be called backminister of fullerene. This has to be read. Our pentagonal ring is X. X = pentagonal ring. So what will be the total number of carbon atoms? What will be the total number of carbon? 5 * X. Because there are five carbons in a ring.

[00:16:51]Five carbons in pentagonal.

[00:16:53]So it will become 5 * X. And Y is the hexagonal ring. So, Y hexagon. So what will be the total number of carbon in it? 6 * Y.

[00:17:02]And Z we have is the total number of carbon atoms. Z What is ours? Carbon atom and fluorine. Now listen a little carefully.

[00:17:09]I am deleting this. I will destroy the one above.

[00:17:13]Now what does the child think? What's a little mistake in this? That Sir, you told me about Z being a carbon atom. Here a five member carbon atom was created. And created a six member carbon atom. So do 5x + 6y.

[00:17:25]And the total number of 5X means the number of carbons in the carbon five member will be 5 * X. Because x is the number of rings. If there are five in one carbon then it is 5 * x and if there are six members then the total number of six member rings is y and if there are six in one carbon then it is 6 * y. And this is the total number of carbons in the following z, so make it z. This is your answer. So 5x + 6y = z will be our option.

[00:17:46]You will choose the first option which will be your wrong option.

[00:17:50]Why does this happen Then listen to this very carefully, Babu. Basically, the fluorine that is formed forms a ball-like structure. There is another concept of this. What is the concept that any five member ring, it is surrounded by only six member ring and every six member ring is surrounded by alternately 5 6 5 6 56 alternatively.

[00:18:11]So just look at it and understand what I am saying.

[00:18:13]Like we can take a six member ring or a five member ring, whichever you wish. Above this there are both five member and six member rings. There is a ring for five members and a ring for six members.

[00:18:21]Inside this there is a five member ring and also a six member ring.

[00:18:26]Say, look, this is six members, five members. In this way, five members will come here again. Six members will come. So these two carbons in this are colliding, right?

[00:18:36]Tell me yes or no? These two carbons are also being counted inside this member. There are a few different rings. This ring of six members, a ring of six members, of six members. There are not 12 such rings, we do not have Y number of rings or X number of rings of pentagonal, they are not different like this.

[00:18:52]Look at this form, they create such mole like structures.

[00:18:54]So there will be some carbon atoms in it which will keep getting attached commonly.

[00:18:58]So the number of carbon atoms will keep decreasing for you. Are you getting the point? So the number of carbon atoms will have to be increased.

[00:19:04]So we work in this, take an example, which we have to study in Buckminster fullerene, we have to study Buckminster fullerene, so we take that and work on it, let us understand a little bit, how will it happen if Buckminster becomes C60, then what is the number of five member rings in C60, five member ring = 12 and number of six member ring is equal to 20, okay, so what will be the number of carbon atoms in five member, so 2 * sorry 5 * 12 five member 12 total number of carbon atoms in five member is five and number of rings is 12, so 12 * 5 plus number of six member rings is 20, so 20 * 6 see how much is coming, it is 60 and plus it is 120, so it comes to 180 and how much total carbon did we take, 60, see, it is coming to 100 180 carbon atoms, so something will decrease, why it will decrease because we have to change something here in it. And the total carbon atoms are 60, meaning 60 carbon atoms together can form 20 six number rings and 12 five number rings. Number of carbon atoms decreased. Why? Because in this which you are making by coinside, some carbon atoms are commonly being used which will reduce the number of carbon atoms in the total number of fluorine. Ok? So how much did it cost? 100 180 came. How much do I have here?

[00:20:19]60. So we can write that as 3 * 60.

[00:20:21]Meaning this z is ours. This is z. And this is our x. This is our y. So basically if we follow this theory then one will come to us. What will come? How much is it coming out to be 5x + 6y = z? 3z which we have worked on the basis of Buck Minister Fullerin.

[00:20:38]Our result table is coming from this, Sir 5x + 6y will not be z, Sir it is sure that it will not be z but it will be 2z or 3z or 4z, we have calculated this on the basis of the Minister of Fullerene, we have a present experimental result of Fullerene, on the basis of that we went to calculate that it was setting us from 3z because the total was coming out to be 180.

[00:20:55]Here it is 60, so if we do 3 * 60 then we will get 180. Ok? This is your result on the basis of what Minister Phoolan has said. You don't have any other way to remove it.

[00:21:04]Okay, right? This is what we have talked about. So that's why A won't happen.

[00:21:08]This option will already be wrong. will not be equal to Z. Some carbon atoms are decreasing in us. Ok? And the result that came was 3Z This is based on the Buck Minister of Florin. We did this by making them sit on 660 because that is what we have to study in our syllabus. I have to read Buck Minister Fullerin only.

[00:21:21]So, the correct option for 5X + 6y = 3Z 54 will be option number three. Let us move on to the next question.

[00:21:33]Question No. 55 Which of the following has peroxy linkage? Look, for peroxy linkages, I repeatedly say calculate the oxidation state. If the oxidation state is greater than the maximum, it means that it contains peroxy linkage. You can find out the oxidation state. H2SH2O3 2 out of 6 will give 4 + 2. But we have already told you that in this the oxidation state of both the sulphurs is not +2. He changes it. S double bond S OH OH This is the structure of H2H2O3 Its oxidation state is -6 and its is +2. When we calculate the average of both, we will get 4/2 to. But both do not equal +2 + 2.

[00:22:08]This thing has to be kept in mind.

[00:22:09]Well, there is no peroxy linkage in it.

[00:22:11]H2S2O5 2 out of 10 becomes 8 in it +8.

[00:22:14]It will have peroxy linkage. S double bud O double bud OOHOH This is the structure this is where the peroxyl linkage comes in which reduces its oxidation state a little bit otherwise it would be at. Sulphur guide is not possible. This means the oxygen you talk about is -2 -2 -2 not -2 everywhere. The one with -1 also has two oxygens in it. Ok? So this is what will contain the peroxy linkage.

[00:22:34]This is oleum. This is +6.

[00:22:36]+6 will have SOH linkage. You will make up the rest of the structure. Double Bud O Double Bud OH OH This is the structure. It does not contain peroxyl linkage. Two out of 12 is 10 + 5, this is less than the maximum. So it has SS linkage. S double bed O double bed O double bed O double bed OOH OH this is the structure.

[00:22:57]It does not contain peroxy linkage.

[00:22:59]Peroxy linkage is contained in H2SO5.

[00:23:02]Its name itself is peroxy sulphuric acid or persulphuric acid.

[00:23:06]Okay, right? So the correct option for 55 will be option number second. Let's move on to the next question.

[00:23:18]56 Product of oxidation of -i with MNO4- in alkaline medium is ok?

[00:23:26]When we react I- with MNO4-, our medium will be alkaline, then I- will convert into IO3 iodate and MNO2 will convert into IO3-, so the correct option for 56 will be option number first. Let us move on to the next question.

[00:23:45]Question No. 57 Larger number of oxidation states are exhibited by actinoids than those by the lanthanoids. Look at The Main Region.

[00:23:54]This is true. The oxidation states in actinides and lanthanoids show only three.

[00:23:59]One is +2, +3 and +4 and +3 is most stable. The one with +2 will also convert to +3. The one with +4 will also convert to +3. But if we talk about actinoids, then in actinoids, if you start with actinoids, then start from thorium and go till lawrencium, then it shows up to +7, then it starts decreasing from here and reaches +3 here, so there are some pyramidal things in it, that is +3 +4 +5 +6 +7, plutonium etc., these people show up to +7.

[00:24:28]Okay, right? So more numbers show in this. The reason for this is that the contraction in actinoids is slightly more powerful. The size of 5F is bigger than 4F and due to its bigger size, the energy gap between 5f and its 6d is less. When the energy gap is small, its electronic transition will occur easily. When the electronic transition occurs easily, then the electrons will move easily in it and its number of oxidation states will keep increasing.

[00:24:57]Therefore, the lesser energy difference between the 5f and 6d is 4f, right, that's it, the energy difference between 5f and 4d is less in comparison to 4f and 5d, that's why the electronic transition happens easily in it, first understand my point, like 5f 1 to 14 sorry 6d 0 to 1 and 7s2, that's what happens, so this is finished, it will remain as it is, now in this the electron moves from here, the more it remains, the more the number of electrons per one will increase. Now if the number of electrons increases then there will be a tendency to form more bonds. His covalency will increase. The oxidation state will also keep increasing.

[00:25:31]So, the smaller the energy difference between 5F and 6D, the easier the electronic transition will be. The lower the energy, the more the number of oxidation states will increase. This is why actinoids show higher oxidation states in comparison to lanthanoids. If there is more energy then there will be no transition.

[00:25:48]So this is given wrongly. More [nasal sound] reactive nature of actinoids than the lanthanoids. This is also not true, nothing happens due to high reactivity. The 4f orbitals are diffuse, so there is no point in this. So energy difference is what works in this. The correct option for 57 will be option number first. Let's move on to the next question.

[00:26:09]Question number 67.

[00:26:12]The standard reduction potential values ​​of three metallic cations x, y and z are 0.5 -1.5 and -2.1 volts respectively the order of oxidising power will be. Ok? Look, this is a good thing.

[00:26:26]We have three methyl cations. There is X, there is Y and there is Z. The reduction potential of all these is like that of any metal, by the way, if we take charge +2 on X. Let's take +3. Let's take +3. Take +3. Take +3. Now its reduction potential means make x to +2, make y to +2, make z to +2. This is the reduction that has happened. And their value for going from +3 to +2 is given as 0.5 volts. Their value is 1.5 volts in negative and their value is -2.1 volts in negative. One thing I tell kids over and over again is that delta g not equals -nf * e not cell. Now note that if delta g of any reaction is negative then our reaction will be spontaneous.

[00:27:07]And the reaction being spontaneous means that our product will be stable. The product is our stable. And delta g not will be negative when e not value is positive. Because this is our number of electrons involved. This will always be positive. And f is the Faraday constant 96500 coulombs. This too will always be positive. So if this note cell becomes positive then its negative value will take negative here.

[00:27:27]So e note cell positive delta g note negative. And as much as these note cells because both of these are constants. Innot cell the more positive the delta g note the more negative and the more negative the delta g note the more spontaneous the reaction and reaction spontaneous means the product is stable means the reaction will move easily in the forward direction.

[00:27:42]So look, this is the most positive thing.

[00:27:44]Both of these are negative. So this will be his.

[00:27:46]It will act as a good oxidizing agent. These people would not want to behave like oxidizing agents. Now if the oxidizing nature is checked in both of these, then the one which is less negative will be better. The one who has more negative things will suffer worse. So the best oxidizing agent will be the reduction.

[00:27:59]Acts as an oxidizing agent.

[00:28:01]The best oxidizing agent will be the cation with the symbol x. Then the y cation and then the z cation. Meaning, the one whose IT value is more positive will have good oxidizing nature. His reduction will be good. The one whose It value is more in negative will have bad reduction. Its oxidizing nature will be bad. So that's why x y z will be our correct option.

[00:28:18]So the correct option for 67 is option number first. Let's move on to the next question. Question No.

[00:28:25]72 Vi among the following is a planar molecule. Ok? This is n. CH3 CH3 CH3 One lone pair of electrons, it will be in pyramidal shape. It is non-planar.

[00:28:38]XCF4 is planar. FF F A lone pair A lone pair. Square is planar geometry hence is planar. X SF4 trigonal bi pyramidal has cisso. It's Seeso. Look, this is non-planar. In this also the molecule will exist in two planes.

[00:28:54]H2SO3 S double bond OH OH one lone pair. We will also talk about SP3 hybridization. If it takes a pyramidal shape then this is also non-planar.

[00:29:06]Planer asked. The XC4 will be. It is a simple question. The correct option for 72 will be option number second. Well, one more thing, if we talk about planar molecule, by the way, it will become tetrahedral. If we talk about this then it will become triangular bipyramidal. If we talk about geometry, then whenever we ask for planar or non- planar molecule, octahedral shape is taken on the basis of hybridization. If we talk about geometry, that is, trigonal bipyramidal of sp3d hybridization, what is its trigonal bipyramidal shape? This is triangular bipyramidal geometry.

[00:29:34]Shep will be the seeso. We will see in the mirror whether it is planar or non-planar. Like it is tetrahedral. H2O has tetrahedral geometry but the molecule will be bent and the bent will be planar. Therefore, we will talk about planar H2O molecule. This thing has to be taken care of. Okay, right? We take the shape consideration planner unplanner. Planar does not take non-planar geometry.

[00:29:52]So the correct option for 72 will be option number second. Let's move on to the next question.

[00:30:05]Question No. 73 Correct order of covariant characteristics of given halides. Well, this is a simple question. Now you have given us only cation beans in everything. When the cation is given, we will see the polarizing potential of the anion.

[00:30:18]What will be the polarizability of anion?

[00:30:20]F- Cl- Br- all have the same charge. The size will increase. What will be the polarity? Will increase.

[00:30:26]If polarity increases, the covent character will increase. Ionic character will decrease. So who will have the most covent character?

[00:30:33]In Br one then in Cl then in AlF3. This is what will happen.

[00:30:36]Albr has the highest amount, then AlCl3, then Alf3, Alf3, it will have more ionic character, less covalent character in it, that is, it is an ionic bond. This is an ionic compound and both of them are covalent compounds. AlF3 is an ionic compound. AlCl3 is a covalent compound. Covalent character is highest in AlBr3.

[00:30:53]Then in AlCl3 then in Alf3.

[00:30:55]What is Reason? Br- has higher probability. The size is big. It will easily polarize the anion. That is why the correct option for 73 will be option number four. Let's move on to the next question.

[00:31:09]Question No. 74 Which among the following complex has highest value of spin only magnetic moment?

[00:31:15]Check the electron on number one in this. Zero charge over g Ni in NiCO4.

[00:31:18]4s2 is 3d8. I repeatedly tell S to empty. The formation of the complex can never occur without S being vacated.

[00:31:26]This becomes 4s0 3d10. It does not have any unpaired electrons. So the value of n will be zero in this. This is a diamagnetic compound. It has tetrahedral geometry. sp3 hybridisation occurs. PT is +2. PT +2 has 4s0 3d sorry not 4s but 6s0 5d8.

[00:31:44]Ok? Is. Platinum is our fifth group.

[00:31:45]Fifth is from the 5D series. Its period number is six. Okay, right? So we'll talk about 6 x 0 5d8. This is a strong field. Platinum All the leagues for platinum are string fields. All of them are stung fields for 5d 4d. This will get the pairing done.

[00:32:04]What happens when the pairing is done in d8?

[00:32:06]This will happen. dsp2 hybridization In this also the number of unpaired electrons will become zero.

[00:32:10]NiCl4 is Ni+2 so 4s0 3d8 and weak field occurs for Nick +2.

[00:32:17]No pairing will do. So it will have two unpaired electrons left. The value of n will be two.

[00:32:20]Copper +2 is 4s0 3d9 in this it will behave like a Weak Field for Copper + 2.

[00:32:27]nobody. The number of electrons per d9 is one. What have you asked? The highest value of spin only magnetic moment will be third. There is zero difference between these two.

[00:32:36]There are two of them. There is one in this. So the value of maximum magnetic moment of two will be.

[00:32:39]The value of magnetic moment will become 8bm. Its formula is n * n + 2. The correct option for 74 will be option number three. Let us move on to the next question.

[00:32:53]Question number 75 is An element with atomic number 52 belongs to block disorder 52.

[00:32:59]54 We know that xenon exists. 53 here 52 here What block would it be in? All of them are in the p block, right? From 13 to 18, they belong to the p block. If we have to find out the group number, it will be 18, 17, 16.

[00:33:13]Okay, it is an element of the p block, so 75 will be the correct option. Option number second. Let's move on to the next question.

[00:33:23]Question number 76. What is the correct order of decreasing field strength of the given legged decreasing field strength? CN is, CO is. So look, understand that CO happens.

[00:33:32]Then it will be CN-. Then it will be EN. This is your spectrochemical series. You will have to remember.

[00:33:37]I- Rain Sunny Class, you guys must remember this. As you move. CNCO is preceded by NO2. - CN-CO EN occurs here.

[00:33:48]Ok? You have to remember this.

[00:33:50]Spectrochemical series is given in NCERT. This has to be remembered. That's what COCN would be.

[00:33:55]COC NN second then the correct option for 76 will be option number second. Let us move on to the next question.

[00:34:06]Question No. 89 The alkali metals form salts like hydrides by the direct synthesis at the elevated temperature. The thermal stability of these hydrides decreases.

[00:34:16]Look listen, if we have to decide the thermal stability of ionic compound, then this salt has been given. Salt-like hydride means the compound is ionic. It is a salt, which means it is ionic. Okay, right? FeSO4 Fe+2O42- So in FeSO4 the bond between Fe+2 and SO42- is ionic bond. Salt is an ionic compound formed in the Okay, right?

[00:34:40]So it is an ionic salt. Salt is a hydride.

[00:34:42]So if you have to decide the thermal stability of ionic compounds, then there are two things for it. Either you decide by lattice energy or you decide by polarizing power.

[00:34:51]Meaning, who has more covent character? The covent character will increase in the ionic character.

[00:34:54]What will happen to stability? It will keep decreasing. So children face problem as to what should we see in what?

[00:34:59]So, in this we teach such children that generally if you have poly atomic anion.

[00:35:03]Polyatomic anions like sulphate, hydroxide, carbonate, all these are polyatomic anions having more than one anion, so we will see polarizing power in them. If we have mono atomic anion, we have mono atomic anion. Mono atomic anions like look hydride is mono atomic, fluoride is mono atomic, oxide is mono atomic, nitride is mono atomic. Then we will see lattice energy in this. And one more thing is that Cl- Br- I- also decides our covalent character.

[00:35:30]Its stability is decided by its covent character. The reason is that after this there is another reason. You have seen the polarizing power.

[00:35:36]Another thing that happens is that the larger anion is happy with the larger cation. The smaller anion is happy with the smaller cation. They are the Cl Br are the bigger anion. They will be more stable with the bigger cation. If you go from top to bottom, you will get a big cation. Therefore their stability will increase. So we're in this this is my H-. H- F- O2- N3- These will be decided by the lattice and all the poly atomic anions plus chloride bromide iodide will be decided by the covalent character polarizing power. Okay, right? What I mean to say is that polarization is better in these.

[00:36:06]When there is a larger anion, the polabilit y will be better. When the polabilit y is good then the covalent character will determine the stability of the ionic compound. Ok?

[00:36:13]This is what it means to say. So look at it all has hydride hydride hydride.

[00:36:17]Lithium plus sodium plus potassium plus rubidium plus cesium plus all pay charge lattice energy is directly proportional to the product of charges. First we will see the product of charge. Inversely proportional to the size of ions. Tell me, the product of charge is same for everyone. The larger the size of the ion, the lower its lattice energy. And if the lattice energy is low, what will happen to the stability? Will reduce.

[00:36:36]So see who owns what? Plus four is the same for everyone. Its size is small.

[00:36:42]Small size, more lattice. More lattice means more style. So that's why this will be the most. This will be our order. The most stable is LiH, then NaH, then KH, then RBH, then CSH. Ok? So this is the correct order of thermal stability of hydrides. LiH is the most stable and CHSH is the least stable. If you are asked to convert Li2CO3 into CH2CO3, if you apply lattice energy to it, then the order will be this which will be wrong. Its order will be A because we will see polarizing power in it or else children are also taught that bigger cation is stable, smaller cation is stable, smaller anion is also stable.

[00:37:16]Carbonate is a large anion. The more polyatomic anions there are, the bigger the anions and they will be more stable with the bigger cations. You can make it even faster than this.

[00:37:23]Ok? This H- F- O2 - N3- gives the smaller anion. They will be more stable with the smaller cation. Ok? So the thermal stability of hydride will be our LiH is the most stable and C-S-H is the least stable. The correct option for 89 will be option number second. Let us move on to the next question.

[00:37:46]Question No. 90 The decreasing value of bond angles from NF3 to SBF3 down the group 15 of the periodic table is due to.

[00:37:55]Ok? Look, let's move on from NF3. If we find out the hybridisation of all these till SBF3, then all have the same hybridisation sp3, there will be no steric repulsion in any of them because our side atom is fluorine. Fluorine has a sterically inactive group. There seems to be no steric repulsion. After that we will see the theory on Wes. So Weiser's first rule number of lone pairs and central atoms will increase. The bond angle will decrease. All Pay Central Atom Pay Loan Payers are Same. is a unit. Second rule electronegativity and central atom will increase. The bond angle will decrease. So therefore the electronegativity of nitrogen is more, then of phosphorus, then of arsenic, then of antimony, this is the order of electronegativity, hence the order of bond angle will also be, electronegativity of central atom will increase, what will happen, bond angle will increase, okay when bond angle increases then this will be the reason for decreasing, someone is asking why it is decreasing, so what will be the reason for decreasing, due to decrease in electronegativity of central atom, see increasing is not all this, there is no bond pair bond pair repulsion, there is no increasing P orbital character in anyone, this is also not decreasing lone pair lone pair repulsion decreasing electronegativity, this is our correct option. Decreasing Electronegativity As you go from top to bottom, electronegativity will decrease.

[00:39:02]Therefore this will be our right option.

[00:39:04]Decreasing electrons causes the bond angle to decrease. This is all about your inorganic equation. Thank you everyone. Thank you so much. All the best for your 21 Zone Examination, Main Examination. Be Calm, Be Quiet and All the Best for Your Examination. Thank you everyone. Thank you so much.