INORGANIC CHEM-TEST-09-VIDEO SOLUTION FOR RE NEET-2026

Added: 2026-06-03

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

[00:00:10]Hello Students, We are going to discuss an organic question of test number nine of test series for RENET 2026 which was held on 1st June 2026. So let us discuss an organic question of this examination. The question number first in organica that is question number second.

[00:00:30]The atomic radius of group 13 elements R is okay boron aluminium gallium indium thallium what was expected what we expected was that going from top to bottom the shell would increase and because of increasing shell the size should have increased so this order should have come but we did not get this order experimentally the order was not found the order which actually came is the smallest size of boron which changed into aluminium and gallium.

[00:01:03]Gallium became smaller. Aluminum grew up.

[00:01:05]Then there's indium and then there's thallium.

[00:01:11]Its reason writes that its size became smaller due to de-contraction.

[00:01:16]Look, understand one thing what I am saying.

[00:01:19]D contraction, lanthanoid contraction or actinoid contraction are all smaller in size. So after the size becomes smaller, there can be three different conditions in it.

[00:01:32]The size will be smaller but the size will contract. After contracting, it may become larger than the previous sale. It may become smaller than the previous cell and it may become equal to the previous cell.

[00:01:47]Contractions Because of these three contractions, anything is possible. If there is contraction, the size will be smaller. How small will it be? It should remain bigger than the previous cell. It should be smaller than the previous cell or equal to the previous cell.

[00:01:58]All three conditions have to be read. For example, look here, thallium will have lanthanoid contraction.

[00:02:02]But due to lanthanoid contraction the size became smaller. But it did not become so small that it would become smaller than indium.

[00:02:08]Still got bigger than indium. So this will be known through experiments. The data will tell us where and what kind of values ​​we have to take. Ok? They cannot predict whether this will happen or not.

[00:02:18]Indium Thallium Thallium has length contraction and it is also known that length contraction is more powerful than D contraction. So then here the deconstriction size became smaller. But here the size of the lenth contraction did not decrease.

[00:02:28]That's why there is complete data. You guys must have studied there in D Block. Remember me in D Block.

[00:02:32]Ah 4D five 3D 4D size big.

[00:02:36]Size of 4D is big and size of 3D is small. In 4D and 5D the size becomes equal due to length and white contraction. So this is the data. This became equal. In this the size became smaller than de contra.

[00:02:44]Now move on to cesium francium or indium thallium, then ah tin and lead.

[00:02:50]Ah move to antimony bismuth. The size is bigger in all of these. But everyone seems to be experiencing lengthy contractions.

[00:02:56]But still the size remains big. Ok? So what I mean to say is that all three conditions are possible. Who will apply where? You have to work on the basis of data.

[00:03:03]Ok? It becomes a data based thing. So among gallium and indium, the size of gallium became smaller. The size of aluminium increased. Due to deconstruction. The rest is absolutely normal.

[00:03:12]So the smallest is boron then gallium then aluminum then indium then thallium the largest is thallium then indium then aluminum then gallium again correct. The first option seems right. The largest boron is wrong. It wo n't matter in size. Boron after Thallium, this is also wrong. The biggest boron, this is also wrong. So our correct option will be option number first. The correct option for second is option number first. Let's move on to the next question.

[00:03:42]Question number five the correct statement about Brf5 and PCL5 is both are ok we have talked a little about its hybridization and shape let us see Brf5 has one lone pair bromine has six-seven electrons valence electrons are seven five bonds have been formed one lone pair will be left with PCL5 phosphorus has five electrons no lone pair will be left if you calculate its steric number then it will be six.

[00:04:08]Its static number will be five. Its hybridisation will be sp3d2. And its hybridization is written as sp3dz² z² so that its geometry is trigonal by pyramidal and its geometry will be octahedral and the shape will be. Now look at the octahedral you make.

[00:04:27]Octahedral has one lone pair.

[00:04:29]In octahedral all positions are equivalent.

[00:04:31]One thing has to be remembered. If octahedral has one lone pair.

[00:04:34]Keep it wherever you want, it won't make any difference.

[00:04:36]Keep it wherever you want, it won't make any difference.

[00:04:38]I am telling this because once a question was asked in CLF3 that which of the following is correct statement, correct structure, in that the child got confused again because when the child makes Clf3, then Clf3 sees that one lone pair here, one lone pair here, fluorine here, fluorine here, fluorine here, so it feels that this is the correct structure.

[00:04:57]If he makes such a structure of Clf3 and gives it to him, one lone pair here, one lone pair here, one fluorine here, one fluorine here, then the child feels that this is a wrong structure.

[00:05:07]This is also correct. Both are equivalent structures. Lone Pair Wants to Occupy the Equatorial Position. So this also equatorial also equatorial went into triangular by pyramidal. So in this both the structures were one and of the same type.

[00:05:16]Loan payer here and loan payer here. Ok? What I mean to say is that all of them have the same structure. This thing has to be remembered. So Brf5 has a lone pair.

[00:05:24]Place a lone pair wherever you want in octahedral. Doesn't matter. All positions are equivalent. But if there are two lone pairs then it will always go diagonally. To bear a 180° angle so that repulsion is reduced. The stability of the molecule should increase. There's a lone pair where there's a fluorine fluorine fluorine fluorine fluorine Br Br Cover up the lone pair to tell the lone shape.

[00:05:45]After covering the lone pair, you will see this square pyramidal.

[00:05:47]This will be square pyramidal.

[00:05:51]Ok? So its shape will be square pyramidal. Above this is the lone pair zero. The molecule in which lone pair is zero, the geometry of the same shape is trigonal bi pyramidal shape is also trigonal bi pyramidal.

[00:06:05]Ok? Let's see. Both are isostructural is wrong, right? One is square pyramidal and one is triangular bi pyramidal. BRF5 is a square pyramidal and PCL5 is trigonal bipyramid. This is absolutely the correct statement.

[00:06:17]PCL5 is a square planer. From where? It is wrong. Brf Is C. So this is also wrong. Brf is sp3d hybridization. This is also wrong. Br5 is sp3d2 hybridisation. So the correct option for five will be option number second. Let's move on to the next question.

[00:06:35]Question number seven. The electronic configuration of californium, neptunium and terbium is respectively. Ok, good.

[00:06:43]Electronic configuration of all three elements is given. Californium, Neptunium and Terbium. So listen to what I say carefully. Whenever it comes to electronic configuration, writing electronic configuration is not a big deal. With respect to noble gases and this is what you people also have to learn. Okay, right?

[00:06:58]This is what you have to say in respect of Nobel Gas.

[00:06:59]So remember one thing that whenever the electronic confirmation of any element is given, then our element is not an exception.

[00:07:08]You should know this. If there is any exception element then it should immediately strike your mind.

[00:07:13]And till date, this question has been asked as many times as possible. Well, it is an exception.

[00:07:17]We will talk now and discuss it. Otherwise, if the electronic configuration of any element is asked individually, then whenever it is asked somewhere, generally only the exception one is asked. Okay, right? So you should remember the exception element. All chromium, copper, Raghunand, see, these are so many elements which are exceptions.

[00:07:59]These are from D Block. Such lanthanum actinium also belong to the d block.

[00:08:04]You can keep it in D block.

[00:08:05]But everything that you will read about lanthanum actinium, electronic conformation size etc., all is discussed in A block. Okay, right? So let's write it in A block.

[00:08:12]You will see the size of the A block in the A block, it has been discussed together. Ok? Some people will definitely read such literature, if they read research books then it is written in them that Lanthanum bacteria are the D block elements but their properties should be discussed in the A block elements. Ok?

[00:08:28]Along with the A block elements. Size etc. are physical properties not the chemical properties. Physical properties should be discussed along with the A block elements.

[00:08:35]This is written in the theory. But anyway these lanthanum actinium r the d block elements. There is no doubt about it.

[00:08:40]What are our elements with exceptions?

[00:08:42]Chromium, Copper, Nibium, Molbedum, Ruthenium. Technetium Look, I have not written Technetium. Technetium is not an exception. Remember. Technetium is not an exception. If you read D NCERT, it has been given as an exception in the DNA block.

[00:08:55]Okay, right? given incorrectly. D5 is given in PT.

[00:08:57]Technetium has D5. Remember. It is 4D5. And it is given in PT.

[00:09:02]You go and check it. This is right.

[00:09:04]This is written in NCERT. If you read it in the DNA block, it has changed. Ok?

[00:09:09]He has transferred one electron to D6. One electron is transferred from S3 to D.

[00:09:12]This thing has to be remembered. Technetium is not an exception. Again I repeat you. Ok? I am not speaking from the Technicium and I am not speaking from any outside book either. Go to PT and see it in PT.

[00:09:23]This technician's 3D 4D5 is written only.

[00:09:25]What has he done in DNF that he has made it six.

[00:09:27]We will not accept that. Why would you agree brother? D5 is my peacock stable half field. We will proceed according to this. And it is written in PT in our NCERT. So we will accept this as it is.

[00:09:33]Why Should We Follow The DNF Electronic Configuration of D6 will not follow.

[00:09:37]We will follow this. Ok?

[00:09:39]Remember this is written in the PT.

[00:09:40]You go and check it. Chromium, copper, niobium, molybdenum, ruthenium, rhodium, palladium, silver, platinum, gold, lanthanum, cerium all belong to this block.

[00:09:48]This one is from A Block. Cerium, Gandonium, Actinium, Thorium, Protactinium, Uranium, Neptunium and Curium. These are the exceptions of the N + Al rule. All these are from D Block. Now look, understand one more thing. Palladium and thorium have a difference of two electrons.

[00:10:04]Palladium and Thorium have a difference of two electrons and all others have a difference of one electron.

[00:10:08]First of all, whenever we have to make a transition in the f block, we will send the electron from s to d and whenever we have to make a transition in the f block, we will send the electron from f to d. It also needs to be kept in mind from where to where it has to be sent.

[00:10:19]So just look at what Neptunium is in exception in California? It is not listed anywhere. He will follow absolutely normal. Neptunium is an exception and belongs to A block. When we make the transition of electron, we will send it from F to D. Terbium is not. Just look at it. Then I have the same exception.

[00:10:34]Where it is written Californium 98. If you find out its period number, its consumption will come out.

[00:10:42]Start with the noble gas radon 86 7s above it. Finish at 7p.

[00:10:48]6d 5f in this two 86 + 2 88 88 + 10 90 this will be 0 this will also be zero, it does not matter to you, we will just do the same thing, Radon 7s2 5f 6d0 5f10 6d0 which one is correct, see 60 6d0 is not written, there is no need to write zero, it is written correctly, Neptunium atomic number 93, its period number will also be seven. Start with the Nobel Gas Red on 86 7 of the one above it. It has to be finished at 7p.

[00:11:26]In the middle 6d 5f it has two 86 + 2 88 88 + 5 93 but in exception.

[00:11:34]Look, neptunium is in the one electron exception.

[00:11:36]Where will you send an electron from? F has to be sent to d. So F will become four. And the forest will go into d.

[00:11:42]So Radon 7s2 6d1 5f4 is written correctly and TB 65 will be its period number sixth. The one above this will become the noble gas xenon.

[00:11:59]Start with 54 six. Finish at 60. 5d 4F Its two 6 52 + 54 + 2 56 56 + 9 65 0 will be 6s2 6s2 4f9 6s2 4f is absolutely correct. The first one seems correct. All the first options are correct. 4f7 is given in this which will be wrong for us. 5f4 is given, this will also come out wrong.

[00:12:23]5f9 This will also come out wrong. This is where it is left out.

[00:12:25]So you can check all the other options.

[00:12:27]So the correct option to consume it would be option number first. Let's move on to the next question.

[00:12:36]Next is question number eight.

[00:12:40]When acidic KMnO4 is treated with potassium iodide the iodine is converted. This question has been asked many times. You all know that when I- is added to MNO4- in acidic medium, Mn+2 will be formed along with I2.

[00:12:54]Iodine will be formed or iodine will be formed. IO- No, not iodized. Iodate is IO3-, neither iodated will be formed. But iodized will also not be formed. Iodine will be formed. So the correct option for 8 will be option number second. Let's move on to the next question.

[00:13:15]The Element Question No. 13 The element does not show variable oxidation state. It happens with bromine, right?

[00:13:20]Bromine will show -1, show +1, show +3, show +5.

[00:13:25]These show -1, +1, +3, +5, +7.

[00:13:30]-1, +1, +3, +5 will show. Fluorine Only -1 Bond has been formed with anyone on earth, sky or underworld.

[00:13:37]Fluorine only exhibits the -1 oxidation state. Apart from this nothing else happens. And this is what makes fluorine the most powerful oxidizing agent.

[00:13:46]So the correct option for 13 will be option number four. Let's move on to the next question.

[00:13:55]Question number nine. The Correct Order of Melting Points of Hydrides of Group 16 Elements H2O H2S H2S H2TE H2O has hydrogen bonding. The rest are due to Vander Waals forces which are directly proportional to molecular weight.

[00:14:15]So that is why the order of melting point and boiling point will be highest for H2O because hydrogen bond is stronger due to Vander Waal force.

[00:14:23]Hydrogen bond is a strong bond. Vander Wall is Force Week.

[00:14:26]Now, Vendor Forcers seem to be involved in all this.

[00:14:28]So which one will incur the most vendor wobble? Its molecular weight is high, then H2T, H2SC and then H2S, most of all H2O, H2T, H2SC, H2S, the first option seems correct. There won't be much of H2T. It is wrong. Not H2SC after H2O. This is also wrong. H2O, H2T, H2, I made a mistake in both of them here. So the correct option for 19 will be option number first. Let's move on to the next question.

[00:15:03]Next is question number 27 in view of the following: The bond order has increased and the magnetic behaviour has changed? Look, MOT is a very good question.

[00:15:12]Okay, right? MOT has a great question. In this, magnetic properties were also asked and you were also asked about bond order. There are two very good things about MOT. Its bond order and magnetic property and on the basis of this bond order you can calculate the stability. Ok? So let's see.

[00:15:26]Asking that the bond order is increasing. The magnetic behavior should change. Take away the C2 number of electrons.

[00:15:31]In C2 there is 12 and in this there is 11, so its bond order is one thing, you guys must know the trick of 14.

[00:15:36]14 has a bond order of three. 13 12 11 10 9 8 2.5 2 1.5 1 0.50 And from here 15 16 17 18 19 20 2.5 2 1.51 0.50 20 becomes zero. Ok?

[00:16:01]You should remember this. And for parametric diamagnetic, all the even diamagnetic ones, excluding which one, accept 10 and 16 electrons and all the odd ones, paramagnetic ones, excluding which one, paramagnetic and also odd number and including which 10 and 16 electrons and taking 10 and 16 along because 10 and 16 are not there in it, not even, all the diamagnetic ones, excluding this, 10 and 16, so where will 10 and 16 go in paramagnetic, okay, so the odd ones plus 10 and 16 are paramagnetic. 12 This will be diamagnetic. First bond order sorry first bond order take out.

[00:16:43]Look at its bond order of 12, two 11 bond orders are 1.5, so the bond order has decreased. He is telling us that bond orders should be increased. So this is where it went wrong. Let's move ahead. The number of electrons of NO is 15, in this 14 its bond order is 2.5. And it has a bond order of 14 with a bond order of three.

[00:17:03]Bond orders increased. 15 will be paramagnetic and 14 will be diamagnetic.

[00:17:10]This is even, right? 10 16 is not. If 10 is excluding 16 then 10 is 16 in even, otherwise it will be 14. is diamagnetic. 15 odd so paramagnetic. So its magnetic behavior also changed and its bond order also increased. This option seems right to us. NO to NO+. Let's move on to the next one.

[00:17:27]O2 O2+ O2 has number of electrons 16 in it 15 its bond order is 16 look 2 2.5 is correct? Bond orders have increased.

[00:17:37]Another condition is that the magnetic behavior should change. So even though the 16 is even, the 16 is also paramagnetic and the 15 is also paramagnetic. This is indeed an odd one.

[00:17:48]16 is paramagnetic despite being even.

[00:17:50]Such paramagnetic diamagnetic M OT has that unpaired electron. You draw the MOT diagram of O2.

[00:17:55]You will get an unpaired electron. That's why it is paramagnetic. Okay, right? And there is more experimental data than that. This is a substance. The best experimental way to make a substance paramagnetic or diamagnetic is to place it in an external magnetic field.

[00:18:09]If it shows deflection after being placed in an external magnetic field, then it is paramagnetic and if it does not show any deflection, then it is diamagnetic. So, from O2 only, this also comes under the exception of MOT.

[00:18:17]Because according to the VBT that was told, it seemed that all the electrons in O2 were paired. It should be diamagnetic. But when O2 was placed in an external magnetic field, it started showing deflection. Then work was done on it on emoti. Then it was seen in MOT, when MOT was made it was seen that it has unpaired electrons.

[00:18:33]That is why it is said that if any property of a diatomic molecule is to be explained, it will be explained through MOT.

[00:18:38]Ok? VBT is not able to explain all the properties to him. So its magnet bond order increased but the magnetic behavior remained the same. So that's why we wo n't have any option. into 14 into plus 13. Now tell me it will be three.

[00:18:51]Its value will be 2.5 13, so if the bond order itself is decreased then this option will also not be available.

[00:18:56]So what will be our correct option?

[00:18:58]Second, as the bond order increases, the magnetic behavior changes. So the correct option for 27 will be option number second. Let's move on to the next question.

[00:19:11]Question No. 28 The correct IPACK name of CON N ka Whole Twice NCS ka Whole Twice. Ok?

[00:19:19]Ethylene diamine comes first. Let's look at ethylene diamine 20 right here. 20 Ethylene diamine is correct. The dye is connected to thiocyanate n.

[00:19:30]M by dentate is legged.

[00:19:32]Charge x0 -2 over cobalt to cobalt across pay charge +1 x k +3 na to no went wrong.

[00:19:39]Cobalt will have a charge of +3. The rest of the name was correct but cobalt would have a charge of +3. Bis ethylene diamine dithiocyanate is absolutely correct. Cobalt three nitrate is the right option. B ethyne diamine di thiocyanato n cobalt four not three gone wrong.

[00:19:57]Bis ethylene diamine dithiocyanato n cobaltate no cobaltate is used for the negatively charged sphere. Coordination spheres are used when they have a negative charge. In this the coordination sphere is positively charged. So in this we will use cobalt and not cobaltate.

[00:20:11]English name is used. So the correct option for 28 will be option number second. Let's move on to the next question.

[00:20:22]Question No. 31 The compound formed in the borax with test of copper +2 ion in the oxidising flame. Absolutely correct. Look Na2B4O7.10H2O this is borax. Heat it. After heating it becomes anhydrous borax.

[00:20:40]Na2B4O7 This is called anhydrous borax. It swells up a little and it puffs up.

[00:20:47]Swells up because when its water crystals are loosened, porous holes are formed there. Due to which fluffiness occurs. It feels bloated. If we heat it further, NaBO2 will be formed and along with it B2O3 will also be formed.

[00:21:00]Ok? Sodium metaborate and it is called boric anhydride.

[00:21:06]Its name is boric anhydride. This is what Transparent Glassy Bud provides. It is a transparent glass bed.

[00:21:17]Just understand the experiment. Experiment: Understand what is Boraxbit test?

[00:21:22]In the borax bead test, you will take one bead in a bowl which is a china dish.

[00:21:28]We will first add borax in it.

[00:21:30]After adding borax, we will add salt to it. Make a paste of both borax and salt.

[00:21:34]Ok? After making the paste, we put it on a loop, so it is best to take a platinum loop so that platinum is unreactive and does not react with anything and does not form any compound. If any other iron is taken, then the iron itself will react with it and form some different compound of its own. So platinum, which is the least reactive, cannot react. That's why it is the best.

[00:21:53]So platinum is very expensive. This is not suitable for use in the lab. There is a fear that theft will happen, this will happen, that will happen.

[00:21:57]So that's why he keeps that glass spatula made. Or make your own out of copper. Make one of it and it is taken like this. Ok? Take that paste in it and heat it on flame. What happens after heating? As you see, if you heat it, then in the end you will get a transparent glassy bud, a complete glass ball which is made of glass, like a glass ball. The whole bud is like a goli type. Now this glassy bud is a transparent glass bud. There is no such color. So glass type is water type. The color that gets deposited on the transparent glass bead is due to the salt of the metal that you have added to it.

[00:22:35]And only for the transition element is done. The borax butt test is performed only for the transition element. This is not done for s and p block metals.

[00:22:43]Ok? So whoever climbs will climb because of that. So sometimes you both face the problem that the same colour gives two cations twice, then in that case you have to test both of them. Do it in oxidizing flame as well as in reducing flame. Do it in both.

[00:22:56]So what we do is we take this and then heat it.

[00:23:00]After heating, if we get the color of the bead which is blue, the oxidizing of copper +2 gives blue color in the flame. And what is the reason behind giving it blue colour? understand. The oxide of copper which is generally taken is your salt, copper +2, you will heat it or copper sulphate, you will get oxide, so the oxide is generally taken, copper +2 reacts with a compound, with NaO2 in B2O3, so first of all you have to remember whose compound is formed, copper +2 will combine with BO2-, this will be its anion, so what does it give in oxidizing flame, the colour it gives in oxidizing flame is copper +2, the compound of copper +2 is obtained and its colour is blue. But if you take its colour in reducing flame, it is only copper. It is available in reddish brown colour. You will have to remember all this. You can see the chart in Salt Analysis. So, in an oxidizing flame, copper +2 compound is formed and what is its colour? It is blue. So in whom did you see copper +2? Meaning in NaO2 it will form a compound of BO2-. This means that the compound that is going to be formed is the whole twice of Cubo2. This will be made and its color will be blue. This has to be remembered.

[00:24:15]Cu is asking the same thing that due to which the compound will be in which form? So Cubo2's Whole Twice. The whole two of Cubo2 will form a compound of +2.

[00:24:24]Copper +2 is formed in an oxidizing flame. And you have to remember all of them, now you ca n't remember how many compounds, right?

[00:24:28]You have to remember which cation is formed in the oxidizing flame?

[00:24:32]What does it charge? Which cation is formed in reducing flame? What does it charge?

[00:24:35]And what will anion always be?

[00:24:38]What will the BO2- anion always be? And BO2- formation of the same compound, the same combination which will form the compound, the same reason for the formation of colour. This thing has to be remembered. Ok? So the correct option for 31 will be option number three. Let's move on to the next question.

[00:24:55]Question number Question number 34. Match column one with column two.

[00:25:01]Everyone knows NH4 plus tetrahedral. Br5 has no lone pair in the square pyramidal SF6 just discussed.

[00:25:08]Octahedral Clf3 will be Clf3. We just talked about Clf3 being a very important compound. FF If you cover it, it becomes T shape. If you want to write it correctly then write bent T. There is no complete T. Remember, if someone asks Bent T that if the child does not make it in Clf3, he will make it with a little bit of consciousness. If someone asks what is the 90° bond angle in Clf3, the child will start counting it as 90°.

[00:25:35]This is not a 90° bond angle.

[00:25:37]90° bond angle in Clf3 will be zero. There is not a single 90° bond angle because by giving repulsion its bond angle is slightly decreased or decreased. Ok?

[00:25:45]So that's why it's called bent T gets a little bent. Bentity. T is also written, there is no problem. We will manage with that only. It is T shaped. Look there's tetrahedral third. Square pyramidal is first. Octahedral is fourth. Anti T shape is second. A's third, A's third, B's first, C's fourth, D's second. First option. The correct option for 34 will be option number first. Let's move on to the next question.

[00:26:20]Question No. 36 Which of the following are isostructural pairs? You can take this out. SO42- S double bond O double bond OO- O- will become tetrahedral. There is no loan payer. Cro- Cr double bed O double bed OO- O- There is no lone pair above this also.

[00:26:38]Hybridisation of both is sp3 sp3 zero lone pair tetrahedral both iso structure. It is absolutely the right option. Sil4 tetrahedral titanium also has four valence electrons.

[00:26:47]4s2 3d2 is titanium. It also has only four valence electrons.

[00:26:51]So this will also make Ti4. No loan payer. So its structure will also be tetrahedral. sp3 hybridization tetrahedral geometry tetrahedral shape is the perfect option. NH3 is pyramidal, do you know that? NO3- is trigonal planar N double bond O-coordinate bond O pyramidal trigonal planar.

[00:27:16]Its trigonal sp2 hybridization will be trigonal planar shape. So these two are not isostructural. It went wrong.

[00:27:20]BCl3 is trigonal planar and BrCl3 is also called BrCl3 is the brother of Clf3. Clclcl lone pair lone pair this will be a T shape bent T shape. So this is also not isostructural. So isostructural are A and B. There are two such pairs. Option A and B will be isostructural.

[00:27:42]A and B will be isostructural. So that is why the fourth option will be correct. The correct option for 36 will be option number four. And B only. Let's move on to the next question.

[00:28:00]Question No. 38 Consider the order O2 - F- Na + Mg + 2 is correct? They all are isoelectronic species. All of them are isoelectronic.

[00:28:13]The correct statement is and are decreasing order of Z effective. Z is the decreasing order of the effective. No, increasing order, right? Z of effective. There is a little decreasing. It is true that increasing and Z effective. Look, the Z effect is directly proportional to the number of protons. This slogan has to be remembered. There is a slogan. And inversely proportional to the number of electrons.

[00:28:33]Increasing the number of electrons will increase shielding. If shielding increases, Z effectiveness will decrease.

[00:28:36]If the number of protons is increased, the nuclear charge will increase. His Z effective will increase.

[00:28:39]In isoelectronic the number of electrons remains the same. So the one which has higher number of protons will have higher Z effective. The number of protons in this is 12, in this is 11, in this is nine, in this is eight. So therefore Z effective will be more in this.

[00:28:51]So there is no decreasing order. There is an increasing order. So it is written wrong here. Increasing order of size. Not even the size. Now what is the size?

[00:28:57]What will be the size if Z effective increases? It will be small. So therefore the order of size will be largest to largest.

[00:29:01]O2- then F- then Na+ and then Mg +2. So, you wrote this order wrong also. There is no increasing order. This is an increasing order, but the increasing order of size is not correct.

[00:29:12]This will be the increasing order of size. So, this also went wrong.

[00:29:14]Increasing order of ionization energy. Absolutely correct. This is a correct statement.

[00:29:19]If you want to remove the electron it is m + mg + 2. The positive charge is greater. The z effective is very high.

[00:29:24]Removing electrons will be difficult. So its ionization will be more. After that it will have its lowest ionization energy. So that's why our third option is correct. Decreasing order of electron affinity.

[00:29:35]Mg +2 will have the highest electron affinity. The lowest will be of O2-. The electron affinity of negative becomes negative. The negative species do not want to take electrons.

[00:29:42]So then its not decreasing, its not decreasing order, it is increasing order, it is increasing order. Wherever you see a decreasing order, just cut it wrong. Because the order written is an increasing order. So A and A cannot exist. There will be one option among these two.

[00:29:54]Increasing order of size will be our A. The increasing order of ionisation energy is written correctly. The highest ionisation energy will be of Mg +2 and the lowest of O2-. Therefore the correct option for 38 will be option number three. Let's move on to the next question.

[00:30:12]Question No. 40 Among the following molecules having the same number of lone pairs of xenon lone pair on xenon xenon pe same number of lone pairs XC double bond O double bond O double bond O one lone pair XC double bond O one fluorine two fluorine three fluorine four fluorine eight electrons. Another loan payer will be saved. 4 2 6 XCF6 1 2 3 4 5 6 Fluorine Fluorine Fluorine Fluorine One lone pair Everyone has one lone pair, you are saying that the number of lone pairs is same. You have asked that everyone has one lone pair. Everyone has the same A B & C all are correct. The fourth option is the correct one. The correct option for 40 will be option number four. Everyone has the same lone pair. Let's move on to the next question.

[00:30:58]Question number 41: The hybridisation in the shape of Brf5. Just now you told me what will be the hybridisation of Brf5?

[00:31:08]Brf has seven electrons. Five fluorines are attached to it and one lone pair is kept.

[00:31:14]Hybridisation will be sp3d2 and the shape will be square pyramidal.

[00:31:21]sp3d2 trigonal bi pyramidal no. This is wrong. There is no sp3d. d sp2 is not.

[00:31:27]sp3d2 Square Pyramidal. So the correct option for 41 will be option number four. Let's move on to the next question.

[00:31:39]Question No. 43 The amphoteric oxide among the following is Cr2O3 is amphoteric.

[00:31:45]Mn2O7 is acidic. V2O5 Complete Bass [sound of clearing throat] CRO Complete Bass. Chromium's +3 is acidic, vanadium 's +5 is acidic. Ok? Cr2O3 is acidic. So the correct option for 43 will be option number first.

[00:32:05]This is all about your inorganic question.

[00:32:08]Thank you everyone. Thank you so much.