INORGANIC CHEM-TEST-06-VIDEO SOLUTION FOR RE NEET-2026

Added: 2026-05-28

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

[00:00:10]Hello Students, We are going to discuss the inorganic question of test number six of test series for the RENET 2026 which was held on 25th May 2026. So, let us discuss the organic questions of this examination. Which is question number first in organica that is question number second. This is an urgent question. In this Assertion is saying Among all the hydrides of group 15 elements BH3 is the strongest oxidizing agent and is giving wrong answer.

[00:00:40]Strongest oxidizing agent will not be the strongest reducing agent. Listen NH PSA SBA & BH as you go top to bottom the bond length increases. As the bond length increases, the bond dissociation energy decreases. And when the bond dissociation energy decreases, it means that the bond between Bi and H can break easily. When broken down easily, it would be easy to give hydrogen to BiH. So that's why loss of hydrogen is called oxidation. Gain of hydrogen is called oxidation. Loss of hydrogen oxidation is called gain of hydrogen reduction.

[00:01:23]and loss of hydrogen oxidation. Gain of hydrogen reduction. So what does loss of hydrogen mean?

[00:01:30]Oxidation will do it. When oxidation occurs it will act as a reducing agent. So the worst reducing agent would be NH3 and the best reducing agent is BH3. Ok? Because of low dissociation energy, the bond of BH will break very easily and when it breaks easily, it will have more tendency to give. It will readily give up hydrogen. Loss of hydrogen would be good.

[00:01:52]And loss of hydrogen is oxidation.

[00:01:54]Gain of hydrogen is reduction. So that is why it is not the strongest oxidizing agent but the strongest reducing agent.

[00:02:00]What's giving here at Reese's? The size of bismuth is largest. That's why the BH bond dissociates more easily than other hydrides of group 15. This is absolutely correct what we were just explaining.

[00:02:12]This is what we were talking about. The bond length of Bih becomes larger because the size of Bi is larger. Hence the bond length will be larger. If the bond length is large then its bond dissociation will be very less. This is written correctly. But it is unable to produce an oxidizing agent. That makes BiH3 a reducing agent. Therefore the assertion was false.

[00:02:28]Reese turned true. Assertion is false.

[00:02:31]Reason is true. So the correct option for second will be option number four. Let's move on to the next question.

[00:02:44]Question number three list some basic radical given in this. Cations are given and their group is given. Which group does he belong to? He is asking questions in every test. You can ask because it is a part of salt analysis.

[00:02:55]Classification of group cations is very important. Ok? You will have to remember this. It belongs to the first group.

[00:03:00]Today you have gone crazy, it contains silver, mercury and lead.

[00:03:04]So, he will have to remember that. Throw the crispy potatoes, throw the crispy potatoes, it is of the third group, is n't it? Throw the crispy potatoes, throw the crispy potatoes, see, I tell you again and again, it has come in almost every test, whatever test series has been given, it has come in all of them, I tell you, it is of the third group, do n't go on Monday, it is the fourth group, do n't go on Monday, it is the fourth group, Calcium Plus2, Government of India, it is the fifth group, Government of India, it is of the fifth group. This copper Punjabi girl will also be imprisoned. He belongs to the second group.

[00:03:44]Punjabi girl also got imprisoned.

[00:03:52]Copper belongs to the second group. Let's see.

[00:03:55]Fourth, this is third. In the third group, the fourth option.

[00:03:57]This will be his fourth. Nick + 2 Fourth Group. The fourth group is in the third option. It will be third here. Calcium is in fifth second and copper is in + two to first. Look at A's fourth A's fourth is in two. B's third B's third is in it. C's second D's first. So the correct option for third will be option number four. Let's move on to the next question.

[00:04:23]Question number four.

[00:04:26]Among the following oxides of p block elements the number of oxides having amphoteric nature. I tell you this again and again. You should know the nature of all oxides.

[00:04:36]What elements make acidic, basic, amphoteric?

[00:04:39]Which elements do not make what? One more thing, it is also known whether amphoteric is predominantly acidic or predominantly basic.

[00:04:45]This means whether the acidic character is more dominant or the base character is more dominant. This should be known. This is a neutral oxide. Do you guys know? There is boron, it is acidic oxide, there is silicon, it is also acidic. The oxides of aluminium are amphoteric.

[00:05:00]But look, aluminum has a charge of +3. The +3 oxidation state of the metal tends to be basic.

[00:05:05]Aluminum is definitely amphoteric but it has more base. The acid is low.

[00:05:09]So we will say Al2O3 is predominantly basic, meaning it will be more basic. It will be a little less acidic. TL2O3 is basic. PbO2 is +4. This is also amphoteric. All oxides of the metal lead are amphoteric. In this also there will be more basic and less acidic. So what you have asked is how many oxides are amphoteric in number? So Al2O3 PBO2 only two oxides are amphoteric. Therefore the correct option for fourth will be option number fourth. Let's move on to the next question.

[00:05:44]Question number six The IPACK official name of the element with atomic number Now this should be known.

[00:05:50]I am telling you again.

[00:05:51]This is a good thing. Atomic numbers 1001 to 118 are given official IPAC names based on the scientist's name. So you will have to remember this, you should remember all the names because in this it has been asked to give you the atomic number and then ask the reverse of it. Like 155 is moscovium. Moscovium has 155. I would have given you what is the eye-catching name of moscobium? He would have given the MC symbol or just written the name. So if you do not know the atomic number of moscopium, then if someone asks you what is the ipack name of 155, then you immediately tell them that it is unpentium, this is a childish question, but if the same thing was asked this time in 2026, the ipack name of 151 was rontium etc., then it would have been an easy question, but if it is asked that tell me the official ipack name of atomic number 155, which is given on the basis of scientists, or its opposite, tell me the ipack name of moscovium, what would it be, okay, so you should know Moscovium by now. So, please remember him once. You must remember all the official IPAC names from 1001 to 118 which were based on the name of the scientist.

[00:06:58]Whose is 115? Moscovium. So the correct option for six will be option number first. Ok? Let's move ahead.

[00:07:10]Question number nine.

[00:07:15]The complex diamine diamine NH3 has a whole twice diamine chlorideo Cl nitrite NNO2 platinum to platinum charge will come automatically two, see it will come -1 -1 +2 can be okay see which one is PTNH3Cl the first one seems right to me, the second one is hydrazine which is wrong we don't have hydrazine in it the name ethyne diamine is also wrong. This diamine difluoride has also been given wrongly.

[00:07:47]Chlorideo is the same, right? Where is the dye in colloid? So the correct option for nine will be option number one. Let's move on to the next question.

[00:08:04]Question No. 10 The density of group 13 elements increases down the group. Move from boron to thallium.

[00:08:12]Its density increases due to increase in molecular mass. There is a very regular increment in density.

[00:08:16]As the molecular mass increases, the density increases. So the density increases as we go from top to bottom in group 50.

[00:08:21]Density increases down the group from boron to gallium. No, this is wrong.

[00:08:27]Increase down the group from bo exactly.

[00:08:29]Boron to thallium increases. Will not decree. It increases. It will not decrease due to increase in molecular mass. He increases it. So the correct option for 10 would be option number second. The density of group 13 elements increases from top to bottom.

[00:08:43]Ok? From boron onwards, as the molecular mass increases, the density will increase. Ok?

[00:08:48]Next 22 22 Question No. 22 Match the complex species given in column one with the possible isomerism given in column two. Ok?

[00:09:06]Look at isomerism, this is a good topic. Please try to understand it once.

[00:09:14]CO NH3 times 4Cl2 plus charge, see this is how I tell the child.

[00:09:23]is an octahedral complex. It has two legged beans.

[00:09:25]Select this. And after selecting it, press it once. The rest will become trans NH3 NH3 NH3.

[00:09:35]Now there will be no other possible pair for it.

[00:09:37]Now once you have selected on She, place it on C and then place it on Trans.

[00:09:41]Clcl NH3 NH3 NH3NH3 This is in it. Now this is trans and this is cis.

[00:09:54]This is also optically inactive because here the trans is the same ligand. It will have POS.

[00:09:58]I have shown it in this. If there will be POS in this also then it will be optically inactive and there will be POS in this also.

[00:10:02]This will also be optically inactive.

[00:10:04]So what will you do? GI: Two GIs will be made for this. One will become cis and one will become trans. It is optically inactive. will not show any optical isomers. So this will show geometrically. So force would be the right option.

[00:10:16]Sis ptcl2 Look, this is a slogan and you have to understand it as well. What? Two by dentate with the cis orientation always optically active. Three bidentate always optically active. Two bidentate with the cis orientation always optically active. But why does this happen, please understand. Now only its cis orientation is given. In two bidentates, we will always select cl. Clcl n This is cis orientation. If we make a trans of this then the trans will be like this.

[00:10:46]nn n This is trans. It is optically inactive.

[00:10:51]This is optically inactive. But it is optically active. Why? Its mirror image will also be optically active.

[00:10:56]What reason do I tell children about it being optically active? I mean, whatever literature I write, I tell them that it is bydentate.

[00:11:04]Both of these are Balki groups. Bidentate will not be a small group but a big group and because it is a big group, when these two remain in cis orientation, then due to their distance being less, there is more repulsion between them.

[00:11:15]When the repulsion is high, these two groups move out of each other's plane.

[00:11:19]They go to another plane. In the part you might have read about steric inhibition resonance, these two groups move out of each other's plane.

[00:11:26]When both of them go out of each other's plane then there is no chance of POS.

[00:11:29]When both are in different planes then no POS. When not POS then it is optically active. And because of this, the three bi-dentate always becomes optically active. Because this orientation of three by dentate will always come. It is not possible that you can escape this orientation. Three by dentate one. This orientation will always come.

[00:11:46]Therefore it becomes optically active.

[00:11:48]He will do this optically active show. Okay, right? Can't think well geometrically. He said Sir, geometrical because it is written in the sys. Do n't think about trans. I have taken an orientation, sis, so you have to talk in this only. There is no need to think about this. Trans has not said sis. So that's why it is optically active. You will be first in this.

[00:12:07]Look, which one is this here? Here's Br and here's SO4. These two will be interchanged with each other.

[00:12:12]Ionization isomer. SO4 will go out and Br will go in. Now it will give the test of Br. When SO4 goes out it will give test of SO4. So therefore the ionization isomers of different ions in the aqueous solution will be second [nasal sound] and a coordination isomer third when both are complex cation and complex anion, then those types of complexes are used to show the coordination isomers, there is an interchange between these two, partially and fully, okay but remember one thing, imagine I am telling a very good thing to a child, you guys should try to understand otherwise you will come back with a mistake. Well, by the way, let me ask what is the total number of isomers that will be formed?

[00:12:54]Six of CrCN will have charge +3 on this and charge -3 on this.

[00:13:01]This one. Ok? Chromium's +3 is the most stable. Therefore its charge will be +3 and this will become -3. Therefore the total will be +3.

[00:13:10]Now you start interchanging. Hole 5 of CO NH3 Hole 4 of CN Hole twice of CO NH3 Hole twice of CN Hole twice of CO NH3 Hole twice of CN Hole 4 of CO NH3 CN call 5 and hole 6 of CO CN When you do it like this then four +3 will come on this +2 will come on this +1 will come on this 0 will come on this you will take out -1 -2 -3 in this and similarly here here -3 so -2 -1 0 +1 +2 +3 you will do it like this what I want to explain is that the complex that will be formed will have zero zero charge on both these complexes and when there will be zero zero charge then we do not count it in isomer. This is not counted as a coordination isomer. So its coordination isomer will be 1 2 3 4 5 6, seventh we will not count this because when we look at both the complexes, this entire complex is one, there are not two complexes, the entire complex is one and what is the bond between these two, ionic, what is the bond between these two, ionic, to be ionic there should be a cation and an anion, this is +3, so what is this -3 cation anion, only then an ionic bond will be formed, ionic bond means the one which dissociates, so we should get ions, when the charge on both is zero, then in this, ionic bond will not be formed between the two, in the coordination sphere, ionization sphere.

[00:14:37]This is our coordination, so this is ionization. This ionization is coordination. Either coordination will be an ionization. Now the charge on both of them is zero. Then is it possible for an ionic bond to form between the two? It is never possible. This means that a complex will separate you. The complexes that follow this will separate both of them.

[00:14:52]Both will separate. There is no relation between the two complexes. Now you can establish it. Therefore, we will not count it among the hydrate isomers.

[00:14:58]Both the complexes will not look like one. There is no bond in both complexes.

[00:15:01]Both are different. It is completely separate. There is no relation between the two.

[00:15:05]That's why this has to be remembered on both the complexes.

[00:15:07]In coordination isomer, when we exchange the ligand between both the complexes, between both the spheres and if zero charge comes on both the spheres, then we will not count it as coordination isomer. Ok?

[00:15:19]Remember this. Well, now we have to find out whether it is a coordination isomer. It is third. So the fourth of A is geometric. B's first, A's fourth, B's first, C's second, D's third.

[00:15:32]Looks like a fourth option. The correct option for 20 seconds will be option number four. Let's move on to the next question.

[00:15:42]Question number 23.

[00:15:46]Which of the following is the correct order of dipole moment? Ok? Look, each one is a small thing. You people have to understand this.

[00:15:53]CH3F, CH3Cl, CH3Br and CH3I, just listen. To find the dipole moment μ. Let's write. q.d q will be more.

[00:16:07]μ will be greater, d will be greater, yet μ will be greater. Generally q is the dominating factor. This is the only case where P/D dominates. Look, if we talk about q in this, then q will increase but d will decrease.

[00:16:21]Sorry d will also increase. No, he must have written it upside down.

[00:16:23]1 minute 1 minute when from here to here q what will happen? Will decrease. The highest q's are delta negative, delta positive, delta negative, delta positive, delta negative, delta positive, delta negative, delta positive.

[00:16:38]Which will have the highest delta negative delta positive? into carbon fluorine. Then carbon chlorine then carbon bromine then carbon iodine. So the value of q, the difference in charge, will decrease. But what will happen to d? Will increase. So now we have a problem.

[00:16:50]But that is why I am telling you that Q dominates almost 95-97% of the space. It happens at some places where D dominates.

[00:16:59]So this is a case where Pe D dominates.

[00:17:01]D dominates between fluorine and chlorine.

[00:17:03]And because of this, due to the larger size of its chlorine, its dipole moment becomes more than CH3F.

[00:17:11]So CH3Cl then CH3F but it has less chlorine. It is not that it will reduce much just by chlorine. Ok? Everything else has a charge. CH3Br followed by CH3I This is the correct order of dipole moment of carbon halides. CH3Cl CH3F CH3Br CH3i sec 23 correct option will be option number sec. Let's move on to the next question.

[00:17:43]Question No. 25 Which of the following complex ion has a magnetic moment same as the whole number of CrH2O. Well, the value of magnetic moment has to be found. Find the number of unpaired electrons.

[00:17:53]It contains chromium +3. 4s0 is 3d3. There will be no pairing. So the number of unpaired electrons is three. Now see in which of these the number of unpaired electrons is three. Cobalt +2 4s0 3d7 and Weak Field is Legend for Cobalt +2. There will be no pairing.

[00:18:09]and How Many Unpaired Electrons Are Present in d7? 1 2 3 4 5 6 7 Three unpaired electrons. The value of n is three. It will be the same. But let's check some more. is mn + 3. mn + 3 has 4s0 3d5 sorry 3d4 mn + 3 right? Behaves like a Weak Field for 3d4 and Mn+3. There will be no pairing in the weak field. When there is no pairing, what will be the number of unpaired electrons? Four. So, it won't happen.

[00:18:37]We have it wrong. Iron is +3, 4s0 will behave like a Weak Field in 3d5 and 3d5. There will be no pairing in this also.

[00:18:46]When pairing does not occur, the number of electrons on the front is five. So its magnetic moment root will also come out to be under 35. So, this will not happen either. Copper + 2 is a d9. It will excite. In this the number of front pair electrons will remain one in every case. So this will also not be equal. So who will be equal to whom? First K. Therefore the correct option for 25 will be option number first. Let's move on to the next question.

[00:19:14]Question No. 26 Match the column one with column two. There are some complex givens in this. Its hybridization is given. Look at NiCO4, hybridisation of NiCO4 becomes sp3.

[00:19:25]In this, the charge on Ni is zero. The electronic configuration is 4s2 3d8.

[00:19:30]Carbonyl is a very strong field legged. He would definitely want to get this pairing done. But when you get the pairing done, when you get the pairing done, then the pairing will happen. Ok? But after that one will vacate D. Just look here, after pairing it, it will have this status and 4s2 is here.

[00:19:47]4p is empty here. This is 4s2. This is 3d8 and 3p empty 4p empty. Now because in complex compounds the same atomic orbital participates in the hybridisation of CMI which is completely vacant because lone pair of electrons has to be accepted. One of them became empty. Now s came after d of s. s is not empty here. And it cannot participate in hybridization by taking full field.

[00:20:11]Because then how will it be able to accept electrons from the ligand? So it is not possible. So that's why he didn't benefit from this pairing. So what does he do? First, s is emptied.

[00:20:20]To vacate it, it sends both its s electrons to 3d. And this becomes 4s0 3d10. Now it becomes 3d10.

[00:20:28]Whether it is a strong field or a weak field, what difference will it make? The hybridisation of all is sp3 at coordination number four.

[00:20:31]And the geometry will be tetrahedral. NiCO4 Look, now there's Ni+2 here.

[00:20:37]NH2 Ni +2 will have D8. It will become 40.

[00:20:40]This will do the pairing here. ds coordination number is four. dsp2 hybridisation will give square plane geometry. Ok? Fe+2 cn will act as a stung field for this. Will get the pairing done. No, when pairing is done, if the coordination number is four or six, then what will happen to hybridization? d2 sp3 will behave as a weak field in mn +2 mn+2.

[00:21:00]No one will do the pairing.

[00:21:01]So what will its hybridization become?

[00:21:02]sp3d2 out orbital complex will be formed. So look, the first one is sp3, this is correct. The first one is correct.

[00:21:10]Its dsp2 seconds is just the second of B.

[00:21:14]Its d2 sp3 is fourth. And its sp3d2 is sp third.

[00:21:22]A's one, A's one, B's second, D's C 's fourth and D's third. Our first option is correct. The correct option for 26 will be option number first. Let's move on to the next question.

[00:21:40]Question No. 31 A solution of FeCl3 when treated with K4CN gives a proton blue precipitate due to formation of hol6. This is an easy question.

[00:21:50]FeCl3 is. Get it reacted.

[00:21:53]6 of K4 FeCN wants to ask you whether you know the formula of Protion Blue or not.

[00:21:59]Fe+3. Here is K4. Will exchange it. So look, the charge on this is 4-. Its remainder is +1. Fe+3 will come here. Will replace K4. So what will happen? Fe 4 FeCN 6 and three will come here. There will be an interchange, right?

[00:22:18]Who is going to replace this K+? Fe+3. And the charge on this will remain just the same. This is not going to change. Four minus. Now write down its formula. Four will come here.

[00:22:26]Three will come here. And this is what Procian Blue's PPT gives. This is Procian blue color or its other name is charge +3 on it, charge +2 on it, so its other name is ferri ferro cyanide, it is also called because there is +3 on the outside and +2 on the inside, so it is also called ferriferrocyanide.

[00:22:47]See what formula it is?

[00:22:49]Hole6 Third option of Fe4 FeCN. So the correct option for this will be option number three. Let's move on to the next question.

[00:23:02]Question No. 35 The set having ions which are coloured and paramagnetic both. Ok?

[00:23:08]Coloured as well as Copper + D10 Colourless as well and is diamagnetic. So this work is finished.

[00:23:13]Zinc +2: This is also colorless. Mn+4 is color colored. Mn +4 is colored. So this is fine. But both of them are also colorless and diamagnetic. is not paramagnetic.

[00:23:23]Ni+2 is colored. Feni is green in colour. Mn +7 is colorless. This is also colorless. D10 is. It is also paramagnetic. It is also diamagnetic.

[00:23:31]Sorry it is diamagnetic. Mercury + 2 is D10. is diamagnetic. So this will also not happen.

[00:23:38]So in this also our option is wrong.

[00:23:40]Copper Plus is D9. It is coloured, right? What colour is it? Blue. Copper cooker is blue in colour. The cooker is blue in colour. Chromium +3 is also a colour. What colour is it?

[00:23:53]Violet. Feel free to do so. Feel free to do so.

[00:24:01]Be a VIP. All these are violet in colour.

[00:24:03]Ok? This is also colored. Scandium Plus has been given. It is also colored because Scandium Plus also has D1.

[00:24:11]We also have this in colour. 4s1 is 3d1. This is also colored. The highly unstable compound scandium +1 never exists. I have given you only one d1, that's why it is colored. Find out this.

[00:24:21]Paramagnetic because the number of electrons per unit is one. So this seems to be your option as well. All of them are paramagnetic. All of them are colours. Copper plus two is blue. A is violet. If this compound itself does not change its colour, then who will be able to decide its colour? It is difficult to tell the color.

[00:24:37]Scandium Plus This Scandium Plus One Hardly Exists in Any Compound. It cannot exist. So its color will be ho colored. Why are you saying colored?

[00:24:46]Because D1 is there. D can transition to D1. It can show color in it.

[00:24:49]This will be coloured. So this seems to be the correct option Scandium +3 is colourless and is also diamagnetic Vanadium +5 is also zero and is also diamagnetic Vanadium +5 is also zero and is also diamagnetic So all of them are not paramagnetic All of them are diamagnetic and all of them are colourless No colour, all of them will be colourless So the one which is coloured and is also paramagnetic is the third option Copper +2 Chromium +3 and Scandium +1 will be the correct option for 35 Option number three Let's move on to the next question Question number 38 When SO2 is passed in an acidified potassium dichromate solution of SO2 When potassium dichromate K2Cr2O7 is reacted in the acidic medium, it automatically converts into chromium +3 and gives SO42- and both of them together give green ppt. Cr2SO4 gives green coloured solution of chromium sulphate. Ok?

[00:25:48]What is? SO2 is passed into the acidified potassium dichromate solution. The oxidation number of sulfur and chromium in the final product. Ok?

[00:25:57]What will be the oxidation state of sulfur in the final product? +6 and +3 of chromium +6 and +3 respectively. +6 and +3 respectfully.

[00:26:06]So the correct option for 38 will be option number second. Let's move on to the next question.

[00:26:15]Question No. 41 Select the incorrect statement. The stability of hydride increases from NH3 to BH3 is given incorrectly, is n't it? It will increase a little. How do you discuss the properties of all hydrides NH3, PH3, ASH3, SBH3, BH3? Bond length will increase top to bottom.

[00:26:36]When the bond length increases, the bond energy will decrease. Thermal stability will also decrease. And this is how our further discussion will proceed.

[00:26:46]So what would NH32BH3 be? Stability decreases when going top to bottom. It does not increase. It is wrong. His first statement itself seems correct. But check further also.

[00:26:56]Nitrogen can not form D Pi. Absolutely correct. Nitrogen does not have any vacancy.

[00:27:00]So nitrogen cannot form a D pi D pi P pi bond. It is written correctly. Single NN bond is weaker than single, this is also written correctly. Just understand. In the triple bond N and P triple bond P, the bond dissociation of nitrogen is more because it does not want to have 3p pi 3p pi formation.

[00:27:17]Its bond dissociation energy is very low. Whereas the bond dissociation energy of 2p pi 2p pi is very high. Hence N2 in triple bonded state is more stable than P triple bond P. But if we talk about nitrogen single bond nitrogen and P single bond P then it has two lone pairs, what is its lone pair is present in the second shell and its lone pair is present in the third shell, so the size of both these second shells is small. So there will be more repulsion between these two. When the repulsion is high then their stability will decrease. If the stability decreases then their bond dissociation energy also decreases.

[00:27:55]And the size of its third shell between the two is bigger. Repulsion will be less.

[00:27:58]Decreasing repulsion will increase stability and result in higher bond dissociation energy.

[00:28:03]So therefore the bond dissociation of nitrogen nitrogen single bond is less than phosphorus same situation here also oxygen double bond oxygen sulfur double bond is more in sulfur only.

[00:28:12]But oxygen single bond oxygen sulfur single bond single has more sulfur.

[00:28:16]This thing has to be remembered. N2O4 has two resonating structures. Correct. Two will be formed from N2O4.

[00:28:21]N double bond O coordinate bond O N double bond O coordinate bond O So do the pi bond here and do the single bond here. Here a single will make two RS. Another RS ​​will be made of this. This is also correct.

[00:28:35][nasal sound] If you have asked incorrectly, then incorrect will be our first choice.

[00:28:39]Ok?

[00:28:42]This is all about your inorganic question.

[00:28:45]Thank you everyone. Thank you so much.