COMEDK Chemistry PYQs (Last 10 Years) | Chemical Bonding and Molecular Structure COMEDK 2026

Added: 2026-05-10

[00:00:00]Hello champions, welcome to the channel and in today's video we are going to solve the last 10 years of comet K PQs of a very very important chapter that is chemical bonding from PU1. So before doing the PQS let's understand the weightage of this chapter. So if you see that in 2023 there were two questions asked in 2024 there were two questions asked but somehow in 2025 one question is asked. So you can easily expect two questions from this chapter because this is a very high weighted chapter. So two questions definitely can come from this chapter and this is very simple chapter you don't have to worry too much just focus on the westward theory formal charge bond order and molecular orbital theory okay so now let's see the first question here which of the following statements about CO3 2 minus is correct okay now whenever you get this question be very very careful whether it is correct or incorrect okay now the CO bond order is 1.5 this is not correct okay bond order varies because this is a if you see CO3 2 minus structure is like Okay. So this is like completely in resonance. So you cannot have like 1.5 of fixed bond order. So that is why this is an incorrect statement. The formal charge on each oxygen atom is -0.67.

[00:01:09]Now if you see the structure on this central oxygen because it is having only two bonds the charge is zero. And if you see on this oxygen you can see that the total valence electron of oxygen is six.

[00:01:19]Now how many unpaired electrons are there or how many non-bonding electrons are there. So if you see already two electrons are present, two pairs of electron and one negative charge is there. So that means it has six unpaired electrons minus half of bonding electrons that is two. So that will give me minus1. So this is minus1 this is also minus1. So total charge is what -1 + -1 + 0 divided by 3 and that gives me -2x 3 which is around 067. So option number B is the correct answer. Next in which of the following the central atom is sp3 hybridized. So for central atom sp3 hybridized you know CH3 plus is sp2 and if you see NH3 NH4 plus okay so here you have four nitrogen groups.

[00:02:03]Okay and all of them are sp3 hybridized.

[00:02:06]So option number B here. Next which of the following is not paramagnetic means it is asking which of the following is diamagnetic. Okay which of the following is diamagnetic. Now diamagnetic means it should have no unpaired electrons.

[00:02:20]Okay. Now you have to actually try this one and do like for each of them. So for CO it is 6 + 8 there are total 14 electrons. Right? And for 14 electrons the configuration is sigma 1s sigma* 1s.

[00:02:35]Then we have sigma 2s and then we have sigma star 2s. Then after that for 14 you'll have after that pi 2px is equal to pi 2p y. Then we have sigma 2pz.

[00:02:48]Okay. And then comes pi star 2px. Then equal to pi star 2p y. And we have sigma star 2p z. So total 14 electrons are there. So 1 2 3 4 5 6 7 8 9 10 11 12 13 14. Oh, I got the answer in the first option itself. So you can see all are paired electron. So there is no unpaired electron. So that is why option number A here. Now if you see O2, O2 has 16 electrons. Okay. And for 16 like more than 14, the configuration is like this.

[00:03:19]You have sigma 1s, then you have sigma star 1s, then you have sigma 2s, then you have sigma star 2s, then here there's a shift. You get sigma 2pz first and then you have pi 2px, then we have pi 2p y, and then we have pi star 2px.

[00:03:38]Then we have pi star 2py.

[00:03:42]And then we have sigma star 2p z. So 16 electrons 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 and 16. Okay. So we have unpaired electrons here. Now B2 is hydrogen, helium, lithium, berium, boron 10 electrons, right? So 10 electrons means you will follow this configuration. So if I just write down here 1 2 3 4 5 6 7 8 9 and 10. So this also has two unpaired electron. Now n if you see it is uh 7 + 8 15 electrons right means you have to follow this configuration means because it is more than 14. So if I write here 1 2 3 4 5 6 7 8 9 10 11 12 13 14 and then 15. You have again one unpaired electron here. So option number A is the correct answer. Now students if you have written the KCT 2026 examination and are curious to know about your rank then me and my team has prepared a very very useful tool for all of you. So in this tool you just have to enter your second PC marks that is out of 300 and your case score out of 180 and then just click on predict rank and it will show you the estimated rank.

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[00:05:25]Now let's see question number four here consider the following statement. Now for this question just remember one thing that if bond bond order okay is inversely proportional to bond length and for bond order we have one trick. So 14 13 12 11 10 and they said we have 15 16 17 and 18. So if the total electrons in a molecule is 14 bond order will be three and if it is 15 it is 2.5 2 1.5 and then 1 and this side also it is 2.5 2 1.5 and 1. Okay. Now bond length in N2+ in N2+ means there are total 13 electrons. Okay. So N2 plus what is the bond order of this one? 2.5.

[00:06:10]Now bond length in N2+ is 002 anstrom greater than that of N2. So in N2 the bond length is bond order is three. So definitely if you see here the bond order is less so the bond length will be more. So this statement is correct. Next the bond length in N plus N plus is again 14 right? N plus is having 14 electron. So the bond order is three is 09 anstrom less than that of N O. Okay.

[00:06:37]So this is 2.5.

[00:06:40]Now if you see that this bond order is more. Okay. Okay, bond order is more then the bond length here will be less.

[00:06:46]So bond length in N plus you is 0.09 angstrom less. Right? You can see here that here since the bond order is more its bond length will be less. So this is also correct statement. O22 minus has a shorter bond length than O2. O2 is 2 minus is 8 + 16 17 18. O2 2 minus is 18 electrons and O2 is 16 electron. So what is the bond order of O2 2 minus? uh for 18 electrons it is one and for O2 it is two that is the bond O22 minus has a shorter bond length no because it is having higher bond uh sorry because it is having lower bond order its bond length will be more so this is the incorrect statement so we have only option number C here next hybridization of ethine ethine and ethane this question was asked in KCT 2026 also ethine is your spine is sp2 and ethane is sp3 So option number B here. Next. In a homonuclear molecule, which of the following set orbitals are degenerate?

[00:07:46]Degenerate means they have same energy.

[00:07:48]So pi2px and pi2py. By specify the coordination geometry and the hybridization of nitrogen and boron atoms in a 1 is to1 complex of BF3 and NH3. So if you see NH3 is there and BF3 is there. NH3 has a lone pair of electron. It donates here. And then you have a complex like this. N H3 is there and then you have B F3 I just draw here. Okay. So if you see now N is also tetraedral and B is also tetrahedral both are sp3. So option number A here. Next which of the following represents zero overlapping.

[00:08:24]So PZ and P and S orbital can overlap.

[00:08:28]px and px they can overlap but they will have lateral overlapping. Here also PX is having uh I can say destructive interference but there is no overlap between PX and S. Okay, PZ and S can overlap but not this one. So option number D here. Next the stability of the species are increasing in the order of okay so stability of the species means you have to just calculate their bond order. Higher the bond order more is the stability. So lithium 2 hydrogen helium lithium lithium 2 has six electrons.

[00:08:59]Lithium 2 minus has seven electrons and this has five electrons. Now if I calculate for lithium 2 okay so that will be sigma 1s sigma star 1 s and sigma 2s. So we have 1 2 3 4 and 5 6.

[00:09:15]Okay. So what is the bond order here?

[00:09:16]Half of bonding minus non-bonding. So 4 - 2 that is equal to 1 here. Now if you see lithium 2 minus means you have one more electron here. So that is sigma uh star 2 s1 so l i 2 minus so that will be we have 3 - 4 3 - 4 / 2 - 1x 2 - 0.5 we'll get okay okay first sorry bonding minus antib-bonding right so 4 - 3x2 I'll get + 0.5 now if I have li2 and then plus means from here you are removing one electron okay like this part is not there 1 2 3 4 5 6 right so here it will become 1 electron so again here also I will have 3 - 2 by 2 that will also give me 0.5 so definitely li2 is the one which is more stable and I have only one option with that so option number C here now if I compare Li 2 minus and Li2 plus in Li2 minus the electron goes to the anti-bonding orbital so that is why it is less stable moving to the next question among LIC BL2 BCL3 and CL4 the coalent character varies Now if the annions are same the coalent character first depends on the charge of the kion.

[00:10:30]If the charge of the kion is same then you should consider the size. Smaller the size more will be the coalent character. But if you see here the charges are different right? So more is the charge more is the coalent character. So obviously CL4 will be more than b3 b2 and l. So option number c here. The relation between the dissociation energy of n2 and n2 plus.

[00:10:50]Now n2 bond order is three. N2 plus bond order is 2.5. That means the dissociation energy of N2 will be more than that of the N2 plus. So dissociation energy of N_sub_2 is more than that of N2 plus. Option number C here. Next a molecule is square planer with no lone pair. What type of hybridization is associated with it? So for square planer it is DSP2 hybridization. A coalent molecule AB3 has a pyramidal structure. Okay. AB3 right has a pyramidal structure. The number of lone pair and bond pair electrons. So parameal structures means you'll have a lone pair. Number of lone pair and bond pair. So one lone pair and three bond pair. Next. The charge to size ratio of an kion determines its polarizing power. Which one of the following sequences represents the increasing order of the polarizing power of the kionic species. Okay. So as I said first of all you have to see the charge. K plus is having less charge. So it will have the least I can say polarizing power and that I have only in one option. So option number D. But now if you check here it depends on the size. Berilium, magnesium and calcium.

[00:11:56]Smaller the size, more will be the polarizing power. So, berium berilium will be highest then magnesium and then calcium. Next, the melting point of orthohydroxybenzihide is lower than that of parah hydroxybenzihide because so in orthohydroxybenzihide you have intramolecular hydrogen bonding but in par hydroxybenzihide you have intermolecular hydrogen bonding. So option number D here which one of the molecular species has unpaired electrons.

[00:12:25]So N2 if you see here you have to definitely test the electronic configuration. So n_sub_2 is 14 and if I write for 14 it is sigma 1s sigma star 1 s then I have sigma 2s sigma* 2s then you have pi 2px pi 2p y okay then we have sigma 2pz then we have pi star 2px pi star 2p y and then we have sigma star 2p z now nitrogen is having 14 electrons so I'll just write all the options here one by one So for option number A it is 1 2 3 4 5 6 7 8 9 10 11 12 13 14. So here there is no unpaired electron.

[00:13:07]Okay. So option number A is incorrect.

[00:13:09]N22 plus means you will remove two electron. Means you will remove this two electron and then also there is no unpaired electron. N22 minus means you will add two electrons here. Right? So N2 is there 2 minus you will add two electrons one and one. Yes. So this has unpaired electron. O2 2 minus will have a different configuration actually. O2 is having 18 electrons. So we have sigma 1 sigma 1s sigma star 1 sigma 2s sigma star 2s pi 2px pi 2p y uh sigma 2pz then we have sorry in case of oxygen sigma 2pz will come first right so after I have sigma* 2s I will have sigma 2p z then we have pi star 2px Okay, pi star 2p y and then we have sigma star 2p z. Okay, so I have 18 electrons. So 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18. So here also no unpad electrons option number C here. Next the formal charge on the central oxygen atom in the O3 molecule.

[00:14:22]So if you see the structure of O3 molecule, it is like this. So if I calculate on the first one. So number of valence electron is six. Number of non-bonding electrons is 2 minus half of bonding electron. So it is 1 2 3 4 5 6.

[00:14:36]Okay. So it is 6 - 2 - 3. So that is + 1. So option number B here. Next. Pick the incorrect statement among those given below. Multiple coalent bonds are shorter than the you have to pick the incorrect statement. Okay. Please remember this one. Multiple coalent car bonds are shorter than single coalent bond between the same set of atoms. This is a correct statement because uh when you have multiple bonds the bond order is more so bond length will be less. Bond strength varies inversely with the bond length. This is also correct statement. Bond order of iso electronic species remains the same.

[00:15:13]Bond enthalpy increases with the increase in bond length. Now when you increase the bond length, bond strength decreases. So bond energy also decreases. option number D here. Next, which of these represents the correct order of increasing of bond order? Okay.

[00:15:28]So for that you need to calculate the number of electrons here. Okay. So if I see first of all carbon is having C2 uh that is 6 + 6 12 + 2. So I'll just write down here C2 2 minus. So it is 6 + 6 is 12 12 13 14. So for 14 the bond order is three. Now H E2 plus right for this one you have to write the electronic configuration because the number of electrons is less than 10. So I have for H E2 plus the like I can write here H E2 plus. So we have sigma 1s sigma* 1s sigma 2s and sigma* 2s. So we have hydrogen helium.

[00:16:09]So 2 to 4 electrons 4 - 1 we have three electrons. So one and this one. So 2 - 1 by 2 that is 0.5.

[00:16:19]So this will give me 0.5.

[00:16:22]Now O2 plus O2 plus is having how much?

[00:16:26]16 uh 16 - 1 will be 15 and 15 will have a bond order of 2 sorry not 2 plus just two it is 2.5 and O2 minus means 17. So if you remember for 14 it is 3 15 it is 2.5 16 it is 2 and 17 is 1.5 and O22 minus is 17 means it is 1.5 so if I now calculate helium 2 plus will be the least so that is there only in one option that after that O2 minus after that O2 plus and after that we have C22 minus option number B here next pick up the incorrect statement again dipole moment in ammonia is due to the orbital dipole resultal dipole in the same direction. That is correct. Here in BF3, bond dipoles are higher but dipole moment is zero because they cancel out each other. Dipole moment is a vector quantity. O2 and H2 show bond dipole due to polarization. No in O2 what happens?

[00:17:25]Because the two atoms are same. There's no electro negativity difference. So there is no bond dipole here. So option number D here. Next the pair of species that has the same bond order in the following is it should have same bond order. For that you need to have iso electronics disease. Now carbon is 6 + 8 14 and this is 7 + 8 15 and 14. So they are having same electron. So option number A here. Now let's check the other option here. 7 8 15 + 1 16. Here carbon 6 and 7 uh 13 + 1 14 uh 16 and 14 16 and 10. So you can see that only in the first option we have same number of electrons. So that is why here the bond order will be same. Next the outer orbitals of C in the ethine molecule can be considered to be hybridized to give you three sp2 orbitals. The total number of sigma and pi bond. So if you see the structure of ethine it is like this right. So the total number of sigma bonds is five and pi bond is 1. Next in which of the following pair both the species have sp3 hybridization. So beh is a linear molecule. NF3 and H2O both of them have sp3 hybridization. Next, in which of the following ionization processes, the bond energy increases and the magnetic behavior changes from paramagnetic to diamagnetic.

[00:18:41]Okay, bond energy increases and the behavior changes from paramagnetic to diiamagnetic. Okay, bond energy increases means bond enthalpy should increase. Okay, or bond order should increase. So, we'll check in which of this bond order is increasing. So, if you see O2 is I just write down the bond order values here.

[00:18:59]16 17 18 13 12 11 and 10. So this is 3 2.5 2 1.51.

[00:19:09]This is 2.5 to 1.51.

[00:19:13]So 16 is there here the bond order will be 2 and 165 will have the bond order of 3.5. So yes, bond energy is definitely increasing in this option. Carbon is 12, 12 bond order is 2, carbon 2 plus is uh like 11 and its bond order is 1.5. So bond order is not increasing. Eliminate this option. 7 + 8 15 15 bond order is 2.5 and 14 bond order is three. So bond order is increasing in both these cases.

[00:19:42]Nitrogen is three uh like 14 and this one is 13 2.5 bond order is decreasing.

[00:19:48]So I can eliminate this one. Now the question is regarding paramagnetic to diamagnetic. Now see oxygen you should remember you have two unpaired electrons in your sigma sorry pi star 2px and pi star 2py and if you rem sorry remove one electron it still remains paramagnetic so it doesn't changes from paramagnetic to diamagnetic so the only option that might be correct is option number c but let's check it so here I have 7 + 8 15 and here I have 14 okay so 15 is there 14 is here so if I write the electronic configuration sigma 1s Sigma star 1 S okay sigma 2s sigma* 2s then I will have nitrogen is there so it will be with pi 2px pi 2p y sigma* 2p z now I have 15 1 2 3 4 5 6 7 8 9 10 11 12 13 14 and in 15 we have pi star 2p x that is equal to pi star 2p Y here one electron is there. Now when you remove this one you can see all of them are completely paired. So it changes from paramagnetic to diamagnetic. So option number C here. Next bond order in nitrogen molecule we know N3 has three 14 electrons. The bond formed between the ammonia and the proton in a ammonium ion is so when you have NH3 okay the lone pair of nitrogen is donate to donated to one H+ electron and here we have this bond order okay and that gives me coordinate bond option number B here. So children these are the last 10 years of comet K pyqs and you can see most of the questions are based on bond order and the molecular orbital theory. to prepare these concepts thoroughly and also the west theory. Okay. So, thank you so much and the case rank predictor link is there in the description section.

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