COMEDK Chemistry PYQs (Last 10 Years) | Structure of Atom Chapter with Solutions

Hinzugefügt: 2026-05-11

[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 PU1 chapter that is structure of atom. Okay. So now let us discuss the weightage of this chapter. So if you see in 2025 three questions were asked from this chapter.

[00:00:16]In 2024 two questions and in 2023 one question. So you can see that the weightage is increasing. So definitely you need to prepare this chapter thoroughly and it is an easy chapter.

[00:00:25]Just remember the basic formulas you will be able to do all the questions.

[00:00:29]Okay. So let's get started with the first question. How many nodal planes are there in the atomic orbital for the principal quantum number n is equal to 3s. Okay. Now this is a very new type of question. So in three we have like and when n is equal to 3 we have 3s 3p and 3d. Right? Now 3s has no nodal plane and 3p has three orbitals. Right? So we have 3 px 3p y and 3p z. And each of them has one nodal plane. So here I will get three nodal plane. Now 3D we have three types of 3D that is 3D uh x - y 3D y - z uh 3D z - x then we have 3D uh x² - y² and then we have 3D z² okay now if you see we have four nal planes here means uh in each of them actually you will get two nodal planes if you see the d orbital it is like this I'll just draw one for you okay so if you see the d orbital It is like this. See, so this is suppose X orbital and this is Y orbital. So dxy will be somewhere like this. Okay. So this one is plus, this is minus. Then you will have this is plus and then this is minus. So you can see here there is one change of sign and there is here one change of sign. So we have two nodal planes. Similarly, all of them has two nodal planes. Okay. And this Z² does not have any nodal planes. So here I will get eight nodal planes. So total it will be 11. So option number C here. Next if the uncertaintity principle is applied to an object of mass 1 mg the uncertaintity value of the velocity and the position will be. So here we have Heisenberg uncertaintity principle delta x into delta p is equal to hx 4 pi.

[00:02:15]Okay. Now when you just expand it delta p you will have m into delta v that is equal to hx 4 pi. Right? Now what do we want to calculate here? We want to calculate the uncert uncertaintity value of velocity and position means this one and this one. So we have delta x into delta v is equal to h by 4 pi m that gives me 6.63 into 10 ^ of - 34 okay jou second. Now for jewel I can write down kg m²ared second inverse and then second okay this is jew second unit of h then four is there and then you have 3.14 and mass is there that is 1 mg that means 1 into 10 ^ of -6 kg okay so kg kg will get cancel here and what is the thing that we have to find out we have to find delta x into delta v okay so that is the unit we have now so here I will m² inverse. Okay. Uh yes. So now if you just calculate this one 4 into 3 means almost 12 and we have 10 ^ of - 6. This will go up. So it will come 10 ^ of - 28. So I have only one option with 10 ^ of - 28. So it will be option number c here. Next the energy required to remove an electron from the x is this much. The calculate the maximum wavelength of light that can photo eject an electron from metal X. Okay. So we have the formula E is equal to HC by lambda. Now again now question is asking me to find the value of lambda. So it will be H C by E. Now H value is 6.63 into 10 ^ of - 34 JW and the C value is 3 into 10 ^ of 8 m inverse. Next we have the energy value which is 3.31 into 10 ^ of -20.

[00:04:12]Again jewel will get cancel. Second and second inverse will get cancel. We have this meter here. Now 3.3 and 6.6 you can round it off to around 2 and 2 into 3 is somewhere around 6 here. And I have the value here that is 10 ^ of - 26 right?

[00:04:28]Oh sorry 120 is there also. So + 28. So this is minus 6 only. So option number A here. So students now before moving ahead if you have written KCT 2026 examination then I have something very very interesting for all of you. So me and my team has prepared a very useful tool for all of you using which you can predict your expected rank in KCT 2026 examination. So it is very simple. You just have to enter your PCM marks here.

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[00:05:43]The number of angular and the radial nodes in the 3p orbital is so angular nodes formula is simply L. So we have SP 01. So uh angular for P the L value is one. So I have angular node is 1 and radial node is n - l - 1. So it is 3 - 1 - 1. So that will give me 1. So 1 and 1 option number b here. Next, what is the maximum number of electrons in the in all the subsells for n + l is equal to 4. Okay. So now if I see for n + l, you have to calculate the maximum number of electrons in all the possible subsells.

[00:06:19]Okay. Now if I see n is equal to 4.

[00:06:22]Okay. then L has to be zero and that will give me four s orbital where I can have maximum two number of electrons.

[00:06:29]Now if I have three here then I will have 3 + 1 that is like three and here it should be one. So that is 3 p. Now 3 p you know can hold six electrons. Now any other configuration I have spd 22 2d is not there. So only these two configurations total eight electrons.

[00:06:47]Next, what will be the mass of a particle if the uncertaintity in its position is this much and the uncertaintity in the velocity is this much. So, we have this formula delta x into delta p is equal to h by 4 pi.

[00:07:01]Right? Now, question is asking regarding mass. Now, we can just modify this equation into delta x into m into sorry yeah m into delta v is equal to h by 4 pi. Now you have to calculate the mass value that is h by 4 pi into delta x into delta v. Okay. Now what is the value of x here? H here 6.63 into 10 ^ of - 34. Okay. Again I'll write jul as kz m² second minus square and second.

[00:07:34]Right? What is the unit of h value? It is jou second. So jewel is this much.

[00:07:38]I'm writing this one so that we can understand how we are canceling the units. Okay. Next we have 4 into pi and then we have delta x value is 10 ^ of minus 8 and delta v is 5.26 into 10 ^ of - 255. Okay. Now this one will give me first of all - 33. Now when it goes up it becomes + 33. And now if I say 6.63 whole divided by uh 4 into 3.14 into 5.26.

[00:08:12]Okay. So this is - 33. This goes up it becomes + 33 and here I have 10 ^ of minus1. Okay. Now see 4 3 is somewhere around 12. 12 5 is 60 and this one is 66. So almost it will be like 1 something you will get here. So if I just calculate this one. So I'll write 4 are 12 12. So I'll just write down here 4 3.14 into 4. Okay. So how much I'll get 4 are 16 4 1 are 4 + 1 5 4 3's are 12. So 12.56 into I have 526 right 12 56 into 526.

[00:08:49]So now I have 6 are 36 6 are 30 + 3 33 6 2 are 12 + 3 15 6 1 are 6 + 1 7 You have calculator in the exam so you don't have to worry about this uh you will have two here then I will have 5 6 are 30 5 are 25 + 3 28 5 2's are 10 11 12 5 1 are 5 + 1 6 now what is there 7 5 3 6 2 5 1 2 and 6 2 8 0 Okay So I will have uh 7 + 5 12 + 8 20. Then I have 6 and 6. Right?

[00:09:26]Now how many decimal places I have here?

[00:09:28]Two decimal places and 2. So total I have 1 2 3 4. Okay. So almost you are getting uh this one as 66 sorry 6.6 divided by 66. Okay. And you have one more 10 ^ of -1 here. So if I just cancel this one will be.1 and then 10 ^ of -1 will be 01. So option number A here. Next the number of radial nodes and angular nodes for d orbital can be represented as so for d orbital the number of radial nodes will be n minus for d the l value you know for s it is zero for p it is 1 and for d it is 2. So n minus lus1 so that is nus 3 radial node and angular node will be two here.

[00:10:12]Okay. And the total nodes will definitely be n minus sorry uh yes n minus 1 that is the total number of nodes. So let's see here n minus2 radial nodes no nus1 radial nodes no nus3 radial nodes two angular nodes and n minus lus1 total nodes no nus3 radial nodes two angular nodes n minus one uh this is one actually okay n minus one total nodes. So that is option number d here. Next what will be the energy of a photon which corresponds to this one.

[00:10:42]This is simple question. Energy formula is HC by lambda. So just substitute here 6.63 into 10 ^ of - 34. Okay. And uh it is also if I write kg m²ared second squared and second and we have 3 into 10 ^ of 8 m/ second divided by 0.5 into 10 ^ of - 10 m.

[00:11:11]Okay. Now here this meter and this meter will get cancel. Okay. And we have here uh I want to calculate what energy.

[00:11:22]Okay. So just we'll calculate this value. So I have here three. So I'll multiply 3 3's are 9. 3 6 are 18. 3 18 + 1 19. So 19.89 first of all I have got here. Then I have this 0.5 in the denominator. And what are the power of 10? I have - 34 is there - 34 this goes up so this is 18 so 18 will be somewhere around I think uh 1 second so I have 34 and 18 right so that will be -6 - 16 somewhere I'll get okay now uh I think this should be the answer more nearly but we'll check here so if I see 19.89 89 I'll just assume it as 20.

[00:12:09]Okay. And I have I can write it like this. So I have 5 4 into 10 ^ of -6. Now if I just make it scientific notation. So it is 4 into 10 ^ of -5. So option number A here. Which of the following statements is incorrect? Okay. The shape of an atomic orbital depends on the asomical quantum number. That is correct. Orientation depends on the magnetic number. Energy of the electron in an atomic orbital of multi-electron depends on the principal quantum number. No, the energy depends on n plus l means principle plus isomeal. So this is the incorrect statement. The probability distribution curve of 2s electron looks like this.

[00:12:47]Select the correct radial distribution curve. So in 2s you know there should be only one node right sorry n minus one.

[00:12:54]So n value is one value. So only one node should be there. So here there is no node. I can eliminate first option number b. Here also you get two nodes.

[00:13:03]So this is also incorrect. Okay. Now if I compare these two here it is starting from zero but here it is starting from a number. So we all know for 2S the value has to be option number A. You can just read this graph from the NCERT. You cannot uh define any logic here because you haven't studied that concept. It is not in your like it is not according to the class 11 syllabus. So just prepare these diagrams for 2s 2p and all. Okay.

[00:13:26]Assuming red constant are equal the ground state energy of the electron in the hydrogen is equal to. So ground state energy of the electron is equal to -3.6 into z² by n². So hydrogen is z² and this is also n². So we'll have -3.6.

[00:13:40]Now if I have the ground state energy of the electron in helium plus so it is -3.6.

[00:13:46]Okay. Z square hydrogen helium. So this is 2² and ground state is it is 1. So this number and this number is not matching. So we can eliminate first excited state energy of the helium atom.

[00:13:57]So we have minus 13.6 six and hydrogen helium 2 squared and first excited state is also second orbital so 2 square so this one will get cancel so option number B here next we have a body of mass 10 mg is moving with a velocity of 100 m/s the wavelength of the drogley wavelength associated with it will be okay so lambda is equal to h by mv 6.63 63 into 10 ^ of -34. Okay. Uh we have to calculate m. What is the mass given? 10 mg. So I'll convert it into 10 ^ of - 6 kg. Okay. And velocity is 100 m. Okay.

[00:14:37]Just 100. So now if I solve this is 10 ^ of 3, right? Then it will minus 3 goes up becomes + 3. So 6.63 into 10 ^ of minus uh 31 it will become because I have three here. So that is option number C. Next for different sets of quantum numbers four electrons are given below the order. So you know n plus l value you have to take. So this two addition is four. This two addition is four. This two addition is five and this one is three. So now E3 will have the highest value. Right? So there's only one option. So option number C here. Now when the N plus L value is same the one which N uh the one where the N value is higher that will have higher energy. So if I compare these two, E1 will have higher energy because its N value is higher. Next, which of the one of the following sets of quantum number represents the highest level? So again, you have to add N plus L. So this is four here. This is four here. This is five and this is three. Okay. So option number C. Next, in an atom, an electron is moving with a speed of 600 m/s with an accuracy of 0 5%. Certainly with which position of the electron can be located? Okay. [snorts] Okay. Now if I see we have to calculate the delta x value. Okay. So we have delta x into m sorry into m into delta v is equal to h by 4 pi. Okay. Now I have to calculate delta v value. What is delta v? It is 600 m/s. But the accuracy is 0.5%.

[00:16:15]Right? So I can just cancel this one. So it is 6 5 are 30 and 1 2 3. Okay. So 03 here. Uh so that is what that is the value of delta v and I have to calculate delta x here. So delta x is equal to h by 4 pi m into delta v which is equal to 6.6 into 10 ^ of - 34. Okay. 4 into 3.14.

[00:16:45]mass given is how much electron mass is 9.1 into 10 ^ of -31 okay and this one will be 03 so if you see 4 3's are 12 12 9 is somewhere around I'll just make it a rough calculation so it will become uh 31 and 3 2.2 2 right this is 10 ^ of -2 -2 this is 10 ^ of - 31 so -2 will become - 33 now this goes up it will become + 33 so I'll have 10 ^ of -1 first of all okay next if you see 4 into 3 is uh okay I'll just multiply this is not that big number I can multiply here so I'll get 4 4's are 16 4 1's are 4 + 1 5 4 3's are 12 and then you have 12.56 into 91. Okay. So 6 5 2 1 9 6 are 54 5 9 5 are 45 + 5 50 9 are 18 + 5 23 9 are 9 10 11. Okay. So 6 9 2 4 1 1. So I have 12.56 that means two decimal and here three decimal. So around 114 something I'll get right. So I have 1 second. Yeah. And I have 2.2 2 divided by 140. Okay, then I can just cancel one more time. So I'll get 1.1 and 2 5 are 10 57 something. So if I just uh this one will be somewhere on 11 divided by 570.

[00:18:25]Okay, then I'll get this one. I'll put one more decimal. Then I'll get this one. And I think 2 0 0 2 7 are 14 2 5 are 10 plus that that is 01 okay so I'll get here see 1 into 10 ^ of - 1 so that will be somewhere around 10 ^ of minus 3 but I have this option incorrect this is 10 ^ of - 3 this is 10 ^ of - 3 but my answer should start with one here so definitely option number C here next the number of electrons protons and neutrons in the species are 18 16 and 16 respectively the appropriate symbol for the element is. So if you see the two extra electrons are there that means 2 minus charge should be there. Okay, we can eliminate this one. The atomic the proton number is written here. So that is why option number A here. An ion with mass number 56 contains three units of positive charge and 30.4% more neutrons than electrons. The ion is this is I'll say a bit tricky question okay or a difficult question. So let us assume that the number of electrons is equal to X. Then what is the number of neutrons here? If you see an ion with the mass number of 56 contains three units of positive charge. Okay. And 30.4% more neutrons than the electrons. So what is the number of neutrons here?

[00:19:46]30.4% more than the number of electrons. So that gives me around this will be 130.4 4 uh by 100. So I'll get 1.304x that is the number of neutrons. Now what is there? Uh now if you know number of the actual number of I can write actual number of electrons will be how much?

[00:20:13]Now see because three units of positive charge is there means three units actual number of electrons is x + 3 and actual number of protons will be same in the case of a neutral atom. So this is the number of protons. Now the atomic number is 56. That means the total number of neutrons plus protons is 56. So I'll have 1.304 of x plus x + 3 is equal to 56. Now if I just uh this is 2.304x and this side it goes it becomes 53. So x will be equal to 53 divided by 2.304.

[00:20:48]Okay. So if you see this if you divide this one you will get around uh I think 28 some 26 something we'll get because you can see 53 right 2.3 times right so 23 if I see 2's are 2 3's are 6 2's are 4 sorry not like this one second so you have to see now that suppose 53 and 2.3 I'm dividing okay so 53 divided by 2.3 means It is 530 divided by 23. So [snorts] roughly I'll get how much? So if I see 23 1 is 23. Can I take 2 3?

[00:21:28]Okay, I can go for two also. So I'll just divide here.

[00:21:33]Yeah, 23 is 46. So 13 - 6 will be uh sorry 13. Uh so I have this I have taken 23 into 2. So it is 46.

[00:21:44]Okay. So I'll get seven here and then I'll have 70. So I'll have 23 into 4 I and 3 3's are 9. 3 2's are 6. So three here. 3 3's are 9.

[00:21:57]Okay. Uh 69. So 1 0. Then I'll have one more zero. I'll have one more zero. Then 4 3's are 12 4 23.04 something I'll get right. So x value I got 23. Okay. Now x is what? The number of electrons. But what is the actual number of electrons?

[00:22:16]The actual number of electrons is x + 3 means that is 26. So answer will be option number b here. Next energy dissociated with the first orbit of H+ is associated with the first orbit of H+. So we have -2.18 into 10 ^ of -8. Okay. Uh then we have Z² by N². Okay. So Z square will be 4.

[00:22:39]Hydrogen atom number is 2. So 2² divided by first orbit. So it is 1 square. So we have 4 are 32 7 4 2 are 12 12.72 sorry 4 2's are 8 8.72 into 10 ^ of -8 - 8.72 into 10 ^ of -8 an orbital with n is equal to 3 l is equal to 1 is designated as so 3 sp 01 sp so option number c here next a 150 watt bulb emits light of the wavelength of 66,000 anstrom and only 8% of the energy is uh emitted as light. So if you see 150 what bulb means it emits 150 jw per second.

[00:23:20]Okay. But only 8% of that is the light energy. So 8% of 150 will be how much?

[00:23:26]Uh so 15 8 is 8 5 are 40. 8 2 are 16 + 4 20 8 5 are 40. 8 1 are 8 + 4 12 sorry 120. 120 divided by 10. So that means 12 jew. Okay. So only 12 jewles of energy is there in the uh in our uh this bulb.

[00:23:46]Okay. Now how many photons are emitted?

[00:23:49]So we know that E is equal to N of H C by lambda. So you have to calculate this N value. Okay. So if I want to calculate the value of N. It will be what? E into lambda divided by H into C. H is E is 12 J. Lambda is uh Okay. So first what we basically doing is we are calculating the energy of one photon and then you have to divide it by 12 to get the number of photons. Okay let's do it in that way that will be more better to understand but yeah calculation will be a one more step. So let's first calculate what is the energy of one photon here. So I have 6.6 6 into 10 ^ of - 34 okay joule second and then I have 3 into 10 ^ of 8 m second and then the denominator we have 6 into 10 ^ of 3 angstrom is there so that will be 10 ^ of - 10 so now if I solve this one is a very simple 1.1 so I will get 3.3 into 10 ^ of this goes up right 18 it will become now 34 and I have 18 so I will have here 16 - 16 this is the energy of 1. I think I have done the calculation correctly. Uh no one second 3 and 10 this is + 3 - 10 this is - 7. This goes up. It is 15 actually. So this one will become 17 I think. I'll just check once again. So in the denominator we have - 7 this goes up. 7 + 8 becomes 15. So I have to subtract 34 and 15. So uh 14 - 5 is 9. So - 19. See I am a bit slow in calculation but yeah for comet K we don't have to worry because we have a savior that is our calculator okay so this is the energy of one photon okay now what I can do 12 / 3.3 into 10 ^ of -9 so you can see almost 3 4 is 12 so 4 into 10 ^ of 19 next we have in the following sets of ions which one is not iso electronic with the rest of the species. So, sodium has 11 electrons.

[00:25:54]All other has 10. So, that is why option number D here. Next, the outer electronic configuration of gadolium is 4F7 5D1 6S2. Next, according to the law of photochemical equivalence, the energy absorbed is given. Okay. So, energy absorbed formula is HC by lambda into N A. Okay. So, HC value is given 6.62 into 10 ^ of - 27. Okay. C is 3 into 10 ^ of 10 cm/s and we have 6.02 into 10 ^ of 23. Okay.

[00:26:28]So 33 27 + 6 is 33 I think. Yeah. So I have 10 ^ of 6 into I have 6 are 36 36 into 3 108.

[00:26:46]Okay. and I have um so I have taken uh here so you'll have decimal places so I have taken um six six so I have to take at least one decimal place right so we'll get the correct answer here so you can see that if you just put the all decimals and everything okay you'll get as 1.196 into 10 ^ of 8 by lambda is the correct answer so please just check this calculation you will have calculator so it won't take much time so I'm not solving it but just multiply this one you'll get the answer. Now moving to the next question. Which of the following electron transitions will have the largest amount of energy? Now remember whenever the gap is very less the energy is higher. So if you see here the gap is 3 to1 means if the at the orbitals are closer to the nucleus then there will be more energy. So n equal to 1 to2 will have more energy. So the transition will also require more energy. So option number b here. Next an electron having a spin quantum number of s is equal to minus 1 by2 and magnetic quantum number of m is equal to +3 can be represented by so +3 comes only in f orbital option number c here next the atomic number of the element with the highest ionization energy among the following is once again the atomic number of the element with the highest energy ionization energy among the following is so if you see here these are all elements so zed is 16 and 16 is sulfur Right. So oxygen below we have sulfur 13 boron aluminium and then 14 is your 14 is which element? Silicon. Right. So so this is 13 and this is 16. So I'll just write it a bit spacious here.

[00:28:29]Yeah. So this is option number A. Then sorry option number A is this one. That is 16. Sulfur is 16. Option number A.

[00:28:38]Then I have B14. So that is your nit not nitrogen that is your silicon.

[00:28:45]Okay, this is option number B. Then we have 13. This is option number C which is aluminium. And then we have 15 that is phosphorus.

[00:28:55]Okay, so now we all know that phosphorus will have an 3p3 configuration or you can say the p orbital is exactly half filled. So it will have more ionization energies. Option number D here. Okay. So students these are the most important PYQs from the ST chapter structure of atom. So just remember the basic formulas and you will be done with this chapter. Okay. So stay tuned, subscribe the channel and uh just for the next chapters and do not forget to check the link of the rank predictor given in the description section.