 Hello, guys, welcome to the session. As you know, we are discussing problems on atomic structure. This is the second session of problem solving atomic structure. So we were solving some questions that has been asked in previous exam. It helps you in both KVVY or J or need exams. So those past year problems we are solving, actually. So let us start with the session. Let us discuss this question. The electrons identified by the quantum number NNL, these values are given, and can be placed in the order of increasing energy from lowest to highest. Easy one. You can easily compare the N plus L rule, compare with N plus L rule, which we also call it as off-bow rule, N plus L rule. According to this rule, what we say that the value of N plus L, the energy, is directly proportional to N plus L value. If N plus L values are equal, then we compare N value. More value of N, more will be the energy. So let us discuss this. The first one, N plus L value for the first one is 5. For the second one, N plus L value is 4. Third one, N plus L value is 5. Again, fourth one, N plus L value is 4. So obviously, second and fourth has less energy than first and third. So first and third, if you compare the N value, third has the minimum N value, less N value than 1. So the third one has maximum energy, sorry. The third one has minimum energy. Then we have first one. And then if you compare, it is other way, the energy of 1 and third is more than to that of second and fourth. 1 and third, the energy of first is more. So first one has maximum. Then we have third, second and fourth N value, if you compare. N value is lesser for the fourth one. So we have second here. And then in the last, we have four. So this is the decreasing order of energy. So let's check the option. First has maximum. Then third, then second, and then fourth option A is described here. OK, now question number 27. The number of nodal planes for XY. PXY has only one nodal plane, that is YZ plane. OK, has only one nodal plane. That is YZ plane. And the answer for this one is option one, that is only one. Perpendicular to the plane of X axis, that is YZ plane. OK, answer is option A. The next one you see. The wavelength associated with a golf ball weighing this much plan, moving at a speed of this, 5 meter per hour, is of the order. OK, so how do we do this? We know wavelength lambda associated with any moving object of mass m is h divided by mv. OK, so h is the plane constant. The value is 6.626 into 10 to the power minus 34 divided by m is 200 gram, the mass that we have. We don't have to change this. 200 gram, sorry, it is in gram. So we have to convert this into kg, the SI unit, because it is in joule, 10 to the power minus 3 kg into, velocity is 5 meter. We have to convert this r into second, so divided by 3600. OK, just to solve this, you'll get the answer. So this is you have to find out in order, so we can do this. OK, suppose we have 6.6 into 3600 into 10 to the power minus 34 divided by 1000 into 10 to the power minus 3. So if you solve this, this will cancel out. And this would be, so we have 1000 into 10 to the power minus 3. This is the calculation mistake. So 5 into 200 is 1000, 1000 into 10 to the power minus 3. And that's right, 1000 into 10 to the power minus 3. So we need to solve this. So we get it here. This will cancel out. And we get 66 into 360 into 10 to the power minus 30. OK, so if you multiply this, what we get is, I'll do this just a second. So we have 66 into 3600, 2376. So we are getting it as 23760 into 10 to the power minus 30. OK, so fine. So we'll write down point here. 40 will come down into the power minus 30. So answer would be 10 to the power minus 30 option C. In this order, we get this. OK, now the next question is, quantum numbers plus half and minus half of the electron spin represent. The rotation of electron in clockwise and anticlockwise direction respectively, rotation of electron in anticlockwise and clockwise direction respectively, magnetic moment of the electron, the electron pointing up and down respectively to quantum mechanical spin state, which have no classical analog. To quantum mechanical spin state, which has no classical analog. So plus half and minus half, we usually say that it is the orientation of electron, about its own axis. If it is said to be clockwise orientation, it is plus half. If it is said to be minus half, then it is anticlockwise orientation. But actually, it represents the magnetic moment. So it is a component of a spin angular momentum. A spin angular momentum. And if it is plus half, then we see the component of this spin angular momentum is along the, along in the direction of magnetic field. In the direction of magnetic field and minus half, it is opposite to the direction of magnetic field. So plus half and minus half, it represents the magnetic moment of electron pointing up and down respectively. Usually for easier way, we say as when the electron enters into the orbital, it can have clockwise or anticlockwise orientation. And that is plus half and minus half. But this actually represents plus half and minus half. This actually represents the spin magnetic moment. Or simply, we can see the magnetic moment when it is along the magnetic field and opposite to the magnetic field. So when it is along the direction of the magnetic field, then it is plus half. And when it is anticlockwise, it is minus half. So answer for this question is option C, magnetic moment of electron pointing up and down respectively. OK, now the next question you see. If nitrogen atom has electronic configuration 1S7, it would have energy lower than that of normal ground state, that is 1S2, 2S2, 2P3. Because the electrons would be closer to the nucleus, yet 1S7 is not observed because it violates Heisenberg uncertainty principle, Wundsruhe, Unstruhe, poly-exclusion principle, and Bohr's postulate of stationary orbits, OK? So you see, Heisenberg has nothing to do with the electronic configuration. It deals with the uncertainty in the position of electrons, correct? Unstruhe, it says that electrons occupy in the orbital, occupies the orbital one by one, right? And then the pairing takes place, right? So pairing is not possible unless all the orbitals belongs to the same sub-chill is singly occupied. That's Unstruhe. Poly-exclusion principle is no two electron could have the same set of quantum numbers, right? But 1S, if it has more than two electrons, then obviously, two electron is this. It's a hypothetical case. Two electron is this. If you have supposed third one, third one will have either clockwise or anti-clockwise orientation. This is the possibility we have. So it means the two electron in the same shell has the same set of quantum numbers. And hence, it is not possible. That's why this particular question, it violates the poly-exclusion principle, right? OK, 33rd. The radius of which of the following orbit is same as that of first Bohr's orbit of hydrogen atom, OK? Radius of which of the following orbit is same, easy one. OK, so what is the first Bohr's orbit of hydrogen atom? We know the formula of radius. The radius formula is in nth orbit. Rn is equals to 0.529n square by z. Hydrogen is the first Bohr's orbit means 529 into 1 square by 1 only, right? Same as the first Bohr's orbit, fine? So if you talk about helium, so Rhe plus is 0.529 into 2 square by 2. Helium atomic number is 2. Obviously, it is not equal. So we need to just compare the n square by z value for all this. OK, if it is equal to 1, you see? The value of this for hydrogen is 1. So for which of these ions, it is 1, OK? Lithium. Lithium n square is 2 square by 3, not equal. Li 2 plus for 3 y. So it is 3 square by 3, not possible. Beryllium 3 plus, right? So it is 2 square by 4, the atomic number of beryllium. Hence, this value is 1. Answer is option 15 for this one. Easy one, just you need to compare the n square by z value. If it is coming out to be equal for hydrogen, that is 1. It is the same orbit as the first orbit of hydrogen has. OK, fine. The next one. The number of radial nodes of 3s and 2p orbitals are respectively easy. See, radial node is equals to n minus l minus 1. So for 3s, it is 3 minus 0 minus 1 equals to 2. For 2p, it is 2 minus 1 minus 1, and that is 0. 2 and 0, the answer is option a. The kinetic energy of an electron in the second bore orbit of a hydrogen atom, OK? So kinetic energy is what? Kinetic energy is half mv square of mv square. And the value of v is in terms of h we need to find out, OK? So if you do not memorize this, you can easily derive this. How do we do this? We know the angular momentum mvr is the integral multiple of h by 2pi. We need to find out mv square, half mv square, right? So what is v from this? It is nh by 2pi into 1 by mr, right? This v will substitute here. So it becomes half mv square is n square by h square by 4pi square m square r square, OK? This square and m will get cancelled. And we'll get here n square h square by 8pi square m r square, OK? Kinetic energy of an electron, a0 is the bore radius. So a0 is equals to what we know the bore radius is for 1, 1 square by 1. This is a0. So r square. So how do we write down r in terms of a? a0 is this. So what we can write r, radius r is equals to, we can write down a0 into any orbit n square by z1. So r we can substitute as a0 square n square. So we'll substitute r here. It becomes n square h square divided by 8pi square m. r square is a0 square n to the power 4. So this square and this 4 will get cancelled. We'll get n to the power 2. OK, so this when you solve, you'll get h square by 8pi square m a0 square n square. h square 8pi square m, is there any mistake? r is a0 n square, no. r square is n square, 4pi square, this is fine. Second bore orbit is given. That's what I was thinking. Second bore orbit is given. So we'll put here 4 square into this. Second words, 2 square into 4. So we'll get h square by 2 square 4 into 8 is 32, pi square m a0 square. Answer is option c. The energy of an electron in the first bore orbit of hydrogen atom is this. So it is given in the first bore orbit. So e1 is given. It is minus 13.6 electron volt. The possible energy values for the excited state of electrons involves orbit of the hydrogen ion. So energy we know it is e is equals to minus 13.6 z square by n square. So this is the formula we have in electron volt. Now, it is for hydrogen atom. So z we don't have to think about because it is 1 only. n can be 1, 2, 3, anything. So if it is 2, if n is 2, we have to check this one by one. If n is 2, then energy for the second orbit is minus 13.6 divided by 4. So we are getting here 2 or 3.4 minus 3.4 electron volt. This is possible. If n is equals to 3, then e3 is equals to minus 13.6 by 9. That is 1.946, 1.5 electron, minus 1.5 electron. What do you see? D is also correct. So minus 3.4, then minus 1.5. Then it means minus 0.4 in between these two it is not possible. If you talk about n is equals to 4, then e4 is equals to, we are getting minus 0. something because 13.6 by 16 would be 0. something which is not the case here. So option would be A and D both possible into this one. It's a multiple correct question. The answer is A and D from this one. Which of the following statements is our correct? This is also multiple correct question. The electronic configuration of chromium is 4S135. This one is correct. The magnetic quantum number may have a negative value. Yes, that is also possible minus L2 plus L. It is there. Silver atom 23 electrons have a spin of one type and 24th of opposite type. The atomic number of AG is 47. So if silver atom, silver is AG. So it is 1S2, OK. 23 electrons have a spin of one type. OK, it means one type we have. So 1, 2, 3. So 23 will have opposite sign. If you see the electronic configuration of copper, so copper zinc we have and here we have AG. If you look at the position of it, right. So copper is, copper electronic configuration is 4S13D10. Hence silver would be, it is 5S1 and 4D10. 5S1 and 4D10. It means if 4D is filled, means 4S is also filled completely. And 3D, 3P3S also filled completely. So the configuration is 1S2, 2S2, 2P6, 3S2, 3P6, 4S2, 3D10, 4P6, 5S, 4P6, 4D, 5S1, and 4D10, OK. So this is the electronic configuration. So it is 6 plus 410 plus 10, 20 plus 10, 30, 40, 47. This one is also correct, 47. All these you see, 20 plus 10, 30. So out of 30, 5 electrons will have the same configuration, and here we have 10, 10 plus 5, 1, 2, 3 plus 2, 5 plus 1, 6, 6 plus 3, 9, 9 plus 1, 10, 10 plus 5, 15, 15 plus 3, 18, 18 plus 5, 23. 23 electron will have the same configuration and same spin. Hence this is the answer. Oxidation is state of nitrogen. Nitrogen is minus 1 by 3 and plus 1, right. Minus 1 by 3, oxidation is state of nitrogen we have. So I think option D is not correct. Oxid, this is plus 1, plus 1, and nitrogen should be minus 1 by 3, so that overall it is OK. So answer is A, D, and C for this question. Next question you see. Decrease in atomic number is observed during what? OK. So this one is information based question. You should know the relation here. What happens in alpha emission? What happens in beta emission, et cetera? Suppose you have an atom. I'm taking XA mass number and atomic number. If it goes under alpha emission, right. Minus alpha means alpha emission. Alpha means what? Helium particle. So helium particles comes out. We have HE 4 and 2. Hence the left here is A minus 4 and Z minus 2. Decrease in atomic number, right. You see the atomic number is decreased. It was Z initially, now it becomes Z minus 2. If this goes under, if this goes under beta emission, then beta emission is electrons comes out from this, right. So we'll get here XA, electrons comes out. So E0 and minus 1, the charge we have here, means it is Z plus 1 to balance the equation. So here we have increase in atomic number. So I think D is wrong. So far this one is correct. Let's talk about the other one that is positron emission, right. In positron emission, means one positron, one positive charge comes out, so plus 1. So that would be simple 1 XA Z minus 1. It means this is also correct, right. And the last one we have, if we have electron capture, means it takes one electron plus E. It takes one electron. So it becomes XA and it takes one electrons, so charge decreases, right, C minus 1, right. So D is also correct, A, C, D are the correct ones. The ground state electronic configuration of nitrogen atom can be represented by, means outermost electron is talking about, it is understood in this question. N has seven electron, 1S2, 2S2, 2B, 3. So 1S has two electron, 2S has two electron, and 2B has three electron, which is this, according to Hans' rule. Hence, the answer for this question, it could be, this is not possible, because either it is clockwise or anticlockwise, correct. So answer for this would be, this is also correct, and this is also correct, A and D, both are correct. So these are the few questions we discussed, okay, very basic questions. The next session we'll have, we'll take some good questions, you know, some high level questions we'll see, but this kind of questions helps you in KVPY exam a lot. Okay, thank you guys, thank you so much.