This video explains key concepts in quantum mechanics including the Pauli Exclusion Principle (no two electrons can have identical quantum numbers), exchange energy (stabilization from electrons with same spin in different orbitals), and how to determine quantum numbers for specific electrons in atoms like oxygen and chlorine, with practical examples of electron configurations for elements like manganese.
Deep Dive
Prerequisite Knowledge
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Deep Dive
11th Chemistry | Chapter 2 Book Back Part 2 | TN State Board | Prakash Sir
Added:Good morning warriors.
Good morning.
Good morning.
Good morning. Good morning.
Good morning.
Holy look. Okay. Okay. Good. Good. Good.
Good morning. Good morning. Yes.
Chapter two.
Chapter two.
So now we will just continue it. So chapter 2 and three.
Just open the description description Library one course all in one library.
So you will get a course standard state will be become zero.
You can just get it.
Go to my morning 11 skateboard.
Join class.
So you can get it.
You can get it bookmark part two part one. Okay.
Further right. So now we'll move on to the question answer.
Any idea?
The half fil.
Main point.
Okay.
The stabilization of half field is more pronounced than the half field of d orbital. Half field of d orbital is more pronounced. D orital half stable short orbital. So suborbital px py pz px p y p for half because it is having five or three orbital. First reason it is having more exchange energy.
Exchange energy.
Exchange energy.
Exchange energy.
Exchange energy.
Symmetry.
symmetry exchange.
Exchange.
It is having four exchange places.
Exchange energy is high. Exchange energy. High exchange energy.
The symmetry is high.
5d orital 5d is more stable. 5d is more stable.
First symmetry.
The half filled orbitals are more symmetrical than partially filled orital. And this symmetry leads to greater stability.
Partly filable.
The electron with the same spin in the different orbital. The electron in the same spin with different orbitals.
Electron in the same spin in different orbitals.
Same spin.
same spin.
The electron with the same spin in different orbitals orital electrons with same orbital of same subshell. Same subd each such exchange releases energy and this is known as exchange energy.
Greater the number of exchanges, greater the number of exchange possibility clear. Is that clear?
I will explain it. Okay. Consider the following electronic arrangement for D5 configuration. Right? So D5 configuration D5 D5 D5 D5 D5 electron So mother rule Hans rule Hans rule exclusion principle poly pairing will not takes place.
Pairing will not takes place until each orbit have at least one electron.
One 2 3 4 Five.
This is wrong.
Huh?
D7.
D7.
increasing order.
Increasing lower energy lower energy dal D7 D4 D4 D4 D4. Any idea what will be D4? D4 D4 D4 D4 option A. Option B. Option C.
[snorts] First one is wrong. Second only last.
So C is the right answer. C is the right answer. Very good. C. Right. Right.
Okay.
Polyexclusion principles right exclusion principle one of the most important question most important question exclusion principle just state polyexclusion principle state polyexclusion principle Very simple.
Polyexclusion principle states that no two electrons electron. No two electrons in an atom can have a same set of value of all four quantum numbers. All four quantum numbers.
Right.
Right. Quantum numbers.
What are the quantum numbers we have ML MS?
Okay.
Correct. Right. Super LS.
So principal quantum number quantum number magnetic quantum number spin only magnetic quantum number two electrons for example helium number atomic number of helium Quantum number is that equal to two configuration 1. First elect.
elect first elect formula L= L= 0 0 to NUS 0= 0= - L 2 0 2 + L + 1 by 2 or - 1 by 2.
Okay.
Right. Second electronig = 1 = 0 = 0 m= - 1 by 2 it is - 1x 1x 2 mus 1x2 mus 1x2.
So number same set of four quantum numbers value.
No two electrons in an atom can have a same set of value of all four quantum numbers. Example hydrogen 1= + 1 by= 1x2 spin direction there is no specification you will do it in bcitative US.
Okay.
Right.
So definition.
So define orital n values of 3 px and 4 dx^ 2us y^2 probability of finding electron is maximum Right.
S orital. P orital. Very good. Very good. S orbital. P orbital. D orbital. F orbital. SPDF.
SPDF. S orbital. P orbital. D orbital. F orbital.
SP DF. S orbital. P orbital. D orbital.
F orbital. Right.
N values. N value for 3 PX and four D. 3 PX 3 PX.
value.
L values values.
L value. S P orital 1. D orital 2. F 1 3 3 P X N = 3 L = 1 5 DX² - Y^2 4 5 4 DX 2 - Y^2 4 D X² - Y^2 Go ahead.
D= 2= 2.
Okay.
value possibility to sir = = 3 0 1 2 0 1 - 3 - 0 1 right 3 px 3 px So that's it. 3x 4 dx²us y^2.
Next question.
Determine the values of all four quantum number.
chapter generation problem.
No worries about it. So eighth electron of oxygen electron of oxygen right of oxygen atomic number. What is the atomic number of oxygen? Oxygen is equal to 8yal.
Okay. Is that equal to 8 configuration?
electronic configuration.
elect 1 S 2 S 3 S 2 P 3 P 4 P first Right.
First, second shell. So first of how will you find the maximum number of electrons in a shell max?
What is the formula for that? 2 n² for n= 1 2 n = 1 maximum highest two electrons.
N= Second 2² is equal to 2 into 4 = 8 electrons.
Second electron Second max P orital orbital 1 s2 2 s2 2 p 4th quantum number four quantum number for e electron of oxygen oxygen e electron 2p Right.
So 1 2 1 X 2 another right super First electron 1 elect 2 p x 2 p y 2p z is this correct Absolutely not correct. Absolutely it is not correct.
We should put one electron in each over orbit.
for right electron.
electron green electron right 2px 2 px electron clear 2px 2 px Four quantum numbers = 2 P value -1 0 + value I'm writing it asus one sorry 2 these three are in same energy level.
Same Energy level= mal.
First, second - 1 by 2. So - 1 by 2 - 1 by 2.
Okay. Right.
Clear. Clear. Is this clear? Have you got it?
Oxygen electron number.
quantum numbers.
15th elect 17th elect write down chlorine atomic number is that equal to 17 electron configuration 1 s2 2 s2 2 p6 3 s2 3 p Right.
Practice.
Clear.
for 35 35 35 33 and 34 33 problem.
Right.
15th elect first electron, second, third, fourth, fifth electron.
for 2 P N = 1 L = 1 M = M = + 1 by 2 M = + 1 by 2 M= + 1 by 2 Clear right 3 PZ uh 3 PZ sorry 3 P 3 P is 3 PZ 3 PZ 3 PZ 3.
Super. Good. Good. Good. Good. You guys are awesome, man. Where level? Super.
3 P is at 3 P. So, yeah. 3 P is it?
Okay. Clear. H.
Right. Super.
For each of the following, give the suble designation the allowable M values.
First let us consider it as -2 -1 0 + 1 + 2 + 1 + N= 5 L= 3 N= 5 L= 3 L= 3 M= - 3 - 2 -1 0 + 1 + 2 + 3 L= 0 m= 0 L= 0 L= 0 right day day. The most most most important UPM question.
One of my favorite question manganese atomic number is that equal to 25 it is going to be give you a different approachal 25.
Okay. Right.
So 1 s 2 elect 2 s 2 electron 2 6 electron 3s 2 electron 3 p 6 electron 36 3p 4s electron confusion in case we have to use N + 3D D + D value 0 1 2 5 4 S value 4 + 0 = Four.
3D5. Right. 3D5. Okay. 3D5.
Huh? 3D. Okay.
They have asked MN2 plus MN2 plus MN2 plus. What is the meaning of MN2 plus?
MN2 plus M2.
MN2.
What is the meaning of MN2 plus?
MN2 plus loss of two electrons outermost.
outer mostn 2 s2 2 p6 3 s2 3p6 3d5 Foreigne.
Yes. Yes.
Good morning.
Okay. So, tomorrow morning we will see the chromium.
Okay. Right. Super.
Right. Bye-bye. See you.
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