Class 11 chap 4 | Chemical Bonding 11 || Molecular Orbital Theory IIT JEE NEET || MOT Part II ||

So now we are talking about how to do electronic configuration in molecular orbital. How to put electron in sigma 1s, sigma 1*1s, sigma 2s. So for that you have to read two rules. First you have to read that if total number of electrons in molecule, in molecule, the total electrons in that molecule, in molecule,

is less than equal to 14, means till 14, we will use 14 also. Then the electronic configuration of the molecule, the order of electronic configuration, now we are following the Bausch principle, sigma 1s, sigma star 1s, sigma 2s, sigma star 2s, pi 2px equal to pi 2py, their energy is equal, sigma 2pz

π*2px = π*2py anti-bonding then σ*2pz So, you must first note this order that we have to follow this order. Pause the video here, note this order or take a screenshot, whatever you want to do. Do it. You have to follow this and you have to learn this thing because there is no n+l rule here like atomic orbital that we have to determine in the exam.

So, you have to remember this thing till when? Till 14 or less than 14 electrons. Sigma 1 is, Sigma star is, Sigma 2 is, Sigma star 2 is, Pi 2 pi x, Pi 2 pi x, Sigma 2 pi x, etc. I will write this on the side somewhere. Then we will start the electron configuration.

So, I will write here on the side, Sigma 1s, Sigma star 1s, Sigma 2s, Sigma star 2s, Pi 2px equal to Pi 2py, Sigma 2pz, Pi star 2px equal to Pi star 2py and Sigma star 2pz. Okay, we will follow this. Let's start. First of all, we will take the molecule H2, Hydrogen.

How many electrons are there in hydrogen? Two electrons. One in hydrogen atom and one in another. Two electrons. In which one will we go? In the first one. How many electrons can go in one orbital? Two. So, sigma 1 is 2. Done. The story of hydrogen is done. Let's take the second one. I took HE2. Let's assume that helium has made its molecule.

Although Helium is a noble gas, let's assume that it is made. So how many electrons will be there in Helium? 4. 2 in 1, 4 in 2. Configuration, Sigma 1s2, Sigma star 1s2. Let's take one more example. Hydrogen plus. What happened? Take two atoms of hydrogen, make a molecule, and take out one electron from it. So what will become? H2 plus. Means how many electrons? 1 electron.

H2+ means, if we take one electron from this, it will be sigma 1s. This will be it. So, this is how you can configure molecules. We will go further. Now, I am going to tell you a very special thing, guys. The special thing is Bond Order. Bond Order. Now, we have to learn a new thing. Bond Order means number of bonds.

The formula for this is number of electrons in bonding minus number of electrons in anti-bonding divided by 2. To find out the number of bonds, we have to divide the number of bonding electrons minus the number of electrons in anti-bonding by 2. For example, if I ask for the bond order, then how many are there in bonding? 2. There is no anti-bonding, so we have 0. Divide by 2, how many? 1. That means, how many bonds are there between H and H? 1 bond.

Here, Bond order, number of electrons in bonding, 2, minus number of in anti-bonding, 2, divided by 2, how many bonds? 0. How many bonds are there in He2? 0. This means this molecule does not exist. It is not a bond. It does not exist. So, this is a good theory. If the bond order of any molecule is 0, then it cannot exist. Or, the smaller the bond order of any molecule, the less stable it is.

That means, the stability of any molecule depends on the bond order. The more the bond order, the more the stability. For example, if we talk about H2+ Bond order. Number of electrons in bonding. 1 Anti-bonding. 0 divided by 2. Bond order is 1/2. Now, I will tell you who is the most stable among these three. So, this can't be explained by the valence bond theory. But, it can be explained. The most stable is the one whose bond order is the most. That is, this. After that, this. After that.

If I ask you to arrange it in stability, then you can do it very easily. You will say the most of hydrogen, then you will say H2+ and then HE. Clear? It will be clear till now. Now I am going to confuse you. Remember, we will make an energy diagram now. Some people are tensed when will that energy diagram come, that long one, that will come. Keep learning. Suppose I give the question of H2-. Now the question of J will come.

It was in lithium, Li2, Li2+ and Li2-. Same question. H2-, how many electrons? H2- means one more electron in the hydrogen molecule, means three electrons. Sigma 1s2, sigma star 1s1. Three electrons will be like this. First two in this, then one in this. Let's talk about the bond order. Number of electrons in bonding, minus number of electrons in antibonding, divided by 2. Is it correct? There will be three electrons in this. Two in this, one in this. Two in bonding, one in antibonding. How much? One by two.

Now I will ask what is the stability order in these three? Now tell me what is their stability order? So it seems that the most stable is this and their stability is equal. It is not like that. Write down the note, if bond order is equal, then stability is checked by seeing which has lesser number of anti-bonding electrons.

If the bond order of two is same, then see in which electron is less in anti-bonding. The bond order of these two is same, half. In which electron is less in anti-bonding? In this. There is no electron in anti-bonding, so this is more stable. When the bond order is equal, then check whose electron is less in anti-bonding. This one does not have electron in anti-bonding, it has, so this is less stable. Because anti-bonding reduces stability.

So, the order of stability will be H2, then H2+ and then H2-. Both of these have the same bond order but it does not have electron in the anti-bonding. Is it clear now? Do you understand how to guess the stability and how to get the bond order? Okay. Now let's talk about the stability. Do you understand? The bond length. The higher the number of bonds, the closer the atoms will be.

Bond length will be smaller. Bond length is inversely proportional to bond order. The more the bond order, the more the number of bonds, the closer the atoms, the smaller the bond length. So, you can ask me to tell the order of bond length. This is the only question that can come up in comparison. Tell the order of bond length.

The order of our bond order will be opposite to that of the other two. For example, the bond length of these two will be the same. Now, don't check the stability here. The bond length of these two is the same and its bond length is the same.

Inversely proportional to bond order. Only bond order. Don't go into stability. Only bond order. So, the bond length of these two is same. And the bond length of this is less than that. Inversely proportional. Whose bond length is less? Sorry, whose bond order is less? Stability is more. Sorry, whose bond order is less? Bond length is more. So, cut, cut. No, let's keep on running. Okay? Okay? So, the more the bond order, the less the bond length. Are all these things clear? Small things. What else comes? What else comes, kids? Magnetic nature. Magnetic.

I have noted how to get the bond order, stability, and bond order. We have to guess the magnetic nature in two ways. One is paramagnetic and the other is diamagnetic. Paramagnetic is the one which is attracted to the magnetic field. Diamagnetic is the one which is repelled in the magnetic field. So paramagnetic is the one which has unpaired electrons.

Unpaired electron means one electron, its magnetic moment is not being cancelled by anyone, so it will show its magnetic property. And the one who has paired electrons, how will our diamagnetic be? It has no difference from the magnetic field, its electrons are paired, it will be a little bit different. This is a little bit of a ripple in the magnetic field, this is a little bit of a crack. So tell me the nature of all three. Here the electrons are paired. Paired electrons means this is diamagnetic.

And they also agree on the experiment. Here, tell me, this is paired and this is unpaired. So, how did this happen? Paramagnetic. Now, the question came in the exam. Which one is paramagnetic among these three? This one. Tell me this one. H2 plus unpaired. There are not two electrons. There is one electron in 1s. Sigma 1s has one electron. So, how did this happen? Unpaired electron. That means, again it is paramagnetic. So, these two are paramagnetic and this is dire.

Where paired electron, there is dia. Where unpaired, there is para. Where paired, there is dia. Where unpaired, there is aaa. Clear? Very good, I think I have told you the theory. Now, do the electronic configuration and the example, your hands will be set. Now, we will tell you everything. Let's move ahead. Let's take the next molecule. H2 is done, helium-2 is done. Now, Li2.

Li2, say it. Li2, 3 electrons in lithium, 6 electrons in Li2. Sigma 1 is 2, Sigma star 1 is 2, Sigma 2 is 2. 6 electrons are formed. Let's talk about bond order. Bond order, number of electrons in bonding minus antibonding divided by 2. 2 in bonding is 4, 2 in antibonding is 2, divided by 2, bond order is 1. That means, if there is a bond between them, then there will be a bond.

Li2+ how many electrons? one out of six went away 5 electrons so Sigma 1s2 Sigma star 1s2 Sigma 2s1 talk about bond order

Say, in bonding 3, in anti-bonding 2, divide by 2, how much happened? 1 by 2. And along with that, talk about Li2 minus, means 7 electrons. Sigma 1 is 2, sigma star 1 is 2, sigma 2 is 2, who is after this? Sigma star 2 is 1. 2, 2, 4, 2, 6, 1, 7. Talk about bond order. 2, 2, 4, 1, 5, minus, anti-bonding.

Sorry, Bond order 2, 2, 4 minus anti-bonding 2, 1, 3 divided by 2, how much is it? 1 by 2. Stability is the most, of this. Then tell, then tell, then tell, then tell, then tell, then tell, then tell, of this. Why? Because there are no electrons in anti-bonding. Then of this, because there is anti-bonding. Bond order is the most according to this. Then equal to these two. But, there are electrons in anti-bonding of this, not in this. So, what will be the stability? Li2, Li2 plus,

Li2 minus. This will be the order. Let's go, brother. Did you understand this too? Let's go. So, lithium is done. Hydrogen is done. Helium is done. Now, you can do beryllium yourself. Try. B2. Beryllium has 4 electrons. This is done. 8 electrons. 8 electrons. 2, 2, 4, 2, 6, 2, 8. So, sigma 1 is 2. Sigma 2.

*1 is 2, Sigma 2 is 2, Sigma*2 is 2 See, the bond order will be 0 because the number of bonds are there, the number of bonds are there. Minus 0, cancel this, cancel this, the bond order is 0. Means do not exist. Do not exist. Next, let's take B2, Boron 2. How many electrons are there? 10 electrons.

Boron 5, so 10 electrons. 10 electrons, 2, 4, 6, 8. Now see how to fill it. Sigma 1 is 2. Sigma star 1 is 2. Sigma 2 is 2. Sigma star 2 is 2. How many are there? 2, 4, 6, 8. Now they will come together. Pi 2px equal to pi 2py. They both open together.

Now, both these have equal energy. How many electrons are left? Two. Eight are left, two are left. So, one will go in this, one in this. If there were eleven, then one in this, twelve, then in that. Don't put two in any one. It is a rule of Hunts. First in this, then in this. Let's talk about the bond order. Tell me the bond order. Tell me the bond order. In bonding, two, two, four, five, one, six. In anti-bonding, two, two, four. Divide by two. Two by two. Bond order one. How much bond is there in between? One.

Magnetic nature, we haven't talked about this for a long time, let's talk about it. So, how are the electrons in this? Unpaired. Unpaired means the moment is not cancelled, single electron, it will be attracted in the magnetic field, means paramagnetic. If you see unpaired electron, it is paramagnetic. Tell me this one, diamagnetic, all electrons are paired. All electrons are paired. It won't exist, but how would it be? Diamagnetic.

Done? Now you show me C2. Pause the video. Do everything by yourself. Bond order, magnetic nature, everything. C2 has 12 electrons. C2 has 12 electrons. 1, 2, 3, 4, 5, 6, 7, 8. Done? Pause. Done? Come on. Carbon 2 has 12 electrons. How will you fill 12? Sigma 1 is 2, Sigma star 1 is 2, Sigma 2 is 2, Sigma star 2 is 2. Oh man, this is very boring. Let's do one thing. Let's remove this whole thing.

Let's remove this whole thing. There are 8 electrons, 2, 2, 4, 2, 6, 2, 8. There is no difference in the bond order. This is getting cancelled by this, this is getting cancelled by this. Brother, is there any difference in the bond order? This is getting cancelled by this, this is getting cancelled by this. So, there must be a difference in the remaining electrons. These must be getting cancelled internally. There is no difference. And there is no point in writing it again and again. So, we will stop writing it further. Pi 2Px = Pi 2Py. Come on, brother. These are the 8 electrons, 12.

So, one in this, nine in this, ten, then one more in this, eleven, then one more in this, twelve. Complete. Let's talk about the bond order. So, it doesn't matter, it's going to be cut. How much bonding? Four. Anti-bonding is zero. Divide by two. How much is there? Two. This will get cancelled. Count it. Two, two, four, two, six, two, eight. Eight bonding is there. Minus. Two, two, four. Anti-bonding. Divide by two. This is what is coming. This can be ignored. When you write long molecules, ignore this. Bond order. Say magnetic nature. Diamagnetic.

Why diamagnetic? Because all electrons are paired. So, we will write its configuration as follows. KK dash. KK means this. KK dash means this. Both of these are filled. After this, pi 2px equal to pi 2py2. Did they make any difference in the bond order? No, they didn't. See, I left them outside. Because this is...

cancel, cancel, how many bonding are left? 4 anti-bonding 0 divided by 2. So now learn to write like this, how many electrons are there? 8 electrons. So, you don't have to write 8 electrons up to sigma*2s. Start directly from pi2px, pi2py. Let's go, carbon is done. Next is nitrogen. Next is nitrogen, N2. Let's see how much bond N2 has. 14 electrons. So, let's stop writing from sigma, KK.

KK dash. How many electrons are finished? 8 are finished. How many are left? 6. Now, pi 2Px equal to pi 2Py. Then, sigma 2Pz. 2, 2. 1 will be added, then 1 will be added. 12 are there. And 2 is added in this. 14 electrons. Okay? Bond order. Number of electrons in bonding. 2, 2, 4, 2, 6. Anti-bonding. 0 divided by 2. All these will be cancelled. 2 bonding, 2 anti-bonding. 2 bonding, 2 anti-bonding. Cancel. So, 6 bonding, 1 anti-bonding.

Bond order 3 means how much bond between nitrogen and nitrogen? 3 bonds. Right? Nice. Paramagnetic, diamagnetic. Diamagnetic. Why diamagnetic? Why? Why? Why did this happen? Because not even one electron is unpaired in this. All electrons are paired. Magnetic field will be a little repelled. Clear? Now see here I am telling you a very clicking point where there is confusion in children. Listen carefully.

I have added one electron in N2. What will be? N2-. Number of electron is 14. Number of electron is 5. How far was this electronic configuration? Less than or equal to 14 electrons. So, will this configuration be applied on N2- or will it be applied on any other configuration? This is a big question. All your videos are flopped here. Here my video explains you the details inside. You get stuck here.

that 14 electrons in N2 and 15 electrons in N2- so for 15 electrons, another configuration will be given. This was upto 14. So what to do in this? Will we apply 15Y or 14Y? You apply 14Y in this. Why? Because it is made of this. So N2- will follow this configuration only. So if I talk about N2- in which how many electrons are there? 15 electrons. It will follow this configuration only. Everything will remain the same. Dash, dash, dash, dash, dash, dash, dash, dash, dash. How many were these? 14. They reached here. After this, it is pi*2px.

Let's write it all. Sigma 1 is 2. Let's have some fun. Sigma star 1 is 2. Sigma 2 is 2. Sigma star 2 is 2. Pi 2 px equal to pi 2 py 2 2. Sigma 2 pz 2. Pi star 2 px. Oh, shit! Pi star 2 px equal to pi star 2 py. Let's go. 2 2 4 2 6 2 8. This is 8 electrons. K K K K dash is over.

Okay? 2 x 10, 2 x 12, 2 x 14, 1 x 15. In this configuration, 15 electrons will work because it is made of N2, there is no new molecule. Now, let's say, for this, there is a bond order. No tension, everyone cancels. Cancel, cancel, this is left. 2 x 2 x 4 x 2 x 6 minus, in anti-bonding, 1 x 2 x 5 x 2, how much is it? 2.5. Who is more stable than these two? Here it is. What was its bond order? 6 x -0 x 2.

How many? 3. This is more stable, this is less stable. If someone asks you about N2+, how many electrons will be there? 13. So, there is no tension. This will be the system for 13. In 14, 13, in 13, reduce one electron. In which? In the last one, remove one electron in sigma 2pz. It will have a paramagnetic nature. Like its nature will also be paramagnetic. It will be a little attracted to the magnet. Because one electron, how is it? Unpaired. Can you see? This one. Unpaired. This 2px, 2py are orbitals. So, there is one electron in this and it is completely filled. No.

N2+13 electrons, here sigma2Pz1 will be there, bond order 2,2,4,1,5,2,2.5 will come and and and and and and paramagnetic nature because it has an electron. Okay, its configuration will be same, at last sigma2Pz1 will be there. How much bond order will come? Say, 2,2,4,1,5, -0,2, how much will come? 2.5. This is the most stable among these three. Who is more stable among these two? Who is more stable among these two? It has electrons in anti-bonding, it will not have. What will happen before this?

This is the configuration of electron. Here, there is only one electron. These two are unpaired electrons. Who has electrons in anti-bonding? This one. This is less stable and this is more stable. So, n2, n2+, n2-. Order n2, n2+, n2-. So, you understood all the story and understood the bond order. Very good. You liked it very much. It was fun. It was fun.

You made the question. Okay, let's say I make you spin and ask you to do another question. For example, I give you carbon monoxide. Or I give you CN. I can give you. Why? I can't give you. I can give you. Or I give you CN minus. Now what to do? Number of electrons have fallen. Let's do the configuration. We can do 6 plus 8. How many are there? 14 electrons. We have written the whole story for 14 electrons and understood things. We can do this. Or let's take a shortcut of this.

Now, understand one shortcut. I am writing here. Understand one shortcut. 14, 13, 12, 11, 10. Bond order. Bond order is 3 on 14. 2.5 on 13. 2.2 on 12. 1.5 on 11. 1.5 on 10. Finished. If number of electrons is 14, Bond order is 3. 13, 2.5. 12, 2. 11, 1.5, 10, 1. Keep 3 on 14. After that, 0.5, 0.5, 0.5.

As I asked you which is more stable? Which is more stable? Yeah. There was a question in J. Which has same bond order as N2? It was a similar question. Now, I gave you four options. Let's say C, N, C, O, O2. I gave you four options. N, O+. Now, I asked whose bond order is similar to N2? So, don't do anything. Count the number of electrons. What did I count? Number of electrons.

How many electrons are there in N2? 14. So, what is the bond order for 14? 3. 14 for 3, 13 for 2.5, 12 for 2. Okay. Come here, CN. 6,

Nitrogen is 13. So its bond order is 2.5. CO is 14. So its bond order is 3. So it will fall here and here. Oxygen is 16. NO+ is 7. So how many electrons are there in oxygen?

Let's count the total electrons. Not the valence shell. 6 on carbon, 7 on nitrogen, 13 on nitrogen. 8 on oxygen, 16 on carbon. 7 and 8, 15. And 1 electron is gone. 14. Yes. What's the plus? 7 and 8, 15. 1 electron is gone. 14. What's the bond order on 14? 3. So, who is following the bond order of nitrogen? B is also following. Now, you understood. So, you can use this trick. Yes, there are some things hidden in this. Whether it is paramagnetic or diamagnetic.

Now see which one is more stable? What should you know about which one is more stable? Which one has electron in anti-bonding? So that thing is missing from here. So it will work in some places. Where there are questions of lightness, it will work there. It will work in stability. Like we ask which is more stable? Which is stable? I have put option A as Cn. I have put option B as No+. Sorry, No. I have put option C as O2. I have put option D.

yin, xin, minus. Now, we are asked which one is more stable. So, now take out the bond order. No of electrons 6 + 7 = 13. Bond order 2.5. 7 + 8 = 15. We don't know about 15. 15 also has 2.5. So, I will change it for now. NO. I will do NO+. So, how much is this? 7 + 8 = 15. -1 = 14. Bond order 3. Come to this.

Oxygen, 16 electrons. Bond order, we don't know about 16. It decreases as we go forward. Write 15, 16. Let me write. Here, 2.5, 2 on 16. 1.5 on 17. 1 on 18. Same order here and there. Clear? Keep 3 on 14. Then, move forward or backward, reduce 0.5. How much will be the bond order for 16 electrons? 2.

then here is carbon 6 nitrogen 7 6 7 13 -1 14 what is the bond order? so you can select the least stable from this least which will be the least stable whose bond order is the lowest so 2.5 3

2.5, 3, 2, 3. The least is 2. So, we will say oxygen is least stable. If you ask the order of stability, then you will have to think a little. See, the stability of these two can be equal. Now, what will have to be checked? Whose electron is present in anti-bonding? Then, it will get confused. Here, if you ask least stable, then you will easily answer J.S.A. If you ask the whole, then you have to do configuration.

So, you have a shortcut to get a bond order. What is the shortcut of a bond order? Assume 14:3, 13:2, 12:2, 11:1, 10:1. If you go down, then the same. 15:2, 16:2, 17:1, 18:1. This is the shortcut for a bond order. You can apply it. Is it clear? Which has the same border? Which has greater bond order? It is possible that the two are not the same. If we are asked to compare only the bond order. Is it clear?

Very good. So, you have understood this shortcut. Now, let's come to the energy library diagram. I am going to write less than or equal to 14 electrons. I have not told the configuration for 15 and 16, but the shortcut. The shortcut is valid. Just remember, 3 bond orders on 14. If the electron is 14, then 3 bond orders. 15 on 2.5. 13 on 2.5. 12 on 2.

16 to 2. Just remember this 14 figure and then reduce it by 0.5. Now, let's make energy level diagram of orbitals. So, let me explain the story of energy level diagram. The lowest energy is sigma 1s. How will it be made? When one is 1s and the other is 1s. So, 1s will be 1s and two orbitals will be made. Sigma 1s and sigma star 1s. Whose energy is more between these two? This one. So, I will

I will make two platforms, one orbital here and one here. This is sigma*1s and this is sigma1s. Its energy is more and its energy is less. We are showing energy from the top, so its energy is more and its energy is less. 1s, 1s, 1s, 1s. Similarly, we will get 2s, 2s. So, from here, 1 2s, from here, 1 2s. Did you get both?

This is sigma*2s and this is sigma2s. Whose energy is more? Sigma*. We are moving in the order of energy. Is it clear? After this, 2p and many more are made. So, let's make 2p. How many orbitals are there in p? Three. So, p has three orbitals. And here also, p has three orbitals. Now, see how to make it. These two orbitals have the same energy. Then, there is a sigma2pz.

Sub bonding. After this, the energy of these two is same in anti-bonding. So, we raised it up. After that, its energy is higher in anti-bonding. So, we raised it higher. It is done. Now, what is pi 2px and pi 2py? So, two orbitals are together here. Which ones? pi 2px and pi 2py. After this, sigma 2pz is on it. So, this is done. After this, sigma 2pz.

On this, there is pi*2px and pi*2py. So, I have put pi*2px and pi*2py on this. And on this, there is sigma*2pz. So, here it is, sigma*2pz. This is the energy level diagram. You can follow this and put electrons in any molecule. You can make an energy level diagram of any molecule's electronic configuration. How did you understand?

No new thing is done. 1 is 1 is, 1 is anti-bonding, 1 is bonding. 2 is 2 is, anti-bonding, bonding. The most confusion is in P. So, look at it again here. Then we will practice it. I have 3 orbitals of P and 3 orbitals of P. See, first, how is π2Px and π2Py? Bonding. So, we have to keep the bonding down. So, we have kept it down. I have made the energy of both equal. π2Px, π2Py. Then, sigma2Pz is bonding. So, we have kept it down.

Sigma 2pz. The anti-bonding of pi*2px and pi*2py went up. Why? Because anti-bonding is unstable. Let me tell you one more thing. You will have to put it in the notes. Pi*2py. After this, sigma 2pz, sigma* anti-bonding, this also went up. Sigma*2p. This is our p-orbital. We don't have to look at px, py, and pz. We have made them. Now, let's go to a very important thing.

This is an atomic orbital and this is an atomic orbital. This is an atom orbital and this is an atom orbital. And the molecules are coming in the middle. So, the bonding molecular orbital, take a note of it, put it in the copy. The energy of the bonding molecular orbital is less than the atomic orbital. That's why it went below it.

And the energy of the anti-bonding molecular orbital is above the atomic orbital. That is why I have made both of them below. I cannot make one above because this is also bonding and this is also bonding. The energy of the bonding molecular orbital is less than the atomic orbital and the energy of the anti-bonding molecular orbital is more than the atomic orbital. This is more stable than this.

less stable than this. That's why it went up. Bonding molecular orbital has lesser energy than atomic. That's why both bonding molecular orbital are down. This will go in the notes. Both bonding molecular orbital are down. The rational reason for one number will ask that this diagram is like this because. Now we have to give the reason. So, in because, you have to write this. In statement 2, you will get this. That bonding molecular orbital has lesser energy than atomic orbitals. This is done. Now, you have to

You can make a diagram like this. Let me tell you. You can do it. H2. Hydrogen. What happens in hydrogen? H is made of H. 1 is 1. 1 is 1. So, one electron is in this. One electron is in this. There are two electrons. What will be the configuration? Sigma 1 is 2. Where did the two electrons go? In bonding. So, combine these two and make it here. That's it. You have to do this. You won't be able to do this. Oh, brother. Okay. Let me tell you. I will tell you to make this. Let's increase it a little.

Let's increase the level a little bit. Suppose you make B2, a system of 10 electrons. So, say, sigma 1 is 2, sigma star 1 is 2, sigma 2 is 2, sigma star 2 is 2, and what else? Say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say, say

2+4+6+8+9+1=10. Okay, what is the configuration of Boron? It has 5 electrons. Configuration? 1 is 2, 2 is 2, 2P1. So, let's make the configuration of Boron on both sides.

both sides are boron configuration, here also boron, here also boron, 1 is 2, 2 is 2, both borons are coming from both sides, and what is 2P1, P has to have one electron, P has only one electron, here also 1, here also 1, clear? Now tell me, what is the configuration, sigma 1 is 2, means 2 electrons in this, sigma star 1 is 2, here also 2 electrons, sigma 2 is 2, 2 electrons,

2 electrons, sigma*2s2, 2 electrons, 1 in pi2px and 1 in pi2py. We put a parallel spin and followed the Hunt's rule. Don't make a mistake here, both are upside down, both have the same energy. You can do any electronic configuration, do the electronic configuration of the atom, the molecule and then come to this diagram. Let's do the nitrogen, the last example, N2, how many electrons? 14 electrons, say for 14 electrons, sigma1s2, sigma2,

1 is 2, star. Sigma 2 is 2, Sigma star 2 is 2. Pi 2Px equal to Pi 2Py. 1, 1, 2, 2. How many are these? 12. After this, Sigma 2Pz2, 14. Do nitrogen configuration here. 7 electron configuration. 1 is 2, 2 is 2, 2P3. 2, 2, 4, 3, 7. 1 is 2, 1 is 2. 2 nitrogen will be there, right?

2s2 2s2 is correct. 3 in 2p. 1 2 3. 1 2 3. From here, 1 nitrogen came. From here also, 1 nitrogen came. Now, sigma 1s2 is correct. Sigma star 1s2 is correct. Sigma 2s2 is correct. Sigma star 2s2 is correct. 2 electrons in pi2px2. 2 electrons in pi2py2. 2 electrons in sigma2pz2. It was fun. You understood how to make an energy level diagram.

It is clear. If you remember the electronic configuration then the energy level diagram will be formed. This is this hole. This hole is the electronic configuration for less than equal to 14 electrons in total of the molecule. Now if it contains N2 minus then remember to follow this configuration. So note this. Now we will move on to electrons greater than 14. That is, we will talk about systems from 15 to 20 electrons. This part is completed.

Let's talk about what will change in the electronic configuration from 15 to 20 and what will change in the energy level diagram. So, greater than 14 or 15 to 20. I had to apply this till 14. So, electronic configuration for total number of electrons greater than 15 and less than 20. 15 to 20 and both limits are included.

What you have to do now? It will remain the same. Sigma 1 is 2. Sigma star 1 is 2. Sigma 2 is 2. Sigma star 2 is 2. Experimentally it has come. Pi 2 Px equal to Pi 2 Py. I will remove 2 2. Now these electrons do not fill. Pi 2 Px Pi 2. Here change will happen.

We were writing this and then what we used to write? Sigma 2pz, then pi*2px = pi*2py and then sigma*2pz. We used to write this. There is only one change here. There is only one interchange. Where was this configuration? In notes, where was this? Less than or equal to 48.

Pick up this and that and that and that is it. This is the only interchange. Put sigma 2pz here. Put pi 2px here equal to pi 2py. That's it. That's it. For 15 to 20 electrons. That's it. Now, the most important example of 15 to 20 electrons is O2. Oxygen. 16 electrons. It will definitely come in the paper.

Sigma 1 is 2, Sigma star 1 is 2, Sigma 2 is 2, Sigma star 2 is 2. That is KKKK. KKKK dash. It doesn't matter. Next, Sigma 2PZ2. Then, Pi 2PX equal to Pi 2PY. 1 in this, 1 in this, 2 in this, 2 in this. How many are there? 2, 2, 4, 2, 6, 2, 8, 2, 10, 2, 12, 2, 14. Next, this will come.

How many electrons are left? 2. Don't put 2 in this. 1 in this and 1 in this because both are equal. It's 16. 2, 2, 4, 2, 6, 2, 8, 2, 10, 2, 12, 2. 2, 2, 4, 2, 6, 2, 8, 2, 10, 2, 12, 2, 14, 2. First, tell me the magnetic nature without thinking anything. What do you want to see in the magnetic nature? Are all the electrons paired or not? So, you can see that here 2 electrons are unpaired.

This is not paired, it is unpaired. So since oxygen has unpaired electron, hence we can say it is paramagnetic. And this question is very common. Show that oxygen is paramagnetic, then do electronic configuration and show that what is in oxygen? It is unpaired electron, that is why our oxygen is paramagnetic. Clear? Is it clear? Bond order?

Bond order is number of electrons in bonding minus anti-bonding. So, leave this, it will cancel bonding, anti-bonding, bonding, anti-bonding. Come from here, 2, 2, 4, 2, 6, 2, 6 are bonding. 1, 2, 2, 4, 2, 6. In anti-bonding, 1, 2. Divide by 2. 6 minus 2, 4 by 2, 2. Means, in O2, how many bonds will be there in oxygen and oxygen? 2. Because look at the bond order, how much is there?

Look at the bonding, 2, 2.4, 2.6 minus, in anti-bonding, 2 divided by 2. Bond order? How much is it? 2. I have told you with a short trick. How much is it at 14? 3. Say it at 15. Say it at 2.5. Say it at 16. Say it at 2. The same thing is coming here by doing the calculation. Clear? Now you will solve a question for me. Following this electronic configuration, that is, for the one above 14, you will calculate the O2 plus.

O2- and O2-2 electronic configuration O2+ 16-1 electron O2- 16-1 electron O2- 2 electron O2- 16-2 electron Their electronic configuration and their bond order. Let's do it. So, this is the configuration O2+ and O2- You won't go for 14 electrons, it's still 15. So, say 2, 2, 4 means KK, KK' Is it correct?

15 electrons are left. Now, we have 2Pz2, 2Py2Py. 14 electrons are left. Now, 2Px2Py.

How many electrons will go in this? 8, 2, 10, 2, 12, 2, 14, 1, 15. 15 electrons. Can you get the bond order from the box? How much will the bond order be? Their will be cancelled. 2, 2, 2, 2, cancelled. Here 2, 2, 4, 2, 6 minus, in anti-bonding 1 divided by 2, 5 divided by 2, how much is it? 2.5. See, it is written as 15.5. 15.5. 17. Same configuration in 17.

15, 16, 17. Did you understand? No? 15 is 16, 17. Means all the configurations will be same. In the end, pi*2px is equal to pi*2py. This was 15. So, 16, 17. This will be the bond order of 17. Rest will be fine. 2, 4, 6.

Minus 3 divided by 2. 3 divided by 2, how much is it? Yes, 3 divided by 2, how much is it? 1.5. See, if you go from 16 to 17, then 1.5. 0.5 is less. Are you understanding the story? The number of points you will go from 14 to 1, this is 0.5, 0.5. 16 to 2 was 15 to 2.5, 17 to 1.5.

What will happen in 18? 17 is done, add one more electron to it, 18. Same here, pi*2px=pi*2py, 2 here, 2 here. What is the bond order? 2, 2, 4, 2, 6 minus, in antibody, 2, 2, 4, divide by 2, 2 by 2, 1. See from here, 17 is 1.5, so how much will come to 18? 1. 17 is 1.5, 18 is 1. We could have done it straight. Clear is very good.

Let's talk about paramagnetic and diamagnetic. One is unpaired paramagnetic, the other is unpaired paramagnetic. Not even one is unpaired, it is diamagnetic. O2 2- is diamagnetic, rest O2+ is paramagnetic, O2- is paramagnetic. Oxygen itself is clear. So, this is how you can do electronic configuration. And you can also make energy level diagram of this. What will be the difference? The difference in this will be reflected in the energy level diagram. Let's do its energy level diagram. Come.

Let us make energy level diagram of this story. Sigma 1s2, Sigma star 1s2, Sigma 2s2, Sigma star 2s2. What is the difference here? Sigma 2pz. Okay. After this, say pi 2px equal to pi 2py. Okay. Then practice. Pi star 2px equal to pi star 2py. Okay.

Then comes sigma star 2p z. Diagram, is taraf energy, is taraf orbitals. To ye 1s 1s, sigma 1s, sigma star 1s. Isi tarah se 2s 2s, sigma star 2s, sigma star 2s.

This is sigma 2s and this is sigma star 2s. The energy of bonding is less than the atomic orbital and the energy of anti-bonding is more than the atomic orbital. What are we writing here and there? Atomic orbitals. And what are we writing in the middle? Molecular orbitals. Then P will come. 3 orbitals in P and 3 orbitals in P. Now look at the order of energy. First, if there is bonding of sigma 2pz, then we will do it below. Sigma 2pz. After this, if there is bonding of pi 2px 2py, then it will be below.

These are both pi 2px and this is pi 2p. After this, pi*2pxy, this is anti-bonding, so from here up, pi*2px, pi*2py. After this, sigma*2pz, anti-bonding, sigma*2pz. You understood the whole method.

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Electronic configuration, 8 electrons, 1s2, 2s2, 2p, 2, 2, 4, 4, 8. So, 2 in 1s, 2 in 2s, and 4 in p. 1, 2, 3, 4. On this side also, 1s2, 2s2, 1, 2, 3, 4. What is here? Oxygen. What is here also? Oxygen.

Clear? Now look here, Sigma 1 is 2, 2 electrons in it. Sigma star 1 is 2, Sigma 2 is 2, Sigma star 2 is 2. In Sigma 2pz, 2 electrons. In Pi 2px 2py, 1 electron. Paramagnetic, diamagnetic, paramagnetic. So, you could do it like this. Now, let's do some good questions left in the last. Let's do it. As we were asked, we have 4 options. Compare all of you. Bond order, Bond length, Stability.

So, I am giving you four options. You have to compare all three. O2, O2-, O2+, O2, 2- So, arrange in all the order. What you have to do is, write all these in the side in the notebook. O2, O2-, O2+, O2- Now, you remember, how much is the bond order on 14? 3. On 15? 2.5.

16 + 2 = 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

So, 1 for 18. So, the bond order is 1. Now, if we want to put the order of the bond order, then whose is the simplest? This, then this, then this, then this. Then, what is being asked? Bond length. So, bond length is inversely proportional to bond order. So, see whose bond order is the most. See whose is the longest. So, whose bond order is the shortest? Whose bond order is the shortest, whose bond order is the longest. That means, reverse the bond order.

then this, then this, then this, then stability. Stability is directly proportional to bond order. So, the most stability is this, then this, then this, then this. There are four questions. How will you ask this question in the GE after turning? Which has arranged in order of bond length

Now see how it will be. First option KO2, second option Na2O2, third option O2. In these three molecules, the bond between oxygen and oxygen, what is its length? So how will KO2 be made? K plus O2, sorry, how will KO2 be made? Yes, K plus O2 minus. So which molecule is here? O2 minus. Na2O2, 2Na plus O2 2 minus, it will be made like this.

How will it be made? 2Na+O2 2- means it has O2 2- and what is this? O2-O2-Bond order 1.5 Bond order O2 2-Bond order 1 O2 Bond order 2 Can you calculate the number of electrons? We have asked about Bond length Bond length is inversely proportional to Bond order The smallest Bond order is the maximum Bond length The maximum Bond length of this

After that, this, after that this. Meaning, order will be B, A, C. Most of it is this, the borne order is the least. Then this, then this. You can ask such questions. So, you have to keep all these things in mind. Like, one kid has another confusion. We have discussed it earlier, now we will discuss it again. It is told us to do electronic configuration of N2 and N2-. So, N2 means 14 electrons, N2- means 15 electrons. This is a big confusion.

Which one will you use? Less than equal to 14 electron method. In which, first comes pi2px equal to pi2py and then comes sigma2pz. And this one is bigger than 14, so should we change it? No. How is n2 minus made? By combining an electron in n2 and making n2 minus. So, first n2 minus will be made or n2? First n2 will be made.

means we have to follow this configuration and move ahead. This will be included in less than equal to 14 electron case. For example, there is a confusion NO + NO. NO is 6, sorry 7 and 8. How many electrons are there? 15. So, this is greater than or equal to 14 case. Means 15 to 20.

What changes in 15 to 20? Sigma 2pz comes first, then pi 2px, pi 2py, then pi* etc. 2px, pi*, 2py. And here, after this comes pi*. So, this went on the case greater than 14. NO+ 7, 8, 15. Plus means electron, 14.

So, will this go on the 14th case? No. First NO will be created, an electron will be produced from it, then NO+. Meaning, you have to treat NO+, too, as greater than or equal to 14, when you do electronic configuration.

If you ask about the bond order, then no problem, just hit with the shortcut and both will get the same answer. But if you ask about the electronic configuration, paramagnetic, diamagnetic nature, then you have to keep in mind that NO+ has 14 electrons, but it is made of NO, which is from 15 to 20 cases, the configuration should be like this. Is it clear? So, these were all the things that we had to study in molecular orbital theory. We have covered all the concepts. Keep studying. All the very best.