 The number of bond present is the number of bond present between the two atoms in a molecule It can be fractional value of bond order is possible in case of resonance. When resonance is possible then fractional value of bond order is different. Integral value will happen there is no resonance ok CH3, CH2, CH3, CH3, CH2, CH3 all the carbon carbon bond if you see it has bond order 1 right because there is no resonance ok. When there is no resonance integral value we have otherwise we have the fractional value. Now calculation of bond order we know also one thing that bond order is inversely proportional to bond length and directly proportional to bond strength. All these comparisons we can do if you know the bond order ok. Calculation of bond order in case of equal resonance in case of equal resonance. Like for example you see the first example Nc double bond O 4 minus and it is resonating that is Nc O minus double bond 4 ok. The bond order in this case will write number of sigma bond plus number of pi bond divided by number of sigma bond sigma plus pi by sigma ok. See in this molecule the resonance is possible from this oxygen to this oxygen right only in this part. So we will count the sigma bond and pi bond in this part where the resonance is possible. How many sigma bond we have here 2 sigma bond so 2 how many pi bond divided by 1. Sir in the formula there is a number of bonds by sigma bond. Yes. In the formula there is a number of bonds by sigma bond. No. There is a number of bonds by sigma bond. Yes. How many sigma bond we have here 1 plus 2 1 plus 1 2 sigma bond and 1 pi bond 2 plus 1 2. Sir how the number of sigma bonds of this number of pi bond was equal to the number of bonding and the bond minus number of pi bond. It is not that in that case we will get integral value. No, not integral value but then no, no but that is different molecule if you have N2 like that same atom present. We are talking about this molecule say the amount. So here we will think we have resonance will use this problem. We have another formula for this also. We can also write number of bonds in all R s, but this one is the easier one, the first one. The mission of number of bonds in all R s divided by the number of R s, there is an ending in the script. Like in this one you see, suppose you have to count the number, the bond order of this bond, bond order of this carbon-oxygen bond you have to find out, bond order of this carbon-oxygen you have to find out. So how many R s do we have? 1 plus 1, 2, right? So we have 2 R s and how many bonds? Two bonds here and one bond here. 2 plus 1, again it is 1. But for this formula you have to draw the resonating structure, the bond resonating structure. But here the advantage is what? With this only you can say 1 plus 2 sigma plus 1, number of sigma plus number of pi by number of s. Second formula is number of bonds in all R s. How many R s do we have? Two. And we have to count the bond order of any carbon-oxygen bond. Suppose this one we are taking, C 1, C 2. We have two bonds here and one bond here. 2 plus 1 by number of R s. Sir, is hydrogen bonding a sigma bond or a pi bond? Hydrogen bonding. It has a sigma bond characteristic. No, because resonance is possible here, no? So take only that part in which the resonance is possible. The first method will be count if there is hydrogen bond. You have resonance here, so number of sigma bond plus pi bond. Yes, we will take only that part in which the resonance is possible. Sir, but is the first case we consider the bond order? Why did we consider both oxygen? We are considering both oxygen only. How many of you understood this? See, we are taking this portion because resonance is possible here only. This edge we are not considering. So how many sigma bonds we have? 1 plus 1, 2 sigma bonds, this we let it be. So 2 sigma bonds, how many pi bonds? 1 pi bonds, so 2 plus 1 by 2. Sir, I didn't get the second one. The formula is exactly same. Either you can use this one or this one. Look at the same answer. But the disadvantage of this is what you have to draw the R s, the entire R s. Then you have to substitute here because number of R s you should know. But if you are using this formula, you can directly say with this one, what do you have? Number of sigma and pi by sigma. Now when you draw the R s, suppose one molecule has 5 R s, 6 R s. And you have to draw all those R s and then you see how many R s possible and then you have to write this. So better use this formula, sigma and pi. Got it? Another example right down, CO3 2 minus. What is the bond order of carbon-oxygen bond? CO3 2 minus. CO3 2 minus. Okay, how many if you understood the second formula? That's that idea. See, number of bond in all R s. First of all you have to draw the resonating structure. We have two resonating structure. Like I said in the beginning, you have to only concentrate the part of the molecule in which the resonance is possible. Now here also you see, in this part the resonance is possible only here. This part is not in resonance. So we won't take this into the count. Okay, this is one thing. Now we have two R s. Formula is this. Number of bond in all R s. Now how many bonds you have here? Number of bonds will what? Suppose we have to find out the bond order. First of all you try to understand. Bond order between two atoms will be fine. It's not defined for the molecule. The molecule is the bond renewal. It's a bond order between the two atoms of a molecule. Like you can find out the bond order of this carbon. This carbon-oxygen bond, this carbon-oxygen bond. You can find out the bond order of carbon-carbon bond. Carbon-oxygen bond like this. So we are taking only one bond here. So I'm taking this C1 and this oxygen bond. This carbon-oxygen bond. This carbon-oxygen bond. Which will be same for this carbon-oxygen bond and this carbon-oxygen bond. Correct? So any one of this bond you can take. Because it is equal resonance it won't affect the answer. Right? So I'm taking this bond. How many bonds we have here? Two. How many bonds we have here? One. One. So summation of number of bonds in all R s. We have two R s. Two bond here. One bond here. Two plus one divided by number of R s. Two. Of so many. Sir, but in the first method we counted the bonds of both oxygens. Because we are here the formula is, that's what I said. The advantage of this formula is you don't have to draw the entire R s of this structure. You have this structure, let it be. How many, since we have resonance possible from here to here? So we'll count the sigma bond in this part and pi bond in this part. So we have two sigma bond and one pi bond. Two plus one. So one point five is the bond for both oxygens. No, both carbon-oxygen bond. See, it's the same thing. When you have this molecule, since it is the equal resonance, so hybrid is what? Hybrid is this. This molecule wants to give this bond R s to. This wants to give one. Half will be given by this. Half will be given by this. So average is what? Two plus one by two. Because it is equal resonance, this and this contributes equally. AB contributed equal to AB contributed. So you'll get the average of that, that is it. So in case of equal resonance, you can find out the average. It is nothing but the average. If you look at the formula, it is nothing but the average. We have to find out the bond. So this is the average value, this is the average value. That's why we have number of R s. Bond divided by R s. It's average only. Understood this. Correct? Yes. Okay. Now what is the bond order here? Carbon-oxygen bond. Which formula we should use? This one or this one? First one. First one will have number of sigma bond. You see, first of all, here is the resonance possible. This part also and this part also. Okay, only this. Resonance possible, this part and this part. How many sigma bonds we have? One, two and three sigma bonds. Three sigma bonds plus one pi bond. Four pi z. With this also, you can do. You'll get three resonating structures. First, you'll get this, comes over here. C o minus double bond o, o minus. And another one from this is what? When this comes here, this goes here, right? C o minus double bond o, o minus. To use the second formula, what you have to do, you have to take any one of the carbon-oxygen bonds. This one. Okay, so total bond is what? One plus one plus one. Four. One plus one plus two. Four divided by R s is what? Three, four by R s. So any one formula. I would suggest this one, I'll give you one. Okay, now this one you tell me. For this one, third one. Resonance is possible in this molecule, entire molecule, right? So it is what? We have sigma bond is three. One pi bond by three. Four pi three. What about this one? This part is not in resonance. So it is three by two. Clear? Okay, so equal contributing structure you won't get. It's very easy and simple. You have to find out the average. Okay, so mostly you'll get unequal contributing structure.