The C2 molecule can be represented by an MO diagram similar
to t...

Question

The C2 molecule can be represented by an MO diagram similar 
to that in Figure 8.22a. 
(b) To increase the bond order of C2, should you add or 
remove an electron?

Accepted Answer

Hello. Everyone in this video. We're given this big diagram here. So this is a demo diagram or molecular orbital diagram of our entire molecule. Were being asked here is if we wanted to increase the bond order of our entire molecule which you add or remove an electron. Let's go ahead and recall the bond order equation. Let's go ahead and write this baby below. So B. O. Which is respond order. This equals to half times the sum of the electrons in our bom, ding orbital's minus our electrons in our anti bonding orbital's. All right. So we'll just do it as is we're doing a neutral one first. So this is for just end too. So we have half electrons in the bonding. Let's go ahead and actually count this. So taking a look at our um oh diagram. Alright, so all the bonding ones is basically all the orbital's without the stars. The first example this one here, this one here and this one here. So we're first looking at how many electrons are in our bonding orbital's and then our anti bonding. So for bonding we have to 4, 6 and eight for anti bonding we just have to so scrolling back down to our equation here. So we have 8 -2 and that'll give us a value of three because when we have eight minus two which is 6 65 by two is three. So let's say we remove an electron from our diagram. So let's start a new calculation here using green. So if we remove an electron we're making it more positive. So we'll have and two plus. So it's gonna again behalf. And let's go ahead and look at the mo diagram again and seeing how much electrons will have in our anti bonding and bonding orbital. And do the calculations. So again we're removing an electron in that case let's just do any green. We'll remove maybe this one here. Doesn't really matter which one. So then if we count everything again and putting it into our calculations, you can see that we have some electrons in the bonding now and we have the same number of anti bonding, which is just two. So doing the calculation here, then we can see that we have 7 -2 which is 5, 5/2 is you go to 2.5. Now, lastly the other question is being asked if we remove an electron with this bond order increase. So if we add electron we're adding more negativity. So the formula for that is going to be n two minus again, it's going to be half minus the electrons in the bonding minus the electrons in the anti bonding. So again let's go ahead and scroll up. So we see here we'll move the screen here. So an electron and electron into the anti bonding. So going back down here to finish off our calculations. So the number of electrons in the bonding is going to be eight and we have increased the number of anti bonding. So that's going to be three. So 8 -3 of course, is five divided by two, is 2.5. So you can see here that the bond or decreases regardless again, what we have in red here. That's just the neutral compound just by itself. But here we try adding or taking out electron here, trying to add electron. But either way the bond order does decrease. So the answer then is neither the bond order decreases. Alright, and this is going to be my final answer for this problem.

Watch next