The C2 molecule has a MO diagram similar to N2
(Figure 8.22a). W...

Question

The C2 molecule has a MO diagram similar to N2 
(Figure 8.22a). What is the bond order of C2 and is it paramagnetic 
or diamagnetic? (LO 8.12) 
(a) Bond order = 2, diamagnetic 
(b) Bond order = 2, paramagnetic 
(c) Bond order = 0, paramagnetic 
(d) Bond order = 3>2, diamagnetic

Accepted Answer

Hello. Everyone in this video we're being told that the M. O diagrams of the molecular orbital diagram for H E two plus is similar to H two. So let's go ahead and determine the bond order for H E two plus. As well as determining if this is going to pick a paramedic or die magnetic. So first step is of course draw out the molecular orbital diagram. So the M. O diagram. So of course we have our increasing energy from bottom to top. So for our H E two plus we only deal with our s orbital. So we'll go ahead and first label that this is going to be our molecular orbital to the right and left of atomic orbital's. We have only one type of atom here so they will all have the same energy level for the atomic orbital's. So you're dealing with of course are to us to us. So then that creates here a sigma bond and a sigma star orbital of course they have interactions with each other. So at that representing with a double or a dotted line. Alright, so we know that helium so H. E. That has two valence electrons according to our trend here for um from appearing table. So go ahead and add two electrons each for our atomic orbital. So we have 1-1 and two. And because we have two helium then that's two times two which is equal to four. However we are dealing with A I. On in this case is I have a plus one charge. It's only positive because we're taking away a negative energy, it's going to be an electron. So if we have originally four electrons, we take away one. That'll be three. So add three electrons into our M. O orbital's here. So then we'll have +12 and three. Now calculating for our bond order will do this in green. Let's go ahead and recall that. The equation for bond order is equal to have multiplied by the sum of the electrons in our bonding orbital, minus our electrons in our anti bonding orbital. Alright. We're calculating for our discounting for the number of electrons in the bonding, that's just the orbital's without the star. So we say we just have to here. So that's two. And then how many we have in our anti bonding. So the one that with the stars we see just one here. So we'll put that into the equation. So 2 -1 is 11 times half is equal to well, half. So 0.5. So the bond order for our ion here is equal to 0.5. Now we're being asked if this is paramedic or diet magnetic. So para magnetic is if the M. O diagram has an unpaid electron and we see from our structure here that the sigma star orbital does have a um paired electron. Therefore we say that HE two plus is a para magnetic. I am. Alright. And this is going to be my final answer for this problem. Thank you all so much for watching.

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