Calcium carbide, CaC2, reacts with water to produce acetylene, C2H2, and is sometimes used as a convenient source of that substance. Use the MO energy diagram in Figure 8.22a to describe the bonding in the carbide anion, C22-. What is its bond order?

Video transcript

Hello. Everyone in this video we're dealing with our M. O diagram of peroxide ion which is 022 minus were also calculated for the bond order. So first step is having to draw the diagram which is molecular orbital diagram of course we have our increasing energy from top to bottom. That's E. So here we're dealing with the molecule ion. Let's actually make this line straight. All right. So again we're dealing with our 02 to minus and then over to the right and left. Of course we just have our auction atoms. So these are atomic orbital is doing the right and left. So of course since we're dealing with auction, that's all the way to the right of the pr table. So we're talking about R. S. Orbital interactions as well as our p orbital interactions. So then starting with the atomic orbital's We of course are dealing with the two S simply to the right now of course because we have these s orbital's we have interactions of sigma and sigma star so go ahead and represent the interactions with this dotted line here. Alright, so again we are dealing with s orbital interactions as well as p orbital interactions. So if it's all right, we'll draw 12342. P. Same thing to the left 123. This is two P. Then we're dealing with now with the bonding and anti bonding orbital's so we'll have one for our sigma two for our pi maybe we'll just make these pure rose higher to make it more accurate. Alright so for the anti bonding now we have to, for the pi star and then one for our sigma star, of course there's still going to be interactions. So we'll just draw this dotted line. It might be a little bit messy but we get the point here. Alright, so what we're doing with auction auction is in group 68, so that's going to be six valence electrons per adam to have 123456. Same thing over to the right? We have 12345 and six for each of our atomic orbital's now for the middle. Since we're dealing with the ion, we need a net total amount of gains electrons. So we're dealing here with two oxygen atoms each giving us six valence electrons. So two times six is 12. Let's go ahead. And also remind ourselves of dealing with a ion in this specific island, we have a two minus charge. So we have two extra electrons. So we have 12 other generally, but adding to now is equal to 14. So go ahead and add 14 total advance electrons in the M. O orbital. So 123456789, 10, 11, 12, 13 and 14. And now that we've completed our M O diagram of our peroxide ion, let's go ahead and calculate for our bond order lot of scrolling down here, give us a little bit more space. So a bond orbital equation is equal to half times the number of electrons in our bonding orbital's minus the number of electrons in our anti bonding. Alright, so filling out these numbers for equations. So our bonding orbital's is going to be all of the orbital's that do not have a star. So this one here, this one and these two. So let's go ahead and count our count how many electrons are in these bonding orbital's? So we have 246 and eight. So go ahead and fill that in our equation here. Alright, so that's eight now for anti bonding we have 24 and six. Alright, so of course 8 -6 is two and then one half or multiplied by two is equal to one. So I bought orbital for our proxy item is equal to one and that's my final answer.