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Ch.9 - Molecular Geometry and Bonding Theories

Chapter 9, Problem 72c

(c) Calculate the bond order in H2-.

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hey everyone in this example, we need to identify the bond order for an atom of helium gas with a plus one caddy in charge. So what we should recall first as our formula to calculate bond order and that is going to be equal to our value of one half, multiplied by the number of electrons in our bonding molecular orbital. Subtracted from the number of electrons in our anti bonding molecular orbital. We should recall that our anti bonding molecular orbital will have an asterisk. So for our helium Adam, when we find that on our periodic tables, we would see that it's in group two a. It's a noble gas, or sorry, it's in group eight a. Which is our noble gas group. And we would recall that helium is going to be a noble gas. So it's going to exist As a molecule HE two. We would recognize that has an atomic number, which we recall is represented by the symbol Z equal to a value of two. And so for a neutral atom of helium, we would say that we therefore have two electrons and two protons. However, we don't have a neutral atom here, we have a plus one charge according to the prompt. And so what that means is that instead of having two electrons each for R two helium Adams because we have that subscript of two, we would instead have instead of four electrons total, we would subtract one electron based on our plus one charge because it's a cat ion. And so that would leave us with three electrons total for our orbital diagram for the helium two plus one catalon. And so to draw out our orbital diagram, we're going to recognize that helium is only going to be Filled in for R one S Sublevel because that's where it's located. So we're going to begin our orbital diagram where we start out at the sigma one s sublevel. And then for our anti bonding molecular orbital above that we would have sigma asterix one of us. So as we confirmed we should have three electrons total for our helium two plus or helium plus one cat ion. And so filling in those electrons, we would start out with our sigma one s bonding molecular orbital. So we would fill in our first electron as follows and then our second electron with the opposite spin according to our poly exclusion principle. And then for our third electron we would move up to our anti bonding molecular orbital where we would fill in our third electron, it can have a spend in either the upward direction as I've drawn in or a downward direction. It's up to you. So now we can go ahead and calculate for our bond order And we would see that we would take 1/2 multiplied by the number of electrons in our bonding molecular orbital which is our orbital our molecular orbital without the asterix we would see that we have a total of two electrons here. So that would be to subtracted from the number of electrons in our anti bonding molecular orbital the orbital with the asterix And that only has one electron. So that would be 2 -1 electron in the anti bonding molecular orbital. To complete our calculation for bond order. So beginning with our brackets, we would take two minus one, that would give us one, One electron left over and we would multiply that by 1/2 for our bond order Equal to a value of 1/2. And so this would be our final answer as the bond order Of our helium to our helium plus one cat ion. So I hope that everything I explained was clear. What's boxed in here in blue is our final answer for this example. So I hope that everything I explained was clear. If you have any questions, please leave them down below and I will see everyone in the next practice video.