Using Figures 9.35 and 9.43 as guides, draw the molecular orbital electron configuration for (d) Ne22 +. In each case indicate whether the addition of an electron to the ion would increase or decrease the bond order of the species.

Video transcript

hey everyone in this example, we need to use the molecular orbital diagram for the beryllium plus one, carry on and identify how the removal of an electron from the ion would change our bond order. So what we want to do is recall our formula for calculating bond order, where we take one half and multiplied by the electrons in the bonding molecular overall subtracted from our number of electrons in our anti bonding molecular orbital. So we want to draw out the molecular orbital diagram for the B two plus or plus one catty on. So we have B plus one. We want to first calculate its total valence electrons. So we would find beryllium on a periodic tables in group to a corresponding to two valence electrons. So we have for our two atoms of beryllium multiplied by sorry, there are two valence electrons. This would give us four electrons total. However, we have a plus one charge here, meaning that we would lose one electron because we recall that a Catalan charge means we lose that number of electrons leaving us with three electrons total for our molecular orbital diagram. And because we recognize that beryllium is across the second period of our periodic tables, we're going to begin our molecular orbital diagram at the sigma two s bonding molecular orbital where we'll place our first two electrons, we're then going to move up to the sigma anti bonding to s molecular orbital where we'll place our third electron there and it can be in any direction spin of your choice. Now according to the prompt, we see that we're going to compare the bond order where we manipulate our molecular orbital diagram and remove an electron but let's go ahead and first calculate the bond order while the electrons are staple here. So we would say we have one half times are electrons in the bonding molecular orbital which again was the molecular orbital without the abstract. So we would count two electrons in the bonding molecular orbital minus our electrons in the anti bonding molecular orbital which is our orbital with the asterisk where we have just one electron here and so we would have two minus one which would give us one and then one half times one which would give us a bond order equal to one half for our B. E plus one cat island. So according to the prompt we would remove one electron from our configuration. So we will just remove this one here meaning we would just be left with our two electrons and our bonding sigma two s molecular orbital. So to calculate the bond order based on this change, we would now have just one half multiplied by our two electrons in our bonding molecular orbital. And now we would say minus zero electrons in our anti bonding molecular orbital since we've removed our electron from the higher energy orbital which was our sigma anti bonding two S molecular orbital where we had that one electron that we removed. So that would give us 2 0 which is two. And then 1/2 times two, which would give us a bond order equal to one. And so we can say that one is greater than one half. And so therefore removing an electron from our P. A. Plus one Catalan increases bond order. And so for our final answer, we would say that removing this electron increases the bond order. So what's highlighted in yellow is our final answer. If you have any questions, please leave them down below. Otherwise, I'll see everyone in the next practice video.