If we assume that the energy-level diagrams for homonuclear
diato...

Question

If we assume that the energy-level diagrams for homonuclear
diatomic molecules shown in Figure 9.43 can be applied
to heteronuclear diatomic molecules and ions, predict the
bond order and magnetic behavior of (b) NO+.

Accepted Answer

hey everyone in this example, we need to predict the bond order and magnetic behavior of flooring oxide as an ion. If we assume that the energy level diagrams for the home, a nuclear di atomic molecules can be applied to hetero nuclear di atomic ions. So our first step is to recall our terms for magnetic behavior. The first being di magnetic, we would recall that a atom or molecule that is dying magnetic has all electrons paired in its configuration and therefore that atom or molecule is capable of repelling a magnet. We want to also recall our term for para magnetic, where all or where at least one electron is UNP aired in the configuration for an atom or molecule and therefore that atom or molecule would be capable of magnetism. So we're focusing on our molecule being the flooring oxide an ion. So we want to begin by drawing out its molecular orbital diagram but we need to calculate total valence electrons before we do so. So we would find flooring on our periodic tables in group seven A. So for our one florian atom in this compound, we would recall that group number and valence electrons are Or correspond with one another. And so because Florida is in group seven a. We would have seven valence electrons Added to this. We want our electrons from oxygen. So we would find oxygen on the periodic table in group six a. We have one oxygen atom here and we're going to multiply by its six valence electrons. So we would have six plus seven. And this would give us a total of 13 valence electrons for our molecular orbital diagram. We want to recognize that oxygen on our product table as well as florian are across the second period in group six A and seven A. And because they're across the second period, we would begin our molecular orbital diagram at the bonding sigma two s molecular orbital where we're going to place our first two electrons, we still have 11 more electrons to fill in. So we're going to move up an energy to the anti bonding sigma two S molecular orbital where we'll place our next two electrons further up in energy. We have our sigma bonding to p molecular orbital where we'll place our next two electrons and then we would have the pie to p bonding molecular orbital where we have two orbital's P Z and P Y. We will place our next four electrons. So we would have 123 and four following our poly exclusion principle. Right now, we have 10 electrons of our 13 filled in. And before we continue, I just want to clarify that because we also have this negative anti uncharged here. This means that we gain one electron and so we actually would to be correct on our total electrons for our molecular orbital Diagram would have a total of 14 electrons total since we have that an ion charge. And so what we need to do is fill in four more electrons in our molecular orbital diagram. And so further up in energy we would have our pie anti bonding to p molecular orbital which has two orbital's PX and py we're both fill in our next four electrons. So we would have 123 and four pairing up our electrons last. And so now this is our molecular orbital diagram for our flooring, oxygen and ion. And what we can see from this diagram is that we have therefore zero um paired electrons. So we can say thus the flooring oxygen and ion is dia magnetic incapable of repelling magnets. The next step in this example is to calculate bond order. So we want to recall that to calculate bond order. We're going to use the following formula where we take one half times the number of electrons in our bonding molecular orbital. And then subtract that from our number of electrons in our anti bonding molecular orbital. And so to calculate our bond order for the flooring oxygen, an ion, we would have one half multiplied by the electrons in the bonding molecular orbital which on our molecular orbital diagram are the orbital's that have no asterix. So we would count 2468. So we have eight electrons in the bonding molecular orbital subtracted from the electrons in the anti bonding molecular orbital which are the orbital's with the asterix. So we would count two, four and 6. So we have six electrons in the anti bonding. And sorry, it's getting cut off. So we'll scoot this over. So we have six electrons in the anti bonding molecular orbital. And so now we would get for our bond order of our flooring oxygen and I on a value 8 -6 would be two and then we would have one half times two, which is going to give us a bond order equal to one. And so for our final answers to complete this example, we've determined that flooring or flooring oxide as an ion is di magnetic and capable of repelling magnets and has a bond order equal to a value of one. So everything highlighted in yellow represents our final answers. If you have any questions, please leave them down below and I will see everyone in the next practice video.

If we assume that the energy-level diagrams for homonuclear diatomic molecules shown in Figure 9.43 can be applied to heteronuclear diatomic molecules and ions, predict the bond order and magnetic behavior of (b) NO+.

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