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 (d) ClF.

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hey everyone in this example, we need to predict the bond order and magnetic behavior of osmium sulfur. If we assume that the energy level diagrams for the homo nuclear di atomic molecules can be applied to hetero nuclear di atomic ions. So because we need to predict magnetic behavior, we want to recall upon the following terms first being die a magnetic, which we should recall is corresponding to the electrons being paired in the configuration for a given atom or molecule, meaning that that atom or molecule is capable of repelling magnets. We also want to think of our second term para magnetic. So we should recall that for a para magnetic atom or molecule the electrons are paired in the configuration. We also need to recall our formula for bond order. So we should recall that's calculated by taking 1/2 and multiplying by the number of electrons in the bonding molecular orbital of a molecular orbital diagram for that molecule minus our number of electrons in the anti bonding molecular orbital for a given molecule. So we're going to take our molecule for oxygen sulfur and we're going to draw out its bonding molecular orbital's or its molecular orbital diagram rather. So we should first calculate the total valence electrons. So looking at oxygen on the periodic table, we find it in group six A. And recall it that corresponds to six valence electrons. So for our one oxygen atom, we're going to multiply it by six electrons, we're then going to add our electrons from sulfur. So recognize that sulfur is also in group six A. So our for our one sulfur atom, we would multiply this by six electrons and so for our total event electrons, this gives us six plus six which is 12 electrons total for our valence molecular orbital diagram. And so we're going to recognize also that for our lowest energy orbital, we would begin at the bonding sigma two S molecular orbital because oxygen is located in the second period of the periodic table and we're going to place our first two electrons here. Moving up in energy. We're going to go to the anti bonding sigma two S molecular orbital where we'll place our next two electrons and I'm just going to scoot this up so that we have enough room to draw our molecular orbital diagram. So next we have our sigma two p bonding molecular orbital where we'll fill in our next two electrons. And then higher up in energy we have our pi two p bonding molecular orbital with two orbital's where we'll fill in 123 and four electrons pairing them up last. So so far we have 2468 10 electrons that we filled in. And now we need to go ahead and fill in two more electrons for our 12 total valence electrons. So higher up an energy we would have our pie anti bonding to p molecular orbital with two orbital's where we'll fill in our last two electrons. So one electron here and one electron here because we want to honor the poly exclusion principle. And so this is our molecular orbital diagram for our oxygen sulfur molecule. And what we can clearly see is that we have here to um paired electrons in our pi two p anti bonding molecular orbital. And so we would say that since we have these two unpaid electrons, therefore our oxygen sulfur molecule is para magnetic and capable of attracting a magnet. Now, we also need to calculate bond order. So we would say for the bond order of the oxygen sulfur molecule, we would have won half multiplied by the electrons in the bonding molecular orbital. So those are our molecular orbital's without the asterix. So we would count too 468. So we would have eight electrons in the bonding molecular orbital subtracted from the electrons and the anti bonding molecular orbital which are the molecular orbital's with the asterix. So we would count too 34. So this would be four electrons in the anti bonding molecular orbital. And so to calculate for bond order, we should have eight minus four, which is four and then one half times four, which would give us a bond order for the oxygen sulfur molecule equal to a value of two. And so for our final answers, we've determined that the bond order for the oxygen sulfur molecule is equal to two, and oxygen silver is a para magnetic molecule. And this would be our final answers here, highlighted in yellow to complete this example. So since the magnetic behavior is para magnetic, we can say that therefore oxygen and sulfur is capable of attracting magnets just to expand on our answer. 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 (d) ClF.

Verified Solution

Key Concepts

Bond Order

Molecular Orbital Theory

Magnetic Behavior