Explain the following: (c) The O22 + ion has a stronger O—O bond ...

Question

Explain the following: (c) The O22 + ion has a stronger O—O bond than O2 itself.

Accepted Answer

hey everyone in this example, we need to compare the bond strength between a sulfur two plus Catalan molecule and a sulfur neutral molecule. So we should recall the concept of bond order and to find bond order. We would calculate it by taking one half, multiplied by our number of electrons and our bonding molecular orbital. And subtracted from this, we would have our number of electrons in our anti bonding molecular orbital. So we end the bracket here and we should recall that there is a direct relationship between bond order and bond strength. So this is what we're going to need to calculate in order to determine which molecule has the stronger or weaker bond. So right now we're going to consider our neutral sulfur molecule and we want to draw out the molecular orbital diagram. But before we can draw this diagram out, we want to recognize how many valence electrons our sulfur molecule has. So to calculate total valence electrons, we would recall that sulfur on our periodic tables is located in group six. A and recall that the group number corresponds to a number of valence electrons. So we would take our two atoms of our sulfur in this molecule and multiply it by six valence electrons. So this is going to give us a total of 12 electrons total in the valence show. And so now we can go ahead and use this to draw out our molecular orbital diagram. So beginning with our first level, we want to recall that sulfur is located across period three on our periodic tables. So we would begin with the sigma. sorry, we would begin With our σ three us molecular orbital bonding orbital And here we would fill in two electrons. So we have 12 electrons of our 12. So we need to fill in 10 more. Our next level up is going to be the anti bonding. So we would have the sigma asterix three s bonding molecular orbital for the anti bonding molecular orbital where we would also fill in two more electrons. So now we have four electrons filled in. We are next going to have our sigma three p bonding molecular orbital where we're going to fill in for two electrons and then we move up and we have our pi three P bonding molecular orbital where we're going to fill in a total of four electrons. And actually before we pair them up, we want to always fill in one in the first orbital one in the second orbital. And then we pair our orbital's up or electrons up in the orbital's. And then right now we're going to count a total of 2468, 10 electrons. We need to fill in 12 for our sulfur molecule here. So we're going to move up to the next orbital where we have our anti bonding pi three p anti bonding molecular orbital, which has to a stroke here. And we would fill in just two more electrons. So one and then two following our poly exclusion principle, which is why we filled them in the orbital In that fashion. And so now we have our 12 electrons filled in our molecular orbital diagram. And we can see that we have based on this diagram. four electrons in the anti bonding molecular orbital's where we have to hear. So we counted two here at the three p pi anti bonding molecular orbital and then we counted two more here in the sigma three s anti bonding molecular orbital. And then we can say we also have in our bonding molecular orbital's 2468. So we have eight electrons in our bonding molecular orbital's. And so now to calculate bond order, we can say that we would have one half. And for a number of electrons in the bonding, we said we have eight -4 in the anti bonding and this gives us a bond order Equal to a value of two. This is our bond order of a sulfur di atomic molecule. And now we're going to compare this to the bond order for a di atomic cat ion of sulfur where we have one half. Now for the number of electrons in the bonding molecular orbital, that will not change. That's still going to be eight electrons minus our number of electrons in the anti bonding molecular orbital. And that is what is going to change due to the fact that we would recognize that we're going to remove these two electrons, we're going to remove these two electrons here in our anti bonding molecular orbital here, the pi three P one, two not include in our orbital diagram for the two plus Catalan of sulfur. And so removing these two electrons here, we would just only be left with the anti bonding electrons in the three s sigma orbital. So that would just be too. And so we would say we just have two electrons in the anti bonding molecular orbital for a two plus cat ion of sulfur. And so this changes our bond order to now, a value equal to three. So just to be clear, we lose these two electrons here and that is why we would say we have only two electrons in the anti bonding molecular orbital. So instead of lose, we would say we would remove we would remove these two electrons here from the highest energy anti bonding molecular orbital which is our three p pi anti bonding molecular orbital. And so we can say therefore the bond order of the sulfur to Pluskat iron is greater than the bond order of the neutral molecule of sulfur. And so thus our cat eye on R. Two plus cosine of sulfur has a stronger bond. Or rather it has a higher bond order and a stronger bond. And so this would be our final answer here. To complete this example is that our two plus Catalan of sulfur is a stronger bond compared to our neutral sulfur molecule. So I hope that everything I reviewed was clear. Everything highlighted in blue represents our final answer, and I will see everyone in the next practice video.

Explain the following: (c) The O22 + ion has a stronger O—O bond than O2 itself.

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