Do you expect a compound with vanadium in the +2 oxidation state to be an oxidizing or a reducing agent? Explain.

Video transcript

Welcome everyone. Let's take a look at our next question. Will the substance having manganese with an oxidation number plus two perform as an oxidizing or reducing agent. Justify your answer. All of our answer. Choices below start with the substance having manganese with an oxidation number plus two. So I'm not going to read that out every time just the first time. So choice A says the substance having manganese with an oxidation number plus two performs as a strong reducing agent because of its high electron affinity. Choice B this substance performs as a reducing agent because it seeks greater stability through an increased Z effective. To I see this substance performs as a reducing agent because of its tendency to donate electrons or choice D this substance has is neither an oxidizing nor a reducing agent because of its stable electron configuration. Well, first, let's remember what oxidizing and reducing agents are. Notice from our answer choices, we know it's not an oxidizing agent because that's not one of our choices. Our choices are basically strong reducing agent, reducing agent or neither one. So reducing agent is something that reduces the oxidation state. If the oxidation state is reduced, its oxidation number becomes more negative. So it gains electrons. So reducing agent is going to give electrons. So I sometimes find that kind of tricky to keep straight. I often start with what it means to be reduced because that's the easiest for me to remember reduced. I think of becoming more negative in its oxidation state. So I usually have to think all the way through that way as I find it a little tricky to keep those straight and you don't want to get mixed up. So we know it won't perform as an oxidizing agent. So it's going to be giving electrons. Well, to think about the likelihood of this happening, we'd want to look at our electron configuration for this element. So manganese neutrally, we look that up on the periodic table. It's in groups seven B, it has seven valence electrons, its closest noble gas is argon. So we put a R in brackets, our seven valence electrons will be four S two and then 3d 5. So that means that manganese two plus which is missing two electrons, we'll have that argon in brackets, we're going to lose our four S two electrons. So it will just be 3d 5. Well, these half fell D orbitals are very stable configurations. Relatively speaking, if we think about our five D orbitals, five electrons, you've got a single electron in each orbital. So they kind of like that sense of each orbital having one, but you don't have that slight repulsion of electron pairs when you start filling them. So it's not as stable as a 3d 10 configuration, but it's pretty stable. So it would not be a strong reducing agent because it definitely doesn't have a high electron affinity due to its relatively stable electron configuration. So let's cross off choice. A but now we just need to figure out is it going to be a reducing agent or neither just how stable is this? Well, let's think about what would happen if it donates an electron, it goes down to 3d 4, which is not enormously less stable. So it's not a huge loss of stability. And what it does have is an increased z effective the subscript eff so let's recall the effective is the effective attractive force that the electrons feel from the positively charged nucleus. Well, as we go from left to right along the periodic table and this is ad block, we have an increased the effective because our electrons are all in the same shell. So they're not getting farther away. So when we have our, when we're comparing our MN with its 3d 5 or excuse me, my MN two plus with its 3d 5, and then we imagine it losing an electron and becoming 3d 4, we have the same number of protons in the nucleus, but their positive charge is sort of divided up amongst fewer electrons. So this essentially increases the Z effect of that, those four electrons feel which increases the stability of that MN three plus form once you've oxidized your MN two plus. So we're actually going to choose Choice B that it can perform as a reducing agent though not a strong one because that 3d 4 configuration gives it greater stability through an increase the effective, the greater attractive force felt by the electrons keeps that form more stable. So we'll go ahead and choose. Choice. B. Uh Choice C says it performs a reducing agent because of its tendency to donate electrons. You wouldn't really describe this as the tendency to donate electrons as an explanation for why it can do this because that isn't going to be the main explanatory factor. And then as we said, we'll cross out choice d neither oxidizing or reducing because that greater the effective does allow it to act as a reducing agent even though it has the stable electron configuration. So once again, will the substance having manganese with an oxidation number plus two perform as an oxidizing or reducing agent. Justify your answer. Choice B it performs as a reducing agent because it seeks greater stability through an increased Z effective. See you in the next video.