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Read the labels of several commercial products and identify monatomic ions of at least four transition elements contained in the products. Write the complete electron configurations of these cations.
Chemistry 101
Chapter 6
Electronic Structure and Periodic Properties of Elements
Electronic Structure
Periodic Table properties
Carleton College
University of Central Florida
Rice University
University of Kentucky
Lectures
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Chemistry is the science o…
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In chemistry and physics, …
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Read the labels of several…
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Write out the full electro…
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Write the electron configu…
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Write electron configurati…
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Partial (valence-level) el…
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Write the condensed ground…
01:03
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Copy and complete the foll…
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On the periodic table belo…
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Use the periodic table to …
All right. So for this problem, we're asked to take a look at some of the labels sitting around our house to find some ionic species and then write out the electron configurations for the ions that we find. So for this, I took a look at a protein powder that was sitting around my house. That's where the zinc oxide comes from. The last three actually come from a can of my dog's food. Um, so we have fair sulphate, manganese sulfate, copper protein eight. Um, the important thing for this problem is we're just going to be looking at the cat I own, so don't worry too much about the lions here. Um so, t give these charges. These, actually, um, conveniently, in a way, are are already all represent the same charge. So they're all going to be the two Pluskat ions for each of these medals. Um, So what we're going to do is just for each of these, walk through with their electron configurations would be in full reading out form, So starting with zinc, actually, all of these medals are also in the top row of the transition metals, which makes us a little bit more straightforward. Um, zinc, for example, is a two plus I on which means that if we walked on our periodic table, it's going to have fully filled one s orbital, fully filled to us orbital fully filled to pee, um, fully filled three s and a fully filled three d. So it's it's in group 12 which means that it has a total of 12 electrons in this Vaillant show spread between the four s and the three D at ground state in an uncharged version. So we know that it's going to have 10 electrons in the three d State and the two electrons that is going to you lose or going to come from this two s orbital shifting now to iron. Same thing is going to hold true. It's going to have the same basic, uh, sort of it's gonna have. The base configuration could be the same as are gone for all of these are goingto have fully fill it. Orbital's from one s through three p using our, um, standard method of writing them out in order. As I've done here, the next thing we need to consider with iron the irons and group ate. That means that there's eat valence electrons, which means that we take two of them away. Um, we're going to be left with a total of six. So again, normally they're going to get pulled from the four s orbital, so we would expect to write is three d six. However, you should recall that the transition metals have this special property where if they're one electron away from forming 1/2 filled their fulfilled orbital, Um, because they have one electron too many. Oftentimes they will take that one electron and push it into a different orbital to accommodate such a change. So this instead of having three d six, this electron is actually going to get pulled from that three d orbital, and it's actually gonna get pushed up into the four s. So what we're going to see is this iron this Ferris cat eye on is going to be falling into this three d five with the four s one. And that's because the three d five is 1/2 filled sub show for the D orbital's, and this is more favorable than having six electrons in the orbital set by itself. These same principles, We're going to apply it to the remaining two cat ions. So we look at manganese again. Uh, I so electric initial configuration toe are grown and then we add in the final valence electrons. So because it lost to the for us is now gone. So we're left with three d five and same thing's going to be true for copper. We take it all the way up through, um, the one s orbital to s to p et cetera. We have fully filled building up to this point again. We're going to remove the two electrons from the four s and we're going to left B B left with three electrons in the three D exceptional.
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