The dichromate ion, Cr2O72-, has neither Cr¬Cr nor O¬O bonds. (a) Taking both 4s and 3d electrons into account, draw an electron-dot structure that minimizes the formal charges on the atoms.

Video transcript

Hello everyone today. We have the following problem. The permanganate ion does not contain any oxygen oxygen bonds. Taking into account the forest and three D. Orbital's draw a lewis structure for the permanganate ion with minimal formal charges. So first we want to take into account how many valence electrons we have to work with. So we have our Manganese component here, and Manganese has seven valence electrons. And then we have oxygen here Which is in group six a. So it has six valence electrons and there are four of them. So we have 24 Electrons total for that one. And then we have this negative charge here and this negative charge is going to contribute to the number of electrons that we have and therefore will give us one electron. And so in total we have valence electrons To work with. So we have 32 valence electrons to work with. What's special about manganese is that it can expand its octet. So manganese can expand its octet, and why is that? It's because it has an empty D orbital. In fact any atom below the Third period or third row in the periodic table can expand its octet. So this is a very useful tool that elements can use when bonding. And so because of this, it can form as many bonds as there are valence electrons. Therefore this can form seven bonds. Total oxygen prefers to form up to two Up to two bonds. So oxygen conform One bond or two bonds depending on its use. And so if we have manganese here and the center and we want to surround it by seven, we cannot surrounded by seven individual single bonds because there are only four oxygen bonds. So what we can do is we can accommodate for that. I'll draw it out. We can draw a double bond here, we can draw a double bond here, We can draw a double bond here and we only have one more left. So we have a single bond there and notice how we fulfilled the rules. We have seven individual bond lines that manganese has and we have up to two bonds for each oxygen. And so now we need to determine how many dots and lines with the formal charge that we have. So we will start with the formal charge of our mega knees. And formal charge is going to be the number of valence electrons minus our bond lines minus. How many electrons we see individual electrons we see. And so for valence electrons, we said that was seven mega knees has seven bond lines on it and there are zero dots around there. So our formal charge for manganese is going to be zero. Next. We need the formal charge for our oxygen. But with our single bond, that oxygen that's going to pertain or that's going to contain the same formula which is valence electrons minus the bond lines minus the number of electrons around that individual oxygen, oxygen is a group six a element. So it has six valence electrons. We see that it has one bond line. And when there's only one bond line We can draw in six individual electrons 6 -1 -6 gives us a formal charge of negative one for our single bonded oxygen. So we're just gonna draw a negative charge there and then we need to calculate our formal charge for oxygen. But the double bonded oxygen's Same equation valence electrons minus our bond lines minus our electrons that we see valence electrons. We have six online. We have to and when we have two different bond lines to oxygen, we can accommodate by drawing four different valence electrons. So we'll say -4, -2 -4 is zero. So our formal charge is going to be zero. You don't need to add any charges to that. And so if we have our compound here, what we can do is we can draw this compound here in brackets with the same connectivity. But instead of placing that negative charge just in an oxygen, we can apply that outside the brackets there. And with that we have drawn our structure overall. I hope that this helped. And until next time