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Iodine forms a series of fluorides (listed here). Write Lewis structures for each of the four compounds and determine the formal charge of the iodine atom in each molecule:(a) IF(b) IF3(c) IF5(d) IF7
a) Thus, in IF, the formal charge on I atom is $[0]$b) Thus, in IF; ithe formal charge on I atom is $[0]$c) Thus, in IF_; the formal charge on I atom is $[0]$d) Thus, in IF_; the formal charge on I atom is $[0]$
00:57
Aadit S.
03:27
David C.
Chemistry 101
Chapter 7
Chemical Bonding and Molecular Geometry
Chemical Bonding
Molecular Geometry
Veronica M.
December 14, 2020
electron pair
Rice University
University of Maryland - University College
Brown University
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Yeah. So for I F we have seven plus seven is 14 electrons. And you want eight plus 8, 16. Mhm. So that's gonna be two electrons that are shared divided by two is going to be mhm one pair. And that means they'll each have three uh lone pairs around them. And then the formal charge on iodine is going to be, so there's seven around it and it has seven valence electrons. So formal charge is zero. Mhm. Next we have IF three. So we have um seven times four is 28 And eight times 4 is 32. So we want to share four electrons which is equal to two pairs of electrons. Okay, okay. Yeah, mm hmm. Each of them wants eight CF 18s was 32, seven times first. Meet four. Okay. Yeah. So the actually this doesn't this is invalid because you can't have two pairs and there's three atoms. So I'm gonna disregard it and assume that there's an expanded architect. So we'll start with single bonds here. And then if we put Yeah, two lone pairs and then we put 1234, 9 10 11 12 13 14 15 16 17 18 1920, 22, Yep. So that adds up to 28. All right, So then the formal charge on iodine is going to be you have 1234567 Electrons around it and seven valence electrons is a formal charge of zero. Okay, then we have I F five and d is I have seven. Both of these are expanded octet, which is calculate the number of electrons. So we have in this case it's going to be six times seven is 42 black trunks. So if we have five bonds to the florian's and then each flooring has okay, It's three lone pairs to satisfy. It's dr That will come out to 40 electrons eight times 5. So then we'll need another lone pair here and then we have 1234567. So again, formal charge of zero on night Lastly we have eight times seven is going to be 56 electrons. Yeah, Yeah, 34567. I'm not going to draw all the lone pairs, but each florida has three lone pairs around. Uh Actually I will because I guess it'll look cute. Why not? All right. So if you have seven florentines And each of them has eight electrons. Okay, around it. That means that's eight times 7 56. And so you don't need any other lone pairs around iodine. And sure enough, if you have one electron from each of the bond belonging to iodine, that's seven around it. And the formal charge is zero
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