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Ch.9 - Chemical Bonding I: The Lewis Model

Chapter 9, Problem 107c

Draw the Lewis structure for each compound. c. H3AsO4

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Hello everyone. So in this video we have to go ahead and draw a lewis dot structure for this compound right here. So first things first, go ahead and calculate the total number of electrons. So we have our hydrogen atoms and oxygen. So for hygiene adam that has one valence electron for a bombing that has seven valence electrons because it's in group 78 For our oxygen atom. That's in group six a. So I have six fans electrons in the compound. We have two oxygen atoms. So multiply that number by two. So six times two is 12. And then adding those three numbers up we get a total of 20 vans electrons in this compound. So next we have to go ahead and draw a lewis dot structure. So are hiding atom that has a duet rule. So I can only form one bond comparing our bromate atom and oxygen atom. Our booming will have a lower electro negativity and therefore will be our central atom. Again we have our bro me connected to two hydrogen or oxygen and then one of the oxygen's will be connected to a hydrogen. Usually our oxygen atoms like to have two bonds. So we'll go ahead and add when we're bond to the terminal oxygen. So to fulfill its octet rule on the left oxygen here, let's go ahead and add two lone pairs for our bro. Mean let's see here we have 123 bonds. Each bond will have two events electrons. So that's six so far it seems that we need to add one more lone pair to a booming atom For oxygen. It needs to lone parents. So let's see here. So far, we have to use 123456789, 10, 11, 12, 13, 14, 15, 16, 17, and 18. So we still have two vans electrons left. Which one can we add the last long pair up onto? So because our oxygen's are very happy and hydrogen can only form one bond. The only adam left that can still hold the remaining last two extra band electrons will be our romy. And so because of that, this will be our final Lewis structure for this acid. All right, thank you all so much for watching.