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Ch.10 - Chemical Bonding I: The Lewis Model

Chapter 10, Problem 57a

Write the Lewis structure for each molecule. a. SF2

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Hello everyone. So in this video we're going to go ahead and write the lewis structure for arson. So The formula for that is going to be given to us in this problem as ASH3. So first we can go ahead and count the number of valence electrons. So for R A. S. H. Three compound, A. S. Has five Total Valence Electrons. And hydrogen has one valence electron per atom in this molecule. We can see that there's going to be three atoms of hydrogen. So multiply one by three to give us total of three vans electrons coming from our hydrogen. So the total amount of vans electrons in our compound, well five plus three is going to be eight. So we have a total of eight valence electrons in our molecule. So hydrogen can only form one bond because of the duet rule. But R A. S. Can go ahead and then be our central item. So we have the A. S. B in the central atom and again hydrant can only have one bond so automatically know that we have three hydrogen is bonded to R. A. S. So let's see. Each bond represents two electrons. We here have 12 and 33 bonds. Three times two is a total of six electrons that we have used. So there's two vance electrons left since we said that the higher regions need two electrons only because of the duet role A. S. Needs to have eight because of the octet rule. So again we have six electrons surrounding it because of the three bonds. The remaining two can go ahead and go on to our central item to go and fulfill its octet. So again, we have three hydrogen attached to our A. S. Central atom, and this two remaining electrons will go as a long pair to fulfill the central items octet, so the Lewis structure, or arson, is going to be this right over here. All right, thank you all so much for watching.