Identifying an element from a given Lewis Structure (3.37)

[Mike Matthews 1D]

Hi. I was hoping someone could explain to me how to do homework problem 3.37. The problem says, "The following Lewis structure was drawn for a Period 3 element. Identify the element." The Lewis Structure shows a central atom "E" surrounded by 3 Cl atoms and 1 O atom. There is a double bond between the central atom and the O atom and single bonds between the central atom and the three Cl atoms. I'm confused because all the atoms in period 3 could have the expanded octet shown in this Lewis structure which means that in order to identify the element one must take formal charge into account. However, I'm not quite sure how to start this problem. Any help would be greatly appreciated.

[Diane Bui 2J]

I'm hoping this is right, but from what I would do is to count up the total number of electrons in the molecule, which seems to me that there are 32 electrons. 21 electrons are from Chlorine due to its 7 valence electrons, and there are 3 Chlorines in the molecule. 32-21 leaves 11 electrons left over. 6 electrons come from Oxygen because Oxygen has 6 valence electrons. That leaves 11-6=5 electrons that should come from E, meaning that it should come from the Group 15 elements. Since the problem states that it is from the 3rd period, the element E should be Phosphorus.
Hope this helps!!

[Camille 4I]

Hi, are you using the formal charge equation to figure out that this is phosphorus?

[Cooper Baddley 1F]

You don't need to use the formal charge equation for this problem just the total number of valence electrons. By finding the that that element accounts for five total valence electrons it points you to group 15 and the problem tells you it's in the third period so you go over to group 15 down to period three and you are on phosphorus.

[Meia Schram 1J]

No formal charge is needed for this problem you just need to figure out how many electrons the element has for this case it is 5e- (found by subtracting the total electrons, 32, by the amount of electrons from known elements 3(Cl) + 1(O) = 3(7)+ 1(6)= 27. Then 32-27= 5 electrons.) We go down the group 15 and across period 3 (given) getting Phosphorus !