Draw a Lewis structure for the following polyatomic ions: b. Sulf...

Question

Draw a Lewis structure for the following polyatomic ions: b. Sulfite, SO²⁻₃

Accepted Answer

Welcome back, everyone provide the correct Lewis structure of the poly atomic ion phosphate P 043 minus. We need to begin by calculating total valence electrons. So beginning with phosphorus recall that phosphorus is located in group five, a corresponding to five valence electrons. We then have our atom oxygen, which we recall is in group six a corresponding to six valence electrons. But notice that we have a subscript of four. So we multiplied by four atoms of oxygen which gives a contribution of 24 electrons from oxygen in our structure. So adding our electrons together valence electrons together, we have a total of 29 valence electrons from just phosphorus and oxygen. But recognize that this minus three anti in charge means that we gain Three electrons. So we would add three electrons to this total, which would give us a total of now 32 valence electrons total for our structure. Next, recall our trend for electro negativity on our periodic table which increases towards the top right of our periodic table. So it increases in the direction of flooring. And based on this trend, we can understand that our electro negativity of phosphorus is going to be less than the electro negativity of oxygen. And so therefore, phosphorus is our central atom because we always place the least electro negative atom in the center. So we have phosphorus surrounded by four oxygen atoms in which we can begin by recalling the bonding preference of oxygen, which is to have two bonds and two lone pairs on itself. And being that phosphorus only has five valence electrons. And we've currently have to directly attached valence electrons in covalin bonds to this first oxygen. We're going to have to use the rest to make single bonds with our three other oxygen's. So in adding these electrons, we've now used a total of 2468, 10, 12, 14 of our 32 valence electrons. So we can subtract that from our total. And now that leaves us with a balance of 18 valence electrons. So now we want to recall that as we stated, oxygen should have six valence electrons around itself. And each of our three other oxygen's only have one directly attached valence electron. So we're going to have to add the remaining 18 valence electrons as lone pairs on oxygen. So we can count 2468, 10, 12, 14, 16 and 18, Using up our last 18 valence electrons leaving us with a balance of zero to complete our structure here. Notice that these oxygen's still have one extra valence electron attached to them because we have three bonds and one of the bonds or sorry, we have three lone pairs I met and one of the valence electrons attached to oxygen is shared in a covalin bond to our phosphorus atom in the center. And so that means that we have one bonding electron and a total of six non bonding electrons. These make up a total of seven directly attached vance electrons to this oxygen atom, which is one more than it would prefer giving a formal charge of minus one for each of these three oxygen atoms bonded. In the same way, we can also use the calculation where we take formal charge and find it from taking the group number of our atom subtracted from the number of bonds to that atom plus non bonding electrons. So for our oxygen with two bonds, which is bonded to its bonding preference, we see that it's group number is six. And again, we're calculating its formal charge. We see that it's group number is six a so six minus its number of bonds being two plus, it's number of non bonding electrons being it's two lone pairs which some to a total of four non bonding electrons. This would give us six -6 which is equal to zero. So that's its formal charge. And then for our formal charge of our oxygen's with just one bond which are all bonded. In the same way we have it's group number six minus its number of bonds. One plus, it's non bonding electrons being a total of three lone pairs. So that gives a contribution of six non bonding electrons. So that would give us six minus seven, which would give us positive one or sorry, negative one as the formal charge. Since again, this is one more than it prefers to have, which is £6 electrons to be stable. And so, in summary, each of these negative one formal charges will add up to a net formal charge of minus three or three minus for our structure. And this would be our correct answer as the best Lewis structure for our phosphate an ion. So I hope that everything I explained was clear, our final answer corresponds to choice, seeing the multiple choice and I I'll see everyone in the next practice video.

Draw a Lewis structure for the following polyatomic ions: b. Sulfite, SO²⁻₃

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