Although I₃^- is a known ion, F₃^- is not. (b) One of your classmates says that F₃^- does not exist because F is too electronegative to make bonds with another atom. Give an example that proves your classmate is wrong.

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welcome back everyone in this example, we have the phosphate an ion. As a stable ion of phosphorus. We need to use lewis structures to explain why our or throw nitrate an ion does not exist. Even the nitrogen and phosphorus or from the same group on the periodic table. Which we would recall is group five A. Or group five on the periodic table. Because we recall that groups are the columns going down our periodic table. So to begin, we're going to start by drawing the lewis structure of our phosphate an ion and we want to figure out total valence electrons first in order to do so. So beginning with our atom of phosphorus, we have just one. We're going to recall that phosphorus on our product table is in group five A. Which we recall corresponds to its five valence electrons. So put a colon and we'll save five valence electrons. Moving on to our atoms of oxygen. We have oxygen which were causing group six A corresponding to six valence electrons. However, because we have that subscript of four next to oxygen, that means we have to multiply this by four atoms of oxygen. And this is going to give us 24 electrons contributed from oxygen. And then we recognize that we have that minus three and ion charge. So we have a minus three and I in charge. And so we would say therefore we would add three electrons. So for our total valence electrons, we would say we have 32 electrons total. And so drawing out our structure, we have our phosphorus surrounded in the middle. We have four oxygen surrounding that. So 123 and four. We're going to make our base connections to our central atom here. So one, 234 bonds where each bond represents two electrons used. So we would count 2468 electrons used so far. And let's go ahead and make this aligned better. Sorry about that. So We have -8 electrons. This is going to then leave us with 24 and sorry about that wrong pen. So 24 electrons next we want to recognize that phosphorus is specifically a period three element and we would recall it can therefore have an expanded octet. And so we can go ahead and make a fifth bond to one of our oxygen's here will go with this top oxygen so that we actually now have 2468, 10 electrons surrounding our phosphorus atom. So that would use up another two electrons. We make that bond leaving us with 22 electrons which we're going to have to fill in as lone pairs. So we would have because we recognize that we have a total of six valence electrons around oxygen and we would have total here, Making this oxygen stable and having a formal charge of zero. And then moving on, we have four electrons. We filled in. So six, 8, 10, 12, 14, 16, 18, 20. And then 22. And as you can see each of these oxygen's here have three sets of lone pairs around themselves, meaning that they now have 1234567 valence electrons making them unstable with a formal charge of minus one on each of these oxygen's. So we'll fill that in for our structure And this uses up our total 22 electrons left. So we have the proper structure. Now we want to base our explanation off of this molecule that we've just drawn. So we would recall that on our periodic table. Nitrogen lies across Period two and so we would say therefore nitrogen cannot have an expanded octet like phosphorus because as we said above, phosphorus is a period three elements. And we recall that period three elements and below can have expanded octet. It's with more than eight electrons around themselves in a lewis structure. And so because we understand that they are in different periods, but rather even though they're in the same group, but they're in different periods. The structure of our anthro or sorry, or throw nitrate and I on here will not exist. So we can say thus our or throw nitrate and ion cannot exist because nitrogen does not have an expanded octet. And so this would be our final answer to complete this example. So I hope that everything I explained was clear if you have any questions, please lead them down below. And I will see everyone in the next practice video

Although I₃^- is a known ion, F₃^- is not. (b) One of your classmates says that F₃^- does not exist because F is too electronegative to make bonds with another atom. Give an example that proves your classmate is wrong.

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Key Concepts

Electronegativity

Covalent Bonding

Polyatomic Ions

Although I3- is a known ion, F3- is not. (b) One of your classmates says that F3- does not exist because F is too electronegative to make bonds with another atom. Give an example that proves your classmate is wrong.

Recommended similar problem, with video answer:

Verified Solution

Key Concepts

Electronegativity

Covalent Bonding

Polyatomic Ions

Although I₃^- is a known ion, F₃^- is not. (b) One of your classmates says that F₃^- does not exist because F is too electronegative to make bonds with another atom. Give an example that proves your classmate is wrong.