Use Lewis structures to explain why Br 3 - and I 3 - are stable, while F 3 - is not.

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

Ch.10 - Chemical Bonding I: The Lewis Model

Tro - Chemistry: A Molecular Approach 5th Edition

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Ch.10 - Chemical Bonding I: The Lewis Model

Problem 100

Chapter 10, Problem 100

Use Lewis structures to explain why Br₃^- and I₃^- are stable, while F₃^- is not.

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insert step 1> Draw the Lewis structures for Br3-, I3-, and F3-. Start by determining the total number of valence electrons for each species.

insert step 2> For Br3- and I3-, note that the central halogen atom can expand its octet due to the availability of d-orbitals, allowing it to accommodate more than 8 electrons.

insert step 3> In the case of F3-, fluorine is in the second period and cannot expand its octet, as it lacks d-orbitals. This makes it difficult to stabilize the extra electron in F3-.

insert step 4> Consider the size of the atoms: larger atoms like bromine and iodine can better accommodate additional electron density due to their larger atomic radii, reducing electron-electron repulsion.

insert step 5> Conclude that the ability to expand the octet and the larger size of Br and I contribute to the stability of Br3- and I3-, while the inability of F to expand its octet and its smaller size make F3- unstable.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Lewis Structures

Lewis structures are diagrams that represent the bonding between atoms in a molecule and the lone pairs of electrons that may exist. They help visualize the arrangement of electrons and the connectivity of atoms, which is crucial for understanding molecular stability and reactivity. By depicting valence electrons, Lewis structures allow chemists to predict the shape and polarity of molecules.

Resonance and Stability

Resonance refers to the phenomenon where a molecule can be represented by two or more valid Lewis structures, which contribute to its overall stability. In the case of Br3- and I3-, resonance structures allow for delocalization of electrons, distributing charge more evenly and enhancing stability. In contrast, F3- lacks sufficient resonance structures to stabilize the negative charge effectively, leading to instability.

Formal Charge

Formal charge is a theoretical charge assigned to an atom in a molecule, calculated based on the number of valence electrons, the number of bonds, and the number of lone electrons. It helps assess the most stable Lewis structure by minimizing formal charges across the molecule. For Br3- and I3-, the formal charges can be distributed favorably, while in F3-, the high electronegativity of fluorine leads to unfavorable formal charge distributions, contributing to its instability.