Account for each of the following observations.
b. SF₄ exists, but OF₄ does not.

Question

Account for each of the following observations.

b. SF₄ exists, but OF₄ does not.

Accepted Answer

All right. Hi, everyone. So for this question, let's go ahead and explain the following oxygen hexafluoride does not form but sulfur hexafluoride does. Now oxygen hexafluoride does not form but sulfur hexafluoride does because a oxygen cannot exhibit an expanded octet. While sulfur can b oxygen has a much larger atomic radius than sulfur C because oxygen is not electron enough to accommodate the bonds formed or d because fluorine can only form bonds with larger atoms like sulfur. So to understand the answer for this question, let's go ahead and start by drawing out what the lowest structure is expected to be for both of these compounds starting off with oxygen hexafluoride or of six. Now recall that oxygen is found in group six A of the periodic table, which means that oxygen is going to contribute six valence electrons to the lowest structure fluorine. On the other hand, is a halogen residing in group seven A which means that it's going to contribute seven valence electrons each now because I happen to have six atoms of fluorine in my structure. I'm going to multiply my seven valence electrons by six. They give me a total of 42 valance electrons. So when I add together my six electrons from oxygen and my 42 from each of my fluorine atoms or all of them combined together, I get a total of 48 valence electrons and recall that sulfur hexafluoride is actually going to have the same number of, of valent electrons because sulfur is also found in group six a of the periodic table, which means that it's also going to contribute six valence electrons to the lowest structure. And in this case, we happen to have the same amount of fluorine atoms which is six multiplied by the number of valence electrons per atom, which is seven to give me a total of 42 electrons. So in the structure of sulfur hexafluoride, I'm also going to have a total of 48 valence electrons. So now I'm going to scroll down briefly to give myself some more space before discussing the structure of the atom more specifically, right, when it comes to the structures of oxygen, hexafluoride and sulfur hexafluoride, recall that both oxygen and sulfur are less electron than fluoride. And additionally, fluorine is a halogen, right, which means that fluorine can only make one bond, which means that in their respective Lewis structures, oxygen and sulfur are going to be the central atom, right. So for 06 oxygen is the central atom and in sf six sulfur is the central atom. So that means that in the proposed Lewis structures for both of these compounds, oxygen and sulfur are expected to have six covalent bonds, one to each individual atom of fluorine, which means that both of them are expected to have a total of 12 electrons or 12 valence electrons. Now, because 12 is not equal to eight, this is going to be in violation of the octet rule. So the question is, can either of these two atoms in this case, oxygen or sulfur accommodate this exception to the octet rule, right? Can they expand their octet to accommodate more valent electrons? And the answer is that only one of these atoms is capable of doing so. So let's go ahead and compare and contrast oxygen and sulfur recall that oxygen is found in period two of the periodic table. Whereas sulfur is found in period three now because sulfur is an element in period three of the periodic table, there is going to be an empty D orbital when it comes to sulfur specifically, right? Because third period elements have this low lying D orbitals that are generally empty, right, or normally empty. So if needed sulfur could potentially use that empty D orbital for bonding, which means that sulfur has the capacity to expand the octet and therefore have more than eight electrons surrounding it. Whereas sulfur or oxygen, excuse me being an element in period two, the periodic table, oxygen cannot expand it octet the way that sulfur can because there is no such thing as an empty D orbital in period two. Now, in addition to this recall that oxygen is highly electron and it is certainly more electron than sulfur, right. So because of this oxygen would not form a compound in which the oxygen atom is going to have an oxidation state that is too highly positive. Because when you consider the structure of of six fluorine has an oxidation state of negative one due to the fact that it is a halogen, which means that oxygen and sulfur respectively must have an oxidation state of positive six to make the structure neutral overall. And so oxygen being highly electron would not be capable of forming a compound in which the oxidation state is so highly positive. So therefore, our answer is going to be option A and the multiple choice right oxygen hexafluoride does not form but sulfur hexafluoride does because oxygen cannot exhibit an expanded octet while sulfur can't. So if you stuck around to the end, thank you so very much for watching. And I hope you found this helpful.

Account for each of the following observations.
b. SF₄ exists, but OF₄ does not.

Verified Solution

Key Concepts

Valence Shell Electron Pair Repulsion (VSEPR) Theory

Electronegativity and Bonding

Molecular Stability and Octet Rule

Account for each of the following observations.b. SF4 exists, but OF4 does not.

Verified Solution

Key Concepts

Valence Shell Electron Pair Repulsion (VSEPR) Theory

Electronegativity and Bonding

Molecular Stability and Octet Rule

Account for each of the following observations.
b. SF₄ exists, but OF₄ does not.