Account for each of the following observations. b. SF 4 exists, but OF 4 does not.

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.

Ch.22 - The Main Group Elements

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McMurry 8th Edition

Ch.22 - The Main Group Elements

Problem 138b

Chapter 22, Problem 138b

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

Verified step by step guidance

insert step 1> Consider the electronic configuration of sulfur (S) and oxygen (O). Sulfur has the electronic configuration [Ne] 3s^2 3p^4, while oxygen has [He] 2s^2 2p^4.

insert step 2> Analyze the ability of sulfur and oxygen to expand their octet. Sulfur, being in the third period, can utilize d-orbitals to expand its octet, allowing it to form more than four bonds.

insert step 3> Consider the valence shell electron pair repulsion (VSEPR) theory. SF_4 has a seesaw shape due to the presence of one lone pair and four bonding pairs around sulfur.

insert step 4> Evaluate the electronegativity and size of the central atom. Oxygen is more electronegative and smaller than sulfur, which limits its ability to form multiple bonds with fluorine atoms.

insert step 5> Conclude that the inability of oxygen to expand its octet and its high electronegativity prevent the formation of OF_4, whereas sulfur can form SF_4 due to its ability to expand its octet and accommodate more bonds.

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Key Concepts

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

Valence Shell Electron Pair Repulsion (VSEPR) Theory

VSEPR theory is used to predict the geometry of molecular structures based on the repulsion between electron pairs around a central atom. In the case of SF4, the sulfur atom has six valence electrons, allowing it to form four bonds with fluorine and maintain a stable structure with a seesaw shape. Conversely, OF4 would require oxygen to accommodate four bonds, which is not feasible due to its limited valence electrons.

Electronegativity and Bonding

Electronegativity refers to the tendency of an atom to attract electrons in a bond. Sulfur is less electronegative than fluorine, allowing SF4 to form stable covalent bonds. In contrast, oxygen is more electronegative than sulfur, leading to instability in OF4 as the oxygen atom would struggle to form four stable bonds without exceeding its octet.

Molecular Stability and Octet Rule

The octet rule states that atoms tend to form bonds in such a way that they have eight electrons in their valence shell, achieving a stable electron configuration. SF4 adheres to this rule, while OF4 does not, as oxygen can only accommodate a maximum of two additional electrons in its valence shell, making the formation of OF4 energetically unfavorable and unstable.