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Chapter 22: Problem 46

The SF \(_{5}^{-}\) ion is formed when \(S F_{4}(g)\) reacts with fluoride salts containing large cations, such as \(\mathrm{CsF}(s) .\) Draw the Lewis structures for \(\mathrm{SF}_{4}\) and \(\mathrm{SF}_{5}^{-}\) , and predict the molecular structure of each.

Short Answer

Expert verified
The Lewis structures for \(\mathrm{SF}_4\) and \(\mathrm{SF}_{5}^{-}\) are: * \(\mathrm{SF}_4 \! \! \! \! \! \! \! \! \! \! :\, \dot{\mathrm{S}} (\text{-}\mathrm{F}\text{-})_{4}\) * \(\mathrm{SF}_5^-\! \! \! \! \! \! \! \! \! \!:\, :\, \mathrm{S}^-\! (\text{-}\mathrm{F}\text{-})_{5}\) The molecular geometry of \(\mathrm{SF}_4\) is "see-saw" or "disphenoidal," and the molecular structure of \(\mathrm{SF}_{5}^{-}\) is octahedral.

Step by step solution

01

Identify the central atom for both molecules.

In both cases, sulfur (S) is the central atom, and fluorine (F) atoms are bonded to it.
02

Count the valence electrons for each molecule.

For \(\mathrm{SF}_4\), count the valence electrons as follows: Sulfur has 6 valence electrons, and each fluorine atom has 7. Since there are 4 fluorine atoms, we have a total of: \( 6 + (4 \times7)=6 + 28 = 34 \) valence electrons. For \(\mathrm{SF}_{5}^{-}\), we count the valence electrons the same way, but remember to account for the extra electron from the negative charge: \( 6 + (5 \times 7) + 1 = 6 + 35 + 1 = 42 \) valence electrons.
03

Distribute valence electrons to create a Lewis structure.

For both cases, follow the Octet Rule and distribute the valence electrons in such a way that each atom achieves a full valence shell (except for Sulfur that can accept more than 8 electrons). For \(\mathrm{SF}_4\): * Place a single bond (2 electrons) between the S and each F atom. * Distribute the remaining electrons around the F atoms to complete their octet and around S to use all 34 valence electrons. * \[ \mathrm{S} - \mathrm{F}_4:\,\, \dot{\mathrm{S}} (\text{-}\mathrm{F}\text{-})_{4} \] For \(\mathrm{SF}_{5}^{-}\): * Place a single bond (2 electrons) between the S and each F atom. * Distribute the remaining electrons around the F atoms to complete their octet and around S to use all 42 valence electrons. * \[ \mathrm{SF}_5^-\! :\, :\, \mathrm{S}^-\! (\text{-}\mathrm{F}\text{-})_{5} \]
04

Use VSEPR theory to predict molecular structures.

For \(\mathrm{SF}_4\): * Sulfur has 5 electron groups (4 bonding F atoms and 1 lone pair), so its electron geometry is trigonal bipyramidal. * In a trigonal bipyramidal molecular geometry, lone pairs occupy equatorial positions. * Thus, the molecular geometry of \(\mathrm{SF}_4\) is "see-saw" or "disphenoidal" because one of the equatorial positions is occupied by a lone pair. For \(\mathrm{SF}_{5}^{-}\): * Sulfur has 6 electron groups (5 bonding F atoms and no lone pairs); therefore, its molecular geometry is octahedral. * Since all the six electron groups are bonding pairs, the molecular structure of \(\mathrm{SF}_{5}^{-}\) is octahedral. In conclusion, the Lewis structures for \(\mathrm{SF}_4\) and \(\mathrm{SF}_{5}^{-}\) have been drawn as: * \(\mathrm{SF}_4 \! \! \! \! \! \! \! \! \! \! :\, \dot{\mathrm{S}} (\text{-}\mathrm{F}\text{-})_{4}\) * \(\mathrm{SF}_5^-\! \! \! \! \! \! \! \! \! \!:\, :\, \mathrm{S}^-\! (\text{-}\mathrm{F}\text{-})_{5}\) The molecular geometry of \(\mathrm{SF}_4\) is "see-saw" or "disphenoidal," and the molecular structure of \(\mathrm{SF}_{5}^{-}\) is octahedral.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Valence Electrons
Valence electrons are the electrons located in the outermost shell of an atom, which play a pivotal role in chemical bonding and reactions. Understanding how to count valence electrons is essential for determining an element's capacity to bond with others.

For example, in sulfur hexafluoride (SF4), sulfur has 6 valence electrons while each fluorine atom has 7. By multiplying the number of fluorine atoms by their valence electrons and adding those to the valence electrons of sulfur, we can calculate the total valence electron count for the molecule. It's also crucial to remember to include any additional electrons resulting from negative charges, as seen in the SF5- ion which has an extra electron due to its negative charge.

Correctly determining the number of valence electrons allows us to predict molecular structures using Lewis structures, where we represent these electrons as dots surrounding the atomic symbols.
VSEPR Theory
VSEPR theory, which stands for Valence Shell Electron Pair Repulsion theory, is a model used to predict the shape of individual molecules based upon the extent of the electrostatic repulsion between electron pairs around a central atom.

The idea behind VSEPR is quite straightforward: electron pairs, whether they be in bonding pairs or lone pairs, will arrange themselves as far apart as possible to minimize repulsion. For instance, for SF4, sulfur's four bonding pairs of electrons and one lone pair adopt a trigonal bipyramidal electron geometry for minimal repulsion. However, its molecular shape is known as the 'see-saw' or 'disphenoidal' due to the placement of the lone pair at an equatorial position, modifying the pure trigonal bipyramidal shape.

Similarly, SF5- follows VSEPR theory whereby its six bonding electron pairs spread out evenly to create an octahedral molecular shape, as there are no lone pairs on the sulfur atom to cause further structural deviations.
Molecular Geometry
Molecular geometry is the three-dimensional arrangement of atoms in a molecule. It's determined primarily by the locations of valence electrons around a central atom, taking into account the differences between bonding pairs and lone pairs of electrons. This concept is crucial for understanding the behavior and properties of molecules, such as reactivity, polarity, phase of matter, color, magnetism, biological activity, and intermolecular interactions.

Using the SF4 and SF5- examples from VSEPR theory, we can visualize that the 'see-saw' shape of SF4 arises from four bonding pairs and one lone pair of electrons, while the perfectly symmetrical octahedral shape of SF5- results from six equivalent bonding pairs. The molecular geometry of each molecule can have profound effects on its physical and chemical properties and influences how it interacts with other molecules.

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Most popular questions from this chapter

Write a balanced net ionic equation for each of the following reactions: (a) Dilute nitric acid reacts with zinc metal with formation of nitrous oxide. (b) Concentrated nitric acid reacts with sulfur with formation of nitrogen dioxide. (c) Concentrated nitric acid oxidizes sulfur dioxide with formation of nitric oxide. (d) Hydrazine is burned in excess fluorine gas, forming \(\mathrm{NF}_{3}\) . (e) Hydrazine reduces \(\mathrm{CrO}_{4}^{2-}\) to \(\mathrm{Cr}(\mathrm{OH})_{4}^{-}\) in base (hydrazine is oxidized to \(\mathrm{N}_{2} )\) .

Name the following compounds and assign oxidation states to the halogens in them: (a) \(\mathrm{KClO}_{3},\) (b) \(\mathrm{Ca}\left(\mathrm{IO}_{3}\right)_{2}\) ,\((\mathbf{c}) \mathrm{AlCl}_{3},(\mathbf{d}) \mathrm{HBrO}_{3},(\mathbf{e}) \mathrm{H}_{5} \mathrm{IO}_{6},(\mathbf{f}) \mathrm{XeF}_{4}\)

Account for the following observations: (a) Phosphorus forms a pentachloride, but nitrogen does not. (b) \(\mathrm{H}_{3} \mathrm{PO}_{2}\) is a monoprotic acid. (c) Phosphonium salts, such as \(\mathrm{PH}_{4} \mathrm{Cl}\) can be formed under anhydrous conditions, but they cannot be made in aqueous solution. (d) White phosphorus is more reactive than red phosphorus.

The solubility of \(\mathrm{Cl}_{2}\) in 100 \(\mathrm{g}\) of water at STP is 310 \(\mathrm{cm}^{3}\) . Assume that this quantity of \(\mathrm{Cl}_{2}\) is dissolved and equilibrated as follows: $$\mathrm{Cl}_{2}(a q)+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{Cl}^{-}(a q)+\mathrm{HClO}(a q)+\mathrm{H}^{+}(a q)$$ (a) If the equilibrium constant for this reaction is \(4.7 \times 10^{-4}\) , calculate the equilibrium concentration of HClO formed. (b) What is the pH of the final solution?

Explain the following observations: (a) For a given oxidation state, the acid strength of the oxyacid in aqueous solution decreases in the order chlorine \(>\) bromine \(>\) iodine.(b) Hydrofluoric acid cannot be stored in glass bottles. (c) HI cannot be prepared by treating Nal with sulfuric acid. ( \(\mathbf{d}\) ) The interhalogen ICl_ is known, but BrCl_ is not.
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