Question

Sulfur tetrafluoride \(\left(\mathrm{SF}_{4}\right)\) reacts slowly with \(\mathrm{O}_{2}\) to form sulfur tetrafluoride monoxide \(\left(\mathrm{OSF}_{4}\right)\) according to the following unbalanced reaction: $$ \mathrm{SF}_{4}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{OSF}_{4}(g) $$ The \(\mathrm{O}\) atom and the four \(\mathrm{F}\) atoms in \(\mathrm{OSF}_{4}\) are bonded to a central \(\mathrm{S}\) atom. (a) Balance the equation. (b) Write a Lewis structure of \(\mathrm{OSF}_{4}\) in which the formal charges of all atoms are zero. (c) Use average bond enthalpies (Table 8.4) to estimate the enthalpy of the reaction. Is it endothermic or exothermic? (d) Determine the electron-domain geometry of \(\mathrm{OSF}_{4}\), and write two possible molecular geometries for the molecule based on this electron-domain geometry. (e) Which of the molecular geometries in part (d) is more likely to be observed for the molecule? Explain.

Short answer

(a) Balanced equation: \(2 SF_4 (g) + O_2 (g) \rightarrow 2 OSF_4 (g)\) (b) Lewis structure of OSF4 with zero formal charges: ``` F | O-S-F / \ F F ``` (c) Enthalpy of the reaction is 24 kJ/mol, and the reaction is endothermic. (d) Electron-domain geometry is trigonal bipyramidal. Two possible molecular geometries are Axial and Equatorial configurations. (e) The most likely molecular geometry of OSF4 is the Equatorial configuration, as it minimizes electron domain repulsion and is more stable.

Step-by-step solution

(a) Balance the reaction equation

We have the unbalanced chemical equation: \( SF_4 (g) + O_2 (g) \rightarrow OSF_4 (g) \) To balance it, we need to balance the number of atoms for each element on both sides of the equation. We can achieve this by putting a coefficient of 2 in front of SF4: \( 2 SF_4 (g) + O_2 (g) \rightarrow 2 OSF_4 (g) \) Now the equation is balanced since there are 4 F atoms and 1 O atom on both sides of the equation.

(b) Lewis Structure of OSF4

In order to draw a Lewis structure for OSF4 with zero formal charges, we first need to count the total number of valence electrons. Oxygen has 6 valence electrons, sulfur has 6 valence electrons, and each fluorine has 7 valence electrons. So, the total number of valence electrons is: 6 (sulfur) + 6 (oxygen) + 4(7) (four fluorine atoms) = 40 electrons. Now, we can draw the Lewis structure: 1. Place the least electronegative atom (sulfur) in the center, and arrange the other atoms (oxygen and fluorine) around it. 2. Connect each atom to the central atom with a single bond (two electrons for each bond). 3. Distribute the remaining electrons to complete the octets of the outer atoms. The Lewis structure that has zero formal charges is: ``` F | O-S-F / \ F F ``` In this structure, sulfur and oxygen are double bonded, while each fluorine atom is single bonded to the sulfur atom. There are no lone pairs on the sulfur atom.

(c) Enthalpy of the reaction and endothermic or exothermic

To determine the enthalpy of the reaction, we can use the average bond enthalpies listed in Table 8.4. We will first calculate the sum of bond enthalpies for the reactants and the products, and then find the difference of enthalpies to get the reaction enthalpy. Reactants: SF bond (4 SF bonds in each SF4 molecule, 2 SF4 molecules) = 4 * 2 * 339 kJ/mol = 2712 kJ/mol O=O bond (1 O2 molecule) = 498 kJ/mol Total energy for the reactants = 2712 + 498 = 3210 kJ/mol Products: OS double bond = 1 * 522 kJ/mol S-F single bond (each OSF4 molecule has 4 SF bonds, 2 OSF4 molecules) = 4 * 2 * 339 kJ/mol = 2712 kJ/mol Total energy for the products = 522 + 2712 = 3234 kJ/mol Enthalpy of the reaction = (Energy of products - Energy of reactants) = 3234 - 3210 = 24 kJ/mol Since the enthalpy of the reaction is positive, the reaction is endothermic.

(d) Electron-domain geometry and two possible molecular geometries

To determine the electron-domain geometry of OSF4, we should first find the number of electron domains around the central sulfur atom. In OSF4, sulfur is bonded to 4 fluorine atoms and 1 oxygen atom via a double bond so we have 5 electron domains. For 5 electron domains, the electron-domain geometry is trigonal bipyramidal. Now, let's find two possible molecular geometries for the molecule based on this electron-domain geometry: 1. The molecular geometry might be that the double-bonded oxygen is in the axial position and the other four fluorine atoms are in the equatorial positions (called Axial configuration). 2. The molecular geometry might be that the double-bonded oxygen is one of the equatorial positions, and the other fluorine atoms occupy the remaining positions (called Equatorial configuration).

(e) Most likely molecular geometry of OSF4

The most likely molecular geometry of OSF4 would be the one with the least repulsion between the electron domains. In the axial configuration, the double-bonded oxygen atom would experience more repulsion from the fluorine atoms than in the equatorial configuration due to the presence of more lone pairs on the oxygen atom. Therefore, the equatorial configuration (with O in the equatorial position) is more likely to be observed for the molecule. This choice minimizes the repulsion between electron domains, making it more stable.

Key concepts

Chemical Reaction Balancing

Balancing chemical reactions ensures that mass is conserved and the same number of each type of atom is present on both sides of the equation. In the case of sulfur tetrafluoride (SF extsubscript{4}) reacting with oxygen (O extsubscript{2}) to form sulfur tetrafluoride monoxide (OSF extsubscript{4}), the unbalanced equation is:

• SF extsubscript{4} (g) + O extsubscript{2} (g) → OSF extsubscript{4} (g)

To balance this equation, we start by identifying the atoms involved: sulfur (S), oxygen (O), and fluorine (F). Initially, there is 1 sulfur, 4 fluorine, and 2 oxygen atoms on the reactant side, but only 1 sulfur, 4 fluorine, and 1 oxygen on the product side. To balance the oxygen atoms, place a coefficient of 2 in front of SF extsubscript{4}:

• 2 SF extsubscript{4} (g) + O extsubscript{2} (g) → 2 OSF extsubscript{4} (g)

Now, each side of the equation has 2 sulfur atoms, 8 fluorine atoms, and 2 oxygen atoms, achieving balance.

Lewis Structures

Lewis structures are diagrams that represent the valence electron distribution in molecules. For OSF extsubscript{4}, a well-drawn Lewis structure shows how atoms are bonded while ensuring the lowest formal charges.
To draw the Lewis structure, first count the total valence electrons:

• Sulfur (S) has 6
• Oxygen (O) has 6
• Each fluorine (F) has 7
• Total = 6 (sulfur) + 6 (oxygen) + 4(7) (fluorine) = 40 electrons

Sulfur, being the least electronegative, sits in the center. Connect S to each surrounding atom with single bonds. Place the remaining electrons to satisfy the octet rule. The most stable Lewis structure for OSF extsubscript{4} features:

• Four single bonds between sulfur and fluorine
• A double bond between sulfur and oxygen
• No formal charges on any atom

Representing it, you'd see sulfur with four fluorine atoms bonded around it, and oxygen double-bonded.

Enthalpy Calculations

Enthalpy calculations help determine if a reaction absorbs or releases energy. The reaction's enthalpy change (\( \Delta H \)) is derived from the difference in bond enthalpies between reactants and products. Using average bond enthalpies available in tables:

• SF bonds in reactants: 4 SF bonds per SF extsubscript{4}, for two molecules: \( 4 \times 2 \times 339 = 2712 \text{kJ/mol} \)
• O=O bond: \( 498 \text{kJ/mol} \)
• Total energy for reactants: \( 3210 \text{kJ/mol} \)
• OS=O double bond in products: \( 522 \text{kJ/mol} \)
• S-F single bonds: \( 4 \times 2 \times 339 = 2712 \text{kJ/mol} \)
• Total energy for products: \( 3234 \text{kJ/mol} \)

Finally, \( \Delta H = 3234 - 3210 = 24 \text{kJ/mol} \), indicating an endothermic reaction since energy input is needed, reflected by the positive value.

Electron-Domain Geometry

The electron-domain geometry provides insights into how the molecule's electron pairs are arranged around a central atom. OSF extsubscript{4} has 5 electron domains around its sulfur atom:

• 4 single bonds with fluorine atoms
• 1 double bond with an oxygen atom

This number, five, corresponds to a trigonal bipyramidal electron-domain geometry. This geometry maximizes the space between electron domains, minimizing repulsion between them. It's crucial for predicting molecular shape and properties.
Understanding this geometry helps us predict possible molecular arrangements and interactions within the molecule.

Molecular Geometry Prediction

Molecular geometry prediction determines the physical shaping of a molecule, building on the electron-domain geometry. For OSF extsubscript{4}, the geometry could take on configurations based on minimizing repulsion:

• **Axial configuration:** Placing the double-bonded oxygen atom in an axial position, with fluorines occupying equatorial positions.
• **Equatorial configuration:** Positioning the double-bonded oxygen atom in an equatorial slot could reduce electron pair repulsion.

The preferred geometry is where electron repulsions are minimized, making the equatorial configuration more stable. In this setup, lone pairs on the oxygen exert less repulsive force on the attached fluorines, leading to a more stable molecular structure.