Sulfur tetrafluoride \(\left(\mathrm{SF}_{4}\right)\) reacts slowly with \(\mathrm{O}_{2}\) to form sulfur tetrafluoride monoxide (OSF_ \(_{4} )\) according to the following unbalanced reaction: \begin{equation}\mathrm{SF}_{4}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{OSF}_{4}(g) \end{equation} The O atom and the four \(\mathrm{F}\) atoms in OSF \(_{4}\) are bonded to a central \(\mathrm{S}\) atom. (a) Balance the equation. (b) Write a Lewis structure of OSF_ in which the formal charges of all atoms are zero.(c) Use average bond enthalpies (Table 8.3 ) 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) For each of the molecules you drew in part (d), state how many fluorines are equatorial and how many are axial.

Step by step solution

01

Balance the chemical equation

To balance the equation, make sure that the same number of each type of atom appears on both sides of the arrow. \[\mathrm{SF}_{4}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{OSF}_{4}(g)\] Balanced equation: \[\mathrm{SF}_{4}(g)+\frac{1}{2}\mathrm{O}_{2}(g) \longrightarrow \mathrm{OSF}_{4}(g)\]

02

Write the Lewis structure of OSF\(_{4}\)

To write the Lewis structure of OSF\(_{4}\): 1. Calculate total number of valence electrons: Sulfur (S) = 6 valence electrons Oxygen (O) = 6 valence electrons 4 Fluorines (F) = 4 x 7 valence electrons = 28 valence electrons Total valence electrons = 6 + 6 + 28 = 40 valence electrons 2. Place the least electronegative atom (Sulfur) in the center and surround it with the other atoms. 3. Connect each of the surrounding atoms to the central atom with a single bond (one electron pair). 4. Fill the octets of the surrounding atoms (Oxygen and Fluorine) by assigning the remaining valence electrons as lone pairs. 5. Assign the remaining valence electrons to the central atom. The resulting Lewis structure is: O || S - F···F | | F In this structure, all atoms have formal charges of zero.

03

Determine the enthalpy of the reaction

The given reaction: \[\mathrm{SF}_{4}(g) + \frac{1}{2}\mathrm{O}_{2}(g) \longrightarrow \mathrm{OSF}_{4}(g)\] Using average bond enthalpies from Table 8.3: \[1 \times \text{S-F bond} + 1 \times \text{O=O bond} \longrightarrow 1 \times \text{S=O bond} + 4 \times \text{S-F bonds}\] \[1 \times (327 \, \mathrm{kJ/mol}) + 1 \times (\frac{1}{2} \times 498 \, \mathrm{kJ/mol}) \longrightarrow 1 \times (532 \, \mathrm{kJ/mol}) + 4 \times (327 \, \mathrm{kJ/mol})\] Calculate the enthalpy of the reaction: \[∆H = \mathrm{Bonds\, Broken\, (\, reactants\, )} - \mathrm{Bonds\, Formed\, (\, products\, )}\] \[∆H = (327 + 249) - (532 + 4 \times 327)\] \[∆H = 576 - 1856 = -1280 \, \mathrm{kJ/mol}\] The reaction is exothermic, as ∆H is negative.

04

Determine the electron-domain geometry and two possible molecular geometries of \(\mathrm{OSF}_{4}\)

To determine the electron-domain geometry of \(\mathrm{OSF}_{4}\), we have to count the electron domains. There are 5 domains (1 oxygen atom, 4 fluorine atoms) around the central sulfur atom, and thus the electron-domain geometry is trigonal bipyramidal. Two possible molecular geometries for the \(\mathrm{OSF}_{4}\) molecule based on the trigonal bipyramidal electron-domain geometry are: 1. Axial: Oxygen axial, 4 Fluorines equatorial 2. Equatorial: Oxygen equatorial, 3 Fluorines equatorial, 1 Fluorine axial These two possible structures can be represented as: 1. Axial oxygen: O | S - F···F | | F 2. Equatorial oxygen: F | S - F···O | | F

05

State the number of equatorial and axial fluorines in each of the molecular geometries

For the two possible molecular geometries of \(\mathrm{OSF}_{4}\) mentioned in step 4: 1. Axial oxygen: - Equatorial fluorines: 4 - Axial fluorines: 0 2. Equatorial oxygen: - Equatorial fluorines: 3 - Axial fluorines: 1

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