Question

An unusual category of acids known as superacids, which are defined as any acid stronger than 100\(\%\) sulfuric acid, can be prepared by seemingly simple reactions similar to the one below. In this example, the reaction of anhydrous HF with SbF produces the superacid \(\left[\mathrm{H}_{2} \mathrm{F}\right]^{+}\left[\mathrm{SbF}_{6}\right]^{-} :\) $$ 2 \mathrm{HF}(l)+\mathrm{SbF}_{5}(l) \longrightarrow\left[\mathrm{H}_{2} \mathrm{F}\right]^{+}\left[\mathrm{SbF}_{6}\right]^{-}(l) $$ a. What are the molecular structures of all species in this reaction? What are the hybridizations of the central atoms in each species? b. What mass of \(\left[\mathrm{H}_{2} \mathrm{F}\right]^{+}\left[\mathrm{SbF}_{6}\right]^{-}\) can be prepared when 2.93 \(\mathrm{mL}\) anhydrous \(\mathrm{HF}\) (density \(=0.975 \mathrm{g} / \mathrm{mL} )\) and 10.0 \(\mathrm{mL}\) SbFs (density \(=3.10 \mathrm{g} / \mathrm{mL}\) ) are allowed to react?

Short answer

The molecular structures and hybridizations of the central atoms in each species are as follows: In HF, Hydrogen (H) has no hybridization. In \(\mathrm{SbF}_{5}\), Antimony (Sb) has \(\mathrm{sp^3d}\) hybridization. In \(\left[\mathrm{H}_{2}\mathrm{F}\right]^{+}\left[\mathrm{SbF}_{6}\right]^{-}\), Hydrogen (H) has \(\mathrm{sp}\) hybridization, and Antimony (Sb) has \(\mathrm{sp^3d^2}\) hybridization. When 2.93 mL of anhydrous HF and 10.0 mL of \(\mathrm{SbF}_{5}\) are allowed to react, approximately 16.89 g of \(\left[\mathrm{H}_{2}\mathrm{F}\right]^{+}\left[\mathrm{SbF}_{6}\right]^{-}\) can be prepared.

Step-by-step solution

Part a: Determine molecular structures and hybridizations

In this part, we will identify the central atoms of \(\mathrm{HF}\), \(\mathrm{SbF}_{5}\), and \(\left[\mathrm{H}_{2}\mathrm{F}\right]^{+}\left[\mathrm{SbF}_{6}\right]^{-}\), and determine the hybridization of these central atoms. - Central atom in \(\mathrm{HF}\): Hydrogen (H) Hybridization: σ bonding results in no hybridization, H is involved only in 1 bond, so no hybridization occurs. - Central atom in \(\mathrm{SbF}_{5}\): Antimony (Sb) Hybridization: Since the electron configuration of Antimony (Sb) is, \([Kr] 5s^2 4d^{10} 5p^3\). When Sb atom bonds with 5 F atoms it forms 5 covalent bonds; 3 single bonds share its 5p orbital, 2 single bonds share its 5s and 5d orbitals. So, the hybridization of Sb in \(\mathrm{SbF}_{5}\) is \(\mathrm{sp^3d}\). - Central atom in \(\left[\mathrm{H}_{2}\mathrm{F}\right]^{+}\left[\mathrm{SbF}_{6}\right]^{-}\): Antimony (Sb) and Hydrogen (H) Hybridization of Sb: As Sb forms a complex with 6 F atoms, the hybridization of Sb in \(\left[\mathrm{SbF}_{6}\right]^{-}\) is \(\mathrm{sp^3d^2}\). Hybridization of H: As H forms a complex with 2 F atoms, the hybridization of H in \(\left[\mathrm{H}_{2}\mathrm{F}\right]^{+}\) is \(\mathrm{sp}\). We have determined the molecular structures and hybridizations of each species in the given reaction.

Part b: Calculate the mass of the superacid produced

In this part, we will calculate the mass of the superacid produced from the given volumes and densities of reactants. 1. Convert volume of reactants to mass: Mass of \(\mathrm{HF} = \text{Volume}\times\text{Density} = 2.93\,\mathrm{mL} \times 0.975\,\frac{\mathrm{g}}{\mathrm{mL}} = 2.85625\,\mathrm{g}\) Mass of \(\mathrm{SbF}_{5} = \text{Volume}\times\text{Density} = 10\,\mathrm{mL} \times 3.1\,\frac{\mathrm{g}}{\mathrm{mL}} = 31\,\mathrm{g}\) 2. Determine the moles of each reactant: Moles of \(\mathrm{HF} = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{2.85625\,\mathrm{g}}{20.01\,\frac{\mathrm{g}}{\mathrm{mol}}} = 0.14273\,\mathrm{mol}\) Moles of \(\mathrm{SbF}_{5} = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{31\,\mathrm{g}}{216.75\,\frac{\mathrm{g}}{\mathrm{mol}}} = 0.14306\,\mathrm{mol}\) 3. Determine the limiting reactant: The stoichiometry of this reaction is such that 2 moles of HF react with 1 mole of SbF5. Using the moles we've calculated, we need to find which reacts first. \(\frac{\text{Moles of HF}}{2} = \frac{0.14273\,\mathrm{mol}}{2} = 0.071365\,\mathrm{mol}\) Since 0.071365 mol is less than the moles of SbF5, HF is the limiting reactant. 4. Calculate moles of the product formed: From the stoichiometry of the reaction, two moles of HF react to produce one mole of \(\left[\mathrm{H}_{2}\mathrm{F}\right]^{+}\left[\mathrm{SbF}_{6}\right]^{-}\). Using this information, the moles of the product formed can be calculated. Moles of the product = 0.071365 mol (from limiting reactant) / 1 5. Calculate mass of the product formed: Mass of the product = Moles × Molar Mass of the product Molar Mass of the product = Molar Mass of \(\left[\mathrm{H}_{2}\mathrm{F}\right]^{+}\) + Molar Mass of \(\left[\mathrm{SbF}_{6}\right]^{-} = 2(20.01) + 216.75 = 236.76\,\frac{\mathrm{g}}{\mathrm{mol}}\) Mass of the product = 0.071365 mol × 236.76 \(\frac{\mathrm{g}}{\mathrm{mol}} = 16.8921\,\mathrm{g}\) The mass of \(\left[\mathrm{H}_{2}\mathrm{F}\right]^{+}\left[\mathrm{SbF}_{6}\right]^{-}\) that can be prepared is approximately 16.89 g.

Key concepts

Molecular Structure

Understanding the molecular structure of a compound helps us know how atoms are connected. In the reaction between anhydrous HF and SbFyou have different species with unique structures.

• Hydrogen fluoride (HF): This is a simple molecule where hydrogen (H) is bonded to fluorine (F). The structure is linear because there are only two atoms involved.
• Antimony pentafluoride (\[\mathrm{SbF}_{5}\]): Here, antimony (Sb) is the central atom surrounded by five fluorine atoms. Its geometry is trigonal bipyramidal because it forms five covalent bonds.
• Superacid \[\left[\mathrm{H}_{2}\mathrm{F}\right]^{+}\left[\mathrm{SbF}_{6}\right]^{-}\]: This product results from HF and SbFy reactions. The \(\left[\mathrm{H}_{2}\mathrm{F}\right]^{+}\) is a unique species resembling a cationic complex where hydrogen bonds with two fluorines. SbF₆⁻ consists of six fluoro groups around antimony, resulting in an octahedral structure.

Understanding these molecular structures allows us to better predict the behavior of the reaction components.

Reaction Stoichiometry

Reaction stoichiometry involves calculating the relative quantities of reactants and products involved in a chemical reaction. This is crucial for understanding the production of a superacid like \[\left[\mathrm{H}_{2}\mathrm{F}\right]^{+}\left[\mathrm{SbF}_{6}\right]^{-}\].

• The given equation shows that two moles of HF react with one mole of SbF₅ to form one mole of the superacid complex.
• To determine how much product forms, you start by calculating the mass of each reactant using their densities and volumes.
• Next, convert these masses to moles using their molar masses. This helps in identifying the limiting reactant, which is the reactant that is entirely used up first and thus limits the amount of product formed.
• Following stoichiometric ratios, you can determine the exact moles of product formed and then calculate the mass of the resulting superacid.

This orderly approach ensures precise calculation of product quantities in chemical reactions.

Molecular Hybridization

Molecular hybridization helps explain the geometry of molecules by describing the mixing of atomic orbitals. In the context of the superacid reaction:

• HF: In HF, hydrogen is bonded to fluorine using a sigma bond. Because there is only one bond, no hybridization occurs for the hydrogen atom.
• SbF₅: Antimony (\(\mathrm{Sb}\)) exhibits \(\mathrm{sp^3d}\) hybridization due to its bonding with five fluorines. This hybridization allows for the formation of a trigonal bipyramidal shape.
• \(\left[\mathrm{H}_{2}\mathrm{F}\right]^{+} \): Here, hybridization of hydrogen is \(\mathrm{sp}\) due to bonding with two fluorine atoms, which leads to a linear arrangement.
• \(\left[\mathrm{SbF}_{6}\right]^{-} \): In this complex, antimony undergoes \(\mathrm{sp^3d^2}\) hybridization, which results in an octahedral structure.

Understanding hybridization allows us to visualize the 3D shape of molecules, which is key for predicting reactions and interactions.