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?

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.

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.