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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 \(\mathrm{SbF}_{5}\) 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 HF (density \(=0.975 \mathrm{~g} / \mathrm{mL}\) ) and \(10.0 \mathrm{~mL} \mathrm{SbF}_{5}\) (density \(=3.10 \mathrm{~g} / \mathrm{mL}\) ) are allowed to react?

Short Answer

Expert verified
The molecular structures and hybridizations of the species involved in the reaction are: 1. Anhydrous HF: linear, no central atom. 2. SbF5: trigonal bipyramidal, central atom Sb with sp³d hybridization. 3. [H2F]+: linear, central atom F with sp hybridization. 4. [SbF6]-: Octahedral, central atom Sb with sp³d² hybridization. When 2.93 mL of anhydrous HF and 10.0 mL of SbF5 are allowed to react, 16.84 g of the superacid [H2F]+[SbF6]- can be prepared.

Step by step solution

01

Determine the molecular structures and hybridizations

To determine the molecular structures, we need to look at the species involved in the reaction. 1. Anhydrous HF: The molecular structure of HF is linear with a single bond between hydrogen and fluorine. There are no central atoms as it's a diatomic molecule. 2. SbF5: The central atom is Sb (antimony), which is surrounded by five fluorine atoms. Each F is bonded to the Sb through a single bond. The molecular structure of SbF5 is trigonal bipyramidal. The hybridization of the central atom Sb is sp³d since it forms five sigma bonds. 3. [H2F]+: In this cation, there is a central fluorine (F) bonded to two hydrogen atoms. The molecular structure is linear. The hybridization of the central atom F is sp. 4. [SbF6]-: The central atom is Sb, which is surrounded by six fluorine atoms. Each F is bonded to the Sb through a single bond. The molecular structure of [SbF6]- is octahedral. The hybridization of the central atom Sb is sp³d² since it forms six sigma bonds.
02

Calculate the moles of reactants

First, we'll find the moles of each reactant using their densities and volumes: Density of anhydrous HF: 0.975 g/mL Volume of anhydrous HF: 2.93 mL Molar mass of HF: 20.01 g/mol Moles of HF = (Density of HF × Volume of HF) / Molar mass of HF Moles of HF = (0.975 g/mL × 2.93 mL) / 20.01 g/mol ≈ 0.1424 moles Density of SbF5: 3.10 g/mL Volume of SbF5: 10.0 mL Molar mass of SbF5: 178.76 g/mol Moles of SbF5 = (Density of SbF5 × Volume of SbF5) / Molar mass of SbF5 Moles of SbF5 = (3.10 g/mL × 10.0 mL) / 178.76 g/mol ≈ 0.1730 moles
03

Determine the limiting reactant and moles of product

The reaction has a stoichiometry of 2 moles of HF reacting with 1 mole of SbF5 to produce 1 mole of superacid. We need to determine which reactant is limiting. Moles of HF required to react with moles of SbF5 = 0.1730 moles of SbF5 × 2 moles of HF/mole of SbF5 = 0.3460 moles Since there are only 0.1424 moles of HF available, HF is the limiting reactant. Now, we determine the moles of the superacid produced: Moles of superacid = (0.1424 moles of HF) × (1 mole of superacid/2 moles of HF) = 0.0712 moles
04

Calculate mass of the superacid

Now that we have the moles of the product, we can calculate its mass using the molar mass: Molar mass of [H2F]+ = 20.01 g/mol Molar mass of [SbF6]- = 216.75 g/mol (including the extra F compared to SbF5) Molar mass of the superacid = 20.01 g/mol + 216.75 g/mol = 236.76 g/mol Mass of superacid produced = Moles of superacid × Molar mass of superacid Mass of superacid produced ≈ 0.0712 moles × 236.76 g/mol ≈ 16.84 g So, 16.84 g of the superacid [H2F]+[SbF6]- can be prepared when 2.93 mL of anhydrous HF and 10.0 mL of SbF5 are allowed to react.

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Key Concepts

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

Chemical Reaction Formulas
Chemical reaction formulas are vital tools in chemistry that represent the substances involved in chemical reactions—the reactants and products—and display the conversion ratios between them. To understand superacid chemistry, it's essential to decipher how molecules such as anhydrous HF (\r\( \r\text{HF} \r\) ) and antimony pentafluoride (\r\( \r\text{SbF}_5 \r\) ) interact to form superacids like \r\( \r\text{[H}_2 \text{F]}^+ \text{[SbF}_6 \text{]}^- \r\).\r

\rIn the provided reaction, we see that two moles of anhydrous HF combine with one mole of \r\( \r\text{SbF}_5 \r\) to produce one mole of the superacid. This stoichiometric relationship is critical for understanding the quantities of reactants needed and the amount of product that can be formed, a fundamental concept in stoichiometry calculations. Ensuring accuracy in writing and interpreting chemical reaction formulas is imperative for students to master the ability to predict the outcomes of chemical reactions and make practical laboratory decisions.
Molecular Hybridization
Molecular hybridization explains how atomic orbitals mix to form new hybrid orbitals, which in turn dictate the geometry and bonding properties of molecules. Delving into superacid chemistry reveals the importance of hybridization in understanding molecular structures.\r

\rFor example, in \r\( \r\text{SbF}_5 \r\), antimony (Sb) undergoes sp\textsuperscript{3}d hybridization to form a trigonal bipyramidal structure. Knowing the hybridization helps us visualize and predict the molecular shapes, which ultimately influence the chemical behavior and reactivity of substances. With the superacid cation \r\( \r\text{[H}_2 \text{F]}^+ \r\), the central fluorine atom is sp hybridized, leading to a linear structure. Students should aim to grasp this concept to interpret molecular geometries and predict how molecules will interact in various chemical reactions.
Stoichiometry Calculations
Stoichiometry calculations are mathematical techniques used to determine the quantities of reactants and products in a chemical reaction. They are based on the mole concept and the principles of mass conservation and are indispensable in chemical engineering and experimental chemistry.\r

\rIn the reaction forming a superacid, we execute stoichiometry calculations to find the mass of the superacid that can be synthesized from given volumes of HF and \r\( \r\text{SbF}_5 \r\). By converting volumes to moles using density and molar mass, and identifying the limiting reactant, we can predict the yield of the reaction—the amount of superacid produced. This calculation is crucial for practical lab work, where efficiency, resource optimization, and safety are of utmost importance. Students are encouraged to hone their skills in stoichiometry to better manage real-world chemical reactions and processes.

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Most popular questions from this chapter

The diatomic molecule OH exists in the gas phase. The bond length and bond energy have been measured to be \(97.06 \mathrm{pm}\) and \(424.7 \mathrm{~kJ} / \mathrm{mol}\), respectively. Assume that the OH molecule is analogous to the HF molecule discussed in the chapter and that molecular orbitals result from the overlap of a lower-energy \(p_{z}\) orbital from oxygen with the higher- energy \(1 s\) orbital of hydrogen (the \(\mathrm{O}-\mathrm{H}\) bond lies along the \(z\) -axis). a. Which of the two molecular orbitals will have the greater hydrogen \(1 s\) character? b. Can the \(2 p_{x}\) orbital of oxygen form molecular orbitals with the \(1 s\) orbital of hydrogen? Explain. c. Knowing that only the \(2 p\) orbitals of oxygen will interact significantly with the \(1 s\) orbital of hydrogen, complete the molecular orbital energy- level diagram for OH. Place the correct number of electrons in the energy levels. d. Estimate the bond order for OH. e. Predict whether the bond order of \(\mathrm{OH}^{+}\) will be greater than, less than, or the same as that of \(\mathrm{OH}\). Explain.

What type of molecular orbital would result from the in-phase combination of two \(d_{x z}\) atomic orbitals shown below? Assume the \(x\) -axis is the internuclear axis.

Which of the following are predicted by the molecular orbital model to be stable diatomic species? a. \(\mathrm{N}_{2}{ }^{2-}, \mathrm{O}_{2}^{2-}, \mathrm{F}_{2}^{2-}\) b. \(\mathrm{Be}_{2}, \mathrm{~B}_{2}, \mathrm{Ne}_{2}\)

The three most stable oxides of carbon are carbon monoxide (CO), carbon dioxide \(\left(\mathrm{CO}_{2}\right)\), and carbon suboxide \(\left(\mathrm{C}_{3} \mathrm{O}_{2}\right)\). The spacefilling models for these three compounds are For each oxide, draw the Lewis structure, predict the molecular structure, and describe the bonding (in terms of the hybrid orbitals for the carbon atoms).

\(\mathrm{FClO}_{2}\) and \(\mathrm{F}_{3} \mathrm{ClO}\) can both gain a fluoride ion to form stable anions. \(\mathrm{F}_{3} \mathrm{ClO}\) and \(\mathrm{F}_{3} \mathrm{ClO}_{2}\) will both lose a fluoride ion to form stable cations. Draw the Lewis structures and describe the hybrid orbitals used by chlorine in these ions.

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