Problem 64 - Chemistry The Science in Context

Question

Draw Lewis structures that show how electron pairs move and bonds form and break in this reaction, and identify the Lewis acid and Lewis base. (Note: HSbF \(_{6}\) is an ionic compound and one of the strongest Bronsted-Lowry acids known.) $$ \mathrm{SbF}_{5}(s)+\mathrm{HF}(g) \rightarrow \mathrm{HSbF}_{6}(s) $$

Step-by-Step Solution

Answer

Question: Analyze the reaction between solid \(\mathrm{SbF_5}\) and gaseous \(\mathrm{HF}\), draw Lewis structures, show electron pair movement and bond formation, and identify the Lewis acid and base. Answer: In this reaction, the Lewis acid is \(\mathrm{SbF_5}\), and the Lewis base is \(\mathrm{HF}\). The Lewis structures of the reactants are as follows: \(\mathrm{SbF_5(s):}\) ``` F | Sb - F - F | F ``` \(\mathrm{HF(g):}\) ``` H - F ``` The electron pair movement during the reaction is: $$ \mathrm{SbF_5}\:\: + \:\:H - F \ \longrightarrow \ H - \mathrm{SbF_6} $$

Step 1: Draw the Lewis structures of the reactants

Draw the Lewis structure of the solid compound \(\mathrm{SbF_5}\), which consists of an antimony (Sb) atom surrounded by five fluorine (F) atoms, each sharing one electron pair with antimony. In addition, draw the Lewis structure of gaseous hydrogen fluoride, \(\mathrm{HF}\), wherein the hydrogen (H) and fluorine atoms share one electron pair. Lewis structure of \(\mathrm{SbF_5}\): ``` F | Sb - F - F | F ``` Lewis structure of \(\mathrm{HF}\): ``` H - F ```

Step 2: Show electron pair movement and bond formation/breaking

Moving forward, we need to demonstrate the electron pair movement in the formation of the product \(\mathrm{HSbF_6}\). The reaction involves breaking the \(\mathrm{H-F}\) bond in the hydrogen fluoride reactant and forming a new \(\mathrm{H-Sb}\) bond in the product. The electron pair movement can be depicted as follows: $$ \mathrm{Dotted\:arrow\:indicating\:electron\:pair\:transfer:}\\ \mathrm{SbF_5}\:\: + \:\:H - F \ \longrightarrow \ H - \mathrm{SbF_6} $$

Step 3: Identify the Lewis acid and Lewis base

In this reaction, the Lewis acid is the compound that accepts an electron pair, while the Lewis base donates an electron pair. As a result, the Lewis base donates an electron pair and forms a new bond through this donation. In our case, \(\mathrm{HF}\) donates a lone pair of electrons, making it the Lewis base; whereas, \(\mathrm{SbF_5}\) accepts the lone pair of electrons, thus acting as the Lewis acid. In summary, the Lewis structures of the reactants, electron pair movement, bond formation and breaking, and the identification of the Lewis acid and Lewis base are as follows: Lewis structures of reactants: $$ \mathrm{SbF_5(s):}\\ F\\ |\\ \mathrm{Sb}\:-\:\mathrm{F}\: - \:\mathrm{F}\\ |\\ F \mathrm{HF(g):}\\ \mathrm{H}\:-\:\mathrm{F} $$ Electron pair movement and bond formation/breaking: $$ \mathrm{SbF_5}\:\: + \:\:H - F \ \longrightarrow \ H - \mathrm{SbF_6}\\ $$ Lewis acid: \(\mathrm{SbF_5}\) Lewis base: \(\mathrm{HF}\)

Key Concepts

Electron Pair Movement

The concept of electron pair movement lies at the heart of understanding chemical reactions, particularly when constructing or interpreting Lewis structures. Electron pair movement refers to the transfer or shift of electrons from one atom or molecule to another. This movement is typically depicted using curved arrows in Lewis structures.
In the given reaction, \[ \mathrm{SbF}_{5}(s) + \mathrm{HF}(g) \rightarrow \mathrm{HSbF}_{6}(s) \]the transfer of electron pairs plays a critical role. The electron pair from the \(\mathrm{H-F}\)molecule in hydrogen fluoride (\(\mathrm{HF}\))moves towards the antimony pentafluoride(\(\mathrm{SbF}_{5}\))to form a new bond. The curved arrow from the lone pair on fluorine in\(\mathrm{HF}\)to antimony in\(\mathrm{SbF}_{5}\)illustrates this electron transfer.
Understanding and visualizing how electrons move helps in predicting product formation. This crucial insight is foundational in the study of reaction mechanisms and helps to anticipate how atoms will interact in chemical reactions.

Lewis Acid and Base

Lewis acids and bases are a pair of chemical species that interact through electron pair transfer. A Lewis acid is a compound that accepts an electron pair, while a Lewis base donates an electron pair.
In many reactions, the identification of acids and bases using the Lewis theory can be crucial for understanding the direction and outcome of the reaction.

• **Lewis Base**: In the reaction under consideration, \(\mathrm{HF}\)acts as a Lewis base. It donates a lone pair of electrons from the fluorine to form a new bond.
• **Lewis Acid**: Conversely, \(\mathrm{SbF}_{5}\)operates as a Lewis acid. It accepts the electron pair offered by \(\mathrm{HF}\),thereby completing its valence shell configuration and forming\(\mathrm{HSbF}_{6}\).

The terms 'acid' and 'base' don't exclusively relate to the familiar hydrogen ion and hydroxide ion concept but extend to include the broader idea of electron pair dynamics. Understanding these interactions allows us to better categorize many reactions beyond the aqueous solutions familiar from Brønsted-Lowry theory.

Bond Formation and Breaking

In any chemical reaction, bond formation and breaking is the core event that results in new products. Bonds can be conceptualized as interactions of shared or exchanged electron pairs between atoms. Whether a reaction is endothermic or exothermic often hinges on this principle.
When analyzing the given reaction, \(\mathrm{SbF}_{5}(s) + \mathrm{HF}(g) \rightarrow \mathrm{HSbF}_{6}(s)\),the two atoms involved in bond formation and breaking are hydrogen and fluorine. Initially, the \(\mathrm{H-F}\)bond in the reactant \(\mathrm{HF}\)will break. Subsequent to this, a new bond forms between the hydrogen from \(\mathrm{HF}\)and the antimony in \(\mathrm{SbF}_{5}\).

• **Breaking Bonds**: The \(\mathrm{H-F}\)bond is considered broken when \(\mathrm{HF}\)donates its electron pair to form the \(\mathrm{H-Sb}\)bond in\(\mathrm{HSbF}_{6}\).
• **Forming Bonds**: The \(\mathrm{H-Sb}\)bond represents the new connection formed through electron pair acceptance by\(\mathrm{SbF}_{5}\).

Every aspect of chemical reactivity involves precision in understanding how bonds break and form; without this understanding, prediction of reaction products becomes significantly less reliable.