Problem 107 - Chemistry Matter and Change

Question

Identify the Lewis acids and bases in the following reactions. \begin{equation} \begin{array}{l}{\text { a. } \mathrm{H}^{+}+\mathrm{OH}^{-} \rightleftharpoons \mathrm{H}_{2} \mathrm{O}} \\ {\text { b. } \mathrm{Cl}^{-}+\mathrm{BCl}_{3} \rightleftharpoons \mathrm{BCl}_{4}^{-}} \\\ {\text { c. } \mathrm{SO}_{3}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{H}_{2} \mathrm{SO}_{4}}\end{array} \end{equation}

Step-by-Step Solution

Answer

a. H⁺ is the acid, OH⁻ is the base; b. BCl₃ is the acid, Cl⁻ is the base; c. SO₃ is the acid, H₂O is the base.

Step 1: Understanding Lewis Acids and Bases

A Lewis acid is a chemical species that can accept an electron pair, whereas a Lewis base is a species that can donate an electron pair. In order to identify Lewis acids and bases in each reaction, we must look at the species that accept or donate electron pairs.

Step 2: Reaction a: Analyzing $H^+$ and OH^-

In the reaction \( ext{H}^+ + ext{OH}^- ightleftharpoons ext{H}_2 ext{O} \), the hydrogen ion \( ext{H}^+ \) is a Lewis acid because it accepts an electron pair from the hydroxide ion \( ext{OH}^- \). The hydroxide ion \( ext{OH}^- \) is a Lewis base as it donates an electron pair to form water.

Step 3: Reaction b: Analyzing $Cl^-$ and $BCl_3$

In the reaction \( ext{Cl}^- + ext{BCl}_3 ightleftharpoons ext{BCl}_4^- \), \( ext{BCl}_3 \) is the Lewis acid as it accepts an electron pair from the chloride ion \( ext{Cl}^- \). \( ext{Cl}^- \) acts as the Lewis base by donating an electron pair to form \( ext{BCl}_4^- \).

Step 4: Reaction c: Analyzing $SO_3$ and $H_2O$

In the reaction \( ext{SO}_3 + ext{H}_2 ext{O} ightleftharpoons ext{H}_2 ext{SO}_4 \), \( ext{SO}_3 \) functions as the Lewis acid because it accepts an electron pair from the water \( ext{H}_2 ext{O} \). \( ext{H}_2 ext{O} \) is the Lewis base as it donates an electron pair, forming sulfuric acid.

Key Concepts

Electron Pair Donation

When we talk about electron pair donation, we're diving into the world of Lewis bases. A Lewis base is any species that donates an electron pair to another species to form a bond. Think of it like a friend lending you a book. The friend is the one donating something, in this case, a book. In chemistry, this "book" is an electron pair. When a molecule or ion has a lone pair of electrons, it can donate these to a Lewis acid.

Let's take some examples from the exercises:

• In the reaction with hydroxide ion (\( \mathrm{OH}^- \)), it donates an electron pair to the hydrogen ion \( \mathrm{H}^+ \) to form water \( \mathrm{H}_2\mathrm{O} \).
• With \( \mathrm{Cl}^- \), it donates an electron pair to the \( \mathrm{BCl}_3 \) molecule to form \( \mathrm{BCl}_4^- \).
• In the case of water \( \mathrm{H}_2\mathrm{O} \), it donates an electron pair to \( \mathrm{SO}_3 \) to make sulfuric acid \( \mathrm{H}_2\mathrm{SO}_4 \).

The donor is crucial in the formation of new chemical bonds, highlighting its role as a Lewis base.

Electron Pair Acceptance

On the flip side, we have the concept of electron pair acceptance, which is essential for understanding Lewis acids. A Lewis acid is a species that accepts an electron pair. It's like being on the receiving end of someone handing you a book. You are taking it, which is exactly what a Lewis acid does with electrons. It creates a new bond by accepting these electrons.

In the exercise examples, Lewis acids show up as:

• \( \mathrm{H}^+ \) accepts an electron pair from \( \mathrm{OH}^- \) to create water molecules.
• \( \mathrm{BCl}_3 \) takes an electron pair from \( \mathrm{Cl}^- \) to form \( \mathrm{BCl}_4^- \).
• \( \mathrm{SO}_3 \) accepts an electron pair from \( \mathrm{H}_2\mathrm{O} \) to convert into \( \mathrm{H}_2\mathrm{SO}_4 \).

Lewis acids play an essential role in many chemical reactions by enabling bond formation through the acceptance of an electron pair.

Chemical Reactions Analysis

Diving into chemical reactions, especially with a Lewis perspective, involves understanding how electron pairs are exchanged. In a typical Lewis acid-base reaction, you'll see an electron pair donated by a base and accepted by an acid. This transfer results in the formation of a new chemical compound. Analyzing such reactions means identifying which species behave as acids or bases from a Lewis viewpoint.

Let's break down the exercise:

• In reaction (a), the formation of water \( \mathrm{H}_2\mathrm{O} \) demonstrates how \( \mathrm{OH}^- \) donates and \( \mathrm{H}^+ \) accepts, achieving a stable product.
• In reaction (b), \( \mathrm{Cl}^- \) transfers its electron pair to \( \mathrm{BCl}_3 \), producing \( \mathrm{BCl}_4^- \). Understanding this transfer is key in chemical synthesis.
• For reaction (c), forming \( \mathrm{H}_2\mathrm{SO}_4 \) highlights the interaction where \( \mathrm{SO}_3 \) accepts the electrons from \( \mathrm{H}_2\mathrm{O} \).

Chemical reactions are the stage where atoms and molecules dramatically change partners through electron pair exchanges, summarized by the Lewis acid-base theory.