Problem 63 - Chemistry: An Atoms-Focused Approach

Question

Draw Lewis structures that show how electron pairs move and bonds form and break in the following reaction, and identify the Lewis acid and Lewis base. $$\mathrm{B}(\mathrm{OH})_{3}(a q)+2 \mathrm{H}_{2} \mathrm{O}(\ell) \rightleftharpoons \mathrm{B}(\mathrm{OH})_{4}^{-}(a q)+\mathrm{H}_{3} \mathrm{O}^{+}(a q)$$

Step-by-Step Solution

Answer

Provide a brief explanation. Answer: In the reaction between B(OH)3 and 2 H2O molecules, B(OH)3 acts as the Lewis acid, and H2O acts as the Lewis base. This is because B(OH)3 accepts a lone pair from an H2O molecule to form B(OH)4-, while H2O provides the electron pair.

Step 1: Identify reactants and products

In this reaction, the reactants are B(OH)3 and 2 H2O molecules, and the products are B(OH)4- and H3O+ ions. It is crucial to analyze how bonds form and break between reactants to form products, focusing on the movement of electron pairs.

Step 2: Draw Lewis structures for reactants and products

Start by drawing the Lewis structures for the reactants, B(OH)3 and H2O, and for the products, B(OH)4- and H3O+. Reactants: B(OH)3: B has three valence electrons, and each O has six valence electrons. Each O shares one electron with B, and each O is single bonded to an H. This results in six unshared electrons on each O(atom). H2O: O has six valence electrons, and each H single bonded to O. This results in four unshared electrons on the O(atom). Products: B(OH)4-: B forms a single bond with each of the four OH- groups (O with one single bonded H and one extra electron), resulting in four unshared electrons on each O(atom). H3O+: O shares one electron with each of the three H atoms (different from H2O where O shares only with two H). This results in only two unshared electrons on O.

Step 3: Observe electron pair movement and bond formation/breaking

In the reaction, B(OH)3 accepts a lone pair from one H2O to form a bond with the O in B(OH)4-. Also, one H from the other H2O molecule transfers to H2O and forms a bond with O to become H3O+.

Step 4: Identify the Lewis acid and Lewis base

Now that we've observed the electron pair movement, we can identify the Lewis acid and Lewis base in the reaction. Since B(OH)3 accepts a lone pair from an H2O molecule to form B(OH)4-, it is the Lewis acid, and H2O, providing the electron pair, is the Lewis base. So, the Lewis acid is B(OH)3, and the Lewis base is H2O.

Key Concepts

Electron Pair Movement

Understanding electron pair movement is crucial in chemical reactions, as it helps us see how bonds break and form. When we examine the given reaction, we need to focus on the transfer and sharing of electron pairs among the atoms involved. In our reaction, the electron pair movement can be seen in two primary changes:

• First, a lone pair from the oxygen in one of the water molecules is donated to the boron atom in B(OH)₃, facilitating the formation of B(OH)₄⁻.
• Secondly, a proton (H⁺) from another water molecule bonds with an oxygen to form the hydronium ion, H₃O⁺.

These movements illustrate the central concept of electron pair transfer, essential for bond making and breaking. Recognizing these shifts can help predict how substances interact in a chemical reaction.
By understanding these changes, students can follow the intricacies of different reactions in chemistry, making complex processes much clearer.

Lewis Acid

A Lewis acid is a species that can accept an electron pair. In our example reaction, B(OH)₃ operates as the Lewis acid.

• Boron in B(OH)₃ has an empty orbital that allows it to accept an electron pair from a water molecule. This characteristic aligns with the definition of a Lewis acid, as boron is looking to complete its octet by acquiring more electrons.

When an atom or molecule accepts electron pairs, it is crucial to realize how these interactions lead to molecule stability. This process forms part of bond formation and is common in reactions involving elements with incomplete electron shells, especially those like boron.

Lewis Base

In contrast to a Lewis acid, a Lewis base is a species that donates an electron pair. In the given reaction, water (H₂O) acts as the Lewis base.

• The oxygen atom in H₂O has two lone pairs of electrons, one of which it donates to boron in B(OH)₃.

This donation results in the transformation of B(OH)₃ into B(OH)₄⁻.
Recognizing a Lewis base helps clarify which molecules act as electron pair donors in chemical processes. Identifying these interactions can make it easier to understand the direction and outcome of chemical reactions. By looking at Lewis structures, students can identify potential Lewis bases by locating unshared electron pairs.

Chemical Bond Formation

Chemical bonds are formed through electron pair interactions. In the reaction we are analyzing, new bonds form between molecules based on electron sharing and transfer.

• The formation of B(OH)₄⁻ occurs when an extra electron pair from a water molecule is accepted by B(OH)₃, creating a new stable bond.
• Additionally, the hydronium ion (H₃O⁺) forms when one of the water molecules acquires an H⁺ ion, making a stronger bond with oxygen.

Bond formation results in the creation of more stable compounds as atoms reach less energetic states. In reactions, this is a balance between bond breaking and forming, driven by the movement of electron pairs.
By constructing Lewis structures, students are better able to 'see' these bonds and grasp how molecular geometry influences reactivity and stability.