Problem 62 - 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. $$ \mathrm{SeO}_{3}(g)+\mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow \mathrm{H}_{2} \mathrm{SeO}_{4}(a q) $$

Step-by-Step Solution

Answer

In the reaction SeO₃(g) + H₂O(l) → H₂SeO₄(aq), the water molecule (H₂O) acts as the Lewis base by donating a lone pair of electrons to selenium in SeO₃. In contrast, SeO₃ acts as the Lewis acid by accepting the electron pair donation from H₂O.

Step 1: Draw the Lewis structures

In this step, we will draw the Lewis structures for the reactants and the product. They are the following: - SeO3: Selenium (Se) forms double bonds with each of the three oxygen (O) atoms. Each oxygen atom has two lone pairs of electrons while selenium has no lone pairs. - H2O: Oxygen (O) forms single bonds with two hydrogen (H) atoms. Oxygen has two lone pairs of electrons, and each hydrogen atom has no lone pairs. - H2SeO4: Oxygen (O) forms a double bond with selenium (Se), and three oxygen atoms form a single bond with selenium. Additionally, one oxygen atom forms a single bond to each of the two hydrogen atoms. Each hydrogen has no lone pairs, two oxygens in the substrate have two lone pairs each, selenium has no lone pairs and oxygen that forms a bond with hydrogen atoms has three lone pairs each.

Step 2: Identify electron pair movements and bond formation/breaking

Next, we will examine how electron pairs move and how bonds form and break in the given reaction. The process occurs as follows: 1. One of the lone pairs of electrons on the oxygen atom in H₂O donates an electron pair to selenium (Se) in SeO₃, creating a bond between selenium (Se) and the oxygen (O) atom in H₂O. 2. After this process, H₂SeO₄ is formed, and the Lewis structure from step 1 reflects the new bonding.

Step 3: Determining the Lewis acid and Lewis base

Now that we've seen the movement of electron pairs and bond formation, let's determine the Lewis acid and Lewis base. A Lewis acid is a species that accepts electron pairs, while a Lewis base is a species that donates electron pairs. In our reaction, the oxygen in water (H₂O) donates a lone pair of electrons to form a bond with selenium in SeO₃. Thus, H₂O is the Lewis base. On the other hand, selenium in SeO₃ accepts the electron pair donation from H₂O, making SeO₃ the Lewis acid.

Key Concepts

Lewis Acid

In chemistry, a Lewis acid is defined as any substance that can accept an electron pair. In the context of the given reaction, selenium trioxide, represented as SeO₃, acts as the Lewis acid.
When SeO₃ comes into contact with water (H₂O), it is capable of accepting an electron pair from the water molecule.
This acceptance forms a new bond between selenium and oxygen, which is the core of understanding Lewis acids in this scenario.

• Lewis acids are key in reactions where bond formation involves electron pair acceptance.
• The usual identifier of a Lewis acid is the presence of an element or compound able to attain a lower energy state by receiving electron pairs.

Understanding the role of Lewis acids helps us see how new molecular structures are created where atoms like selenium can take on new partners, utilizing available electrons.

Lewis Base

A Lewis base is a chemical species that donates an electron pair to a Lewis acid to form a covalent bond. In our reaction, the water molecule, H₂O, is the Lewis base.
Water donates a lone pair of electrons from its oxygen atom to the selenium atom in selenium trioxide (SeO₃). This donation is what helps facilitate a new bond creation.
Knowing how Lewis bases operate in reactions is crucial because:

• They often initiate reactions by donating electrons.
• They can stabilize positive charges, which leads to greater molecular stability.

As we see in the example of H₂O, the readiness to donate electrons plays a pivotal role in molecular interactions and transformations.

Electron Pair Movement

The movement of electron pairs is an essential part of understanding chemical reactions, as it dictates how bonds are formed or broken. Electron pair movement in our example starts with a lone pair on the oxygen atom in water.
This lone pair gets attracted to the selenium atom in SeO₃, initiating a chemical bond. As the electrons move from the water to selenium, a new coordinated bond forms between these two species.
This movement of electron pairs is fundamental because:

• It leads to the stabilization of the reactive species by forming new bonds.
• These movements can also result in the breaking of existing bonds, facilitating the transformation of reactants to products.

Grasping electron pair movement offers insights into the dynamic nature of molecules and their tendency to reach more stable configurations.

Bond Formation and Breaking

Bond formation and breaking are two sides of the same coin in chemical reactions. In this reaction between SeO₃ and H₂O, a critical transformation occurs.
Upon donation of electrons from the oxygen in water to the selenium in SeO₃, a new S-O bond forms. This bond formation is a clear result of electron pair movement from one atom to another.
Concurrently, the restructuring of electron arrangements can lead to the breaking of weaker bonds to accommodate the new ones.

• Bond formation provides structural stability and energy release.
• Bond breaking requires energy but allows molecules to transform and adapt to new configurations.

Understanding the intricacies of these processes is crucial for predicting reaction outcomes and visualizing molecular shifts.