Question

Lewis structures can be used to understand why some molecules react in certain ways. Write the Lewis structure for the reactants and products in the reactions described below. a. Nitrogen dioxide dimerizes to produce dinitrogen tetroxide. b. Boron trihydride accepts a pair of electrons from ammonia, forming \(\mathrm{BH}_{3} \mathrm{NH}_{3}\). Give a possible explanation for why these two reactions occur.

Short answer

In reaction (a), two NO₂ molecules dimerize to form N₂O₄, resulting in a more stable molecule without any unpaired electrons. This occurs due to nitrogen atoms sharing their unpaired electrons. In reaction (b), the electron-deficient boron atom in BH₃ achieves a complete octet by accepting a lone pair of electrons from the nitrogen atom in NH₃, forming a more stable complex molecule. This is an example of a Lewis acid-base reaction.

Step-by-step solution

To draw the Lewis structure for NO₂, start by counting the valence electrons. Nitrogen (N) has 5 valence electrons and each oxygen (O) has 6 valence electrons. Thus, we have a total of 5 + 2(6) = 17 valence electrons. Place the nitrogen atom in the center and draw single bonds to the two oxygen atoms. This uses up 4 valence electrons (2 for each bond). Distribute the remaining 13 valence electrons as lone pairs on the oxygen atoms (each oxygen atom should have 6 valence electrons). At this point, we still have one valence electron left, which belongs to the nitrogen atom. Consequently, nitrogen has an unpaired electron, making NO₂ a free radical. Lewis structure for NO₂: O = N - O #Step 2: Draw the Lewis structure for dinitrogen tetroxide (N₂O₄)#

To draw the Lewis structure for N₂O₄, start by counting the valence electrons. We have two nitrogen atoms and four oxygen atoms. Thus, we have a total of 2(5) + 4(6) = 34 valence electrons. Place the two nitrogen atoms in the center and draw single bonds to the four oxygen atoms. This uses up 8 valence electrons (2 for each bond). Distribute the remaining 26 valence electrons as lone pairs on the oxygen atoms and the nitrogen atoms (each oxygen atom and nitrogen atom should have 6 valence electrons). Finally, dimerize the nitrogen dioxide molecules by forming a bond between the unpaired electrons on the nitrogen atoms. Lewis structure for N₂O₄: O = N - O - N = O #Step 3: Draw the Lewis structure for boron trihydride (BH₃)#

To draw the Lewis structure for BH₃, start by counting the valence electrons. Boron (B) has 3 valence electrons and each hydrogen (H) has 1 valence electron. Thus, we have a total of 3 + 3(1) = 6 valence electrons. Place the boron atom in the center and draw single bonds to the three hydrogen atoms. This uses up 6 valence electrons (2 for each bond). Lewis structure for BH₃: H - B - H | H #Step 4: Draw the Lewis structure for ammonia (NH₃)#

To draw the Lewis structure for NH₃, we start by counting the valence electrons. Nitrogen (N) has 5 valence electrons and each hydrogen (H) has 1 valence electron. Thus, we have a total of 5 + 3(1) = 8 valence electrons. Place the nitrogen atom in the center and draw single bonds to the three hydrogen atoms. This uses up 6 valence electrons (2 for each bond). Distribute the remaining 2 valence electrons as a lone pair on the nitrogen atom. Lewis structure for NH₃: H - N - H | H #Step 5: Draw the Lewis structure for the complex \(\mathrm{BH}_{3} \mathrm{NH}_{3}\) and provide a possible explanation for the reaction#

According to the reaction, BH₃ accepts a pair of electrons from the lone pair of NH₃. Hence, we form a coordinate bond between the nitrogen atom of NH₃ and the boron atom of BH₃. Lewis structure for \(\mathrm{BH}_{3} \mathrm{NH}_{3}\): H - N - H | | B - H | H Explanation: The reaction occurs because boron in BH₃ has an incomplete octet. By forming a coordinate bond with nitrogen in NH₃, boron can achieve a complete octet, making the complex more stable. #Step 6: Provide a possible explanation for reactions (a) and (b)#

Possible explanation for reaction (a): In the case of NO₂, it exists as a free radical due to its unpaired electron in the nitrogen atom. When two NO₂ molecules dimerize to form N₂O₄, they form a more stable molecule without any unpaired electrons by sharing the unpaired electron of each nitrogen atom. Possible explanation for reaction (b): The reaction between BH₃ and NH₃ is an example of a Lewis acid-base reaction. The boron atom in BH₃ has only six valence electrons, making it electron-deficient. The nitrogen atom in NH₃ has a lone pair of electrons, making it an electron donor. Thus, the reaction occurs because boron in BH₃ achieves a complete octet by accepting a lone pair of electrons from the nitrogen atom in NH₃, leading to the formation of a more stable complex molecule.

Key concepts

Molecular Reactions

Molecular reactions involve the transformation of reactants into products, often driven by the stability that molecules seek in achieving favorable electron configurations. In the case of dimerization, such as the conversion of nitrogen dioxide (NO₂) into dinitrogen tetroxide (N₂O₄), the primary driving force is stability. NO₂ molecules, which are unstable due to their free radicals, combine to form a more stable N₂O₄ without unpaired electrons, thus completing a stable molecular structure.

These reactions can often be visualized and better understood through Lewis structures, which illustrate how electrons are distributed among the atoms within a molecule. By representing both bonding and non-bonding electrons in a molecule, Lewis structures help predict the stability and reactivity of molecules involved in chemical reactions.

• Stability seeks: Molecules seek to achieve a state of lower energy and greater stability.
• Predicting reactions: Lewis structures provide insight into possible reaction pathways.
• Dimerization: A process where two identical simpler molecules combine to form a complex molecule.

Valence Electrons

Valence electrons are the outermost electrons of an atom that are involved in chemical bonding. They play a crucial role in determining how atoms interact and bond with each other. In the context of our example problems, understanding the valence electrons is key to drawing the correct Lewis structures.

For instance, nitrogen dioxide (NO₂) has one nitrogen atom with five valence electrons and two oxygen atoms with six valence electrons each. Understanding this distribution allows us to predict the formation of bonds and lone pairs. Similarly, in boron trihydride (BH₃), boron has three valence electrons which form three bonds with hydrogen atoms, each contributing one valence electron.

Mastering the concept of valence electrons enables students to predict molecular structures and potential reactivity. Correctly counting and assigning these electrons is fundamental to understanding chemistry at the molecular level.

• Outer shell: Valence electrons are found in an atom's outermost electron shell.
• Bonding power: They help determine how atoms bond and form structures.
• Lewis structures: These diagrams use valence electrons to illustrate molecule structures.

Free Radicals

Free radicals are molecules or atoms containing an unpaired electron, making them highly reactive. In chemical reactions, the presence of free radicals can significantly affect the behavior and stability of molecules. In the Lewis structure of NO₂, the nitrogen atom holds one unpaired electron, marking it as a free radical.

The unpaired electron in a free radical seeks pairing to achieve stability, thus driving reactions such as dimerization. This process can make a molecule more stable through electron sharing. Understanding the nature of free radicals can help explain why certain reactions occur and predict the existence of highly reactive species in a chemical mixture.

• Unpaired electrons: A characteristic feature of free radicals, leading to high reactivity.
• Instability: Free radicals are often unstable and seek stability through reactions.
• Dimerization: A mechanism by which free radicals generally achieve stability.

Lewis Acid-Base Reactions

Lewis acid-base reactions involve the transfer of a pair of electrons from a donor to an acceptor. In this context, a Lewis acid is an electron acceptor, while a Lewis base is an electron donor. Such reactions can be visualized through the creation of coordinate covalent bonds, where both electrons in the bond are provided by the Lewis base.

An exemplary case of this reaction type is the formation of the \(\text{BH}_3\text{NH}_3\) complex from boron trihydride (BH₃) and ammonia (NH₃). Boron, with only six valence electrons, acts as a Lewis acid needing more electrons to achieve an octet. Ammonia, with a lone pair of electrons on nitrogen, acts as a Lewis base, donating this pair to form a stable complex with boron.

• Electron donor: The role of a Lewis base in providing electron pairs in reactions.
• Electron acceptor: The role of a Lewis acid in seeking additional electrons.
• Stability: Stabilization occurs as the Lewis acid achieves a complete octet.