Question

Lewis structures can be used to understand why some molecules react in certain ways. Write the Lewis structures 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

The nitrogen dioxide molecules dimerize to form dinitrogen tetroxide because the oxygen atoms in the NO2 molecules are highly electronegative, which leads to an unequal sharing of electrons and the creation of a radical in the structure. By dimerizing to form N2O4, the unpaired electrons are shared and a more stable molecule is formed. Boron trihydride accepts a pair of electrons from ammonia to form BH3NH3 because boron is electron-deficient, having only 3 valence electrons in its outer shell. Ammonia can donate its lone pair of electrons to boron to form a dative bond, thereby satisfying the octet rule for boron and creating a more stable compound.

Step-by-step solution

Reaction a: Nitrogen dioxide dimerization

Step 1: Draw Lewis structures for the reactants - Nitrogen dioxide (NO2): Nitrogen has 5 valence electrons, while each Oxygen atom has 6. So, the total number of valence electrons is 17. Lewis structure: \( O=N-O^{\bullet} \) Step 2: Draw Lewis structures for the products - Dinitrogen tetroxide (N2O4): Nitrogen has 5 valence electrons and each Oxygen has 6, so the total number of valence electrons is 34. Lewis structure: \( O_{2}N-O-N_{2}O \) Step 3: Provide a possible explanation for the reaction The nitrogen dioxide molecules dimerize to form dinitrogen tetroxide because the oxygen atoms in the NO2 molecules are highly electronegative, which leads to an unequal sharing of electrons and the creation of a radical in the structure. By dimerizing to form N2O4, the unpaired electrons are shared and a more stable molecule is formed.

Reaction b: Boron trihydride accepting electrons from ammonia

Step 1: Draw Lewis structures for the reactants - Boron trihydride (BH3): Boron has 3 valence electrons, while each hydrogen has 1, making a total of 6 valence electrons. Lewis structure: \( H-B-H \) | H - Ammonia (NH3): Nitrogen has 5 valence electrons, and each hydrogen has 1, making a total of 8 valence electrons. Lewis structure: \( H \) | N-H | H Step 2: Draw Lewis structures for the products - BH3NH3: The total number of valence electrons is 14. Lewis structure: \( H-B-H-N-H \) | | H H Step 3: Provide a possible explanation for the reaction Boron trihydride accepts a pair of electrons from ammonia to form BH3NH3 because boron is electron-deficient, having only 3 valence electrons in its outer shell. Ammonia can donate its lone pair of electrons to boron to form a dative bond, thereby satisfying the octet rule for boron and creating a more stable compound.

Key concepts

Chemical Bonding

Chemical bonding forms the fundamental basis of understanding how molecules interact, form, or even break apart. In chemistry, bonds are the forces that hold atoms together.
One main type of chemical bonding is covalent bonding. Here, atoms share electrons. Hydrogen (H) and carbon (C) in methane ( \( \text{CH}_4 \)) showcase covalent bonds, where each hydrogen shares electrons with carbon.

• Covalent bonds are often observed in organic compounds and molecules where nonmetals are present, such as water ( \( \text{H}_2\text{O} \)).
• Another form of bonding is ionic bonding, which involves the complete transfer of electrons from one atom to another. This usually occurs between metals and nonmetals, as seen in sodium chloride ( \( \text{NaCl} \)).

Understanding the types of chemical bonds in a molecule helps predict its behavior, reactivity, and physical properties.

Reactivity of Molecules

The reactivity of molecules strongly depends on their structure, including the types of bonds and any lone pairs of electrons that might be present.
Reactivity often involves breaking old bonds and forming new ones. For example, in a dimerization reaction like that of nitrogen dioxide ( \( \text{NO}_2 \)) to form dinitrogen tetroxide ( \( \text{N}_2\text{O}_4 \)), radical reactions occur. The unpaired electron in \( \text{NO}_2 \) makes it more reactive, leading it to pair up and stabilize into \( \text{N}_2\text{O}_4 \).

• Molecules with unpaired electrons or radicals are often highly reactive because these electrons tend to pair up to achieve a lower energy state.
• Those that have fulfilled their valence shell (octet) tend to be less reactive as they are already stable.
• Lone pairs on atoms, like in nitrogen in ammonia ( \( \text{NH}_3 \)), can also affect reactivity by forming dative covalent bonds as seen with boron trihydride ( \( \text{BH}_3 \)).

Valence Electrons

Valence electrons are the outermost electrons of an atom and play a key role in chemical bonding and reactions. These electrons are the ones involved in forming chemical bonds, whether they be covalent or ionic.
For example, nitrogen in ammonia ( \( \text{NH}_3 \)) has 5 valence electrons. Its valence electrons allow it to bond with hydrogen atoms and can provide an additional lone pair to form a dative bond with boron in \( \text{BH}_3 \).

• Understanding the number of valence electrons helps predict how atoms will bond and the type of bonds they will form.
• Elements strive to achieve a full valence shell, following the octet rule for stability, as seen in noble gases like neon ( \( \text{Ne} \)), which are rarely reactive due to having a complete valence shell.
• Reactions occur when atoms or molecules interact to either share, gain, or lose electrons to stabilize their electronic configuration.

Recognizing valence electrons also aids in drawing Lewis structures, which provide a visual representation of molecule formation.