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.

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.

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.