This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
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Alright — let's do this. 1. Hybrid Orbitals and Bonding a) Methane (CH₄) Step 1: Determine the electron configuration of the central carbon atom. Carbon has an electron configuration of 1s^2 2s^2 2p^2. To form four bonds, one electron from the 2s orbital is promoted to an empty 2p orbital, resulting in 1s^2 2s^1 2p^3. Step 2: Identify the hybridization. The carbon atom mixes one 2s orbital and three 2p orbitals to form four equivalent sp^3 hybrid orbitals. Step 3: Describe the bonding. Each of the four sp^3 hybrid orbitals on carbon contains one electron and overlaps head-on with the 1s orbital of a hydrogen atom (which also contains one electron) to form a strong sigma () bond. This results in four C-H sigma bonds. The sp^3 hybridization leads to a tetrahedral geometry around the carbon atom with bond angles of 109.5^. b) Ethyne (H-C≡C-H) Step 1: Determine the electron configuration of each carbon atom. Each carbon atom has an electron configuration of 1s^2 2s^2 2p^2. For bonding, one electron from the 2s orbital is promoted to an empty 2p orbital, resulting in 1s^2 2s^1 2p^3. Step 2: Identify the hybridization. Each carbon atom is bonded to one hydrogen and one other carbon atom, forming two electron domains. Therefore, each carbon atom undergoes sp hybridization, mixing one 2s orbital and one 2p orbital to form two sp hybrid orbitals. Two 2p orbitals remain unhybridized. Step 3: Describe the bonding. • One sp hybrid orbital on each carbon overlaps with the 1s orbital of a hydrogen atom to form a C-H sigma () bond. • The other sp hybrid orbital on each carbon overlaps head-on with the sp hybrid orbital of the adjacent carbon atom to form a C-C sigma () bond. • The two unhybridized 2p orbitals on each carbon (oriented perpendicular to each other and to the sp orbitals) overlap side-by-side with the corresponding unhybridized 2p orbitals on the adjacent carbon atom to form two pi () bonds. This results in a triple bond between the carbon atoms (one and two bonds) and two C-H sigma bonds. The sp hybridization leads to a linear geometry around each carbon atom with bond angles of 180^. 2. Valence Bond Theory and Bonding in NH₃ Step 1: Determine the electron configuration of the central nitrogen atom. Nitrogen has an electron configuration of 1s^2 2s^2 2p^3. It has one lone pair in the 2s orbital and three unpaired electrons in the 2p orbitals. Step 2: Apply the concept of hybridization within Valence Bond Theory. While the 2p orbitals could theoretically overlap with hydrogen 1s orbitals to form three bonds, this would predict bond angles of 90^. However, the experimentally observed bond angle in ammonia is approximately 107^. To account for this, Valence Bond Theory uses the concept of hybridization. The nitrogen atom undergoes sp^3 hybridization. Step 3: Describe the formation of hybrid orbitals and bonding. The nitrogen atom mixes its one 2s orbital and three 2p orbitals to form four equivalent sp^3 hybrid orbitals. • One of these sp^3 hybrid orbitals contains the lone pair of electrons. • The other three sp^3 hybrid orbitals each contain one unpaired electron and overlap head-on with the 1s orbital of a hydrogen atom (each containing one electron) to form three N-H sigma () bonds. Step 4: Explain the molecular geometry. The four sp^3 hybrid orbitals arrange themselves in a tetrahedral electron geometry. However, because one position is occupied by a lone pair, the molecular geometry of NH₃ is trigonal pyramidal. The lone pair exerts greater repulsion on the bonding pairs, compressing the H-N-H bond angles from the ideal 109.5^ to approximately 107^.