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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Let's find the hybridization of ethene (C₂H₄). Step 1: Identify the central atom. In ethene, we have two carbon atoms, and we'll determine the hybridization for each. Step 2: Determine the steric number for one of the carbon atoms. Each carbon atom is bonded to two other carbon atoms and two hydrogen atoms. There are no lone pairs on the carbon atoms. So, the steric number is 4 (2 atoms bonded to C + 2 H atoms + 0 lone pairs). Step 3: Relate the steric number to hybridization. A steric number of 4 corresponds to sp³ hybridization. However, ethene has a double bond between the carbon atoms. This means one of the bonds is a sigma (σ) bond and the other is a pi (π) bond. The hybridization we calculated above (sp³) is for sigma bonds and lone pairs. For atoms involved in double bonds, one of the p orbitals remains unhybridized to form the pi bond. Let's re-evaluate considering the double bond: Step 1: For each carbon atom in ethene, count the number of sigma bonds and lone pairs. Each carbon atom forms: One sigma bond with the other carbon atom. Two sigma bonds with two hydrogen atoms. Zero lone pairs. Total sigma bonds + lone pairs = 1 + 2 + 0 = 3. Step 2: Relate this count to hybridization. A count of 3 corresponds to sp² hybridization. This means one 2s orbital and two 2p orbitals hybridize to form three sp² hybrid orbitals. The remaining 2p orbital is unhybridized and forms the pi bond. The hybridization state of each carbon atom in ethene is sp². That's 2 down. 3 left today — send the next one.