Given the molecule $C_2{H}_4$ (ethene), determine the hybridization and geometry around each carbon atom.

• A. Each carbon is ${\mathrm{sp}}_3$ hybridized with tetrahedral geometry.
• B. Each carbon is ${\mathrm{sp}}_2$ hybridized with trigonal planar geometry.
• C. Each carbon is ${\mathrm{sp}}_1$ hybridized with linear geometry.
• D. Each carbon is ${\mathrm{sp}}_1$ hybridized with trigonal planar geometry.

Answer: (B)

Verified step by step guidance

1. Identify the molecular formula given: C\_2H\_4, which corresponds to ethene, a molecule with a carbon-carbon double bond.
2. Recall that the hybridization of a carbon atom depends on the number of regions of electron density (bonds and lone pairs) around it. For ethene, each carbon forms three sigma bonds: two with hydrogen atoms and one with the other carbon atom.
3. Since each carbon has three sigma bonds and one pi bond (from the double bond), the hybridization is sp\_2. This involves mixing one s orbital and two p orbitals to form three sp\_2 hybrid orbitals.
4. The geometry around each sp\_2 hybridized carbon is trigonal planar, meaning the atoms bonded to the carbon lie in a plane with approximately 120° bond angles.
5. Summarize that each carbon in ethene is sp\_2 hybridized and has trigonal planar geometry, which explains the planar structure and bond angles observed in the molecule.