Which best describes any unhybridized p orbitals on the central carbon atom of ethene (C2H4)?

• A. They participate in the formation of the pi bond in the C=C double bond.
• B. They are involved in forming lone pairs on the carbon atom.
• C. They are used to form sigma bonds with hydrogen atoms.
• D. They remain non-bonding and do not interact with other atoms.

Answer: (A)

Verified step by step guidance

1. Identify the hybridization of the central carbon atoms in ethene (C2H4). Each carbon in ethene forms three sigma bonds: two with hydrogen atoms and one with the other carbon atom, which suggests sp2 hybridization.
2. Recall that sp2 hybridization involves mixing one s orbital and two p orbitals to form three sp2 hybrid orbitals, which lie in a plane and form sigma bonds.
3. Understand that the remaining unhybridized p orbital on each carbon atom is perpendicular to the plane of the sp2 orbitals and is not involved in sigma bonding.
4. Recognize that these unhybridized p orbitals on each carbon overlap side-by-side to form a pi (π) bond, which is part of the double bond (C=C) in ethene.
5. Conclude that the unhybridized p orbitals on the central carbon atoms in ethene participate in the formation of the pi bond in the C=C double bond, rather than forming lone pairs, sigma bonds with hydrogen, or remaining non-bonding.