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Q.Diagram for bonding in ethene with sp2 hybridisation.

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Step 1. Hybridize each carbon. Each carbon's one 2s and two 2p orbitals mix into three equivalent sp2 hybrid orbitals, 120° apart, all lying in one plane; the third, unhybridized 2p orbital on each carbon is left standing perpendicular to that plane (section 5.4.5).

Step 2. Form the C-H bonds. Two of each carbon's three sp2 hybrids overlap axially with a hydrogen 1s orbital, giving 4 C-H sigma bonds total (2 per carbon).

Step 3. Form the C-C sigma bond. The remaining (third) sp2 hybrid orbital on each carbon overlaps axially with its counterpart on the other carbon, giving one C-C sigma bond (sp2-sp2 overlap).

Step 4. Form the C-C pi bond. The two unhybridized p orbitals (one per carbon), both perpendicular to the molecular plane, overlap sideways (laterally) to give one C-C pi bond.

Step 5. Assemble. The C=C double bond of ethene is therefore one sigma + one pi bond; the whole molecule is planar with H-C-H and H-C-C angles close to 120°.

✓Final answer

C2H4: 4 C-H σ (sp2-s) + 1 C-C σ (sp2-sp2) + 1 C-C π (p-p), planar, angles ≈120°

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