Concept:Hybridization is determined by the number of sigma bonds and lone pairs on the central atom.
Formula:$$ \text{Steric Number} = (\sigma \text{ bonds}) + (\text{lone pairs}) $$
Solution:- In methane (\( \text{CH}_4 \)), Carbon forms 4 sigma bonds and has 0 lone pairs.
- Steric number = 4, which perfectly corresponds to \( \text{sp}^3 \) hybridization (mixing one s and three p orbitals).
Why other options are incorrect:- Option B: Ethyne (\( \text{C}_2\text{H}_2 \)) has a triple bond, making its carbons \( \text{sp} \) hybridized.
- Option C: Ethene (\( \text{C}_2\text{H}_4 \)) has a double bond, making its carbons \( \text{sp}^2 \) hybridized.
- Option D: Carbon dioxide (\( \text{CO}_2 \)) has two double bonds on the central carbon, making it \( \text{sp} \) hybridized.
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