The hybridization scheme in ethyne is

1. Analyze the question You are asked for the hybridization of the carbon atoms in ethyne (acetylene), C₂H₂.

2. Explain the concept Hybridization depends on the number of sigma-bonded regions (steric number) around each carbon:

• 4 regions → sp³ (e.g., ethane, tetrahedral)
• 3 regions → sp² (e.g., ethene, trigonal planar)
• 2 regions → sp (e.g., ethyne, linear)

3. Solution Ethyne has the structure H–C≡C–H.

• Each carbon forms one sigma bond to hydrogen and one sigma bond to the other carbon, so each carbon has 2 regions of electron density.
• Two regions of electron density means one s orbital and one p orbital mix, giving two sp hybrid orbitals at 180° (linear).
• The two remaining unhybridized p orbitals on each carbon overlap sideways to form two pi bonds. Together with the one sigma bond, this makes the C≡C triple bond.

So the molecule is linear, with H–C–C bond angles of 180°, and each carbon is sp hybridized.

Checking the options:

• A. sp: correct.
• B. sp³: this is for alkanes such as ethane.
• C. sp²d: not a hybridization scheme for ethyne.
• D. sp²s: not a valid scheme for ethyne.

4. Answer: The correct option is A. sp.

5. Exam tip Remember the pattern with the "bond type" trick:

• Single bond (alkane) → sp³ → 109.5°
• Double bond (alkene) → sp² → 120°
• Triple bond (alkyne) → sp → 180°

The more s-character, the shorter and stronger the bond. Keep going, you are doing well!