Question

which of these is the correct lewis structure for ethyne (acetylene), c₂h₂? h—c(=c)—h (with lone pairs), :h—c=c—:h (with lone pairs), :c—h—h—:c (with lone pairs), h—c≡c—h

Explanation:

Step1: Recall Lewis Structure Rules

Lewis structures show valence electrons. For \( \text{C}_2\text{H}_2 \) (ethyne), each C has 4 valence electrons, each H has 1. Total valence electrons: \( 2\times4 + 2\times1 = 10 \).

Step2: Analyze Carbon Bonding

Carbon in ethyne forms a triple bond (\( \ce{C\equiv C} \)) to satisfy octet (each C shares 6 electrons in triple bond, plus 1 bond to H: \( 6 + 2 = 8 \) electrons for C; H has 2 electrons from single bond).

Step3: Evaluate Each Option

• First option: Double bond (\( \ce{C=C} \)) – incorrect, needs triple bond for 10 valence electrons.
• Second option: H with lone pairs – H can't have lone pairs (only 1 valence electron, forms 1 bond, no lone pairs). Incorrect.
• Third option: H atoms between C atoms – incorrect structure (H can only form 1 bond, can't be between two C).
• Fourth option: \( \ce{H - C\equiv C - H} \) – Triple bond between C, each C bonded to H. Valence electrons: Triple bond (6) + 2 single bonds (2 each) = 10. Correct.

Answer:

The correct option is the fourth one: \( \boldsymbol{\ce{H - C\equiv C - H}} \) (the last option with the triple bond between the two carbon atoms).