Reaction 1: CaC2(s) + 2 H2O(l) ⇄ C2H2(g) + Ca(OH)2(s) Reaction 2: NaOCl(aq) + 2 HCl(aq) ⇄ Cl2(g) + N — Bonding Chemistry Question

STEPS:

1. Analyze the molecular structure of the reactant, acetylene ($\text{C}_2\text{H}_2$): In $\text{C}_2\text{H}_2$, the two carbon atoms share a carbon-carbon triple bond, and each carbon is also bonded to one hydrogen atom via a single bond ($\text{H}-\text{C}\equiv\text{C}-\text{H}$).
2. Determine the steric number of carbon in $\text{C}_2\text{H}_2$: The steric number is the sum of the number of bonded atoms and the number of lone pairs on the central atom. Each carbon atom has 0 lone pairs and is bonded to exactly two atoms (one $\text{H}$ and one $\text{C}$), which gives it a steric number of 2.
3. Determine the hybridization of carbon in $\text{C}_2\text{H}_2$: A steric number of 2 corresponds to $\text{sp}$ hybridization (resulting in a linear molecular geometry).
4. Analyze the molecular structure of the product, dichloroethene ($\text{C}_2\text{H}_2\text{Cl}_2$): During Reaction 3, the addition of chlorine gas breaks one of the $\pi$ bonds of the triple bond, converting it into a carbon-carbon double bond. Each carbon is now bonded to one hydrogen atom, one chlorine atom, and the other carbon atom ($\text{Cl}-\text{CH}=\text{CH}-\text{Cl}$).
5. Determine the steric number of carbon in $\text{C}_2\text{H}_2\text{Cl}_2$: Each carbon atom still has 0 lone pairs but is now bonded to exactly three atoms (one $\text{H}$, one $\text{Cl}$, and one $\text{C}$), which increases its steric number to 3.
6. Determine the hybridization of carbon in $\text{C}_2\text{H}_2\text{Cl}_2$: A steric number of 3 corresponds to $\text{sp}^2$ hybridization (resulting in a trigonal planar geometry).
7. Conclude the hybridization change: Comparing the reactant to the product, the hybridization of the carbon atoms changes from $\text{sp}$ to $\text{sp}^2$, which directly confirms Option A as the correct answer.

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WHY_OTHERS_WRONG:

• Option B is incorrect: This option claims the hybridization changes to $\text{sp}^3$. $\text{sp}^3$ hybridization corresponds to a steric number of 4, which would occur if the carbon atoms formed four single bonds (such as in tetrachloroethane, $\text{C}_2\text{H}_2\text{Cl}_4$). Because dichloroethene ($\text{C}_2\text{H}_2\text{Cl}_2$) retains a double bond, each carbon has a steric number of only 3 ($\text{sp}^2$).
• Option C is incorrect: This option claims the starting hybridization of carbon in acetylene ($\text{C}_2\text{H}_2$) is $\text{sp}^2$. Acetylene contains a triple bond, making it $\text{sp}$ hybridized. $\text{sp}^2$ would only be the starting hybridization if we were reacting an alkene (such as ethene, $\text{C}_2\text{H}_4$), rather than an alkyne.
• Option D is incorrect: This option claims there is no hybridization change. The addition reaction decreases the carbon-carbon bond order from 3 to 2, increasing the steric number of each carbon from 2 to 3. This change in molecular geometry requires a corresponding change in orbital hybridization.