Question

Calcium carbide \(+\,D_2O \rightarrow X + Ca(OD)_2\). The hybridization of carbon atom/s in \(X\)

• A. \(sp^2\)
• B. \(sp\)
• C. \(sp^3\)
• D. \(dsp^2\)

Hint:

Carbon atoms involved in a triple bond always show \[ sp \] hybridization and possess linear geometry with bond angle \[ 180^\circ \]

Answer 1

The Correct Option is B

Step 1: Write the reaction of calcium carbide with heavy water.
Calcium carbide reacts with $D_2O$ analogously to its reaction with ordinary water: \[ CaC_2 + 2D_2O \rightarrow C_2D_2 + Ca(OD)_2 \] Product X = $C_2D_2$ (deuterated acetylene/ethyne).
Step 2: Write the structure of $C_2D_2$.
Structure: $D-C \equiv C-D$ with a carbon-carbon triple bond.
Step 3: Determine hybridization based on sigma bonds formed.
Each carbon forms 2 sigma bonds: one with D and one in the C-C triple bond. Two sigma bonds require 2 hybrid orbitals, characteristic of $sp$ hybridization. The remaining 2 pure p-orbitals form 2 pi bonds of the triple bond.
Step 4: Verify geometry implied by $sp$ hybridization.
$sp$ hybridization gives linear geometry with bond angle $180^\circ$. The molecule $D-C \equiv C-D$ is indeed linear.
Step 5: Distinguish from $sp^2$ and $sp^3$ hybridization.
$sp^3$ forms 4 sigma bonds (tetrahedral); $sp^2$ forms 3 sigma bonds (trigonal planar). Only 2 sigma bonds per carbon confirms $sp$ hybridization uniquely.
Step 6: State the final answer.
\[ \boxed{sp} \]