Question

Determine the hybridization and geometry of carbon in ethyne \((C_2H_2)\).

• A. \(sp^3\), Tetrahedral
• B. \(sp^2\), Trigonal planar
• C. \(sp\), Linear
• D. \(sp^3d\), Trigonal bipyramidal

Hint:

Hybridization shortcut: 2 regions → \(sp\) (linear) 3 regions → \(sp^2\) (trigonal planar) 4 regions → \(sp^3\) (tetrahedral)

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The question asks for the orbital mixing (hybridization) and spatial arrangement (geometry) of carbon atoms in an alkyne.
Step 2: Detailed Explanation:
In ethyne (\(H-C \equiv C-H\)), each carbon atom is bonded to one hydrogen atom (sigma bond) and one other carbon atom (sigma bond).
The triple bond consists of one sigma (\(\sigma\)) bond and two pi (\(\pi\)) bonds.
Number of sigma bonds per carbon = 2.
Number of lone pairs per carbon = 0.
Steric Number = 2 + 0 = 2.
A steric number of 2 corresponds to \(sp\) hybridization.
The VSEPR theory predicts that two electron domains will arrange themselves as far apart as possible (\(180^\circ\)), resulting in a linear geometry.
Step 3: Final Answer:
The hybridization is \(sp\) and the geometry is Linear.