Describe the molecular geometry of the central atoms in; a) CH3COOH b) CH3CHCH2 c) CH3CCCH2OH
The Correct Answer and Explanation is :
Here’s the molecular geometry of the central atoms in the given molecules:
a) CH₃COOH (Acetic acid)
- CH₃ group: The central atom (C) in CH₃ is tetrahedral due to four sigma bonds (three C-H bonds and one C-C bond).
- COOH group: The central atom (C) in the carboxylic acid group is trigonal planar because it is sp² hybridized (one C=O double bond and two sigma bonds to the hydroxyl group and CH₃ group).
b) CH₃CH=CH₂ (Propene)
- CH₃ group: The central atom (C) in CH₃ is tetrahedral for the same reason as in acetic acid: it has four sigma bonds.
- CH group: The second carbon (in CH=CH₂) is trigonal planar due to sp² hybridization (one C=C bond and two sigma bonds to the CH₃ group and one hydrogen).
- CH₂ group: The last carbon in CH₂ is also trigonal planar because it is sp² hybridized with one C=C bond and two sigma bonds to hydrogen atoms.
c) CH₃C≡CCH₂OH (Butyne-1-ol)
- CH₃ group: As before, the CH₃ carbon is tetrahedral.
- C≡C group: The central carbon in the triple bond is linear due to sp hybridization (two sigma bonds and two π bonds).
- CH₂ group: The carbon in CH₂ is tetrahedral (sp³ hybridized, with three sigma bonds to H atoms and one sigma bond to the OH group).
Explanation:
The molecular geometry is determined by the hybridization of the central atoms. Hybridization arises from the number of sigma bonds and lone pairs on a central atom. For instance:
- Tetrahedral geometry (sp³ hybridized) occurs when the atom forms four sigma bonds.
- Trigonal planar geometry (sp² hybridized) results from three sigma bonds.
- Linear geometry (sp hybridized) arises when there are two sigma bonds.
These geometries are derived from VSEPR theory, which minimizes electron-pair repulsion around the central atom. Each geometry optimizes the spatial arrangement for stability.