Building a model of the 1,2-ethanediol molecule (C2H6O2) can help us understand its structure and geometry. Let’s break this down into two parts.
a. Identifying the Electron Pair and Molecular Geometry
In 1,2-ethanediol, there are two carbon atoms and two oxygen atoms. Each carbon (C) atom is bonded to two hydrogen (H) atoms and one hydroxyl group (-OH). The two oxygen atoms are also bonded to one hydrogen atom each, forming the hydroxyl groups.
1. **Carbon Atoms:** Each carbon atom is sp3 hybridized due to four regions of electron density: two C-H bonds, one C-O bond, and one non-bonding pair, leading to a tetrahedral arrangement. The molecular geometry around each carbon is tetrahedral.
2. **Oxygen Atoms:** Each oxygen atom is bonded to a carbon atom and a hydrogen atom. The molecular geometry around each oxygen is bent or angular due to the presence of two bond pairs (one C-O bond and one O-H bond) and two lone pairs of electrons.
b. Bond Angles for H-C-H
The bond angles in 1,2-ethanediol can vary slightly due to steric effects and the presence of lone pairs. However, when focusing on the H-C-H bond angle around the carbon atom:
The typical H-C-H bond angle in a tetrahedral geometry is approximately 109.5 degrees. Thus, in the case of 1,2-ethanediol, the H-C-H bond angles are close to this value, indicating an ideal tetrahedral arrangement.
In summary, building a model of 1,2-ethanediol reveals its tetrahedral geometry around carbon atoms and bent geometry around oxygen atoms, with H-C-H bond angles near 109.5 degrees.