Answer
In biacetyl, there is: C-C-O bond angles of 120º, sp2 hybridisation, the four carbon atoms and two oxygen atoms lie in the same plane, and it contains 6 σ bonds and 2 π bonds.
In acetoin, there is: C-C-O bond angle of 120º, sp2 hybridisation and it contains 8 σ bonds and 1 π bond.
Work Step by Step
To determine the Lewis structures, bond angles, hybridization and planarity of biacetyl and acetoin molecules, we first need to understand their basic chemical structures.
1. Biacetyl: Biacetyl, also known as butane-2,3-dione, possesses two carbonyl (CO) groups. Its Lewis structure would normally be represented with a carbon chain backbone with double-bonded oxygen atom on each ends. Biacetyl has two carbon-carbon-oxygen (CCO) angles, both of which are found within the carbon-oxygen double bond, implying angles of 120 degrees. This is consistent with the sp2 hybridization of the carbon atoms. Answering whether atoms lie in the same plane requires looking at their hybridization, as sp2 hybridized atoms are planar.
Then, consider the number of σ (single covalent) and π (pi, those that comprise double and triple bonds) bonds. Biacetyl has 6 σ bonds and 2 π bonds.
2. Acetoin: Acetoin, also known as 3-hydroxybutan-2-one, is a molecule with a hydroxyl (OH) and carbonyl groups. In the Lewis structure of acetoin, the carbon chain serves as the backbone with an oxygen double-bonded to a terminal carbon, and a OH group attached to the neighbouring carbon atom. The CCO bond angle is determined by the carbonyl carbon's hybridization. With the double bond and the remaining two regions occupied by singular atoms, the carbonyl carbon is sp2 hybridized meaning that the CCO bond angle would be 120 degrees.
Likewise, we can evaluate the number of σ and π bonds. Acetoin has 8 σ bonds and 1 π bond.
