Question

consider the atoms that are bonded to two or more other atoms. highlight the non - hydrogen atoms with an sp hybridization in red, an sp² hybridization in green, and an sp³ hybridization in blue.

Explanation:

Step1: Recall hybridization rules

• $sp$ hybridization: Atom has 2 electron - group around it (linear geometry).
• $sp^{2}$ hybridization: Atom has 3 electron - group around it (trigonal planar geometry).
• $sp^{3}$ hybridization: Atom has 4 electron - group around it (tetrahedral geometry).

Step2: Analyze the given molecule

• In the given molecule, the oxygen in the $-O - CH_3$ group has 4 electron - groups (2 bonding pairs and 2 lone pairs), so it is $sp^{3}$ hybridized (highlight in blue).
• The carbon atoms in the aromatic rings have 3 electron - groups around them (3 $\sigma$ - bonds), so they are $sp^{2}$ hybridized (highlight in green).
• The carbonyl carbon ($C = O$) has 3 electron - groups (2 $\sigma$ - bonds and 1 $\pi$ - bond), so it is $sp^{2}$ hybridized (highlight in green).
• The carbon attached to the carbonyl and the aromatic ring and the methyl group has 4 electron - groups, so it is $sp^{3}$ hybridized (highlight in blue).
• There are no $sp$ hybridized non - hydrogen atoms in this molecule.

Answer:

Highlight the oxygen in the $-O - CH_3$ group and the carbon attached to the carbonyl, aromatic ring and methyl group in blue (for $sp^{3}$ hybridization). Highlight the carbon atoms in the aromatic rings and the carbonyl carbon in green (for $sp^{2}$ hybridization). Do not highlight any atoms in red as there are no $sp$ hybridized non - hydrogen atoms.