Question

indicate the hybridization of the atoms labeled a, b, c, and d (in that order) for the molecule below. you may want to draw the non - bonding electrons (lone pairs of electrons) and/or hidden hydrogens to give each atom an octet before making your determination.
a
b
c
d
ch3
h3co
oh
o
o
o
h
n - ch3
ch3
h3c
ho
oh
ho
sp2, sp2, sp, sp3
sp2, sp2, sp2, sp2
sp2, sp2, sp, sp2
sp, sp2, sp, sp2
sp, sp2, sp, sp3

Explanation:

Step1: Analyze atom A

Atom A is a carbonyl - carbon. It has 3 bonding regions (double - bond to oxygen and single - bonds to two other carbons), so it is $sp^{2}$ hybridized.

Step2: Analyze atom B

Atom B is a carbon in an aromatic ring. It has 3 bonding regions (two single - bonds to carbons and one double - bond to a carbon), so it is $sp^{2}$ hybridized.

Step3: Analyze atom C

Atom C is a carbon in a triple - bond. It has 2 bonding regions (triple - bond to another carbon and a single - bond to an oxygen), so it is sp hybridized.

Step4: Analyze atom D

Atom D is a nitrogen atom. It has 4 electron - pair regions (3 bonding regions to hydrogen, carbon, and lone - pair of electrons), so it is $sp^{3}$ hybridized.

Answer:

$sp^{2}, sp^{2}, sp, sp^{3}$