indicate the hybridization of the atoms labeled a, b, c, and d (in that order) for the molecule below. you…

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.\na\nb\nc\nd\nch3\nh3co\noh\no\no\no\nh\nn - ch3\nch3\nh3c\nho\noh\nho\nsp2, sp2, sp, sp3\nsp2, sp2, sp2, sp2\nsp2, sp2, sp, sp2\nsp, sp2, sp, sp2\nsp, sp2, sp, sp3

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.\na\nb\nc\nd\nch3\nh3co\noh\no\no\no\nh\nn - ch3\nch3\nh3c\nho\noh\nho\nsp2, sp2, sp, sp3\nsp2, sp2, sp2, sp2\nsp2, sp2, sp, sp2\nsp, sp2, sp, sp2\nsp, sp2, sp, sp3

Answer

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:

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