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Chapter 11: Problem 12

Propose a hybridization scheme to account for bonds formed by the central carbon atom in each of the following molecules: (a) hydrogen cyanide, HCN; (b) methyl alcohol, \(\mathrm{CH}_{3} \mathrm{OH} ;\) (c) acetone, \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CO}\) (d) carbamic acid,

Short Answer

Expert verified
Hybridizations for each molecule are: HCN - sp hybridization, CH3OH - sp3 hybridization, (CH3)2CO - sp2 hybridization, NH2COOH - sp2 hybridization.

Step by step solution

01

Determine Hybridization for Hydrogen Cyanide (HCN)

In HCN, the central carbon atom is bonded to a hydrogen atom and a nitrogen atom. Therefore, it is forming two sigma bonds which will use two of the carbon atom's valence electrons. This suggests that carbon's 2s and one of the 2p orbitals will hybridize into two sp hybrid orbitals.
02

Determine Hybridization for Methyl Alcohol (CH3OH)

In CH3OH, the central carbon atom is bonded to three hydrogen atoms and one oxygen atom. The carbon atom therefore forms four sigma bonds using all its four valence electrons. This indicates that carbon's 2s and all the three 2p orbitals will hybridize into four sp3 hybrid orbitals.
03

Determine Hybridization for Acetone (CH3COCH3)

In CH3COCH3 or (CH3)2CO, the central carbon atom bonded to two other carbon atoms and one oxygen atom. This results in three sigma bonds with other atoms and one pi bond with the oxygen atom, therefore using all its four valence electrons. Hence, it can be said that carbon's 2s and two of the 2p orbitals will hybridize into three sp2 hybrid orbitals.
04

Determine Hybridization for Carbamic Acid (NH2COOH)

In NH2COOH, the central carbon atom is bonded to one nitrogen atom, one oxygen atom and one hydroxyl group. This results in three sigma bonds with other atoms and one pi bond with one of the oxygen atoms, using all its four valence electrons. As such, it can be inferred that carbon's 2s and two of the 2p orbitals will hybridize into three sp2 hybrid orbitals.

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Most popular questions from this chapter

Consider the molecules \(\mathrm{NO}^{+}\) and \(\mathrm{N}_{2}^{+}\) and use molecular orbital theory to answer the following: (a) Write the molecular orbital configuration of each ion (ignore the 1 s electrons). (b) Predict the bond order of each ion. (c) Which of these ions is paramagnetic? Which is diamagnetic? (d) Which of these ions do you think has the greater bond length? Explain.

Which of these diatomic molecules do you think has the greater bond energy, \(\mathrm{Li}_{2}\) or \(\mathrm{C}_{2} ?\) Explain.

Describe the molecular geometry of \(\mathrm{H}_{2} \mathrm{O}\) suggested by each of the following methods: (a) Lewis theory; (b) valence-bond method using simple atomic orbitals; (c) VSEPR theory; (d) valence-bond method using hybridized atomic orbitals.

A substance in which the valence and conduction bands overlap is (a) a semiconductor; (b) a metalloid; (c) a metal; (d) an insulator.

Represent bonding in the carbon dioxide molecule, \(\mathrm{CO}_{2},\) by \((\mathrm{a})\) a Lewis structure and \((\mathrm{b})\) the valencebond method. Identify \(\sigma\) and \(\pi\) bonds, the necessary hybridization scheme, and orbital overlap.
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