Chapter 9, Problem 36QP

Interpretation Introduction

Interpretation:

The hybrid orbitals of carbon atoms in the given molecules are to be determined.

Concept introduction:

Hybridization is the combining of atomic orbitals to form hybrid orbitals.

To determine hybridization of an atom, first draw the Lewis structure of the molecule.

Find the number of electron domains around an atom so as to get the number of hybrid orbitals used by the atom for bonding.

When atomic orbitals combine, they form equal number of hybrid orbitals.

The s orbital combines with one, two, or three p orbitals to form $\text{sp,}{\text{sp}}^{\text{2}},{\text{ or sp}}^{\text{3}}$ hybrid orbitals, respectively.

Answer to Problem 36QP

Solution:

a) $s{p}^3,s{p}^3$

b) $s{p}^3,s{p}^2,s{p}^2$

c) $s{p}^3,sp,sp,s{p}^3$

d) $s{p}^3,s{p}^2$

e) $s{p}^3,s{p}^2$

a)

Explanation of Solution

The Lewis structure of ${\text{CH}}_{\text{3}}-{\text{CH}}_{\text{3}}$ molecule is as follows:

In this, the two carbon atoms are bonded to each other and the three hydrogen atoms by single bonds. Thus, there are four electron domains around each carbon atom. Four electron domainsdepict thepresence of four hybrid orbitals. Thus, in both carbon atoms $s{p}^3$ hybrid orbitals are present.

b)

The Lewis structure of ${\text{CH}}_{\text{3}}{\text{CH=CH}}_{\text{2}}$ molecule is as follows:

The carbon on the left is bonded to three hydrogen atoms by single bonds and to a carbon atom by a single bond. Thus, there are four electron domains around each carbon atom. Four electron domainsdepict thepresence of four hybrid orbitals. Thus, in this carbon $s{p}^3$ hybrid orbitals are present.

The central carbon is bonded to one carbon atom by a single bond, to another by double bond and to a hydrogen by a single bond. Thus, there are three electron domains around this carbon atom. Three electron domainsdepict thepresence of three hybrid orbitals. Thus, hybridization of the central carbon is $s{p}^2$.

The carbon on the right is bonded to two hydrogen atoms by a single bond and to a carbon by a double bond. Thus, there are three electron domains around this carbon atom. Three electron domainsdepict thepresence of three hybrid orbitals. Thus, hybridization of the carbon on the right is $s{p}^2$.

Thus, the hybridizations of the carbon atoms, from left to right in the molecule, are $s{p}^3,s{p}^2,\text{ and }s{p}^2$.

c)

The Lewis structure of ${\text{CH}}_{\text{3}}{\text{-C=C-CH}}_{\text{2}}\text{OH}$ molecule is as follows:

The carbon on left is bonded to three hydrogen atoms by single bonds and to a carbon atom by single bond. Thus, there are four electron domains around each carbon atom. Four electron domainsdepict thepresence of four hybrid orbitals. Thus, in this carbon, ${\text{sp}}^{\text{3}}$ hybrid orbitals are present.

The two carbon atoms in the center are bonded to each other by double bonds and to a carbon by a single bond. Thus, there are two electron domains around each carbon atom. Two electron domains depict the presence of two hybrid orbitals. Thus, in these two carbons, sp hybrid orbitals are present.

The carbon on the right is bonded to two hydrogen atoms by single bonds, to a carbon atom by a single bond and to an oxygen by a single bond. Thus, there are four electron domains around this carbon atom. Four electron domainsdepict the presence of four hybrid orbitals. Thus, in this carbon, ${\text{sp}}^{\text{3}}$ hybrid orbitals are present.

Thus, the hybridizations of the carbon atoms from left to right in the molecule are $s{p}^3,sp,sp,\text{ and }s{p}^3$.

d)

The Lewis structure of ${\text{CH}}_{\text{3}}\text{CH=O}$ molecule is as follows:

The carbon on the left is bonded to three hydrogen atoms by single bonds and to a carbon atom by a single bond. Thus, there are four electron domains around each carbon atom. Four electron domainsdepict the presence of four hybrid orbitals. Thus, in this carbon $s{p}^3$ hybrid orbitals are present.

The carbon on the right is bonded to a hydrogen atom by a single bond and to an oxygen atom by a double bond. Thus, there are three electron domains around this carbon atom. Three electron domainsdepict the presence of three hybrid orbitals. Thus, the hybridization of the carbon atom on the right is $s{p}^2$.

Thus, the hybridizations of carbon atoms from left to right in the molecule are $s{p}^3\text{ and }s{p}^2$.

e)

The Lewis structure of ${\text{CH}}_{\text{3}}\text{COOH}$ molecule is as follows:

The carbon on left is bonded to three hydrogen atoms by single bonds and to carbon atom by single bond. Thus, there are four electron domains around each carbon atom. Four electron domainsdepict the presence of four hybrid orbitals. Thus, in this carbon $s{p}^3$ hybrid orbitals are present.

The carbon on the right is bonded to an oxygen atom by a double bond and to another oxygen atom by a single bond. Thus, there are three electron domains around this carbon atom. Three electron domainsdepict the presence of three hybrid orbitals. Thus, the hybridization of the carbonatomon the right is $s{p}^2$.

Thus, the hybridizations of carbon atoms from left to right in the molecule are $s{p}^3\text{ and }s{p}^2$.