Chapter 3, Problem 3.34P

Interpretation Introduction

Interpretation:

The electron geometry, hybridization, and molecular geometry of each nonhydrogen atoms in levomenol are to be determined.

Concept introduction:

According to the Valence Shell Electron Pair Repulsion (VSEPR) theory, the geometry is identified based on groups attached to the central atom. Electron geometry depends on the total number of electrons groups attached to the central atom. Molecular geometry depends on the total number of bonded atoms or groups to the central atom.

The number of electron groups and geometry is determined on the basis of the Lewis structure of the molecule/ion.

According to the Valence bond (VB) theory, the hybridization of atomic orbitals is used to account for electron and molecular geometry around the central atom. A hybrid orbital is a combination of one or more atomic orbitals from the valence shell of an atom.

The number of hybrid orbitals required is the same as the number of electron groups. If the number of electron groups is two, two-hybrid orbitals are needed. These are formed by a combination of s and one p orbital, giving $\text{sp}$ hybridization. For three electron groups, three hybrid orbitals are required, s and two of the p orbitals giving ${\text{sp}}^2$ hybridization. Four groups require four orbitals giving ${\text{sp}}^{\text{3}}$ hybridization.

Answer to Problem 3.34P

The electron geometry, hybridization, and molecular geometry of each nonhydrogen atoms in levomenol are

Type	Number of C atom on the chain	Electron Geometry	Molecular Geometry	Hybridization
${\text{CH}}_{\text{3}}$	$\text{7, 8, 12, and 15}$	Tetrahedral	Tetrahedral	${\text{sp}}^{\text{3}}$ for each C
${\text{CH}}_{\text{2}}$	$\text{3, 4, 9, 13, and 14}$	Tetrahedral	Tetrahedral	${\text{sp}}^{\text{3}}$ for each C
$\text{OH}$	$\text{2}$	Tetrahedral	Bent	${\text{sp}}^{\text{3}}$ for O
C	$\text{1}$	Tetrahedral	Tetrahedral	${\text{sp}}^{\text{3}}$
C	$\text{2}$	Tetrahedral	Tetrahedral	${\text{sp}}^{\text{3}}$
C	$\text{10}$	Trigonal planar	Trigonal planar	${\text{sp}}^{\text{2}}$
C	$\text{11}$	Trigonal planar	Trigonal planar	${\text{sp}}^{\text{2}}$
C	$5$	Trigonal planar	Trigonal planar	${\text{sp}}^{\text{2}}$
C	$6$	Trigonal planar	Trigonal planar	${\text{sp}}^{\text{2}}$

Explanation of Solution

The Lewis structure of levomenol is drawn based on the total valence electrons in the structure. The carbon atom has four valence electrons, hydrogen has one valence electron, and oxygen has six valence electrons. We need to add lone pairs on the oxygen atom to complete the octet.

The electron geometry, hybridizations, and molecular geometry of nonhydrogen atoms of each carbon and oxygen atoms are determined on the basis of the number of electron groups around each as:

1. C in each ${\text{CH}}_{\text{3}}$ group:

There are four groups of electrons around the carbon atom - three single bonds with hydrogen and one single bond with carbon. No lone pair is present on the central carbon atom. According to VSEPR theory, electron geometry and molecular geometry of C in each ${\text{CH}}_{\text{3}}$ group is tetrahedral.

Each carbon has four electron groups around it and needs four hybrid orbitals. Therefore, the hybridization of carbon in each ${\text{CH}}_{\text{3}}$ group is ${\text{sp}}^3$.

1. C in each ${\text{CH}}_{\text{2}}$ group:

There are four groups of electrons around the carbon atom - two single bonds with hydrogen and two single bonds with carbon. No lone pair is present on the central carbon atom. According to VSEPR theory, electron geometry and molecular geometry of C in each ${\text{CH}}_{\text{2}}$ group is tetrahedral.

Each carbon has four electron groups around it and needs four hybrid orbitals. Therefore, the hybridization of carbon in each ${\text{CH}}_{\text{2}}$ group is ${\text{sp}}^3$.

1. O in $\text{-OH}$ group.

There are four groups of electrons around the oxygen atom - one single bond with hydrogen, one single bond with carbon, and two lone pairs on it. According to VSEPR theory, its electron geometry is tetrahedral, and its molecular geometry is bent.

Oxygen has four electron groups around it and needs four hybrid orbitals. Therefore, the hybridization of the oxygen atom is ${\text{sp}}^3$.

1. C in ${\text{C}}_{\text{1}}\text{and}{\text{C}}_{\text{2}}$ groups.

There are four groups of electrons around the carbon atom. The carbon of ${\text{C}}_{\text{1}}$ position has three single bonds with C and one single bond with hydrogen. The carbon of ${\text{C}}_{\text{2}}$ position has three single bonds with C and one single bond with oxygen. No lone pair is present on the central carbon atom. With four electron groups, the electron geometry and molecular geometry of both carbon is tetrahedral.

Each carbon has four electron groups around it and needs four hybrid orbitals. Therefore, the hybridization of both carbon is ${\text{sp}}^3$.

1. C in $\text{C=C}$ groups.

There are three groups of electrons around the carbon atom in $\text{C=C}$ bond, two single bonds and one double bond, therefore three electron groups around it. According to VSEPR theory, electron geometry and molecular geometry of each C in $\text{C=C}$ groups are trigonal planar.

Each carbon has three electron groups around it and needs three hybrid orbitals. Therefore, the hybridization of each carbon is ${\text{sp}}^2$.

Conclusion

According to VSEPR theory, the electron geometry and molecular geometry is identified from valance bond theory, and hybridization is determined.