Chapter 3, Problem 3.33P

Interpretation Introduction

(a)

Interpretation:

The electron geometries of all nonhydrogen atoms in the six listed species 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. The electron geometry depends on the total number of electrons groups attached to the central atoms. Electron geometry describes the orientation of the electron groups in an atom’s valence shell. An electron group is a lone pair or a bond between the two atoms. The bond, whether a single, double or triple, is considered as just one electron group.

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

Answer to Problem 3.33P

The electrons geometries of nonhydrogen atoms in the listed species are:

1. ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{}^{\text{+}}$: Tetrahedral for C and trigonal planar for ${\text{C}}^{\text{+}}$ carbon.
2. ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{}^{\text{-}}$: Tetrahedral for both C.
3. ${\text{CH}}_{\text{2}}{\text{=OH}}^{\text{+}}$: Trigonal planar for both C and O.
4. ${\text{CH}}_{\text{6}}{\text{N}}^{\text{+}}$: Tetrahedral for both C and N.
5. ${\text{CH}}_{\text{5}}{\text{O}}^{\text{+}}$: Tetrahedral for both C and O.
6. ${\text{C}}_{\text{3}}{\text{H}}_{\text{3}}{}^{\text{-}}$: Tetrahedral for left C, linear for both triple bonded C.

Explanation of Solution

The Lewis structure is drawn based on the total valence electrons in the structure. The carbon atom has four valence electrons, hydrogen has one valence electrons and oxygen has six valence electrons.

1. The Lewis structure of ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{}^{\text{+}}$ is:

Left side carbon has four electron groups around it, all bonding groups. With four electron groups, the electron geometry around carbon is tetrahedral.

Right side carbon has three electron groups around it, all bonding groups and no lone pair on it. With three electron groups, the electron geometry around carbon is trigonal planar.

1. The Lewis structure ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{}^{\text{-}}$ of is:

Left side carbon has four bonds, and therefore four electron groups around it. With four electron groups, the electron geometry around carbon is tetrahedral.

Right side carbon has four electron groups around it, three bonding groups and one lone pair on it. According to the VSEPR theory, the electron geometry around carbon is tetrahedral.

1. The Lewis structure of ${\text{CH}}_{\text{2}}{\text{=OH}}^{\text{+}}$ is:

In this species, the carbon atom has two single bonds and one double bond, therefore three electron groups around it. According to the VSEPR theory, its electron geometry is trigonal planar.

Oxygen has one single bond with hydrogen, one double bond with carbon and one lone pair on it, therefore three electron groups around it. According to the VSEPR theory, its electron geometry is trigonal planar.

1. The Lewis structure of ${\text{CH}}_{\text{6}}{\text{N}}^{\text{+}}$ is:

In this species, there are four groups of electrons around the carbon atom, one single bond with nitrogen and three single bonds with hydrogen atoms. No lone pair is present on the central carbon atom. According to the VSEPR theory, its electron geometry is tetrahedral.

Nitrogen has four electrons groups, three single bonds with hydrogen and one single bond with carbon. Therefore, the electron geometry around the nitrogen atom is tetrahedral.

1. The Lewis structure of ${\text{CH}}_{\text{5}}{\text{O}}^{\text{+}}$ is:

In this species, there are four groups of electrons around the carbon atom, one single bond with oxygen and three single bonds with hydrogen atoms. No lone pair is present on the central carbon atom. According to the VSEPR theory, its electron geometry is tetrahedral.

Oxygen has four electrons groups, two single bonds with hydrogen, one single bond with carbon and one lone pair on it. Therefore, the electron geometry around the oxygen atom is tetrahedral.

1. The Lewis structure of ${\text{C}}_{\text{3}}{\text{H}}_{\text{3}}{}^{\text{-}}$ is:

Left side carbon has four groups of electrons around it, one single bond with carbon and three single bonds with hydrogen atoms. According to the VSEPR theory, its electron geometry is tetrahedral.

Middle carbon has one single bond with left side carbon and triple bond with right side carbon. So, middle carbon has two electron groups around it. Therefore, the electron geometry of the middle carbon is linear.

Right side carbon has a triple bond with middle carbon and one lone pair electrons. So, the right side carbon has two electron groups around it. Therefore, the electron geometry of the middle carbon is linear.

Conclusion

The electron geometry around an atom is determined by the number of electron groups in its valence shell.

Interpretation Introduction

(b)

Interpretation:

The hybridizations of all nonhydrogen atoms in the six listed species are to be determined.

Concept introduction:

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.

Typically, the s and the p orbitals are involved in the hybridization from the valence shell, resulting in the same total number of hybrid orbitals of the same energy and shape. The number of hybrid orbitals required is 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 the s and one p orbital, giving $\text{sp}$ hybridization. For three electron groups, three hybrid orbitals are required, the 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 3.33P

The hybridizations of nonhydrogen atoms in the six listed species are:

1. ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{}^{\text{+}}$: ${\text{sp}}^{\text{3}}$ for C and ${\text{sp}}^2$ for ${\text{C}}^{\text{+}}$ carbon.
2. ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{}^{\text{-}}$: ${\text{sp}}^{\text{3}}$ for both C.
3. ${\text{CH}}_{\text{2}}{\text{=OH}}^{\text{+}}$: ${\text{sp}}^2$ for both C and O.
4. ${\text{CH}}_{\text{6}}{\text{N}}^{\text{+}}$: ${\text{sp}}^{\text{3}}$ for both C and N.
5. ${\text{CH}}_{\text{5}}{\text{O}}^{\text{+}}$: ${\text{sp}}^{\text{3}}$ for both C and O.
6. ${\text{C}}_{\text{3}}{\text{H}}_{\text{3}}{}^{\text{-}}$: ${\text{sp}}^{\text{3}}$ for left C, $\text{sp}$ for both triple bonded C.

Explanation of Solution

The hybridizations of nonhydrogen atoms in each of the four species are determined on the basis of the number of electron groups around each as:

1. The Lewis structure of ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{}^{\text{+}}$ is:

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

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

1. The Lewis structure ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{}^{\text{-}}$ of is:

Both carbons have four electron groups around them and each atom needs four hybrid orbitals. Therefore the hybridization of both atoms is ${\text{sp}}^3$.

1. The Lewis structure of ${\text{CH}}_{\text{2}}{\text{=OH}}^{\text{+}}$ is:

Carbon and oxygen both have three electron groups around them and each atom needs three hybrid orbitals. Therefore, the hybridization of both atoms is ${\text{sp}}^2$.

1. The Lewis structure of ${\text{CH}}_{\text{6}}{\text{N}}^{\text{+}}$ is:

Carbon and nitrogen both have four electron groups around them and each atom needs four hybrid orbitals. Therefore, the hybridization of both atoms is ${\text{sp}}^3$.

1. The Lewis structure of ${\text{CH}}_{\text{5}}{\text{O}}^{\text{+}}$ is:

Carbon and oxygen both have four electron groups around them and each atom needs four hybrid orbitals. Therefore, the hybridization of both atoms is ${\text{sp}}^3$.

1. The Lewis structure of ${\text{C}}_{\text{3}}{\text{H}}_{\text{3}}{}^{\text{-}}$ is:

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

The middle and right side both carbon have two electron groups around them and each atom needs two hybrid orbitals. Therefore, the hybridization of both atoms is $\text{sp}$.

Conclusion

Hybridization of an atom is determined by the number of electron groups around it.