Chapter 3, Problem 3.27P

Interpretation Introduction

(a)

Interpretation:

Among molecules mentioned in the question, it is to be determined which has shorter $\text{C-C}$ distance. Also the molecule which has stronger $\text{C-C}$ bonds is to be determined.

Concept introduction:

Hybridization affects both bond length and bond strength. The two $\text{sp}$ hybrid atomic orbitals are formed by mixing one $\text{s}$ orbital and one of the three $\text{p}$ orbitals from the valence shell. Each $\text{sp}$ hybrid orbital formed has $\text{50\%}$ $\text{s}$ character and $\text{50\%}$ $\text{p}$ character. Three ${\text{sp}}^{\text{2}}$ hybrid atomic orbitals are formed by mixing one $\text{s}$ orbital and two of the three $\text{p}$ orbitals from the valence shell. Each ${\text{sp}}^{\text{2}}$ hybrid orbital formed has $\text{33}\text{.33\%}$$\text{s}$ character and $\text{66}\text{.67\%}$ $\text{p}$ character. Four ${\text{sp}}^{\text{3}}$ hybrid atomic orbitals are formed by mixing the one $\text{s}$ orbital and three $\text{p}$ orbitals from the valence shell. Each ${\text{sp}}^{\text{3}}$ hybrid orbital formed has $\text{25\%}$ $\text{s}$ character and $\text{75\%}$ $\text{p}$ character. As the $\text{s}$ character increases, the hybrid atomic orbital becomes more compact. The bonds formed by shorter atomic orbitals are shorter in length. The shorter bonds connecting the same atoms are stronger. Therefore, as the hybridization of an atom changes from ${\text{sp}}^{\text{3}}$ to ${\text{sp}}^{\text{2}}$ to $\text{sp}$, its bonds become shorter and stronger.

Answer to Problem 3.27P

Explanation of Solution

The structure of the molecules mentioned in the question is shown below:

The two $\text{sp}$ hybrid atomic orbitals are formed by mixing one $\text{s}$ orbital and one of the three $\text{p}$ orbitals from the valence shell. Each $\text{sp}$ hybrid orbital formed has $\text{50\%}$ $\text{s}$ character and $\text{50\%}$ $\text{p}$ character. Three ${\text{sp}}^{\text{2}}$ hybrid atomic orbitals are formed by mixing one $\text{s}$ orbital and two of three $\text{p}$ orbitals from the valence shell. Each ${\text{sp}}^{\text{2}}$ hybrid orbital formed has $\text{33}\text{.33\%}$$\text{s}$ character and $\text{66}\text{.67\%}$ $\text{p}$ character. Four ${\text{sp}}^{\text{3}}$ hybrid atomic orbitals are formed by mixing the one $\text{s}$ orbital and three $\text{p}$ orbitals from the valence shell. Each ${\text{sp}}^{\text{3}}$ hybrid orbital formed has $\text{25\%}$ $\text{s}$ character and $\text{75\%}$ $\text{p}$ character. As the hybridization of an atom changes from ${\text{sp}}^{\text{3}}$ to ${\text{sp}}^{\text{2}}$ to $\text{sp}$, its bonds become shorter and stronger.

In the first molecule, according to VSEPR theory, the ${\text{C}}_{\text{1}}$ carbon atom is surrounded by two electron groups – a triple bond and a single bond. The electron geometry of ${\text{C}}_{\text{1}}$ carbon atom is linear. Therefore, the ${\text{C}}_{\text{1}}$ carbon atom must be $\text{sp}$ hybridized. In the first molecule, according to VSEPR theory, the ${\text{C}}_{\text{2}}$ carbon atom is surrounded by four electron groups – four single bonds. The electron geometry of ${\text{C}}_{\text{2}}$ carbon atom is tetrahedral. Therefore, the ${\text{C}}_{\text{2}}$ carbon atom must be ${\text{sp}}^{\text{3}}$ hybridized.

In the second molecule, according to VSEPR theory, the ${\text{C}}_{\text{1}}$ carbon atom is surrounded by four electron groups – four single bonds. The electron geometry of ${\text{C}}_{\text{1}}$ carbon atom is tetrahedral. Therefore, the ${\text{C}}_{\text{1}}$ carbon atom must be ${\text{sp}}^{\text{3}}$ hybridized. In the second molecule, according to VSEPR theory, the ${\text{C}}_{\text{2}}$ carbon atom is surrounded by three electron groups – a double bond and two single bonds. The electron geometry of ${\text{C}}_{\text{2}}$ carbon atom is trigonal planar. Therefore, the ${\text{C}}_{\text{2}}$ carbon atom must be ${\text{sp}}^{\text{2}}$ hybridized. In second molecule, according to VSEPR theory, the ${\text{C}}_{\text{3}}$ carbon atom is surrounded by four electron groups – four single bonds. The electron geometry of ${\text{C}}_{\text{3}}$ carbon atom is tetrahedral. Therefore, the ${\text{C}}_{\text{3}}$ carbon atom must be ${\text{sp}}^{\text{3}}$ hybridized.

This shows that in the first molecule, the percentage s character in $\text{C-C}$ bond is more than in the second molecule. Therefore, the $\text{C-C}$ bond is shorter and stronger in the first molecule.

Conclusion

Bond length and bond strength is determined from the hybridization involved in the molecule.

Interpretation Introduction

(b)

Interpretation:

Among the molecules mentioned in the question, it is to be determined which has shorter $\text{C-N}$ distance. Also the molecule which has stronger $\text{C-N}$ bonds is to be determined.

Concept introduction:

Hybridization affects both bond length and bond strength. The two $\text{sp}$ hybrid atomic orbitals are formed by mixing one $\text{s}$ orbital and one of the three $\text{p}$ orbitals from the valence shell. Each $\text{sp}$ hybrid orbital formed has $\text{50\%}$ $\text{s}$ character and $\text{50\%}$ $\text{p}$ character. Three ${\text{sp}}^{\text{2}}$ hybrid atomic orbitals are formed by mixing one $\text{s}$ orbital and two of the three $\text{p}$ orbitals from the valence shell. Each ${\text{sp}}^{\text{2}}$ hybrid orbital formed has $\text{33}\text{.33\%}$$\text{s}$ character and $\text{66}\text{.67\%}$ $\text{p}$ character. Four ${\text{sp}}^{\text{3}}$ hybrid atomic orbitals are formed by mixing the one $\text{s}$ orbital and three $\text{p}$ orbitals from the valence shell. Each ${\text{sp}}^{\text{3}}$ hybrid orbital formed has $\text{25\%}$ $\text{s}$ character and $\text{75\%}$ $\text{p}$ character. As the $\text{s}$ character increases, the hybrid atomic orbital becomes more compact. The bonds formed by shorter atomic orbitals are shorter in length. The shorter bonds connecting the same atoms are stronger. Therefore, as the hybridization of an atom changes from ${\text{sp}}^{\text{3}}$ to ${\text{sp}}^{\text{2}}$ to $\text{sp}$, its bonds become shorter and stronger.

Answer to Problem 3.27P

Explanation of Solution

The structure of the molecules mentioned in the question is shown below:

The two $\text{sp}$ hybrid atomic orbitals are formed by mixing one $\text{s}$ orbital and one of the three $\text{p}$ orbitals from the valence shell. Each $\text{sp}$ hybrid orbital formed has $\text{50\%}$ $\text{s}$ character and $\text{50\%}$ $\text{p}$ character. Three ${\text{sp}}^{\text{2}}$ hybrid atomic orbitals are formed by mixing one $\text{s}$ orbital and two of the three $\text{p}$ orbitals from the valence shell. Each ${\text{sp}}^{\text{2}}$ hybrid orbital formed has $\text{33}\text{.33\%}$$\text{s}$ character and $\text{66}\text{.67\%}$ $\text{p}$ character. Four ${\text{sp}}^{\text{3}}$ hybrid atomic orbitals are formed by mixing the one $\text{s}$ orbital and three $\text{p}$ orbitals from the valence shell. Each ${\text{sp}}^{\text{3}}$ hybrid orbital formed has $\text{25\%}$ $\text{s}$ character and $\text{75\%}$ $\text{p}$ character. As the hybridization of an atom changes from ${\text{sp}}^{\text{3}}$ to ${\text{sp}}^{\text{2}}$ to $\text{sp}$, its bonds become shorter and stronger.

In the first molecule, according to VSEPR theory, the ${\text{C}}_{\text{1}}$ carbon atom is surrounded by two electron groups – a triple bond and a single bond. The electron geometry of ${\text{C}}_{\text{1}}$ carbon atom is linear. Therefore, the ${\text{C}}_{\text{1}}$ carbon atom must be $\text{sp}$ hybridized. In the first molecule, according to VSEPR theory, the nitrogen atom is surrounded by two electron groups – a lone pair and a triple bond. The electron geometry of nitrogen atom is linear. Therefore, the nitrogen atom must be $\text{sp}$ hybridized.

In the second molecule, according to VSEPR theory, the ${\text{C}}_{\text{2}}$ carbon atom is surrounded by three electron groups – a double bond and two single bonds. The electron geometry of ${\text{C}}_{\text{2}}$ carbon atom is trigonal planar. Therefore, the ${\text{C}}_{\text{2}}$ carbon atom must be ${\text{sp}}^{\text{2}}$ hybridized. In the second molecule, according to VSEPR theory, the nitrogen atom is surrounded by three electron groups – a lone pair, a single bond, and a double bond. The electron geometry of nitrogen atom is trigonal planar. Therefore, the nitrogen atom must be ${\text{sp}}^{\text{2}}$ hybridized.

This shows that in the first molecule, the percentage s character in C-N bond is more than that in the second molecule. Therefore, the C-N bond is shorter and stronger in the first molecule.

Conclusion

Bond length and bond strength is determined from the hybridization involved in the molecule.