Chapter 10, Problem 10.76P

(a)

Interpretation Introduction

Interpretation:

A Lewis structure for ${\text{C}}_{\text{3}}{\text{H}}_{\text{4}}$ is to be drawn and if resonance possible it is to be identified. Also, the carbon-carbon bond order is to be determined.

Concept introduction:

Lewis structure is generally considered as a simplified structure of any molecule or atom. Lewis structure for any atom or molecule depicts the valence electrons as dots around the element’s symbol present in the molecule along with the bonds that connect them. Every element tries to complete an octet except the hydrogen atom.

Every element in the Lewis structure tries to attain eight electrons in its valence shell by transfer or share of electrons. This rule is known as the octet rule.

To draw the Lewis structure of the molecule there are following steps:

Step 1: Find the central atom and place the other atoms around it. The atom in a compound which has the lowest group number or lowest electronegativity considered as the central atom.

Step 2: Calculate the total number of valence electrons.

Step 3: Connect the other atoms around the central atoms to the central atom with a single bond and lower the value of valence electrons by 2 of every single bond.

Step 4: Allocate the remaining electrons in pairs so that each atom can get 8 electrons.

Step 5: Convert the lone pair into bond pair.

Answer to Problem 10.76P

The Lewis structure for ${\text{C}}_{\text{3}}{\text{H}}_{\text{4}}$ is as follows:

The carbon-carbon single bond has the bond order 1 and carbon-carbon double bond has the bond order 2.

Explanation of Solution

The total number of valence electrons of ${\text{C}}_{\text{3}}{\text{H}}_{\text{4}}$ is calculated as,

  $\text{Total valence electrons}=\left[\begin{array}{l}\left[\begin{array}{l}\left(\text{Total number of C atom}\right)\\ \left(\text{Valence electrons of C}\right)\end{array}\right]\\ +\left[\begin{array}{l}\left(\text{Total number of H atom}\right)\\ \left(\text{Valence electrons of H}\right)\end{array}\right]\end{array}\right]$        (1)

Substitute 3 for the total number of $\text{C }$ atom, 4 for valence electrons of $\text{C }$, 1 for the total valence electrons of $\text{H}$, and 4 for the total number of $\text{H}$ atom in equation (1).

  $\begin{array}{c}\text{Total valence electrons}=\left[\left(3\right)\left(4\right)+\left(4\right)\left(1\right)\right]\\ =16.\end{array}$

In ${\text{C}}_{\text{3}}{\text{H}}_{\text{4}}$, the three $\text{C }$ atom use 6 electrons to form single bond with each other and form three-membered ring and the two carbon atom in the ring also form one double bond with each other. The remaining electrons are used by the carbon atoms to form bond with the hydrogen atom.

The Lewis structure of ${\text{C}}_{\text{3}}{\text{H}}_{\text{4}}$ is,

The single bond has the bond order 1 and double bond has the bond order 2. Therefore, the carbon-carbon single bond has the bond order 1 and carbon-carbon double bond has the bond order 2.

Conclusion

The single bond has the bond order 1, the double bond has the bond order 2 and triple bond has the bond order 3.

(b)

Interpretation Introduction

Interpretation:

A Lewis structure for ${\text{C}}_{\text{3}}{\text{H}}_6$ is to be drawn and if resonance possible it is to be identified. Also, the carbon-carbon bond order is to be determined.

Concept introduction:

Lewis structure is generally considered as a simplified structure of any molecule or atom. Lewis structure for any atom or molecule depicts the valence electrons as dots around the element’s symbol present in the molecule along with the bonds that connect them. Every element tries to complete an octet except the hydrogen atom.

Every element in the Lewis structure tries to attain eight electrons in its valence shell by transfer or share of electrons. This rule is known as octet rule.

To draw the Lewis structure of the molecule there are following steps:

Step 1: Find the central atom and place the other atoms around it. The atom in a compound which has the lowest group number or lowest electronegativity considered as the central atom.

Step 2: Calculate the total number of valence electrons.

Step 3: Connect the other atoms around the central atoms to the central atom with a single bond and lower the value of valence electrons by 2 of every single bond.

Step 4: Allocate the remaining electrons in pairs so that each atom can get 8 electrons.

Step 5: Convert the lone pair into bond pair.

Answer to Problem 10.76P

The Lewis structure of ${\text{C}}_{\text{3}}{\text{H}}_6$ is,

There is only single bond in ${\text{C}}_{\text{3}}{\text{H}}_6$ so the carbon-carbon bond order is 1.

Explanation of Solution

The total number of valence electrons of ${\text{C}}_{\text{3}}{\text{H}}_6$ is calculated as,

  $\text{Total valence electrons}=\left[\begin{array}{l}\left[\begin{array}{l}\left(\text{Total number of C atom}\right)\\ \left(\text{Valence electrons of C}\right)\end{array}\right]\\ +\left[\begin{array}{l}\left(\text{Total number of H atom}\right)\\ \left(\text{Valence electrons of H}\right)\end{array}\right]\end{array}\right]$        (2)

Substitute 3 for the total number of $\text{C }$ atom, 4 for valence electrons of $\text{C }$, 1 for the total valence electrons of $\text{H}$, and 6 for total number of $\text{H}$ atom in equation (2).

  $\begin{array}{c}\text{Total valence electrons}=\left[\left(3\right)\left(4\right)+\left(6\right)\left(1\right)\right]\\ =18.\end{array}$

In ${\text{C}}_{\text{3}}{\text{H}}_6$, the three $\text{C }$ atom use 6 electrons to form single bond with each other and form three-membered ring. The remaining electrons are used by the carbon atoms to form bond with hydrogen atom. Each carbon atom forms two carbon-hydrogen bonds respectively to complete the valence electrons.

The Lewis structure of ${\text{C}}_{\text{3}}{\text{H}}_6$ is,

The single bond has the bond order 1 and double bond has the bond order 2. Therefore, there is only single bond in ${\text{C}}_{\text{3}}{\text{H}}_6$ so the carbon-carbon bond order is 1.

Conclusion

The single bond has the bond order 1, double bond has the bond order 2 and triple bond has the bond order 3.

(c)

Interpretation Introduction

Interpretation:

A Lewis structure for ${\text{C}}_{\text{4}}{\text{H}}_6$ is to be drawn and if resonance possible it is to be identified. Also, the carbon-carbon bond order is to be determined.

Concept introduction:

Lewis structure is generally considered as a simplified structure of any molecule or atom. Lewis structure for any atom or molecule depicts the valence electrons as dots around the element’s symbol present in the molecule along with the bonds that connect them. Every element tries to complete an octet except the hydrogen atom.

Every element in the Lewis structure tries to attain eight electrons in its valence shell by transfer or share of electrons. This rule is known as the octet rule.

To draw the Lewis structure of the molecule there are following steps:

Step 1: Find the central atom and place the other atoms around it. The atom in a compound which has the lowest group number or lowest electronegativity considered as the central atom.

Step 2: Calculate the total number of valence electrons.

Step 3: Connect the other atoms around the central atoms to the central atom with a single bond and lower the value of valence electrons by 2 of every single bond.

Step 4: Allocate the remaining electrons in pairs so that each atom can get 8 electrons.

Step 5: Convert the lone pair into bond pair.

Answer to Problem 10.76P

The Lewis structure of ${\text{C}}_{\text{4}}{\text{H}}_6$ is,

The carbon-carbon single bond has the bond order 1 and carbon-carbon double bond has the bond order 2.

Explanation of Solution

The total number of valence electrons of ${\text{C}}_{\text{4}}{\text{H}}_6$ is calculated as,

  $\text{Total valence electrons}=\left[\begin{array}{l}\left[\begin{array}{l}\left(\text{Total number of C atom}\right)\\ \left(\text{Valence electrons of C}\right)\end{array}\right]\\ +\left[\begin{array}{l}\left(\text{Total number of H atom}\right)\\ \left(\text{Valence electrons of H}\right)\end{array}\right]\end{array}\right]$        (3)

Substitute 4 for the total number of $\text{C }$ atom, 4 for valence electrons of $\text{C }$, 1 for the total valence electrons of $\text{H}$, and 6 for total number of $\text{H}$ atom in equation (3).

  $\begin{array}{c}\text{Total valence electrons}=\left[\left(4\right)\left(4\right)+\left(6\right)\left(1\right)\right]\\ =22.\end{array}$

In ${\text{C}}_{\text{4}}{\text{H}}_6$, the four $\text{C }$ atom use 8 electrons to form single bond with each other and form a four-membered ring and the two carbon atom in the ring also form one double bond with each other. The remaining electrons are used by the carbon atoms to form bond with the hydrogen atom.

The Lewis structure of ${\text{C}}_{\text{4}}{\text{H}}_6$ is,

The single bond has the bond order 1 and double bond has the bond order 2. Therefore, the carbon-carbon single bond has the bond order 1 and carbon-carbon double bond has the bond order 2.

Conclusion

The single bond has the bond order 1, the double bond has the bond order 2 and triple bond has the bond order 3.

(d)

Interpretation Introduction

Interpretation:

A Lewis structure for ${\text{C}}_{\text{4}}{\text{H}}_4$ is to be drawn and if resonance possible it is to be identified. Also, the carbon-carbon bond order is to be determined.

Concept introduction:

Lewis structure is generally considered as a simplified structure of any molecule or atom. Lewis structure for any atom or molecule depicts the valence electrons as dots around the element’s symbol present in the molecule along with the bonds that connect them. Every element tries to complete an octet except the hydrogen atom.

Every element in the Lewis structure tries to attain eight electrons in its valence shell by transfer or share of electrons. This rule is known as the octet rule.

To draw the Lewis structure of the molecule there are following steps:

Step 1: Find the central atom and place the other atoms around it. The atom in a compound which has the lowest group number or lowest electronegativity considered as the central atom.

Step 2: Calculate the total number of valence electrons.

Step 3: Connect the other atoms around the central atoms to the central atom with a single bond and lower the value of valence electrons by 2 of every single bond.

Step 4: Allocate the remaining electrons in pairs so that each atom can get 8 electrons.

Step 5: Convert the lone pair into bond pair.

Answer to Problem 10.76P

The Lewis structure of ${\text{C}}_{\text{4}}{\text{H}}_4$ is,

The average bond order in ${\text{C}}_{\text{4}}{\text{H}}_4$ is $1.5$.

Explanation of Solution

The total number of valence electrons of ${\text{C}}_{\text{4}}{\text{H}}_4$ is calculated as,

  $\text{Total valence electrons}=\left[\begin{array}{l}\left[\begin{array}{l}\left(\text{Total number of C atom}\right)\\ \left(\text{Valence electrons of C}\right)\end{array}\right]\\ +\left[\begin{array}{l}\left(\text{Total number of H atom}\right)\\ \left(\text{Valence electrons of H}\right)\end{array}\right]\end{array}\right]$        (4)

Substitute 4 for the total number of $\text{C }$ atom, 4 for valence electrons of $\text{C }$, 1 for the total valence electrons of $\text{H}$, and 4 for total number of $\text{H}$ atom in equation (4).

  $\begin{array}{c}\text{Total valence electrons}=\left[\left(4\right)\left(4\right)+\left(4\right)\left(1\right)\right]\\ =20\end{array}$

In ${\text{C}}_{\text{4}}{\text{H}}_4$, the four $\text{C }$ atom use 8 electrons to form single bond with each other and form four-membered ring and the four $\text{C }$ atoms in the ring also forms two alternate double bond with other carbon other in the ring. The remaining electrons are used by the carbon atoms to form bond with the hydrogen atom.

There is alternate single and double present in the ring so the molecule will show resonance.

The Lewis structure of ${\text{C}}_{\text{4}}{\text{H}}_4$ is,

There are 4 single bonds and 2 double bonds in ${\text{C}}_{\text{4}}{\text{H}}_4$.

The formula to calculate the average bond order in ${\text{C}}_{\text{4}}{\text{H}}_4$ is,

  $\text{Average bond order}=\frac{\text{Number of shared electron pair}}{\text{Number of bonded atoms}}$        (5)

Substitute 6 for the number of shared electron pair and 4 for the number of bonded atoms in equation (5).

  $\begin{array}{c}\text{Average bond order}=\frac{6}{4}\\ =\mathrm{1.5.}\end{array}$

Conclusion

The single bond has the bond order 1, the double bond has the bond order 2 and triple bond has the bond order 3.

(e)

Interpretation Introduction

Interpretation:

A Lewis structure for ${\text{C}}_{\text{6}}{\text{H}}_6$ is to be drawn and if resonance possible it is to be identified. Also, the carbon-carbon bond order is to be determined.

Concept introduction:

Lewis structure is generally considered as a simplified structure of any molecule or atom. Lewis structure for any atom or molecule depicts the valence electrons as dots around the element’s symbol present in the molecule along with the bonds that connect them. Every element tries to complete an octet except the hydrogen atom.

Every element in the Lewis structure tries to attain eight electrons in its valence shell by transfer or share of electrons. This rule is known as the octet rule.

To draw the Lewis structure of the molecule there are following steps:

Step 1: Find the central atom and place the other atoms around it. The atom in a compound which has the lowest group number or lowest electronegativity considered as the central atom.

Step 2: Calculate the total number of valence electrons.

Step 3: Connect the other atoms around the central atoms to the central atom with a single bond and lower the value of valence electrons by 2 of every single bond.

Step 4: Allocate the remaining electrons in pairs so that each atom can get 8 electrons.

Step 5: Convert the lone pair into bond pair.

Answer to Problem 10.76P

The Lewis structure of ${\text{C}}_{\text{6}}{\text{H}}_6$ is,

The average bond order in ${\text{C}}_{\text{6}}{\text{H}}_6$ is $1.5$.

Explanation of Solution

The total number of valence electrons of ${\text{C}}_{\text{6}}{\text{H}}_6$ is calculated as,

  $\text{Total valence electrons}=\left[\begin{array}{l}\left[\begin{array}{l}\left(\text{Total number of C atom}\right)\\ \left(\text{Valence electrons of C}\right)\end{array}\right]\\ +\left[\begin{array}{l}\left(\text{Total number of H atom}\right)\\ \left(\text{Valence electrons of H}\right)\end{array}\right]\end{array}\right]$        (6)

Substitute 6 for the total number of $\text{C }$ atom, 4 for valence electrons of $\text{C }$, 1 for the total valence electrons of $\text{H}$, and 6 for total number of $\text{H}$ atom in equation (6).

  $\begin{array}{c}\text{Total valence electrons}=\left[\left(6\right)\left(4\right)+\left(6\right)\left(1\right)\right]\\ =30\end{array}$

In ${\text{C}}_{\text{6}}{\text{H}}_6$, the four $\text{C }$ atom use 12 electrons to form single bond with each other and form a six-membered ring and the six $\text{C }$ atoms in the ring also forms three alternate double bonds with other carbon other in the ring. The remaining electrons are used by the carbon atoms to form bond with the hydrogen atom.

There is alternate single and double present in the ring so the molecule will show resonance.

The Lewis structure of ${\text{C}}_{\text{6}}{\text{H}}_6$ is,

There are 6 single bonds and 3 double bonds in ${\text{C}}_{\text{6}}{\text{H}}_6$.

The formula to calculate the average bond order in ${\text{C}}_{\text{6}}{\text{H}}_6$ is,

  $\text{Average bond order}=\frac{\text{Number of shared electron pair}}{\text{Number of bonded atoms}}$        (7)

Substitute 9 for the number of shared electron pair and 6 for the number of bonded atoms in equation (7).

  $\begin{array}{c}\text{Average bond order}=\frac{9}{6}\\ =\mathrm{1.5.}\end{array}$

Conclusion

The single bond has the bond order 1, the double bond has the bond order 2 and triple bond has the bond order 3.