Chapter 10, Problem 10.61P

(a)

Interpretation Introduction

Interpretation:

The bond strength, bond length, and bond order of $\text{N}-\text{N}$ bond in ${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}$, ${\text{N}}_{\text{2}}{\text{H}}_2$ and ${\text{N}}_{\text{2}}$ is to be compared.

Concept introduction:

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.

Bond strength is directly proportional to the bond order. While the bond length is inversely proportional to the bond order thus higher bond order means smaller bond length.

A single bond is weaker and longer than a double bond and a triple bond is most stronger and shorter.

Answer to Problem 10.61P

The bond order of the $\text{N}-\text{N}$ bond in ${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}$, ${\text{N}}_{\text{2}}{\text{H}}_2$ and ${\text{N}}_{\text{2}}$ are 1,2 and 3 respectively.

Therefore the order of $\text{N}-\text{N}$ bond strength is as follows:

  ${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}<{\text{N}}_{\text{2}}{\text{H}}_2<{\text{N}}_2$

And the order of $\text{N}-\text{N}$ bond length is as follows:

  ${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}>{\text{N}}_{\text{2}}{\text{H}}_2>{\text{N}}_2$

Explanation of Solution

Both $\text{N}$ atoms are central atoms and other atoms are evenly distributed.

The total number of valence electrons of ${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}$ is calculated as,

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

Substitute 2 for the total number of $\text{N}$ atom, 5 for valence electrons of $\text{N}$, 4 for the total number of $\text{H}$ and 1 for the valence electrons of $\text{H}$ in equation (1).

  $\begin{array}{c}\text{Total valence electrons}\left(\text{TVE}\right)=\left[\left(2\right)\left(5\right)+\left(4\right)\left(1\right)\right]\\ =14\end{array}$

${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}$ has 14 valence electrons, from which ten are used in the formation of all single bonds. The remaining 4 electrons are used in completing the octets of both the $\text{N}$ atoms.

Lewis structure of ${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}$,

Therefore the bond order of the $\text{N}-\text{N}$ bond in ${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}$ is 1.

The total number of valence electrons of ${\text{N}}_{\text{2}}{\text{H}}_2$ is calculated as follows:

Substitute 2 for the total number of $\text{N}$ atom, 5 valence electrons of $\text{N}$, 2 for the total number of $\text{H}$ and 1 for the valence electrons of $\text{H}$ in equation (1).

  $\begin{array}{c}\text{Total valence electrons}\left(\text{TVE}\right)=\left[\left(2\right)\left(5\right)+\left(2\right)\left(1\right)\right]\\ =12\end{array}$

${\text{N}}_{\text{2}}{\text{H}}_2$ has 12 valence electrons, from which six are used in the formation of all single bonds. From the 12 valence electrons remaining 6 electrons are not able to fulfill the octets of both $\text{N}$ atoms due to which formation of $\text{N}-\text{N}$ double bonds takes place.

Lewis structure of ${\text{N}}_{\text{2}}{\text{H}}_2$,

Therefore the bond order of the $\text{N}-\text{N}$ bond in ${\text{N}}_{\text{2}}{\text{H}}_2$ is 2

The total number of valence electrons of ${\text{N}}_{\text{2}}$ is calculated as,

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

Substitute 2 for the total number of $\text{N}$ atom, and 5 valence electrons of in equation (2).

  $\begin{array}{c}\text{Total valence electrons}\left(\text{TVE}\right)=\left[\left(2\right)\left(5\right)\right]\\ =10\end{array}$

${\text{N}}_{\text{2}}$ has 10 valence electrons, from which two electrons are used to place a single bond between both $\text{N}$ atoms. Remaining four pairs of electrons are not enough to complete the octets of both $\text{N}$ atoms the requirement is of six electrons. Hence two lone pairs become bonding pair and a triple bond will form.

Lewis structure of ${\text{N}}_{\text{2}}$ is,

Therefore the bond order of the $\text{N}-\text{N}$ bond in ${\text{N}}_{\text{2}}$ is 3

In Hydrazine $\left({\text{N}}_{\text{2}}{\text{H}}_{\text{4}}\right)$, both $\text{N}$ atoms are attached with single bonds and a single bond is weaker and longer. In Diazene $\left({\text{N}}_{\text{2}}{\text{H}}_2\right)$ both $\text{N}$ atoms are connected with a double bond and its strength is more and bond length is shorter than the Hydrazine. Nitrogen has a triple bond. The triple bond is most stronger and shorter among all three.

Thus the order of $\text{N}-\text{N}$ bond strength is ${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}<{\text{N}}_{\text{2}}{\text{H}}_2<{\text{N}}_2$.

And the order of $\text{N}-\text{N}$ bond length is ${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}>{\text{N}}_{\text{2}}{\text{H}}_2>{\text{N}}_2$.

Conclusion

Bond strength is directly proportional to the bond order and bond length is inversely proportional to bond order. Hydrazine has a weakest and longest $\text{N}-\text{N}$ bond and nitrogen has the strongest and $\text{N}-\text{N}$ weakest bond.

Interpretation Introduction

Interpretation:

The Lewis structure for tetrazene $\left({\text{H}}_{\text{2}}{\text{NNNNH}}_{\text{2}}\right)$ is to be drawn. Also $\text{Δ}{H}_{\text{rxn}}^{°}$ for the decomposition of tetrazene is to be calculated.

Concept introduction:

To draw the Lewis structure of the molecule following steps are used.

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.

The heat of the reaction $\left(\Delta H_{\text{rxn}}^{°}\right)$ is defined as the heat released or absorbed during a chemical reaction as a result of the difference in the bond energies (BE) of reactant and product in the reaction. $\Delta H_{\text{rxn}}^{°}$ is negative for exothermic reaction and $\Delta H_{\text{rxn}}^{°}$ is positive for an endothermic reaction.

The formula to calculate $\Delta H_{\text{rxn}}^{°}$ of reaction is as follows:

  $\Delta H_{\text{rxn}}^{°}=\sum \Delta H_{\text{reactant bond broken}}^{°}+\sum \Delta H_{\text{product bond formed}}^{°}$

Or,

  $\Delta H_{\text{rxn}}^{°}=\sum {\text{BE}}_{\text{reactant bond broken}}-\sum {\text{BE}}_{\text{product bond formed}}$

The bond energy of reactants is positive and the bond energy of products is negative.

Answer to Problem 10.61P

The Lewis structure for Tetrazene is

The heat of reaction $\left(\Delta H_{\text{rxn}}^{°}\right)$ for the decomposition of Tetrazene is $\text{ –367 kJ}$.

Explanation of Solution

The total number of valence electrons of ${\text{N}}_4{\text{H}}_4$ is calculated as follows:

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

Substitute 4 for the total number of $\text{N}$ atom, 5 for valence electrons of $\text{N}$, 4 for the total number of $\text{H}$ and 1 for the valence electrons of $\text{H}$ in equation (3).

  $\begin{array}{c}\text{Total valence electrons}\left(\text{TVE}\right)=\left[\left(4\right)\left(5\right)+\left(4\right)\left(1\right)\right]\\ =24\end{array}$

${\text{N}}_4{\text{H}}_4$ has 24 total number of valence electrons. 7 pairs of electrons are used in the formation of single bonds between the atoms. Remaining 7 pairs are used to complete the octets of all the N atoms but one N remain short of two electrons due to which one lone pair is used to make a double bond.

The Lewis structure of Tetrazene is,

The given chemical equation for the decomposition of Tetrazene is as follows:

The number of broken bonds is $\text{4 N}-\text{H}$ bonds, $\text{2 N}-\text{N}$ bonds, and $\text{1 N}=\text{N}$ bond.

The number of bonds formed is $\text{4 N}-\text{H}$ bonds, $\text{1 N}-\text{N}$ bond, and $\text{1 N}\equiv \text{N}$ bond.

The formula to calculate the enthalpy of the given reaction is as follows:

  $\Delta H_{\text{rxn}}^{°}=\left({\text{4BE}}_{\text{N}-\text{H}}+2{\text{BE}}_{\text{N}-\text{N}}+1{\text{BE}}_{\text{N=N}}\right)-\left({\text{4BE}}_{\text{N}-\text{H}}+1{\text{BE}}_{\text{N}-\text{N}}+1{\text{BE}}_{\text{N}\equiv \text{N}}\right)$        (4)

Substitute $391\text{ kJ/mol}$ for ${\text{BE}}_{\text{N}-\text{H}}$, $160\text{kJ/mol}$ for ${\text{BE}}_{\text{N}-\text{N}}$, $418\text{kJ/mol}$ for ${\text{BE}}_{\text{N=N}}$ and $965\text{kJ/mol}$ for ${\text{BE}}_{\text{N}\equiv \text{N}}$ in the equation (4).

  $\begin{array}{c}\Delta H_{\text{rxn}}^{°}=\left[\begin{array}{l}\left(\left(\text{4 mol}\right)\left(391\text{kJ/mol}\right)+\left(\text{2 mol}\right)\left(160\text{kJ/mol}\right)+\left(\text{1 mol}\right)\left(418\text{kJ/mol}\right)\right)-\\ \left(\text{4 mol}\right)\left(391\text{kJ/mol}\right)+\left(\text{1 mol}\right)\left(160\text{kJ/mol}\right)+\left(\text{1 mol}\right)\left(965\text{kJ/mol}\right)\end{array}\right]\\ =-367\text{kJ}\end{array}$

Conclusion

Only one Lewis structure is possible for tetrazene.