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.

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.