Q10.61CP - Chemistry: Molecular Nature Of Matter And Change

Question

In addition to ammonia, nitrogen forms three other hydrides: hydrazine $\left({\mathbf{N}}_2{\mathbf{H}}_4\right)$, diazene $\left({\mathbf{N}}_2{\mathbf{H}}_2\right)$, and tetrazene $\left({\mathbf{N}}_4{\mathbf{H}}_4\right)$.

1. Use Lewis structures to compare the strength, length, and order of nitrogen-nitrogen bonds in hydrazine, diazene, and ${\mathbf{N}}_{\mathbf{2}}$.
2. Tetrazene (atom sequence ${\mathbf{H}}_2{\mathbf{NNNNH}}_2$) decomposes above ${\mathbf{0}}^{\mathbf{o}}$ to hydrazine and nitrogen gas. Draw a Lewis structure for tetrazene, and calculate ${\mathbf{\Delta H}}_{\mathrm{rxn}}^{\mathbf{o}}$ for this decomposition.

Step-by-Step Solution

Answer

1. The Lewis structures of hydrazine, diazene, and ${\mathrm{N}}_2$ are:

The Bond strength of nitrogen-nitrogen bonds in hydrazine, diazene and nitrogen are:   Nitrogen > Diazene > Hydrazine

The Bond length of nitrogen-nitrogen bonds in hydrazine, diazene and nitrogen are:   Hydrazine > Diazene > Nitrogen

The Bond order of nitrogen-nitrogen bonds in hydrazine, diazene and nitrogen are: Nitrogen = 3, Diazene = 2, Hydrazine =1

 

       b. The Lewis structure of Tetrazene:

The ${\mathrm{\Delta H}}_{\mathrm{rxn}}^{\mathrm{o}}$ for the decomposition reaction is -367 kJ.

Step 1: Definition of Lewis Structure, Standard Heat of Reaction, Bond Order

Lewis’s structure is the two-dimensional structure which contains electron dot symbols of every atom present and the bonding pairs that hold them together. Standard warmness of response is the sum of the usual heats of formation of the goods minus the sum of the usual heats of formation of the reactants.

 

${\mathrm{\Delta H}}_{\mathrm{rxn}}^{\mathrm{o}}=\sum \mathrm{m}{\mathrm{\Delta H}}_{\mathrm{f}\left(\mathrm{products}\right)}^{\mathrm{o}}-\sum \mathrm{n}{\mathrm{\Delta H}}_{\mathrm{f}\left(\mathrm{products}\right)}^{\mathrm{o}}$

Where,

$\sum$ is sum of m, n are amounts of products and reactants (in mol) represented by coefficients from the balanced equation.

 

The number of chemical bonds or the bonded electrons between the pair of atoms in a molecule is called bond order. 

E.g., for C-H bond, the bond order is 1, for diatomic nitrogen ${\mathrm{N}}_2$,

• The bond length depends on the bond order and they are inversely proportional to each other.
• As bond order increases the bond length decreases and as bond order decreases the bond length increases. Therefore, the order of bond length would be:

                     triple bond < double bond < single bond.

• The strength of the bond increases the bond length decreases. The stronger the bond shorter will be the length of the bond. The order of strength of the bond is:

                     triple bond > double bond > single bond   

• Therefore, a triple bond is the strongest bond with the shortest bond length compared to a double bond and a double bond is stronger with a shorter bond length than a single bond.

Step 2: Drawing Lewis structure and Comparing the Strength, Length and Order of nitrogen-nitrogen bonds in hydrazine, diazene, and N 2 .

The Lewis structures of hydrazine, diazene, and ${\mathrm{N}}_2$ are:

The triple nitrogen-nitrogen bond (in nitrogen), with bond order 3 has stronger bond with shorter bond length than others. The single nitrogen-nitrogen bond (in hydrazine), with bond order 1 has weaker bond with longer bond length than others. The double nitrogen-nitrogen bond (in diazene), with bond order 2 has intermediate bond strength and length. 

Bond Strength: Nitrogen > Diazene > Hydrazine                                                                    

                           Strongest                     Weakest                                              

Bond Length: Hydrazine > Diazene > Nitrogen                                                         

                            Longer                         Shorter

 

Step 3: Lewis Structure of Tetrazene and Calculation of &#916;H rxn o for its Decomposition Reaction

Tetrazene decomposes above $0°$ to hydrazine and nitrogen gas.

We know that:

${\mathrm{\Delta H}}_{\mathrm{rxn}}^{\mathrm{o}}=\sum \mathrm{m}{\mathrm{\Delta H}}_{\mathrm{f}\left(\mathrm{products}\right)}^{\mathrm{o}}-\sum \mathrm{n}{\mathrm{\Delta H}}_{\mathrm{f}\left(\mathrm{products}\right)}^{\mathrm{o}}$

Substituting the values, we get;

$\begin{array}{c}{\mathrm{\Delta H}}_{\mathrm{rxn}}^{\mathrm{o}}=\sum \mathrm{m}{\mathrm{\Delta H}}_{\mathrm{f}\left(\mathrm{products}\right)}^{\mathrm{o}}-\sum \mathrm{n}{\mathrm{\Delta H}}_{\mathrm{f}\left(\mathrm{products}\right)}^{\mathrm{o}}\\ =\left[{\mathrm{\Delta H}}_{\mathrm{f}}^{\mathrm{o}}\left({\mathrm{N}}_2{\mathrm{H}}_4\right)+{\mathrm{\Delta H}}_{\mathrm{f}}^{\mathrm{o}}\left({\mathrm{N}}_2\right)\right]-\left[{\mathrm{\Delta H}}_{\mathrm{f}}^{\mathrm{o}}\left({\mathrm{N}}_4{\mathrm{H}}_4\right)\right]\\ =\left[\left(50.63\mathrm{kJ}+0\right)\right]\text{\hspace{0.17em}\hspace{0.17em}}-\left[417.63\mathrm{kJ}\right]\\ {\mathrm{\Delta H}}_{\mathrm{rxn}}^{\mathrm{o}}=-367\mathrm{kJ}\end{array}$