Question

How does the bond energy for the formation of NaCl(s) take place?

Answer 1

Hint: In the above question, we have to find out how the bond energy for the formation of NaCl takes place. We have to write a series of balanced chemical equations along with heat released in each step to get the desired lattice enthalpy.

Complete step by step solution:
Sodium is metal and chlorine is a diatomic gas in room temperature and pressure but in NaCl, sodium and chlorine ions are present. So, we have to basically find out the following reaction:
 $\text{Na(s)}\stackrel{\text{sublimation}}{\to }\text{Na(l)}\stackrel{\text{1st ionization}}{\to }\text{N}{\text{a}}^{\text{ + }}\text{(g)}$
 ${\displaystyle \frac{\text{1}}{\text{2}}}\text{C}{\text{l}}_{\text{2}}\text{(g)}\stackrel{{\displaystyle \frac{\text{1}}{\text{2}}}\text{ bond breaking}}{\to }\text{Cl(g)}\stackrel{\text{1st electron affinity}}{\to }\text{C}{\text{l}}^{\text{ - }}\text{(g)}$
The standard heat of formation of NaCl is given as:
 $\text{NaCl(s)}\to \text{Na(s) + }{\displaystyle \frac{\text{1}}{\text{2}}}\text{C}{\text{l}}_{\text{2}}{\mathrm{\Delta }H}_{\text{f,NaCl}}^{\text{0}}\text{ = 411}$ kJ …………..(1)
The enthalpy change during sublimation of sodium is given by:
 $\text{Na(s)}\to Na(g),\mathrm{\Delta }{\text{H}}_{\text{sub,Na}}\text{ = 107}$ kJ …………..(2)
The first ionization energy of Na, that is, removal of one electron from sodium atom is given by:
 $\text{Na(g)}\to \text{N}{\text{a}}^{\text{ + }}\text{(g) + }{\text{e}}^{\text{ - }}\text{,I}{\text{E}}_{\text{1,Na(g)}}\text{ = 502}$ KJ…………..(3)
Now let us look at the enthalpy change when chlorine molecule breaks down to chlorine atom which is given by:
 ${\displaystyle \frac{\text{1}}{\text{2}}}\text{C}{\text{l}}_{\text{2}}\text{(g)}\to \text{Cl(g),}{\displaystyle \frac{\text{1}}{\text{2}}}\mathrm{\Delta }{\text{H}}_{\text{bond,C}{\text{l}}_{\text{2}}\text{(g)}}\text{ = }{\displaystyle \frac{\text{1}}{\text{2}}}\times 242=121$ kJ…………..(4)
The enthalpy change when chlorine gains an electron is given by:
 $\text{Cl(g) + }{\text{e}}^{\text{ - }}\to \text{C}{\text{l}}^{\text{ - }}\text{(g),E}{\text{A}}_{\text{1,Cl(g)}}\text{ = - 355}$ kJ…………..(5)
We have to find the enthalpy change ( $\mathrm{\Delta }{\text{H}}_{\text{lattice}}$ ) during-
 $\text{N}{\text{a}}^{\text{ + }}\text{(g) + C}{\text{l}}^{\text{ - }}\text{(g)}\to \text{NaCl(s)}$ …………..(6)
If we add equation 1, 2, 3,4 and 5, we will get:
 $\text{NaCl(s) + Na(s) + Na(g) + }{\displaystyle \frac{\text{1}}{\text{2}}}\text{C}{\text{l}}_{\text{2}}\text{(g) + Cl(g) + }{\text{e}}^{\text{ - }}\to \text{Na(s) + }{\displaystyle \frac{\text{1}}{\text{2}}}\text{C}{\text{l}}_{\text{2}}(g)+\text{Na(g)}+\text{N}{\text{a}}^{\text{ + }}\text{(g) + }{\text{e}}^{\text{ - }}+\text{Cl(g)}+\text{C}{\text{l}}^{\text{ - }}\text{(g)}$
Simplifying the above equation, we get:
 $\text{NaCl(s)}\to \text{N}{\text{a}}^{\text{ + }}\text{(g) + C}{\text{l}}^{\text{ - }}\text{(g)}$
 $\mathrm{\Delta }H=411+107+502+121-355=786$ kJ
Hence, for the reaction
 $\text{N}{\text{a}}^{\text{ + }}\text{(g) + C}{\text{l}}^{\text{ - }}\text{(g)}\to \text{NaCl(s)}$
 $\mathrm{\Delta }{\text{H}}_{\text{lattice}}=-\mathrm{\Delta }H=-786$ kJ
Hence, the bond energy for the formation of NaCl(s) is $\text{ - 786}$ kJ.

Note:
The lattice energy of a crystalline solid is a measure of the energy released when ions are combined to make a compound. It is a measure of the cohesive forces that bind ions. Lattice energy is relevant to many practical properties which includes solubility, hardness, and volatility.