Question

What is the $\vartriangle {H_f}$ of $MgC{l_2}$ if the lattice enthalpy involved is C KJ / mol?
$Mg \to \to \to M{g^{2 + }}$ ; Energy = A KJ /mol.
 $C{l_2}(g) \to \to \to 2C{l^ - }$ ; Energy = B KJ / mol

Answer 1

Hint: As we know that Heat of formation which is also known as the standard heat of formation is defined as the amount of heat absorbed or evolved when one mole of a compound is formed from its constituent elements. As lattice enthalpy is involved here so we will consider Born-Haber cycle.

Complete step by step answer:
The question arises that from an energetic point of view, magnesium chloride is $MgC{l_2}$ rather than $MgCl$ and $MgC{l_3}$. It turns out that $MgC{l_2}$ has the most negative enthalpy change of formation or in other words we can say that it is the most stable one relative to the element magnesium and chlorine.
Let us look at this in terms of Born-Haber cycle:
We need to add in the second ionization energy of magnesium because we are making a $ + 2$ ion. We also need to multiply the atomization enthalpy of chlorine by $2$, because we need $2$ moles of gaseous chlorine atoms. We also need to multiply the electron affinity of chlorine by $2$, because we are making $2$ moles of chloride ions. So, we obviously need a different value for lattice enthalpy.
$Mg\xrightarrow{{S.E( + )}}M{g_{(g)}}\xrightarrow{{I.{E_1}( + )}}M{g^ + }_{(g)}\xrightarrow{{I.{E_2}( + )}}M{g^{2 + }};$ Energy=$A\;kJ\;mo{l^{ - 1}}$
$C{l_2}\xrightarrow{{E.A.( - )}}C{l^ - }\xrightarrow{{E.A.( + )}}2C{l^ - }$; Energy =$B\;kJ\;mo{l^{ - 1}}$
Now, adding these two equations we will get:
$Mg + C{l_2} \to MgC{l_2}\;;{E_{L.E}} = C\;kJ\;mo{l^{ - 1}}$
Thus, the heat of formation will become:
$\Delta {H_f} = \Delta {H_{S.E}} + I.E + \dfrac{1}{2}\Delta {H_{Diss.}} + E.A - {E_{L.E}}$
Or we can write it as:
$\Delta {H_f} = A + B - C$
As we can see that lattice enthalpy is in negative form because a lot more energy is released as lattice enthalpy. That is because there are stronger ionic attractions between $1 - $ ions and $2 + $ions than between the $1 - $and $1 + $ in $MgCl$

Note: So, we can see that much more energy is released when we make $MgC{l_2}$ than when we make $MgCl$. As we need to put in more energy to ionize the magnesium to give a $2 + $ion. Using the Born-Haber cycle for lattice enthalpy is a good way of judging how purely ionic is any crystal.