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Question: What is the \(\vartriangle {H_f}\) of \(MgC{l_2}\) if the lattice enthalpy involved is C KJ / mol? ...

What is the △Hf\vartriangle {H_f} of MgCl2MgC{l_2} if the lattice enthalpy involved is C KJ / mol?
Mg→→→Mg2+Mg \to \to \to M{g^{2 + }} ; Energy = A KJ /mol.
Cl2(g)→→→2Cl−C{l_2}(g) \to \to \to 2C{l^ - } ; Energy = B KJ / mol

Explanation

Solution

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 MgCl2MgC{l_2} rather than MgClMgCl and MgCl3MgC{l_3}. It turns out that MgCl2MgC{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 + 2 ion. We also need to multiply the atomization enthalpy of chlorine by 22, because we need 22 moles of gaseous chlorine atoms. We also need to multiply the electron affinity of chlorine by 22, because we are making 22 moles of chloride ions. So, we obviously need a different value for lattice enthalpy.
Mg→S.E(+)Mg(g)→I.E1(+)Mg+(g)→I.E2(+)Mg2+;Mg\xrightarrow{{S.E( + )}}M{g_{(g)}}\xrightarrow{{I.{E_1}( + )}}M{g^ + }_{(g)}\xrightarrow{{I.{E_2}( + )}}M{g^{2 + }}; Energy=A  kJ  mol−1A\;kJ\;mo{l^{ - 1}}
Cl2→E.A.(−)Cl−→E.A.(+)2Cl−C{l_2}\xrightarrow{{E.A.( - )}}C{l^ - }\xrightarrow{{E.A.( + )}}2C{l^ - }; Energy =B  kJ  mol−1B\;kJ\;mo{l^{ - 1}}
Now, adding these two equations we will get:
Mg+Cl2→MgCl2  ;EL.E=C  kJ  mol−1Mg + C{l_2} \to MgC{l_2}\;;{E_{L.E}} = C\;kJ\;mo{l^{ - 1}}
Thus, the heat of formation will become:
ΔHf=ΔHS.E+I.E+12ΔHDiss.+E.A−EL.E\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:
ΔHf=A+B−C\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 1−1 - ions and 2+2 + ions than between the 1−1 - and 1+1 + in MgClMgCl

Note: So, we can see that much more energy is released when we make MgCl2MgC{l_2} than when we make MgClMgCl. As we need to put in more energy to ionize the magnesium to give a 2+2 + ion. Using the Born-Haber cycle for lattice enthalpy is a good way of judging how purely ionic is any crystal.