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Question: Given: **Reaction**| **Energy change (in \(kJmo{{l}^{-1}}\)) \(kJmo{{l}^{-1}}\)** ---|--- ...

Given:

ReactionEnergy change (in kJmol−1kJmo{{l}^{-1}}) kJmol−1kJmo{{l}^{-1}}
Li(s)→Li(g)L{{i}_{(s)}}\to L{{i}_{(g)}}161161
Li(g)→Li+(g)L{{i}_{(g)}}\to L{{i}^{+}}_{(g)}520520
12F2(g)→F(g)\dfrac{1}{2}{{F}_{2(g)}}\to {{F}_{(g)}}7777
F(g)+e−→F−(g){{F}_{(g)}}+{{e}^{-}}\to {{F}^{-}}_{(g)}Electron gain enthalpy
Li(g)+F(g)−→LiF(s)L{{i}_{(g)}}+F_{(g)}^{-}\to Li{{F}_{(s)}}−1047-1047
Li(s)+12F2(g)→LiF(s)L{{i}_{(s)}}+\dfrac{1}{2}{{F}_{2(g)}}\to Li{{F}_{(s)}}−617-617

Based on data provided, the value of electron gain enthalpy of fluorine would be:
A. −300 kJmol−1-300\text{ kJmo}{{\text{l}}^{-1}}
B. −350 kJmol−1-350\text{ kJmo}{{\text{l}}^{-1}}
C. −328 kJmol−1-328\text{ kJmo}{{\text{l}}^{-1}}
D. −228 kJmol−1-228\text{ kJmo}{{\text{l}}^{-1}}

Explanation

Solution

To solve this question, determine which reaction is undergoing what process i.e. either the reaction is undergoing sublimation, etc. Then consider the Born Haber cycle equation, where the heat of formation energy equals all types of energies.

Complete step by step solution:
Before proceeding to the calculation, let us have some knowledge about the Born Haber cycle. The Born-Haber cycle is an approach to measure the reaction energies. The cycle is mostly related to the formation of ionic solid from various metals that belong to the first and second group when it reacts with a halogen or a non-metallic element. So, here as per the question let’s see each reaction and determine what type of energy is involved.
So, in the first reaction Li(s)→Li(g)L{{i}_{(s)}}\to L{{i}_{(g)}} we see that solid is changed directly to a gas. So, it is a sublimation reaction and involves sublimation energy (S) i.e. 161 kJmol−1161\text{ kJmo}{{\text{l}}^{-1}}.
In the next reaction, Li(g)→Li+(g)L{{i}_{(g)}}\to L{{i}^{+}}_{(g)} there is a charge gain in the product. So, this involves ionization energy (I) i.e. given as 520 kJmol−1520\text{ kJmo}{{\text{l}}^{-1}}
Moving to next reaction, 12F2(g)→F(g)\dfrac{1}{2}{{F}_{2(g)}}\to {{F}_{(g)}} it is a dissociation reaction and thus involves dissociation energy (D) i.e. 77 kJmol−1\text{77 kJmo}{{\text{l}}^{-1}}.
F(g)+e−→F−(g){{F}_{(g)}}+{{e}^{-}}\to {{F}^{-}}_{(g)} involves gain of electron and thus involve electron gain enthalpy (E) which we have to find out.
In the reaction, Li(g)+F(g)−→LiF(s)L{{i}_{(g)}}+F_{(g)}^{-}\to Li{{F}_{(s)}} involves lattice energy (L) which is given as −1047 kJmol−1-1047\text{ kJmo}{{\text{l}}^{-1}}.
While in the reaction, Li(s)+12F2(g)→LiF(s)L{{i}_{(s)}}+\dfrac{1}{2}{{F}_{2(g)}}\to Li{{F}_{(s)}} the energy that is involved in formation energy (F) and it is given as −617 kJmol−1-617\text{ kJmo}{{\text{l}}^{-1}}.
So, according to Born Haber cycle equation the heat of formation energy equals to the sum total of the energies involved in the reaction and the formula can be represented as:
F=S+I+D−E+LF=S+I+D-E+L
So, the electron gain enthalpy will be:
E=(S+I+D+L)−FE=(S+I+D+L)-F
By placing the values, we get:
E=161+520+77+(−1047)−(−617)=328E=161+520+77+(-1047)-(-617)=328
So, the value of electron gain enthalpy of fluorine is 328 kJmol−1328\text{ kJmo}{{\text{l}}^{-1}}

Hence, the correct option is C.

Note: It is important to note that; the Born Haber cycle is primarily used in calculating the lattice energy which is difficult to measure in general. You should know, the lattice energy is that energy change involved in the formation of ionic solids from gaseous ions.