Question

Given:

Reaction	Energy change (in $kJmo{{l}^{-1}}$) $kJmo{{l}^{-1}}$
$L{{i}_{(s)}}\to L{{i}_{(g)}}$	$161$
$L{{i}_{(g)}}\to L{{i}^{+}}_{(g)}$	$520$
$\dfrac{1}{2}{{F}_{2(g)}}\to {{F}_{(g)}}$	$77$
${{F}_{(g)}}+{{e}^{-}}\to {{F}^{-}}_{(g)}$	Electron gain enthalpy
$L{{i}_{(g)}}+F_{(g)}^{-}\to Li{{F}_{(s)}}$	$-1047$
$L{{i}_{(s)}}+\dfrac{1}{2}{{F}_{2(g)}}\to Li{{F}_{(s)}}$	$-617$

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

Answer 1

Hint: 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 $L{{i}_{(s)}}\to L{{i}_{(g)}}$ we see that solid is changed directly to a gas. So, it is a sublimation reaction and involves sublimation energy (S) i.e. $161\text{ kJmo}{{\text{l}}^{-1}}$.
In the next reaction, $L{{i}_{(g)}}\to L{{i}^{+}}_{(g)}$ there is a charge gain in the product. So, this involves ionization energy (I) i.e. given as $520\text{ kJmo}{{\text{l}}^{-1}}$
Moving to next reaction, $\dfrac{1}{2}{{F}_{2(g)}}\to {{F}_{(g)}}$ it is a dissociation reaction and thus involves dissociation energy (D) i.e. $\text{77 kJmo}{{\text{l}}^{-1}}$.
${{F}_{(g)}}+{{e}^{-}}\to {{F}^{-}}_{(g)}$ involves gain of electron and thus involve electron gain enthalpy (E) which we have to find out.
In the reaction, $L{{i}_{(g)}}+F_{(g)}^{-}\to Li{{F}_{(s)}}$ involves lattice energy (L) which is given as $-1047\text{ kJmo}{{\text{l}}^{-1}}$.
While in the reaction, $L{{i}_{(s)}}+\dfrac{1}{2}{{F}_{2(g)}}\to Li{{F}_{(s)}}$ the energy that is involved in formation energy (F) and it is given as $-617\text{ kJmo}{{\text{l}}^{-1}}$.
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:
$F=S+I+D-E+L$
So, the electron gain enthalpy will be:
$E=(S+I+D+L)-F$
By placing the values, we get:
$E=161+520+77+(-1047)-(-617)=328$
So, the value of electron gain enthalpy of fluorine is $328\text{ kJmo}{{\text{l}}^{-1}}$

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.