Question

The electron gain enthalpy of chlorine is $3.7eV$ How much energy in kcal is released when $~2g~$ of chlorine is completely converted to $C{{l}^{-}}~$ ions in the gaseous state? $\left( 1eV=23.06kcal \right)$
A. $5.8kcal$
B. $3.52kcal$
C. $4.8kcal$
D. None of these

Answer 1

The amount of energy released when an electron is added to a neutral atom to form an anion is known as the electron affinity of an element. It is the potential energy required to change a neutral atom into anion. More negative the electron affinity of an element, the more favorable the addition process.

Complete step by step answer:
The electron affinity is also known as electron gain enthalpy. It is represented as EA.
${{A}_{(g)}}+{{e}^{-}}\to A_{(g)}^{-}$
Electron affinity depends on the atomic size as well as the nuclear charge. The energy value of electron gain enthalpy can be endothermic or exothermic depending on the nature of the element. The electron gain enthalpy is measured in $kJ/mol$or $kcal/mol.$
Group $16$ has a high electron affinity. As we move down the group from Oxygen to sulphuric, the electron affinity value increases and then decreases from $S-Po$ The oxygen has lower electron affinity and this is an exception.
The electron density in the $2p$ energy shell in oxygen is high due to the small size of oxygen. Due to this, there is some resistance to the incoming electron and thus the electron affinity.
Here, we have, Electron affinity in $eV=3.7$
Electron affinity in $kcal/mol=3.7\times 23.06=85.322~kcal/mol$
$\therefore Number(moles)=\dfrac{2}{35.5}=0.05633$
Since we know that the one moles gives $85.322~kcal$
Therefore $0.056338~\left( gives \right)~85.322\times 0.056338=4.806~kcal$

Therefore, the correct answer is option C.

Note: Electron affinity will be low when the atomic size will be very less. Electron affinity will be low when the last orbital is too far from the nucleus due to less effective nuclear charge.