Question

What is the value (including the sign) of the electron gain enthalpy or electron affinity of $N{a^ + }$ ion in electron -volts? The first ionization potential for $Na$ is $118\;kcal\;mol{e^{ - 1}}$.

Answer 1

In this question we have to find the value of the electron gain enthalpy. We have to state about the ionization energy and electron gain enthalpy and also about the relationship between them. Ionization energy is the opposite of electron gain enthalpy.

Complete answer:
Firstly, we should know about the ionization energy and electrons gain enthalpy.
Ionization energy: The change in enthalpy associated with the removal of the electron from a gaseous atom in its gaseous state is called Ionization energy. The factors affecting the ionization energy are:
Penetration effect
Shielding effect
Electronic configuration
It is also denoted by the symbol $IE$.
Electron gain enthalpy: when an electron is added to an isolated gaseous atom, the amount of energy released during the addition is termed as electron gain enthalpy. It is also denoted by the symbol $H$.
Here it is clear that both are opposite to each other. Thus, if the ionization energy is given then the electron gain enthalpy would have an opposite sign. According to this question,
$Na \to N{a^ + } + {e^ - }$
Here, $IE = 118\;Kcal\,mo{l^{ - 1}}$
Whereas,
$N{a^ + } + {e^ - } \to Na$
Here, $EA = - 118\,kcal\,mo{l^{ - 1}}$
$E,A = \left( {\dfrac{{ - 118 \times {{10}^3} \times 4.18}}{{6.023 \times {{10}^{23}} \times 1.6 \times {{10}^{ - 19}}}}} \right)eV\;ato{m^{ - 1}}$
$\Rightarrow - 5.12\;eV$
Thus, this is the required value of electron gain enthalpy.

Note:
The strength of extra electrons can be found by the value of electron gain enthalpy of the element. The amount of energy released in the chemical reaction depends upon the electron gain enthalpy. In simple words, they are directly proportional to each other.