Question

What is the value of electron gain enthalpy of $N{a^ + }$ if $I{E_1}$ of$Na = 5.1eV$?
A) $ - 5.1eV$
B) $ - 10.2eV$
C) $ + 2.55eV$
D) $ + 10.2eV$

Answer 1

Hint: We know that Electron gain enthalpy is characterized as the enthalpy change related with a segregated vaporous molecule (X) when it increases an electron to shape its comparing anion. The response can be given as underneath:
\[X\left( g \right) + {e^ - } \to {X^ - }\left( g \right)\]

Complete step by step answer:
As we realize that ionization energy is the base measure of energy needed to eliminate an electron from the valence shell of a confined vaporous molecule in its ground state bringing about the arrangement of particle though, electron gain enthalpy is the measure of energy related with the increase of electrons by the segregated gaseous atom. The electron gain enthalpy of sodium cation is equivalent in size and inverse in sign to the primary ionization energy of sodium molecule. It is equivalent to $ - 5.1eV$.

So, the correct answer is Option A.

Note: Now we can discuss about the variety in electron gain enthalpy in the period as,
In the advanced occasional table, on moving from left to directly over a period, the nuclear size of components diminishes and the successful atomic charge increments. Consequently, the power of fascination between the core and added electron increments. Consequently, electron gain enthalpy turns out to be more negative while moving right over a period.
Let’s we know that the variety of electron gain enthalpy in a group as,
Generally, when we descend in a group turns out to be more positive. This is on the grounds that, as we go down the group both the nuclear size and atomic charge increments, however the impact of nuclear size is more conspicuous than that of atomic charge. Henceforth, the power of fascination between the core and the additional electron diminishes and in this way enthalpy turns out to be more positive.