Question

The ionization potential of Na is 5.1 eV . The value of electron gain enthalpy of $N{{a}^{+}}$ will be:
A. $-5.1eV$
B. $-10.2eV$
C. $+2.55eV$
D. $-2.55eV$

Answer 1

Hint: Electron gain enthalpy is basically the amount of energy released when an electron is added to an isolated gaseous atom. It is found that during addition of an electron, energy can neither be released or absorbed. The amount of energy required to remove the electron will be the same amount of energy released to add an electron for the same atom in the same condition. Only the sign will be different.

Complete step by step answer:
-Ionization potential is also called ionization energy, it is actually the energy required to remove an electron from an isolated molecule or an atom.

- We can write the equation of ionization potential as:
The energy release here is:
\[Na\to N{{a}^{+}}+{{e}^{-}}\]
We can consider this equation as first ionization enthalpy
Here, the value of $\Delta H$ is= 5.1 eV , that is the first ionization enthalpy

- Now, have to find the value of electron gain enthalpy of $N{{a}^{+}}$, so we can write the equation as:
\[N{{a}^{+}}+{{e}^{-}}\to Na\]
We can consider this equation as second ionization enthalpy.

- We can see here that the second equation is the reverse of the first equation, and we know the value of first equation, so the value of second equation will be:
Here, the value of $\Delta H$ is = - 5.1 eV , that is the electron gain enthalpy. This value is equal in magnitude and opposite in sign to the first ionization energy of Na.

-Hence, we can say that the correct option is (A), that is the value of electron gain enthalpy of $N{{a}^{+}}$ will be - 5.1 eV .
So, the correct answer is “Option A”.

Note: - Electron gain enthalpy is measured in the unit of kJ per mol or electron volts per atom.
-There are several factors that affect the electron gain enthalpy like: Atomic size, nuclear charge symmetry of electronic configuration.