Question

The first ionization potential of sodium $ \left( {Na} \right) $ is $ 5.1eV $ . The value of electron gain enthalpy of $ N{a^ + } $ will be:
(A) $ - 2.55eV $
(B) $ - 5.1eV $
(C) $ - 255eV $
(D) $ - 51eV $

Answer 1

Hint :The ionization potential is the amount of energy required to remove an electron from its outermost shell. The electron gain enthalpy is the energy released when an electron is added to a neutral gaseous atom. The ionization potential value is opposite to the electron gain enthalpy.

Complete Step By Step Answer:
Sodium is a chemical element with atomic number $ 11 $ and has only one electron in its outermost shell. Sodium is a metal and can easily lose an electron from its outermost orbital or shell as it tries to attain the neon gas configuration. Neon is a noble gas with completely filled electronic configuration and also a stable element.
Ionization potential is the amount of energy required to remove an electron from its outermost shell. The reaction will be as follows:
 $ Na \to N{a^ + } + {e^ - } - - - - 5.1eV $
The electron gain enthalpy is the energy released or absorbed when an electron is added to a neutral gaseous atom or an ion. The reaction will be as follows:
 $ N{a^ + } + {e^ - } \to Na - - - - - \left( { - 5.1eV} \right) $
The value of ionization potential is exactly opposite to the value of electron gain enthalpy in case of sodium. As the two reactions are opposite, it leads to the opposite value.
Option B is the correct one.

Note :
As the reactions are opposite to each other, the value of electron gain enthalpy of $ N{a^ + } $ is exactly opposite to the first ionization potential of sodium $ \left( {Na} \right) $ . The energy can sometimes be absorbed when an electron is added to a negative ion like anions.