Question

Arrange $Cl,F,{{F}^{-}},C{{l}^{-}}$ in increasing order of ionization potential.
A. ${{F}^{-}}$ < $C{{l}^{-}}$ < $ Cl$ < $ F$
B. $C{{l}^{-}}$ < ${{F}^{-}}$ < $F$ < $Cl$
C. $C{{l}^{-}}$ < $F$ < ${{F}^{-}}$ < $Cl$
D. ${{F}^{-}}$ < $C{{l}^{-}}$ < $F$ < $ Cl$

Answer 1

Hint: The amount of energy which is required to remove an electron from an isolated gaseous atom or a molecule is called ionization energy or ionization potential. It gives ideas about the chemical reactivity of atoms or molecules.

Complete Solution:
- Ionization energy or ionization potential is the minimum amount of energy which is required from an electron to come out from the influence of the nucleus. Ionization energy increases with the increase in the atomic number when we move from left to right in a period and it decreases when we move down the group.
- First ionization potential is the energy required to remove the first electron or one electron from the valence orbital of the atom or an ion. Let's consider the electronic configuration of all the given options.
Electronic configuration of chlorine: $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{5}}$
Electronic configuration of chloride ion: $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}$
Electronic configuration of fluorine: $1{{s}^{2}}2{{s}^{2}}2{{p}^{5}}$
Electronic configuration of fluoride ion: $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}$

Fluorine will have its valence electrons in the 2p orbital while the valence electrons of chlorine are present in the 3p orbital. There is more attractive force in fluorine hence the ionization potential of fluorine will be more than that of chlorine.

So, the correct answer is “Option A”.

Note: This order is only valid for the first ionization enthalpy or energy. For the second ionization enthalpy the order changes because losing an electron makes some of the compound stable and some unstable. This affects the order of ionization enthalpy. The element having the half-filled or fully filled electronic configurations are most stable.