What is the electronic configuration for a Fe (III) ion in its ground state?
(A) $\left[ Ar \right]3{{d}^{5}}$
(B) $\left[ Ar \right]3{{d}^{6}}$
(C) $\left[ Ar \right]4{{s}^{2}}3{{d}^{3}}$
(D) $\left[ Ar \right]4{{s}^{2}}3{{d}^{6}}$

Hint: Start by writing the atomic number of Iron followed by its ground state electronic configuration. Now, we have to find the electronic configuration for Fe (III) ion in its ground state. So, the oxidation state of Fe is +3, accordingly, write the electronic configuration and get the answer.

Complete step by step solution:
-Iron is a part of the d-block in the periodic table. It is a transition metal.
-Iron has atomic number 26.
-The ground state electronic configuration of iron, Fe is \[1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}4{{s}^{2}}3{{d}^{6}}\]
-It can also be represented as, Fe $=\left[ Ar \right]4{{s}^{2}}3{{d}^{6}}$
-Now, we have to find the ground state electronic configuration of Fe (III) ions. Fe (III) ion indicates iron is in its ferric state. So, its oxidation state is +3.
-This implies, Fe has lost three electrons. So, it will lose electrons from its outermost orbitals.
-Therefore first, it will lose two electrons from 4s orbital to form Fe (II) ion, that is, ferrous state and then it will lose one more electron from 3d orbital to form Fe (III) ion, that is, ferric state.
-Therefore, the electronic configuration of Fe (III) ion in its ground state will be \[1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}3{{d}^{5}}4{{s}^{0}}\] which can also be represented as, $\left[ Ar \right]3{{d}^{5}}$

Therefore, the correct answer is option (A).

Note: Remember that iron having +3 oxidation state is known as ferric ion and the one having +2 oxidation state is known as a ferrous ion. Remember removal of electrons will always take place from the last filled orbital regardless of whether it is half-filled, partially filled or completely filled.