Question

The valence shell electronic configuration of the $F{{e}^{2+}}$ is:
A) $3{{s}^{2}}3{{d}^{6}}$
B) $3{{s}^{1}}3{{d}^{7}}$
C) $3{{s}^{0}}3{{d}^{8}}$
D) $3{{s}^{2}}3{{d}^{5}}$

Answer 1

The valence electrons are those electrons present in the outermost shell of the atom.
- The atomic number of Fe is 26.

Complete Solution :
So in the question we are asked to write the valence shell electronic configuration of $F{{e}^{2+}}$.
We know that the electronic configuration of an atom is a way of representing the electrons present in each orbitals in each shell of the atom. And the electrons are filled according to the increasing order of energy of the orbitals.
- The valence shell is the outermost shell of an atom having the greatest value for ‘n’ which is the principal quantum number and it represents the number of shells in an atom of an element.
Now we should know how to write the electronic configuration of an element. For that we should know the periodic table thoroughly so that we could easily know the atomic number of the given element.
Here the given element is Fe and we know it is a transition element.
- The special characteristics of the metals that belong to transition series is that they show variable oxidation state due to the less energy gap between the outermost s and d orbitals.
Iron possess two oxidation states, $F{{e}^{2+}}$ and $F{{e}^{3+}}$
The atomic number of Fe is 26, hence it possesses 26 electrons in a Fe atom.
Now let’s write the configuration of Fe.
$\text{Electronic}\,\text{configuration}\,\text{of}\,\text{Fe=1}{{\text{s}}^{\text{2}}}\text{2}{{\text{s}}^{\text{2}}}\text{2}{{\text{p}}^{\text{6}}}\text{3}{{\text{s}}^{\text{2}}}\text{3}{{\text{p}}^{\text{6}}}\text{4}{{\text{s}}^{\text{2}}}\text{3}{{\text{d}}^{\text{6}}}$
$\text{Electronic}\,\text{configuration}\,\text{of}\,\text{Fe=}\left[ \text{Ar} \right]\text{4}{{\text{s}}^{\text{2}}}\text{3}{{\text{d}}^{\text{6}}}$
For $F{{e}^{2+}}$ there are only 24 electrons, since the elemental Fe loses 2 electrons and forms $F{{e}^{2+}}$.
And the 2 electrons are lost from the 4s orbitals, even though it is first filled with electrons prior to the 3d orbitals, the electrons from the 4s are removed to get $F{{e}^{2+}}$.
Hence the electronic configuration of $F{{e}^{2+}}$ is:
$\text{=1}{{\text{s}}^{\text{2}}}\text{2}{{\text{s}}^{\text{2}}}\text{2}{{\text{p}}^{\text{6}}}\text{3}{{\text{s}}^{\text{2}}}\text{3}{{\text{p}}^{\text{6}}}\text{3}{{\text{d}}^{\text{6}}}$
So, the correct answer is “Option A”.

Note: If the electronic configuration of $F{{e}^{3+}}$was asked, then it has only 23 electrons and the electronic configuration is:
$\text{Electronic}\,\text{configuration}\,\text{of}\,\text{F}{{\text{e}}^{\text{3+}}}\text{=1}{{\text{s}}^{\text{2}}}\text{2}{{\text{s}}^{\text{2}}}\text{2}{{\text{p}}^{\text{6}}}\text{3}{{\text{s}}^{\text{2}}}\text{3}{{\text{p}}^{\text{6}}}\text{3}{{\text{d}}^{\text{5}}}$
- The energy gap between the ns orbital and (n-1) d orbitals is comparatively less and hence the electrons can be filled and removed from the both the orbitals and electrons from both the orbitals actively involved in the chemical bond formation.
- But there is an exceptional case, when the d-orbital is empty, half-filled or completely filled. The d-orbital possesses extra stability for these three cases.