Question

Write the electronic configuration of: \[M{n^{2 + }},F{e^{2 + }}\] .

Answer 1

Hint: We must know the total number electrons present in the particular element which is equal to their respective atomic number. Using the Aufbau principle, we can write the electronic configuration as per their increasing energy levels. And in case of their cationic species, we remove electrons from their electronic configuration of ground state.

Complete step-by-step answer:

Electronic configurations are the summary of where the electrons are positioned around a nucleus. We know that each neutral atom has a number of electrons equal to its number of protons. What we will do now is place those electrons into an arrangement around the nucleus as per their energy levels and the shape of the orbital in which they are located.

Orbitals and Electron Capacity of the First Four Principle Energy Levels
Principle Energy Level(n)	Type of sublevel	Number of orbitals per type	Number of orbitals per level	Maximum number of electrons
1	s	1	1	2
2	s	1	4	8
	p	3
3	s	1	9	18
	p	3
	d	5
4	s	1	16	32
	p	3
	d	5
	f	7

The order in which electrons are placed into the orbitals is based on the order of their energy levels. This is called the Aufbau principle. The lowest energy orbitals are filled first.
Let us now write the electronic configuration of neutral manganese and iron atoms which have atomic numbers 25 and 26 respectively which means they will have electrons equal to their atomic numbers.
Electronic configuration of Mn: \[1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^2}3{d^5}\]
Electronic configuration of Fe: \[1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^2}3{d^6}\]
But we have to find the electronic configuration of their cationic species: \[M{n^{2 + }},F{e^{2 + }}\] . Here, +2 charge indicates that two electrons have been removed from the electronic configuration. Therefore, we can remove two electrons from 4s orbitals as they are easy to remove. This is because removing an electron from 3d will require more energy as half-filled configuration is highly stable in case of manganese and same with iron.
Electronic configuration of \[M{n^{2 + }}\] : \[1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^0}3{d^5}\]
Electronic configuration of \[F{e^{2 + }}\] : \[1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^0}3{d^6}\]

Note: We can also write the electronic configuration in short form by involving previous noble gas element’s configuration and substituting it in the main configuration. In these two cases, we can use electronic configuration of argon, which is a noble gas having atomic number 18. So, we can now write these electronic configurations as \[[Ar]4{s^2}3{d^5}\] and \[[Ar]4{s^2}3{d^6}\] for ground state.