Question

What is the electronic configuration of iron(ii) ion?

Answer 1

Hint : Divide the periodic table into parts that display the atomic orbitals, or the regions where electrons are found, to calculate an electron structure. The s-block is made up of groups one and two, the d-block is made up of groups three to twelve, the p-block is made up of groups thirteen to eighteen, and the f-block is made up of the two rows at the right. The energy ranges that form the orbitals and electrons are represented by the rows one through seven

Complete Step By Step Answer:
We know that the atomic number of $Fe$is 26. Then, the electronic configuration of $Fe$is $[Ar]3{d^6}4{s^2}$. When iron cools to a molten state, it takes on three allotropic forms at varying temperatures. One of them is $F{e^{2 + }}$. In this oxidation state two electrons are being donated by the ion.
In the electronic configuration, these 2 electrons are being removed from the last orbital that is $4s$. Hence, the electronic configuration of $F{e^{2 + }}$is $[Ar]3{d^6}$ or $1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^2}3{d^6}$.
Metals are naturally attracted to iron because of its unique crystalline form and electronic configuration. Ferromagnetic structures are what they're called. Even though iron does not have a single crystalline structure, it has a variety of allotropic shapes.

Note :
Iron compounds are mostly formed at oxidation states of $+ 2$ and $+ 3$. They're also possible at a higher oxidation state of $+ 6$. Potassium ferrate is an exemplar of this. Iron $(4)$serves as an intermediate in a variety of biochemical oxidation reactions. Iron is one of the first elements in its group which cannot achieve an oxidation state of $+ 8$.