Question

What is the electron configuration of $C{r^{2 + }}$ ?

Answer 1

Hint : An element's electronic configuration is a symbolic representation of how the atoms' electrons are spread through various atomic orbitals. When writing electron configurations, a generic notation is used, with the energy level and form of orbital written first, then the number of electrons in the orbital written in superscript.

Complete Step By Step Answer:
$C{r^{2 + }}$ ion is formed when a $Cr$ atom loses two electrons. When losing electrons, the electrons in the outermost (valence) shells are lost first.
When it comes to electron configurations, chromium and copper are two special cases, with just one electron in the $4s$ orbital compared to the other transition metals in the first row, which have a filled $4s$ orbital.
This is because electron repulsion is minimised in this configuration. For $Cr$ in particular, half-filled orbitals are the most stable configuration.
As a result, elemental Chromium's electron configuration is
$1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^1}3{d^5}$
The electrons in the $4s$ orbital are separated first because they are further away from the nucleus and hence easier to remove during ionisation.
To form the $C{r^{2 + }}$ ion, we must eliminate two electrons, one $4s$ and one $3d$ , leaving us with:
$1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^4}$
Therefore, the electron configuration of $C{r^{2 + }}$ is $1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^4}$

Note :
Always remember that when filling up orbitals, we must follow the Aufbau theory and fill up the orbitals with the least energy first. However, since the electrons in the valence shells are easily lost when withdrawing electrons, we must separate them from the valence shells first, then the pen-ultimate shells, and so on.