Question

How many ‘ $d$ ’electrons are present in $C{r^{2 + }}$ ion?
(A) $4$
(B) $5$
(C) $6$
(D) $3$

Answer 1

Aufbau principle given a sequence in which various subshells are filled up depending on the relative order of the energies of various subshells. The orbitals of the subshell in such a way as to give the maximum number of unpaired electrons with parallel spin.

Complete step by step answer:
In order to write the chromium electron configuration we first need to know the number of electrons for the $Cr$ atom (there are $24$ electrons)
Once we have the configuration for $Cr$ , the ions are simple. When we write the configuration we shall put all $24$ electrons in orbitals around the nucleus of the chromium atom.
In writing the electron configuration for chromium the first two electrons will go in the $1s$ orbital. Since $1s$ can only hold two electrons the next $2$ electrons for chromium go in the $2s$ orbital. The next six electrons will go in the $2p$ orbital. The $p$ orbital can hold up to six electrons. We will put six in the $2p$ orbital and then put the next two electrons in the $3s$ . Since the $3s$ if now full we’ll move to the $3p$ where we will place the next six electrons. We now shift to the $4s$ orbital where we place the remaining two electrons after the $4s$ is full we put the remaining four electrons in the $3d$ orbital and end with$3{d^4}$.
Therefore the expected electron configuration for chromium will be
$1{s^2}2{s^2}2{p^6}3{s^2}3{p^4}4{s^2}3{d^9}$
Correct electron configuration for chromium ( $Cr$ )
Half-filled and fully-filled subshell have got extra stability. Therefore, one of the $4{s^2}$ electrons jumps to the $3{d^5}$ so that it is half-filled. This gives us the (correct) configuration of
$1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^5}4{s^1}$
For the $C{r^{2 + }}$ ion we remove one electron from $4{s^1}$ and one from $3{d^5}$ leaving us with:
$1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^4}$
So, the correct answer is A.
Additional Information:
The configuration notation helps scientists to write and communicate how electrons are arranged around the nucleus of an atom. This makes it much easier to understand and predict how atoms will interact to form chemical bonds.
For the $C{r^{3 + }}$ ion we remove a total of three electrons (one from the $4{s^1}$ and two from the $3{d^5}$ ) leaving with
$1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^3}$

Note:
Chromium is an exception to the rules for writing electron configurations when writing the electron configuration for an atom like $Cr$,$3d$ is usually written before the $4s$ . Both the configurations have the correct numbers of electrons in each orbital, it is just a matter of how the electronic configuration notation is written.