Question

How many “d” electrons are present in $C{r^{2 + }}$ ion?
A. 4
B. 5
C. 6
D. 3

Answer 1

D-block elements are those elements in which the last electron enters in (n-1) d subshell i.e. penultimate shell. There are four series in the d block corresponding to the filling up of 3d, 4d, 5d or 6d orbitals.

Complete step by step answer:
As we know the atomic number $Cr$ is 24 and its electronic configuration is: $1{s^2},2{s^2},2{p^6},3{s^2},3{p^6},3{d^5},4{s^1}$
Here, Chromium exists in the +2 oxidation state that means there is a loss of 2 electrons. As we know during the filling of an electron in d orbital, the electrons enter in 4s orbital first then in 3d orbital but after the filing of the electron the energy of 4s orbital is quite high hence the electron is removed from 4s first then 3d. So, electronic configuration of $C{r^{2 + }}$ ion will be:
$1{s^2},2{s^2},2{p^6},3{s^2},3{p^6},3{d^4}$
Hence in d orbital, there are 4 unpaired electrons.
Due to the unpaired electrons, $C{r^{2 + }}$ is paramagnetic and forms a coloured compound.

Hence the correct answer is option A .

Additional Information:
d block elements are also known as transition elements and have partially filled d subshell either in the ground state or most common oxidation state. Zn, Cd, Hg are not considered to be transition elements because they have fulfilled d orbital in the ground state as well as in the most common oxidation states.

Note:
Due to paramagnetic nature, $C{r^{2 + }}$ ions have some Dipole moment. The value of dipole moment is found by the formula $\sqrt {n(n + 2)}$
Where n is the number of unpaired electrons
In the case of $C{r^{2 + }}$ ion, the number of unpaired electrons is 4.
Hence the value of dipole moment is calculated as:
$\sqrt {4(4 + 2)} \\\ \Rightarrow \sqrt {4 \times 6} \\\ \Rightarrow \sqrt {24} = 4.89{\text{B}}{\text{.M}} \\\ \\\$