Question

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

Answer 1

Hint : Electronic configuration of an atom is the specific arrangement of electrons in the various orbit levels around the nucleus. Here, there is a positive charge on the given atom. This means that the number of electrons present in the ion are less than the neutral atom by the difference equal to the charge of the ion.

Complete Step By Step Answer:
To be able to write the electronic configuration of Cadmium, we must know the number of electrons present in it.
From the periodic table we obtain the atomic number of Cadmium as $\;48$ and is present in the $d$ -block of the periodic table. Hence, the last electron entering will be in the $d$ orbit.
Hence, there are $\;48$ electrons present in the Cadmium atom.
Now, beginning with the lowest energy level we will start arranging the electrons till we get a total of $\;48$ electrons.
We must remember here to arrange the electrons in $4{{s}^{2}}$ orbit before the $3{{d}^{10}}$ orbit because of relatively low energy level and similar method for subsequent higher energy levels.
Thus the electronic configuration of Cadmium is
$Cd=1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}4{{s}^{2}}3{{d}^{10}}4{{p}^{6}}5{{s}^{2}}4{{d}^{10}}$
Arranging them as per the distance from the nucleus,
$Cd=1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}3{{d}^{10}}4{{s}^{2}}4{{p}^{6}}4{{d}^{10}}5{{s}^{2}}$
This is the expanded method for the electronic configuration of Cadmium.
We can compress it by replacing the completely filled energy levels with the previous nearest inert gas.
From the periodic table, the nearest previous inert gas is Krypton whose electronic configuration is written as,
$Kr=1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}3{{d}^{10}}4{{s}^{2}}4{{p}^{6}}$
Hence, the energy levels can be replaced by inert gas as,
$Cd=\left[ Kr \right]4{{d}^{10}}5{{s}^{2}}$
This is the electronic configuration of Cadmium with inert gas core.
Now, here we are given $C{{d}^{2+}}$ which means that Cadmium ion has two electrons less as compared to the neutral element.
Now, as the orbit $5{{s}^{2}}$ is the farthest from the nucleus the electrons will be eliminated from that orbit. Hence, the electronic configuration of $C{{d}^{2+}}$ is,
$C{{d}^{+2}}=\left[ Kr \right]4{{d}^{10}}$

Note :
For cadmium, we can see that the electrons enter the orbit $5{{s}^{2}}$ before the orbit $4{{d}^{10}}$ because its energy level is lower. Hence, the last electron in Cadmium enters in $4{{d}^{10}}$ . However, when an electron is removed it will be removed first from the $5{{s}^{2}}$ orbit, as it is further away from the nucleus, and hence the attraction due to nucleus is less compared to $4{{d}^{10}}$ orbit. This applies to all the elements in the transition metals in the $d$ block of the periodic table.