Question

The number of d-electrons in $\text{ F}{{\text{e}}^{\text{2+}}}\text{ }$$\text{ ( Z = 26 ) }$ is not equal to the number of electrons in which one of the following?
A) d - electrons in $\text{ Fe }$$\text{ ( Z = 26 ) }$
B) p - electrons in $\text{ Ne }$$\text{ ( Z = 10 ) }$
C) s - electrons in $\text{ Mg }$$\text{ ( Z = 12 ) }$
D) p - electrons in $\text{ Cl }$$\text{ ( Z = 17 ) }$

Answer 1

Hint: the electronic configuration of the $\text{ F}{{\text{e}}^{\text{2+}}}\text{ }$ ions is as follows:
$\text{ F}{{\text{e}}^{\text{2+}}}\text{= 1}{{\text{s}}^{\text{2}}}\text{2}{{\text{s}}^{\text{2}}}\text{2}{{\text{p}}^{\text{6}}}\text{3}{{\text{s}}^{\text{2}}}\text{3}{{\text{p}}^{\text{6}}}\text{3}{{\text{d}}^{\text{6}}}\text{ }$
There are 6 d-electrons in the $\text{ F}{{\text{e}}^{\text{2+}}}\text{ }$ ions. Write down the electronic configuration of each element and determine the number of electrons in the respective orbitals.


Complete step by step answer:
We have given that the $\text{ F}{{\text{e}}^{\text{2+}}}\text{ }$ion has an atomic number equal to 26. We are interested in the $\text{ F}{{\text{e}}^{\text{2+}}}\text{ }$ ion. The $\text{Fe}$ atom loses its two electrons from the $\text{ 4s }$ orbital. Therefore, the electronic configuration of the $\text{ F}{{\text{e}}^{\text{2+}}}\text{ }$is as follows,
$\text{ F}{{\text{e}}^{\text{2+}}}\text{= 1}{{\text{s}}^{\text{2}}}\text{2}{{\text{s}}^{\text{2}}}\text{2}{{\text{p}}^{\text{6}}}\text{3}{{\text{s}}^{\text{2}}}\text{3}{{\text{p}}^{\text{6}}}\text{3}{{\text{d}}^{\text{6}}}\text{ }$
Here, the $\text{ F}{{\text{e}}^{\text{2+}}}\text{ }$ ions the $\text{ 3d }$ orbital accommodated the 6 electrons.
Now, we are interested to find out a species which does not contain the 6 electrons in the valence shell.

Let's have a look at the option.
A) d- electrons in the $\text{Fe}$atom:
The electronic configuration of $\text{Fe}$ (zero states) is as shown below,
$\text{ Fe = 1}{{\text{s}}^{\text{2}}}\text{2}{{\text{s}}^{\text{2}}}\text{2}{{\text{p}}^{\text{6}}}\text{3}{{\text{s}}^{\text{2}}}\text{3}{{\text{p}}^{\text{6}}}\text{3}{{\text{d}}^{\text{6}}}\text{4}{{\text{s}}^{2}}\text{ }$
The $\text{Fe}$atom has the six electrons in the$\text{ 3d }$. On comparing it with the electronic configuration of $\text{ F}{{\text{e}}^{\text{2+}}}\text{ }$ions, we can say that the number of electrons in the d-orbital of $\text{ F}{{\text{e}}^{\text{2+}}}\text{ }$and $\text{Fe}$ have the 6 electrons.

B) p electrons in the $\text{ Ne }$ atom:
The electronic configuration of $\text{ Ne }$ is as shown below,
$\text{ Ne = 1}{{\text{s}}^{\text{2}}}\text{2}{{\text{s}}^{\text{2}}}\text{2}{{\text{p}}^{\text{6}}}\text{ }$
The p – orbital of the$\text{ Ne }$atom contains the 6 electrons. That is the same number of electrons which are present in the d-orbital of the $\text{ F}{{\text{e}}^{\text{2+}}}\text{ }$ ions.

C) s- electrons in the $\text{ Mg }$ :
The electronic configuration of the magnesium is as shown below:
$\text{ Mg = 1}{{\text{s}}^{\text{2}}}\text{2}{{\text{s}}^{\text{2}}}\text{2}{{\text{p}}^{\text{6}}}\text{3}{{\text{s}}^{\text{2}}}\text{ }$
In magnesium, the $\text{ 1s }$ orbital contains the 2 electrons .Similarly, $\text{ 2s }$and $\text{ 3s }$orbitals contain 2 electrons each. Therefore, the total number of s orbitals would be equal to sum of the electrons in the$\text{ 1s }$, $\text{ 2s }$and $\text{ 3s }$orbitals.
$\begin{align}
 & \text{ s}-\text{electrons = 1s }{{\text{e}}^{-}}\text{ + 2s }{{\text{e}}^{-}}\text{ + 3s }{{\text{e}}^{-}} \\
 & \therefore \text{ s}-\text{electrons }=\text{ 2 + 2 + 2 = 6 }{{\text{e}}^{-}}\text{ } \\
\end{align}$
Therefore, there are a total 6 s electrons. That is the same number of electrons that are present in the d-orbital of the $\text{ F}{{\text{e}}^{\text{2+}}}\text{ }$ ions.

D) p – electrons in the $\text{ Cl }$ atom:
The electronic configuration of $\text{ Cl }$ an atom is as follows:
$\text{ Cl = 1}{{\text{s}}^{\text{2}}}\text{2}{{\text{s}}^{\text{2}}}\text{2}{{\text{p}}^{\text{6}}}\text{3}{{\text{s}}^{\text{2}}}\text{3}{{\text{p}}^{\text{5}}}\text{ }$
The p orbital of the chlorine atom accommodates the $(6+5\text{ = 11) }$ electrons. The d-orbital of the $\text{ F}{{\text{e}}^{\text{2+}}}\text{ }$ atom contains the 6 electrons which are not the same as the p-electrons in the chlorine atom.
So, the correct answer is “Option D”.

Note: The atoms or the ions which contain the equal number of electrons are known as the isoelectronic species. For example, \[\text{ }{{\text{O}}^{\text{2}-}}\text{ , }{{\text{F}}^{-}}\text{ , M}{{\text{g}}^{\text{2+}}}\text{ }\] have the 10 electrons.