
The number of d-electrons in $F{{e}^{2+}}$ (Z = 26) is not equal to the number of electrons in which one of the following?
(a)- d-electrons in Fe (Z = 26)
(b)- p-electrons in Ne (Z = 10)
(c)- s-electrons in Mg (Z = 12)
(d)- p-electrons in Cl (Z = 17)
Answer
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Hint: First write the configuration of Fe in the ground state and then in $F{{e}^{2+}}$, there is a removal of two electrons. Now compare all the configuration of the elements Fe, Ne, Mg, and Cl.
Complete answer:
Iron is the element that belongs to d-block so its configuration in ground state is given below:
$1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}4{{s}^{2}}3{{d}^{6}}$
When there is an excitation of two electrons then the $Fe$ converts into $F{{e}^{2+}}$. So from the above configuration, two electrons will be removed from the $4s$orbital. So the configuration becomes:
$1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}3{{d}^{6}}$
Now, the electronic configuration of $Fe$ having atomic number 26 is:
$1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}4{{s}^{2}}3{{d}^{6}}$
So, it has the same number of electrons in d-orbital as that of an electron in d-orbital of $F{{e}^{2+}}$, i.e., 6 electrons.
In $Ne$ having atomic number 10, the electronic configuration is:
$1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}$
So, it has the same number of electrons in p-orbital as that of an electron in p-orbital of $F{{e}^{2+}}$, i.e., 6 electrons.
In $Mg$ having atomic number 12, the electronic configuration is:
$1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}$
So, it has the same number of electrons in s-orbital as that of an electron in s-orbital of $F{{e}^{2+}}$, i.e., (2+2+2) 6 electrons.
In $Cl$ having atomic number 17, the electronic configuration is:
$1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{5}}$
So, the number of p-electrons in $Cl$is: $6+5=11$. But in $F{{e}^{2+}}$, the number of p-electrons is: $6+6=12$. So they are not equal.
Therefore, the correct answer is an option (d)- p-electrons in Cl (Z = 17).
Note:
Magnesium is the element of s-block having atomic number 12 and its last electron comes in the s-orbital. Neon and chlorine are the elements of p-block having atomic numbers 10 and 17 respectively and their last electron comes in the p-orbital.
Complete answer:
Iron is the element that belongs to d-block so its configuration in ground state is given below:
$1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}4{{s}^{2}}3{{d}^{6}}$
When there is an excitation of two electrons then the $Fe$ converts into $F{{e}^{2+}}$. So from the above configuration, two electrons will be removed from the $4s$orbital. So the configuration becomes:
$1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}3{{d}^{6}}$
Now, the electronic configuration of $Fe$ having atomic number 26 is:
$1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}4{{s}^{2}}3{{d}^{6}}$
So, it has the same number of electrons in d-orbital as that of an electron in d-orbital of $F{{e}^{2+}}$, i.e., 6 electrons.
In $Ne$ having atomic number 10, the electronic configuration is:
$1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}$
So, it has the same number of electrons in p-orbital as that of an electron in p-orbital of $F{{e}^{2+}}$, i.e., 6 electrons.
In $Mg$ having atomic number 12, the electronic configuration is:
$1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}$
So, it has the same number of electrons in s-orbital as that of an electron in s-orbital of $F{{e}^{2+}}$, i.e., (2+2+2) 6 electrons.
In $Cl$ having atomic number 17, the electronic configuration is:
$1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{5}}$
So, the number of p-electrons in $Cl$is: $6+5=11$. But in $F{{e}^{2+}}$, the number of p-electrons is: $6+6=12$. So they are not equal.
Therefore, the correct answer is an option (d)- p-electrons in Cl (Z = 17).
Note:
Magnesium is the element of s-block having atomic number 12 and its last electron comes in the s-orbital. Neon and chlorine are the elements of p-block having atomic numbers 10 and 17 respectively and their last electron comes in the p-orbital.
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