Question

Total spin of $M{n^{2 + }}(Z = 25)$ ion will be:
A. $\dfrac{3}{2}$
B. $\dfrac{1}{2}$
C. $\dfrac{5}{2}$
D. $\dfrac{7}{2}$

Answer 1

Hint: We know that electrons in the orbitals are filled up first by $ + \dfrac{1}{2}$ and when all the orbitals of a shell are filled then orbitals are filled with $ - \dfrac{1}{2}$ spin. So the total spin of any atom or ion can be calculated by multiplying the unpaired electrons to $\dfrac{1}{2}$.. Transition elements show magnetic moment due to spin of their electron particles.

Complete answer:
> We can calculate the total spin of any atom or ion with the help of following step;
1. First we should write the electronic configuration of the atom or ion.
2. In the second step we check whether the atom or ion possesses any unpaired electrons or not.
3. Then at last we multiply the number of total unpaired electrons to $\dfrac{1}{2}$.
> We know that $Mn$ has atomic number $25$ and after removing two electron from its outer shell we get $M{n^{2 + }}$ ion so its electronic configure of manganese and its ion are given below;
$Mn = $$1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^2}3{d^5}$ and \[M{n^{2 + }} = 1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}3{d^5}\] . Here we can see that manganese ions in d- orbital there are five unpaired electrons when two electrons are removed from s- orbital of manganese atom.
- Hence the total spin of $M{n^{2 + }}$ ion $ = 5 \times \dfrac{1}{2} = \dfrac{5}{2}$.

So, the correct answer is “Option C”.

Note:
1-We have calculated the total spin of manganese ions by calculating unpaired electrons. With the help of their electronic configuration.
2-We know that electrons and protons can rotate at their space that is why the term spin is used to determine their direction of rotation whether it is clockwise or anticlockwise. Hence the total spin gives the direction of rotation of the atom or ion.