Question

An element belonging to 3d series in the modern periodic table has spin magnetic moment = 5.92 B.M. in +3 oxidation state. Determine the atomic number of the element.

Answer 1

Hint: To solve this, you can sue the formula $\mu =\sqrt{n\left( n+2 \right)}\text{ BM}$ to determine n, where n is the number of unpaired electrons and BM is the unit of magnetic moment. After finding n, find the total number of electrons in the outermost shell of the element. You should know that the general configuration of the 3d-series is $\left[ Ar \right]3{{d}^{1-10}}4{{s}^{2}}$.

Complete step by step solution:
We know that magnetic moment is the tendency of a certain pattern in which the atoms are arranged in an orbital.
To solve this, firstly we have to know the formula of spin magnetic moment.
The magnetic moment is given by $\mu =\sqrt{n\left( n+2 \right)}\text{ BM}$ where n is the number of unpaired electrons.
In the question, the value of spin magnetic moment for +3 oxidation state of the element is given to us as 5.92 BM.
We know 3d elements have an outermost orbital as $3{{d}^{1-10}}4{{s}^{2}}$ .
Let us assume that the element is X.
Therefore, we can write that ${{X}^{+3}},\mu =\sqrt{n\left( n+2 \right)}\text{ BM = 5}\text{.92 BM}$
Now, if we solve the above equation for n, we can write that-
$\begin{align}
& \sqrt{n\left( n+2 \right)}\text{ BM = 5}\text{.92 BM} \\
& \text{n(n+2) BM = (5}\text{.92}{{\text{)}}^{2}}\text{ BM } \\
& \text{n(n+2) BM = 35}\text{.04 BM}\simeq \text{35 BM} \\
\end{align}$
From here we can assume that n is 5. Because, if n is 5 then n+2 is 7 and 5 times 7 is 35 which matches the value obtained from the above equation.
Therefore, we can say n is 5. We have mentioned earlier that n is the number of unpaired electrons.
It means that the element X has 5 unpaired electrons.
Also, the oxidation state of the element is given to us as +3, which means it has total of 5+3 i.e. 8 electrons in the valence shell.
Now, we know that the general configuration of 3d elements is $\left[ Ar \right]3{{d}^{1-10}}4{{s}^{2}}$ .
So, if the element has 8 electrons in the valence shell, it means there are 2 electrons in the s-orbital and the remaining 6 electrons in d-orbital.
Therefore, for the element the electronic configuration is $\left[ Ar \right]3{{d}^{6}}4{{s}^{2}}$ .
Therefore, we can write that the atomic number of the element is 18+6+2 = 26.

Therefore, the answer will be, the atomic number of the element is 26.

Note: Here, the unit of magnetic moment we have written BM which is Bohr magneton. It is the moment of the electrons. We have a similar unit for moments of nuclei, protons and neutrons known as the neutron magneton. We can also solve the above question using the spin-only formula of magnetic moment including spin number. The formula is $\mu =\sqrt{4S\left( S+1 \right)}\text{ BM}$ ,where s is the spin of the electrons in the orbital.