Question

Calculate the 'spin only' magnetic moment of $${M^{2 + }}\left( {aq} \right)\;$$ ion ($$Z = 27$$).

Answer 1

Hint:To calculate the magnetic moment number of the given ion we have to find the total number of unpaired electrons present in the ion. For that, we have to write the electronic configuration of this species.
Formula Used:
The spin only magnetic moment is represented by the formula:
$\mu = \sqrt {n(n + 2)} $
Where, $n = $ Number of Unpaired electrons.

Complete step by step answer:
The magnetic moment is the term used with the electronic spectra to achieve the details about the oxidation state and the stereochemistry of the central metal ion present in a coordination complex. It is calculated with the help of unpaired electrons present in the ion. The unit of Magnetic moment is Bohr magneton. The net magnetic moment of an electron is calculated using the vector sum of orbital magnetic moment and the spin only magnetic moment
Now, we will calculate the number of unpaired electrons present in the given ion. The atomic number is $27$. Its electronic configuration will be $$\left[ {Ar} \right]{\;^{18}}3{d^7}4{s^2}$$. The ${M^{2 + }}$ ion will form after the removal of two electrons from the s orbital. The E.C. of ${M^{2 + }}$ ion will be $$\left[ {Ar} \right]{\;^{18}}3{d^7}4{s^0}$$. Now we have $7$ electrons in the d-orbital. So, the total number of unpaired electrons in the d-orbital will be $3$.
On putting this value of $n$in the formula we will get the magnetic moment. It will be;
$\mu = \sqrt {n(n + 2)} $
$\mu = \sqrt {3(3 + 2)} = \sqrt {15} = 3.87BM$
Hence, the 'spin only' magnetic moment of $${M^{2 + }}\left( {aq} \right)\;$$ ion will be $3.87\,BM$

Note:
 The orbital angular momentum of an electron of an object about a fixed origin is called the angular momentum of the center of mass about the origin. The orbital angular momentum is given by the formula $ = \dfrac{h}{{2\pi }}\sqrt {l(l + 1} $ here $'l'$ is azimuthal quantum number.