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

$\text{Number of unpaired electrons}$

Given:
Atomic number $\left(Z\right)=27$

The valence electronic configuration of cobalt $\left(Co\right)$ is $3d^7,4s^2$.
$M^{2+}\left(aq\right)$ ion means, it loses two electrons.

Hence, valence electronic configuration becomes $d^7$.
Now, Number of unpaired electrons $=3$

$\text{Spin only magnetic moment}$

We know that $\mu =\sqrt{n(n+2)}$
where $n=$ number of unpaired electrons

Putting $n=3$
we get, $\mu =\sqrt{3(3+2)}=3.87BM$

$\text{Final answer:}$

$\mu =3.87BM$