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

(N/A) The atomic number of the element $M$ is $Z=27$,which corresponds to Cobalt $(Co)$.
The electronic configuration of $Co$ is $[Ar] 3d^7 4s^2$.
For the $M^{2+}$ ion,two electrons are removed from the $4s$ orbital:
$M^{2+} = [Ar] 3d^7$.
The $3d^7$ configuration has $3$ unpaired electrons as shown below:
$3d^7 = \boxed{\uparrow\downarrow} \boxed{\uparrow\downarrow} \boxed{\uparrow} \boxed{\uparrow} \boxed{\uparrow}$
Therefore,the number of unpaired electrons $n = 3$.
The formula for 'spin only' magnetic moment is $\mu = \sqrt{n(n+2)} \ BM$.
Substituting $n=3$:
$\mu = \sqrt{3(3+2)} = \sqrt{3 \times 5} = \sqrt{15} \ BM$.
$\mu \approx 3.87 \ BM$.