Question

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

Answer 1

Given ion is $M^{2+}$ with the atomic no. 27.

So, electronic configuration of $M^{2+}$ is [Ar] $3d^{7}4s^0$

[Ar] $\upharpoonleft$$\downharpoonright$ $\upharpoonleft$ $\downharpoonright$ $\upharpoonleft$ $\upharpoonleft$ $\upharpoonleft$

3d 4s

No. of unpaired $e^{-}$ s =3

So, the spin only magnetic moment is given by,

$\mu$=$\sqrt{n(n+2)}$ =$\sqrt{3(3+2)}$ =$\sqrt{3X5}$ =$\sqrt{15}$ BM

=3.87 BM