Calculate the ‘spin only’ magnetic moment of

Z = 27

 [Ar] 3d7 4s2

 M2+ = [Ar] 3d7

3d7 = 

i.e., 3 unpaired electrons

 n = 3

$\Rightarrow \sqrt{n(n+2)}=\mu$

$\Rightarrow \sqrt{3(3+2)}=\mu$

$\Rightarrow \sqrt{15}=\mu$

μ ≈ 4 BM