Q. The ‘spin-only’ magnetic moment [in units of Bohr magneton,$\left(\mu_{\beta}\right)$ ] of $Ni^{2+}$ in aqueous solution would be (Atomic number of Ni = 28):
AIEEEAIEEE 2008
Solution:
$Ni^{2+}=\left[Ar\right]3d^{8}$
Number of unpaired electrons = 2
Hence, magnetic moment =$\sqrt{n \left(n+2\right)}$
$=\sqrt{8}=2.84$
