Question

The ??pin-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):

• A. 2.84
• B. 4.9
• C. 0
• D. 1.73

Answer 1

Answer: (A)

2.84

Answer 2

$Ni^{2+}=\left[Ar\right]3d^{8}$ Number of unpaired electrons = 2 Hence, magnetic moment =$\sqrt{n \left(n+2\right)}$ $=\sqrt{8}=2.84$