Question

The ‘spin-only’ magnetic moment [in units of Bohr magneton,$\left({\mu }_{\beta }\right)$ ] of $N{i}^{2+}$ in aqueous solution would be (Atomic number of Ni = 28):

• A. 2.84
• B. 4.90
• C. 0
• D. 1.73

Answer 1

Answer: (A)

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