Question

The 'spin only' magnetic moment of $N{i}^{2+}$ in aqueous solution would be :
[Atomic number of Ni = 28]

• A. $\sqrt{6}$ BM
• B. $\sqrt{15}$ BM
• C. $\sqrt{2}$ BM
• D. $\sqrt{8}$ BM

Answer 1

The correct option is D $\sqrt{8}$ BM

$N{i}^{2+}$ has $3d^8$ as outer electronic configuration. It has 2 unpaired electrons.
The 'spin only' magnetic moment of $N{i}^{2+}$ in aqueous solution would be $\sqrt{n(n+2)}=\sqrt{2(2+2)}=\sqrt{8}$ BM.