Question

The spin-only magnetic moment [in units of Bohr magneton, ${\mu }_B$] of $N{i}^{2+}$ in aqueous solution would be:

[Atomic number of Ni = 28]

• A. 1.73
• B. 2.84
• C. 4.9
• D. 0

Answer 1

The correct option is B 2.84

In aqueous solution, as aqua is poor ligand so high spin complex forms.

Electronic configuration of $Ni=\left[Ar\right]4s^{2}3d^8$

Electronic configuration of $N{i}^{2+}=\left[Ar\right]4s^{0}3d^8$

$N{i}^{+2},d^8$ configuration, have 2 unpaired electrons so the magnetic moment is:

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

$=\sqrt{2\times 4}=2.84$ BM

Hence, option B is correct.