Question

The “spin-only” magnetic moment, in units of Bohr magnetron $\left({\mu }_B\right)$, of $N{i}^{2+}$in aqueous solution would be

• A. 4.9
• B. \N
• C. 1.73
• D. 2.84

Answer 1

The correct option is D 2.84

$N{i}^{2+}$ in aqueous solution exists as $\left[Ni\right(H_{2}O{)}_6{]}^{2+}$. It has
$s{p}^3{d}^2$ hybridisation as
$N{i}^{2+}\to 3d^{8}4s^0$

$\therefore$ No. of unpaired electron = 2
$\therefore$ ${\mu }_s=\sqrt{n(n+2)}=\sqrt{2(2+2)}=2.84$