Question

Magnetic moment 2.83 BM is given by which of the following ions?
(At. nos. Ti = 22, Cr = 24, Mn = 25, Ni = 28)

• A. $T{i}^{3+}$
• B. $N{i}^{2+}$
• C. $C{r}^{3+}$
• D. $M{n}^{2+}$

Answer 1

The correct option is B $N{i}^{2+}$

Magnetic moment is given by $\mu =\sqrt{n(n+2)}B.M$
[where n = no. of unpaired electrons]
When n = 2, then $\mu$ = 2.83 B.M.
For $T{i}^{3+}\left(3d^1\right),n=1;C{r}^{3+}\left(3d^3\right),n=3$
$N{i}^{2+}\left(3d^8\right),n=2M{n}^{2+}\left(3d^5\right),n=5$
Hence, $N{i}^{2+}$ has two unpaired electrons, with magnetic moment 2.83 B.M.