Question

Consider the following ions :
(i) $N{i}^{2+}$
(ii) $C{o}^{2+}$
(iii) $C{r}^{2+}$
(iv) $F{e}^{3+}$
(Given, at. no. $Cr=24,Fe=26,Co=27,Ni=28$)
The correct order of magnetic moment of these ions is

• A. (i) < (ii) < (iii) < (iv)
• B. (iv) < (ii) < (iii) < (i)
• C. (i) < (iii) < (ii) < (iv)
• D. (iii) < (iv) < (ii) < (i)

Answer 1

Answer: (A)

Magnetic moment, $\mu =\sqrt{n(n+2)}BM$

where, $n=$ number of unpaired electrons

Higher the number of unpaired electrons, more the value of $\mu$.

The electronic configuration of given species is

$N{i}^{2+}=[Ar]3d^8$ (two unpaired electrons)

$C{o}^{2+}=[Ar]3d^7$ (three unpaired electrons)

$C{r}^{2+}=[Ar]3d^4$ (four unpaired electrons)

$F{e}^{3+}=[Ar]3d^5$ (five unpaired electrons)

Thus, the order of magnetic moment is

$\underset{(i)}{N{i}^{2+}}<\underset{(ii)}{C{o}^{2+}}<\underset{(iii)}{C{r}^{2+}}<\underset{(iv)}{F{e}^{3+}}$