Question

Out of the following, how many of them have spin-only magnetic moment value of $\sqrt{24}$ B.M.
$F{e}^{2+},C{o}^{2+},C{u}^{2+},T{i}^{2+},C{r}^{2+}$

Answer 1

Answer: 2

Ion	Electronic configuration	Unpaired electrons	$\mu (B.M.)$
$F{e}^{2+}$	$[Ar]3d^6$	4	$\sqrt{4(4+2)}=\sqrt{24}$
$C{o}^{2+}$	$[Ar]3d^7$	3	$\sqrt{3(3+2)}=\sqrt{15}$
$C{u}^{2+}$	$[Ar]3d^9$	1	$\sqrt{1(1+2)}=\sqrt{3}$
$T{i}^{2+}$	$[Ar]3d^2$	2	$\sqrt{2(2+2)}=\sqrt{8}$
$C{r}^{2+}$	$[Ar]3d^4$	4	$\sqrt{4(4+2)}=\sqrt{24}$

Alternate method
$\sqrt{n(n+2)}=\sqrt{24}$
$\therefore n(n+2)=24$
$n^2+2n=24$
$\Rightarrow n\ne 1,2,3$
If $n=4$, then
$4^2+2\times 4=24$
$\therefore$ No. of unpaired electrons $=4$
Among the given options, $F{e}^{2+}$ and $C{r}^{2+}$
have $4$ unpaired electrons and hence,
their magnetic moment $=\sqrt{24}B.M$.