Q.
Out of the following, how many of them have spin-only magnetic moment value of $\sqrt{24}$ B.M.
$Fe ^{2+}, Co ^{2+}, Cu ^{2+}, Ti ^{2+}, Cr ^{2+}$
The d-and f-Block Elements
Solution:
Ion
Electronic configuration
Unpaired
electrons
$\mu( B . M .)$
$Fe ^{2+}$
$[Ar] 3d ^{6}$
4
$\sqrt{4(4+2)}=\sqrt{24}$
$Co ^{2+}$
$[Ar] 3d ^{7}$
3
$\sqrt{3(3+2)}=\sqrt{15}$
$Cu ^{2+}$
$[Ar] 3d ^{9}$
1
$\sqrt{1(1+2)}=\sqrt{3}$
$Ti ^{2+}$
$[Ar] 3d ^{2}$
2
$\sqrt{2(2+2)}=\sqrt{8}$
$Cr ^{2+}$
$[Ar] 3d ^{4}$
4
$\sqrt{4(4+2)}=\sqrt{24}$
Alternate method
$\sqrt{n(n+2)}=\sqrt{24}$
$\therefore n(n+2)=24$
$n^{2}+2 n=24$
$\Rightarrow n \neq 1,2,3$
If $n=4$, then
$4^{2}+2 \times 4=24$
$\therefore $ No. of unpaired electrons $=4$
Among the given options, $Fe ^{2+}$ and $Cr ^{2+}$
have $4$ unpaired electrons and hence,
their magnetic moment $=\sqrt{24}B.M$.
| Ion | Electronic configuration | Unpaired electrons | $\mu( B . M .)$ |
|---|---|---|---|
| $Fe ^{2+}$ | $[Ar] 3d ^{6}$ | 4 | $\sqrt{4(4+2)}=\sqrt{24}$ |
| $Co ^{2+}$ | $[Ar] 3d ^{7}$ | 3 | $\sqrt{3(3+2)}=\sqrt{15}$ |
| $Cu ^{2+}$ | $[Ar] 3d ^{9}$ | 1 | $\sqrt{1(1+2)}=\sqrt{3}$ |
| $Ti ^{2+}$ | $[Ar] 3d ^{2}$ | 2 | $\sqrt{2(2+2)}=\sqrt{8}$ |
| $Cr ^{2+}$ | $[Ar] 3d ^{4}$ | 4 | $\sqrt{4(4+2)}=\sqrt{24}$ |