Q. Which one of the following has a magnetic moment of $1.75\, B.M.$ ?
Solution:
Magnetic moment, $\mu=\sqrt{n(n+2)} BM$
where $n=$ number of unpaired electrons.
Ion
Outermost electronic configuration
No. of unpaired $e^-$
Magnetic moment (BM)
$V ^{3+}$
2
$\sqrt{2(2+2)}=2.82$
$Cr ^{3+}$
3
$\sqrt{3(3+2)}=3.87$
$Fe ^{3+}$
5
$\sqrt{5(5+2)}=5.91$
$Ti ^{3+}$
1
$\sqrt{1(1+2)}=1.73$
Thus, it is $Ti ^{3+}$ ion, magnetic moment which is $1.75 \,BM$
| Ion | Outermost electronic configuration | No. of unpaired $e^-$ | Magnetic moment (BM) |
|---|---|---|---|
| $V ^{3+}$ | |
2 | $\sqrt{2(2+2)}=2.82$ |
| $Cr ^{3+}$ | |
3 | $\sqrt{3(3+2)}=3.87$ |
| $Fe ^{3+}$ | |
5 | $\sqrt{5(5+2)}=5.91$ |
| $Ti ^{3+}$ | |
1 | $\sqrt{1(1+2)}=1.73$ |