Question

If hydrogen atoms (in the ground state) are passed through an homogeneous magnetic field, the beam is split into two parts. This interaction with the magnetic field shows that the atoms must have magnetic moment. However, the moment cannot be due to the orbital angular momentum since $\phi =0$. Hence one must assume existence of intrinsic angular momentum, which as the experiment shows, has only two permitted orientations.
Spin of the electron produces angular momentum equal to $s=\sqrt{s(s+1)}\frac{h}{2\pi }$ where $s=+\frac{1}{2}$
Total spin of an atom $=+\frac{n}{2}or-\frac{n}{2}$
where n is the number of unpaired electron.
The Substance which contain species with unpaired electrons in their orbitals behave as paramagnetic substances. The paramagnetism is expressed in terms of magnetic moment.
The magnetic moment of an atom
$\mu =\sqrt{s(s+1)}\frac{eh}{2\pi mc}=\sqrt{\frac{n}{2}(\frac{n}{2}+1)}\frac{eh}{2\pi mc}$ $s=\frac{n}{2}$
${\mu }_s=\sqrt{n(n+2)}$B.M
n-number of upaired electrons
1.B.M. (Bohr magneton)$=\frac{eh}{4\pi mc}$
If magnetic moment is zero the substance is di-magnetic.

Which of the following ion has highest magnetic moment?

• A. $F{e}^{2+}$
• B. $M{n}^{2+}$
• C. $C{r}^{3+}$
• D. $V3+$

Answer 1

Answer: (B)

$M{n}^{2+}$

E.C. of $M{n}^{2+}=1s^2,2s^2,2p^6,3s^2,3p^6,3d^5,$
$M{n}^{2+}$ has 5 unpaired electrons i.e, $\left[3d^5\right]$.
All others have less than 5 unpaired electrons.
So, $M{n}^{2+}$ has highest magnetic moment.