Q2 / 120 · VITEEE 2006 · Chemistry
The magnetic moment of lanthanide ions is determined from which one of the following relation?
- A$\mu=\sqrt{\mathrm{n}(\mathrm{n}+2)}$
- Correct answerB$\mu=g \sqrt{J(J+1)}$
- C$\mu=g \sqrt{\mathrm{n}(\mathrm{n}+1)}$
- D$\mu=2 \sqrt{\mathrm{n}(\mathrm{n}+1)}$
Solution
In case of lanthanoids, 4 f orbitals lie too deep and hence the magnetic effect of the motion of the electron in its orbital is not quenched out. Here spin contribution $\mathrm{S}$ and orbital contribution $\mathrm{L}$ couple together to give a new quantum number $\mathrm{J}$.
Thus magnetic moment of lanthanoids is given by, $\mu=g \sqrt{J(J+1)}$
where $\mathrm{J}=\mathrm{L}-\mathrm{S}$ when the shell is less than half fill $\mathrm{J}=\mathrm{L}+\mathrm{S}$ when the shell is more than half fill and
$\mathrm{g}=1 \frac{1}{2}+\frac{\mathrm{S}(\mathrm{S}+1)-\mathrm{L}(\mathrm{L}+1)}{2 \mathrm{~J}(\mathrm{~J}+1)}$
Hence, the correct answer is option: 'B'.