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Question

Pair of lanthanide ions which show significant deviation between the experimental and calculated magnetic moments, considering contribution from the ground state only (given μ eff = g[J(J + 1)]1/2 , is

The correct answer is Sm3+ and Eu3+

Lanthanide Magnetic Moments Explained

Lanthanide ions are known for their magnetic properties, which primarily arise from the unpaired electrons in their 4f orbitals. For most lanthanide ions, the orbital angular momentum (L) and spin angular momentum (S) are strongly coupled (Russell-Saunders coupling), and the total angular momentum (J) is the good quantum number, especially in the ground state.

Calculating Magnetic Moments (Ground State)

The effective magnetic moment (μeff) for most lanthanide ions can be calculated by considering the contribution from the total angular momentum (J) of the ground state. The formula used is:

\(\mu_{\text{eff}} = g\sqrt{J(J+1)}\)

Where:

  • μeff is the effective magnetic moment in Bohr magnetons (BM).
  • J is the total angular momentum quantum number for the ground state.
  • g is the Landé g-factor, calculated as \(g = 1 + \frac{J(J+1) + S(S+1) - L(L+1)}{2J(J+1)}\).
  • S is the total spin quantum number.
  • L is the total orbital quantum number.

For most lanthanide ions, this formula provides a value that is in reasonably good agreement with experimental measurements at room temperature.

Ions Showing Significant Deviation

However, there are exceptions to this rule. For certain lanthanide ions, the energy difference between the ground state and the first excited state is very small. When this energy gap is small, the excited states can be significantly populated at room temperature due to thermal energy.

If excited states are populated, their contribution to the magnetic moment becomes important. The magnetic moment is then not solely determined by the ground state J value, and the simple formula μeff = g√J(J+1) no longer accurately predicts the experimental value. This phenomenon leads to a significant deviation between the calculated magnetic moment (based on ground state only) and the experimental magnetic moment.

The lanthanide ions where this effect is most pronounced, leading to significant deviation, are Samarium(III) (Sm3+) and Europium(III) (Eu3+). For Sm3+ and Eu3+, the energy difference between the ground state and the low-lying excited states is much smaller compared to other lanthanide ions, making the contribution from excited states significant even at moderate temperatures.

Conclusion

Therefore, the pair of lanthanide ions that show significant deviation between the experimental and calculated magnetic moments, when considering contribution only from the ground state, is Sm3+ and Eu3+.

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Important Questions from Transition Elements and Inner Transition Elements

  1. Which actinide, discovered by Glenn T Seaborg in 1940, is used as a heat source for sensitive electrical components in satellites as well as a power source for satellites?

  2. The known oxidation state(s) of Eu in aqueous solution is/are

  3. The effective magnetic moment (in BM) for a lanthanide f10 ion is approximately

  4. The ore (X) gives a d‐block metal (M) in the elemental form, following a chemical process. Which of the sets X / M / Chemical process below is correct?

  5. The red colour of the gem, ruby is predominantly due to

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