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

Concept:

The effective magnetic moment of lanthanoids is calculated by:

\(g\sqrt{J(J+1)} \)

where g is called Lande's splitting factor and calculated as:

\(g= 1+ \frac{J(J+1)+S(S+1)-L(L+1)}{2J(J+1)} \)

Explanation:

For f¹⁰, the magnetic moment is calculated as below:

electronic configuration of f¹⁰ is:

Now, the total orbital angular momentum quantum number is calculated as:

\(L = \sum m_{l} \)

= +3+3+2+2+1+1-1-2-3

= 6

Total spin angular momentum quantum number is:

\(S = \frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2} \)

= 2

Now, the total angular momentum quantum number is calculated as:

\(J = \left | L\pm S \right | \)

\(J = L + S \)  when orbitals are more than half filled

 \(J = L - S \)  when orbitals are less than half filled.

Here orbitals are more than half filled therefore the total angular momentum quantum number is

\(J = L + S \)

  \(=6+2\)

   \(=8\)

Now to calculate magnetic moment first we have to find the value of g which is calculated as:

\(g= 1+ \frac{J(J+1)+S(S+1)-L(L+1)}{2J(J+1)} \)

 \(= 1 + \frac{8(8+1) + 2(2+1)-6(6+1)}{2\times 8(8+1)} \)

 \(= 1 + \frac{8\times 9 + 2\times 3 - 6\times 7}{2\times 8\times 9}\)

 \(= 1 + \frac{72 + 6 -42}{144}\)

 \(= 1+0.25\)

  \( = 1.25 \)

Now the magnetic moment is:

\(\mu = g\sqrt{J(J+1)} \)

\(= 1.25\sqrt{8(8+1)}\)

 \(= 1.25\sqrt{72}\)

 \(= 1.25\times 8.4 \)

 \(= 10.6 \)

Conclusion:

Hence, the The effective magnetic moment (in BM) for a lanthanide f10 ion is approximately 10.6