Question

The calculated magnetic moment of [Ce(NO)5]2- is _____ BM (rounded off to two decimal places).
(Given: atomic number of Ce is 58) 
 

Hint:

For transition metals, the spin-only formula \(\mu = \sqrt{n(n+2)}\) BM is usually sufficient.
For lanthanides (4f block), orbital contribution is significant, so one must use the L-S coupling approach.
Ce\(^ {3+}\) (4f\(^1\)) has \(\mu_{\text{eff}} \approx 2.54\) BM, a classic textbook value.

Answer 1

Correct Answer: 2.51

Step 1: Determine oxidation state of Ce.
The complex is \([Ce(NO_3)_5]^{2-}\). Each nitrate group (\(\ce{NO3^-}\)) has a charge of \(-1\). With 5 nitrates, total = \(-5\). The overall charge is \(-2\). Let Ce oxidation state = \(x\).
\[ x + (-5) = -2 \quad \Rightarrow \quad x = +3 \] So Ce is in +3 oxidation state. Step 2: Write electronic configuration.
Cerium (Z = 58): ground state = [Xe] \(4f^1 5d^1 6s^2\).
For Ce\(^ {3+}\): remove 3 electrons (first from 6s and 5d). Configuration = [Xe] \(4f^1\).
Thus, Ce\(^ {3+}\) has 1 unpaired electron. Step 3: Apply spin-only formula for magnetic moment.
The spin-only magnetic moment is: \[ \mu = \sqrt{n(n+2)} \ \text{BM} \] where \(n =\) number of unpaired electrons.
Here, \(n = 1\).
\[ \mu = \sqrt{1(1+2)} = \sqrt{3} = 1.732 \] However, for \emph{lanthanides}, orbital contribution is significant. The experimental effective moment is calculated using the \emph{L-S coupling formula}: \[ \mu_{\text{eff}} = g \sqrt{J(J+1)} \ \text{BM} \] with \(g = 1 + \dfrac{J(J+1) + S(S+1) - L(L+1)}{2J(J+1)} Step 4: Term symbol for Ce\(^ {3+}\) (4f\(^1\)).
- For \(4f^1\): \(L = 3\) (since f-orbital).
- \(S = \tfrac{1}{2}\) (1 unpaired electron).
- \(J = L - S = \tfrac{5}{2}\) (because for less than half-filled shells, \(J = |L - S|\)).
Step 5: Calculate Landé g-factor.
\[ g = 1 + \frac{J(J+1) + S(S+1) - L(L+1)}{2J(J+1)} \] \[ = 1 + \frac{\tfrac{5}{2}\left(\tfrac{7}{2}\right) + \tfrac{1}{2}\left(\tfrac{3}{2}\right) - 3(4)}{2 \times \tfrac{5}{2}\times \tfrac{7}{2}} \] \[ = 1 + \frac{\tfrac{35}{4} + \tfrac{3}{4} - 12}{2 \times \tfrac{35}{4}} = 1 + \frac{\tfrac{38}{4} - 12}{\tfrac{35}{2}} = 1 + \frac{9.5 - 12}{17.5} = 1 - \tfrac{2.5}{17.5} = 1 - 0.1428 = 0.857 \] Step 6: Calculate magnetic moment.
\[ \mu_{\text{eff}} = g \sqrt{J(J+1)} = 0.857 \times \sqrt{\tfrac{5}{2}\left(\tfrac{7}{2}\right)} = 0.857 \times \sqrt{\tfrac{35}{4}} = 0.857 \times \sqrt{8.75} \] \[ = 0.857 \times 2.958 = 2.54 \ \text{BM} \] Thus, the calculated magnetic moment = \(\boxed{2.54 \ \text{BM}}\).