Question

Calculate magnetic moment of \(Ti^{3+}\) by using spin only formula. (Z of Ti = 22)

Hint:

A quick shortcut for BM values:
\(n=1 \rightarrow \sim 1.7-1.8\)
\(n=2 \rightarrow \sim 2.7-2.8\)
\(n=3 \rightarrow \sim 3.8-3.9\)
Basically, the first digit is the number of unpaired electrons.

Answer 1

Step 1: Understanding the Concept:
The magnetic moment of transition metal ions is calculated based on the number of unpaired electrons in their \(d\)-orbitals.
Step 2: Key Formula or Approach:
Spin-only formula: \(\mu = \sqrt{n(n+2)}\) Bohr Magnetons (BM), where \(n\) is the number of unpaired electrons.
Step 3: Detailed Explanation:
Atomic number of Titanium (\(Ti\)) is 22.
Electronic configuration of \(Ti\): \( [Ar] \, 3d^2 \, 4s^2 \).
In \(Ti^{3+}\) ion, three electrons are removed (two from \(4s\) and one from \(3d\)).
Configuration of \(Ti^{3+}\): \( [Ar] \, 3d^1 \).
Number of unpaired electrons, \(n = 1\).
\[ \mu = \sqrt{1(1+2)} = \sqrt{3} \]
\[ \mu \approx 1.732 \text{ BM} \]
Step 4: Final Answer:
The magnetic moment of \(Ti^{3+}\) is 1.73 BM.