Question:medium

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

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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.
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Solution and Explanation

Step 1: Understanding the Concept:
Magnetic moment (\(\mu\)) of transition metal ions depends on the number of unpaired electrons (\(n\)).
Step 2: Key Formula or Approach:
Spin-only formula: \(\mu = \sqrt{n(n+2)} \text{ BM}\)
Step 3: Detailed Explanation:
1. Write configuration of \(Ti\) (\(Z=22\)): \([Ar] 3d^2 4s^2\).
2. Write configuration of \(Ti^{3+}\): Remove 2 electrons from \(4s\) and 1 from \(3d\).
Configuration: \([Ar] 3d^1\).
3. Number of unpaired electrons, \(n = 1\).
4. Calculate \(\mu\):
\[ \mu = \sqrt{1(1+2)} = \sqrt{3} \]
\[ \mu \approx 1.732 \text{ BM} \]
Step 4: Final Answer:
The magnetic moment of \(Ti^{3+}\) is \(1.73 \text{ BM}\).
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