Question

The calculated spin-only magnetic moment of \(Ti^{2+}\) is:

• A. \(3.87\text{ BM}\)
• B. \(5.92\text{ BM}\)
• C. \(4.90\text{ BM}\)
• D. \(2.82\text{ BM}\)

Hint:

Spin-only magnetic moment is \(\mu=\sqrt{n(n+2)}\), where \(n\) is the number of unpaired electrons.

Answer 1

The Correct Option is D

Step 1: Write atomic number and electronic configuration of titanium.
Titanium has atomic number: \[ Z=22 \] Its electronic configuration is: \[ Ti=[Ar]\,3d^2\,4s^2 \]

Step 2: Write configuration of \(Ti^{2+}\).
When titanium forms \(Ti^{2+}\), two electrons are removed. Electrons are removed first from the \(4s\) orbital. So: \[ Ti^{2+}=[Ar]\,3d^2 \]

Step 3: Find number of unpaired electrons.
The \(3d^2\) configuration has: \[ 2 \] unpaired electrons. Thus: \[ n=2 \]

Step 4: Apply spin-only magnetic moment formula.
Spin-only magnetic moment is: \[ \mu=\sqrt{n(n+2)} \] Substitute: \[ n=2 \] \[ \mu=\sqrt{2(2+2)} \] \[ \mu=\sqrt{8} \] \[ \mu=2.82\text{ BM} \] Therefore, the magnetic moment of \(Ti^{2+}\) is: \[ 2.82\text{ BM}. \]