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: Understanding the Question:
The question asks for the spin-only magnetic moment of the titanium ion in its $+2$ oxidation state. Magnetic moments are a result of the orbital and spin motions of electrons. In transition metals, the "spin-only" formula is widely used because the orbital contribution is often "quenched" by the surrounding ligands or crystal field. This calculation relies entirely on the number of unpaired electrons in the $d$-subshell of the ion.
Step 2: Key Formula or Approach:
The Spin-only magnetic moment ($\mu$) formula is: \[ \mu = \sqrt{n(n+2)} \text{ Bohr Magnetons (BM)} \] where $n$ is the number of unpaired electrons. To find $n$, we must write the electronic configuration of the ion. Step 3: Detailed Explanation:

Neutral Atom Configuration: Titanium ($Ti$, Atomic Number $Z = 22$) has the configuration $[Ar] 3d^2 4s^2$.
Ion Configuration: For $Ti^{2+}$, two electrons are removed. Electrons are always removed from the outermost shell ($4s$) first. Thus, $Ti^{2+}$ has the configuration $[Ar] 3d^2$.
Unpaired Electrons (n): In the $3d$ subshell, there are five orbitals. Following Hund's Rule, the two electrons will occupy separate orbitals with parallel spins. Thus, there are $n = 2$ unpaired electrons.
Calculating Magnetic Moment: \[ \mu = \sqrt{2(2+2)} \] \[ \mu = \sqrt{2 \times 4} = \sqrt{8} \]
Final Value: $\sqrt{8}$ is approximately $2.8284$. Looking at the provided options, $2.84$ BM is the closest and correct value.
Step 4: Final Answer:
The calculated spin-only magnetic moment for $Ti^{2+}$ is $2.84 BM$.