The calculated 'spin-only' magnetic moment $\mathrm{Ti}^{2+}\left(3 \mathrm{~d}^2\right)$ is :

Question

Accepted Answer

C. 1. The spin-only magnetic moment of an ion depends only on the number of **unpaired electrons** it has.
Given that magnetic moment arises from electron spin,
$$
\mu = \sqrt{n(n+2)} \ 	ext{BM}
$$
where $n$ is the number of unpaired electrons.

2. For $\mathrm{Ti}^{2+}$, the electronic configuration is $[\mathrm{Ar}]3d^2 4s^0$, since two electrons are removed from the neutral titanium atom.

3. In the $3d^2$ configuration, by Hund's rule the two electrons occupy two different d-orbitals with parallel spins, so there are 2 unpaired electrons.
$$
n = 2
$$

4. Substituting $n = 2$ into the formula,
$$
\mu = \sqrt{2(2+2)} = \sqrt{8} \approx 2.84 \ 	ext{BM}
$$

5. Hence, the spin-only magnetic moment of $\mathrm{Ti}^{2+}$ is approximately 2.84 BM.
Hence, the answer is **C**.

The calculated 'spin-only' magnetic moment Ti2+(3 d2)\mathrm{Ti}^{2+}\left(3 \mathrm{~d}^2\right)Ti2+(3 d2) is :

The calculated 'spin-only' magnetic moment $\mathrm{Ti}^{2+}\left(3 \mathrm{~d}^2\right)$ is :