Consider the hydrated ions of Ti$$^{2+}$$, V$$^{2+}$$, Ti$$^{3+}$$ and Sc$$^{3+}$$. The correct order of their spin-only magnetic moments is:

For any transition‐metal ion the spin-only magnetic moment is calculated by the formula

$$\mu_{\text{spin}}=\sqrt{n(n+2)}\ \text{Bohr magnetons (BM)}$$

where $$n$$ is the number of unpaired (parallel-spin) electrons present in the ion.

Because the question speaks of “hydrated ions”, we assume the ligand is water, a weak-field ligand; thus the complexes are high-spin, so the number of unpaired electrons is exactly the number expected from the gas-phase $$3d$$ configuration of the ion.

We therefore begin by writing the ground-state electron configurations of the neutral atoms and then remove electrons equal to the charge of each ion, always removing the $$4s$$ electrons before the $$3d$$ electrons.

Scandium: $$\text{Sc}\;(Z=21):\;[\,\text{Ar}\,]\,3d^{1}4s^{2}$$

Titanium: $$\text{Ti}\;(Z=22):\;[\,\text{Ar}\,]\,3d^{2}4s^{2}$$

Vanadium: $$\text{V}\;(Z=23):\;[\,\text{Ar}\,]\,3d^{3}4s^{2}$$

Now we create each required ion.

1. Sc$$^{3+}$$

We must remove three electrons: two from $$4s$$ and one from $$3d$$.

$$[\,\text{Ar}\,]\,3d^{1}4s^{2}\;-\;2(4s)\;-\;1(3d)= [\,\text{Ar}\,]\,3d^{0}$$

Thus $$n=0$$ unpaired electrons.

$$\mu_{\text{spin}}=\sqrt{0(0+2)}=0\ \text{BM}$$

2. Ti$$^{3+}$$

Remove three electrons from $$[\,\text{Ar}\,]\,3d^{2}4s^{2}$$: two from $$4s$$, one from $$3d$$.

$$[\,\text{Ar}\,]\,3d^{2}4s^{2}\;\longrightarrow\;[\,\text{Ar}\,]\,3d^{1}$$

So $$n=1$$ unpaired electron.

$$\mu_{\text{spin}}=\sqrt{1(1+2)}=\sqrt{3}\ \text{BM}$$

3. Ti$$^{2+}$$

Remove two electrons, both from $$4s$$.

$$[\,\text{Ar}\,]\,3d^{2}4s^{2}\;\longrightarrow\;[\,\text{Ar}\,]\,3d^{2}$$

Here $$n=2$$ unpaired electrons.

$$\mu_{\text{spin}}=\sqrt{2(2+2)}=\sqrt{8}\ \text{BM}$$

4. V$$^{2+}$$

Remove two electrons from $$[\,\text{Ar}\,]\,3d^{3}4s^{2}$$, again both from $$4s$$.

$$[\,\text{Ar}\,]\,3d^{3}4s^{2}\;\longrightarrow\;[\,\text{Ar}\,]\,3d^{3}$$

So $$n=3$$ unpaired electrons.

$$\mu_{\text{spin}}=\sqrt{3(3+2)}=\sqrt{15}\ \text{BM}$$

Comparing the four values, we clearly have

$$0\;{\text{BM}}\lt\sqrt{3}\;{\text{BM}}\lt\sqrt{8}\;{\text{BM}}\lt\sqrt{15}\;{\text{BM}}$$

That is, in terms of the ions themselves,

$$\text{Sc}^{3+}\lt\text{Ti}^{3+}\lt\text{Ti}^{2+}\lt\text{V}^{2+}$$

This sequence exactly matches Option C.

Hence, the correct answer is Option C.