Consider the hydrated ions of $T{i}^{2+}, V^{2+}, T{i}^{3+} and S{c}^{3+}$. The correct order of their spin-only magnetic moments is

$\mathrm{\mu }=\sqrt{\mathrm{n}(\mathrm{n}+2)}\mathrm{BM}$

Were, n = no. of unpaired electrons.

and BM = Bohr magneton

The electronic configurations of a given transition metal ion are listed.

${Sc}^{3+}(Z=21)1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{0}4s^0$

${\mathrm{Ti}}^{3+}(\mathrm{Z}=22)1{ \mathrm{s}}^{2}2{ \mathrm{s}}^{2}2{\mathrm{p}}^{6}3{ \mathrm{s}}^{2}3{\mathrm{p}}^{6}3{ \mathrm{d}}^{1}4{ \mathrm{s}}^0$

${\mathrm{Ti}}^{2+}(\mathrm{Z}=22)1{ \mathrm{s}}^{2}2s^{2}2{\mathrm{p}}^{6}3{ \mathrm{s}}^{2}3p^{6}3{ \mathrm{d}}^{2}4S^0$

$V^{2+}(Z=23)1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{3}4s^0$

X = Atomic number.

Because the magnetic moment is proportional to the number of unpaired electrons.

${\mathrm{Sc}}^{3+}\to 3{\mathrm{d}}^0$

$\mathrm{n}=0\Rightarrow \mu =\sqrt{0(0+2)}=0BM$

${\mathrm{Ti}}^{3+}\to 3{\mathrm{d}}^1$

$\mathrm{n}=1\Rightarrow \mathrm{\mu }=\sqrt{1(1+2)}=\sqrt{1\times 3}=\sqrt{3}\mathrm{Bm}$

${\mathrm{Ti}}^{2+}\to 3{\mathrm{d}}^2$

$\mathrm{n}\text{=2⇒μ=}\sqrt{2(2+2)}=\sqrt{2\times 4}=\sqrt{8}\mathrm{BM}$

${\mathrm{V}}^{2+}\to 3{\mathrm{d}}^3$

$\mathrm{n}=3\Rightarrow \mu =\sqrt{3(3+2)}=\sqrt{3\times 5}=\sqrt{15Bm}$

Because they have 0, 1, 2, and 3 unpaired electrons, the correct increasing order of magnetic moment is ${\mathrm{Sc}}^{3+}<{\mathrm{Ti}}^{3+}<{\mathrm{Ti}}^{2+}<{\mathrm{V}}^{2+}$