The 'spin-only' magnetic moment ($\mu$) is determined by the number of unpaired electrons ($n$) using the formula $\mu = \sqrt{n(n+2)} \text{ BM}$. Let us find the number of unpaired electrons for each given ion:

• $Ti^{3+} (Z=22)$: Electronic configuration is $[Ar] 3d^1$. It has 1 unpaired electron ($n=1$).
• $Cr^{2+} (Z=24)$: Electronic configuration is $[Ar] 3d^4$. It has 4 unpaired electrons ($n=4$).
• $Mn^{2+} (Z=25)$: Electronic configuration is $[Ar] 3d^5$. It has 5 unpaired electrons ($n=5$).
• $Fe^{2+} (Z=26)$: Electronic configuration is $[Ar] 3d^6$. It has 4 unpaired electrons ($n=4$).
• $Sc^{3+} (Z=21)$: Electronic configuration is $[Ar] 3d^0$. It has 0 unpaired electrons ($n=0$).

Since both $Cr^{2+}$ and $Fe^{2+}$ have 4 unpaired electrons, they have the same 'spin-only' magnetic moment of $\sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \text{ BM}$ . Thus, B and D is the correct pair.