To solve this problem, we need to calculate the 'spin only' magnetic moment for each of the given ions.$\newline$The 'spin only' magnetic moment $(\mu)$ is given by the formula:$\newline \mu = \sqrt{n(n+2)} \, \mu_B \newline$where $n$ is the number of unpaired electrons and $\mu_B$ is the Bohr magneton.$\newline$Let's determine the number of unpaired electrons for each ion:$\newline$• Ti$^{3+}:$ Titanium has an atomic number of 22. The electronic configuration is $[$Ar$] \, 3d^1. \newline$ Therefore, Ti$^{3+}$ has 1 unpaired electron.$\newline$• Cr$^{2+}:$ Chromium has an atomic number of 24. The electronic configuration is $[$Ar$] \, 3d^4. \newline$ Therefore, Cr$^{2+}$ has 4 unpaired electrons.$\newline$• Mn$^{2+}:$ Manganese has an atomic number of 25. The electronic configuration is $[$Ar$] \, 3d^5. \newline$ Therefore, Mn$^{2+}$ has 5 unpaired electrons.$\newline$• Fe$^{2+}:$ Iron has an atomic number of 26. The electronic configuration is $[$Ar$] \, 3d^6. \newline$ Therefore, Fe$^{2+}$ has 4 unpaired electrons.$\newline$• Sc$^{3+}:$ Scandium has an atomic number of 21. The electronic configuration is $[$Ar$] \, 3d^0. \newline$ Therefore, Sc$^{3+}$ has 0 unpaired electrons.$\newline$Now, calculate the magnetic moment for ions with unpaired electrons:$\newline$• For Cr$^{2+}$ and Fe$^{2+}$ (both have 4 unpaired electrons):$\newline \mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90 \, \mu_B \newline$• For Ti$^{3+}: \mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \, \mu_B \newline$• For Mn$^{2+}: \mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 \, \mu_B \newline$• For Sc$^{3+}: \mu = \sqrt{0(0+2)} = 0 \, \mu_B \newline$The ions with the same 'spin only' magnetic moment are Cr$^{2+}$ and Fe$^{2+}. \newline$Therefore, the correct option is Option 3: B and D only.