To solve this problem, we need to determine which ion corresponds to a magnetic moment of 2.84 BM (Bohr Magnetons).$\newline$The magnetic moment $(\mu)$ is given by the formula:$\newline \mu = \sqrt{n(n+2)}$ BM, where $n$ is the number of unpaired electrons.$\newline$Given: $\mu = 2.84$ BM$\newline$We can equate and solve for $n: \newline 2.84 = \sqrt{n(n+2)} \newline$Squaring both sides:$\newline (2.84)^2 = n(n+2) \newline 8.0656 = n(n+2) \newline$Testing integer values for $n: \newline$• If $n = 2,$ then $n(n+2) = 2(2+2) = 8$ (close to 8.0656)$\newline$Thus, $n = 2$ unpaired electrons.$\newline$Now, let's determine which ion has 2 unpaired electrons:$\newline$1. $Ti^{3+}:$ Electronic configuration: $[Ar] \, 3d^1 \newline$(1 unpaired electron)$\newline$2. $Cr^{2+}:$ Electronic configuration: $[Ar] \, 3d^4 \newline$(4 unpaired electrons)$\newline$3. $Co^{2+}:$ Electronic configuration: $[Ar] \, 3d^7 \newline$(3 unpaired electrons)$\newline$4. $Ni^{2+}:$ Electronic configuration: $[Ar] \, 3d^8 \newline$(2 unpaired electrons)$\newline$Therefore, the ion with 2 unpaired electrons is $Ni^{2+}. \newline$This corresponds to Option 4.