To solve this problem, we need to determine which ion has a magnetic moment of 2.83 BM (Bohr Magneton).$\newline$The magnetic moment $(\mu)$ is given by the formula:$\newline \mu = \sqrt{n(n+2)} \,$BM$\newline$where $n$ is the number of unpaired electrons.$\newline$Given: $\mu = 2.83 \,$BM$\newline$Let's calculate $n$ using the formula:$\newline 2.83 = \sqrt{n(n+2)} \newline$Squaring both sides, we get:$\newline (2.83)^2 = n(n+2) \newline 8.0089 = n(n+2) \newline$Solving the quadratic equation $n^2 + 2n - 8.0089 = 0,$ we find $n \approx 2. \newline$Therefore, the ion should have 2 unpaired electrons.$\newline$Let's analyze each option:$\newline$• $Ti^{3+}:$ Atomic number of Ti is 22. Electronic configuration: $[Ar] \, 3d^1 \, 4s^0. \newline$ Ti$^{3+}$ loses 3 electrons: $[Ar] \, 3d^1.$ Number of unpaired electrons $= 1. \newline$• $Ni^{2+}:$ Atomic number of Ni is 28. Electronic configuration: $[Ar] \, 3d^8 \, 4s^2. \newline$ Ni$^{2+}$ loses 2 electrons: $[Ar] \, 3d^8.$ Number of unpaired electrons $= 2. \newline$• $Cr^{3+}:$ Atomic number of Cr is 24. Electronic configuration: $[Ar] \, 3d^5 \, 4s^1. \newline$ Cr$^{3+}$ loses 3 electrons: $[Ar] \, 3d^3.$ Number of unpaired electrons $= 3. \newline$• $Mn^{2+}:$ Atomic number of Mn is 25. Electronic configuration: $[Ar] \, 3d^5 \, 4s^2. \newline$ Mn$^{2+}$ loses 2 electrons: $[Ar] \, 3d^5.$ Number of unpaired electrons $= 5. \newline$The ion with 2 unpaired electrons is $Ni^{2+}. \newline$Therefore, the correct option is Option 2: $Ni^{2+}.$