To solve this problem, we need to calculate the spin-only magnetic moment of the $Cr^{2+}$ ion.$\newline$The spin-only magnetic moment $(\mu_{spin})$ is given by the formula:$\newline \mu_{spin} = \sqrt{n(n+2)} \,$BM$\newline$where $n$ is the number of unpaired electrons.$\newline$Let's determine the electronic configuration of $Cr^{2+}: \newline$• The atomic number of chromium (Cr) is 24.$\newline$• The electronic configuration of a neutral Cr atom is $[Ar] \, 3d^5 \, 4s^1. \newline$• For $Cr^{2+},$ two electrons are removed, typically from the 4s and 3d orbitals.$\newline$• The electronic configuration of $Cr^{2+}$ is $[Ar] \, 3d^4. \newline$Now, determine the number of unpaired electrons:$\newline$• In the $3d^4$ configuration, there are 4 electrons in the 3d subshell.$\newline$• These electrons are distributed as: $\uparrow \uparrow \uparrow \uparrow$ (4 unpaired electrons).$\newline$Substitute $n = 4$ into the formula for $\mu_{spin}: \newline \mu_{spin} = \sqrt{4(4+2)} = \sqrt{24} \,$BM$\newline \mu_{spin} \approx 4.90 \,$BM$\newline$Therefore, the calculated spin-only magnetic moment of $Cr^{2+}$ ion is $4.90 \,$BM.$\newline$This corresponds to Option 1.