To solve this problem, we need to analyze the electronic configuration of the complex $[$Co(CN)$_6]^{3-}. \newline$1. Determine the oxidation state of cobalt in $[$Co(CN)$_6]^{3-}: \newline$• Let the oxidation state of Co be $x. \newline$• Each CN ligand carries a charge of $-1. \newline$• The overall charge of the complex is $-3. \newline$• Therefore, the equation is: $x + 6(-1) = -3. \newline$• Solving for $x: \newline x - 6 = -3 \newline x = +3 \newline$2. Electronic configuration of Co$^{3+}: \newline$• The atomic number of cobalt (Co) is 27.$\newline$• The electronic configuration of neutral Co is $[$Ar$] \, 3d^7 \, 4s^2. \newline$• For Co$^{3+}, remove three electrons: two from 4s and one from 3d. \newline$• The configuration becomes $[$Ar$] \, 3d^6. \newline$3. Determine the spin state:$\newline$• CN$^-$ is a strong field ligand, which causes pairing of electrons.$\newline$• In a low-spin configuration, the 3d electrons will pair up.$\newline$• The 3d^6 configuration in a low-spin state results in all electrons being paired.$\newline$4. Analyze the options:$\newline$• Option 1: Four unpaired electrons and low-spin configuration. This is incorrect.$\newline$• Option 2: Four unpaired electrons and high-spin configuration. This is incorrect.$\newline$• Option 3: No unpaired electrons and high-spin configuration. This is incorrect.$\newline$• Option 4: No unpaired electrons and low-spin configuration. This is correct.$\newline$Therefore, the correct option is Option 4.