To solve this problem, we need to determine the hybridization and geometry of the complex $[$Mn(CN)$_6]^{3-}. \newline$Step-by-step analysis:$\newline$1. Determine the oxidation state of Mn:$\newline$• The complex ion is $[$Mn(CN)$_6]^{3-}. \newline$• Let the oxidation state of Mn be $x. \newline$• Each CN ligand has a charge of $-1. \newline$• The overall charge of the complex is $-3. \newline$• Therefore, the equation is: $x + 6(-1) = -3. \newline x - 6 = -3 \newline x = +3 \newline$• Mn is in the +3 oxidation state.$\newline$2. Determine the electronic configuration of Mn in the +3 state:$\newline$• The atomic number of Mn is 25.$\newline$• The electronic configuration of Mn is $[$Ar$] \, 3d^5 \, 4s^2. \newline$• In the +3 oxidation state, Mn loses 3 electrons: 2 from 4s and 1 from 3d.$\newline$• The configuration becomes $[$Ar$] \, 3d^4. \newline$3. Determine the hybridization:$\newline$• CN is a strong field ligand and causes pairing of electrons in the 3d orbitals.$\newline$• After pairing, the configuration is $t_{2g}^4 \, e_g^0. \newline$• The hybridization involves the use of 2 paired 3d orbitals, 1 4s orbital, and 3 4p orbitals.$\newline$• This results in $d^2sp^3$ hybridization.$\newline$4. Determine the geometry:$\newline$• The $d^2sp^3$ hybridization corresponds to an octahedral geometry.$\newline$Conclusion:$\newline$The correct statement is: It is $d^2sp^3$ hybridized and octahedral.$\newline$This corresponds to Option 2.