To solve this problem, we need to determine which complex shows zero crystal field stabilization energy (CFSE).$\newline$CFSE depends on the distribution of electrons in the d-orbitals of the metal ion in the complex.$\newline$Let's analyze each option:$\newline$• Option 1: $[Mn(H_2O)_6]^{3+} \newline$ - $Mn^{3+}$ has an electronic configuration of $3d^4. \newline$ - In an octahedral field, the electrons will be distributed as $t_{2g}^3 e_g^1. \newline$ - This results in a non-zero CFSE.$\newline$• Option 2: $[Fe(H_2O)_6]^{3+} \newline$ - $Fe^{3+}$ has an electronic configuration of $3d^5. \newline$ - In an octahedral field, the electrons will be distributed as $t_{2g}^3 e_g^2. \newline$ - This results in zero CFSE because the stabilization from $t_{2g}$ is balanced by the destabilization from $e_g. \newline$• Option 3: $[Co(H_2O)_6]^{2+} \newline$ - $Co^{2+}$ has an electronic configuration of $3d^7. \newline$ - In an octahedral field, the electrons will be distributed as $t_{2g}^5 e_g^2. \newline$ - This results in a non-zero CFSE.$\newline$• Option 4: $[Co(H_2O)_6]^{3+} \newline$ - $Co^{3+}$ has an electronic configuration of $3d^6. \newline$ - In an octahedral field, the electrons will be distributed as $t_{2g}^4 e_g^2. \newline$ - This results in a non-zero CFSE.$\newline$Therefore, the complex that shows zero CFSE is $[Fe(H_2O)_6]^{3+}. \newline$This corresponds to Option 2.