Match List-I with List-II. Choose the correct answer from the...

d-orbital is degenerate which split into two levels ${e}_{g}$ and ${t}_{2}g$ in the presence of ligands. This splitting is due to the presence of ligands. This is called the crystal-field splitting and the energy difference between the two levels is called the crystal-field splitting energy, $\Delta0 .$

(A) ${Ti}^{2 +}: 4{s}^{0}3{d}^{2}$

${t}_{2g}{e}^{–} = 2{e}_{g}{e}^{–} = 0$

${\Delta}_{0} = 2\left(- . 4{\Delta}_{0}\right) = - . 8{\Delta}_{0}$

(B) ${V}^{2 +}: 4{s}^{0}3{d}^{3}$

${t}_{2g}{e}^{–} = 3{e}_{g}{e}^{–} = 0$

${\Delta}_{0} = 3\left(- . 4{\Delta}_{0}\right)$

$= - 1 . 2{\Delta}_{0}$

(C) ${Mn}^{3 +} = 4{s}^{0}3{d}^{4}$

${t}_{2g}{e}^{–} = 3{e}_{g}{e}^{–} = 1$

${\Delta}_{0} = 3\left(- . 4{\Delta}_{0}\right) + 1\left(. 6{\Delta}_{0}\right)$

$= - 1 . 2{\Delta}_{0} + 0 . 6{\Delta}_{0}$

$= - 0 . 6{\Delta}_{0}$

(D) ${Fe}^{3 +} = 4{s}^{0}3{d}^{5}$

${t}_{2g}{e}^{–} = 3{e}_{g}{e}^{–} = 2$

${\Delta}_{0} = 3\left(- . 4{\Delta}_{0}\right) + 2\left(. 6{\Delta}_{0}\right)$

$= - 1 . 2{\Delta}_{0} + 1 . 2{\Delta}_{0} = 0$