Chemistry - Coordination Chemistry Question with Solution | TestHub

d-orbital is degenerate which split into two levels ${\mathrm{e}}_{\mathrm{g}}$ and ${\mathrm{t}}_2\mathrm{g}$ 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, $\mathrm{\Delta }0.$

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

$$\begin{gathered} {\mathrm{t}}_{2\mathrm{g}} {\mathrm{e}}^{–}=2 \\ {\mathrm{e}}_{\mathrm{g} }{\mathrm{e}}^{–}=0 \end{gathered}$$

${\mathrm{\Delta }}_0=2\left(-.4{\mathrm{\Delta }}_0\right)=-.8{\mathrm{\Delta }}_0$

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

$$\begin{gathered} {\mathrm{t}}_{2\mathrm{g}} {\mathrm{e}}^{–}=3 \\ {\mathrm{e}}_{\mathrm{g} }{\mathrm{e}}^{–}=0 \end{gathered}$$

${\mathrm{\Delta }}_0=3\left(-.4{\mathrm{\Delta }}_0\right)$

$= -1.2{\mathrm{\Delta }}_0$

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

$$\begin{gathered} {\mathrm{t}}_{2\mathrm{g}} {\mathrm{e}}^{–}=3 \\ {\mathrm{e}}_{\mathrm{g} }{\mathrm{e}}^{–}=1 \end{gathered}$$

${\mathrm{\Delta }}_0=3\left(-.4{\mathrm{\Delta }}_0\right)+1\left(.6{\mathrm{\Delta }}_0\right)$

$=-1.2{\mathrm{\Delta }}_0+0.6{\mathrm{\Delta }}_0$

$= -0.6{\mathrm{\Delta }}_0$

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

$$\begin{gathered} {\mathrm{t}}_{2\mathrm{g}} {\mathrm{e}}^{–}=3 \\ {\mathrm{e}}_{\mathrm{g} }{\mathrm{e}}^{–}=2 \end{gathered}$$

${\mathrm{\Delta }}_0=3\left(-.4{\mathrm{\Delta }}_0\right)+2\left(.6{\mathrm{\Delta }}_0\right)$

$=-1.2{\mathrm{\Delta }}_0+1.2{\mathrm{\Delta }}_0=0$