Chemistry - Coordination Chemistry Question with Solution | TestHub

Crystal field splitting energy $(\Delta )\propto$ charge or oxidation state of central metal atom.

Crystal field splitting energy $(\Delta )\propto$ Field strength of ligand (and chelation)
Crystal field stabilisation energy (CFSE) $=[-0.4{\mathrm{t}}_{{\mathrm{g}}_g}$ +0.6 eg] ${\Delta }_{\circ }$ (for octahedral)
For,
$\begin{array}{rl}& {\left[\mathrm{Co}(\mathrm{en}{)}_3\right]}^{3+}:{\mathrm{Co}}^{3+}:{\mathrm{t}}_{2\mathrm{g}}^6{\mathrm{eg}}^0;\mathrm{CFSE}=-2.4{\left({\Delta }_0\right)}_1\\ & {\left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_6\right]}^{3+}:{\mathrm{Co}}^{3+}:{\mathrm{t}}_{2\mathrm{g}}^6{\mathrm{eg}}^0;\mathrm{CFSE}=-2.4{\left({\Delta }_0\right)}_2\\ & {\left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_6\right]}^{2+}:{\mathrm{Co}}^{2+}:{\mathrm{t}}_{2g}^5{\mathrm{eg}}^2;\mathrm{CFSE}=-0.8{\left({\Delta }_0\right)}_3\\ & {\left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_4\right]}^{2+}:{\mathrm{Co}}^{2+}:{\mathrm{e}}^4{\mathrm{t}}_2^3;\mathrm{CFSE}=-1.2{\Delta }_{\mathrm{t}}\\ & \left(\mathrm{as}:\left({\Delta }_{\mathrm{t}}\right)=\frac{4}{9}{\left({\Delta }_0\right)}_3\right)\\ & \because {\Delta }_{\mathrm{t}}<{\left({\Delta }_0\right)}_3<{\left({\Delta }_0\right)}_2<{\left({\Delta }_0\right)}_1\end{array}$