Coordination Compounds Part 4

Concept:Isoelectronic species have the same total number of electrons. The moles of AgCl formed equal the number of ionisable chloride ions in the complex.Explanation:First, calculate the total electrons for each ion:From the table, $\mathrm{Cr}^{2+}$ and $\mathrm{Mn}^{3+}$ both have 22 electrons. So the number of isoelectronic species is $n=2$.Now consider the complex $\mathrm{CoCl}_3(\mathrm{en})_2\mathrm{NH}_3$. Since $n=2$ moles of AgCl are formed with excess $\mathrm{AgNO}_3$, it means 2 chloride ions are outside the coordination sphere (ionisable). Hence the complex is $[\mathrm{Co}(\mathrm{en})_2\mathrm{NH}_3\mathrm{Cl}]\mathrm{Cl}_2$.Inside the coordination sphere, $\mathrm{en}$ and $\mathrm{NH}_3$ are neutral ligands, and one coordinated $\mathrm{Cl}^-$ contributes $-1$. The complex ion has a charge of $+2$ (due to two external $\mathrm{Cl}^-$). Let oxidation state of Co be $x$. Then $x - 1 = +2$ gives $x = +3$. So cobalt is $\mathrm{Co}^{3+}$.Electronic configuration of $\mathrm{Co}^{3+}$ is $3d^6$. In an octahedral field, $d$ orbitals split into $t_{2g}$ and $e_g$. With strong field ligands ($\mathrm{en}$, $\mathrm{NH}_3$), electrons fill $t_{2g}$ first: $t_{2g}^6$, $e_g^0$. Therefore, total electrons in $t_{2g}$ orbitals is $6$.Answer:6