What is the coordination entity formed when excess of aqueous \(KCN\) is added to an aqueous solution of copper sulphate? Why is it that no precipitate of copper sulphide is obtained when \(H_2S(g)\) is passed through this solution?
\(CuSO_{4(aq)}+4KCN_{(aq)}→K_2[Cu(CN)_4]_{(aq)}+K_2SO_4\)
i.e,\([Cu(H_2O)_4]^{2+}+4CN^{-}→[Cu(CN)_4]^{2-}+4H_2O\)
Thus, the coordination entity formed in the process is \(K_2[Cu(CN)_4]\). \(K_2[Cu(CN)_4]\) is a very stable complex, which does not ionize to give \(Cu^{2+}\) ions when added to water. Hence, \(Cu^{2+}\) ions are not precipitated when \(H_2S_{(g)}\)is passed through the solution.
The correct IUPAC name of \([ \text{Pt}(\text{NH}_3)_2\text{Cl}_2 ]^{2+} \) is:
The Crystal Field Theory (CFT) of coordination compounds is based on the effect of different crystal fields (provided by the ligands taken as point charges) on the degeneracy of d-orbital energies of the central metal atom/ion. The splitting of the d-orbitals provides different electronic arrangements in strong and weak crystal fields. In tetrahedral coordination entity formation, the d-orbital splitting is smaller as compared to the octahedral entity.
What is crystal field splitting energy?
The Crystal Field Theory (CFT) of coordination compounds is based on the effect of different crystal fields (provided by the ligands taken as point charges) on the degeneracy of d-orbital energies of the central metal atom/ion. The splitting of the d-orbitals provides different electronic arrangements in strong and weak crystal fields. In tetrahedral coordination entity formation, the d-orbital splitting is smaller as compared to the octahedral entity.
On the basis of CFT, explain why [Ti(H$_2$O)$_6$]Cl$_3$ complex is coloured? What happens on heating the complex [Ti(H$_2$O)$_6$]Cl$_3$? Give reason.

There are three applications of coordination compounds: