Question

Write the oxidation states, distribution of d-orbitals and coordination number of central metal of following complexes:

Hint:

For coordination complexes, the coordination number is determined by the number of donor atoms (ligands) attached to the central metal ion. A bidentate ligand, like oxalate (C₂O₄²⁻), contributes 2 donor atoms, whereas a monodentate ligand, like water (H₂O), contributes only 1.

Answer 1

(x) \(K_3[Co(C_2O_4)_3]\):
- Oxidation state of Co: +3 (since each oxalate \(C_2O_4^{2-}\) contributes -2 charge, and there are three oxalates, the net charge is -6. The overall charge of the complex is -3, so Co must be +3 to balance this).
- Distribution of d-orbitals: Co\(^{3+}\) (d⁶ configuration) will have 6 electrons in its d-orbitals.
- Coordination number of Co: 6 (since there are three oxalate ions, each of which is a bidentate ligand, contributing 2 donor atoms per ligand, for a total coordination of 6).
(y) \([Mn(H_2O)_6]SO_4\):
- Oxidation state of Mn: +2 (the complex ion \([Mn(H_2O)_6]^{2+}\) is neutralized by the sulfate ion \(SO_4^{2-}\), indicating Mn is in the +2 oxidation state).
- Distribution of d-orbitals: Mn\(^{2+}\) (d⁵ configuration) will have 5 electrons in its d-orbitals.
- Coordination number of Mn: 6 (since there are six water molecules, each contributing one donor atom, making the coordination number 6).