Question

 How would you account for the following? 

(a) [Ti (H₂O)₆]³⁺ is coloured while [Sc (H₂O)₆]³⁺ is colourless. 

(b) [Fe (CN)₆]^3- is weakly paramagnetic while Fe (CN)₆]^4- is diamagnetic. 

(c) [Ni (CO)₄ ] possess tetrahedral geometry while is [Ni (CN)₄]^2- is square planner.

Answer 1

(a) In [Ti (H₂O)₆ ]³⁺, metal ion has d¹ configuration. Because of this unpaired electron, d-d transition is possible and hence, it appears coloured.In [Sc (H₂O)₆] ³⁺, no unpaired electron is present in the d-subshell of metal ion hence, d-d transition is not possible, therefore, it would be colourless. 

 (b) In [Fe (CN)₆]^3- ,Fe is in +3 oxidation state and has 5 unpaired electrons in its d-orbitals. This complex is formed due to d² Sp³ hybridisation, after the pairing of d- electrons (because of strong ligand field of CNions).These six hybridised orbitals receive six electron pairs from six molecules of CNbut one electron in d-subshell remains unpaired , hence it is weakly paramagnetic. But in [Fe (CN)₆]^4-, the oxidation state of iron is +2 and has d⁶ configuration. These six electrons get paired under the influence of ligand molecules and no electron remains unpaired in d-orbitals, therefore, it is diamagnetic. 

(C) In [Ni (CO)₄], Ni is in zero oxidation state and is Sp³ oxidised. As a result, it possess tetrahedral geometry. While in [Ni (CN)₄]^2- , Ni is in +2 oxidation state and is dsp² hybridised. As a result, it is square planner