Write the state of hybridization, the shape and the magnetic behaviour of the following complex entitles :

(i) [Cr(NH₃)₄Cl₂]Cl

(ii) [Co(en)₃]Cl₃

(iii) K₂[Ni(CN)₄]

(i) Cr in [Cr(NH₃)₄Cl₂]Cl has +3 oxidation state which is d³. As –NH₃ is a weak field ligand, the electrons of Cr doesnt pair up inside the orbital. So, the first 3 orbitals of the ‘d’ remains intact with the electrons of Cr. The remaining 2 gets filled by the incoming electrons of –NH₃. 1 s orbital is also filled by –NH₃. 1 p orbitals is also filled by –NH₃ and the remaining 2 is filled by Cl (As there are 4 NH₃ atoms and 2 Cl atoms). So, we get the hybridisation of d²sp³. The geometry is octahedral and as three unpaired electrons (of Cr³⁺) are present, it is paramagnetic.

(ii) Co³⁺ in [Co(en)₃]Cl₃ is in 3d⁶ oxidation state. As -en is a strong field ligand, the electrons pair up inside the orbital. Due to which, the first 3 orbitals of the ‘d’ gets occupied and the remaining 2 gets filled by the incoming electrons of –en. 1 s orbital is also filled by –en and 3 p orbitals are filled by –Cl. So, we get the hybridisation of d²sp³. The geometry is octahedral and as no unpaired electrons are present, it is diamagnetic.

(iii) Ni has oxidation state of +2 in K₂[Ni(CN)₄] and its configuration is d⁸.As -CN is a strong field ligand, the electrons pair up inside the orbital. Due to which, the first 4 orbitals of the ‘d’ gets occupied and the remaining 1 orbital gets filled by the incoming electrons of –CN. 1 ‘s’ orbital is also filled by –CN and 2 ‘p’ orbitals are filled by –CN as there are 4 CN atoms in the complex. So, we get the hybridisation of dsp². The geometry is tetrahedral and as no unpaired electrons are present, it is diamagnetic.