(i) Draw the geometrical isomers of [Co(en)2Cl2]+
(ii) On the basis of CFT, write electronic configuration of d4 ion ∆° >P
(iii) [NiCl4]2- is paramagnetic whereas [Ni(C0)4]is a diamagnetic, though both are tetrahedral, why?(atomic no of Ni= 28)
(i)
When same groups are adjacent to each other it is a cis conformation. And when they are opposite to each other it is a trans conformation.
(ii) in CFT if ∆° .> P, then it is a low spin complex, due to presence of a strong ligand, so the fourth electron will come back and get paired in t2g orbital. So the electronic configuration will be t2g4 eg0.
(iii) [NiCl4]2- is paramagnetic as Cl is weak field ligand, and it can’t undergo pairing, due to which, it has unpaired electrons which are responsible for its paramagnetic behavior. On the other hand, CO is a strong field ligand, which can undergo pairing and in that there will be absence of unpaired electrons resulting in diamagnetic behavior.
3d orbital on Ni in [NiCl4]
3d orbital of Ni in [Ni(CO)4]
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