Question

show cft diagram

5) (20 Pts) a) Ni(II) tends to form a variety of complexes w square-planar geometries, while Pd(IID) and PtII) form predominately square-planar complexes, explain using crystal field theory. b) Another striking difference between these three metals is that the NiIV) oxidation state is very rare, while Pd(IV) is more common and Pt(IV) is very common, explain. Show the CFT diagrams for full credit. ith octahedral, tetrahedral and

Answer 1

5) a) Nickel uses it's 3d orbital for complex formation , so not that much electron density can be stabilized by those d orbitals and due lower in energy of d orbital , the crystal field spilitting is not that much. For Pt(+2) or Pd(+2) , the size of the d orbitals is bigger than that of Nickel , which can stabilize much more electron density than Nickel.They use 4d orbitals for complexation.Moving from up to down across the periodic table , it is seen that the crystal field splitting gets higher up and so , due to Jahn Teller distotion , only square planar complexes are formed for Paladium and Plainum in comparison with Nickel though they all have d8 configufration.

b) We know , lower the size of atom and higher the charge , lesser the stability. Hence Nickel has lower size than that of Paladium or Platinum. Being bigger in size , Paladium and Platinum can afford higher charge.

_ dus az den du square-plorer cugtal iead sputing spliting