The sums of first and second ionization enthalpies and those of third and fourth ionization enthalpies of nickel and platinum are given below: Metal E 1 + E 2 E 3 + E 4 Ni 2 . 49 8 . 80 Pt 2 . 66 6 . 70 Taking these values into account write the name of the metal ( Ni or Pt ) which can form compounds in + 4 oxidation state more easily and why?

The +4 oxidation state is more stable for platinum. It is a consequence of the lower ionization energies as significant from the differences seen by summation of all the values from table. Pt ( IV ) Is more easily attained while more energy would be required for obtaining Ni ( IV ) Ion. Hence, Pt ( IV ) Compounds are more stable than Ni ( IV ) Compounds. For example, K 2 PtCl 6 ( Pt in IV oxidation state ) Is a well-known compound while corresponding nickel salt is not known. Note-Another reason that supports the fact is about the difference in atomic sizes of the two element ions.Platinum has large size with vacant orbital which can form complex of coordination number. 6 , while the size of nickel is small, which cannot hold 6 Ligand at a time therefore does not show K 2 NiCl 6

Write down the electronic configuration of Co 2 +

Co 2 + is formed by the loss of two electrons from the 4 s orbital of the cobalt atom. It has now a total of 25 electrons. Therefore, the electronic configuration can be given as 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 3 d 7 or Ar 3 d 7 .

Name the members of the lanthanoid series which exhibit + 4 oxidation state and those which exhibit + 2 oxidation states. Try to correlate this type of behaviour with the electronic configuration of these elements.

The lanthanoids that exhibit + 4 oxidation state are cerium, terbium, promethium and neodymium. Cerium and terbium shows + 4 oxidation. This can be explained in the terms of extra stability of half filled and completely filled 4 f sub-shell respectively in their + 4 oxidation state. Promethium and neodymium shows + 4 oxidation stable because both these tetra positive ions are excess of one or two electrons respectively from the necessary 4 f -sub-shell stable configuration. The lanthanoids that exhibit + 2 oxidation states are europium, ytterbium, samarium and thulium. The + 2 oxidation state of europium and ytterbium can be explained in terms of the extra stability of half filled and completely filled 4 f -sub-shell. Similarly, the + 2 oxidation states of samarium and thulium is because these ions are short of one electron from the necessary half filled and completely filled stable configurations.