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


The members of lanthanide series along with their electronic configuration are given in the below table:


Name



Symbol



Atomic number



Electron configuration



lanthanum



La



57



(Xe)5d1 6s2



Cerium



Ce



58



(Xe)4f15d06s2



Praseodymium



Pr



59



(Xe)4f35d06s2



Neodymium



Nd



60



(Xe)4f45d06s2



Promethium



Pm



61



(Xe)4f55d06s2



Samarium



Sm



62



(Xe)4f65d06s2



Europium



Eu



63



(Xe)4f75d06s2



Gadolinium



Gd



64



(Xe)4f7 5d16s2



Terbium



Tb



65



(Xe)4f95d06s2



Dysprosium



Dy



66



(Xe)4f105d06s2



Holmium



Ho



67



(Xe)4f115d06s2



Erbium



Er



68



(Xe)4f125d06s2



Thulium



Tm



69



(Xe)4f135d06s2



Ytterbium



Yb



70



(Xe)4f145d06s2



Lutetium



Lu



71



(Xe)4f14 5d16s2



The typical oxidation state of the lanthanides is +3. The oxidation state of +2 and +4 are exhibited by some of the elements. These are shown by those elements which by losing 2 or 4 elements acquire a stable configuration.


+2 oxidation state is exhibited when the lanthanide has the configuration 5d06s2 so that 2 electrons are lost easily. Hence the members which will show +2 oxidation state are:


+2 = 60Nd, 62Sm, 63Eu, 69Tm, 70Yb


+4 oxidation state is exhibited when the configuration left (by losing 2 electrons) is close to 4f0 (example 4f0 , 4f1, 4f2, 4f3) or close to 4f7 (example 4f7or 4f8) or close to 4f14 ( 4f13 or 4f14)


Hence, the members which will show +4 oxidation state are:


+4 = 58Ce, 59Pr, 60Nd, 65Tb, 66Dy


Note: Each case tends to revert to the more stable oxidation state of +3 by loss or gain of an electron. That is why Sm2+, Eu2+ and Yb2+ ions in solutions are good reducing agents and aqueous solution of Ce4+ and Tb4+ are good oxidizing agents.


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