Name the members of the lanthanide series which exhibit +4 oxidation states and those which exhibit +2 oxidation state.

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)4f1 5d06s2

 

Praseodymium

Pr

59

(Xe)4f3 5d06s2

 

Neodymium

Nd

60

(Xe)4f4 5d06s2

 

Promethium

Pm

61

(Xe)4f5 5d06s2

 

Samarium

Sm

62

(Xe)4f6 5d06s2

 

Europium

Eu

63

(Xe)4f7 5d06s2

 

Gadolinium

Gd

64

(Xe)4f7 5d16s2

 

Terbium

Tb

65

(Xe)4f9 5d06s2

 

Dysprosium

Dy

66

(Xe)4f10 5d06s2

 

Holmium

Ho

67

(Xe)4f11 5d06s2

 

Erbium

Er

68

(Xe)4f12 5d06s2

 

Thulium

Tm

69

(Xe)4f13 5d06s2

 

Ytterbium

Yb

70

(Xe)4f14 5d06s2

 

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.