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

<p><strong>8.42&nbsp;</strong>The members of lanthanide series along with their electronic configuration are given in the below table:</p><table><tbody><tr><td width="142"><p><strong>Name</strong></p></td><td width="78"><p><strong>Symbol</strong></p></td><td width="144"><p><strong>Atomic number</strong></p></td><td width="198"><p><strong>Electron configuration</strong></p></td></tr><tr><td width="142"><p><strong>lanthanum</strong></p></td><td width="78"><p>La</p></td><td width="144"><p>57</p></td><td width="198"><p>(Xe)5d¹ 6s²</p></td></tr><tr><td width="142"><p><strong>Cerium</strong></p></td><td width="78"><p>Ce</p></td><td width="144"><p>58</p></td><td width="198"><p>(Xe)4f¹5d⁰6s²</p></td></tr><tr><td width="142"><p><strong>Praseodymium</strong></p></td><td width="78"><p>Pr</p></td><td width="144"><p>59</p></td><td width="198"><p>(Xe)4f³5d⁰6s²</p></td></tr><tr><td width="142"><p><strong>Neodymium</strong></p></td><td width="78"><p>Nd</p></td><td width="144"><p>60</p></td><td width="198"><p>(Xe)4f⁴5d⁰6s²</p></td></tr><tr><td width="142"><p><strong>Promethium</strong></p></td><td width="78"><p>Pm</p></td><td width="144"><p>61</p></td><td width="198"><p>(Xe)4f⁵5d⁰6s²</p></td></tr><tr><td width="142"><p><strong>Samarium</strong></p></td><td width="78"><p>Sm</p></td><td width="144"><p>62</p></td><td width="198"><p>(Xe)4f⁶5d⁰6s²</p></td></tr></tbody></table><table><tbody><tr><td width="142"><p><strong>Europium</strong></p></td><td width="78"><p>Eu</p></td><td width="144"><p>63</p></td><td width="198"><p>(Xe)4f⁷5d⁰6s²</p></td></tr><tr><td width="142"><p><strong>Gadolinium</strong></p></td><td width="78"><p>Gd</p></td><td width="144"><p>64</p></td><td width="198"><p>(Xe)4f⁷ 5d¹6s²</p></td></tr><tr><td width="142"><p><strong>Terbium</strong></p></td><td width="78"><p>Tb</p></td><td width="144"><p>65</p></td><td width="198"><p>(Xe)4f⁹5d⁰6s²</p></td></tr><tr><td width="142"><p><strong>Dysprosium</strong></p></td><td width="78"><p>Dy</p></td><td width="144"><p>66</p></td><td width="198"><p>(Xe)4f¹⁰5d⁰6s²</p></td></tr><tr><td width="142"><p><strong>Holmium</strong></p></td><td width="78"><p>Ho</p></td><td width="144"><p>67</p></td><td width="198"><p>(Xe)4f¹¹5d⁰6s²</p></td></tr><tr><td width="142"><p><strong>Erbium</strong></p></td><td width="78"><p>Er</p></td><td width="144"><p>68</p></td><td width="198"><p>(Xe)4f¹²5d⁰6s²</p></td></tr><tr><td width="142"><p><strong>Thulium</strong></p></td><td width="78"><p>Tm</p></td><td width="144"><p>69</p></td><td width="198"><p>(Xe)4f¹³5d⁰6s²</p></td></tr><tr><td width="142"><p><strong>Ytterbium</strong></p></td><td width="78"><p>Yb</p></td><td width="144"><p>70</p></td><td width="198"><p>(Xe)4f¹⁴5d⁰6s²</p></td></tr><tr><td width="142"><p><strong>Lutetium</strong></p></td><td width="78"><p>Lu</p></td><td width="144"><p>71</p></td><td width="198"><p>(Xe)4f¹⁴ 5d¹6s²</p></td></tr></tbody></table><p>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.</p><p>+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:</p><p>+2 = ₆₀Nd, ₆₂Sm, ₆₃Eu, ₆₉Tm, ₇₀Yb</p><p>+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)</p><p>Hence, the members which will show +4 oxidation state are:</p><p>+4 = ₅₈Ce, ₅₉Pr, ₆₀Nd, ₆₅Tb, ₆₆Dy</p><p>Each case tends to revert to the more stable oxidation state of +3 by loss or gain of an 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.</p>

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