Question
Europium and Ytterbium behave as good reducing agents in $+2$ oxidation state explain.

Answer

  • The most stable oxidation state of lanthanoids is $+ 3$.
  • Hence, $Eu^{2+}$​​​​​​​ and $Yb^{2+}​​​​​​​$​​​​​​​ tend to get $+ 3$ oxidation states by losing one electron.
  • Since they get oxidised, they are good reducing agents in $+ 2$ oxidation state.

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