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Which of the following oxidation states is the most common among the lanthanoids?
A
4
B
2
C
5
D
3
Detailed Solution
The general configuration of the lanthanoids is $[Xe]4f^{1-14}5d^{0-1}6s^2$.
All of them readily lose the two 6s electrons and one 5d or 4f electron to form $Ln^{3+}$ ions.
The +3 state is the most common and most stable oxidation state for every lanthanoid.
+2 and +4 states are shown only by a few elements where an empty, half-filled or completely filled 4f subshell results, e.g., $Ce^{4+}$ ($4f^0$), $Eu^{2+}$ ($4f^7$), $Yb^{2+}$ ($4f^{14}$), $Tb^{4+}$ ($4f^7$); +5 is not shown.
Hence the most common oxidation state is +3.
All of them readily lose the two 6s electrons and one 5d or 4f electron to form $Ln^{3+}$ ions.
The +3 state is the most common and most stable oxidation state for every lanthanoid.
+2 and +4 states are shown only by a few elements where an empty, half-filled or completely filled 4f subshell results, e.g., $Ce^{4+}$ ($4f^0$), $Eu^{2+}$ ($4f^7$), $Yb^{2+}$ ($4f^{14}$), $Tb^{4+}$ ($4f^7$); +5 is not shown.
Hence the most common oxidation state is +3.