1.

Although +3 oxidation state is the characteristic oxidation state of lanthanoids but cerium shows+4 oxidation state also. Why ?

Answer»

Solution :`._(58) Ce= [Xe]^(54) 4f^(2) 5d^(0) 6s^(2) . Ce^(3+) = [Xe]^(54) 4f^(1) `. Further lossof ONE electron GIVES stable `4f^(0)` configuration. Hence, it SHOWS `+4` oxidation state ALSO.


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