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The outer electronic configuration of two members of the lanthanoid series are as follows : 4f^(1) 5d^(1) 6s^(2) and 4f^(7) 5d^(-) 6s^(2) What are their atomic numbers ? Predict the oxidation states exhibited by these elements in their compounds. |
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Answer» Solution :Complte E.C. of 1ST lanthanoid `= [Xe]^(54) 4f^(1) 5d^(1) 6 s^(1) `. Atomic No. ` = 58 (C e) ` COMPLETE E.C. of 2nd lanthanoid `= [ Xe]^(54) 4f^(7) 5d^(0) 6s^(2)` . Atomic No. `=63 (E u)` OXIDATIO states of 1st lanthanoid `= + 2 ( 4f^(2)) , + 3( 4f^(1)) , + 4 ( 4f^(0))` Oxidation states of 2nd lanthanoid ` = +2 (4f^(7)) ` and `+ 3 ( 4f^(6))` |
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