Question

The outer electronic configuration of two members of the lanthanoid series are as follows:

4f¹ 5d¹ 6s² and 4f⁷ 5d⁰ 6s².

What are their atomic numbers? Predict the oxidation states exhibited by these elements in their compounds.

Answer 1

1. From lanthanum (La) to lutetium (Lu), there are fifteen elements in the lanthanoid series. Because of their electrical structures, several members of this series exhibit the oxidation states of +4 and +2. Cerium (Ce), praseodymium (Pr), and terbium (Tb) are commonly found in the +4 oxidation state.
2. The elements lose four electrons in these situations, mostly from the 4f and 6s orbitals. Samarium (Sm), europium (Eu), and ytterbium (Yb) frequently exhibit the +2 oxidation state, in which these elements lose two electrons, typically from the 6s orbital.
3. The electronic configurations of these oxidation states can be linked to their stability; for instance, Eu has a highly stable half-filled 4f shape in the +2 state.
4. Likewise, the +4 state is preferred in elements such as Ce because it can stabilize higher oxidation states by involving 4f electrons.