Question

Which of the following oxidation states is the most common among the
lanthanide?
(a) 4
(b) 2
(c) 5
(d) 3

Answer 1

Lanthanides are the members of the f-blocks elements which involves the filling of the (n-2) f-subshell and the general electronic configuration of the lanthanide is $\text{4}{{\text{f}}^{1-14}}\text{5}{{\text{d}}^{0-1}}\text{6}{{\text{s}}^{2}}$and loses the electrons in order to gain the nearest noble gas configuration. Now identify its oxidation state.

Complete step by step solution:
- Lanthanoids both belong to the family of f-block elements. In f-block elements, the elements involve the filling of f-orbitals i.e. in which the last electron enters the last but two i.e.(n-2) f-subshell of the atom. Their general electronic configuration is: $\text{(n-2)}{{\text{f}}^{1-14}}\text{(n-1)}{{\text{d}}^{0-1}}\text{n}{{\text{s}}^{2}}$. They are also called inner transition elements because they are occurring in between the transition elements.
- f-block elements occur in two series: lanthanides and actinides. In lanthanides, the 4f subshell is progressively being filled. It consists of 14 elements from Ce (58) to Lu (71) after La (57) and their general electronic configuration is: $\text{4}{{\text{f}}^{1-14}}\text{5}{{\text{d}}^{0-1}}\text{6}{{\text{s}}^{2}}$. On the other hand, in cases of actinides, the 5f subshell is progressively being filled. It also consists of 14 elements from Th (90) to Lr (103) after Ac (89) and their general electronic configuration is: $\text{5}{{\text{f}}^{1-14}}\text{6}{{\text{d}}^{0-1}}\text{7}{{\text{s}}^{2}}$.
- In lanthanide , the most common oxidation state is +3 because they lose the electrons to acquire the nearest noble gas configuration or the half-filled electronic configuration.

Hence, option (d) is correct.

Note: In Lanthanides ,there is steady decrease in the ionic radii with the increase in the atomic number i.e. the nuclear charge increases as we move from left to right but the electrons are also added in the n-2 subshell and the effect of increased nuclear charge dominates over the imperfect shielding by the 4f electrons of the lanthanides and thus, there is decrease in their ionic radii and this is known as lanthanide contraction.