Question

Name the members of the lanthanoid series which exhibit $ + 4$ oxidation state and those which exhibit $ + 2$ oxidation state. Try to correlate this type of behavior with the electronic configuration of these elements.

Answer 1

Hint:The lanthanide series mostly show an oxidation state $ + 3$. Few metals in the lanthanide series show $ + 4$ oxidation states. This uneven distribution of oxidation states among the metals is attributed to the high stability of empty, half-filled, or filled subshells.

Complete step by step answer:
The elements in the lanthanide series show an oxidation state of $ + 3$ as mentioned.
The lanthanides that exhibit $ + 4$ oxidation state are Ce (50), Pr (59), Nd (60), Tb(65), Dy (66)
Their electronic configuration is given below:
$Ce = \left[ {Xe} \right]4{f^1}5{d^1}6{s^2}$
$C{e^{4 + }}$ ion is a strong oxidizing agent because the $ + 3$ oxidation state is more stable than $ + 4$ oxidation, and also due to its noble gas electronic configuration.
The $ + 4$ oxidation state of cerium is favored as it acquires a noble gas configuration but it reverts to a $ + 3$ oxidation state and thus acts as a strong oxidant.
$Tb = \left[ {Xe} \right]4{f^9}6{s^2}$
$T{b^{ + 4}}$ the state has $4{f^7}$ a half-filled stable configuration but it readily gets reduced into a common state and hence acts as an oxidant.
The lanthanides that exhibit a $ + 2$ oxidation state are samarium, europium, and ytterbium.
Europium electronic configuration $\left[ {Xe} \right]4{f^7}6{s^2}$ loses two electrons from 6s energy level and attains the highly stable, half-filled $4{f^7}$ configuration and hence it readily forms $E{u^{2 + }}$ ion. $E{u^{2 + }}$ then changes to the common oxidation states of lanthanides ($ + 3$) and forms $E{u^{ + 3}}$ , acting as a strong reducing agent.
Ytterbium is a strong reducing agent, in the $Y{b^{2 + }}$ state, it is filled f-orbital.


Note:
The oxidation state of these elements and their properties due to the presence of f-subshell plays the main role.
The energy gap between 4f and 5d orbitals is large and so the number of oxidation states is limited, unlike the d-block elements.
Lanthanides show variable oxidation states. They show $ + 2$, $ + 3$ and $ + 4$ oxidation states.
But the most stable oxidation state of lanthanides is $ + 3$