Question

Name an element of lanthanide series which is well known to show +4 oxidation state. Is it a strong oxidizing agent or reducing agent?

Answer 1

Hint: Lanthanides: These are the ${ 4f }$ series elements of f-block elements in a periodic table are called lanthanides. Its ammonium nitrate salt is a one-electron oxidizing agent and used for oxidizing addition reaction of electrophilic radicals to alkenes.

Complete step-by-step answer:
Cerium (Ce) and Terbium (Tb) show the ${ +4 }$ oxidation state in the lanthanide series. The electronic configurations of these elements are given below:
${ Ce=[Xn]4f }^{ 1 }{ 5d }^{ 1 }{ 6s }^{ 2 }$
${ Tb=[Xn]4f }^{ 0 }{ 6s }^{ 2 }$

As it is clear from the electronic configuration of Ce that ${ Ce }^{ +4 }$ is favoured by its noble gas configuration, that is,${ [Xn]4f }^{ 0 }{ 5d }^{ 0 }{ 6s }^{ 0 }$ and it can be easily converted into ${ Ce }^{ +3 }$ ${ [Xn]4f }^{ 1 }{ 5d }^{ 0 }{ 6s }^{ 0 }$ Hence, ${ Ce }^{ +4 }$ is a strong oxidizing agent due to common ${ Ce }^{ +3 }$ oxidation state.

Additional Information:
Lanthanoid Contraction: The steady decrease in the atomic and ionic radii of the transition metals as the atomic number increases. This is because of the filling of 4f orbitals before the 5d orbitals. This contraction is quite regular. This is called a lanthanide contraction.
It is because of lanthanoid contraction that the atomic radii of the second row of transition elements are almost similar to those of the third row of transition elements.

Consequences of Lanthanide contraction:
There is a similarity in the properties of the second and third transition series.
Separation of lanthanides is possible due to lanthanide contraction.
It is due to lanthanide contraction that there is variation in the basic strength of lanthanide hydroxides.

Note: The possibility to make a mistake is that ${ +3 }$ oxidation state of Cerium is more stable than ${ +4 }$ although it attains noble gas configuration. This is because it requires a huge amount of energy to extract ${ 4 }$ electrons and the increase in charge leads to reducing stability.