Question

Write general oxidation state of Lanthanides.

Answer 1

Lanthanides form the compound which are mainly ionic and are trivalent in nature. Ionization energies are fairly low and very much comparable with alkaline earth metals. They are also very good reducing agents.

Complete solution.
The $14$ elements that fill up the antepenultimate $4f$ energy levels are popularly known as the lanthanides. They lie between lanthanum ( $La$ ) and Hofmium ( $Hf$ ).Their general electronic configuration of this group can be represented by –
$[Xe]4{f^{1 - 14}}5{d^{0 - 1}}6{s^2}$
The atomic number of the elements starts from the Cerium ( $58$ ) to Lutetium ( $71$ ).
These elements have a special kind of contraction known as the lanthanide contraction which can be defined as the decrease in atomic radius of lanthanides on moving from Ce to Lu due to the increase in the effective nuclear charge experienced by the outermost electrons.
As we know that metallic bonds are the bonds which are formed by the electrons present in the metals, So in Lanthanides only three electrons takes part in metallic bonding, except
$Eu(Z = 63)andYb(Z = 70)$ . This is because in their case one electron from $5d$ orbital moves to $4f$ orbital to get half-filled and fully filled stable electronic configuration. Hence only two electrons participate in metallic bonding. The oxidation state of $Eu$ and $Yb$ is $+ 2$ (exception).
The $4f$ electrons are shielded from the chemical environment by the $5s$ and $5p$ orbital electrons. Thus these electrons do not take part in any chemical reactions.
These elements are mostly stable in their $+ 3$ oxidation state. Generally three electrons are used in the Valency as two electrons from $6s$ orbitals and one electron from either $5d$ orbital or from $4f$ orbital when they are not substantially half filled and fully filled.
This is because the half-filled $4{f^7}$ orbitals and fully filled $4{f^{14}}$ orbitals are exceptionally very stable and do not take part in reaction in normal conditions.

Note:
Lanthanide elements have unpaired electrons in $f$ orbitals, hence they exhibit colour by $f - f$ transition. The colour exhibited by these elements depends upon the number of unpaired electrons. The ions having configuration $4{f^n}$ and configuration $4{f^{14 - n}}$ are expected to have similar colour. The elements having and are colourless as they are diamagnetic and thus no $f - f$ is possible.