Question

The element from atomic number 58 to 71 are called:
A) Transition elements
B) Lanthanides
C) Actinides
D) Alkali metals

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 step by step 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 are mostly stable in their $+ 3$ oxidation state. The $4f$ electrons are shielded from the chemical environment by the $5s$ and $5p$ electrons. Thus these electrons do not take part in any chemical reactions.
The lanthanides contain only one radioactive element named as Promethium ( $Pm$ ).
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.
Lanthanide elements have unpaired electrons in $f$ orbitals, hence they exhibit colour by $f - f$ transition.
$\therefore$ The correct option is (B).

Note:
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.