Question

Write the general oxidation state of lanthanides.

Answer 1

Hint: Lanthanides are the rare elements of modern periodic table i.e the elements with atomic numbers from 58 to 71.

Complete step by step answer:
Oxidation number, also called Oxidation State, the total number of electrons that an atom either gains or loses in order to form a chemical bond with another atom. Generally, Lanthanides show variable oxidation states. They also show $+2,+3,$ and $$+4$$ oxidation states. But the most stable oxidation state of Lanthanides is $$+3$$ Elements in other states try to lose or gain electrons to get $$+3$$ state. samarium, europium, and ytterbium are the lanthanides which have $$+2$$ oxidation state. Hence the most stable oxidation state of lanthanides is $$+3$$.

Additional information:
 Lanthanides are the rare earth elements of the modern periodic table i.e. the elements with atomic numbers from 58 to 71 following the element Lanthanum. They are called rare earth metals since the occurrence of these elements is very small ($3\times {10}^{-4}$of Earth’s crust).They are available in ‘monazite’ sand’ as lanthanide orthophosphates. The term ‘lanthanide’ was first introduced by the Norwegian mineralogist Victor Goldschmidt in 1925. The lanthanide family consists of fifteen metallic elements (from lanthanum to lutetium), all but one of which are f-block elements. The valence electrons of these elements lie in the 4f orbital.


Note:
- Lanthanum, however, is a d-block element with an electronic configuration of $[\text{ Xe }]\text{ 5}{\text{d}}^{\text{1}}\text{6}{\text{s}}^{\text{2}}$. (from lanthanum to lutetium), all but one of which are f-block elements. The valence electrons of these elements lie in the 4f orbital. Lanthanum, however, is a d-block element with an electronic configuration of $[\text{ Xe }]\text{ 5}{\text{d}}^{\text{1}}\text{6}{\text{s}}^{\text{2}}$.
-The oxidation state of an uncombined element is zero.But for lanthanide it can be $+2,+3,+4$.