Question

The most common oxidation state of lanthanides is:
A) $+ 2$
B) $+ 3$
C) $+ 4$
D) $+ 5$

Answer 1

The very high energy difference between $6s$ and $4f$ subshell, it is very hard to remove more than one electron from the $4f$ subshell. Because of this $4f$ electrons will experience a more effective nuclear charge than $6s$ electrons.

Complete step by step answer:
We know that in the modern periodic table, lanthanides are placed in the 6th period and 3rd group.
As Lanthanides comprises 14 elements consisting of partially, half or fully filled $f$ orbitals, it is kept in the $f$ block.
The general representation of lanthanides is given as:
$4{f^n}5{d^{0 - 1}}6{s^2}$
Here,$n$ in the $f$ orbitals means the electrons are filling in the $f$ orbitals linearly from 1 to 14.
If we remove two electrons from $6s$ orbital, we will get the oxidation state as $+ 2$ which is also shown by some of the lanthanides. But if we remove two electrons from the $6s$ orbital and one electron from the $4s$ orbital, we get a $+ 3$ oxidation state which is predominant in most of the lanthanides. Lanthanides in $+ 2$ oxidation state is a strong reducing agent changing to the most common oxidation state which is $+ 3$. Some lanthanides are completely half-filled and fully filled $f$orbitals. So, in this case, the oxidation state of metal remains $+ 2$.

Therefore, we can conclude that the correct answer to this question is option B.

Note:
$+ 3$ oxidation state, the decrease in atomic radii is regular and lanthanides in other oxidation states have an irregular decrease in atomic radii. So, $+ 3$ is the most common oxidation state of lanthanides.