Question

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

Answer 1

Hint: 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.