Question

${\text{L}}{{\text{n}}^{3 + }}$(trivalent lanthanide ion) have EC:
A. $\left[ {{\text{Xe}}} \right]\,{\text{4}}{{\text{f}}^{\text{1}}}$ to $\left[ {{\text{Xe}}} \right]\,{\text{4}}{{\text{f}}^{{\text{14}}}}$
B. $\left[ {{\text{Xe}}} \right]\,4{{\text{d}}^1}{\text{4}}{{\text{f}}^{\text{1}}}$ to $\left[ {{\text{Xe}}} \right]\,4{{\text{d}}^1}{\text{4}}{{\text{f}}^{{\text{14}}}}$
C. $\left[ {{\text{Xe}}} \right]\,4{{\text{d}}^2}{\text{4}}{{\text{f}}^0}$ to $\left[ {{\text{Xe}}} \right]\,4{{\text{d}}^1}{\text{4}}{{\text{f}}^{{\text{14}}}}$
D. $\left[ {{\text{Xe}}} \right]\,{\text{4}}{{\text{f}}^0}$ to $\left[ {{\text{Xe}}} \right]\,{\text{4}}{{\text{f}}^{{\text{14}}}}$

Answer 1

Lanthanides are placed in 5d- transition series after the lanthanum.
The atomic number 58 - 71 represents the lanthanides. The atomic number of xenon is
54.

Complete answer:
The atomic number of the lanthanum is $57$ and electronic configuration of the lanthanum is as follows:
$\left[ {{\text{Xe}}} \right]\,5{{\text{d}}^1}6{{\text{s}}^2}$
Lanthanides are ${\text{f}}$-block elements so, their valence electrons will be present in
${\text{f}}$-orbital.
So, the electronic configuration of the lanthanides is as follows:
$\left[ {{\text{Xe}}} \right]\,4{{\text{f}}^1}5{{\text{d}}^1}6{{\text{s}}^2}$ to $\left[ {{\text{Xe}}} \right]\,4{{\text{f}}^{14}}5{{\text{d}}^1}6{{\text{s}}^2}$
It is given that the lanthanides are trivalent so, the three electrons (one of ${\text{d}}$-orbital and two of ${\text{s}}$-orbital) will be removed. The stability of $4{\text{f}}$-orbital is very high then -orbital so it is hard to remove electrons from $4{\text{f}}$-orbital. So, the electronic configuration of the lanthanides trivalent ions is as follows:
$\left[ {{\text{Xe}}} \right]\,4{{\text{f}}^1}$ to $\left[ {{\text{Xe}}} \right]\,4{{\text{f}}^{14}}$
Therefore, option (A) $\left[ {{\text{Xe}}} \right]\,4{{\text{f}}^1}$ to $\left[ {{\text{Xe}}} \right]\,4{{\text{f}}^{14}}$is correct.
In lanthanides the ${\text{4d}}$-orbital is fully filled, so the option (B) and (C) are incorrect.
The electronic configuration for trivalent lanthanum is as follows:
$\left[ {{\text{Xe}}} \right]\,{\text{4}}{{\text{f}}^0}$
So, the electronic configuration$\left[ {{\text{Xe}}} \right]\,{\text{4}}{{\text{f}}^0}$ to $\left[ {{\text{Xe}}} \right]\,{\text{4}}{{\text{f}}^{{\text{14}}}}$ stands for ${\text{La}} - \,{\text{Lu}}$ but the lanthanum is not lanthanides. So, the option (D) is incorrect

So the option (A) is correct.

Note: Tri-positive is the most stable oxidation number shown by the lanthanides. Simply the electronic configuration of an atom is shown by the noble gas’s electron configuration with valence shell electronic configuration. For example, the $54$ electrons of lanthanum are shown by symbol $\left[ {{\text{Xe}}} \right]$ , xenon that is a noble gas with the valence shell electronic configuration$5{{\text{d}}^1}6{{\text{s}}^2}$.