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Question: General electronic configurations of lanthanides are: A) \[\left( {n - 2} \right){f^{1 - 14}}\lef...

General electronic configurations of lanthanides are:
A) (n−2)f1−14(n−1)s2p6d0−1ns2\left( {n - 2} \right){f^{1 - 14}}\left( {n - 1} \right){s^2}{p^6}{d^{0 - 1}}n{s^2}
B) (n−2)f10−14(n−1)d0−1ns2\left( {n - 2} \right){f^{10 - 14}}\left( {n - 1} \right){d^{0 - 1}}n{s^2}
C) (n−2)f0−14(n−1)d10ns2\left( {n - 2} \right){f^{0 - 14}}\left( {n - 1} \right){d^{10}}n{s^2}
D) (n−2)f0−1(n−1)f1−14ns2\left( {n - 2} \right){f^{0 - 1}}\left( {n - 1} \right){f^{1 - 14}}n{s^2}

Explanation

Solution

Lanthanides (rare earth elements) contain fourteen elements from atomic numbers 58 to 71. Lanthanum is a d-block element whereas cerium to lutetium are f-block elements. In the periodic table, lanthanides are present in the sixth period and third group.

Complete answer:
In lanthanides, the electrons enter the penultimate (n−1)d\left( {n - 1} \right)d subshell and pre-penultimate (n−2)f\left( {n - 2} \right)f subshell. The general electronic configuration of lanthanides is (n−2)f1−14(n−1)d0−1ns2\left( {n - 2} \right){f^{1 - 14}}\left( {n - 1} \right){d^{0 - 1}}n{s^2} . But the penultimate subshell also contains (n−1)s2\left( {n - 1} \right){s^2} and (n−1)p6\left( {n - 1} \right){p^6} electrons. Hence, the general electronic configuration of lanthanides will be (n−2)f1−14(n−1)s2(n−1)p6(n−1)d0−1ns2\left( {n - 2} \right){f^{1 - 14}}\left( {n - 1} \right){s^2}\left( {n - 1} \right){p^6}\left( {n - 1} \right){d^{0 - 1}}n{s^2}.

It can also be written as (n−2)f1−14(n−1)s2p6d0−1ns2\left( {n - 2} \right){f^{1 - 14}}\left( {n - 1} \right){s^2}{p^6}{d^{0 - 1}}n{s^2}.

The option B ) is an incorrect answer as (n−2)f10−14\left( {n - 2} \right){f^{10 - 14}} should be written as (n−2)f1−14\left( {n - 2} \right){f^{1 - 14}}.

The option C ) is an incorrect answer as (n−2)f10−14\left( {n - 2} \right){f^{10 - 14}} should be written as (n−2)f1−14\left( {n - 2} \right){f^{1 - 14}}.
Also (n−1)d10\left( {n - 1} \right){d^{10}} should be written as (n−1)d0−1\left( {n - 1} \right){d^{0 - 1}}.

The option D ) is an incorrect answer as (n−2)f0−1\left( {n - 2} \right){f^{0 - 1}} should be written as (n−2)f1−14\left( {n - 2} \right){f^{1 - 14}}.
Also (n−1)f1−14\left( {n - 1} \right){f^{1 - 14}} should be written as (n−1)d0−1\left( {n - 1} \right){d^{0 - 1}}.

Hence, option A) is the correct answer.

Note: Lanthanides are called rare earth elements as they have very small occurrences (around 3×10−4%3 \times {10^{ - 4}}\% of earth’s crust). In ‘monazite sand’ lanthanides are available as lanthanide orthophosphates.