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Question: The correct configuration of f-block elements is: A. \[(n - 2){f^{1 - 14}}(n - 1){d^{0 - 1}}n{s^2}...

The correct configuration of f-block elements is:
A. (n−2)f1−14(n−1)d0−1ns2(n - 2){f^{1 - 14}}(n - 1){d^{0 - 1}}n{s^2}
B. (n−1)f1−14(n−1)d0−1ns2(n - 1){f^{1 - 14}}(n - 1){d^{0 - 1}}n{s^2}
C. (n−3)f1−14(n−2)d0−1ns2(n - 3){f^{1 - 14}}(n - 2){d^{0 - 1}}n{s^2}
D. (n−2)f0−1(n−1)d0−1ns2(n - 2){f^{0 - 1}}(n - 1){d^{0 - 1}}n{s^2}

Explanation

Solution

The elements in which the additional electron enters in (n−2)f(n - 2)f orbitals are called inner transition elements or the f-block elements. The f-block elements are classified into Lanthanides and Actinides based on the fact that the outermost electron of an atom enters in the f-orbital.

Complete step by step answer:
The general electronic configuration of the f-block elements is represented by:
(n−2)f1−14(n−1)d0−1ns2(n - 2){f^{1 - 14}}(n - 1){d^{0 - 1}}n{s^2}
In this electronic configuration, we can see that if the outermost shell has a number equal to ′n′'n' then the electrons that accommodate the d and f orbital of the atom of the element are represented by (n−1)(n - 1) and (n−2)(n - 2). The f-subshell consists of seven orbitals in which the electrons are present. The f-block elements are divided into two groups and they are Lanthanides and Actinides. The f-block elements are placed separately below the periodic table in two groups of lanthanide and actinide and consist of 14 elements each based on the number of electrons the f-orbital can accommodate. We can notice here that the d-orbital also comprises a range of 0 and 1 electron. This means that due to the introduction of half-filled and fully-filled stable electronic configuration, the outermost electron can even enter into the d-orbital in order to maintain the stability of the electronic configuration of the f-block element. For lanthanides the value of ′n′'n' is equal to 6 and for actinides, the value of ′n′'n' is equal to 7.
Thus, the correct option is A. (n−2)f1−14(n−1)d0−1ns2(n - 2){f^{1 - 14}}(n - 1){d^{0 - 1}}n{s^2} .

Note:
There is some resemblance in the lanthanides and actinides such as both of the series show a dominant oxidation state of +3 and both are electropositive and act as strong reducing agents. The distinction between them is that the lanthanides show less tendency towards the complex formation whereas the actinides have a stronger tendency towards complex formation.