Question

The electronic configuration of lanthanides is:
(A) $(n-2){{f}^{1-14}}(n-1){{d}^{0-1}}n{{s}^{2}}$
(B) $(n-2){{f}^{1-14}}(n-1){{d}^{10-1}}n{{s}^{2}}$
(C) $(n-2){{f}^{1-14}}(n-1){{d}^{10}}n{{s}^{2}}$
(D) $(n-2){{d}^{0-1}}(n-1){{f}^{1-14}}n{{s}^{2}}$

Answer 1

Hint: Lanthanides are separately located in the bottom of the periodic table. The lanthanide compounds are widely used in industries as a catalyst, reducing agents etc. Monazite is the chief rock from which lanthanides are extracted.

Complete answer:
The elements from $C{{e}_{58}}$ to $L{{u}_{71}}$ and from $T{{h}_{90}}$ to $L{{w}_{103}}$ are known as inner transition elements. These elements are separately placed at the bottom of the periodic table. The first series that is from $C{{e}_{58}}$ to $L{{a}_{71}}$ lies in the 6th period and is known as lanthanides. The name lanthanides are given since these elements follow $L{{a}_{57}}$. The second series that is from $T{{h}_{90}}$ to $L{{w}_{103}}$ lies in the 7th period and is known as actinides. The name actinides are given since these elements follow$A{{c}_{89}}$.
$L{{a}_{57}}$ and $A{{c}_{89}}$ show similarities in its properties. The last electrons in the atoms of these enters the f subshell belonging to an anti-penultimate shell that is ${{(n-2)}^{th}}$ shell, these elements are called f-block elements.
The valence shell electronic configuration of the atoms of the atoms of f-block elements can be represented as $(n-2){{f}^{0-14}}.(n-1){{d}^{0-2}}n{{s}^{2}}$ which shows that in these elements the outermost three shells are partially filled while the remaining inner shells are completely filled.
The lanthanides are 4f-block elements and actinides are 5f-block elements.
The electronic configuration of $L{{a}_{57}}$ is ${{[Xe]}_{54}}4{{f}^{0}}5{{d}^{1}}6{{s}^{2}}$. When we move across the period, the additional electron should occupy the vacant 4f orbitals and 5f orbitals should remain singly filled.
So, the electronic configuration of lanthanides is $(n-2){{f}^{0-14}}.(n-1){{d}^{0-1}}n{{s}^{2}}$.

From above discussion we can conclude that option A is the correct answer.

Note:
 If we carefully observe the options we can figure out that option B and D is wrong, that is its representation itself wrong. As discussed above the electronic configuration of f-block is $(n-2){{f}^{0-14}}.(n-1){{d}^{0-2}}n{{s}^{2}}$ so, option B is wrong. The electronic configuration of actinides is $(n-2){{f}^{0-14}}.(n-1){{d}^{0-2}}n{{s}^{2}}$.