
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
457.8k+ views
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 $(n-2){{f}^{0-14}}.(n-1){{d}^{0-2}}n{{s}^{2}}$.
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 $(n-2){{f}^{0-14}}.(n-1){{d}^{0-2}}n{{s}^{2}}$.
Recently Updated Pages
Glucose when reduced with HI and red Phosphorus gives class 11 chemistry CBSE

The highest possible oxidation states of Uranium and class 11 chemistry CBSE

Find the value of x if the mode of the following data class 11 maths CBSE

Which of the following can be used in the Friedel Crafts class 11 chemistry CBSE

A sphere of mass 40 kg is attracted by a second sphere class 11 physics CBSE

Statement I Reactivity of aluminium decreases when class 11 chemistry CBSE

Trending doubts
10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

State and prove Bernoullis theorem class 11 physics CBSE

What organs are located on the left side of your body class 11 biology CBSE

How many valence electrons does nitrogen have class 11 chemistry CBSE
