Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

General electron configuration of lanthanides is:
A. ${\text{(n}} - {\text{2)}}{{\text{f}}^{1 - 14}}\left( {{\text{n}} - 1} \right){d^{0 - 1}}{\text{n}}{{\text{s}}^{\text{2}}}$
B. ${\text{(n}} - 2){{\text{f}}^{10 - 14}}\left( {{\text{n}} - 1} \right){{\text{d}}^{10 - 1}}{\text{n}}{{\text{s}}^{\text{2}}}$
C. $\;\left( {{\text{n}} - 2} \right){{\text{f}}^{0 - 14}}\left( {{\text{n}} - 1} \right){{\text{d}}^{{\text{10}}}}{\text{n}}{{\text{s}}^{\text{2}}}$
D. ($({\text{n}} - {\text{2)}}{{\text{d}}^{0 - 1}}\left( {{\text{n}} - 1} \right){{\text{f}}^{1 - 14}}{\text{n}}{{\text{s}}^{\text{2}}}$

seo-qna
Last updated date: 09th Apr 2024
Total views: 408k
Views today: 8.08k
MVSAT 2024
Answer
VerifiedVerified
408k+ views
Hint: Lanthanides fall in the atomic number range of 57 to 71 in the periodic table. They exhibit similar properties to the first element in the group, Lanthanum.

Step by step answer:
We should keep in mind that though lanthanum is the first element of the lanthanides group and it also has similar properties with all other elements in the group, it’s electron configuration varies with the rest of the elements in the group.
The general electron configuration of lanthanide elements are in the form of $\text{(Xe)4}{{\text{f}}^{\text{n}}}\text{6}{{\text{s}}^{\text{2}}}$. Generally, lanthanum is not included in this generalization but it is considered a part in the lanthanides group due to its tri-valency and many other similarities.
In three cases, exceptions are found to the \[\] pattern, in Cerium, Gadolinium and Lutetium. We can hence stabilize the solution obtained by the group with the equation \[\left( \text{n-2} \right){{\text{f}}^{\text{1-14}}}\left( \text{n-1} \right){{\text{d}}^{\text{0-1}}}\text{n}{{\text{s}}^{\text{2}}}\] which satisfies each case in the lanthanides group.
Hence, we can conclude that electronic configuration of lanthanides is \[\left( \text{n-2} \right){{\text{f}}^{\text{1-14}}}\left( \text{n-1} \right){{\text{d}}^{\text{0-1}}}\text{n}{{\text{s}}^{\text{2}}}\]. Therefore, the correct answer is Option A.

Note: The f-block elements appear in two series in the periodic table characterized by the filling of the 4f and 5f orbitals. Electron configuration of lanthanides follow Aufbau rules except three of the elements. Lanthanides show variable oxidation states like +2, +3 and +4 oxidation states. The most stable oxidation state of lanthanides is +3 oxidation state since elements in other states try to gain or lose electrons to get +3 oxidation state.
Recently Updated Pages