Although $(+3)$ oxidation state is the characteristic oxidation state of lanthanoids,cerium shows $(+4)$ oxidation state also. Why?
(N/A) Cerium ($Ce$,atomic number $58$) has the electronic configuration $[Xe] \ 4f^{1} \ 5d^{1} \ 6s^{2}$.
By losing four electrons,it attains the stable noble gas configuration of Xenon $([Xe])$,which is $[Xe] \ 4f^{0} \ 5d^{0} \ 6s^{0}$.
This extra stability associated with the empty $4f$ subshell allows cerium to exhibit the $(+4)$ oxidation state.
By losing four electrons,it attains the stable noble gas configuration of Xenon $([Xe])$,which is $[Xe] \ 4f^{0} \ 5d^{0} \ 6s^{0}$.
This extra stability associated with the empty $4f$ subshell allows cerium to exhibit the $(+4)$ oxidation state.
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Identify the isoelectronic pair of ions from the following.
The most common oxidation state of Lanthanoid elements is $+3$. Which of the following is likely to deviate easily from $+3$ oxidation state?
Most acidic among the given compounds is:
Medium
View SolutionObserve the following $f$-block elements: $Eu (Z=63)$; $Pu (Z=94)$; $Cf (Z=98)$; $Sm (Z=62)$; $Gd (Z=64)$; $Cm (Z=96)$. How many of the above have half-filled $f$-orbitals in their ground state?
DifficultTS EAMCET 2024
View SolutionWhich of the following ions is the strongest oxidizing agent? [ Atomic Number of $Ce = 58, Eu = 63, Tb = 65, Lu = 71$ ]
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