More number of oxidation states are exhibited by the actinoids than by the lanthanoids. The main reason for this is
- Amore active nature of the actinoids
- Bmore energy difference between $5f$ and $6d$ orbitals than that between $4f$ and $5d$ orbitals
- Clesser energy difference between $5f$ and $6d$ orbitals than that between $4f$ and $5d$ orbitals
- Dgreater metallic character of the lanthanoids than that of the corresponding actinoids.
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Similar Questions
$Ln$ (Lanthanide) $\xrightarrow{\text{With Halogen}} X$
$Ln$ (Lanthanide) $\xrightarrow{\text{Burn With } O_2} Y$
$Ln$ (Lanthanide) $\xrightarrow{\text{Heated with } N_2} Z$
$X, Y$ and $Z$ are respectively:-
$Ln$ (Lanthanide) $\xrightarrow{\text{Burn With } O_2} Y$
$Ln$ (Lanthanide) $\xrightarrow{\text{Heated with } N_2} Z$
$X, Y$ and $Z$ are respectively:-
Difficult
View SolutionWhich lanthanoid element has the electronic configuration $[Xe]4f^7 5d^1 6s^2$?
Easy
View SolutionThe atomic number of cerium $(Ce)$ is $58$. The correct electronic configuration of $Ce^{3+}$ ion is
Identify the element with the smallest ionic radius in the $+3$ oxidation state from the following:
State the $E^o$ values of lanthanoids and provide their reactions with the following: $(i)$ $H_2$,$(ii)$ Carbon,$(iii)$ Dilute acid,$(iv)$ Halogens,$(v)$ $O_2$,and $(vi)$ $H_2O$.
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