The common oxidation state of f-block element | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The common oxidation state of $f$-block elements is $+3$. $Eu$ $(Z=63)$ has the electronic configuration $[Xe] 4f^7 6s^2$. It shows a stable $+2$

Vedclass · Accepted Answer

$+2, +4$

Vedclass · Answer

(A) The common oxidation state of $f$-block elements is $+3$. $Eu$ $(Z=63)$ has the electronic configuration $[Xe] 4f^7 6s^2$. It shows a stable $+2$ oxidation state due to the stable half-filled $4f^7$ configuration. $Tb$ $(Z=65)$ has the electronic configuration $[Xe] 4f^9 6s^2$. It shows a stable $+4$ oxidation state due to the stable half-filled $4f^7$ configuration after losing four electrons. Therefore,the other stable oxidation states for $Eu$ and $Tb$ are $+2$ and $+4$ respectively. Hence,option $A$ is the correct answer.

The common oxidation state of $f$-block elements is $+3$. The other stable oxidation states of $^{63}Eu$ and $^{65}Tb$ are respectively:

Similar Questions

The pair of lanthanides in which both elements have high third-ionization energy is:

Discuss the trends in oxidation states shown by actinoids.

Identify the correct statements from the following :
$I$. The ionic radius of $Pr^{3+}, Dy^{3+}$ and $Sm^{3+}$ follow the order,$Sm^{3+} > Pr^{3+} > Dy^{3+}$.
$II$. $Eu^{2+}$ acts as a strong reducing agent.
$III$. $Pu$ exhibits $+7$ oxidation state.

$(i) La(OH)_3$ is the least basic among lanthanide hydroxides. $(ii) Zr^{4+}$ and $Hf^{4+}$ have almost identical ionic radii. $(iii) Ce^{4+}$ can act as an oxidizing agent. Which of the above statements are correct?

The point of dissimilarity between lanthanides and actinides is

Vedclass Test Series

Exam Paper Generator

Online Exam Module