What is the number of unpaired electrons pres | vedclass

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

Question

(A) The atomic number of Lutetium $(Lu)$ is $71$. The electronic configuration of $Lu$ is $[Xe] 4f^{14} 5d^1 6s^2$. In the $+3$ oxidation state,$Lu$ l

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(A) The atomic number of Lutetium $(Lu)$ is $71$. The electronic configuration of $Lu$ is $[Xe] 4f^{14} 5d^1 6s^2$. In the $+3$ oxidation state,$Lu$ loses three electrons (two from $6s$ and one from $5d$). The electronic configuration of $Lu^{3+}$ is $[Xe] 4f^{14}$. Since the $4f$ subshell is completely filled with $14$ electrons,there are no unpaired electrons present. Therefore,the number of unpaired electrons is $0$.

What is the number of unpaired electrons present in $f$-orbital at $+3$ oxidation state of $Lu (Z=71)$?

Similar Questions

Which of the following statements is incorrect with reference to $Ce$ $(Z=58)$?

The most common oxidation state of Lanthanoid elements is $+3$. Which of the following is likely to deviate easily from $+3$ oxidation state?

What is the correct order of ionic radii for the $+3$ oxidation state in $Ce, La, Pm,$ and $Yb$?

What is lanthanoid contraction? What are the consequences of lanthanoid contraction?

The atomic number of cerium $(Ce)$ is $58$. The correct electronic configuration of $Ce^{3+}$ ion is

Vedclass Test Series

Exam Paper Generator

Online Exam Module