What is the number of unpaired electrons in f | vedclass

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

Question

(D) The atomic number of lutetium $(Lu)$ is $71$. The electronic configuration of neutral $Lu$ is $[Xe] 4f^{14} 5d^1 6s^2$. In its $+3$ oxidation stat

Vedclass · Accepted Answer

$0$

Vedclass · Answer

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

What is the number of unpaired electrons in $f$ orbitals of lutetium in its $+3$ oxidation state?

Similar Questions

Which of the following pairs of ions have the same paramagnetic moment?

Which of the following electronic configurations represents an atom that can exhibit the highest oxidation state?

Which of the following elements has the maximum density?

In the modern periodic table,which elements represent the transition between the most electropositive and the least electropositive elements?

Assertion: Transition metals and their complexes show catalytic activity.
Reason: The activation energy of a reaction is lowered by the catalyst.

Vedclass Test Series

Exam Paper Generator

Online Exam Module