What is the number of unpaired electrons pres | vedclass

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(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

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$0$

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(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 $Lu (Z=71)$?

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