What is the number of unpaired electrons in f | vedclass

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(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

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(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?

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