Search any question & find its solution
Question:
Answered & Verified by Expert
What is the number of unpaired electrons present in $\mathrm{f}$-orbital at +3 oxidation state of $\mathrm{Lu}(\mathrm{Z}=71)$ ?
Options:
Solution:
1833 Upvotes
Verified Answer
The correct answer is:
0
$$
\mathrm{Lu}=[\mathrm{Xe}] 4 \mathrm{f}^{14} 5 \mathrm{~d}^1 6 \mathrm{~s}^2
$$
or $\mathrm{Lu}^{+3}=[\mathrm{Xe}] 4 \mathrm{f}^{14}$
\mathrm{Lu}=[\mathrm{Xe}] 4 \mathrm{f}^{14} 5 \mathrm{~d}^1 6 \mathrm{~s}^2
$$
or $\mathrm{Lu}^{+3}=[\mathrm{Xe}] 4 \mathrm{f}^{14}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.