JEE Mains · Chemistry · STD 12 - 4. d and f- block elements
Number of electrons present in \(4 \mathrm{f}\) orbital of \(\mathrm{Ho}^{3+}\) ion is \(.....\) (Given Atomic No. of \(\mathrm{Ho}=67\) )
- A \(10\)
- B \(15\)
- C \(20\)
- D \(36\)
Answer & Solution
Correct Answer
(A) \(10\)
Step-by-step Solution
Detailed explanation
\(\mathrm{Ho}=[\mathrm{Xe}] 4 \mathrm{f}^{11} 6 \mathrm{~s}^{2}\) \(\mathrm{Ho}^{3+}=[\mathrm{Xe}] 4 \mathrm{f}^{10}\) So number of \(\mathrm{e}^{-}\)present \(4 \mathrm{f}\) is \(10 .\)
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