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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 \(\mathrm{Ho}=67\) )

  1. A \(10\)
  2. B \(15\)
  3. C \(20\)
  4. D \(36\)
Verified Solution

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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