Question:
Number of electrons present in $4 \mathrm{f}$ orbital of $\mathrm{Ho}^{3+}$ ion is . (Given Atomic No. of Ho $=67$ )
Solution:
$\mathrm{Ho}=[\mathrm{Xe}] 4 \mathrm{f}^{11} 6 \mathrm{~s}^{2}$
$\mathrm{Ho}^{3+}=[\mathrm{Xe}] 4 \mathrm{f}^{10}$
so number of $\mathrm{e}^{-}$present in $4 \mathrm{f}$ is 10 .