Question

Number of electrons present in $4f$ orbital of $H{o}^{3+}$ ion is______. (Given Atomic No. of $Ho=67$ )

Answer 1

Answer: 10

$Ho=[Xe]4f^{11}6s^2$
$H{o}^{3+}=[Xe]4f^{10}$
so number of $e^{-}$ present in $4f$ is $10$