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Chemistryf-blockf-blockEasy2 minPYQ_2021

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

Answer:

10

Solution:

$Ho = [Xe] 4 f^{11} 6 s^{2}$

${Ho}^{3 +} = [Xe] 4 f^{10}$

so number of $e^{-}$ present in $4 f$ is $10 .$

Stream:JEESubject:ChemistryTopic:f-blockSubtopic:f-block

⏱ 2mℹ️ Source: PYQ_2021

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