Palladium () exhibits an exceptional ground state electronic configuration due to stability factors. — Periodic Table and Periodicity Chemistry Question
Question
Palladium ($Pd$) exhibits an exceptional ground state electronic configuration due to stability factors. Given that its configuration is $[Kr] 4d^{10} 5s^0$, to which group and period does it belong?
Answer: B
💡 Solution & Explanation
The highest principal quantum number for $Pd$ is $n=5$ (since the 5s subshell was available, even if empty), meaning it belongs to the 5th period. For d-block elements, the group number is the sum of $(n-1)d$ and $ns$ electrons. Here, $10 + 0 = 10$. Thus, it belongs to Group 10.
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