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Question
- what does the electron configuration for pd end with? b. $5d^{8}$ c. $4p^{8}$ d. $4d^{8}$
Palladium (Pd) has an atomic number of 46. Its electron configuration is \([Kr]4d^{10}\). So the last - filled sub - shell is \(4d\) with 10 electrons. But looking at the options, if we consider some common misunderstandings or mis - representations (assuming a typo in the options where it might be related to the \(d\) - block filling near its position), the correct sub - shell related to its valence electrons (in a more general sense of the outer - shell configuration for transition metals) among the given options is \(4d\).
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d. \(4d^{10}\) (assuming the option \(d\) was supposed to be \(4d^{10}\) as the correct electron - configuration ending for Pd. If the options are as they are with no typo, there is an error in the problem setup, but based on the sub - shell type (\(d\) - subshell relevant for Pd as a transition metal), \(4d\) is the relevant sub - shell among the given options)