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This question already has an answer here:

The p block elements generally have the general valence shell configuration as $$n\mathrm{p}^{1-6}\,n\mathrm{s}^{1-2}.$$ Expected electronic configuration for palladium is $$\ce{^46Pd} = \mathrm{1s^2\, 2s^2\, 2p^6\, 3s^2\, 3p^6\, 4s^2\, 3d^{10}\, 4p^6\, 5s^2\, 4d^8}.$$ But I find that it's valence shell instead is $$\mathrm{4d^{10}\, 5s^0}$$ I first thought, for achieving fully filled d-subshell, 2 electrons go to the d-subshell from s-subshell. This electron transfer should happen for all members of its group, yet only palladium demonstrates this. Why is this so?

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marked as duplicate by Gaurang Tandon, M.A.R., pentavalentcarbon, Todd Minehardt, Jon Custer Mar 31 '18 at 23:40

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ It's double electron promotion. $\endgroup$ – Mithoron Sep 16 '15 at 20:22
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    $\begingroup$ You may check the answer given here: chemistry.stackexchange.com/questions/2469/… $\endgroup$ – RBW Sep 23 '15 at 12:52
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    $\begingroup$ I strongly suggest the linked question's title be renamed from being about niobium to a more general one, because the answer to that question is a very general answer, and covers all cases neatly. Most probably, any answer to this question will almost repeat what has already been said. Hence, I've cast a close vote as duplicate. $\endgroup$ – Gaurang Tandon Mar 31 '18 at 12:43
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The Aufbau principle states that, hypothetically, electrons orbiting one or more atoms fill the lowest available energy levels before filling higher levels (e.g., 1s before 2s). In this way, the electrons of an atom, molecule, or ion harmonize into the most stable electron configuration possible.

Electron behavior is elaborated by other principles of atomic physics, such as Hund's rule and the Pauli exclusion principle. Hund's rule asserts that even if multiple orbitals of the same energy are available, electrons fill unoccupied orbitals first, before reusing orbitals occupied by other electrons. But, according to the Pauli exclusion principle, in order for electrons to occupy the same orbital, they must have different spins.

The order in which these orbitals are filled is given by n+l rule known as Madelung Rule or diagonal rule.Orbitals with a lower (n+l) value are filled before those with higher (n+l) values where n represents principal quantum number and l represents azimuthal quantum number.

The fact that most of the ground state configurations of neutral atoms fill orbitals following this n + ℓ, n pattern was obtained experimentally, by reference to the spectroscopic characteristics of the elements.

The Madelung energy ordering rule applies only to neutral atoms in their ground state,even in some cases there are examples which show different electronic configuration than those of experimentally determined.Copper,Chromium and Palladium are common examples of this rule.

For Palladium,according to Madelung's rule, 5s^2 has lower (n+l) value i.e. (5+2=7) and 4d^8 has higher (n+l) value i.e. (4+8=12),so the 5s orbital fills up first before 4d orbital.

Hence, Palladium shows anomalous electronic configuration.

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    $\begingroup$ For 5s isn't the n+l value 5+0=5? As the l value of s subshell is 0? $\endgroup$ – Aneek Sep 23 '15 at 15:05
  • $\begingroup$ The answer fails to clear my question, read it carefully. It is the only element besides Niobium to transfer 2 electrons from the s subshell to d subshell..but the link by @Marko cleared my doubt a bit, and it seems I have to study it little more. Thanks @Marko! But, you would have liked to answer it, as it has a bounty! $\endgroup$ – Aneek Sep 23 '15 at 15:08
  • $\begingroup$ @Aneek, if you think that Marko's answer helped you than that of mine, then go for it! :) $\endgroup$ – Soundarya Sep 23 '15 at 19:07

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