While studying about d block elements I came across this table showing outer shell electronic configuration of group 10 elements

$\ce{Ni}$ $ $ $ $ $ $ $ $ $ $ $3d^8$ $4s^2$

$\ce{Pd}$ $ $ $ $ $ $ $ $ $4d^{10}$ $5s^0$

$\ce{Pt}$ $ $ $ $ $ $ $ $ $ $ $5d^9$ $6s^1$

(Source Concise Inorganic Chemistry, JD Lee, adapted by S. Guha Pg 563)

Now I cannot understand the reason behind this. If we consider that $\ce{Pd}$ has changed its config. to attain stability then what's wrong with $\ce{Ni}$ and $\ce{Pt}$? And should the outer shell electronic configuration of the ground state not be same in elements down a group?


To those who have marked my question as a duplicate of the one mentioned, I humbly request to explain me which answer to the above mentioned question (of which mine is marked as a duplicate of) tells about the anomalous behavior of group 10 elements and why $\ce{Zn}$ assumes $d^{10}$ config and none of the others in the group do the same and again $\ce{Pt}$ assumes a $5d^9 6s^1$ config, while $\ce{Ni}$ does not. Also please do point out where can I find an answer to why in this particular group there is no similarity in ground state electronic configuration b/w elements of the group. Thank You.

  • $\begingroup$ There are relativistic effects in Pt, leading to stabilisation of 6s and destabilisation of 5d $\endgroup$ Commented Sep 25, 2017 at 10:29
  • $\begingroup$ @orthocresol if $\ce{Pd}$ can show $d^{10}$ config due to increased stability why is this not the case with $\ce{Ni}$. Moreover isn't the energy required to pair electrons in d orbitals greater than pairing electrons in s. $\endgroup$
    – jyoti proy
    Commented Sep 25, 2017 at 10:54
  • $\begingroup$ related chemistry.stackexchange.com/questions/2469/… $\endgroup$
    – Mithoron
    Commented Sep 25, 2017 at 15:30
  • $\begingroup$ possible duplicate of chemistry.stackexchange.com/questions/2660/… and of chemistry.stackexchange.com/questions/35487/… $\endgroup$
    – Mithoron
    Commented Sep 25, 2017 at 15:32
  • $\begingroup$ @Mithoron but why then $\ce{Ni}$ is different, it should also have followed $\ce{Pd}$ or $\ce{Pt} $. Moreover we see that elements belonging to a certain group show similar electronic configuration, but why in this particular group different elements take different paths to attain stability. Don't mind but I have asked about the group as a whole not the single exceptions. Btw the links u provided are helpful 😊. $\endgroup$
    – jyoti proy
    Commented Sep 25, 2017 at 18:27

2 Answers 2


I think the relevant comparison should be with the earlier transition metal groups where $s^2$ configurations are more predominant.

In multielectron atoms, electrons interact in such a way that for a given shell quantum number $n$, the orbitals with higher total angular momentum $l$ are raised, thus for example $3d$ ($n=3, l=2$) becomes higher than $3p$ ($n=3, l=1$). This angular momentum effect is most important when the orbitals are well shielded with a low effective nuclear charge, where the electron-electron interactions don't get swamped by electron-nucleus interactions.

In earlier transition groups the valence shells have this low effective nuclear charge and the angular momentum effect is so strong by comparison that the $d$ valence subshell is raised above the $s$ subshell even with one higher $n$ quantum number. So, for instance, in the fourth period we see $3d$ orbitals filled only after $4s$ in neutral atoms. Not so much in the ions, however, where ionization leaves the remaining electrons with more effective nuclear charge, hence we often see $d$ instead if $s$ subshell electrons remaining in the early transition ions.

But by the time we get to the later transition metals the effective nuclear charge has increased even in the neutral atoms, the lower-$n$ $d$ subshell drops relative to the higher-$n$ $s$ subshell, and we begin to see a greater preference for the $d$ subshell being filled first. By the time we get to Group 12, this transition is complete and only the $s$ electrons remain valence electrons except, possibly, in extreme circumstances.


Here Platinum shows d9 s1 configuration due to the relativistic effect of 6s orbital. Which is inert towards oxidation.
But in case of Pd it is d10 configuration due to the full filled highly stable 4d orbital. But Ni cannot show this configuration. When it comes to the case of Ni , it has 3d orbital in its valence shell so it is not so much large and cannot supply spin pairing energy. So the configuration show d8 s2.That's why this group element show analomous behavior.

  • 1
    $\begingroup$ Welcome to Chemistry.se! From these statement, I cannot really connect the dots. What do you mean by 'cannot supply spin pairing energy'? $\endgroup$ Commented Mar 3, 2019 at 9:51

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