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Palladium follows an exception to normal electron filling-up rules and so has the electron configuration of $\ce{[Kr] 4d^{10} 5s^0}$

The oxidation states of palladium are $\ce{+II}$ and $\ce{+IV}$. How can these oxidation states be explained by looking at the special electron configuration of $\ce{Pd}$?

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Ions of the transition metals do not often occur "naked" the way one might think of a sodium ion. Of course, sodium ions cannot occur by themselves either, but let's compare the two for a moment.

Sodium nitrate, $\ce{NaNO3}$

When $\ce{NaNO3}$ dissolves in water, the two ions dissociate.

$$\ce{NaNO3 -> Na+(aq) + NO3- (aq)}$$

Both ions are solvated by the solvent. The water molecules (which are polar) surround the ions. The partial charges due to the dipoles in the water molecules stabilize the charges on the ions. However, the interactions between the sodium ions and the water molecules are fleeting. These interactions are not strong enough to be consider "bonds" (there is no covalent electron sharing between water and sodium ion). Thus, we often consider sodium ions in solution (despite their solvation) to be "just" sodium ions.

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Palladium nitrate, $\ce{Pd(NO3)2}$

Transition metal ions are different. They tend to form complex ions by coordinating solvent molecules. These dative interactions are far stronger than the ion-dipole interactions in solvated sodium ions.

Palladium nitrate does dissociate, but the palladium ion picks up water molecules that it holds tightly. Thus, we do not worry about the electron configuration of $\ce{Pd^{2+}}$, since it never exists on its own. It is always part of a complex ion or coordination compound, and we care about the electron configuration of that complex ion.

$$\ce{Pd(NO3)2 + nH2O -> Pd(H2O)6^{2+}(aq) + 2NO3- (aq)}$$

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  • $\begingroup$ But he didn't say solvated in water- Of course the ion in salt going to have some covalent nature. $\endgroup$ – user2617804 Jan 9 '14 at 6:29
  • $\begingroup$ The OP did not say solvated in water, but then there will always be a counterion. $\endgroup$ – Ben Norris Jan 9 '14 at 13:07

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