# How does Potassium Hexfluoronickelate(IV) exist in a stable state?

Nickel generally does not exist in a $$+4$$ Oxidation state. Wikipedia has the following to say regarding Nickel(IV):

Ni(IV) is present in the mixed oxide $$\ce{BaNiO3}$$. ....... Ni(IV) remains a rare oxidation state of nickel and very few compounds are known to date.

This article directly contradicts the first statement, as it says:

the photoelectron spectroscopy data suggest the description of $$\ce{BaNiO3}$$ in terms of trivalent Ni and oxygen anion holes.In terms of a local ionic model the idealized description of this high-valent oxometalate would correspond to $$\ce{Ba^2+Ni^3+(O2)^2O^•-}$$

Now, we end up coming to the curious case of Potassium Hexfluoronickelate(IV), which decomposes at 350°C, meaning that it is relatively stable. Not much literature is available on this compound online, except this article, which is unfortunately behind a paywall.

Could someone explain how nickel exists in a (IV) state in this compound instead of degenerating to $$\ce{K4[NiF6]^2-}$$? Also, are there any more Nickel compounds exhibiting the (IV) state? Explanations regarding the formation and stability of this compound are also welcome!

To quote from the comments: you can get things into an amazing variety of oxidation state given the right environment.

In the case of maximal oxidation states, particularly above $$+2$$ or $$+3$$ in the $$3d$$ transition series, the right environment (at least in simple species) involves complexation with ligands that are highly electronegative, capable of pulling electrons even from a strongly positive central atom, and resistant to oxidation themselves in an electron-poor setting. Meaning, in practice, oxide or fluoride ligands.

We can therefore roughly classify high oxidation state complexes into oxo and fluoro complexes. Oxo complexes are more likely to be seen with early transition metals, where the central metal ion contributes few $$d$$ electrons and there are a lot of vacant orbitals with which the strongly $$\pi$$-donating oxygen can form back-bonds. The best known examples being chromate/dichromate ($$\ce{Cr(VI)}$$), manganate and permanganate ($$\ce{Mn(VI), Mn(VII)}$$), and ferrate (most commonly, $$\ce{Fe(VI)}$$). Even cobalt reaches its apparent maximum $$+5$$ oxidation state with an oxidizing species, $$\ce{CoO4^{3-}}$$.

Later transition metals, which have lower maximum oxidation states due to increased electronegativity/ionization energy, have more $$d$$ electrons coming from the cationic shells and thus are no longer so stabilized by $$\pi$$ back-donation. Then the fluoro complexes, with a relatively $$\pi$$-inactive ligand, take over. In the $$3d$$ series cobalt, cited above as forming an oxide complex with the formula $$\ce{CoO4^{3-}}$$, also forms a +4 fluoride complex $$\ce{CoF6^{2-}}$$[1]. Fluoride complexes are also seen with both nickel and copper in oxidation states above $$+2$$, including the $$\ce{NiF6^{2-}}$$ ion mentioned in this question and the copper(III) species $$\ce{CuF6^{3-}}$$. Zinc in the $$3d$$ series is limited to oxidation states of $$+2$$ or less, but if mercury has tetrafluoride $$\ce{HgF4}$$ (disputed), it's no coincidence that this too is a fluoride.

Reference

1. J.W. Quail and G.A. Rivett, "Fluirude Complexes of Tetravalent Cobalt", Can J Chem 50:2447 (1972).

• Interesting. A lot of these fluoro complexes also adopt the $\ce{K2[PtCl6]}$ lattice structure and also seem to be similar in their properties. I'm assuming that all these fluoro complexes are low spin, given the high charge on the central metal atom. Could you please confirm? – Aniruddha Deb Jan 20 at 1:56
• At least for the $\ce{NiF6^{2-}}$ ion it is low-spin. – Oscar Lanzi Jan 20 at 2:16