Why is hexafluoridonickelate(IV) diamagnetic?
This is a common mistake to make when analysing transition metal compounds. The undergraduate or the non-coordination chemist looks at a complex, identifies a ligand, puts it into their box of weak-field or strong-field ligands, deduces an electronic structure and is then completely perplexed if something does not work the way the simple picture would assume.

It is then helpful to remember what the molecular orbital scheme of am octahedral complex looks like; refer to scheme 1. Note that scheme 1 ignores &#x3c0; contributions of the ligands but fret not for I shall briefly explain where (and how) they come in.


Scheme 1: Octahedral $\ce{[ML6]}$ complex with no &#x3c0; interactions. Image copied from this answer and originally taken from Professor Klüfers’ internet scriptum to his coordination chemistry course.

On the left-hand side of the scheme you can see the metal’s orbitals, from top to bottom they are (in nickel’s case) $\mathrm{4p\ (t_{1u}), 4s\ (a_{1g})}$ and $\mathrm{3d\ (e_g + t_{2g})}$. On the right-hand side, we have the six ligand orbitals which transform as $\mathrm{a_{1g} + e_g + t_{1u}}$. Only orbitals with identical irreducible representation interact constructively (or destructively) leading to the corresponding molecular orbitals shown in the middle.

What is commonly termed field split, $10~\mathrm{Dq}$ or $\Delta_\mathrm{o}$ is the difference between $\mathrm{t_{2g}}$ and $\mathrm{e_g^*}$ — the former being entirely metal-centred if &#x3c0; interactions are not considered, the latter being formed by both the metal and the ligand in a bonding-antibonding interaction.

Adding the ligands’ &#x3c0; orbitals, realise that they would transform as $\mathrm{t_{1u} + t_{1g} + t_{2u} + t_{2g}}$. Most of these interactions are either strictly non-bonding or contribute only weakly to bonding since the corresponding symmetry $\mathrm{t_{1u}}$ is already being used to make &#x3c3; interactions. Only the $\mathrm{t_{2g}}$ p orbitals introduce a new interaction with the central metal by turning what are nonbonding orbitals in scheme 1 into a bonding/antibonding pair of orbitals. Since the interaction is &#x3c0; type, it will be a weak splitting, but it contributes to the decrease of $10~\mathrm{Dq}$ by raising the bottom layer.

I have talked a lot but so far I did not mention the point why hexafluoridonickelate(IV) is a low-spin complex. However, the basis has been laid. It is important to visualise that the field split is — because it derives from the strength (or weakness) of bonding/antibonding interactions in the molecular orbital scheme — dependent both on the metal’s and the ligands’ properties. This can be visualised in the very simple (and thus not satisfying; when approaching a higher level) equation $(1)$:

$$10~\mathrm{Dq} = g_\mathrm{M} \times f_\mathrm{L}\tag{1}$$

In this formula, $g_\mathrm{M}$ represents the contribution of the central metal while $f_\mathrm{L}$ that of the ligands. Obviously, if we modify the metal in a certain way, we can significantly influence the value of $10~\mathrm{Dq}$, i.e. the field split, i.e. whether the complex prefers a high or a low-spin state. Enough circling the issue, let’s get to the point.

The preference for high or low spin can be greatly influenced by the metal’s oxidation state. Remember that a positive charge attracts electrons, so that their respective orbitals will (all!) be lowered in energy by oxidation. Thus, the energies of the metal’s d orbitals and those of the ligands’ &#x3c3;-symmetric orbitals will move closer together creating a larger stabilisation/destabilisation. At the same time, the &#x3c0; symmetric ligand orbitals will be more removed energetically and thus the weak &#x3c0; interaction will be further diminished. These effects in combination lead to a greater field split for high central metal oxidation states. In hexafluoridonickelate(IV), the oxidation state is $\mathrm{+IV}$ — pretty high for nickel which is probably most commonly found as nickel(II) in coordination complexes. Thus, it is not surprising that this higher oxidation state leads to a low-spin complex. Low-spin $\mathrm{d^6}$ is, of course, diamagnetic.

For further exemplification, I quote two values from a table of $g_\mathrm{M}$ values Professor Klüfers included in the linked internet scriptum (in German):

$$\begin{array}{cc}\hline
\ce{M^n+} &amp; g_\mathrm{M}/ 1000~\mathrm{cm^{-1}}\\ \hline
\ce{Co^2+} &amp; 9 \\
\ce{Co^3+} &amp; 18.2 \\ \hline\end{array}$$

As can be seen from these values, increasing cobalt’s oxidation state from $\mathrm{+II}$ to $\mathrm{+III}$ doubles the metal’s contribution to $10~\mathrm{Dq}$. And that was only an increase of one.