Can a high-spin complex be diamagnetic?

Question

There are two nickel complexes: one is octahedral and the other square planar. One complex is diamagnetic, the other is paramagnetic. Both are high-spin.

Now, $\ce{Ni^2+}$ that is octahedral should be paramagnetic, as should $\ce{Ni^2+}$ that is square planar. Actually any high-spin complex should be paramagnetic. Is it possible that the octahedral $\ce{Ni^2+}$ is paramagnetic, and square planar $\ce{Ni^2+}$ diamagnetic (achieved by not filling the last orbital even tho that goes against high-spin concept)?

Answer 1

A high-spin complex cannot be diamagnetic. By definition, a high-spin system has multiple unpaired electrons. The names "high-spin" and "low-spin" refer to the total electron spin. In the following example ($d^6$ octahedral), the high spin state $S = 2$ and the low spin state $S = 0$.

In an electron configuration such as $d^8$ the outcome is independent of whether electrons were filled by pairing in the $t_{2g}$ orbitals first, or by filling all orbitals and pairing last. You end up with the same configuration. In this case, there is no such thing "high-spin" or "low-spin" states.

The same is true for $d^0$, $d^1$, $d^2$, $d^3$, $d^7$, $d^9$ and $d^{10}$. Octahedral is really the only geometry with both high- and low-spin states, as this is the only geometry where the splitting energy $\Delta$ is comparable to the spin-pairing energy.

Answer 2

Both are high-spin.

No, square planar complexes are nearly always low-spin. With a $\mathrm{d^8}$ configuration, the lowest four orbitals would be filled. That would make the complex diamagnetic.

For the octahedral $\mathrm{d^8}$ complex, regardless of whether it's high- or low-spin, it will have the same electronic configuration $\mathrm{(t_{2g})^6 (e_g)^2}$. By Hund's first rule, the two electrons in the $\mathrm{e_g}$ orbital will both be unpaired, so the complex will be paramagnetic.

Here's a quick sketch (energy levels not to scale):