Can a high-spin complex be diamagnetic? 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)?
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):