- I recently added an answer to Why is [Ni(en)3]2+ paramagnetic?
Atomic number of Ni is 28, $\ce{Ni^{2+}}$ has configuration $\mathrm{[Ar]3d^8}$. It can not form lower spin octahedral complex due to unavailability of 2 vacant d orbitals. Thus, will form $\mathrm{sp^3d^2}$ complex. Similar is the case for $\ce{[Ni(NH3)6]^2+}$...
I just wish to know, is my explanation wrong? It might be too narrow, but still, I feel it's right...
Also I earlier asked What exactly are frontier orbitals?, why was it closed? This question troubles me, thus asked... How shall I make it more clear?
I also didn't get answer for Reason behind more prioritising of normal spinels, is the question not at all good to be answered or even commented upon? Anything that I shall do for improving my question?
Also I got half the bounty amount for Is the number of tetrahedral voids always equal to double the number of octahedral voids in any crystal?, any particular reason?