Is nickel tetracarbonyl a square planar complex or a tetrahedral complex?

According to the crystal field theory, $\ce{CO}$ is a strong field ligand and should hence cause the $t_{2g}$ electrons of nickel to pair up, hence leaving 1 $d$ orbital free, which must pair up with empty $s$ and $2p$ orbitals forming 4 $dsp_2$ orbitals yielding a square planar complex. But my textbook says its tetrahedral, with the electrons not pairing up and the outer d orbitals being used.

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    $\begingroup$ the Ni atom in the molecule is d10. $\endgroup$ – permeakra Sep 2 '16 at 18:13
  • $\begingroup$ So you're saying that the 4s2 electrons are demoted to the t2g orbitals, pairing them up and hence resulting in the formation of the tetrahedral complex? $\endgroup$ – user122143 Sep 2 '16 at 18:15
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    $\begingroup$ Yes! In the formation of a metal complex the 4s electrons are always "demoted" to the 3d orbitals. I would not quite put it that way, but in effect, that is what happens. d-electron count always includes the electrons that were originally from the 4s orbitals (unless they were ionised, of course). I've written an answer on it before: chemistry.stackexchange.com/questions/41962 $\endgroup$ – orthocresol Oct 3 '16 at 19:17
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    $\begingroup$ And now, with a $\mathrm{d^{10}}$ transition metal, no matter how strong-field the ligand is, you won't be able to derive any ligand-field stabilisation energy. So, electronic effects can't dictate the geometry here. What other factors might favour a tetrahedral complex over a square planar complex? $\endgroup$ – orthocresol Oct 3 '16 at 19:25
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    $\begingroup$ nitpick: $t_{2}$ not $t_{2g}$ for tetrahedral molecules $\endgroup$ – getafix Oct 4 '16 at 8:31

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