# Shape of nickel tetracarbonyl

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.

• the Ni atom in the molecule is d10. Sep 2 '16 at 18:13
• 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? Sep 2 '16 at 18:15
• 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 Oct 3 '16 at 19:17
• 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? Oct 3 '16 at 19:25
• nitpick: $t_{2}$ not $t_{2g}$ for tetrahedral molecules Oct 4 '16 at 8:31

According to the work by Ladell et al., the structure of $$\ce{Ni(CO)4}$$ in the solid state ($$\pu{-55 \pm 5 ^\circ{}C}$$, ambient pressure) is tetrahedral with refined $$\ce{Ni-C}$$ distances of $$\pu{(1.84 \pm 0.03) Å}$$, and $$\ce{C-O}$$ distances of $$\pu{(1.15 \pm 0.03) Å}$$. You find the corresponding .cif file for example in the COD database filed there as entry COD 2310876:
Ladell, J.; Post, B.; Fankuchen, I. The Crystal Structure of Nickel Carbonyl, $$\ce{Ni(CO)4}$$. Acta Cryst. 1952, 5, 795-800; doi: 10.1107/S0365110X52002148.
Technical detail: Initially, the reference card of the entry displays only the unit cell and the crystallographic motif. To view the packing in the unit cell as shown in the illustration above, put your mouse over the interactive display provided by JSmol, and click once with the right-hand mouse button to open a new pull down menu. From here, choose entry Symmetry, followed by Reload {1 1 1} to display any molecule with at least one atom within the unit cell. (In the present case, the multiplicity equates to eight molecules per unit cell.) You may zoom-in and out with a mouse wheel, or walk around the model by click-and-drag with the left-hand mouse button.