# What is the shape of [Cr(NO)₄]?

I was learning about crystal field theory recently. I am quite confused about predicting the shape of $$\ce{[Cr(NO)4]}$$ complex because I am not able to conclude whether it will be square planar or tetrahedral.

How to predict the shape of $$\ce{[Cr(NO)4]}$$ using crystal field theory?

Will $$\ce{NO}$$ act as a 3-e- donor here?

Will all the four $$\ce{NO}$$ that are in coordination with $$\ce{Cr}$$ be of the type nitrosyl or will three of them be nitroso and one nitrosyl (because $$\ce{Cr^3+}$$ is one of the common oxidation state of $$\ce{Cr})?$$

• The oxidation state is zero, and not +3. Are you sure that this compound exists ? I have not voted down. – Maurice May 29 '20 at 14:07
• @Maurice Thanks for answering one part of my question.Yes,I am sure that the compound exists.You can check that here en.wikipedia.org/wiki/Metal_nitrosyl_complex – Chem-Learner May 29 '20 at 14:09
• $\ce{[Cr(NO)4]}$ is very bad for CF theory. CF works well with complexes with strongly charged components. CO, NO and PR3 complexes do not really fit. – permeakra May 29 '20 at 14:52
• I quickly googled and found a paper claiming that the system is best viewed as complex of $\ce{Cr^+}$, but actually is very similar (and isoelectronic) to $\ce{Ni(CO)4}$ – permeakra May 29 '20 at 14:54
• Both are 18 electron compounds (en.wikipedia.org/wiki/18-electron_rule). The CO ligands donate 4x2=8 electrons to Ni make 8+10=18. The NO ligands donate 4x3=12 electrons to make 12+6=18. They are therefore isoelectronic. The nickel compound is tetrahedral. So therefore is the Chromium compound. – Ian Bush May 29 '20 at 15:24

Based on available theoretical considerations and a available literature, a tetrahedral geometry appears to be a good assumption for $$\ce{Cr(NO)4}$$.
1. The NO ligand can be rendered as a three-electron donor, if we take it as uncharged. It is actually similar to CO, really, in that it interacts through its $$\pi^*$$ orbitals as well as its $$\sigma$$ orbital; but NO unlike CO has an electron to offer from its $$\pi^*$$. That plus the $$\sigma$$ electrons makes three.
2. We then may identify the central metal atom as having 18 valence electrons. When this happens with multiple ligands we typically see an arrangement with highest order symmetry, thus tetrahedral for four-coordination. This gives the best possible bonding to the entire set of valence orbitals involved in the 18-electron structure. This is seen in the tetrahedral four-coordinationn of $$\ce{Ni(CO)4}$$, and in the octahedral six-coordination of the series $$\ce{[V(CO)6]-, Cr(CO)6, [Mn(CO)6]+, [Fe(CO)6]^{2+}}$$. So, we should expect $$\ce{Cr(NO)4}$$ to be tetrahedral.
3. Heberhold and Razavi found evidence for a tetrahedral structure in $$\ce{Cr(NO)4}$$ from its IR spectrum.