The cis and trans isomers of $\ce{[CoCl2(NH3)4]Cl}$ are quite famous as one is purple and the other one is green which is a huge difference in color. Now I can't find any information to why this is the case?
I remember they told me in the lecture on organometallic chemistry when we were discussing substituted carbonylmetallates that a trans-effect occurs where a π-donor trans to the carbonyl (which binds to the metal's d-Orbitals via its antibonding π*-orbitals) increases the electron density in the metal and therefore also increases the the bond's strength ($\ce{M-CO}$) which then changes the bond length. A π-acceptor ligand on the other hand decreases the e-density.
In case of an octahedron the trans isomer would only possess same ligands trans positioned to each other while the cis isomer has mixed ones, too.
But is this the explanation I am looking for or is there another effect I just can't think of right now?
Answer
If you search a library for a copy of Inorganic Chemistry by Miessler and Tarr, they have an excellent diagram showing which explains the phenomenon.
$\ce{Cl-}$ ligands have lone electrons that can donate into the metal, as you said yourself. The trick is to recognize what effect this has on the $t_\mathrm{2g}^*$ and $t_\mathrm{2g}$ orbitals. Remember that the five frontier orbitals in an octahedral configuration are the $t_\mathrm{2g}$ and $e_\mathrm{g}^*$ orbitals.
When a $\ce{Cl-}$ ligand pi-donates into the metal center, it can only interact with the $\mathrm{d}_{xy}$, $\mathrm{d}_{xz}$, $\mathrm{d}_{yz}$ $(t_\mathrm{2g})$ orbitals because the $\mathrm{d}_{x^2-y^2}$, $\mathrm{d}_{z^2}$ $(e_{g}^*)$ orbitals have only a sigma-type interaction with the ligand (for them, there is no node, draw this to ensure yourself). When these ligands like $\ce{Cl-}$ pi-donate, they fill the $t_\mathrm{2g}$ bonding orbitals, widening the gap between them and the open the $T_\mathrm{2g}^*$ antibonding orbitals. These antibonding orbitals become much closer in energy to the $e_{g}^*$ orbitals, so you reduce the octahedral field splitting energy $\Delta_\mathrm{O}$ which redshifts the absorbance of the compound.
The reason the trans isomer is more redshifted (green as opposed to purple) is due to the fact that in the trans isomer, the ligand-metal pi overlapping does not clash. On the other hand, if you draw the ligand-metal pi overlapping generated by two cis $\ce{Cl-}$ ligands, you will find that there is always one orbital that contains a lobe where the ligand overlap does clash. When this clash does exist, it becomes less favorable for the $\ce{Cl-}$ ligands to pi-donate into the metal via that clashing orbital -- so, the $t_\mathrm{2g}^*$ antibonding energy level is closer in energy to the $t_\mathrm{2g}$ bonding energy level, and further in energy from the $e_\mathrm{g}^*$ energy level. In the cis compound, field splitting $\Delta_\mathrm{O}$ is higher and the color should be purple instead of green.
If you are having trouble seeing the clash, draw a $\mathrm{d}_{xy}$ orbital on a metal $\ce{M}$, and then draw two $\ce{Cl-}$ ligands such that $\ce{Cl-M-Cl}$ is a right angle (cis ligands). When you try to show pi-donation from both ligands, you will find that there is one lobe of the $\mathrm{d}_{xy}$ orbital that is used by both; that is the clash that is disfavored.
No comments:
Post a Comment