12

We sometimes call this type of complex 'pseudotetrahedral' since there is an isomerism from a tetrahedral to a square planar complex possible. I was unable to find the original work here but this link gives some information. As you already mentioned there are two strong and two weak ligands so it's hard to tell how strong the ligand field splitting will be. ...


10

Dichlorobis(triphenylphosphine)nickel(II), or $\ce{NiCl2[P(C6H5)3]2}$ in square planar form is red and diamagnetic. The blue form is paramagnetic and features tetrahedral Ni(II) centers. Both tetrahedral and square planar isomers coexist in solutions. Weak field ligands, favor tetrahedral geometry and strong field ligands favor the square planar isomer. Both ...


5

To determine what orbitals do, it is often a good idea to look at the symmetry group as it can provide a rough overview over what can (symmetry-allowed) and cannot (symmetry-forbidden) happen. In the case of a quadruple bond such as $\ce{[Re2Cl4]^2-}$, we can approximate the rhenium(III) centres as being close to $C_\mathrm{4,v}$. A quick glance at the ...


5

Chelation therapy, which is a very effective way to treat heavy-metal poisoning, is a chemical process in which a synthetic solution is injected into the bloodstream to remove heavy metals and/or minerals from the body. British Anti-Lewisite (BAL) was one of the first chelating agents to be developed as an anti-dote for war gas, dichlorovinyl arsine (Lewsite)...


5

Lead, mercury and other heavy metals form complexes with both monodentate and polydentate ligands. But the complexes made with polydentate ligands like EDTA are much more stable than with monodentates. Usually metallic complexes become toxic if the central metallic ion (non-complexed) is released or not entirely surrounded by ligands. Monodentate complexes ...


3

Based on available theoretical considerations and a available literature, a tetrahedral geometry appears to be a good assumption for $\ce{Cr(NO)4}$. 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; ...


3

If you inspect Table II in Ref. 1, which you will find referenced in the link you provide to the WebElements website, you'll see that AU units are used (also in the linked page), not pm, so the radius of the 3d Cu orbital in Ref. 1 is actually $$ r_{\textrm{Mann}} = 0.613 \times 0.529 Å = 0.324 Å $$ or 32.4 pm. As for Slaters approach, it is not enough ...


3

Octahedral complexes are somewhat of a ‘standard’ for transition-metal complexes. Not only can they happen if there are too few electrons to satisfy the 18-electron rule (e.g. $\ce{[Ti(H2O)6]^3+}$, a $\mathrm d^1$ complex) but also if there are too many electrons to satisfy the 18-electron rule (Jahn-Teller distorted $\ce{[Cu(H2O)6]^2+}$, a $\mathrm d^9$ ...


1

I made both isomers as a PhD student many years ago. The red diamagnetic form crystallises out when a concentrated CH2Cl2 solution is cooled in a dry ice-acetone bath, and you can quickly filter off the red product. But on leaving the solid at room temperature, it reverts in a couple of days at room temperature to the blue, paramagnetic form. So it's a ...


1

This is no Iron(I)! The idea that Fe(I) is present in this famous brown ring test is outdated. NO is a non-innocent ligand and will take the form of $NO^-$ here while the Fe is in the oxidation state +III. Even if you think about how you prepare it. There are no real Fe(I) compounds (I mean simple ones). And Fe(II) oxidizes really easily to Fe(III) and you ...


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