New answers tagged

2

Concise Inorganic Chemistry1 has the following to say about this tetra(cyclopentadienyl) titanium complex: Tetra(Cyclopentadienyl) compounds such as $\ce{Ti(C5H5)4}$ may be made from $\ce{TiCl4}$ and $\ce{Na(C5H5)}$. The formula may be written as $\ce{[Ti(\eta_5-C5H5)2(\eta_1-C5H5)2]}$ where two cyclopentadienyl rings are attached by 5 C-atoms ($\pi$ - ...


8

This depends a lot on the compound in question. In general, a bidentate acetate would lead to a four-membered ring with the central metal; since the $\ce{O-C-O}$ angle is relatively fixed at approximately $120^\circ$ and since we expect the resulting compound to be at least somewhat symmetric, it would typically lead to unusually small bond angles or a $\ce{...


3

I have found the answer: In the following case: $$\ce{[Fe(NH_3)_3(CN)_3][Co(CN)_3(NH_3)_3]}$$ Both the constituent entities are neutral and hence will exist independently, and thus will not form isomers of the original compound.


4

Py and Cl both ligands are monodentate . Therefore, the coordination number of Pt in both the cases are 4 and Pt in general shows +2 oxidation state for 4 coordination number. Again py is neutral ligand whereas Cl is anionic (-1). So, $\ce{[Pt(py)4]^2+}$ is positively charged and $\ce{[PtCl4]^2-}$ is negatively charged. In IUPAC nomenclature, for a ...


6

When a ligand has multiple donor sites, but uses only say $x$ of them at a particular time, then it's called ambidentate ligand. However, if it uses different number of sites in different compounds (say $x$ in one and $y$ in the other where $x\ne y$), then flexidentate is a more appropriate term as said in the other answer. But this doesn't mean all ...


3

The simple answer for your query: the lone pair is practically immobile, and can't orient itself to chelate. To understand this, we first have to understand the structure of DABCO. The rings force a conformational rigidity on both the nitrogens, and therefore the fluxional process of nitrogen inversion is inhibited in these structures. Now, as the accepted ...


5

My copy of Concise Inorganic Chemistry, 5th edition (J.D.Lee, Wiley publishers)1 has nothing contrary to say about hydrazine or 1,4-diazabicyclo[2.2. 2]octane (henceforth referred to as DABCO). It is most likely that the author of your adaptation has used his liberty to add these facts in your copy. This then raises the question: is the author correct? ...


2

The bromide ions contribute an electron pair on each side of the square they form with the nickel centers (thus the bromine has a positive formal charge like a bridged bromonium ion), so in your fraction you need $4×2$ in the numerator where you have $2×2$. Then it will come out to $16$.


13

Although the question is a bit old, I think it still hasn't been answered yet. And by looking at the given answers it seems like the discussion went into a different direction at some point. So let's compare the two compounds, here I plotted the $\ce{Cu(II)}$ centers of $\ce{CuSO4.5H2O}$ and $\ce{CuSO4}$ from their crystal structure data. As you can see, ...


3

If the ligand has 3 donor sites and it is able to show multiple denticity in different compounds, then it will be called as flexidentate ligand. But as per your question, if it is only showing denticity of 2 only it will be called as bidentate ligand. Actually, this concept arose when arguments started for selecting donor sites for the " monodentate ligands ...


1

There are several complexes where methanol acts as ligand. Some examples are: methanol coordinated vanadium(V) chalcogenido complexes: $\ce{V(Chal)Cl2(OMe)(HOMe)2}$ where $\ce{Chal = O, S, Se}$ bis(μ-methoxo)dichromium(III) complexes, $\ce{[(L^{Se})2Cr2(μ-OCH3)2(CH3OH)2]}$ and $\ce{[(L^{Se})2Cr2(μ-OCH3)2(CH3OH)(CH3O)]}$, where $\ce{L^{Se}}$ represents the ...


7

There is an iron hexacarbonyl, not as a neutral compound but as a dication which satisfies the 18-electron rule. This along with its heavier Group 8 congeners is reported by Finze et al. [1]. The authors who report the tetrafluoroborate salts, also reference earlier work with the corresponding fluoroantimonates. The reference was found in Wikipedia. ...


3

Holleman/Wiberg states (translation by me) $\ce{Co^{3+}}$ usually forms octahedral low-spin complexes, because it is the only way to achieve a high ligand-field stabilization energy. Octahedral high-spin complexes are typically an exception; they are formed with the "weakest" ligands (fluoride): $\ce{[CoF6]^{3-}}$, $\ce{[CoF3(H2O)3]}$ (the ion $\ce{[Co(...


2

Ok, after having poked around a bit I doubt that $\ce{Pb(CN)2}$ forms. I'm guessing that you're collecting the froth to get a concentrate of the galena $\ce{PbS}$ out of some finely crushed ore. An aspect of the operation seems to be chemically reducing the concentration of $\ce{ZnS}$ by dissolving via the following reaction: $$\ce{ZnS(s) + 4NaCN(l) -&...


4

Aromatic rings that complex "face on" to a transition element are acting as pi acceptors all the time, including cyclopentadienide complexes. And they act simultaneously as pi donors too. Let's look at the familiar cyclopentadiende pi orbitals from a different perspective. We know that there are three bonding orbitals that should be occupied and two ...


Top 50 recent answers are included