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On negative oxidation states, in general Although it's usually a topic that's covered relatively late in a chemistry education, negative oxidation states for transition metals[1] are actually quite alright. On the Wikipedia list of oxidation states, there are quite a number of negative oxidation states. Some textbooks have tables which only show positive ...


17

Your book was correct that a five coordinate metal complex is able to adopt both square pyramidal and trigonal bipyramidal geometries, and in both cases the sp3d hybridisation scheme applies (if you believe in hybridisation...). Which geometry is adopted depends upon a combination of steric and electronic factors, and isn't necessarily trivial to predict, ...


9

In this complex there are two different 31P environments which are not related by symmetry: The two green phosphorus nuclei can be interconverted by a $C_2$ rotation (the rotation axis bisects the OC–Mo–CO angle), and so can the purple ones, but green and purple cannot be interconverted. As extra proof, consider that the green P is cis to both carbonyl ...


9

tl;dr: There probably isn’t an Fe–Fe bond. We should probably stop drawing an Fe-Fe bond. Textbooks and lecturers should really stop teaching students that there is an Fe-Fe bond. But all that said, there isn’t one single satisfactory explanation for why there isn’t one… Despite a lot of studies (and despite $\ce{[Fe2(CO)9]}$ being one of the earliest ...


7

The name of this compound is carbonyl(η5-cyclopentadienyl)(trimethylphosphane)cobalt. Note that, generally, the prefixes denoting the organic groups and any other ligands are placed in alphabetical order before the name of the metal. The current version of Nomenclature of Inorganic Chemistry – IUPAC Recommendations 2005 (Red Book) presents the rules for ...


7

To supplement the accepted answer, here is something else I found. Bonding in bridging carbonyls An alternative explanation for the bonding in $\ce{[Fe2(CO)9]}$ is presented in a 2012 article by Green1 (which also contains numerous references to earlier computational studies indicating the absence of a bond). In it, the authors describe Fe–C(O)–Fe bonds as ...


7

The most recent single crystal structure investigation (high precision, $R_1 = 1.9\%$, $wR_2 = 5.7\%$) of molecular iron pentacarbonyl [1] suggests trigonal bipyramidal coordination environment (also mentioning anti-Berry pseudorotation), as shown by the calculated geometry index $\tau_5$: \begin{align} \begin{cases} \tau_5 &= 0 \qquad &\text{square ...


6

Ignore list of "allowed oxidation states" for metal complexes. "allowed oxidation states" might be the most common, but definitely not the only ones the atom is allowed to be in. For example, for $\ce{Fe}$ all oxidation state from -2 to +6 are known, for $\ce{Mn}$ -3 and from -1 to +7 are known, for $\ce{Ir}$ it is -3 and from -1 to +8 with tentative +9 ...


6

Unfortunately, you don't give an indication of what you expect the chemical shift for the -Me to be, other than higher than is observed. However, you are correct in your assumption; the π back donation from the Mn centre to CO ligands does decrease the shielding, and therefore does increase the chemical shift, but this is a good case of how chemical ...


5

The hypothetical $\ce{Fe(CO)4}$ complex is isoelectronic to the observed $\ce{[Ni(CN)4]^{2-}}$ complex and therefore would also be square planar.[1,2] Both are 16 electron complexes, i.e. one of the valence orbitals of the central atom remains unoccupied. Your assessment for $\ce{Fe^{\pm0}}$ as $\mathrm{d^8}$ is absolutely correct, also that carbonyl is a ...


4

It is possible for carbon monoxide to coordinate through oxygen, as reported here. However, there is indeed a strong tendency for carbon monoxide to coordinate through carbon, essentially due to the electronegativity difference noted by the OP. Cyanide is more likely to coordinate through N than carbon monoxide through O, simply because the ...


4

The main point is to stop the heat flow from the lamp, while allowing UV through. You want to do photochemistry, not thermochemistry.


4

Some months later but here we go... I agree with most of the comments that multiconfigurational effects should be present for this system, and maybe they are very strong. But I tend to think that this is not the cause of the huge difference with experimental values. The fact that B3LYP returns reasonable values should be taking into account. Surely it "...


3

From the abstract of J. Chem. Theory Comput., 2011, 7, 2112-2125 (DOI) is seems that niobium forms a dinuclear $\ce{Nb2(CO)12}$ complex, in which two $\ce{Nb(CO)6}$ units are connected by a 340 pm $\ce{Nb-Nb}$ bond.


3

I think the mistake you're making is this conclusion: "Higher atomic charge. . . -> higher electron density -> more pi-backbonding". With higher (ie more positive or less negative) atomic charge, there is less electron density donation towards the ligand, because the electron density is pulled to the metal center more strongly. Thus, $\pi$ back-bonding is ...


2

Yes there are more carbonylate ions than metal carbonyl cations, but the latter do exist. A salt $\ce{[Mn(CO)6^+][BF4^-]•SO2}$ has been synthesized in a superacid medium. Metal carbonyl cations are often disfavored because they have to contend with counterions that are anions, which may compete with the carbon monoxide ligands. Such a competition is ...


2

Going to assume the solvent is CDCl3. It is true that pi Back-donation withdraws electron density from the Metal Center to deshield it, but also note that carbonyl groups are still electron donating and the Methyl group can be treated as negatively charged (additional shielding).


2

Also we need some empty orbitals on the metal cation, since we're looking at Lewis adducts. Ahem. Nope. $\pi$-bonding in $\ce{CO/NO+}$ complexes is from back-donation, from occupied metal d-orbitals onto unoccupied $\pi^*$ orbitals of the ligand. Looks like homework question, so no further details provided.


1

Aluminium chloride ($\ce{AlCl_3}$) is mainly used for this Friedel-Crafts alkylation or acylation, where this compound acts as catalyst, and also controls the production of the desired compound. For example, in this case, the reactants are anisole ($\ce{Ph -OMe}$) and succinic anhydride. So, the main reaction which drives these reactants towards product is ...


1

Wikipedia (primary reference) suggests a possible reason for $\ce{Fe2(CO)9}$ dissolving preferentially in THF versus nonpolar solvents: it reacts according to the scheme $\ce{Fe2(CO)9 + THF <=> Fe(CO)5 + Fe(CO)4 \cdot THF}$ Such a reaction is invoked to account for the dinuclear complex giving mononuclear products with various ligands in THF. A ...


1

The reason is thus: For aldoses (like glucose), the copper sulfate in the reagent easily oxidizes it into the carboxylic acid forming copper oxide (a precipitate). For ketoses (like fructose), the ketone cannot be oxidized, so there is no color change. Over time at elevated temperature, the α-hydroxy ketone isomerizes to a α-hydroxy aldehyde. This aldehyde ...


1

First off, be aware that oxidative addition occurs by several mechanisms which are governed by different factors.[1] As you move down a period, the outer electrons become more easily accessed, as the outer electrons are increasingly screened from the nuclear charge by the inner electrons.[2] This is why both ionization energy and electron affinity decreases ...


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