14

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. ...


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, ...


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 ...


9

They are equivalent in energy because symmetry says so, and because the calculations give that as an answer. Okay, that wasn’t exactly satisfactory, was it? The next complicated answer on the road from over-simplified to reality is that the $\mathrm{d_{z^2}}$ extends to both ligands, too: Do not forget the ‘ring’ in the $xy$ plane. By extending towards the ...


8

The chemical property that creates colour is the ability to absorb light of a specific visible wavelength. There is more than one way to do this. Mostly colour is caused by the existence of electronic transitions in substances that match the energy of some wavelength of light so when light hits the substance, some is absorbed by exciting electrons from a ...


8

It is a good question, it is a gemstone hiding in the mud. I have searched the Cambridge database, all the bis-acetylacetonatonickel complexes which have four coordinate nickel centres are square planar. There are three common coordination geometries for nickel(II) which we need to consider. Tetrahedral, square plannar and octahedral. Here is a diagram (...


7

The labels you are talking about refer to the orbitals’ irreducible representations. Which irreducible representations can occur is dictated by the point group of the entire compound. For octahedral compounds, the point group is the very symmetric $O_\mathrm{h}$, which basically includes every element of symmetry you can have in a cube. Tetrahedral complexes ...


7

The bond between $\ce {Pt} $ and ethene is not purely a $\sigma $ bond and also not a pure $\pi $ bond. The bonding actually consists of a combination of both bonding and back-bonding effect. If you look into the Molecular Orbital Diagram of Ethene, the picture will be more clear. For the formation of bond between the metal and the ligand, electrons from ...


6

The answer to this lies in the symmetry of the orbitals involved. Naturally, an octahedral complex will have the point group $O_\mathrm{h}$ — the octahedral point group. In molecular orbital theory, we will first need to work out how the ligand group orbitals transform; then we can find out which metal orbitals transform identically, i.e. belong to the same ...


6

This is a very good question, I must say. It requires the understanding of the very fundamentals. You're right, if the electrons from the ligand pair up with the electrons of the metal, the electrons cannot undergo $d$-$d$ transitions, and no color would be seen. But do they pair up? Think carefully. Recall one of the most basic assumptions of the ...


6

You nailed it, this is almost a chicken-and-egg problem. As long as we don’t know which complex will form, we cannot say which orbitals are populated. This is why it is often and regularly repeated that you need to know the geometry before you can make a call. In an octahedral complex, the $\mathrm{t_{2g}}$ orbitals will always be populated first as they ...


6

As the user above states, if you mix concentrated solutions of $\ce{Cu(II)}$ salts and $\ce{NCS}$ you get $\ce{Cu(NCS)_2}$ which is a black solid. I'd disagree with them that it's not longer a complex because it 100% is - and you are correct to assume that the local coordination is $\ce{Cu(NCS)_4}$. To be more precise, each copper is coordinated by $\ce{Cu(...


6

Theoretically, you cannot predict a priori whether a compound is high- or low-spin. For some reason, a lot of people seem to think that it depends only on the ligand and that it is possible to unambiguously use the position of the ligand in the spectrochemical series to figure out whether a complex is high- or low-spin. This is a very narrow viewpoint and ...


6

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; but ...


5

An alternative to the crystal field theory approach presented in Pritt Balagopal's answer is to make use of ligand field theory. Neither theory is perfect, and crystal field theory can explain many properties of coordination compounds (even if not always for the right reason). Ligand Field Theory Ligand field theory (LFT from here on in) is a newer and more ...


5

The passage you quoted is perfectly correct and contains one word which you may have overlooked: the weighted mean of these two sets of perturbed orbitals is taken as zero [...] In this case the weight corresponds to the degeneracy of the set. The lower-energy $\mathrm{t_{2g}}$ set contains three orbitals and the higher-energy $\mathrm{e_g}$ set contains ...


5

Books have been written on the topic of whether a certain nitrosyl ligand is the nitrosyl cation, the nitrosyl radical or — you forgot this one — the nitrosyl anion $\ce{NO-}$. This ligand is termed a non-innocent ligand, meaning we cannot just look at it and tell what it is. You asked for an answer that does not require IR or Raman spectroscopy. Well sadly,...


5

Your example is a little questionable. I would think you are talking about $\ce{[Fe(H2O)6]^2+}$, in which case you are actually supposed to have an $\ce{Fe^2+}$ ion with a $[\ce{Ar}]\mathrm{(3d)^6}$ configuration. There are no $\mathrm{4s}$ electrons in the $\ce{Fe^2+}$ ion. Nevertheless, it is true that in crystal field theory, electrons are assumed to be ...


5

In the abstract to the original Jahn-Teller paper it is stated:[1] We shall show that stability and degeneracy are not possible simultaneously unless the molecule is a linear one. Thus the Jahn-Teller effect is specifically about the removal of a degeneracy, and that that will always lower the total energy of the system. Thus you can not have a Jahn-...


4

Both elements are stable under certain conditions in various oxidation states. In chromium they range from -2 to +6 and in manganese from 0 to +7. However, some of them are more common than others, especially in aqueous solution. In chromium the most common are +3 and +6. As a "free" ion $\ce{Cr^{3+}}$ is most stable due to a filled s orbital and only one ...


4

The easiest way to do it is to look at what other people before you have done. If you look at the $D_\mathrm{3h}$ character table, you see that $(x^2 - y^2, xy)$ transform together as $E'$. I don't think there is a way of looking at diagrams of the orbitals and concluding that they are necessarily degenerate. As you said, there isn't "enough information to ...


4

This is a question that the crystal field theory cannot answer because of its inherent shortcomings. The crystal field theory assumes point-type charges approaching the central metal from the coordinate axes leading to the destabilisation of those $\mathrm d$ orbitals that do not feature a nodal plane in the coordinate axes: $\mathrm{d}_{x^2-y^2}$ and $\...


4

When electrons come back down they emit the same wavelength of light and that's the colour we see. I think this is a major misconception held by students. The color of these transition metal complexes is due to the absorption of light in the visible region. Very crudely, imagine you have six colors in the rainbow, if you remove a certain color portion from ...


4

First and hopefully obvious things first: sulphate ions do form (weak) coordinate bonds to metal centres and thus by definition cause some d orbital splitting. In a simplified crystal field model, you might imagine them as weaker negative charges but nonetheless the do interact or we would not see the structures we see. Comparing the structures of anhydrous ...


4

If the compound is really nickel(IV), then there are only two choices: a high-spin $\mathrm d^6$ or a low-spin $\mathrm d^6$ configuration, corresponding either to 4 or 0 unpaired electrons. As higher oxidation states make low-spin configurations more likely (due to the lower energy of metal orbitals which means the energy is more similar to the ...


3

Yes, the methodology is the same that is used to derive the crystal field theory a priori. Your course should have taught you that when it introduced crystal field theory, otherwise it is not worth its money. Crystal field theory assumes that the ligands will approach the central metal in a certain manner and that these ligands will be point-shaped negative ...


3

The colour will depend on several factors. You've identified the oxidation state (+5 in both cases) as one. Other factors are: Ligands Complex shape With $\ce{[VO2]}^{+}$, the actual complex ion is $\ce{[VO2(H2O)4]}^{+}$ (Taken from ChemGuide). So our ligand is water, and the shape is tetrahedral (or square-planar). The $\ce{[VO4]^{3-}}$ ion usually only ...


3

You would expect a sharp peak for a well-defined structure which has bond lengths and angles that do not significantly change. In a crystal structure at low temperatures you may be able to observe sharp peaks rather than broad bands. In solution, however, there is a number of effects that broaden the peaks, making the entire thing less sharp: aqua ligands ...


3

You're on the right track. It has to do with the energies of the frontier orbitals. As you rightly said, both species are isoelectronic, and the orbital energies in CO are lower than those in CN−. The lower HOMO energy means that CO is a poorer σ donor orbital towards the metal than CN−. Likewise the lower LUMO makes it a better π acceptor. These two ...


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 $\mathrm d^9$ ...


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