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16 votes

Why is anhydrous copper(II) sulfate white while the pentahydrate is blue, even though both have one unpaired electron?

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, it changes from a $\ce{[Cu(H2O)4[SO4]...
Justanotherchemist's user avatar
16 votes

Why is Ni[(PPh₃)₂Cl₂] tetrahedral?

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 ...
Justanotherchemist's user avatar
13 votes

Why is Ni[(PPh₃)₂Cl₂] tetrahedral?

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 ...
Chakravarthy Kalyan's user avatar
12 votes
Accepted

Is cobalt(II) in the hexaamminecobalt(II) complex high spin or low spin?

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 ...
orthocresol's user avatar
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10 votes
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What chemical properties that allow for colour exist in the dark?

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 ...
matt_black's user avatar
9 votes
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In an octahedral complex, what happens to the electrons donated by the ligand?

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 ...
Pritt says Reinstate Monica's user avatar
9 votes
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Geometry of Ni(acac)2

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 ...
Nuclear Chemist's user avatar
8 votes

Why is hexafluoridocobaltate(III) the only common high-spin cobalt(III) complex?

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 ...
Jan's user avatar
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8 votes
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Why is a [Cu(SCN)2] complex black?

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 ...
Matt Cliffe's user avatar
8 votes
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Bonding in Zeise's salt

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 ...
Soumik Das's user avatar
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8 votes
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What is the shape of [Cr(NO)₄]?

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 ...
Oscar Lanzi's user avatar
7 votes
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Identifying a Positive or Neutral Ligand Inside Coordination Sphere?

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 ...
Jan's user avatar
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7 votes
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Crystal Field Splitting of d-Orbitals in Octahedral and Tetrahedral Ligand Fields

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 ...
Jan's user avatar
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6 votes
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Theoretically, Which d orbital participates in sp3d and sp3d2?

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, ...
Jan's user avatar
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6 votes

What's the configuration of chromium(III) in aqueous medium according to crystal field theory?

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 ...
Jan's user avatar
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6 votes

In an octahedral complex, what happens to the electrons donated by the ligand?

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 ...
NotEvans.'s user avatar
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6 votes
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Colour of diamagnetic compounds using Crystal Field Theory

As Porphyrin already stated in his answer, diamagnetism/paramagnetism has nothing to do with the colour of the compound. Are d-d transitions possible in diamagnetic complexes? It depends on the ...
Infinite's user avatar
  • 1,644
6 votes
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Colour properties of nickel complexes

OP's argument: According to Crystal Field Theory(CFT), complexes with stronger ligands must absorb light with higher frequency hence would transmit corresponding complementary color. First of all, ...
Mathew Mahindaratne's user avatar
6 votes
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Color change in coordination compounds

The choice of «intensity» is not a good one here, because the change from (pale) yellow to red relates to different positions of the absorption band(s) in terms of an energy split between orbitals / ...
Buttonwood's user avatar
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5 votes
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Why is CO ligand higher than CN- in the spectrochemical series?

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−. ...
orthocresol's user avatar
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5 votes
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Energies of d orbitals relative to the barycentre

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 ...
orthocresol's user avatar
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5 votes
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About Fe(III) and Fe(II) coordination complexes

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 $...
Jan's user avatar
  • 68.3k
5 votes

Sulfate and carbonate ions as ligands

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 ...
Jan's user avatar
  • 68.3k
5 votes
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Why must there be degeneracy in the orbitals to have a Jahn-Teller effect?

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-...
Ian Bush's user avatar
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5 votes

What is the magnetic moment of tris(oxalato)nickelate(IV)?

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 ...
Jan's user avatar
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5 votes
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Why are electrons paired up in hexaaquacobalt(III)?

Whether a complex is high or low spin does not solely depend on the ligands attached. There are other factors such as the size of the metal, the charge on the metal, the coordination of ligands, and ...
M.L's user avatar
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4 votes

How can the electronic structure of pentaaquanitrosyliron be explained?

The presence (or absence) of a single strong field ligand is rarely enough to tip the entire complex’ balance from high to low spin (or vice-versa). I have often come across questions here that attemt ...
Jan's user avatar
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4 votes
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Deducing orbital degeneracy in geometries apart from octahedral or tetrahedral

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 ...
orthocresol's user avatar
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4 votes
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Why do ligands approach along the axes?

You basically said it. They can approach from anywhere. Well, I can define the axes to be anywhere. An isolated atom in vacuum with degenerate d-orbitals does not know where $x,y$ or $z$ are. ...
Jan's user avatar
  • 68.3k
4 votes

Colour of metal complexes with the same type and number of ligands

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 ...
AChem's user avatar
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