# Linked Questions

1answer
4k views

### Why are compounds of V5+ and Cr6+ colored? [duplicate]

As far as my knowledge goes, transition metal complex ions show a color due to a transfer of electrons between the levels of split $\mathrm d$-orbitals. In the case of $\ce{V^5+}$ and $\ce{Cr^6+}$ ...
0answers
401 views

### Color of potassium permanganate and potassium dichromate [duplicate]

In $\ce{KMnO4}$, the Oxidation state of $\ce{Mn}$ is +7, which results in an electronic configuration as $\ce{[Ar] 4s^0 3d^0}$ In $\ce{K2Cr2O7}$,the Oxidation state of $\ce{Cr}$ is +6, which results ...
0answers
50 views

### How is Mn+7 in the Permanganate ion considered as momentarily reduced to +6 in LMCT? [duplicate]

I saw this passage on a book and it says that Ligand to Metal Charge Transfer (LMCT) overpowers d-d transitions thus the permanganate ion has an intense purple color. However, it was mentioned that ...
0answers
45 views

### Why is potassium permanganate coloured? [duplicate]

$\ce{Mn}$ in $\ce{KMnO4}$ is in $+7$ oxidation state, has no d electrons so d-d transition is not possible. Then, how is $\ce{KMnO4}$ coloured?
9answers
7k views

### What are oxidation states used for?

When I took high school chemistry many years ago, considerable effort was spent on teaching us to compute oxidation states of atoms in various compounds, following a set of rules that looked somewhat ...
3answers
8k views

### Why is there a difference in the colours of K₂Cr₂O₇ and K₂CrO₄?

In both the compounds the oxidation state of Chromium is +6 so why is there a difference in the colours of their aqueous solutions. $\ce{K2Cr2O7}$ is red-yellow and $\ce{K2CrO4}$ is bright yellow I ...
2answers
3k views

### Color of chromate and permanganate

I've heard quite a few times that the chromate and permanganate have a $d^3s$ configuration. Also, their colors arise due to a rapid switching of electrons between the oxygen and metal atoms. I don't ...
3answers
4k views

### Can any atom donate more than 1 lone pair as a coordinate bond to the same metal?

This would mean the formation of a σ and one or two π bonds, all coordinate. I haven't heard of this happening, like a fluorine anion or water with two lone pairs on oxygen will donate ...
2answers
7k views

### Why do octahedral metal ligand complexes have greater splitting than tetrahedral complexes?

Octahedral complexes have greater splitting in the d orbitals. Is it because octahedral complexes have more atoms and thus more interactions?
1answer
692 views

### How are the π bonds arranged in Osmium Tetroxide?

The structure of $\ce{OsO4}$ has puzzled me greatly for a while. I know that it has a tetrahedral geometry. I am quite piqued at the thought of how the $π$ bonds were arranged in the molecule. My ...
1answer
2k views

### How potassium ferrocyanide is coloured?

The colour of a complex compound is due to unpaired electrons. As per crystal field theory, $\ce{K4[Fe(CN)6]}$ has no unpaired electrons so it has to be colourless. But then why is it coloured?
1answer
872 views

### Why do better π-acceptor ligands cause smaller Δ(T) d-orbital splitting?

Consider a formazanate ligand. Then imagine an electron-withdrawing group added to it. I expect a small σ-donating effect, and good π-accepting properties. My supervisor tells me that an ...
1answer
1k views

### How does complex ion differ from ligand complex?

As I understand, in terms of "colouring", ligand complex cause a transition metal cation centre's d-shell to split, allowing the electrons in that region to absorb specific frequencies of light, and ...
0answers
396 views

### Difference in color between the chromate and permanganate ions

As the title states, I'm wondering why the $\ce{[CrO4]^2-}$ and $\ce{[MnO4]-}$ ions display different colors in solution. Both of the metal ions have a $\mathrm{d}^0$ electron configuration, have the ...
1answer
197 views

### Why is chromate coloured but tungstate colourless?

I have to try and work this out using the tetrahedral molecular orbital diagram for $\ce{[WO4]^2-}$ but I am confused as both are $\mathrm d^0$ right?

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