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Colour of solid


Primarygun

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Many compounds have colour. Is it essential for them all containing moving ion in the body?

We perceive colours when only a selection of wavelengths of the visible light hit our eye. Chemicals can look coloured when they absorb certain frequencies of light. They do this when the light that hits them can be absorbed by an electron and promote it to a higher energy state. Only exact frequencies of light can do this, so a certain chemical will only absorb these frequencies. The resulting light it reflects won't contain these absorbed frequencies, and For many chemicals, like alot of the main group ones, they often don't have electrons that are happy to be promoted to a higher energy state. So they are colourless.

And the ion must be tranisitional metal ion? :confused:

Transitional metals are commonly colourful because they have electrons that can happily be promoted to higher energy states but they aren't the only things that can be coloured.

 

You're right to be confused, it confuses me plenty.

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"Crystal Field Splitting"

Its used to explain colours of metal comlpexes. The field of unpaired electrons splis into two levels, and for most metals the difference b/w these levels corresponds to visible light. Hence the colour. Complexes are coloured only if there are unpaired electrons.

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"Crystal Field Splitting"

Its used to explain colours of metal comlpexes. The field of unpaired electrons splis into two levels' date=' and for most metals the difference b/w these levels corresponds to visible light. Hence the colour. Complexes are coloured only if there are unpaired electrons.[/quote']

 

I'm afraid you've got that a bit wrong. CFST is about the divergence of d-orbital energies in complexes of certain geometries (usally octahedral and tetrahedral). In order to get a transition you need to have empty orbitals (i.e. <10 d electrons), the electron does not have to be unpaired.

 

d-d transitions are often Laporte disallowed (I think) so colours are often weak. Very strong colours in transition metals often come from Metal to ligand charge transfer transitions, especially if the ligands are highly conjugated.

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In order to get a transition you need to have empty orbitals (i.e. <10 d electrons), the electron does not have to be unpaired

I'm sorry if I got it wrong..........but.........

<10 d orbitals would mean unpaired electrons (Hund's rule of maximum multiplicity)

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I'm sorry if I got it wrong..........but.........

<10 d orbitals would mean unpaired electrons (Hund's rule of maximum multiplicity)

 

what about 6d in octahedral and 4d in tetrahedral? if the crystal field splitting energy is sufficiently large it is greater than the electron-electron repulsion energy (Hunds rule only applies for degenerate orbitals, which these are not) and thus all the electrons will reside in the lower-lying orbitals thus all being paired up. d-d transitions are still possible to the higher-lying, empty orbitals.

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so what properties do items which apear white, have, which items which appear black dont have. or vice versa

 

basically;

 

whats the difference in items which appear different colours, as far as the properties of the atoms are concerned? on a sub-atomic scale?

 

i know its all about the different wavelengths being absorbed, and the frequency/wavelength of the incoming light waves... read post #2 for that, what happens on a sub-atomic scale? and with the original properties of different coloured items?

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so what properties do items which apear white' date=' have, which items which appear black dont have. or vice versa

 

basically;

 

whats the difference in items which appear different colours, as far as the properties of the atoms are concerned? on a sub-atomic scale?

 

i know its all about the different wavelengths being absorbed, and the frequency/wavelength of the incoming light waves... read post #2 for that, what happens on a sub-atomic scale? and with the original properties of different coloured items?[/quote']

As far as I know, there's not going to be any categorical differences. So a red compounds aren't going to have any uniform differences in chemical properties to blue compounds. But obviously they react differently to light, and if you have light driven reactions this will only affect those chemicals that react to whatever wavelengths you use (so you can choose what chemicals in a solution will react).

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pulkit' date=' transition elements have excited electrons

and thus lose an electron (or more) in their outermost D orbitals.

[/quote']

 

Only if they are excited by an energy source. by losing an electron from the D orbital I presume you mean excitation to a higher orbital, not ionisation.

 

yes, that means there are unpaired electrons.

 

Not neccesarily. In a sufficiently large ligand field, the electron repulsion energy is smaller than the crystal field splitting, thus for certain electron configurations for certain geometries (such as 6d for octahedral field and 4d for tetrahedral field) it is possible to have fully paired electrons in a non-full d shell.

 

example of a transition metal: 4D^9 5S^1

 

That is an example of an excited state of a 4d10 metal, 4d10 is not a transition metal electron configuration (the d shell is full). Also note that, in transition metal complexes, the simple atomic orbitals no longer exist, so your excited state configuration is incorrect for a complex (which is what the discussion is concerned with).

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As far as I know, there's not going to be any categorical differences. So a red compounds aren't going to have any uniform differences in chemical properties to blue compounds. But obviously they react differently to light, and if you have light driven reactions this will only affect those chemicals that react to whatever wavelengths you use (so you can choose what chemicals in a solution will react).

 

You can also get quantum confinement of electrons too (in so-called "quantum dots"). Compound particles often have band structures rather than totally discreet levels, as the particulate size is decreased to the nano-scale, the number of atoms in the particles becomes smaller and the band structure slowly becomes more discreet (i.e. the electrons in the particle become more locally associated with one atom) this leads to different light scattering behaviour as the excitation energies become more discreet. This manifests itself through the materials often going from black to white (through the visible spectrum of colours) as particulate sizes is decrease. It is a very interesting phenomenon.

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Ok, you don't need less than 10 d electrons to have color. For every molecular bonding orbital there is an antibonding orbital. If the transition to that orbital is low enough in energy to be promoted by a visible photon then you will see color. There are plenty of d10 complexes with plenty of color. You also don't need the antibonding orbital if there is an available orbital on the ligand. Or an electron can go from an orbital on the ligand to one on the metal. MLCT (metal to ligand charge transfer) and LMTC (ligand to metal charge transfer).

 

You don't need transition metals for color, obviously, as there are many colored substances that do not contain metal. All you need is a an electronic transition in the visible region. I think there is another thread about colored smokes. He asked about p-nitroaniline, which is red and there's no metal in sight.

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Oh! I forgot plasmon resonance! If you have a colloid of a certain metal, usually silver or gold, you will see different colors in the solution. The color of the solution depends on the resonance of electrons on the surface (plasmons) of the metal particle. The frequency of the resonance is dependent on the size (and shape) of the particle.

 

You can make a rainbow of different colored colloids of, say, gold. From pink to red to purple to blue, all depending on the size of the particle.

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