Colour in elements and compounds

[hale2bopp]

Ok. Let me start by saying Hi!! I am new to this forum.

Ok, so I am doing a project on colour in elements and compounds, and I came upon a few problems:

1) Ok, so transition metals are coloured because they have unpaired electrons which get excited and move to higher energy levels when light hits it. It is coloured because the energy emitted by the electron when it goes back to ground state is within the range of visible light.

But what about non-metals? Most of them are coloured. Also, non-transition metals like caesium (goldish), magnesium (greyish -black) and francium (reddish) in group 1 are coloured. What is the reason for this?

2) Coloured gases : Some gases like NO2(brownish-red), Cl2(pale-greenish), I2(deep purple) etc are coloured. Does this have anything to do with unpaired electrons? Or the fact that the molecules are further apart, since Iodine in solid form is white?

3) I tried doing a numerical to compute the wavelength of colour of copper (II) ions. ie, Cu 2+ ions. Here it is.

According to one theory, colour is caused when radiation falls on an electron. The electron gains Energy from the radiation and becomes excited and increases its Energy level (as I understand, Energy levels are denoted by 'n' and essentially denote shell number)

In such a case, why cannot one just use the formula ( delta E = -2.18 * (10^-18) ( (1/(nf^2))-(1/(ni^2))) where nf stands for the final value of n (the Energy level at which the atom reaches after excitation) and ni stands for the energy level of atom at ground state?

As in questions like these are there in my normal textbook, but when I later apply the formula for frequency of this emitted radiation , i.e. ( frequency (nu) = [-2.18 * (10 ^-18)*((1/(nf^2))-(1/(ni^2)))] / (6.626* (10^-34)) ]

Let's take Cu2+ ion. It is radiated with visible light (so we can see it-the normal kind of visible light). The atomic number of copper is 29.ground state value of n for copper is 4. But it has lost two electrons (as it is an ion) so its n value is now 3.

Let's check the energy for one copper atom and the frequency of energy emitted.

frequency= [-2.18 * (10 ^-18)*(1/(3^2))] / (6.626* (10^-34)) ]

Which gives a frequency of -3.65563269 × 10^14 hz.

Or, approximately 820 nm wavelength which is outside visible light.

But we know that copper ions lend a blue colour; copper flame is blue and copper sulphate is blue.

Ok, the difference in type size is due to the fact that I have copy-pasted the question I asked in another forum a while ago. I didn't get any replies.

 

Anyway. Hoping that someone can help me with this! Thanks in advance. I want to make a good project.

Thanks,

Hale2bopp

[Mr Skeptic]

Things are colored because they absorb a specific range of photon frequencies (colors). This relates to photon energy because it is directly related to the frequency. Certain molecules can absorb a certain photon energy. This does relate to the electron configuration. Whatever colors the molecule cannot absorb, it reflects. This is the color you see it as.

[mississippichem]

Copper ions are blue in aqueous solution. Copper (II) in aqueous solution takes on the form [ce] [Cu(H_2 O)_6 ]^{2+} [/ce]. This coordination complex is a member of the [ce] O_h [/ce] point group which corresponds to a specific set of symmetry operations and therefore a specific charcter table matrix. One must then use a table of microstates [set of possible total quantum numbers for a given electronic configuration] to decide which transitions are spin allowed and which transitions are Laporte allowed. Some of the forbidden transitions still occur through a process known as vibronic coupling [cascades through quantized vibrational levels that allow transitions to happen without a change in parity].

After all that..one must use a Tanabe-Sugano diagram to approximate the wavelengths of the allowed transitions, of which for members of the [ce] O_h [/ce] point group there are many.

In this specific case, a Tanabe-Sugano diagram is not neccessary because there are no electron-electron repulsions. This case holds true for the [ce] d^1, d^9, d^0 , d^{10} [/ce] cases. In order to calculate this correctly though, you have to take into Crystal Field Theory; much of which is summed up in the following diagram