# Tag Info

164

Yes, this is a beautiful question. As you said, in lower rows of the periodic table, there are relativistic effects for the electrons. That is, for core electrons in gold, the electrons are traveling at a significant fraction of the speed of light (e.g., ~58% for Au $\ce{1s}$ electrons). This contracts the Bohr radius of the 1s electrons by ~22%. Source: ...

114

See the footnotes I've included if you would like to see more of the detail behind a specific statement. The figure below compares the reflectance spectrum for silver and gold (let's forget about aluminum, it's not relevant to this discussion; also keep in mind that where reflectance is low, absorbance is high and vice-versa). The absorption (reduced ...

74

Based on your description, I may have found the article you originally saw, or at least one very similar. Researchers from Dartmouth College published a paper$\mathrm{^1}$ in which they report, among other things, the results of viewing sunlit white paper through two 3 meter lengths of plexiglass; one filled with $\ce{H2O}$ and one with $\ce{D2O}$. Sure ...

65

This does seem to be the case. I don't have images of the different types of water, but I did find this overlaid IR-visible spectrum of water and heavy water: As you stated, the presence of deuterium shifts the absorbance spectrum of heavy water further into the IR region, rendering it colorless. The website I found this on (http://www.webexhibits.org/...

49

This effect comes from the band structure of the metal, not atoms. In simplest terms (see, e.g. Ashcroft & Mermin, Solid State Physics, chapter 1) the electrons in the conduction band have a plasma frequency $\omega_{p}^{2} = 4\pi n e^{2}/m$ - this is a collective excitation of the metal. For Cu (reddish) you start exciting the plasmons around 2eV (red)...

32

The difference between snow and ordinary ice cubes is mainly about the size of the particles. Snow is made from small, irregular crystals with many edges at a very small scale. Light is refracted or scattered by the edges (or the interface between air and the edges). Snow is white because the scattering effect of those edges dominates what happens to light ...

26

Though @DHMO's answer is quite interesting, you shouldn't take it without a pinch of salt. I've to disagree with it in some aspects. Soap colorants range from various kinds of Mica to dyes found naturally in plants. Now typically when you dye anything the quantity of dye used in comparison to the quantity of the substance to be dyed is very small. Too much ...

26

The main substances that cause the yellowish color of the milk are carotenoids [1]. The main carotene involved is the beta-carotene coming from the feed that cows eat. Some studies have been carried on and it has been noticed that the milk with a more yellow tinge was collected during late spring and early summer when carotene levels are at a maximum [2] ...

24

Background $\ce{O2}$ exists as a paramagnetic, triplet since the two electrons in its two (degenerate) HOMO orbitals are unpaired. There are 6 known phases of solid oxygen with color ranging from pale blue to red to black. In the liquid phase it has a light blue color. This color is due to light absorption by the ground state triplet according to the ...

24

What you're specifically describing is typically known as UV degradation. The color change is due to the oxidation of the polymers that make up the plastic. These chemical reactions are demonstrated in satisfying detail here. Under the section titled "Photodegradation" is where you can find the intricacies of the polymer-oxidation process. And yes, such ...

22

You're right--it's got to do with them being transition metals (usually). Transition metal ions form coordination complexes. Their empty $d$ orbitals accept lone pairs from other molecules (called "ligands") and form larger molecules (though we don't call them that--we call them "complexes"). When put in water, the ligand is $\ce{H2O}$, and you get complexes ...

22

The colour is from a thin film of bismuth(III) oxide that forms on the surface if the crystals are formed in air. At the elevated temperatures used to melt bismuth, the oxide forms quite quickly. The iridescence is a result of thin film interference—light waves constructively or destructively interfere as they bounce off the bismuth-oxide and oxide-air ...

21

You are absolutely correct, it all about the metal's electrons and also about their d orbitals. Transition elements are usually characterised by having d orbitals. Now when the metal is not bonded to anything else, these d orbitals are degenerate, meaning that they all have the same energy level. However when the metal starts bonding with other ligands, ...

21

Selection rules The intensity of the transition from a state $\mathrm{i}$ to a state $\mathrm{f}$ is governed by the transition dipole moment $\mu_{\mathrm{fi}}$ (strictly, it is proportional to $|\mu_{\mathrm{fi}}|^2$): $$\iint \Psi_\mathrm{f}^*\hat{\mu}\Psi_\mathrm{i}\,\mathrm{d}\tau \,\mathrm{d}\omega \tag{1}$$ where $\mathrm{d}\tau$ is the usual ...

21

This video shows how a black flame is achieved. If you illuminate the fire with a monochromatic light source (sodium vapor lamp) and introduce a species in the fire that absorbs that wavelength (sodium ions) then the fire will in fact appear black under the illumination. A screen shot is shown below: Click image for video.

20

First, let's think about what makes a substance coloured: a substance will appear to be coloured if it absorbs light in the visible spectrum. Beta carotene: absorbs blue/green light (400–500 nm), so it looks red-orange when white light is shined on it as the blue/green light is subtracted from the white and what remains is what reaches our eyes. The ...

19

Due to Chappuis absorption, ozone does have a bluish color. To determine exactly what kind of blue it is, let's first look at the spectrum of absorption in the Chappuis band. The following plot was done using these data for 293K. This is spectral cross-section of absorption. To determine color from this spectrum, we need to choose some parameters: Number ...

18

Osmium has a bluish-gray tint. Well; slightly. Cesium is silvery-golden!, But don't wear it.

18

The question is really badly worded. For starters, let’s look at solutions of nickel(II): Figure 1: Nickel(II) solutions. From left to right: $\ce{[Ni(NH3)6]^2+}$, $\ce{[Ni(en)3]^2+}$, $\ce{[NiCl4]^2-}$, $\ce{[Ni(H2O)6]^2+}$. Image taken from Wikipedia, where a full list of authors is available. You can ignore the left two but the rightmost is a standard ...

17

Absorption of a photon typically results in a vibrationally excited higher electronic state of the same multiplicity. $$\ce{S_0 ->[h\nu_\mathrm{ex}] S_1}$$ In most cases, the excited state deactivates through internal conversion in a radiationless process via vibrational energy exchange with solvent molecules. No light is emitted here, but the ...

17

Yes, it is all about the absorption of light at specific wavelength. Azobenzene, the parent compound has an absorption maximum around $\lambda$= 430 nm in the visible spectrum. The interesting part is: The absorption can be tuned by substitution of the arenes. This is done before the azo coupling. Some examples are Allura Red (1), Chrysoine Resorcinol (2),...

17

You might have observed that large soap bubbles are transparent. This is because the color has been diluted to a great extent that it basically became transparent. Now, the foam is just a multitude of tiny bubbles that are transparent. You would observe that the unorganized mass of transparent particles (the bubble is quite organized) would appear white. ...

16

The partially full d-orbitals in transition metals have energy splittings that happen to lie in the visible range. Depending on the arrangement of substituents (known as ligands) that attach to them, the electron energies split according to crystal field theory. Similar splitting in the s or p orbitals produce gaps in the ultraviolet, and any visible light ...

16

Under basic condtions, acetone does condense with itself, first to form mesityl oxide, which is colorless, and then to form isophorone, which has a yellow color. Both mesityl oxide and isophorone are still reactive carbonyl compounds likely to be attacked by deprotonated acetone carbanions. That leads to bigger and more conjugated condensation products, ...

16

What you describe as pink is the actual colour of copper. The colour of a clean, solid surface of high-purity copper is typically salmon red. In your case, the appearance may be further affected by the particle size of the precipitate. The familiar reddish-brown colour of copper is caused by copper(I) oxide, which is slowly formed on the metal surface.

16

Let’s take a look at a qualitative MO scheme for a tetrahedric transition metal complex whose ligands have three p-type orbitals each. On the left of figure 1 you have the metal orbitals ($\mathrm{3d}$, $\mathrm{4s}$ and $\mathrm{4p}$) and on the right the twelve degenerate ligand p-orbitals (transform as $\mathrm{a_1 + e + t_1 + 2t_2}$). Only orbitals of ...

16

There is no "pigment" that makes milk white. According to Wikipedia: A pigment is a material that changes the color of reflected or transmitted light as the result of wavelength-selective absorption. Going by that definition of a "pigment", then there is no (white) pigment in milk. In fact (as @Ivan points out in the comments) there is no such thing as ...

15

The thermochromism of $\ce{ZnO}$ results from a minor loss of oxygen upon heating to temperatures around 800 °C, i.e. a non-stoichiometric $\ce{Zn$_{1+x}$O}$ with $x = 7 \times 10^{-5}$ is formed. Under air, this effect is reversible. Heating (and cooling) of the material while hooked up to a vacuum pump might result in a more persistent colour change.

15

The $\ce{Fe}$-porphyrin complex is a vastly conjugated $\pi$-system, in which the oxidation state of the $\ce{Fe}$ plays a central role. The wavelength of light absorbed by hemoglobin exactly corresponds to the the difference in energy between the highest occupied molecular orbital (HOMO) and lowest unoccupied molecular orbital (LUMO) of the heme $\pi$-...

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