Why do we see complementary colors in UV/Vis Spectroscopy? [closed]

Question

From what I read, a compound that absorbs visible light will produce a complementary color (using the color wheel), that can be seen. Why is this the case? Is it because, for example, if a molecule absorbs red, for instance, all the other colors will reflect and combine (?) to give off the color green that we see? I thought if we excite an electron to its excited state, with it absorbing a certain wavelength, when it drops back down to its ground state, it will give off a photon of that same wavelength (and same color). Also, when a molecule absorbs UV radiation and reflects all of the visible wavelengths, I read that it will appear white or colorless (isn't black the absence of all colors?), is this just two ways of saying the substance appears white or does that mean that it can be white or clear?

Answer 1

There are several points

From what I read, a compound that absorbs visible light will produce a complementary color (using the color wheel), that can be seen. Why is this the case?

This has to do with the physiology of the eye and how it has been designed to precept the colors in the visible range. As you stated, if we remove green from the visible spectrum, the transmitted light appears reddish or purplish to our eye. May be a cat or dog sees it differently. Color is a human perception. This complementary chart is just an approximation. If we take potassium permanganate solution, and if you look at what colors does it allow to pass through, you will see that it absorbs mostly in the green. You can easily do such experiments with a cardboard box and a diffraction grating made from CDs and a white LED.

I thought if we excite an electron to its excited state, with it absorbing a certain wavelength, when it drops back down to its ground state, it will give off a photon of that same wavelength (and same color).

This does not happen in majority of the cases i.e, in solutions and solids. The phenomenon you are referring to is called resonance fluorescence and it only happens to atoms in the gas phase. You shine 589 nm to Na atoms in gas phase, and you will see emission of 589 nm from Na atoms. This is accomplished in a flame. See Atomic Fluorescence Spectroscopy.

Normally, once light is absorbed, the absorbing material does not emit light. Instead, the absorbed energy is dissipated as heat.

Also, when a molecule absorbs UV radiation and reflects all of the visible wavelengths, I read that it will appear white or colorless (isn't black the absence of all colors?), is this just two ways of saying the substance appears white or does that mean that it can be white or clear?

Clear and white are two different things. Something clear like water does not absorb much in the visible range, perhaps a very very small portion from the red region. Therefore when you have huge amount of water you start to see the effects of absorbing red portion of the light. Do you recall that ocean water looks bluish? Even liquid oxygen is light blue!

You can say that water transmits most of the visible light. Majority of the substances will also absorb UV. How deep UV do you want? Most of the substance will start absorbing deep UV. Even air does, otherwise we would have been fried by the Sun.

A white piece of plastic or a cotton cloth, is reflecting most of the visible light, hence it appears white to us. If you look the same piece in red light, it will look red, and in blue light it will appear blue and so on.