![]() Red light has the lowest energy in the visible region. When white light is passed through a solution of this ion, some of the energy in the light is used to promote an electron from the lower set of orbitals into a space in the upper set.Įach wavelength of light has a particular energy associated with it. The size of the energy gap between them (shown by the blue arrows on the diagram) varies with the nature of the transition metal ion, its oxidation state (whether it is 3+ or 2+, for example), and the nature of the ligands. Whenever 6 ligands are arranged around a transition metal ion, the d orbitals are always split into 2 groups in this way - 2 with a higher energy than the other 3. The diagram shows the arrangement of the d electrons in a Cu 2 + ion before and after six water molecules bond with it. However, because of the way the d orbitals are arranged in space, it doesn't raise all their energies by the same amount. That raises the energy of the d orbitals. When the ligands bond with the transition metal ion, there is repulsion between the electrons in the ligands and the electrons in the d orbitals of the metal ion. The argument is not really any different if you have multidentate ligands. Technically, the Sc 3 + ion does not count as a transition metal ion because its 3d level is empty.įor simplicity we are going to look at the octahedral complexes which have six simple ligands arranged around the central metal ion. ![]() Although there is a partially filled d level in the metal, when it forms its ion, it loses all three outer electrons. At the other end of the row, scandium ( 3d 14s 2 ) does not really counts as a transition metal either. When it forms an ion, the 4s electrons are lost - again leaving a completely full 3d level. Zinc with the electronic structure 3d 104s 2 does not count as a transition metal whichever definition you use. The usual definition of a transition metal is one which forms one or more stable ions which have incompletely filled d orbitals. This shortened version of the Periodic Table shows the first row of the d block, where the 3d orbitals are being filled. We often casually talk about the transition metals as being those in the middle of the Periodic Table where d orbitals are being filled, but these should really be called d block elements rather than transition elements (or metals). Copper(II) sulfate solution is pale blue (cyan) because it absorbs light in the red region of the spectrum. What this all means is that if a particular color is absorbed from white light, what your eye detects by mixing up all the other wavelengths of light is its complementary color. Mixing together two complementary colors of light will give you white light. ![]() Blue and yellow are complementary colors red and cyan are complementary and so are green and magenta. The diagram shows one possible version of this.Ĭolors directly opposite each other on the color wheel are said to be complementary colors. If you arrange some colors in a circle, you get a "color wheel". You can, however, sometimes get some estimate of the color you would see using the idea of complementary colors. Mixing different wavelengths of light doesn't give you the same result as mixing paints or other pigments. Sometimes what you actually see is quite unexpected. You wouldn't have thought that all the other colors apart from some red would look cyan, for example. Working out what color you will see is not easy if you try to do it by imagining "mixing up" the remaining colors. The diagram gives an impression of what happens if you pass white light through copper(II) sulfate solution. We see this mixture of wavelengths as pale blue (cyan). The light which passes through the solution and out the other side will have all the colors in it except for the red. Copper(II) ions in solution absorb light in the red region of the spectrum. ![]() If white light (ordinary sunlight, for example) passes through copper(II) sulfate solution, some wavelengths in the light are absorbed by the solution. \)Įxample 1: Blue Color of Copper (II) Sulfate in Solution ![]()
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