There is a specific order for filling electron for transition metals, but I don't know if it applies to other elements too?enter image description here

Some thing to take note of in this picture is the order of 3d and 4s. My belief is: 3d has a higher energy level than 4s, and but 4s is still the outermost shell.

The reason that transition metal would form colorful ions is that there are empty or half filled d orbitals, so when excited electrons drop back to the d orbitals, they emit energy in form of light. I am a bit confused about this.( it is from a book) what electron is it referring to? Is it the d electron or electron of lower energy level orbitals, as for instance 3p?


2 Answers 2


This is the order of electrons filling for all elements but typically this diagram is only used for transition elements because when writing electron configurations for any other element its simple and intuitive but transition metals are somewhat irregular so a helpful diagram like that is useful.

Its transition metal compounds or complexes that are coloured not the metal on its own. Its coloured because when it forms compounds or complexes the orbitals in the 3d subshell aren't the same energy as they would be if it wasn't part of a compound or complex. The electrons make transitions between these different energy 3d orbitals when they absorb photons of visible light. The colour of the compound will be the complementary colour of the colour absorbed.

The 3d orbital is typically seen as the higher energy orbital but when a transition metal is ionized the 4s electrons are taken first.

  • $\begingroup$ So which orbital has the valence electrons? $\endgroup$ Jun 23, 2014 at 17:47
  • $\begingroup$ Then what is the change in electron configuration when a transition metal forms compound with another element? Why it produces complementary colors? Aren't they supposed to release the amount of energy it absorbs? $\endgroup$ Jun 23, 2014 at 17:49
  • $\begingroup$ Do the photons light come from the forming of the compound? $\endgroup$ Jun 23, 2014 at 17:54

In response to your comments on my last answer: the term valence electrons isn't really used for compounds - especially ionic compounds. The valence electrons of the element will be the 4s electrons or electron in the case of Chromium and Copper. It's not so much a change in electron configuration its more about the energy or the orbitals. Each 4s orbital will have a different energy level. The electrons make transitions between these orbitals - this is why it is key that the transition metal ion has a partially filled d sub-shell (so there's room for the electrons to go to and from). It's not about the colour it emits, its about the colour it absorbs when visible light is passed through it. A good website to explain is "http://www.chemguide.co.uk/inorganic/complexions/colour.html". It talks about complexes but its the same idea for compounds.

  • $\begingroup$ By the virtue of your answer, if it does not absorb any light at all, doesn't it emit any light? So if this is performed in total darkness, we will see no color, right? $\endgroup$ Jun 23, 2014 at 22:08
  • $\begingroup$ It does absorb light - the light that passes through it. You're right about total darkness. Take a look at the link in my last answer. It has some very useful diagrams. $\endgroup$
    – RobChem
    Jun 24, 2014 at 8:49

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