Color of Fe(II) complexes

  • #1
lee403
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1
There is something that I am not understanding about the visible colors of coordination complexes. I did a lab where I prepared [Fe(NH2trz)3]Br2 from FeBr2.
Fe(NH2trz)3 was a violet color and FeBr2 was red. My understanding of colors in coordination complexes is that reds are high energy because they absorb the complementary color most strongly which would be purplish. Purple is high energy light because it has a shorter wavelength. So if FeBr2 is red that means there is a large d-d orbital splitting or a large energy difference between the t2g and eg levels of the d-d orbitals. However, Br is a weak field ligand (for another reason [poor metal-ligand overlap] that I don't quite understand) and NH2trz since its bonding through the N is an intermediate ligand. I thought that since weak field ligands have small d-d orbital splitting they would absorb low energy (long wavelength) light, which would be red and would therefore appear purple. But the complexes I have are the opposite of how I have interpreted color so far. What part of my reasoning is incorrect?
 
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Likes Pratyaya Ghoshal Das

Answers and Replies

  • #2
DrDu
Science Advisor
6,258
906
There is something that I am not understanding about the visible colors of coordination complexes. I did a lab where I prepared [Fe(NH2trz)3]Br2 from FeBr2.
Fe(NH2trz)3 was a violet color and FeBr2 was red. My understanding of colors in coordination complexes is that reds are high energy because they absorb the complementary color most strongly which would be purplish. Purple is high energy light because it has a shorter wavelength. So if FeBr2 is red that means there is a large d-d orbital splitting or a large energy difference between the t2g and eg levels of the d-d orbitals. However, Br is a weak field ligand (for another reason [poor metal-ligand overlap] that I don't quite understand) and NH2trz since its bonding through the N is an intermediate ligand. I thought that since weak field ligands have small d-d orbital splitting they would absorb low energy (long wavelength) light, which would be red and would therefore appear purple. But the complexes I have are the opposite of how I have interpreted color so far. What part of my reasoning is incorrect?
d-d transitions are dipole forbidden and therefore have only weak intensity. Mostly charge transfer transitions are more intense and dominate the colour impressio of the complex.
 
  • #3
HAYAO
Science Advisor
Gold Member
370
233
There is something that I am not understanding about the visible colors of coordination complexes. I did a lab where I prepared [Fe(NH2trz)3]Br2 from FeBr2.
Fe(NH2trz)3 was a violet color and FeBr2 was red. My understanding of colors in coordination complexes is that reds are high energy because they absorb the complementary color most strongly which would be purplish. Purple is high energy light because it has a shorter wavelength. So if FeBr2 is red that means there is a large d-d orbital splitting or a large energy difference between the t2g and eg levels of the d-d orbitals. However, Br is a weak field ligand (for another reason [poor metal-ligand overlap] that I don't quite understand) and NH2trz since its bonding through the N is an intermediate ligand. I thought that since weak field ligands have small d-d orbital splitting they would absorb low energy (long wavelength) light, which would be red and would therefore appear purple. But the complexes I have are the opposite of how I have interpreted color so far. What part of my reasoning is incorrect?
You might want to be careful with Fe2+ compounds (d6 electron) and crystal field. Ever heard of "Tanabe-Sugano diagram"?
 

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