2
$\begingroup$

This question comes specifically from an IB Chemistry HL Paper 1 in May 2018 TZ1, namely question 8.

Which complex has the greatest d orbital splitting?

It gives 4 Complexes $\ce{[Fe(H_2O)_6]^{2+}}$, $\ce{[Fe(H_2O)_6]^{3+}}$, $\ce{[Co(H_2O)_6]^{3+}}$, $\ce{[Cr(NH_3)_6]^{3+}}$ and it says that they give the colours green, orange, blue and violet respectively.

Initially I thought that the answer would be $\ce{[Cr(NH_3)_6]^{3+}}$ because it gives the highest energy light, being violet. However, the answer is given as $\ce{[Fe(H_2O)_6]^{3+}}$, why is this?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
7
$\begingroup$

The colour at which the complex absorbs reflects the wavelength of the d–d* electronic transitions. However, this colour is not the same as the transmitted colour (which you see), but is instead complementary to the transmitted colour. Therefore, a complex that appears purple is actually absorbing lower-energy light than a complex that appears red.

Improve this answer
$\endgroup$
3
  • $\begingroup$ Please explain how is it 'complementary'. $\endgroup$
    – Pan
    Commented Mar 19, 2019 at 8:55
  • $\begingroup$ From what I understand, $\ce{[Cr(NH_3)_6]^{3+}}$ would be absorbing the complementary colour of violet being yellow, which is low energy. However, how does this make any sense because $\ce{NH_3}$ is a strong field ligand which should create a large d orbital splitting, therefore being able to absorb higher energy lights? $\endgroup$
    – ASP
    Commented Mar 20, 2019 at 2:06
  • $\begingroup$ @AnthonyP The splitting of the d orbitals is not only a function of the ligand, but also the metal (the atom type as well as the oxidation state). It therefore isn’t correct to only look at the ligand. Apart from that, NH3 isn’t even particularly strong-field; it’s only marginally higher than H2O in the spectrochemical series. $\endgroup$
    – orthocresol
    Commented Mar 20, 2019 at 2:40

Not the answer you're looking for? Browse other questions tagged or ask your own question.