3
$\begingroup$

I know that the colour in coordination compounds can be explained by CFT in terms of d-d transitions. Are d-d transitions possible in diamagnetic complexes? Why does $\ce{[Co(NH3)6]^3+}$ show colour?

It is given that $\ce{[Co(NH3)6]^3+Cl3}$ has a yellow colour and that $\ce{[Co(NH3)5Cl]^2+Cl2}$has a purple colour in my textbook. Can the cause of this variation be explained using CFT?

$\endgroup$
2
  • 3
    $\begingroup$ I think this is similar to this one chemistry.stackexchange.com/a/60331/115228 $\endgroup$
    – Infinite
    Apr 28, 2022 at 5:07
  • 2
    $\begingroup$ Not quite. I was looking for an explanation on coordination complexes using CFT. It would be appreciated if it was explained using [Co(NH3)6]^3 (if that is possible). There are 3 parts to my question. $\endgroup$
    – Aja
    Apr 28, 2022 at 14:20

1 Answer 1

6
$\begingroup$

As Porphyrin already stated in his answer, diamagnetism/paramagnetism has nothing to do with the colour of the compound.

Are d-d transitions possible in diamagnetic complexes?

It depends on the energy of the photon used. If it has sufficient energy then excitation could be possible whether the electrons are paired or not. If the wavelength of the absorbed photon lies in the visible light region then the compound could exhibit colour. So this answers why those two compounds are coloured.

It is given that $\ce{[Co(NH3)6]Cl3}$ has a yellow colour and that $\ce{[Co(NH3)5Cl]Cl2}$ has a purple colour in my textbook. Can the cause of this variation be explained using CFT?

$\ce{NH3}$ is a stronger ligand than $\ce{Cl-}$. So, $\ce{NH3}$ would produce more splitting energy,$\Delta$ (the energy difference between $\ce{t_{2g}}$ and $\ce{e_g}$ set of orbitals) than $\ce{Cl-}$. Therefore, the splitting is maximum in case of $\ce{[Co(NH3)6]^{3+}}$ (since it has a greater number of $\ce{NH3}$ ligands when compared to $\ce{[Co(NH3)5Cl]^{2+}}$).

So more energetic photon is required to excite an electron in $\ce{[Co(NH3)6]Cl3}$ as compared to $\ce{[Co(NH3)5Cl]Cl2}$. But according to the colour wheel, if the observed colour is yellow then absorbed colour could be violet similarly if observed colour is red-violet (purple) then absorbed colour could be yellow-green. So $\ce{[Co(NH3)6]Cl3}$ would absorb violet photon which is more energetic than yellow-green photon. So the splitting energy is more in the case of $\ce{[Co(NH3)6]Cl3}$, which is in agreement with our theory.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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