5
$\begingroup$

I study coordination chemistry at high school. I have read about valence bond theory (VBT) and crystal field theory (CFT) as well.

My teacher taught me that VBT is based on the experimental values of magnetic moment, whereas CFT focuses on orbitals splitting in the presence of weak and strong field ligands (WFL and SFL, respectively). Can we explain VBT configuration on a basis of SFL and WFL?

The prerequisite for this question was of my exams where we were asked to write the configuration of $\ce{[Fe(NH3)6]^3+}$ on the basis of VBT with no magnetic moment data provided. How should I have written the configuration?

$\endgroup$
1
  • $\begingroup$ That actually is a tough nut to crack. CFT is a somewhat outdated theory. It was superseded by ligand field theory. This might be a bit too much though at the highschool level. Also, you're probably taught a very crude approximation to VBT. There is more to that, too. Now, I realise that this isn't very helpful, but -at least to me- there is no simple way to answer your question (which I think is a good one). $\endgroup$ Commented Apr 16 at 0:12

1 Answer 1

0
$\begingroup$

For $[Fe(NH_3)_6]^3+$, you can answer as following:

  1. Oxidation state of Fe in this complex is '+3', making it a $3d^5$ configuration, with 5 unpaired electrons occupying each degenerate orbital.
  2. Since $NH_3$ is a strong field ligand, electron pairing occurs disobeying Hund's rule. Now, the configuration becomes $3d^5$ with two orbitals containing 4 paired electrons, and 1 one unpaired in one degenerate orbitals, leaving two vacant 3d orbitals.
  3. Now, hybridisation occurs with two 3d, one 4s, and three 4p vacant atomic orbitals forming six $d^2sp^3$ hybrid orbitals arranged in an octahedral geometry.
  4. The lone pair on each $NH_3$ molecule will now form coordinate covalent bond with six hybrid orbitals.

While ligand strength can be applied in VBT, it might bring some incorrect results sometimes regarding magnetic moment.

This answer might seem correct given it agrees with CFT but this doesn't with few other complexes.

Take, for example, Hexacyannonickelate (II) $[Ni(CN)_6]^4-$ for example. VBT argues that it is a diamagnetic species because all the 8 electrons undergo pairing. While CFT argues that the configuration will be $t_g^6 e_g^2$ with 2 unpaired electrons in $e_g$ orbitals, meaning it is paramagnetic in nature.

VBT fails to explain magnetic properties of few complexes and geometry using hybridisation, and hence CFT is preferred over VBT.

$\endgroup$

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.