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My answer: The inert pair effect in $\ce{Pb}$ causes it to pull back the electrons, resulting in polarisation.

My teacher's answer: An ionisation state of $+4$ is too difficult to achieve, and it is not gained even when 4 chlorine atoms get reduced, as the later ionisation enthalpies of lead (or any metal) are high.

My point is, if we don't think of the inert pair effect, we are generalising this, and moreover, high ionisation energies of lead should be explained by the inert pair effect.

Which explanation do you think is better?

As a side note, I would like to add that my teacher pointed out that the inert pair effect is an outdated concept. Is it so? If yes, why is it rejected?

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The high covalent characted is just explained by the fact of $\ce{Pb^{4+}}$ being very polarizing and $\ce{Cl-}$ fairly polarizable.

The inert pair effect would give an explanation about why it is difficult to reach that oxidation state, and why it has high tendency to get reduced, but there is no need to require to this concept to explain the covalent character of the compound.

Regarding your last question, I do not see why the inner pair concept should be rejected. Is just another model that can explain in a simple way properties or behaviors of certain elements. There are better models and orbital calcualtions that can also do the job? Of course, but that is no reason for avoiding the use of the model. Models that are still useful to explain stuff easily, can be employed even if more accurate ones do exist. One good example is ligand field theory, which is still useful even when molecular orbitals theory is more accurate and can still explain the same (and many more) things than the former. Even though, a theorem or model HAS to be rejected if a new one is found that explains properties better and is incompatible with the old one.

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    $\begingroup$ +1, Agreed, I see absolutely no reason why the inert pair effect should be abandoned/ should be outdated, it is a wonderful example of relativistic effects. The crystal field theory on the other hand should be avoided, since the superior ligand field theory is not much more complex. $\endgroup$
    – Martin - マーチン ♦
    Nov 20, 2014 at 12:47
  • $\begingroup$ Yes, I intended to write LFT, I will edit that. Thanks for the correction. $\endgroup$
    – Altered State
    Nov 20, 2014 at 13:47
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Halogenides with metals in high oxidation state are rarely truly ionic on their own, as a strong positive charge on the cation must be stabilized. To do so, many metal-halogen contacts are required, and the metal cation is usually too small to coordinate many halogen ions. So, usually higher halogenides are either molecules with highly polar bonds or (di/oligo/poly)mers with highly polar bonds where some halogen atoms participate in two metal-halogen bonds.

In case of $\ce{PbCl4}$ NBO analysis reveals -0.4 NBO charges on chlorine atoms, so it is a highly polar bond, even though it is a molecular compound. So, I would say that due to unfavorable ionic radiuses, $\ce{Pb^{4+}}$ is incapable to form enough contacts to stabilize effectively, forming instead a tetrahedral polar-covalent molecule.

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In $\ce{PbCl4}$ the lead is in $\ce{Pb^{IV}}$ state and $\ce{Cl}$ is having a considerable, not very large - but enough to be polarised easily, and the high charge density on lead polarizes the anion much more efficiently, so that the electron density is shifted and partially shared, in effect the compund is much covalent.You would like to see Fajan's rule, if you didn't knew them before.

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