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I would like to find out more about what makes a compound ionic and what makes one covalent. However, I felt this was too general a question so I thought that asking about the chlorides of lead would be more appropriate. I have already come up with an explanation of my own, which I have provided below. However, I would also like to hear the opinions of others.

My explanation is as follows: I consider the interaction between the atomic orbitals of the lead (II) and lead (IV) cations with the chloride ions. Firstly, we must note that their valence electronic configurations are different: lead(II) has its valence electrons in 6s orbital while the valence electrons of lead (IV) reside in the 5d orbital. On the other hand, the valence electrons of the chloride ions reside in the 3p orbital. Based on what I have read from Clayden's Organic Chemistry, I understand that atomic orbitals of similar energy interact strongly to produce strong covalent bonds. I was thinking that since the interacting orbitals of the lead (IV) cation and the chloride ions are closer in energy, a favourable covalent interaction would occur. However, in the case of lead (II), covalent bond formation is not favoured since the orbital energies are considered to be too far apart. I am not confident that this is correct. Please feel free to point out any inaccuracies.

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  • $\begingroup$ At its simplest it is just more positive charge -> more polarising -> more covalent bond. Pb(IV) can't form bonds with a fully filled 5d subshell, btw, unless you invoke some kind of d-s hybridisation. It's probably best represented by 4x dative bonds from chloride into 6s + 6p orbitals of Pb, which have covalent character because of strong polarisation. Same applies to Pb(II) and its fully filled 6s orbital. $\endgroup$ – orthocresol Aug 12 '17 at 17:37
  • $\begingroup$ @orthocresol Ok. That sounds simple enough. On a side note, how would we explain that titanium (IV) iodide is covalent? $\endgroup$ – Tan Yong Boon Aug 12 '17 at 17:49
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  • $\begingroup$ @orthocresol Actually, if you think about it, the argument using positive charge seems to fail when we consider neutral atoms in the first step. This argument only works if you first assume that the interacting species are all ions, not neutral atoms. Then only can we say that it is the high polarising power of lead (IV) cation etc. $\endgroup$ – Tan Yong Boon Aug 12 '17 at 18:03
  • $\begingroup$ Is it possible to explain this in a "less ionic" manner, meaning that we do not invoke the concept of charges and polarisation. I mean the electronegativity of lead is invariant in whatever compound it forms. So why would believe that the lead atom in lead (IV) compounds is more polarising than in lead (II) compounds. $\endgroup$ – Tan Yong Boon Aug 12 '17 at 18:08

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