Group 4 oxidation states

Hey guys, why does the +2 oxidation state in group 4 become more stable than the +4 as the atomic no' increases?
Also how does this relate to the chemistry of tin and lead? Thanks

Reply 1

Sycren

11

ok, well if atomic number increases then the number of protons increases, and so the number of electrons increases falling into the same shell, greater number of delocalised electrons in metal lattice so greater forces of attraction so more stable....right?

Reply 2

theredsox

15

If you're doing Edexcel, I've noticed that they never actually ask why tin is more stable in +4 than +2 and lead is more stable in +2 than +4. You just have to say that they are.
If you're not doing Edexcel, then I can't really help you :smile:

The question is a little more complex than suggested. This is one of those areas where I feel uncomfortable with the answers given. The textbooks suggest that there is an 'inert pair effect' due to the s2 electrons that allows lead and tin (to a lesser extent) to form 2+ ions.
To my mind this inert pair theory is simply an artifice to explain the observation wthout any specific precedent or basis... but that's just my opinion.

Reply 4

oceane

10

charco

The question is a little more complex than suggested. This is one of those areas where I feel uncomfortable with the answers given. The textbooks suggest that there is an 'inert pair effect' due to the s2 electrons that allows lead and tin (to a lesser extent) to form 2+ ions.
To my mind this inert pair theory is simply an artifice to explain the observation wthout any specific precedent or basis... but that's just my opinion.

I haven't come across the term 'inert pair effect' but I imagine it's referring to the fact that as u go down the group the valence s orbital becomes more contracted towards the nucleus relative to the valence p orbital. i.e the separation in energy between the s and p subshells is much larger. (due to the electrostatic forces when u have a heavier proton-rich nucleus)

Therefore it becomes less energetically favourable to use all 4 electrons (to form sp3 hybrids, ie 75% p character - higher energy, 25% s character - lower energy) than to just the use the 2 outer p electrons (to form sp hybrids ie 50% s character, 50% p character so overall lower in energy).

yes, but the II oxidation state in lead and tin is predominantly ionic so there is no hybridisation involved.

The basic logic of the argument, however, is sound, with the likelihood of the 's' pair being involved in bonding decreasing as you descend the group

Reply 6

Cities

16

This is an answer, with a few mark-scheme elements added

Why the equations which lead to different oxidation states?
Stable state for lead 2+ compard to 4+
Stable state for tin 4+ compared to 2+
lead(II)/Pb2+/PbCl2 is more/most stable (than lead(IV)/Pb4+/PbCl4)
+2 is the stable oxidation state of lead
PbCl4 would oxidise the HCl.
Bond strength of Pb-Cl is less than that of tin as lead is a larger atom
Thus the energy to promote electrons to its d-orbital and hybridise to form the 4+ state
Is not compensated by the new bonds formed in the 4+ state
Thus 2+ state is favoured and
Therefore lead 4+ oxidses chloride ion to chloride: relative oxidising powers

Group 4 oxidation states

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