Oxidation state trends in group 4

The inert pair effect in the formation of ionic bonds
If the elements in Group 4 form 2+ ions, they will lose the p electrons, leaving the s² pair unused. For example, to form a lead(II) ion, lead will lose the two 6p electrons, but the 6s electrons will be left unchanged - an "inert pair".
You would normally expect ionisation energies to fall as you go down a Group as the electrons get further from the nucleus. That doesn't quite happen in Group 4.
This first chart shows how the total ionisation energy needed to form the 2+ ions varies as you go down the Group. The values are all in kJ mol^-1.

Notice the slight increase between tin and lead.
This means that it is slightly more difficult to remove the p electrons from lead than from tin.
However, if you look at the pattern for the loss of all four electrons, the discrepancy between tin and lead is much more marked. The relatively large increase between tin and lead must be because the 6s² pair is significantly more difficult to remove in lead than the corresponding 5s² pair in tin.

Again, the values are all in kJ mol^-1, and the two charts are to approximately the same scale.
The reasons for all this lie in the Theory of Relativity. With the heavier elements like lead, there is what is known as a relativistic contraction of the electrons which tends to draw the electrons closer to the nucleus than you would expect. Because they are closer to the nucleus, they are more difficult to remove. The heavier the element, the greater this effect.
This affects s electrons much more than p electrons.
In the case of lead, the relativistic contraction makes it energetically more difficult to remove the 6s electrons than you might expect. The energy releasing terms when ions are formed (like lattice enthalpy or hydration enthalpy) obviously aren't enough to compensate for this extra energy. That means that it doesn't make energetic sense for lead to form 4+ ions.