Why does $\ce{Pb}$ have an oxidation number of $2^{+}$ while carbon and other elements in the same group have an oxidation number of $4^{+}$?

Furthermore, why doesn't carbon also have a $2^{+}$ oxidation number, as can't it also give up its two electrons in the p-suborbital to have a full s-suborbital? Why does it prefer to gain four electrons instead to fill the p-suborbital?

Similarly, why can't $\ce{Pb}$ gain four electrons just like carbon instead of giving up two?

Finally, doesn't this mean that $\ce{Pb}$ forms ionic bonds? Why doesn't carbon do the same?

  • 1
    $\begingroup$ This is usually called the inert pair effect, but there is more to it than just that. It is a consequence of relativistic corrections to the behavior of atoms, which occur more strongly for atoms far down in the periodic table. $\endgroup$ Commented Mar 30, 2014 at 16:47
  • $\begingroup$ For other answers discussing some aspects of relativistic effects, check here, here, and here, and for a very thorough analysis of relativistic effects, check here and here. $\endgroup$ Commented Mar 30, 2014 at 16:52

3 Answers 3


Why can't lead gain four electrons? The lower you go on the periodic table, the lesser you will find elements that have a more non-metallic character (so with this I mean elements that will have a lower tendency to gain electrons). Just visualize a lead atom with a very small nucleus and a large area surrounding the nucleus, filled with electrons. Now, the more you go away from the nucleus, the weaker the positively charged nucleus can 'pull' on the outer , negatively charged, valence electrons. Elements that haven't got a very tight grip on these valence electrons, will lose their valence electrons. However, elements with a small atomic radius (such as Fluorine) will have such a tight grip on its valence electrons, that it will even be able to snatch electrons from other atoms.

Of course the electron configuration of Radon is stable, because it has fully filled orbitals and it has a noble gas configuration, but you need to keep the nuclear forces in mind. This you can also see by looking at the electronegative value of lead : it is 1.87 (Pauling scale). Now compare it to carbon, which has 2.5. The lower you go in the same group, the lower the electronegativity goes and the more metallic character you will get. Elements with more metallic properties (such as lead) will mostly form ionic bonds and elements with more non-metallic properties (such as carbon) will mostly form covalent bonds. There exists a rule to predict (however it isn't always correct) if a compound will be covalently/ionically bonded : if the difference in electronegative value is higher than 1.66, then you have an ionic bond and if it is lower than 1.66 then you have a covalent bond.

Why does lead have an oxidation number of 2+ while carbon and other elements in the same group have an oxidation number of 4+? Just to correct you on this : lead does have an oxidation state of 4+ and elements above such as carbon and tin also have 2+ as an oxidation state. Just look at the electron configuration : the outer $\ce{s^2 p^2}$ orbitals will be able to lose 2 or even 4 electrons, or even gain 4 electrons. In fact these are all possible oxidation states of the elements in the carbon-group:

   Carbon 4, 3, 2, 1, 0, −1, −2, −3, −4
   Silicon 4, 3, 2, 1 -1, -2, -3, -4
   Germanium 4, 3, 2, 1, 0, -1, -2, -3, -4
   Tin  4, 3, 2, 1, -4
   Lead   4, 3, 2, 1
   (source : wikipedia)

Please do note the tendency to have more positive oxidation states, the lower you go in the group. (the reason why? see the beginning of the answer) Note : not all of these oxidation states are (very) stable.

Also have a look at this picture: Size of atoms and ions/.. Notice the metallic (shown by red) character increases the lower you go on the periodic table.


The preference of $\ce{Pb}$ for the lower oxidation state is an example for the so-called inert pair effect. Due to relativistic contraction of the outermost $s$ orbital, its 2 electrons are closer to the nucleus and thus more tightly bound than the electrons in the $p$ orbitals. More energy is required to use the $s$ electrons for covalent or ionic bonds. Stable compounds with $\ce{Pb}$ in the oxidation state $\ce{+IV}$ are therefore mostly formed with very electronegative elements like $\ce{O}$ or $\ce{F}$.


[Carbon cannot shift the electrons from the S orbital to the P orbital because there is a significant difference in energy, unlike lead.] - oops, I was wrong

Lead also has a +4 oxidation state. Look at the chemistry of a lead acid battery. Lead doesn't form ionic bonds since the difference in electronegativity between lead and the anions is too small.

NaCl - large difference in electronegativity
CO2 - small difference in electronegativity
PbO2 - small difference in electronegativity

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