# Why is lead sulfide found in nature, whereas lead oxide is less common?

This question probably applies to other heavier metals as well. The only rationalization I can figure is that lead in the +2 oxidation state (which is most common) is a borderline soft acid and prefers to take electrons from a soft sulfur instead of trying to react with a hard oxygen.

Is this the right direction? Thank you in advance.

Lead(IV) and sulfide ions are large and polarizable, making them soft in the context of hard/soft acid/base theory (HSAB), which states that ions with similar hardness will form stronger bonds. The oxide ion is hard and therefore has a tendency to form weaker interactions with Pb(IV).

HSAB theory has been discussed on Chem.SE in this question and this answer.

I guess that a long time ago there was a lot of $\ce{H2S}$ in the atmosphere and then the $\ce{H2S}$ could have reacted with that lead oxide to form $\ce{PbS}$?

On wikipedia it says (under Lead(II)oxide,reactions) :

The red and yellow forms of this material are related by a small change in enthalpy:

PbO(red) → PbO(yellow) ΔH = 1.6 kJ/mol

PbO is amphoteric, which means that it reacts with both acids and with bases. With acids,it forms salts of $\ce{Pb2+}$ via the intermediacy of oxoclusters such as $\ce{[Pb6O(OH)6]4+}$. With strong base, $\ce{PbO}$ dissolves to form plumbite(II) salts:[6]

$\ce{PbO + H2O + OH- -> [Pb(OH)3]-}$

It could have reacted with $\ce{H2S}$ following this reaction : (in water)

$\ce{Pb2+ + H2S -> PbS + 2 H+}$

• Ores were/are formed deep in the earth crust. Oxygen has only a minor influence in case of ores brought in contact to air by mining. Exaples are cinnabar oxidized to elementary mercury in ores close to surface and oxidation of pyrite leading to acid effluents fron abandoned coal mines. – Georg Aug 25 '13 at 17:59