Lead (Pb) is a heavy post-transition metal with an atomic number of 82, characterized by its softness, density, and low melting point. When lead forms a chemical compound, it loses or shares electrons to achieve a stable arrangement, resulting in an electrical charge known as an oxidation state. This oxidation state represents the charge the lead atom would have if the compound were fully ionic. Lead exhibits more than one oxidation state, but it favors a specific charge in most of its commonly encountered compounds.
Lead’s Most Common Oxidation State
The most prevalent and stable charge lead carries in its compounds is the \(+2\) oxidation state, represented as the lead(II) ion, \(\text{Pb}^{2+}\). This divalent state means the lead atom has lost its two outermost p-orbital electrons. The \(\text{Pb}^{2+}\) ion is found in the vast majority of lead’s inorganic chemistry.
Compounds like lead(II) chloride (\(\text{PbCl}_2\)) and lead(II) sulfate (\(\text{PbSO}_4\)) are characteristic examples where lead exhibits the \(+2\) charge. Lead(II) oxide (\(\text{PbO}\)), also known as litharge, is another common compound in the \(+2\) state, and it readily forms salts when reacting with acids.
The stability of the \(+2\) state is high; even strong oxidizing agents often only convert elemental lead to a lead(II) compound, such as the reaction of lead with fluorine to form \(\text{PbF}_2\). The lead(II) ion is typically colorless in solution. Many lead(II) compounds are sparingly soluble, identified by the tendency of their salts to precipitate out of water.
The Less Stable Oxidation State
Lead also exhibits a secondary, less stable oxidation state of \(+4\), found in the lead(IV) ion, \(\text{Pb}^{4+}\). This tetravalent state means the lead atom has lost all four of its valence electrons. Inorganic compounds where lead is in the \(+4\) state are less common and are often formed only under specific, oxidizing conditions.
The best-known example is lead dioxide (\(\text{PbO}_2\)), a dark-brown solid used commercially as the positive plate in lead-acid batteries. Lead dioxide is a powerful oxidizing agent because the \(\text{Pb}^{4+}\) ion readily accepts two electrons to convert to the more stable \(\text{Pb}^{2+}\) state.
This tendency to revert to the lower charge makes lead(IV) compounds reactive. Lead dioxide reacts with concentrated hydrochloric acid, oxidizing the chloride ions to chlorine gas while the lead is reduced from \(+4\) to \(+2\). Few inorganic lead(IV) compounds are known, as many decompose to the \(+2\) state under normal conditions.
The Chemical Reason for Lead’s Variable Charge
The existence of both the \(+2\) and \(+4\) oxidation states is a consequence of lead’s electron configuration, which is \([\text{Xe}]4f^{14}5d^{10}6s^26p^2\). A lead atom has four valence electrons in its outermost shell: two in the \(6s\) orbital and two in the \(6p\) orbital. The expected oxidation state for an element in Group 14, which includes carbon and silicon, is \(+4\), reflecting the loss of all four valence electrons.
Lead’s stability is governed by the “inert pair effect,” a phenomenon pronounced in the heavier elements of the \(p\)-block. This effect describes the reluctance of the outermost \(s\)-electrons, the \(6s^2\) pair, to participate in chemical bonding. The \(6s\) electrons are held much more tightly by the nucleus due to poor shielding by the intervening \(d\) and \(f\) electrons, which makes them less available for chemical reactions.
When lead reacts, it preferentially loses only the two \(6p\) electrons, resulting in the more stable \(+2\) oxidation state. The \(+4\) state occurs only when the energy of the bond formation is sufficient to overcome the energy required to involve the \(6s^2\) electron pair in bonding. The greater stability of the \(+2\) state compared to the \(+4\) state is a direct demonstration of the inert pair effect in lead chemistry.