Question

Rationalise the given statements and give chemical reactions: \- Lead(II) chloride reacts with \(\mathrm{Cl}_{2}\) to give \(\mathrm{PbCl}_{4}\). \- Lead(IV) chloride is highly unstable towards heat. \- Lead is known not to form an iodide, \(\mathrm{PbI}_{4}\).

Short answer

PbCl₂ reacts with Cl₂ to form PbCl₄, a compound that is unstable to heat, and lead does not form PbI₄ due to the instability of Pb⁴⁺ with iodides.

Step-by-step solution

Understanding the Reaction of Lead(II) Chloride

When lead(II) chloride (PbCl₂) reacts with chlorine gas (Cl₂), it undergoes a redox reaction. The oxidation state of lead increases from +2 in PbCl₂ to +4 in PbCl₄. The balanced chemical equation is: \[\text{PbCl}_2 + \text{Cl}_2 \rightarrow \text{PbCl}_4\]Here, chlorine acts as an oxidizing agent and converts Pb to a higher oxidation state.

Analyzing the Stability of Lead(IV) Chloride

Lead(IV) chloride (PbCl₄) is thermally unstable because PbCl₄ tends to decompose at elevated temperatures, reverting to PbCl₂ and releasing chlorine gas. This instability is attributed to the inert pair effect, where the 2 s² electrons in lead are not involved in bonding and prefer to remain as a pair, thereby reducing the stabilization of the +4 oxidation state. The decomposition reaction can be represented as:\[\text{PbCl}_4 \rightarrow \text{PbCl}_2 + \text{Cl}_2\]

Understanding the Formation Issues of Lead(IV) Iodide

Lead does not form a stable PbI₄ primarily due to the large size and low oxidizing power of the iodide ion (I⁻), which is unable to sufficiently stabilize the +4 oxidation state of lead. Moreover, the high tendency of iodide ions to reduce and the weak bonding nature with lead further prevent the formation of PbI₄. As a result, lead is stable in the +2 oxidation state with iodides, forming PbI₂ instead.

Key concepts

Redox Reactions

Redox reactions are a type of chemical reaction where the oxidation states of atoms are changed. These reactions involve the transfer of electrons between substances. For a redox reaction to occur, one substance must lose electrons (oxidized) while another gains electrons (reduced). In the case of lead(II) chloride reacting with chlorine, we have a classic redox reaction:

• Lead(II) chloride (\[\text{PbCl}_2\]) reacts with chlorine (\[\text{Cl}_2\]), resulting in the formation of lead(IV) chloride (\[\text{PbCl}_4\]).
• The oxidation state of lead changes from +2 to +4, indicating that lead has been oxidized.
• Chlorine acts as an oxidizing agent, meaning it gains electrons (reducing its oxidation state), facilitating the oxidation of lead.

Understanding redox reactions helps predict the behavior of substances in chemical processes, explaining why certain reactions occur as they do.

Oxidation States

Oxidation states denote the degree of oxidation of an element in a compound. They help in identifying how electrons are being transferred in redox reactions. In our example:

• The oxidation state of lead in lead(II) chloride (\[\text{PbCl}_2\]) is +2.
• When lead(II) chloride reacts with chlorine, its oxidation state increases to +4, forming lead(IV) chloride (\[\text{PbCl}_4\]).
• Knowing the oxidation states is crucial because it helps us understand the stability and possible chemical reactions of compounds.

Changes in oxidation states indicate either the gaining or losing of electrons, which forms the basis of redox reaction mechanisms. In essence, the oxidation state offers a practical guide to the electronic features of a molecule.

Inert Pair Effect

The inert pair effect is an important concept in the chemistry of heavier p-block elements, especially lead. It explains the reluctance of the s-electrons in elements like lead to participate in bonding. In the context of lead:

• Lead has outer s-electrons (\[6s^2\]) which are often not involved in forming compounds, contributing to the stability of lower oxidation states (+2).
• This effect is why lead(IV) chloride (\[\text{PbCl}_4\]) is less stable than lead(II) chloride (\[\text{PbCl}_2\]).
• The inert pair effect leads to a preference for lead remaining in the +2 oxidation state, which is energetically more favorable.

Understanding this effect is crucial for rationalizing why lead compounds behave the way they do, particularly their stability and formation tendency.

Stability of Compounds

The stability of a compound is influenced by various factors such as atomic structure, the nature of bonding, and external conditions like temperature. In terms of lead compounds:

• Lead(IV) chloride (\[\text{PbCl}_4\]) is thermally unstable and decomposes at higher temperatures into lead(II) chloride (\[\text{PbCl}_2\]) and \[\text{Cl}_2\] gas.
• This instability stems from the lead's tendency to adopt the more stable +2 oxidation state, influenced by the inert pair effect.
• Lead(IV) iodide (\[\text{PbI}_4\]) does not form stably because the iodide ion is too large and weak of an oxidizing agent to stabilize lead in the +4 state.

In essence, the stability of lead compounds reflects the interplay of atomic characteristics and chemical principles, determining how different leads will react or decompose.