What is the formula for lead (IV) carbonate?
The Correct Answer and Explanation is :
The formula for lead(IV) carbonate is ( \text{Pb(CO}_3\text{)}_2 ).
Explanation
Lead(IV) carbonate is a compound consisting of lead in its +4 oxidation state, represented as lead(IV) or ( \text{Pb}^{4+} ). The carbonate ion, ( \text{CO}_3^{2-} ), carries a -2 charge. In order to balance the charges between lead and carbonate, we need two carbonate ions to neutralize the charge from one lead(IV) ion. Therefore, the formula is derived as follows:
- Understanding Oxidation States: In lead(IV) carbonate, lead has an oxidation state of +4. Carbonate (( \text{CO}_3^{2-} )) consists of one carbon atom bonded to three oxygen atoms, and it has an overall charge of -2.
- Balancing Charges: To balance the +4 charge of lead, we need enough carbonate ions. Since each carbonate ion has a -2 charge, two carbonate ions are required to balance the +4 charge from lead:
[
\text{Charge from lead} (+4) + 2 \times \text{Charge from carbonate} (-2) = 0
] - Writing the Formula: Combining the lead ion and the two carbonate ions, the resulting chemical formula becomes ( \text{Pb(CO}_3\text{)}_2 ).
Properties and Uses
Lead(IV) carbonate is less commonly encountered than lead(II) carbonate, which is more stable and prevalent. However, lead(IV) compounds can be synthesized under specific conditions. Lead compounds, in general, are known for their use in various applications, including pigments and ceramics, although many have fallen out of favor due to toxicity concerns associated with lead exposure.
In summary, ( \text{Pb(CO}_3\text{)}_2 ) represents lead(IV) carbonate, highlighting the relationship between lead’s oxidation state and the balancing of charges with carbonate ions. Understanding these fundamentals is crucial in chemistry, especially when dealing with ionic compounds.