Question

Write the formula for each of the following compounds: g. lead(IV) sulfide

Answer 1

The formula for lead(IV) sulfide is PbS2.

To determine the formula, we need to understand the valencies of the elements involved. Lead (Pb) in its +4 oxidation state, also known as lead(IV), has a valency of 4, meaning it can form four bonds. Sulfur (S) typically has a valency of 2, indicating that it can form two bonds.

In lead(IV) sulfide, one lead(IV) ion combines with two sulfide ions to form the compound. The formula PbS2 represents this combination, indicating one lead(IV) ion and two sulfide ions.

It's important to note that the subscripts in the formula represent the number of ions or atoms present in the compound. In this case, the subscript 2 indicates that there are two sulfide ions bonded to one lead(IV) ion. If the formula were written as PbS or Pb2S, it would imply a different number of ions or atoms, which would be incorrect.

Remember, when writing chemical formulas, it's crucial to consider the valencies and the number of ions or atoms involved to accurately represent the compound.

Answer 2

Answer: PbS₂

Step-by-step explanation:

When we have a Roman numeral after an element, that indicates the charge or oxidation state of that element; the elements that are usually followed by a Roman numeral are transition metals, and lead, Pb, is a transition metal.

Lead (IV) indicates that the charge of Pb is 4+ because metals tend to be cations (positive ions), and IV is 4.

Sulfide is an anion of sulfur, and its charge is 2- because it's a nonmetal, and nonmetals tend to be anions (negative ions). It's also in group 16, and elements in group 16 have 2- as their charge.

So, we know the charges of both our elements:

Pb⁴⁺ and S²⁻

So, we balance the charges by having a subscript of 2 on S (2 * -2 = -4) so that the sum of their charges is 0:

Pb⁴⁺S₂²⁻

Our final formula is PbS₂.

I hope this helps! :)