Question

name the compound shown below. pbo₂ lead (ii) oxide check you have correctly selected 0.

Explanation:

Step1: Determine the oxidation state of lead

In \(PbO_2\), oxygen has an oxidation state of \(- 2\). Let the oxidation state of \(Pb\) be \(x\). Using the formula for the sum of oxidation states in a neutral compound (\(x + 2\times(-2)=0\)), we solve for \(x\).

$$x-4 = 0\Rightarrow x = + 4$$

Step2: Name the compound

For ionic compounds with metals that have variable oxidation states (like \(Pb\)), we use Roman numerals. The compound \(PbO_2\) has \(Pb^{4+}\) and \(O^{2 -}\) ions. So the name is Lead(IV) oxide.

Answer:

Lead(IV) oxide