Question

name the compound shown below.
pb3p2
lead (i phosphide

Explanation:

Step1: Determine the oxidation state of lead (Pb)

In the compound \(Pb_3P_2\), phosphorus (\(P\)) has an oxidation state of \(-3\) (as it forms phosphide ions \(P^{3 -}\)). Let the oxidation state of \(Pb\) be \(x\). Using the charge - neutrality principle (sum of oxidation states in a neutral compound is zero):

$$3x+2\times(- 3)=0$$

$$3x - 6 = 0$$

$$3x=6$$

$$x = 2$$

Step2: Name the compound

The compound is an ionic compound. For ionic compounds, we name the cation first and then the anion. The cation is lead (\(Pb\)) with an oxidation state of \(+2\) (so it is lead (II)) and the anion is phosphide (\(P^{3-}\)).

Answer:

Lead (II) Phosphide