Liquid IF5 conducts electricity. — p Block Elements Chemistry Question

💡 Solution & Explanation

Step 1: Lead (Pb) exhibits both +2 and +4 oxidation states, but Pb(IV) is highly oxidizing due to the inert pair effect. Step 2: Iodide (I-) has a large size and low electronegativity, making it a very strong reducing agent. Step 3: When Pb(IV) and I- are brought together, a redox reaction occurs where Pb4+ is reduced to Pb2+ and I- is oxidized to I2. Consequently, PbI4 is thermodynamically unstable and does not exist, matching option (b).