The correct code for stability, of oxidation states for given cations is :<br>(i) Pb2+ > Pb4+, Tl+ < — p Block Elements Chemistry Question

💡 Solution & Explanation

Step 1: Tin (Sn) belongs to Group 14, where the stability of the +4 oxidation state is higher than that of +2 for the lighter members. Step 2: Because Sn4+ is highly stable, Sn2+ readily acts as a reducing agent by losing two electrons to convert into Sn4+: Sn^2+ -> Sn^4+ + 2e^-. Step 3: Therefore, the reducing nature of SnCl2 is due to Sn4+ being more stable than Sn2+, matching option (c).