Question

The incorrect statement about the oxidation states of group 14 elements is

• A. In addition to +4, +2, carbon also shows negative oxidation states
• B. Tin in +2 state acts as a reducing agent
• C. Lead in +2 state acts as good reducing agent
• D. Lead in +4 state acts as a good oxidising agent

Hint:

Remember the inert pair effect trend for Group 14: The stability of the +2 oxidation state increases as you go down the group, while the stability of the +4 state decreases. For Pb, the +2 state is the most stable. This makes Pb(II) a poor reducing agent and Pb(IV) a strong oxidizing agent.

Answer 1

The Correct Option is C

Step 1: Understanding Inert Pair Effect: In Group 14 (C, Si, Ge, Sn, Pb), the stability of the +4 oxidation state decreases down the group, and the stability of the +2 oxidation state increases due to the inert pair effect. - For Tin (Sn), +4 is more stable than +2. So, \( \text{Sn}^{2+} \) wants to become \( \text{Sn}^{4+} \) (oxidation). Hence, \( \text{Sn}^{2+} \) is a good reducing agent. - For Lead (Pb), +2 is more stable than +4. So, \( \text{Pb}^{4+} \) wants to become \( \text{Pb}^{2+} \) (reduction). Hence, \( \text{Pb}^{4+} \) is a good oxidising agent.
Step 2: Analyzing Statement C: Statement (C) says "Lead in +2 state acts as good reducing agent". This would imply \( \text{Pb}^{2+} \) oxidises to \( \text{Pb}^{4+} \). However, \( \text{Pb}^{2+} \) is the stable state. It does not want to lose electrons. Therefore, this statement is incorrect.
Step 3: Analyzing other statements: (A) Correct. Carbon shows -4 in \( \text{CH}_4 \), etc. (B) Correct. \( \text{Sn}^{2+} \) reduces to \( \text{Sn}^{4+} \). (D) Correct. \( \text{Pb}^{4+} \) oxidizes to \( \text{Pb}^{2+} \).