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 the Concept:
This question assesses knowledge of the trends in oxidation states and the stability of those states for Group 14 elements (C, Si, Ge, Sn, Pb). A key concept here is the "inert pair effect," which becomes significant for the heavier elements in the p-block, like lead (Pb). The inert pair effect is the reluctance of the two s-orbital electrons in the valence shell to participate in bonding.
Step 2: Analyzing Each Statement:
- (A) In addition to +4, +2, carbon also shows negative oxidation states: This is correct. Carbon shows a +4 state (e.g., in CCl₄), a +2 state (e.g., in CO), and various negative oxidation states in its compounds with less electronegative elements (e.g., -4 in CH₄).
- (B) Tin in +2 state acts as a reducing agent: This is correct. For tin (Sn), the +4 oxidation state is more stable than the +2 state. Therefore, Sn²⁺ can be easily oxidized to Sn⁴⁺, meaning it readily loses electrons and acts as a reducing agent (e.g., SnCl₂ is a common laboratory reducing agent).
- (C) Lead in +2 state acts as good reducing agent: This is incorrect. Due to the pronounced inert pair effect in lead (Pb), the +2 oxidation state is significantly more stable than the +4 oxidation state. It is difficult to oxidize Pb²⁺ to Pb⁴⁺. Therefore, Pb²⁺ does not act as a good reducing agent. In fact, the reverse is true: Pb⁴⁺ is a strong oxidizing agent.
- (D) Lead in +4 state acts as a good oxidising agent: This is correct. As the +2 state is more stable for lead, compounds with Pb in the +4 state (like PbO₂) have a strong tendency to gain two electrons and get reduced to the +2 state. This makes them strong oxidizing agents.
Step 3: Final Answer:
The statement that is incorrect is that "Lead in +2 state acts as good reducing agent". Therefore, option (C) is the correct answer.