Question

Which from following statements is true for group 16 elements?

• A. All elements of this group form $\text{EO}_2$ type oxides.
• B. It includes all the nonmetals.
• C. Oxides of all elements of this group are gaseous at room temperature.
• D. Reducing properties of dioxides of this group element decreases form $\text{SO}_2$ to $\text{TeO}_2$.

Hint:

Remember the classic d-block and p-block oxide rule: For a given oxidation state down a group, the heavier element's oxide tends to be a solid and behaves more as an oxidizing agent, while the lighter element's oxide ($\text{SO}_2$) behaves as a reducing agent.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The problem asks us to evaluate four statements regarding the trends and properties of Group 16 elements (the chalcogens) and identify the single true statement.

Step 2: Key Formula or Approach:
Analyze the periodic trends of Group 16 (O, S, Se, Te, Po). As you move down the group, atomic size increases, metallic character increases, and the stability of the higher $+6$ oxidation state decreases due to the inert pair effect, which directly impacts the redox properties of their dioxides ($\text{EO}_2$).

Step 3: Detailed Explanation:
Let's analyze each option systematically to verify its correctness: 1.

Option (A): Incorrect. Oxygen itself is part of Group 16 and cannot form a dioxide of the type $\text{EO}_2$ with itself under standard configurations ($\text{O}_3$ is ozone, an allotrope).
2.

Option (B): Incorrect. Group 16 does not contain all nonmetals in the periodic table (Groups 14, 15, and 17 also contain nonmetals). Furthermore, Group 16 includes metalloids (Te) and metals (Po), so it does not consist exclusively of nonmetals.
3.

Option (C): Incorrect. While $\text{SO}_2$ is a gas at room temperature, selenium dioxide ($\text{SeO}_2$) is a crystalline solid and tellurium dioxide ($\text{TeO}_2$) is a solid.
4.

Option (D): Correct. In Group 16 dioxides, sulfur in $\text{SO}_2$ is in a $+4$ oxidation state and readily wants to be oxidized to the more stable $+6$ state ($\text{SO}_3$), making it a strong reducing agent. Moving down the group, the $+4$ oxidation state becomes progressively more stable than the $+6$ state due to the inert pair effect. Therefore, $\text{TeO}_2$ is an oxidizing agent rather than a reducing agent. Thus, the reducing property of the dioxides decreases from $\text{SO}_2$ to $\text{TeO}_2$.

Step 4: Final Answer:
The true statement is given by option (D).