Question

Which is the correct thermal stability order for\( H_2E\) (E=O, S, Se, Te and Po)? 

• A. \(H_2S< H_2O<H_2Se<H_2Te<H_2Po \)
• B. \(H_2O< H_2S<H_2Se<H_2Te<H_2Po\)
• C. \(H_2Po<H_2Te<H_2Se < H_2S <H_2O\)
• D. \(H_2Se<H_2Te<H_2Po< H_2O <H_2S \)

Answer 1

The Correct Option is C

The question asks about the thermal stability order of the hydrides of Group 16 elements, specifically \( H_2E \) where \( E \) represents the elements Oxygen (O), Sulfur (S), Selenium (Se), Tellurium (Te), and Polonium (Po). Let's analyze the factors affecting their thermal stability and determine the correct order.

Concept: Thermal stability of hydrides decreases down the group in the periodic table. This is because, as we move down the group, the size of the central atom increases, which results in a longer and weaker bond with hydrogen, making the compound less stable thermally.

• H_2O: Oxygen is at the top of Group 16, hence it forms the strongest bond with hydrogen, making water the most thermally stable.
• H_2S: Sulfur is below oxygen, the bond is weaker than in water, making it less stable.
• H_2Se: Selenium forms a weaker bond than sulfur, further reducing thermal stability.
• H_2Te: Tellurium forms an even weaker bond with hydrogen, reducing stability.
• H_2Po: Polonium, being the heaviest, has the weakest bond with hydrogen and hence is the least thermally stable.

Based on this understanding, the thermal stability order from least to most stable should be:

H_2Po < H_2Te < H_2Se < H_2S < H_2O

Conclusion: The correct answer is:

H_2Po<H_2Te<H_2Se < H_2S <H_2O

This order reflects the decreasing bond strength and increasing atom size as we move down the group from oxygen to polonium.