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 thermal stability of hydrides like \( H_2E \) (where E = O, S, Se, Te, Po) is influenced by the bond strength between hydrogen and the element E. As we move down the group in the periodic table, the atomic size increases, causing the bond strength to decrease. Consequently, the thermal stability of the hydrides decreases because the weaker bonds break more easily under thermal stress.

1. Bond Strength: The bond strength between H and E is a key factor. The smaller the atom, the stronger the bond it forms with hydrogen.
2. Bond Length: As the size of the atom E increases down the group, the bond length of \( H-E \) increases, leading to weaker bonds because the overlap between the hydrogen's 1s orbital and E's orbitals becomes less effective.

Applying these principles to the group 16 (oxygen family) hydrides:

• H_2O has the strongest bond due to the small size of O, making it the most thermally stable.
• H_2S is less stable than water but more stable than the others down the group.
• H_2Se is more stable than H_2Te and H_2Po.
• H_2Po is the least stable due to its large atomic size and weak bond with hydrogen.

Therefore, the correct thermal stability order for the hydrides \( H_2E \) is:

H_2Po < H_2Te < H_2Se < H_2S < H_2O

This correctly matches with the third option.