Which is the correct thermal stability order of `H_(2)E(E=O,S,Se,Te` and `Po)`?

Options

A. `H_(2)SeltH_(2)TeltH_(2)PoltH_(2)OltH_(2)S`
B. `H_(2)SltH_(2)OltH_(2)SeltH_(2)TeltH_(2)Po`
C. `H_(2)OltH_(2)SltH_(2)SeltH_(2)TeltH_(2)Po`
D. `H_(2)PoltH_(2)TeltH_(2)SeltH_(2)SltH_(2)O`

Explanation

To determine the correct thermal stability order of the hydrides \( H_2E \) where \( E \) can be \( O, S, Se, Te, \) and \( Po \), we need to analyze the bond dissociation energies of these compounds. The thermal stability of these hydrides generally decreases as we move down the group in the periodic table due to the following reasons:

1. **Bond Strength**: As we move from oxygen to polonium, the size of the atoms increases. Larger atoms have longer bond lengths, which typically results in weaker bonds. Weaker bonds are easier to break, leading to lower thermal stability.

2. **Electronegativity**: The electronegativity of the elements decreases down the group. Higher electronegativity results in stronger bonds due to better overlap of atomic orbitals, contributing to higher bond dissociation energies.

Now, let’s analyze the hydrides one by one:

1. **\( H_2O \)**: Water has a high bond dissociation energy due to the small size of oxygen and its high electronegativity. This makes \( H_2O \) very stable thermally.

2. **\( H_2S \)**: The bond dissociation energy is lower than that of \( H_2O \) because sulfur is larger than oxygen, resulting in a longer and weaker bond.

3. **\( H_2Se \)**: The bond dissociation energy continues to decrease as we move to selenium, which is larger than sulfur, leading to even weaker bonds.

4. **\( H_2Te \)**: Tellurium is larger than selenium, and thus \( H_2Te \) has an even lower bond dissociation energy compared to \( H_2Se \).

5. **\( H_2Po \)**: Polonium is the largest and least electronegative among these elements, resulting in the weakest bond and thus the lowest thermal stability.

Putting this all together, the order of thermal stability based on bond dissociation energies is:

\[ H_2O > H_2S > H_2Se > H_2Te > H_2Po \]

Thus, the correct thermal stability order is:

**Answer**: \( H_2O > H_2S > H_2Se > H_2Te > H_2Po \)