Correct decreasing order of volatility is

Options

A. `H_(2)O gt H_(2)S gt H_(2)Se`
B. `H_(2)S gt H_(2)O gtH_(2)Se`
C. `H_(2)Se gt H_(2)O gt H_(2)S`
D. `H_(2)S gt H_(2)Se gt H_(2)O`

Answer

To determine the correct decreasing order of volatility among the compounds H2O, H2S, and H2Se, we need to analyze their molecular weights and the presence of intermolecular forces.

### Step-by-Step Solution:

1. **Understanding Volatility**:
- Volatility refers to how readily a substance vaporizes. Generally, lower molecular weight compounds are more volatile than higher molecular weight compounds.

2. **Identifying Molecular Weights**:
- Calculate the molecular weights of the compounds:
- H2O (Water): H = 1 g/mol, O = 16 g/mol → Total = 2(1) + 16 = 18 g/mol
- H2S (Hydrogen Sulfide): H = 1 g/mol, S = 32 g/mol → Total = 2(1) + 32 = 34 g/mol
- H2Se (Hydrogen Selenide): H = 1 g/mol, Se = 78 g/mol → Total = 2(1) + 78 = 80 g/mol

3. **Comparing Molecular Weights**:
- H2O = 18 g/mol
- H2S = 34 g/mol
- H2Se = 80 g/mol
- From the weights, we can see that H2O has the lowest molecular weight, followed by H2S, and then H2Se.

4. **Considering Intermolecular Forces**:
- H2O has strong hydrogen bonding due to the presence of highly electronegative oxygen, which significantly affects its volatility.
- H2S and H2Se have weaker van der Waals forces compared to the hydrogen bonding in H2O.

5. **Establishing the Order of Volatility**:
- Despite H2O having the lowest molecular weight, its strong hydrogen bonds make it less volatile than H2S and H2Se.
- Therefore, the order of volatility based on molecular weight and intermolecular forces is:
- H2S > H2Se > H2O

6. **Final Answer**:
- The correct decreasing order of volatility is H2S > H2Se > H2O.