B.P of `H_(2)O (100^(@)C) `and `H_(2)S( -= 42^(@)C)` is explained by

Options

A. Vander Waal's forces
B. Colvalent bonding
C. Hydrogen bonding
D. Ionic bonding

Answer

To explain the boiling points of \( H_2O \) (100°C) and \( H_2S \) (−42°C), we can analyze the intermolecular forces present in each compound.

### Step 1: Identify the compounds and their boiling points
- **Water (\( H_2O \))** has a boiling point of **100°C**.
- **Hydrogen sulfide (\( H_2S \))** has a boiling point of **−42°C**.

### Step 2: Determine the types of intermolecular forces present
- **Water (\( H_2O \))**: The molecule contains oxygen, which is highly electronegative. This leads to the formation of **hydrogen bonds** between water molecules.
- **Hydrogen sulfide (\( H_2S \))**: The molecule contains sulfur, which is less electronegative than oxygen. Therefore, \( H_2S \) primarily exhibits **Van der Waals forces** (also known as London dispersion forces), which are weaker than hydrogen bonds.

### Step 3: Explain the significance of hydrogen bonding in water
- **Hydrogen bonding** is a strong type of dipole-dipole interaction that occurs when hydrogen is bonded to highly electronegative atoms like oxygen, nitrogen, or fluorine. In water, each molecule can form up to four hydrogen bonds with neighboring molecules, creating a strong network of interactions. This results in a higher boiling point due to the increased energy required to break these bonds during the phase transition from liquid to gas.

### Step 4: Compare the boiling points based on intermolecular forces
- The strong hydrogen bonds in water lead to a significantly higher boiling point (100°C) compared to the weaker Van der Waals forces in \( H_2S \), which results in a much lower boiling point (−42°C).

### Conclusion
The boiling point of \( H_2O \) is higher than that of \( H_2S \) primarily due to the presence of hydrogen bonding in water, which is absent in hydrogen sulfide.

### Final Answer
The boiling point of \( H_2O \) (100°C) and \( H_2S \) (−42°C) is explained by **hydrogen bonding**.

---