The molecule having the largest bond angle is :

Options

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

Answer

To determine which molecule has the largest bond angle among water (H₂O), hydrogen sulfide (H₂S), hydrogen selenide (H₂Se), and hydrogen telluride (H₂Te), we can follow these steps:

### Step 1: Understand the molecular geometry
- Water (H₂O) has a bent molecular geometry due to the presence of two lone pairs of electrons on the oxygen atom.
- Hydrogen sulfide (H₂S), hydrogen selenide (H₂Se), and hydrogen telluride (H₂Te) also have bent geometries due to their respective central atoms (S, Se, Te) having two lone pairs as well.

**Hint:** Remember that the presence of lone pairs affects the bond angles due to repulsion between electron pairs.

### Step 2: Analyze the electronegativity of the central atoms
- The electronegativity of the central atoms decreases as we move down the group in the periodic table: Oxygen (O) > Sulfur (S) > Selenium (Se) > Tellurium (Te).
- Higher electronegativity means that the central atom can attract the bonding electrons more strongly, which influences the bond angle.

**Hint:** Consider how electronegativity affects the electron cloud distribution around the central atom.

### Step 3: Compare the bond angles
- In water (H₂O), the bond angle is approximately 104.5°.
- In hydrogen sulfide (H₂S), the bond angle is about 92°.
- In hydrogen selenide (H₂Se), the bond angle is slightly larger than H₂S but still less than that of water.
- In hydrogen telluride (H₂Te), the bond angle is even smaller due to the lower electronegativity of tellurium.

**Hint:** Look up the typical bond angles for each of these molecules to compare them directly.

### Step 4: Conclusion
Based on the analysis, water (H₂O) has the largest bond angle due to the high electronegativity of oxygen, which causes greater repulsion between the bonding pairs of electrons, resulting in a larger angle.

**Final Answer:** The molecule having the largest bond angle is **water (H₂O)**.