Consider the following molecules :
`H_(2)O H_(2)S H_(2) Se H_(2)Te `
`I II III IV`
Arrange these molecules in increasing order of bond angles

Options

A. `I lt II lt III lt IV`
B. `IV lt III lt II lt I`
C. `I lt II lt IV lt III`
D. `I lt IV lt III lt I`

Explanation

To determine the increasing order of bond angles for the molecules \( H_2O \), \( H_2S \), \( H_2Se \), and \( H_2Te \), we can follow these steps:

### Step 1: Identify the central atom and its group
All four molecules have a central atom from Group 16 of the periodic table:
- \( H_2O \) (Oxygen)
- \( H_2S \) (Sulfur)
- \( H_2Se \) (Selenium)
- \( H_2Te \) (Tellurium)

### Step 2: Understand the trend in electronegativity
Electronegativity decreases as we move down the group in the periodic table. The electronegativity values for these atoms are approximately:
- Oxygen (O): 3.44
- Sulfur (S): 2.58
- Selenium (Se): 2.55
- Tellurium (Te): 2.1

### Step 3: Analyze the bond angles
The bond angle in a molecule is influenced by the repulsion between the electron pairs around the central atom. Higher electronegativity results in stronger attraction of the bonding electrons towards the central atom, which leads to greater repulsion between the bonding pairs, resulting in larger bond angles.

- \( H_2O \): The bond angle is approximately 104.5°.
- \( H_2S \): The bond angle is approximately 92°.
- \( H_2Se \): The bond angle is approximately 88°.
- \( H_2Te \): The bond angle is approximately 90°.

### Step 4: Arrange the molecules in increasing order of bond angles
Based on the bond angles identified:
1. \( H_2Se \) (88°)
2. \( H_2Te \) (90°)
3. \( H_2S \) (92°)
4. \( H_2O \) (104.5°)

Thus, the increasing order of bond angles is:
\[ H_2Se < H_2Te < H_2S < H_2O \]

### Final Answer
The increasing order of bond angles is:
\[ H_2Se < H_2Te < H_2S < H_2O \]

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