Which of the following is the correct reducing order of bondangle

Options

A. `NH_3 lt CH_4 lt C_2H_2 lt H_2O`
B. `C_2H_2 gt NH_3 gt H_2O lt CH_4`
C. `NH_3 gt H_2O gt CH_4 gt C_2H_4`
D. `H_2O lt NH_3 gt CH_4 lt C_2H_2`

Explanation

To determine the correct reducing order of bond angles for the given molecules, we will analyze the bond angles of each compound step by step.

### Step 1: Identify the bond angles of each molecule

1. **Ammonia (NH₃)**:
- Ammonia has a bond angle of approximately **107.3°**. This is due to the presence of one lone pair of electrons on the nitrogen atom, which causes some repulsion and reduces the bond angle from the ideal tetrahedral angle.

2. **Methane (CH₄)**:
- Methane has a bond angle of **109.5°**. It has a tetrahedral geometry with no lone pairs, leading to the ideal tetrahedral bond angle.

3. **Water (H₂O)**:
- Water has a bond angle of approximately **104.5°**. The two lone pairs on the oxygen atom exert repulsion on the hydrogen atoms, reducing the bond angle from the tetrahedral angle.

4. **Ethyne (C₂H₂)**:
- Ethyne has a bond angle of **180°**. This is due to its linear geometry resulting from sp hybridization.

### Step 2: Arrange the bond angles in decreasing order

Now that we have the bond angles, we can arrange them from the largest to the smallest:

1. **Ethyne (C₂H₂)**: 180°
2. **Methane (CH₄)**: 109.5°
3. **Ammonia (NH₃)**: 107.3°
4. **Water (H₂O)**: 104.5°

### Step 3: Write the final order

Thus, the correct reducing order of bond angles is:

**C₂H₂ > CH₄ > NH₃ > H₂O**

### Conclusion

The correct answer is that the bond angles decrease in the order of **C₂H₂ > CH₄ > NH₃ > H₂O**.

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