Question

In which of the following options, the molecules are correctly arranged in the increasing order of their bond angles?

• A. \( \text{NH}_3<\text{O}_3<\text{H}_2\text{O}<\text{SO}_2 \)
• B. \( \text{H}_2\text{O}<\text{O}_3<\text{NH}_3<\text{SO}_2 \)
• C. \( \text{H}_2\text{O}<\text{NH}_3<\text{SO}_2<\text{O}_3 \)
• D. \( \text{H}_2\text{O}<\text{NH}_3<\text{O}_3<\text{SO}_2 \)

Hint:

VSEPR Theory General Rules for Bond Angles: 1. Start with the ideal angle based on the number of electron domains (Linear: 180°, Trigonal Planar: 120°, Tetrahedral: 109.5°). 2. Lone pair - lone pair repulsion>lone pair - bonding pair repulsion>bonding pair - bonding pair repulsion. 3. Lone pairs and multiple bonds occupy more space than single bonds and thus compress the angles between other bonds.

Answer 1

The Correct Option is D

Let's determine the approximate bond angle for each molecule using VSEPR theory.
1. \(\text{H}_2\text{O}\) (Water): The central oxygen atom has 2 bonding pairs and 2 lone pairs. The geometry is bent, based on a tetrahedral electron arrangement. The two lone pairs repel more strongly than bonding pairs, pushing the H-O-H bonds closer together. The angle is approximately 104.5°.
2. \(\text{NH}_3\) (Ammonia): The central nitrogen atom has 3 bonding pairs and 1 lone pair. The geometry is trigonal pyramidal, based on a tetrahedral electron arrangement. The single lone pair repels the bonding pairs, reducing the angle from the ideal 109.5°. The angle is approximately 107°.
3. \(\text{O}_3\) (Ozone): The central oxygen atom has one single bond, one double bond, and one lone pair. This gives a total of 3 electron domains (treating the double bond as one domain). The geometry is bent, based on a trigonal planar electron arrangement. The ideal angle would be 120°. The lone pair repulsion will compress this angle slightly. The actual bond angle is about 116.8°.
4. \(\text{SO}_2\) (Sulfur Dioxide): The central sulfur atom has two double bonds (in resonance structures) and one lone pair. This also gives 3 electron domains. The geometry is bent, based on a trigonal planar arrangement. The ideal angle is 120°. The lone pair compresses the angle, but the repulsion between the double bonds is greater than between single bonds in ozone, so the angle is larger than in ozone. The actual bond angle is about 119°.
Based on these approximate values:
H\(_2\)O (104.5°)<NH\(_3\) (107°)<O\(_3\) (116.8°)<SO\(_2\) (119°)
This corresponds to the increasing order \( \text{H}_2\text{O}<\text{NH}_3<\text{O}_3<\text{SO}_2 \).