Question

The correct order of bond angles of the following is
I. \( \text{H}_2\text{O} \)
II. \( \text{NH}_3 \)
III. \( \text{CH}_4 \)
IV. \( \text{SO}_2 \)

• A. IV>III>II>I
• B. IV>III>I>II
• C. I>II>III>IV
• D. I>II>IV>III

Hint:

Use VSEPR theory to predict the geometry and bond angles. The number of lone pairs and bond pairs around the central atom determines the electron pair geometry and molecular geometry. Lone pair-lone pair repulsion>lone pair-bond pair repulsion>bond pair-bond pair repulsion. Multiple bonds also cause greater repulsion than single bonds.

Answer 1

The Correct Option is A

The bond angles in molecules are determined by the repulsion between electron pairs around the central atom, according to the VSEPR (Valence Shell Electron Pair Repulsion) theory.
I.
\( \text{H}_2\text{O} \): The central atom is oxygen (O).
It has 6 valence electrons.
In \( \text{H}_2\text{O} \), it forms two bonds with hydrogen atoms and has two lone pairs of electrons.
The electron pair geometry is tetrahedral (4 electron pairs), and the molecular geometry is bent.
The bond angle is approximately 104.
5°.
The two lone pairs cause greater repulsion, reducing the bond angle from the ideal tetrahedral angle of 109.
5°.
II.
\( \text{NH}_3 \): The central atom is nitrogen (N).
It has 5 valence electrons.
In \( \text{NH}_3 \), it forms three bonds with hydrogen atoms and has one lone pair of electrons.
The electron pair geometry is tetrahedral (4 electron pairs), and the molecular geometry is trigonal pyramidal.
The bond angle is approximately 107°.
The single lone pair causes repulsion, reducing the bond angle from 109.
5°.
The repulsion is less than in \( \text{H}_2\text{O} \) due to only one lone pair.
III.
\( \text{CH}_4 \): The central atom is carbon (C).
It has 4 valence electrons.
In \( \text{CH}_4 \), it forms four bonds with hydrogen atoms and has no lone pairs of electrons.
The electron pair geometry is tetrahedral (4 electron pairs), and the molecular geometry is also tetrahedral.
The bond angle is 109.
5°.
There are no lone pairs to cause additional repulsion.
IV.
\( \text{SO}_2 \): The central atom is sulfur (S).
It has 6 valence electrons.
In \( \text{SO}_2 \), it forms two double bonds with oxygen atoms and has one lone pair of electrons.
The electron pair geometry is trigonal planar (3 electron pairs), and the molecular geometry is bent.
The ideal bond angle for trigonal planar is 120°.
The presence of a lone pair causes repulsion, reducing the bond angle to approximately 119°.
The double bonds also contribute to greater electron density and thus more repulsion than single bonds, which can affect the bond angle.
However, the dominant factor here is the trigonal planar electron pair geometry and the effect of the lone pair.
Comparing the approximate bond angles: \( \text{SO}_2 \) (\( \sim 119^\circ \))>\( \text{CH}_4 \) (\( 109.
5^\circ \))>\( \text{NH}_3 \) (\( \sim 107^\circ \))>\( \text{H}_2\text{O} \) (\( \sim 104.
5^\circ \)) The correct order of bond angles is IV>III>II>I.